url
stringlengths 90
342
| html
stringlengths 602
98.8k
| text_length
int64 602
98.8k
| __index_level_0__
int64 0
5.02k
|
|---|---|---|---|
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/08%3A_Phase_Equilibrium/8.02%3A_Single_Component_Phase_Diagrams |
The stability of phases can be predicted by the chemical potential, in that the most stable form of the substance will have the minimum chemical potential at the given temperature and pressure. This can be summarized in a phase diagram like the one shown below. In this diagram, the phase boundaries can be determined by measuring the rate of cooling at constant temperature. A typical cooling curve is shown below. The temperature will decrease over time as a sample is allowed to cool. When \(\kappa_T\) the substance undergoes a phase change, say from liquid to solid, the temperature will stop changing while heat is extracted due to the phase change. The temperature at which the halt occurs provides one point on the boundary at the temperature of the halt and the pressure at which the cooling curve was measured. The same data can be obtained by heating the system using a technique such as scanning calorimetry. In this experiment, heat is supplied to a sample at a constant rate, and the temperature of the sample is measured, with breaks occurring at the phase change temperatures. | 1,104 | 2,674 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/08%3A_Phase_Equilibrium/8.05%3A_The_Clausius-Clapeyron_Equation |
The Clapeyron equation can be developed further for phase equilibria involving the gas phase as one of the phases. This is the case for either sublimation (\(\text{solid} \rightarrow \text{gas}\)) or vaporization (\(\text{liquid} \rightarrow \text{gas}\)). In the case of vaporization, the change in molar volume can be expressed \[ \Delta V = V_{gas} -V_{liquid} \nonumber \] Since substances undergo a very large increase in molar volume upon vaporization, the molar volume of the condensed phase (liquid in this case) is negligibly small compared to the molar volume of the gas (i.e., \(V_{gas} \gg V_{liquid}\)). So, \[\Delta V \approx V_{gas} \nonumber \] And if the vapor can be treated as an ideal gas, \[V_{gas} = \dfrac{RT}{p} \nonumber \] Substitution into the Claperyron equation yields \[\dfrac{dp}{dT} = \dfrac{p\Delta H_{vap}}{RT^2} \nonumber \] Separating the variables puts the equation into an integrable form. \[dp = \dfrac{p\Delta H_{vap}}{R} \dfrac{dT}{T^2} \label{diffCC} \] Noting that \[\dfrac{dT}{T^2} =- d\left(\dfrac{1}{T} \right) \nonumber \] makes the integration very easy. If the enthalpy of vaporization is of temperature over the range of conditions, \[ \int_{p_1}^{p_2} \dfrac{dp}{p} = - \dfrac{\Delta H_{vap}}{R} \int_{T_1}^{T_2} d\left(\dfrac{1}{T} \right) \nonumber \] \[ \ln \left( \dfrac{p_2}{p_1}\right) = - \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2} -\dfrac{1}{T_2} \right) \label{CC} \] This is the . It can also be used to describe the boundary between solid and vapor phases by substituting the enthalpy of sublimation (\(\Delta H_{sub}\)) The vapor pressure of a liquid triples when the temperature is increased from 25 °C to 45 °C. What is the enthalpy of vaporization for the liquid? The problem can be solved using the Clausius-Clapeyron equation (Equation \ref{CC}). The following values can be used: Substitution into the Clausius-Clapeyron equation yields \[ \ln \left( \dfrac{3p_1}{p_1}\right) = - \dfrac{\Delta H_{vap}}{9.314 \dfrac{J}{mol\,K}} \left( \dfrac{1}{318\,K} -\dfrac{1}{298\,K} \right) \nonumber \] \[ \Delta H_{vap} = 43280 \,\dfrac{J}{mol} = 43.28 \, \dfrac{kJ}{mol} \nonumber \] The Clausius-Clapeyron equation also suggests that a plot of \(\ln(p)\) vs. \(1/T\) should yield a straight line, the slope of which is \(–\Delta H/R\) (provided that \(\Delta H_{vap}\) is independent of temperature over the range of temperatures involved.. \[ \ln(p) = - \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T} \right) + const. \nonumber \] This approach is very useful when there are several pairs of measurements of vapor pressure and temperature. Such a plot is shown below for water. For water, which has a very large temperature dependence, the linear relationship of \(\ln(p)\) vs. \(1/T\) holds fairly well over a broad range of temperatures. So even though there is some curvature to the data, a straight line fit still results in a reasonable description of the data (depending, of course, on the precision needed in the experiment.) For this fit of the data, \(\Delta H_{vap}\) is found to be 43.14 kJ/mol. For systems that warrant it, temperature dependence of \(\Delta H_{vap}\) can be included into the derivation of the model to fit vapor pressure as a function of temperature. For example, if the enthalpy of vaporization is assumed to take the following empirical form \[ \Delta H_{vap} = \Delta H_o + aT + bT^2 \nonumber \] and substituting it into the differential form of the Clausius-Clapeyron equation (Equation \ref{diffCC}) generates \[ \dfrac{dp}{p} = \dfrac{\Delta H_o + aT + bT^2}{R} \dfrac{dT}{T^2} \nonumber \] or \[ \dfrac{dp}{p} = \dfrac{\Delta H_o}{R} \dfrac{dT}{T^2} + \dfrac{a}{R} \dfrac{dT}{T} + \dfrac{b}{R} dT \nonumber \] And so the integrated form becomes \[ \ln (p) = - \dfrac{\Delta H_o}{R} \left(\dfrac{1}{T}\right) + \dfrac{a}{R} \ln T + \dfrac{b}{R} T + constant \nonumber \] The results of fitting these data to the temperature dependent model are shown in the table below. This results in calculated values of \(\Delta H_{vap}\) of 43.13 kJ/mol at 298 K, and 43.15 kJ/mol at 373 K. The results are a little bit skewed since there is no data above 100 C included in the fit. A larger temperature dependence would be found if the higher-temperature data were included in the fit. | 4,301 | 2,675 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.03%3A_Units_of_Concentration |
There are several different ways to quantitatively describe the concentration of a solution. For example, molarity was introduced in Chapter 4 as a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions, introduced in Chapter 10, are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example 4 reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known. Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL. : mass of substance and mass and density of solution : molarity and mole fraction : : A The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. The volume of the solution equals its mass divided by its density. The calculations follow: \[moles\; CH_3CO_2H=\dfrac{3.78\; \cancel{g}\; CH_3CO_2H}{60.05\; \cancel{g}/mol}=0.0629 \;mol\] \[volume=\dfrac{mass}{density}=\dfrac{100.0\; \cancel{g}\; solution}{1.00\; \cancel{g}/mL}=100\; mL\] \[ molarity\; of\; CH_3CO_2H=\dfrac{moles\; CH_3CO_2H }{\text{liter solution}}=\dfrac{0.0629\; mol\; CH_3CO_2H}{(100\; \cancel{mL})(1\; L/1000\; \cancel{mL})}=0.629\; M \;CH_3CO_2H \] This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than \(\frac{1}{2}\) mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL. B To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have \[moles\; H_2O=\dfrac{96.2\; \cancel{g}\; H_2O}{18.02\; \cancel{g}/mol}=5.34\; mol\; H_2O\] The mole fraction \(X\) of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present: \[X_{CH_3CO_2H}=\dfrac{moles\; CH_3CO_2H}{moles \;CH_3CO_2H + moles\; H_2O}=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol}=0.0116=1.16 \times 10^{−2}\] This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01. A solution of \(HCl\) gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of \(HCl\) per 100.0 g of solution, and its density is 1.10 g/mL. : The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent: \[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \tag{13.5}\] Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of \(H_2O\) under these conditions is very close to 1.0 L, and a 0.50 M solution of \(KBr\) in water, for example, has approximately the same concentration as a 0.50 m solution. Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb): \[\text{mass percentage}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \tag{13.6}\] \[\text{parts per million (ppm)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \tag{13.7}\] \[\text{parts per billion (ppb)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \tag{13.8}\] In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution. The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of \(H_2SO_4\) per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL). Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL. : volume of sample, solute concentration, and density of solution : molarity of solute and mass of solute in 250 mL : : a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore \[ molarity=\dfrac{moles}{liter solution}=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L}=1.63 \times 10^{-4} M\] b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is The maximum allowable concentration of lead in drinking water is 9.0 ppb. What is the molarity of \(Pb^{2+}\) in a 9.0 ppb aqueous solution? Use your calculated concentration to determine how many grams of \(Pb^{2+}\) are in an 8 oz glass of water. : 4.3 × 10−8 M; 2 × 10−6 g How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature. Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as we saw in Chapter 10. As you will learn in Section 13.5, mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies. Table 13.5 summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example 6. Different Units for Expressing the Concentrations of Solutions* Vodka is essentially a solution of pure ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka. : volume percent and density : mass percentage, mole fraction, molarity, and molality : : The key to this problem is to use the density of pure ethanol to determine the mass of ethanol (\(CH_3CH_2OH\)), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density: \[mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\] If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution. B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage: \[\%EtOH=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100)=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) = 34.5\%\] C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL: \[ moles\; EtOH=(31.6\; \cancel{g\; EtOH}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; EtOH}}\right)=0.686 \;mol\; CH_3CH_2OH\] Similarly, the number of moles of water is The mole fraction of ethanol is thus \[ X_{EtOH}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\] D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward: \[ m_{EtOH}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m \] A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene. : | 11,964 | 2,676 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/22%3A_Helmholtz_and_Gibbs_Energies/22.01%3A_Helmholtz_Energy |
We have answered the question: what is entropy, but we still do not have a general criterion for spontaneity, just one that works in an isolated system. We will consider what happens when we hold volume and temperature constant. As discussed previously, the expression for the change in internal energy: \[dU = TdS -PdV \nonumber \] is only valid for changes. Let us consider a spontaneous change. If we assume constant volume, the \( -PdV \) work term drops out. From the \(dS>\dfrac{δq}{T}\) we get: \[\underset{ \text{constant V} }{dU \le TdS } \nonumber \] \[\underset{ \text{constant V} }{ dU-TdS \le 0 } \nonumber \] Consider a new state function, , : \[ A ≡ U -TS \nonumber \] \[dA = dU -TdS - SdT \label{diff1} \] If we also set \(T\) constant, we see that Equation \(\ref{diff1}\) becomes \[\underset{ \text{constant V and T} }{ dA=dU-TdS \le 0 } \nonumber \] This means that the Helmholtz energy, \(A\), is a for spontaneous processes (regardless of isolation!) when \(T\) and \(V\) are held constant. \(A\) becomes constant once a reversible equilibrium is reached. A good example is the case of the mixing of two gases. Let's assume isothermal conditions and keep the total volume constant. For this process, \(\Delta U\) is zero (isothermal, ideal) but the \[\Delta S_{molar} = -y_1R\ln y_1-y_2 R \ln y_2 \nonumber \] This means that \[\Delta A_{molar} = RT (y_1\ln y_1+y_2\ln y_2). \nonumber \] This is a negative quantity because the mole ratios are smaller than unity. So yes this spontaneous process has a negative \(\Delta A\). If we look at \(\Delta A = \Delta U - T\Delta S\) we should see that the latter term is the same thing as \(-q_{rev}\) So we have : \[\Delta A = \Delta U - q_{rev} = w_{rev} \nonumber \] This is however the and so the Helmholtz energy is a direct measure of how much work one can get out of a system. \(A\) is therefore often called the Helmholtz energy. Interestingly this work be volume work as volume is constant. so it stands for the maximal work (e.g. electrical work) that can be obtained under the unlikely condition that volume is constant. Because \(A≡ U-TS\) we can write \[dA = dU -TdS -SdT \nonumber \] \[dA = TdS -PdV -TdS -SdT = -PdV - SdT \nonumber \] The natural variables of \(A\) are volume \(V\) and temperature \(T\). | 2,320 | 2,677 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/12%3A_Solids/12.06%3A_Metals_and_Semiconductors |
To explain the observed properties of metals, a more sophisticated approach is needed than the electron-sea model described in . The molecular orbital theory we used in to explain the delocalized π bonding in polyatomic ions and molecules such as NO , ozone, and 1,3-butadiene can be adapted to accommodate the much higher number of atomic orbitals that interact with one another simultaneously in metals. In a 1 mol sample of a metal, there can be more than 10 orbital interactions to consider. In our molecular orbital description of metals, however, we begin by considering a simple one-dimensional example: a linear arrangement of metal atoms, each containing a single electron in an orbital. We use this example to describe an approach to metallic bonding called band theory , which assumes that the valence orbitals of the atoms in a solid interact, generating a set of molecular orbitals that extend throughout the solid. If the distance between the metal atoms is short enough for the orbitals to interact, they produce bonding, antibonding, and nonbonding molecular orbitals. The left portion of shows the pattern of molecular orbitals that results from the interaction of orbitals as increases from 2 to 5. As we saw in , the lowest-energy orbital is the completely bonding molecular orbital, whereas the highest-energy orbital is the completely antibonding molecular orbital. Molecular orbitals of intermediate energy have fewer nodes than the totally antibonding molecular orbital. The energy separation between adjacent orbitals decreases as the number of interacting orbitals increases. For = 30, there are still discrete, well-resolved energy levels, but as increases from 30 to a number close to Avogadro’s number, the spacing between adjacent energy levels becomes almost infinitely small. The result is essentially a continuum of energy levels, as shown on the right in , each of which corresponds to a particular molecular orbital extending throughout the linear array of metal atoms. The levels that are lowest in energy correspond to mostly bonding combinations of atomic orbitals, those highest in energy correspond to mostly antibonding combinations, and those in the middle correspond to essentially nonbonding combinations. The continuous set of allowed energy levels shown on the right in is called an energy band . The difference in energy between the highest and lowest energy levels is the bandwidth and is proportional to the strength of the interaction between orbitals on adjacent atoms: the stronger the interaction, the larger the bandwidth. Because the band contains as many energy levels as molecular orbitals, and the number of molecular orbitals is the same as the number of interacting atomic orbitals, the band in contains energy levels corresponding to the combining of orbitals from metal atoms. Each of the original orbitals could contain a maximum of two electrons, so the band can accommodate a total of 2 electrons. Recall, however, that each of the metal atoms we started with contained only a single electron in each orbital, so there are only electrons to place in the band. Just as with atomic orbitals or molecular orbitals, the electrons occupy the lowest energy levels available. Consequently, only the lower half of the band is filled. This corresponds to filling all of the molecular orbitals in the linear array of metal atoms and results in the strongest possible bonding. The previous example was a one-dimensional array of atoms that had only orbitals. To extrapolate to two- or three-dimensional systems and atoms with electrons in and orbitals is straightforward in principle, even though in practice the mathematics becomes more complex, and the resulting molecular orbitals are more difficult to visualize. The resulting energy-level diagrams are essentially the same as the diagram of the one-dimensional example in , with the following exception: they contain as many bands as there are different types of interacting orbitals. Because different atomic orbitals interact differently, each band will have a different bandwidth and will be centered at a different energy, corresponding to the energy of the parent atomic orbital of an isolated atom. Because the 1 , 2 , and 2 orbitals of a period 3 atom are filled core levels, they do not interact strongly with the corresponding orbitals on adjacent atoms. Hence they form rather narrow bands that are well separated in energy ( ). These bands are completely filled (both the bonding and antibonding levels are completely populated), so they do not make a net contribution to bonding in the solid. The energy difference between the highest level of one band and the lowest level of the next is the band gap . It represents a set of forbidden energies that do not correspond to any allowed combinations of atomic orbitals. Because they extend farther from the nucleus, the valence orbitals of adjacent atoms (3 and 3 in ) interact much more strongly with one another than do the filled core levels; as a result, the valence bands have a larger bandwidth. In fact, the bands derived from the 3 and 3 atomic orbitals are wider than the energy gap between them, so the result is overlapping bands . These have molecular orbitals derived from two or more valence orbitals with similar energies. As the valence band is filled with one, two, or three electrons per atom for Na, Mg, and Al, respectively, the combined band that arises from the overlap of the 3 and 3 bands is also filling up; it has a total capacity of eight electrons per atom (two electrons for each 3 orbital and six electrons for each set of 3 orbitals). With Na, therefore, which has one valence electron, the combined valence band is one-eighth filled; with Mg (two valence electrons), it is one-fourth filled; and with Al, it is three-eighths filled, as indicated in . The partially filled valence band is absolutely crucial for explaining metallic behavior because it guarantees that there are unoccupied energy levels at an infinitesimally small energy above the highest occupied level. Band theory can explain virtually all the properties of metals. Metals conduct electricity, for example, because only a very small amount of energy is required to excite an electron from a filled level to an empty one, where it is free to migrate rapidly throughout the crystal in response to an applied electric field. Similarly, metals have high heat capacities (as you no doubt remember from the last time a doctor or a nurse placed a stethoscope on your skin) because the electrons in the valence band can absorb thermal energy by being excited to the low-lying empty energy levels. Finally, metals are lustrous because light of various wavelengths can be absorbed, causing the valence electrons to be excited into any of the empty energy levels above the highest occupied level. When the electrons decay back to low-lying empty levels, they emit light of different wavelengths. Because electrons can be excited from many different filled levels in a metallic solid and can then decay back to any of many empty levels, light of varying wavelengths is absorbed and reemitted, which results in the characteristic shiny appearance that we associate with metals. For a solid to exhibit metallic behavior, it must have a set of delocalized orbitals forming a band of allowed energy levels, and the resulting band must be only partially filled (10%–90%) with electrons. Without a set of delocalized orbitals, there is no pathway by which electrons can move through the solid. Metallic behavior requires a set of delocalized orbitals and a band of allowed energy levels that is partially occupied. Band theory explains the correlation between the valence electron configuration of a metal and the strength of metallic bonding. The valence electrons of transition metals occupy either their valence , ( − 1) , and orbitals (with a total capacity of 18 electrons per metal atom) or their and ( − 1) orbitals (a total capacity of 12 electrons per metal atom). These atomic orbitals are close enough in energy that the derived bands overlap, so the valence electrons are not confined to a specific orbital. Metals with 6 to 9 valence electrons (which correspond to groups 6–9) are those most likely to fill the valence bands approximately halfway. Those electrons therefore occupy the highest possible number of bonding levels, while the number of antibonding levels occupied is minimal. Not coincidentally, the elements of these groups exhibit physical properties consistent with the presence of the strongest metallic bonding, such as very high melting points. In contrast to metals, electrical insulators are materials that conduct electricity poorly because their valence bands are full. The energy gap between the highest filled levels and the lowest empty levels is so large that the empty levels are inaccessible: thermal energy cannot excite an electron from a filled level to an empty one. The valence-band structure of diamond, for example, is shown in part (a) in . Because diamond has only 4 bonded neighbors rather than the 6 to 12 typical of metals, the carbon 2 and 2 orbitals combine to form two bands in the solid, with the one at lower energy representing bonding molecular orbitals and the one at higher energy representing antibonding molecular orbitals. Each band can accommodate four electrons per atom, so only the lower band is occupied. Because the energy gap between the filled band and the empty band is very large (530 kJ/mol), at normal temperatures thermal energy cannot excite electrons from the filled level into the empty band. Thus there is no pathway by which electrons can move through the solid, so diamond has one of the lowest electrical conductivities known. What if the difference in energy between the highest occupied level and the lowest empty level is intermediate between those of electrical conductors and insulators? This is the case for silicon and germanium, which have the same structure as diamond. Because Si–Si and Ge–Ge bonds are substantially weaker than C–C bonds, the energy gap between the filled and empty bands becomes much smaller as we go down group 14 (part (b) and part (c) of ). (For more information on bond strengths, see .) Consequently, thermal energy is able to excite a small number of electrons from the filled valence band of Si and Ge into the empty band above it, which is called the conduction band . Exciting electrons from the filled valence band to the empty conduction band causes an increase in electrical conductivity for two reasons: Consequently, Si is a much better electrical conductor than diamond, and Ge is even better, although both are still much poorer conductors than a typical metal ( ). Substances such as Si and Ge that have conductivities between those of metals and insulators are called semiconductors . Many binary compounds of the main group elements exhibit semiconducting behavior similar to that of Si and Ge. For example, gallium arsenide (GaAs) is isoelectronic with Ge and has the same crystalline structure, with alternating Ga and As atoms; not surprisingly, it is also a semiconductor. The electronic structure of semiconductors is compared with the structures of metals and insulators in . Because thermal energy can excite electrons across the band gap in a semiconductor, increasing the temperature increases the number of electrons that have sufficient kinetic energy to be promoted into the conduction band. The electrical conductivity of a semiconductor therefore increases rapidly with increasing temperature, in contrast to the behavior of a purely metallic crystal. In a metal, as an electron travels through the crystal in response to an applied electrical potential, it cannot travel very far before it encounters and collides with a metal nucleus. The more often such encounters occur, the slower the motion of the electron through the crystal, and the the conductivity. As the temperature of the solid increases, the metal atoms in the lattice acquire more and more kinetic energy. Because their positions are fixed in the lattice, however, the increased kinetic energy increases only the extent to which they vibrate about their fixed positions. At higher temperatures, therefore, the metal nuclei collide with the mobile electrons more frequently and with greater energy, thus the conductivity. This effect is, however, substantially smaller than the increase in conductivity with temperature exhibited by semiconductors. For example, the conductivity of a tungsten wire decreases by a factor of only about two over the temperature range 750–1500 K, whereas the conductivity of silicon increases approximately 100-fold over the same temperature range. These trends are illustrated in . The electrical conductivity of a semiconductor with increasing temperature, whereas the electrical conductivity of a metal with increasing temperature. Doping is a process used to tune the electrical properties of commercial semiconductors by deliberately introducing small amounts of impurities. If an impurity contains valence electrons than the atoms of the host lattice (e.g., when small amounts of a group 15 atom are introduced into a crystal of a group 14 element), then the doped solid has more electrons available to conduct current than the pure host has. As shown in part (a) in , adding an impurity such as phosphorus to a silicon crystal creates occasional electron-rich sites in the lattice. The electronic energy of these sites lies between those of the filled valence band and the empty conduction band but closer to the conduction band. Because the atoms that were introduced are surrounded by host atoms, and the electrons associated with the impurity are close in energy to the conduction band, those extra electrons are relatively easily excited into the empty conduction band of the host. Such a substance is called an -type semiconductor , with the indicating that the added charge carriers are negative (they are electrons). If the impurity atoms contain valence electrons than the atoms of the host (e.g., when small amounts of a group 13 atom are introduced into a crystal of a group 14 element), then the doped solid has fewer electrons than the pure host. Perhaps unexpectedly, this also results in conductivity because the impurity atoms generate holes in the valence band. As shown in part (b) in , adding an impurity such as gallium to a silicon crystal creates isolated electron-deficient sites in the host lattice. The electronic energy of these empty sites also lies between those of the filled valence band and the empty conduction band of the host but much closer to the filled valence band. It is therefore relatively easy to excite electrons from the valence band of the host to the isolated impurity atoms, thus forming holes in the valence band. This kind of substance is called a -type semiconductor , with the standing for positive charge carrier (i.e., a hole). Holes in what was a filled band are just as effective as electrons in an empty band at conducting electricity. -Type semiconductors are egative charge carriers; the impurity has valence electrons than the host. -Type semiconductors are ositive charge carriers; the impurity has valence electrons than the host. The electrical conductivity of a semiconductor is roughly proportional to the number of charge carriers, so doping is a precise way to adjust the conductivity of a semiconductor over a wide range. The entire semiconductor industry is built on methods for preparing samples of Si, Ge, or GaAs doped with precise amounts of desired impurities and assembling silicon chips and other complex devices with junctions between - and -type semiconductors in varying numbers and arrangements. Because silicon does not stand up well to temperatures above approximately 100°C, scientists have been interested in developing semiconductors made from diamonds, a more thermally stable material. A new method has been developed based on vapor deposition, in which a gaseous mixture is heated to a high temperature to produce carbon that then condenses on a diamond kernel. This is the same method now used to create cultured diamonds, which are indistinguishable from natural diamonds. The diamonds are heated to more than 2000°C under high pressure to harden them even further. Doping the diamonds with boron has produced -type semiconductors, whereas doping them with boron and deuterium achieves -type behavior. Because of their thermal stability, diamond semiconductors have potential uses as microprocessors in high-voltage applications. A crystalline solid has the following band structure, with the purple areas representing regions occupied by electrons. The lower band is completely occupied by electrons, and the upper level is about one-third filled with electrons. band structure variations in electrical properties with conditions Based on the occupancy of the lower and upper bands, predict whether the substance will be an electrical conductor. Then predict how its conductivity will change with temperature. After all the electrons are removed from the upper band, predict how the band gap would affect the electrical properties of the material. Determine whether you would use a chemical oxidant or reductant to remove electrons from the upper band. Predict the effect of a filled upper band on the electrical properties of the solid. Then decide whether you would use an oxidant or a reductant to fill the upper band. Exercise A substance has the following band structure, in which the lower band is half-filled with electrons (purple area) and the upper band is empty. assumes that the valence orbitals of the atoms in a solid interact to generate a set of molecular orbitals that extend throughout the solid; the continuous set of allowed energy levels is an . The difference in energy between the highest and lowest allowed levels within a given band is the , and the difference in energy between the highest level of one band and the lowest level of the band above it is the . If the width of adjacent bands is larger than the energy gap between them, result, in which molecular orbitals derived from two or more kinds of valence orbitals have similar energies. Metallic properties depend on a partially occupied band corresponding to a set of molecular orbitals that extend throughout the solid to form a band of energy levels. If a solid has a filled valence band with a relatively low-lying empty band above it (a ), then electrons can be excited by thermal energy from the filled band into the vacant band where they can then migrate through the crystal, resulting in electrical conductivity. are poor conductors because their valence bands are full. have electrical conductivities intermediate between those of insulators and metals. The electrical conductivity of semiconductors increases rapidly with increasing temperature, whereas the electrical conductivity of metals decreases slowly with increasing temperature. The properties of semiconductors can be modified by , or introducing impurities. Adding an element with more valence electrons than the atoms of the host populates the conduction band, resulting in an with increased electrical conductivity. Adding an element with fewer valence electrons than the atoms of the host generates holes in the valence band, resulting in a that also exhibits increased electrical conductivity. Can band theory be applied to metals with two electrons in their valence orbitals? with no electrons in their valence orbitals? Why or why not? Given a sample of a metal with 10 atoms, how does the width of the band arising from orbital interactions compare with the width of the band arising from orbital interactions? from orbital interactions? Diamond has one of the lowest electrical conductivities known. Based on this fact, do you expect diamond to be colored? Why? How do you account for the fact that some diamonds are colored (such as “pink” diamond or “green” diamond)? Why do silver halides, used in the photographic industry, have band gaps typical of semiconducting materials, whereas alkali metal halides have very large band gaps? As the ionic character of a compound increases, does its band gap increase or decrease? Why? Why is silicon, rather than carbon or germanium, used in the semiconductor industry? Carbon is an insulator, and silicon and germanium are semiconductors. Explain the relationship between the valence electron configuration of each element and their band structures. Which will have the higher electrical conductivity at room temperature—silicon or germanium? How does doping affect the electrical conductivity of a semiconductor? Draw the effect of doping on the energy levels of the valence band and the conduction band for both an -type and a -type semiconductor. The low electrical conductivity of diamond implies a very large band gap, corresponding to the energy of a photon of ultraviolet light rather than visible light. Consequently, diamond should be colorless. Pink or green diamonds contain small amounts of highly colored impurities that are responsible for their color. As the ionic character of a compound increases, the band gap will also increase due to a decrease in orbital overlap. Remember that overlap is greatest for orbitals of the same energy, and that the difference in energy between orbitals on adjacent atoms increases as the difference in electronegativity between the atoms increases. Thus, large differences in electronegativity increase the ionic character, decrease the orbital overlap, and increase the band gap. Of Ca, N, B, and Ge, which will convert pure silicon into a -type semiconductor when doping? Explain your reasoning. Of Ga, Si, Br, and P, which will convert pure germanium into an -type semiconductor when doping? Explain your reasoning. Thumbnail from | 22,036 | 2,678 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.03%3A_Thin_Layer_Chromatography_(TLC)/2.3B%3A_Uses_of_TLC |
TLC is a common technique in the organic chemistry laboratory because it can give quick and useful information about the purity of a sample and whether or not a reaction in progress is complete. When low polarity solvents are used, a TLC plate can be complete in less than 5 minutes. One of the uses of TLC is to assess the purity of a sample. In Figure 2.7 are TLC plates of acetophenone and cinnamaldehyde: samples that were diluted from their reagent bottle, run, and visualized with UV light. Acetophenone appeared as only one spot on the TLC plate, indicating the reagent is likely pure. Conversely, cinnamaldehyde is unquestionably impure as its TLC showed two large spots, and had a few fainter spots as well. Aldehydes are prone to air oxidation, and it is common for aldehydes to be found in their reagent bottles alongside with their corresponding carboxylic acids. TLC is one method that can be used to determine how much an aldehyde has degraded. TLC can be used to analyze a chemical reaction, for example to determine if the reactants have been consumed and a new product has formed. A pure sample of the reactant can be spotted in one lane of a TLC, and the product mixture in another lane. Often the central lane is used for reference, where both reactant and product mixture are spotted over top of one another, in what is called the " ." For example, the reaction shown in Figure 2.8a is analyzed by TLC in Figure 2.8b. In the first lane of Figure 2.8b (labeled F) is spotted a pure sample of the reactant ferrocene. In the last lane (labeled AF) is spotted the product mixture, which is assumed to be acetylferrocene. In the central lane (labeled co) is spotted both pure ferrocene and the product mixture. The right-most "AF" lane shows that the reaction appears to be a success: the higher spot of ferrocene is absent (meaning it has been consumed), and a new product spot is present. More tests would have to be done to confirm that the lower spot is the expected product of acetylferrocene, but the TLC results look promising. The co-spot is used for reference, and can be useful in interpreting certain situations. For example, at times the solvent may run with a slight diagonal, causing identical components to elute to slightly different heights. The co-spot can be useful in tracking these variations. It can also be helpful in identifying spots as different if they have similar \(R_f\) values (a feature difficult to be certain of in Figure 2.9a). Two compounds with slightly different eluted over top of one another in the co-spot lane may produce a spot with an elongated shape (Figure 2.9b), making it more obvious they are different compounds. Furthermore, a cospot can be helpful in deciphering a TLC plate if the reaction solvent persists in the product mixture lane (for example if the solvent has a relatively high boiling point like DMF or DMSO), such that the residual solvent has an effect on the component's \(R_f\) values. TLC can be used to monitor the progress of a reaction. This method is often used in research, and one such journal article reporting its usage is shown in Figure 2.10. To use TLC in this manner, three lanes are spotted on a TLC plate: one for the limiting reactant, one for the co-spot, and one for the reaction mixture. The goal is to note the disappearance of the limiting reactant in the reaction mixture lane and the appearance of a new product spot. When the limiting reactant has completely disappeared, the chemist deduces that the reaction is complete, and can then be "worked up". To demonstrate how TLC can be used to monitor a reaction, the transesterification reaction in Figure 2.11 was studied over time (Figure 2.12). In the first lane of each TLC plate (marked "BA") was spotted a dilute sample of the reactant benzyl acetate, while in the third lane of each was spotted the reaction mixture (marked "Pr") at different times. In the central lane (marked "Co" for the co-spot), both benzyl acetate and the reaction mixture were delivered over top of one another. Figure 2.12 shows the progress of the reaction over time. In Figure 2.12a, with the third lane representing the reaction mixture after one minute of mixing, a faint spot near the middle of the plate corresponds to unreacted benzyl acetate. A new spot appears below it, representing the benzyl alcohol product. Over time (Figure 2.12b-e), the top benzyl acetate spot disappears in the reaction mixture lane, and the lower benzyl alcohol spot intensifies. It is apparent from the TLC plates that the reaction was nearing completion at 10 minutes, and was complete at 20 minutes. The TLC demonstrates that the reaction mixture could be worked up after 20 minutes of mixing. To monitor a reaction's progress by TLC, an " " (or tiny sample) of the reaction mixture is necessary.
If the reaction is run at room temperature or with only mild heating, and the concentration of reactants is conducive to TLC, a capillary spotter can be directly inserted into the flask where the reaction is taking place (Figure 2.13a). A long spotter is ideal if one is available. The aliquot can then be directly spotted on the TLC plate.
If the sample is expected to be UV active, it is a good idea to view the TLC plate under UV light before eluting the plate. If the sample spot is not visible before elution it will not be visible afterwards, as compounds diffuse during elution. If the sample is expected to be UV active, and only a faint hint of material is seen on the baseline, the material can be deposited multiple times before elution (Figure 2.13b): deliver a small spot of sample on the baseline, and let it before delivering another spot over top of the first. If the spots are not allowed to dry in between applications, the spot will be too large. Check the plate under UV light again, and if necessary spot more times.
It is important to fully allow a spot to dry on the TLC plate before placement in the TLC chamber. In Figure 2.13c the spot for the reaction mixture (labeled Pr) was not fully dry before elution, and the ethanol solvent likely affected the appearance of the lower \(R_f\) spot. In Figure 2.13d the sample was allowed to fully dry before elution, and the lower \(R_f\) spot was more distinct. Furthermore, residual ethanol was likely the reason why the compound spotted in the reagent lane (labeled "BA") had a different \(R_f\) than it did in the central co-spot lane ("Co"). To obtain an aliquot of a refluxing solution, briefly remove the condenser and insert a spotter into the reaction mixture (Figure 2.13e). Immediately re-connect the condenser and adjust the clamps while holding the aliquot. Alternatively, lift the flask from the heat source to temporarily cease the reflux before inserting the spotter. The sample may be able to be spotted directly on the TLC plate, but if too concentrated it can be first diluted by running an appropriate solvent (e.g. acetone) through the pipette and into a small vial. | 6,996 | 2,682 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.2%3A_Group_18_-_The_Noble_Gases |
The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O , CO , and water vapor from air and showed that this residual gas was slightly denser than pure N obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens. Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University. Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh. In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N , so scuba divers often use gas mixtures that contain these gases, rather than N , to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N (g) that can occur when a diver returns to the surface too rapidly. Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source. The elements of group 18 all have closed-shell valence electron configurations, either ns np or 1s for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in Table \(\Page {1}\) indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements. For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (Figure \(\Page {1}\)). Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved. In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century. The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF , a compound used in the Manhattan Project, could oxidize O . Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O , Bartlett recognized that PtF should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF vapor, yellow-orange crystals immediately formed (Figure \(\Page {3}\)). Although Bartlett initially postulated that they were \(\ce{Xe^{+}PtF6^{−}}\), it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the \(\ce{XeF^{+}}\) ion: \[\ce{Xe(g) + PtF6(g) -> [XeF^{+},PtF5^{−}](s)} \label{Eq1}\] Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF , XeF , or XeF , depending on conditions; one such reaction is as follows: \[\ce{Xe(g) + 2F2(g) -> XeF4(s)} \label{Eq2}\] The ionization energies of helium, neon, and argon are so high (Table \(\Page {1}\)) that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F and Cl . Although \(\ce{XeCl2}\) and \(\ce{KrF2}\) can be prepared directly from the elements, they are substantially less stable than the xenon fluorides. The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known. Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF produces CsXeF , which gives Cs XeF when heated: \[\ce{XeF6(s) + CsF(s) -> CsXeF7(s)} \label{Eq3}\] \[\ce{2CsXeF7(s) ->[\Delta] Cs2XeF8(s) + XeF6(g)} \label{Eq4}\] The \(\ce{XeF8^{2-}}\) ion contains eight-coordinate xenon and has the square antiprismatic structure, which is essentially identical to that of the IF ion. Cs XeF is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs XeF are almost certainly the formation of a stable ionic lattice and the of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF .) For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between \(O\) and \(Xe\). Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either \(XeF_4\) or \(XeF_6\) produces \(XeO_3\), an explosive white solid: \[\ce{XeF6(aq) + 3H2O(l) -> XeO3(aq) + 6HF(aq)} \label{Eq5}\] Treating a solution of XeO with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO , a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO ), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO is isoelectronic with the iodate ion (IO ), and XeF is isoelectronic with the IF ion. Xenon has a high affinity for both fluorine and oxygen. Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored. On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon? nonradioactive isotope of radon summary of its chemistry Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon. We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF , RnF , RnF , and possibly RnF (due to its large radius). The ion RnF should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO and possibly RnO . The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl and possibly even RnCl . Predict the stoichiometry of the product formed by reacting XeF with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion. \(\ce{KXeF7}\); the xenon atom in XeF has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s electrons behave like an inert pair that does not participate in bonding. The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8. | 12,826 | 2,683 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4F%3A_Steam_Baths |
A steam bath (Figure 1.49) is a relatively safe way to heat flammable organic liquids. They are designed to heat beakers, Erlenmeyer flasks, and round-bottomed flasks, and have a series of concentric rings that can be removed to adjust to the size of the flask. Many science buildings have in house steam lines in their labs, allowing for this convenient and safe method to heat various solvents. The steam line should be connected to the upper arm of the steam bath, and condensation should be allowed to drain from the lower arm of the bath to the sink (Figure 1.50a). The steam tap should be adjusted so that a moderate amount of steam can be seen coming only from the central opening in the bath, then a flask should be set atop the opening to warm the flask (Figure 1.50b). Steam should only warm the bottom of the flask, and steam should not be visible coming out anywhere else in the bath. When warming highly volatile solvents, the flask may need to be held the opening of the bath to control the rate of heating (Figure 1.50c), or the steam rate lowered. | 1,077 | 2,684 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.05%3A_Electron_Affinities |
The electron affinity (\(EA\)) of an element \(E\) is defined as the energy change that occurs when an electron is added to a gaseous atom or ion: \[ E_{(g)}+e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA \label{7.5.1} \] Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with a negative value corresponded to the energy change for an exothermic process, which is one in which heat is released (Figure \(\Page {1}\)). The chlorine atom has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element: \[ \ce{ Cl(g) + e^- \rightarrow Cl^- (g)} \;\;\; =-346\; kJ/mol \label{7.5.2} \] In contrast, beryllium does not form a stable anion, so its effective electron affinity is \[ \ce{ Be(g) + e^- \rightarrow Be^- (g)} \;\;\; EA \ge 0 \label{7.5.3} \] Nitrogen is unique in that it has an electron affinity of approximately zero. Adding an electron neither releases nor requires a significant amount of energy: \[ \ce{ N(g) + e^- \rightarrow N^- (g)} \;\;\; EA \approx 0 \label{7.5.4} \] Generally, electron affinities become more negative across a row of the periodic table. In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as increases, the extra electrons enter orbitals that are increasingly far from the nucleus. Atoms with the largest radii, which have the lowest ionization energies (affinity for their own valence electrons), also have the lowest affinity for an added electron. There are, however, two major exceptions to this trend: The equations for second and higher electron affinities are analogous to those for second and higher ionization energies: \[E_{(g)} + e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA_1 \label{7.5.5} \] \[E^-_{(g)} + e^- \rightarrow E^{2-}_{(g)} \;\;\; \text{energy change=}EA_2 \label{7.5.6} \] As we have seen, the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons. For example, the first electron affinity of oxygen is −141 kJ/mol, but the second electron affinity is +744 kJ/mol: \[O_{(g)} + e^- \rightarrow O^-_{(g)} \;\;\; EA_1=-141 \;kJ/mol \label{7.5.7} \] Thus the formation of a gaseous oxide (\(O^{2−}\)) ion is energetically quite (estimated by adding both steps): \[O_{(g)} + 2e^- \rightarrow O^{2-}_{(g)} \;\;\; EA=+603 \;kJ/mol \label{7.5.9} \] Similarly, the formation of all common dianions (such as \(S^{2−}\)) or trianions (such as \(P^{3−}\)) is energetically unfavorable in the gas phase. While first electron affinities can be negative, positive, or zero, second electron affinities are always positive. Electron Affinity: If energy is required to form both positively charged cations and monatomic polyanions, why do ionic compounds such as \(MgO\), \(Na_2S\), and \(Na_3P\) form at all? The key factor in the formation of stable ionic compounds is the favorable electrostatic interactions between the cations and the anions . Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity? three elements element with most negative electron affinity We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements. Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion? Rb The ( of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table. | 4,970 | 2,685 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/15%3A_Equilibria_of_Other_Reaction_Classes/15.E%3A_Equilibria_of_Other_Reaction_Classes_(Exercises) |
Complete the changes in concentrations for each of the following reactions: \(\begin{alignat}{3}
&\ce{AgI}(s)⟶&&\ce{Ag+}(aq)\,+\,&&\ce{I-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{CaCO3}(s)⟶&&\ce{Ca^2+}(aq)\,+\,&&\ce{CO3^2-}(aq)\\
& &&\underline{\hspace{45px}} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg(OH)2}(s)⟶&&\ce{Mg^2+}(aq)\,+\,&&\ce{2OH-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg3(PO4)2}(s)⟶&&\ce{3Mg^2+}(aq)\,+\,&&\ce{2PO4^3-}(aq)\\
& && &&x\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ca5(PO4)3OH}(s)⟶&&\ce{5Ca^2+}(aq)\,+\,&&\ce{3PO4^3-}(aq)\,+\,&&\ce{OH-}(aq)\\
& &&\underline{\hspace{45px}} &&\underline{\hspace{45px}} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{AgI}(s)⟶&&\ce{Ag+}(aq)\,+\,&&\ce{I-}(aq)\\
& &&x &&\underline{x}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{CaCO3}(s)⟶&&\ce{Ca^2+}(aq)\,+\,&&\ce{CO3^2-}(aq)\\
& &&\underline{x} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg(OH)2}(s)⟶&&\ce{Mg^2+}(aq)\,+\,&&\ce{2OH-}(aq)\\
& &&x &&\underline{2x}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg3(PO4)2}(s)⟶&&\ce{3Mg^2+}(aq)\,+\,&&\ce{2PO4^3-}(aq)\\
& &&\underline{3x} &&2x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ca5(PO4)3OH}(s)⟶&&\ce{5Ca^2+}(aq)\,+\,&&\ce{3PO4^3-}(aq)\,+\,&&\ce{OH-}(aq)\\
& &&\underline{5x} &&\underline{3x} &&x
\end{alignat}\) Complete the changes in concentrations for each of the following reactions: \(\begin{alignat}{3}
&\ce{BaSO4}(s)⟶&&\ce{Ba^2+}(aq)\,+\,&&\ce{SO4^2-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ag2SO4}(s)⟶&&\ce{2Ag+}(aq)\,+\,&&\ce{SO4^2-}(aq)\\
& &&\underline{\hspace{45px}} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Al(OH)3}(s)⟶&&\ce{Al^3+}(aq)\,+\,&&\ce{3OH-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Pb(OH)Cl}(s)⟶&&\ce{Pb^2+}(aq)\,+\,&&\ce{OH-}(aq)\,+\,&&\ce{Cl-}(aq)\\
& &&\underline{\hspace{45px}} &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ca3(AsO4)2}(s)⟶&&\ce{3Ca^2+}(aq)\,+\,&&\ce{2AsO4^3-}(aq)\\
& &&3x &&\underline{\hspace{45px}}
\end{alignat}\) How do the concentrations of Ag and \(\ce{CrO4^2-}\) in a saturated solution above 1.0 g of solid Ag CrO change when 100 g of solid Ag CrO is added to the system? Explain. There is no change. A solid has an activity of 1 whether there is a little or a lot. How do the concentrations of Pb and S change when K S is added to a saturated solution of PbS? What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised? The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO , CuI, PbCO , PbCl , Tl S, KClO ? Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO , CaF , Hg I , MnCO , ZnS, PbS? CaF , MnCO , and ZnS Write the ionic equation for dissolution and the solubility product ( ) expression for each of the following slightly soluble ionic compounds: Write the ionic equation for the dissolution and the expression for each of the following slightly soluble ionic compounds: The gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. The gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a)1.77 × 10 ; 1.6 × 10 ; 2.2 × 10 ; 7.91 × 10 Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF , Hg Cl , PbI , or Sn(OH) . Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: 2 × 10 ; 1.3 × 10 ; 2.27 × 10 ; 2.2 × 10 Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. 7.2 × 10 = [Ag ], [Cl ] = 0.025 Check: \(\dfrac{7.2×10^{−9}\:M}{0.025\:M}×100\%=2.9×10^{−5}\%\), an insignificant change; 2.2 × 10 = [Ca ], [F ] = 0.0013 Check: \(\dfrac{2.25×10^{−5}\:M}{0.00133\:M}×100\%=1.69\%\). This value is less than 5% and can be ignored. 0.2238 = \(\ce{[SO4^2- ]}\); [Ag ] = 2.30 × 10 Check: \(\dfrac{1.15×10^{−9}}{0.2238}×100\%=5.14×10^{−7}\); the condition is satisfied. [OH ] = 2.8 × 10 ; 5.7 × 10 = [Zn ] Check: \(\dfrac{5.7×10^{−12}}{2.8×10^{−3}}×100\%=2.0×10^{−7}\%\); is less than 5% of [OH ] and is, therefore, negligible. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions. Check: \(\dfrac{7.6×10^{−3}}{0.025}×100\%=30\%\) This value is too large to drop . Therefore solve by using the quadratic equation: Check: \(\dfrac{1.7×10^{−3}}{0.0313}×100\%=5.5\%\) This value is too large to drop x, and the entire equation must be solved. \(\ce{[C2O4^2- ]}=3.5×10^{−3}\) Check: \(\dfrac{3.5×10^{−3}}{0.02444}×100\%=14\%\) This value is greater than 5%, so the quadratic equation must be used: Check: \(\dfrac{3.15×10^{−3}}{0.050}×100\%=6.28\%\) This value is greater than 5%, so a more exact method, such as successive approximations, must be used. Explain why the changes in concentrations of the common ions in can be neglected. Explain why the changes in concentrations of the common ions in cannot be neglected. The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change. Calculate the solubility of aluminum hydroxide, Al(OH) , in a solution buffered at pH 11.00. Refer to for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter. CaSO ∙2H O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L. Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract. This use of BaSO is possible because of its low solubility. Calculate the molar solubility of BaSO and the mass of barium present in 1.00 L of water saturated with BaSO . Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 × 10 ) of \(\ce{SO4^2-}\) because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO (“gyp” water) as a result or passing through soil containing gypsum, CaSO •2H O, meet these standards? What is \(\ce{SO4^2-}\) in such water? 4.9 × 10 = \(\ce{[SO4^2- ]}\) = [Ca ]; Since this concentration is higher than 2.60 × 10 , “gyp” water does not meet the standards. Perform the following calculations: The solubility product of CaSO •2H O is 2.4 × 10 . What mass of this salt will dissolve in 1.0 L of 0.010 \(\ce{SO4^2-}\)? Mass (CaSO •2H O) = 0.34 g/L Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products). Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products). [Ag ] = [I ] = 1.2 × 10 ; [Ag ] = 2.86 × 10 , \(\ce{[SO4^2- ]}\) = 1.43 × 10 ; [Mn ] = 2.2 × 10 , [OH ] = 4.5 × 10 ; [Sr ] = 4.3 × 10 , [OH ] = 8.6 × 10 ; [Mg ] = 1.6 × 10 , [OH ] = 3.1 × 10 . The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate for each of the slightly soluble solids indicated: The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate for each of the slightly soluble solids indicated: 2.0 × 10 ; 5.1 × 10 ; 1.35 × 10 ; 1.18 × 10 ; 1.08 × 10 Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for values.) Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for values.) Calculate the concentration of Tl when TlCl just begins to precipitate from a solution that is 0.0250 in Cl . Calculate the concentration of sulfate ion when BaSO just begins to precipitate from a solution that is 0.0758 in Ba . 1.42 × 10 Calculate the concentration of Sr when SrF starts to precipitate from a solution that is 0.0025 in F . Calculate the concentration of \(\ce{PO4^3-}\) when Ag PO starts to precipitate from a solution that is 0.0125 in Ag . 9.2 × 10 Calculate the concentration of F required to begin precipitation of CaF in a solution that is 0.010 in Ca . Calculate the concentration of Ag required to begin precipitation of Ag CO in a solution that is 2.50 × 10 in \(\ce{CO3^2-}\). [Ag ] = 1.8 × 10 What [Ag ] is required to reduce \(\ce{[CO3^2- ]}\) to 8.2 × 10 by precipitation of Ag CO ? What [F ] is required to reduce [Ca ] to 1.0 × 10 by precipitation of CaF ? 6.2 × 10 A volume of 0.800 L of a 2 × 10 - Ba(NO ) solution is added to 0.200 L of 5 × 10 Li SO . Does BaSO precipitate? Explain your answer. Perform these calculations for nickel(II) carbonate. 2.28 L; 7.3 × 10 g Iron concentrations greater than 5.4 × 10 in water used for laundry purposes can cause staining. What [OH ] is required to reduce [Fe ] to this level by precipitation of Fe(OH) ? A solution is 0.010 in both Cu and Cd . What percentage of Cd remains in the solution when 99.9% of the Cu has been precipitated as CuS by adding sulfide? 100% of it is dissolved A solution is 0.15 in both Pb and Ag . If Cl is added to this solution, what is [Ag ] when PbCl begins to precipitate? What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the values given in .) A solution contains 1.0 × 10 mol of KBr and 0.10 mol of KCl per liter. AgNO is gradually added to this solution. Which forms first, solid AgBr or solid AgCl? A solution contains 1.0 × 10 mol of KI and 0.10 mol of KCl per liter. AgNO is gradually added to this solution. Which forms first, solid AgI or solid AgCl? AgI will precipitate first. The calcium ions in human blood serum are necessary for coagulation. Potassium oxalate, K C O , is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC O •H O. It is necessary to remove all but 1.0% of the Ca in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca per 100 mL of serum, what mass of K C O is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the K value for CaC O in serum is the same as in water.) About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca (PO ) . The normal mid range calcium content excreted in the urine is 0.10 g of Ca per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form? 4 × 10 The pH of normal urine is 6.30, and the total phosphate concentration (\(\ce{[PO4^3- ]}\) + \(\ce{[HPO4^2- ]}\) + \(\ce{[H2PO4- ]}\) + [H PO ]) is 0.020 . What is the minimum concentration of Ca necessary to induce kidney stone formation? (See for additional information.) Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions: \(\ce{Mg^2+}(aq)+\ce{Ca(OH)2}(aq)⟶\ce{Mg(OH)2}(s)+\ce{Ca^2+}(aq)\) \(\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)⟶\ce{MgCl2}(s)+\ce{2H2O}(l)\) \(\ce{MgCl2}(l)\xrightarrow{\ce{electrolysis}}\ce{Mg}(s)+\ce{Cl2}(g)\) Sea water has a density of 1.026 g/cm and contains 1272 parts per million of magnesium as Mg ( ) by mass. What mass, in kilograms, of Ca(OH) is required to precipitate 99.9% of the magnesium in 1.00 × 10 L of sea water? 3.99 kg Hydrogen sulfide is bubbled into a solution that is 0.10 in both Pb and Fe and 0.30 in HCl. After the solution has come to equilibrium it is saturated with H S ([H S] = 0.10 ). What concentrations of Pb and Fe remain in the solution? For a saturated solution of H S we can use the equilibrium: \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\) (Hint: The \(\ce{[H3O+]}\) changes as metal sulfides precipitate.) Perform the following calculations involving concentrations of iodate ions: 3.1 × 10 ; [Cu ] = 2.6 × 10 ; \(\ce{[IO3- ]}\) = 5.3 × 10 Calculate the molar solubility of AgBr in 0.035 NaBr ( = 5 × 10 ). How many grams of Pb(OH) will dissolve in 500 mL of a 0.050- PbCl solution ( = 1.2 × 10 )? 1.8 × 10 g Pb(OH) Use the from the earlier Link to Learning to complete the following exercise:. Using 0.01 g CaF , give the K values found in a 0.2- solution of each of the salts. Discuss why the values change as you change soluble salts. How many grams of Milk of Magnesia, Mg(OH) ( ) (58.3 g/mol), would be soluble in 200 mL of water. = 7.1 × 10 . Include the ionic reaction and the expression for in your answer. ( = 1 × 10 = [H O ,OH ]) \[\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}+\ce{2OH-} \] \[K_\ce{sp}=\ce{[Mg^2+,OH- ]^2}\] \[1.14 × 10 g Mg(OH) Two hypothetical salts, LM and LQ, have the same molar solubility in H O. If for LM is 3.20 × 10 , what is the value for LQ? Which of the following carbonates will form first? Which of the following will form last? Explain. SrCO will form first, since it has the smallest value it is the least soluble. BaCO will be the last to precipitate, it has the largest value. How many grams of Zn(CN) ( ) (117.44 g/mol) would be soluble in 100 mL of H O? Include the balanced reaction and the expression for in your answer. The value for Zn(CN) ( ) is 3.0 × 10 . Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water? when the amount of solid is so small that a saturated solution is not produced Explain why the addition of NH or HNO to a saturated solution of Ag CO in contact with solid Ag CO increases the solubility of the solid. Calculate the cadmium ion concentration, [Cd ], in a solution prepared by mixing 0.100 L of 0.0100 Cd(NO ) with 1.150 L of 0.100 NH ( ). 2.35 × 10 Explain why addition of NH or HNO to a saturated solution of Cu(OH) in contact with solid Cu(OH) increases the solubility of the solid. Sometimes equilibria for complex ions are described in terms of dissociation constants, . For the complex ion \(\ce{AlF6^3-}\) the dissociation reaction is: \[\ce{AlF6^3- ⇌ Al^3+ + 6F-}\) and \(K_\ce{d}=\ce{\dfrac{[Al^3+,F- ]^6}{[AlF6^3- ]}}=2×10^{−24}\] Calculate the value of the formation constant, , for \(\ce{AlF6^3-}\). 5 × 10 Using the value of the formation constant for the complex ion \(\ce{Co(NH3)6^2+}\), calculate the dissociation constant. Using the dissociation constant, = 7.8 × 10 , calculate the equilibrium concentrations of Cd and CN in a 0.250- solution of \(\ce{Cd(CN)4^2-}\). [Cd ] = 9.5 × 10 ; [CN ] = 3.8 × 10 Using the dissociation constant, = 3.4 × 10 , calculate the equilibrium concentrations of Zn and OH in a 0.0465- solution of \(\ce{Zn(OH)4^2-}\). Using the dissociation constant, = 2.2 × 10 , calculate the equilibrium concentrations of Co and NH in a 0.500- solution of \(\ce{Co(NH3)6^3+}\). [Co ] = 3.0 × 10 ; [NH ] = 1.8 × 10 Using the dissociation constant, = 1 × 10 , calculate the equilibrium concentrations of Fe and CN in a 0.333 M solution of \(\ce{Fe(CN)6^3-}\). Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 × 10 mol of silver cyanide, AgCN. 1.3 g Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 × 10 mol of silver bromide. A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na S O •5H O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as \(\ce{Ag(S2O3)2^3-}\) ( = 4.7 × 10 )? 0.80 g We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H ) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each: (a) ; (b) ; (c) ; (d) ; (e) Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each: Using Lewis structures, write balanced equations for the following reactions: (a) ; \(\ce{H3O+ + CH3- ⟶ CH4 + H2O}\) ; \(\ce{CaO + SO3 ⟶ CaSO4}\) ; \(\ce{NH4+ + C2H5O- ⟶ C2H5OH + NH3}\) Calculate \(\ce{[HgCl4^2- ]}\) in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100- HgCl solution. In a titration of cyanide ion, 28.72 mL of 0.0100 AgNO is added before precipitation begins. [The reaction of Ag with CN goes to completion, producing the \(\ce{Ag(CN)2-}\) complex.] Precipitation of solid AgCN takes place when excess Ag is added to the solution, above the amount needed to complete the formation of \(\ce{Ag(CN)2-}\). How many grams of NaCN were in the original sample? 0.0281 g What are the concentrations of Ag , CN , and \(\ce{Ag(CN)2-}\) in a saturated solution of AgCN? In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 °C) is a strong acid. In liquid HF, HNO acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F ion (for example, BF or SbF ). Write balanced chemical equations for the reaction of pure HNO with pure HF and of pure HF with BF . \(\ce{HNO3}(l)+\ce{HF}(l)⟶\ce{H2NO3+}+\ce{F-}\); \(\ce{HF}(l)+\ce{BF3}(g)⟶\ce{H+}+\ce{BF4}\) The simplest amino acid is glycine, H NCH CO H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH , and a carboxylic acid group, –CO H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH CO H, and the base strength of the amino group is slightly greater than that of ammonia, NH . Write the Lewis structures of the ions that form when glycine is dissolved in 1 HCl and in 1 KOH. Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH and \(\ce{−CO2-}\) groups.) Boric acid, H BO , is not a Brønsted-Lowry acid but a Lewis acid. \(\ce{H3BO3 + H2O ⟶ H4BO4- + H+}\); The electronic and molecular shapes are the same—both tetrahedral. The tetrahedral structure is consistent with hybridization. A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium? Calculate the equilibrium concentration of Ni in a 1.0- solution [Ni(NH ) ](NO ) . 0.014 Calculate the equilibrium concentration of Zn in a 0.30- solution of \(\ce{Zn(CN)4^2-}\). Calculate the equilibrium concentration of Cu in a solution initially with 0.050 Cu and 1.00 NH . 1.0 × 10 Calculate the equilibrium concentration of Zn in a solution initially with 0.150 Zn and 2.50 CN . Calculate the Fe equilibrium concentration when 0.0888 mole of K [Fe(CN) ] is added to a solution with 0.0.00010 CN . 9 × 10 Calculate the Co equilibrium concentration when 0.100 mole of [Co(NH ) ](NO ) is added to a solution with 0.025 NH . Assume the volume is 1.00 L. The equilibrium constant for the reaction \(\ce{Hg^2+}(aq)+\ce{2Cl-}(aq)⇌\ce{HgCl2}(aq)\) is 1.6 × 10 . Is HgCl a strong electrolyte or a weak electrolyte? What are the concentrations of Hg and Cl in a 0.015- solution of HgCl ? 6.2 × 10 = [Hg ]; 1.2 × 10 = [Cl ]; The substance is a weak electrolyte because very little of the initial 0.015 HgCl dissolved. Calculate the molar solubility of Sn(OH) in a buffer solution containing equal concentrations of NH and \(\ce{NH4+}\). Calculate the molar solubility of Al(OH) in a buffer solution with 0.100 NH and 0.400 \(\ce{NH4+}\). [OH ] = 4.5 × 10 ; [Al ] = 2.1 × 10 (molar solubility) What is the molar solubility of CaF in a 0.100- solution of HF? for HF = 7.2 × 10 . What is the molar solubility of BaSO in a 0.250- solution of NaHSO ? for \(\ce{HSO4-}\) = 1.2 × 10 . What is the molar solubility of Tl(OH) in a 0.10- solution of NH ? What is the molar solubility of Pb(OH) in a 0.138- solution of CH NH ? A solution of 0.075 CoBr is saturated with H S ([H S] = 0.10 ). What is the minimum pH at which CoS begins to precipitate? \(\ce{CoS}(s)⇌\ce{Co^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.5×10^{−27}\) \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\) A 0.125- solution of Mn(NO ) is saturated with H S ([H S] = 0.10 ). At what pH does MnS begin to precipitate? \(\ce{MnS}(s)⇌\ce{Mn^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.3×10^{−22}\) \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\) 3.27 Calculate the molar solubility of BaF in a buffer solution containing 0.20 HF and 0.20 NaF. Calculate the molar solubility of CdCO in a buffer solution containing 0.115 Na CO and 0.120 NaHCO To a 0.10- solution of Pb(NO ) is added enough HF( ) to make [HF] = 0.10 . Calculate the concentration of Cd resulting from the dissolution of CdCO in a solution that is 0.010 in H CO . 1 × 10 Both AgCl and AgI dissolve in NH . Calculate the volume of 1.50 CH CO H required to dissolve a precipitate composed of 350 mg each of CaCO , SrCO , and BaCO . 0.0102 L (10.2 mL) Even though Ca(OH) is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH) ? What mass of NaCN must be added to 1 L of 0.010 Mg(NO ) in order to produce the first trace of Mg(OH) ? 5 × 10 g Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.) The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: \(\ce{MgF2}(s)⇌\ce{Mg^2+}(aq)+\ce{2F-}(aq)\) In a saturated solution of MgF at 18 °C, the concentration of Mg is 1.21 × 10 . The equilibrium is represented by the preceding equation. = [Mg ,F ] = (1.21 × 10 )(2 × 1.21 × 10 ) = 7.09 × 10 ; 7.09 × 10 Determine the concentration of Mg and F that will be present in the final volume. Compare the value of the ion product [Mg ,F ] with . If this value is larger than , precipitation will occur. 0.1000 L × 3.00 × 10 Mg(NO ) = 0.3000 L × Mg(NO ) Mg(NO ) = 1.00 × 10 0.2000 L × 2.00 × 10 NaF = 0.3000 L × NaF NaF = 1.33 × 10 ion product = (1.00 × 10 )(1.33 × 10 ) = 1.77 × 10 This value is smaller than , so no precipitation will occur. MgF is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Chatelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic. Which of the following compounds, when dissolved in a 0.01- solution of HClO , has a solubility greater than in pure water: CuCl, CaCO , MnS, PbBr , CaF ? Explain your answer. Which of the following compounds, when dissolved in a 0.01- solution of HClO , has a solubility greater than in pure water: AgBr, BaF , Ca (PO ) , ZnS, PbI ? Explain your answer. BaF , Ca (PO ) , ZnS; each is a salt of a weak acid, and the \(\ce{[H3O+]}\) from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations What is the effect on the amount of solid Mg(OH) that dissolves and the concentrations of Mg and OH when each of the following are added to a mixture of solid Mg(OH) and water at equilibrium? What is the effect on the amount of CaHPO that dissolves and the concentrations of Ca and \(\ce{HPO4-}\) when each of the following are added to a mixture of solid CaHPO and water at equilibrium? Effect on amount of solid CaHPO , [Ca ], [OH ]: increase, increase, decrease; decrease, increase, decrease; no effect, no effect, no effect; decrease, increase, decrease; increase, no effect, no effect Identify all chemical species present in an aqueous solution of Ca (PO ) and list these species in decreasing order of their concentrations. (Hint: Remember that the \(\ce{PO4^3-}\) ion is a weak base.) A volume of 50 mL of 1.8 NH is mixed with an equal volume of a solution containing 0.95 g of MgCl . What mass of NH Cl must be added to the resulting solution to prevent the precipitation of Mg(OH) ? 7.1 g | 27,632 | 2,686 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX11._Protein_Modifications |
After proteins are synthesized via transcription of DNA, their structures may be modified in a number of ways. This process, in general, is referred to as "post-translational modification." It might involve the attachment of small organic groups to the protein, attachment of other biomolecules, or even the complete alteration of an amino acid residue into another form, so that the protein actually contains a residue that does not exist in the basic beginner's amino acid alphabet. Minor modifications of proteins can be very important in regulating enzyme activity. Modest changes in structure can have a big impact on the conformation of the protein, which in turn might open or close access into the enzyme's interior or alter the shape of the active site. Alternatively, these changes might affect the location of a protein by moving it from the cell membrane into the cell interior or vice versa. They might also result in the formation or dissociation of supramolecular assemblies, by increasing or decreasing the attraction between two separate biomolecules. The physical reasons for these structure-property changes are based in simple intermolecular attractions or, in the case of a large, complicated protein, intramolecular attractions between different parts of the same protein. A hydrogen bonding interaction or ion-dipole interaction that holds the protein in one conformation may be disrupted if a key player in that interaction is suddenly masked. For example, a cationic ammonium side chain may react to become an amide group; the absence of a full positive charge on this group significantly alters its intermolecular attractions. There are probably hundreds of ways in which proteins are modified. We will look at just a few different modifications that occur in proteins, some of which are closly tied to carboxylate substitution. Note that any of these modifications might act to turn an enzyme "off" or "on"; the details depend on the individual case. Acetylation is the attachment of an acetyl group, CH C=O, to another compound such as an amino acid residue. Serine and threonine groups might be acetylated to make esters, cysteine side chains might be acetylated to make thioesters, or amino groups. Frequently, acetylation refers specifically to reaction at an amino group, such as the N-terminus of a protein or in a lysine side chain, forming an amide. The source of the acetyl group is acetyl coenzyme A (AcSCoA, below), the thioester workhorse of the cell. This structure may seem a little bit complicated, but at the most basic level it is just a carboxyl electrophile (CH C=O) attached to a very large thiolate leaving group. Of course, lysine side chains and the N-termini of proteins are usually in a protonated state under biological conditions. That means that there must be a deprotonation step along the reaction pathway. One example of the role of acetylation is seen in histones. Histones are proteins found in , the mixture of DNA and proteins in the cell nucleus. The already-coiled DNA helix is further wrapped around histones, bundling it up into a smaller package. DNA has many, many negative charges all along it because the phosphate units in its phosphate-sugar copolymer backbone are negatively charged at typical biological pH. The DNA interacts easily with the histones because they are rich in positively charged lysine residues. Storing DNA in smaller bundles lets you keep more junk in your nucleus, but you don't want to just let the DNA sit there. Once in a while you want to take it out and do something with it, but how can you do that when it's stuck to those darned histones? The answer is, you just have to turn off the histones' force field. Get rid of that positive charge, and the DNA won't be stuck anymore. It's easy to do that by acetylating the histones. This change in charge results because, although amines are easily protonated, amides are not. Thus, amines are likely to carry positive charge under biological conditions, whereas amides are likely to be neutral. Explain why an amine is easily protonated but an amide is not. Provide a mechanism for the acetylation of a lysine side chain. Assume the presence of histidine and its conjugate acid, histidinium ion; these species are common proton shuttles in biological reactions. Serine, tyrosine and threonine residues are frequently modified by phosphorylation, which is the formation of a phosphate ester. The phosphorylation of these groups results in a change from a neutral side chain to an anionic side chain. The phosphate donor in these cases is another ubiquitous cellular performer, ATP. The complete structure of ATP is shown here. It is a little less complicated than AcSCoA. Once again, at the most basic level of reactivity, it provides a phosphate electrophile, attached to a phosphate leaving group. The anionic nature of the phosphate may not be apparent in the above drawing of ATP, but in reality, at biological pH, a phosphate would be deprotonated. Phosphates actually undergo multiple equilibria involving the loss of two protons. Provide a mechanism for phosphorylation of serine in the presence of an appropriate biological proton shuttle. There are other modifications that occur via different mechanisms. In a farnesylation, reaction does not even take place at a carbonyl. It does not take place at a phosphonate or a sulfonate. Sulfur is the nucleophile in the farnesylation shown above. Identify the electrophile and show a mechanism with curved arrows. Palmitoylation is closely related to acetylation, but the electrophile in this case is a palmitoyl group instead of an acetyl. , | 5,653 | 2,687 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.02%3A_The_VSEPR_Model |
The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds. The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach. We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the , ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form , which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figures \(\Page {1}\) and \(\Page {2}\). In the VSEPR model, the molecule or polyatomic ion is given an AX E designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and and are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the and interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the in a molecule or polyatomic ion. This VESPR procedure is summarized as follows: We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure \(\Page {2}\) and Figure \(\Page {3}\), which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups. Our first example is a molecule with two bonded atoms and no lone pairs of electrons, \(BeH_2\). 1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is 3. Both groups around the central atom are bonding pairs (BP). Thus BeH is designated as AX . 4. From Figure \(\Page {3}\) we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH is . 1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is 2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH , the arrangement that minimizes repulsions places the groups 180° apart. 3. Once again, both groups around the central atom are bonding pairs (BP), so CO is designated as AX . 4. VSEPR only recognizes groups around the atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO is linear (Figure \(\Page {3}\)). The structure of \(\ce{CO2}\) is shown in Figure \(\Page {1}\). 1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is 3. All electron groups are bonding pairs (BP), so the structure is designated as AX . 4. From Figure \(\Page {3}\) we see that with three bonding pairs around the central atom, the molecular geometry of BCl is , as shown in Figure \(\Page {2}\). 1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as 3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX . 4. We see from Figure \(\Page {3}\) that the molecular geometry of CO is trigonal planar with bond angles of 120°. In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time. 1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below. 3. There are two bonding pairs and one lone pair, so the structure is designated as AX E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure \(\Page {4}\)). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO , we have one BP–BP interaction and two LP–BP interactions. 4. The molecular geometry is described only by the positions of the nuclei, by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is , or , which can be viewed as a trigonal planar arrangement with a missing vertex (Figures \(\Page {2}\) and \(\Page {3}\)). The O-S-O bond angle is expected to be 120° because of the extra space taken up by the lone pair. As with SO , this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom. Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH O (AX ), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°). One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions. 1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is 2. There are four electron groups around the central atom. As shown in Figure \(\Page {2}\), repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°. 3. All electron groups are bonding pairs, so the structure is designated as AX . 4. With four bonding pairs, the molecular geometry of methane is (Figure \(\Page {3}\)). 1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure 2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. 3. With three bonding pairs and one lone pair, the structure is designated as AX E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. 4. There are three nuclei and one lone pair, so the molecular geometry is . In essence, this is a tetrahedron with a vertex missing (Figure \(\Page {3}\)). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure \(\Page {3}\) and Figure \(\Page {4}\)). 1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure 3. With two bonding pairs and two lone pairs, the structure is designated as AX E with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles. 4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is , or , with an H–O–H angle that is even less than the H–N–H angles in NH , as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices. In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups. 1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl is 3. All electron groups are bonding pairs, so the structure is designated as AX . There are no lone pair interactions. 4. The molecular geometry of PCl is , as shown in Figure \(\Page {3}\). The molecule has three atoms in a plane in positions and two atoms above and below the plane in positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example. 1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is With an expanded valence, this species is an exception to the octet rule. 2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure \(\Page {2}\). 3. We designate SF as AX E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the axial position, we have three LP–BP repulsions at 90°. If we place it in the equatorial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the . We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair. At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions. 4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a . The F –S–F angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane. 1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is Once again, we have a compound that is an exception to the octet rule. 2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid. 3. With three bonding pairs and two lone pairs, the structural designation is AX E with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°: Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons. 4. The three nuclei in BrF determine its molecular structure, which is described as . This is essentially a trigonal bipyramid that is missing two equatorial vertices. The F –Br–F angle is 172°, less than 180° because of LP–BP repulsions (Figure \(\Page {2}\).1). 1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is 2. There are five electron groups about the central atom in I , two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid. 3. With two bonding pairs and three lone pairs, I has a total of five electron pairs and is designated as AX E . We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions. The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles. 4. With three nuclei and three lone pairs of electrons, the molecular geometry of I is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected. Six electron groups form an , a polyhedron made of identical equilateral triangles and six identical vertices (Figure \(\Page {2}\).) 1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is With an expanded valence, this species is an exception to the octet rule. 2. There are six electron groups around the central atom, each a bonding pair. We see from Figure \(\Page {2}\) that the geometry that minimizes repulsions is . 3. With only bonding pairs, SF is designated as AX . All positions are chemically equivalent, so all electronic interactions are equivalent. 4. There are six nuclei, so the molecular geometry of SF is octahedral. 1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is With its expanded valence, this species is an exception to the octet rule. 2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure: 3. With five bonding pairs and one lone pair, BrF is designated as AX E; it has a total of six electron pairs. The BrF structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all F –Br–F angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs. 4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is . The F –B–F angles are 85.1°, less than 90° because of LP–BP repulsions. 1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is 2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is 3. ICl is designated as AX E and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles. 4. With five nuclei, the ICl4− ion forms a molecular structure that is , an octahedron with two opposite vertices missing. The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure \(\Page {6}\). Using the VSEPR model, predict the molecular geometry of each molecule or ion. two chemical species molecular geometry All electron groups are bonding pairs, so PF is designated as AX . Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal. The PF molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal. There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH , repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. With three bonding pairs and one lone pair, the structure is designated as AX E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. There are three nuclei and one lone pair, so the molecular geometry is , in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions: Using the VSEPR model, predict the molecular geometry of each molecule or ion. trigonal pyramidal octahedral linear Predict the molecular geometry of each molecule. two chemical compounds molecular geometry Use the strategy given in Example\(\Page {1}\). There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid. From B, XeF is designated as AX E and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial: The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I . All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle. With two nuclei about the central atom, the molecular geometry of XeF is linear. It is a trigonal bipyramid with three missing equatorial vertices. There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other. From B we designate SnCl as AX E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions. With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl is bent, like SO , but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing. Predict the molecular geometry of each molecule. trigonal planar square planar The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AX E fragments. We will demonstrate with methyl isocyanate (CH –N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops. We can treat methyl isocyanate as linked AX E fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX . We can therefore predict the CH –N portion of the molecule to be roughly tetrahedral, similar to methane: The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair: Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°. The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure: Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO , so its geometry, like that of CO , is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules. Use the VSEPR model to predict the molecular geometry of propyne (H C–C≡CH), a gas with some anesthetic properties. chemical compound molecular geometry Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(\Page {3}\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole. Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°. Predict the geometry of allene (H C=C=CH ), a compound with narcotic properties that is used to make more complex organic molecules. The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°. You previously learned how to calculate the of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are ; they possess both a and a . The dipole moment of a molecule is therefore the of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO , a linear molecule (Figure \(\Page {8a}\)). Each C–O bond in CO is polar, yet experiments show that the CO molecule has no dipole moment. Because the two C–O bond dipoles in CO are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO molecule has no dipole moment even though it has a substantial separation of charge. In contrast, the H O molecule is not linear (Figure \(\Page {8b}\)); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H O to hydrogen-bond to other polarized or charged species, including other water molecules. Other examples of molecules with polar bonds are shown in Figure \(\Page {9}\). In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment. Molecules with asymmetrical charge distributions have a net dipole moment. Which molecule(s) has a net dipole moment? three chemical compounds net dipole moment For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment. Which molecule(s) has a net dipole moment? \(\ce{CH3Cl}\) and \(\ce{XeO3}\) Lewis electron structures give no information about , the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AX E designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and and are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the . From this we can describe the . The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds. Molecules with polar covalent bonds can have a , an asymmetrical distribution of charge that results in a tendency for molecules to align themselves in an applied electric field. Any diatomic molecule with a polar covalent bond has a dipole moment, but in polyatomic molecules, the presence or absence of a net dipole moment depends on the structure. For some highly symmetrical structures, the individual bond dipole moments cancel one another, giving a dipole moment of zero. | 28,146 | 2,688 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.1%3A_Periodic_Trends_in_Bonding |
Periodic trends affect bonding, because of how the elements are arranged on the periodic table. For example elements can be arranged by their electronegative, electron affinity, atomic radius, or ionization energy. Electronegative is the atoms ability to attract other bonded atoms. Electron affinity is an atoms ability to attract another atom. The atomic radius is the radius of an elements atom. Ionization energy is the energy it takes to remove an atom from another atom. Other periodic trends are when the attraction of the atoms for the pair of bonding electrons is different, this is polar covalent bonds. Properties in compounds are used to determine the type of bonding and structure, not just the elements being used. These different properties help group elements to make them either more available or less available for bonding. It may seem counterintuitive to say that HF is the weakest hydrohalic acid because fluorine has the highest electronegativity. However, the H-F bond is very strong; if Another important trend to note in main group chemistry is the chemical similarity between the lightest element of one group and the element immediately below and to the right of it in the next group, a phenomenon known as the (Figure \(\Page {1}\) ) There are, for example, significant similarities between the chemistry of Li and Mg, Be and Al, and B and Si. Both BeCl and AlCl have substantial covalent character, so they are somewhat soluble in nonpolar organic solvents. In contrast, although Mg and Be are in the same group, MgCl behaves like a typical ionic halide due to the lower electronegativity and larger size of magnesium. Oxides are binary compounds of oxygen with another element, e.g., CO , SO , CaO, CO, ZnO, BaO , H O, etc. These are termed as oxides because here, oxygen is in combination with only one element. Based on their acid-base characteristics oxides are classified as acidic, basic, or neural: There are different properties which help distinguish between the three types of oxides. The term ("without water") refers to compounds that assimilate H O to form either an acid or a base upon the addition of water. Acidic oxides are the oxides of non-metals ( ) and these form acids with water: \[ SO_2 + H_2O \rightarrow H_2SO_3 \label{1}\] \[ SO_3 + H_2O \rightarrow H_2SO_4 \label{2}\] \[ CO_2 + H_2O \rightarrow H_2CO_3 \label{3}\] Acidic oxides are known as acid anhydrides (e.g., sulfur dioxide is sulfurous anhydride and sulfur trioxide is sulfuric anhydride) and when combined with bases, they produce salts, e.g., \[ SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O \label{4}\] Generally and elements form bases called base anhydrides or basic oxides, e.g., \[ K_2O \; (s) + H_2O \; (l) \rightarrow 2KOH \; (aq) \label{5}\] Basic oxides are the oxides of metals. If soluble in water, they react with water to produce hydroxides (alkalies) e.g., \[ CaO + H_2O \rightarrow Ca(OH)_2 \label{6}\] \[ MgO + H_2O \rightarrow Mg(OH)_2 \label{7}\] \[ Na_2O + H_2O \rightarrow 2NaOH \label{8}\] These metallic oxides are known as basic anhydrides. They react with acids to produce salts, e.g., \[ MgO + 2HCl \rightarrow MgCl_2 + H_2O \label{9}\] \[ Na_2O + H_2SO_4 \rightarrow Na_2SO_4 + H_2O \label{10}\] An amphoteric solution is a substance that can chemically react as either acid or base. For example, when HSO reacts with water it will make both hydroxide and ions: \[ HSO_4^- + H_2O \rightarrow SO_4^{2^-} + H_3O^+ \label{11}\] \[ HSO_4^- + H_2O \rightarrow H_2SO_4 + OH^- \label{12}\] Amphoteric oxides are metallic oxides, which show both basic as well as acidic properties. When they react with an acid, they produce salt and water, showing basic properties. While reacting with alkalies they form salt and water showing acidic properties, e.g., \[ZnO + 2HCl \rightarrow \underset{\large{zinc\:chloride}}{ZnCl_2}+H_2O\,(basic\: nature) \label{13}\] \[ZnO + 2NaOH \rightarrow \underset{\large{sodium\:zincate}}{Na_2ZnO_2}+H_2O\,(acidic\: nature) \label{14}\] \[Al_2O_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2O\,(basic\: nature) \label{15}\] \[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2+H_2O\,(acidic\: nature) \label{16}\] Amphoteric oxides have both acidic and basic properties. A common example of an amphoteric oxide is aluminum oxide. In general, amphoteric oxides form with metalloids. (see chart below for more detail). Example with acidic properties: \[ Al_2O_3 + H_2O \rightarrow 2 Al(OH) + 2H^+ \label{17}\] Example with basic properties: \[ Al_2O_3 + H_2O \rightarrow 2Al^{3+} + 3OH^- \label{18}\] Neutral oxides show neither basic nor acidic properties and hence do not form salts when reacted with acids or bases, e.g., carbon monoxide (CO); nitrous oxide (N O); nitric oxide (NO), etc., are neutral oxides. : metals react rapidly with oxygen to produce several different ionic oxides, usually in the form of \( M_2O \). With the oyxgen exhibiting an of -2. \[ 4 Li + O_2 \rightarrow 2Li_2O \label{19} \] : Often and reacts with oxygen to produce the peroxide, \( M_2O_2 \). with the oxidation number of the oxygen equal to -1. \[ H_2 + O_2 \rightarrow H_2O_2 \label{20}\] : Often , Rubidium, and Cesium react with excess oxygen to produce the superoxide, \( MO_2 \). with the oxidation number of the oxygen equal to -1/2. \[ Cs + O_2 \rightarrow CsO_2 \label{21}\] A peroxide is a metallic oxide which gives hydrogen peroxide by the action of dilute acids. They contain more oxygen than the corresponding basic oxide, e.g., sodium, calcium and barium peroxides. \[BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 \label{22}\] \[Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2 \label{23}\] Dioxides like PbO and MnO also contain higher percentage of oxygen like peroxides and have similar molecular formulae. These oxides, however, do not give hydrogen peroxide by action with dilute acids. Dioxides on reaction with concentrated HCl yield Cl and on reacting with concentrated H SO yield O . \[PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O \label{24}\] \[2PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O + O_2 \label{25}\] Compound oxides are metallic oxides that behave as if they are made up of two oxides, one that has a lower oxidation and one with a higher oxidation of the same metal, e.g., \[\textrm{Red lead: } Pb_3O_4 = PbO_2 + 2PbO \label{26}\] \[\textrm{Ferro-ferric oxide: } Fe_3O_4 = Fe_2O_3 + FeO \label{27}\] On treatment with an acid, compound oxides give a mixture of salts. \[\underset{\text{Ferro-ferric oxide}}{Fe_3O_4} + 8HCl \rightarrow \underset{\text{ferric chloride}}{2FeCl_3} + \underset{\text{ferrous chloride}}{FeCl_2} + 4H_2O \label{28}\] Oxides can be generated via multiple reactions. Below are a few. Many metals and non-metals burn rapidly when heated in oxygen or air, producing their oxides, e.g., \[2Mg + O_2 \xrightarrow{Heat} 2MgO\] \[2Ca + O_2 \xrightarrow{Heat} 2CaO\] \[S + O_2 \xrightarrow{Heat} SO_2\] \[P_4 + 5O_2 \xrightarrow{Heat} 2P_2O_5\] At higher temperatures, oxygen also reacts with many compounds forming oxides, e.g., \[2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2\] \[2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2\] \[C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\] \[CaCO_3 \xrightarrow{\Delta} CaO + CO_2\] \[2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2\] \[Cu(OH)_2 \xrightarrow{\Delta} CuO + H_2O\] \[2Cu + 8HNO_3 \xrightarrow{Heat} 2CuO + 8NO_2 + 4H_2O + O_2\] \[Sn + 4HNO_3 \xrightarrow{Heat} SnO_2 + 4NO_2 + 2H_2O\] \[C + 4HNO_3 \rightarrow CO_2 + 4NO_2 + 2H_2O\] The oxides of elements in a period become progressively more acidic as one goes from left to right in a period of the periodic table. For example, in third period, the behavior of oxides changes as follows: \[\underset{\large{Basic}}{\underbrace{Na_2O,\: MgO}}\hspace{20px}
\underset{\large{Amphoteric}}{\underbrace{Al_2O_3,\: SiO_2}}\hspace{20px}
\underset{\large{Acidic}}{\underbrace{P_4O_{10},\: SO_3,\:Cl_2O_7}}\hspace{20px}\] If we take a closer look at a specific period, we may better understand the acid-base properties of oxides. It may also help to examine the , but it is not necessary (Table \(\Page {1}\)). Metal oxides on the left side of the periodic table produce basic solutions in water (e.g. Na O and MgO). Non-metal oxides on the right side of the periodic table produce acidic solutions (e.g. Cl O, SO , P O ). There is a trend within acid-base behavior: basic oxides are present on the left side of the period and acidic oxides are found on the right side. Be B Aluminum oxide shows acid and basic properties of an oxide, it is amphoteric. Thus Al O entails the marking point at which a change over from a basic oxide to acidic oxide occurs. It is important to remember that the trend only applies for oxides in their highest oxidation states. The individual element must be in its highest possible oxidation state because the trend does not follow if all oxidation states are included. Notice how the amphoteric oxides (shown in blue) of each period signify the change from basic to acidic oxides. | 9,040 | 2,689 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.5%3A_Molecular_Orbital_Theory |
delocalized Previously, we described the electrons in isolated atoms as having certain spatial distributions, called , each with a particular . Just as the positions and energies of electrons in can be described in terms of (AOs), the positions and energies of electrons in can be described in terms of molecular orbitals (MOs) —a spatial distribution of electrons that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory is a approach to bonding. Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins. We begin our discussion of molecular orbitals with the simplest molecule, H , formed from two isolated hydrogen atoms, each with a 1 electron configuration. As discussed , electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called . The 1 atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the of the two H 1 wave functions, and the other produced by taking their : \[ \begin{matrix}
MO(1)= & AO(atom\; A) & +& AO(atomB) \\
MO(1)= & AO(atom\; A) & -&AO(atomB)
\end{matrix} \label{Eq1} \] The molecular orbitals created from Equation \(\ref{Eq1}\) are called linear combinations of atomic orbitals (LCAOs) . A molecule must have as many molecular orbitals as there are atomic orbitals. Adding two atomic orbitals corresponds to interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is . The molecular orbital corresponding to the sum of the two H 1 orbitals is called a σ combination (pronounced “sigma one ess”) (part (a) and part (b) in Figure \(\Page {1}\)). In a sigma (σ) orbital, , the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1 denotes the atomic orbitals from which the molecular orbital was derived: \[ \sigma _{1s} \approx 1s\left ( A \right) + 1s\left ( B \right) \label{Eq2} \] Conversely, subtracting one atomic orbital from another corresponds to interference between two waves, which reduces their intensity and causes a in the internuclear electron probability density (part (c) and part (d) in Figure \(\Page {1}\)). The resulting pattern contains a where the electron density is zero. The molecular orbital corresponding to the difference is called \( \sigma _{1s}^{*} \) (“sigma one ess star”). In a sigma star (σ*) orbital , there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis: \[ \sigma _{1s}^{\star } \approx 1s\left ( A \right) - 1s\left ( B \right) \label{Eq3} \] A molecule must have as many molecular orbitals as there are atomic orbitals. The electron density in the σ molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ orbital represents a bonding molecular orbital. . In contrast, electrons in the \( \sigma _{1s}^{\star } \) orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the \( \sigma _{1s}^{\star } \) orbital is an antibonding molecular orbital (a . Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not. Because electrons in the σ orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ molecular orbital has a energy than either of the hydrogen 1 atomic orbitals. Conversely, electrons in the \( \sigma _{1s}^{\star } \) orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1 atomic orbitals. Consequently, the \( \sigma _{1s}^{\star } \) molecular orbital has a energy than either of the hydrogen 1 atomic orbitals. The σ (bonding) molecular orbital is relative to the 1 atomic orbitals, and the \( \sigma _{1s}^{\star } \) (antibonding) molecular orbital is . The relative energy levels of these orbitals are shown in the energy-level diagram in Figure \(\Page {2}\) A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable). To describe the bonding in a homonuclear diatomic molecule such as H , we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (Figure \(\Page {2}\)). We fill the orbitals according to the and : each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds. In the Lewis electron structures, the number of electron pairs holding two atoms together was called the . In the molecular orbital approach, bond order is defined as one-half the number of bonding electrons: \[ bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{Eq4} \] To calculate the bond order of H , we see from Figure \(\Page {2}\) that the σ (bonding) molecular orbital contains two electrons, while the \( \sigma _{1s}^{\star } \) (antibonding) molecular orbital is empty. The bond order of H is therefore \[ \dfrac{2-0}{2}=1 \label{Eq5} \] This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3. We can use energy-level diagrams such as the one in Figure \(\Page {2}\) to describe the bonding in other pairs of atoms and ions where = 1, such as the H ion, the He ion, and the He molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the or . Part (a) in Figure \(\Page {3}\) shows the energy-level diagram for the H ion, which contains two protons and only one electron. The single electron occupies the σ bonding molecular orbital, giving a (σ ) electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is (1-0)/2=1/2 Because the bond order is greater than zero, the H ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H . With a bond order of only 1/2 the bond in H should be weaker than in the H molecule, and the H–H bond should be longer. As shown in Table \(\Page {1}\), these predictions agree with the experimental data. Part (b) in Figure \(\Page {3}\) is the molecular orbital energy-level diagram for He . This ion has a total of three valence electrons. Because the first two electrons completely fill the σ molecular orbital, the Pauli principle states that the third electron must be in the \( \sigma _{1s}^{\star} \) antibonding orbital, giving a \( \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{1} \) electron configuration. This electron configuration gives a bond order of (2-1)/2=1/2. As with H , the He ion should be stable, but the He–He bond should be weaker and longer than in H . In fact, the He ion can be prepared, and its properties are consistent with our predictions (Table \(\Page {1}\)). Finally, we examine the He molecule, formed from two He atoms with 1 electron configurations. Part (c) in Figure \(\Page {3}\) is the molecular orbital energy-level diagram for He . With a total of four valence electrons, both the σ bonding and \( \sigma _{1s}^{\star } \) antibonding orbitals must contain two electrons. This gives a \( \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{1} \) electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He molecule has no net bond and is not a stable species. Experiments show that the He molecule is actually stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions. In molecular orbital theory, . Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H ion. Molecular Orbital Theory: In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons. Use a molecular orbital energy-level diagram, such as those in Figure \(\Page {3}\), to predict the bond order in the He ion. Is this a stable species? chemical species molecular orbital energy-level diagram, bond order, and stability Two He 1 atomic orbitals combine to give two molecular orbitals: a σ bonding orbital at lower energy than the atomic orbitals and a \( \sigma _{1s}^{\star } \) antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram: The He ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He as being formed from two He ions, each of which has a single valence electron in the 1 atomic orbital. We can now fill the molecular orbital diagram: The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ ) orbital, giving a (σ ) electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. So the bond order is \( \dfrac{2-0}{2} =1 \) He is therefore predicted to contain a single He–He bond. Thus it should be a stable species. Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H ion. Is this a stable species? H has a valence electron configuration of \( \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{2} \) with a bond order of 0. It is therefore predicted to be unstable. So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals. A molecular orbital diagram that can be applied to any with two identical alkali metal atoms (Li and Cs , for example) is shown in part (a) in Figure \(\Page {4}\), where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σ bonding molecular orbital and a σ antibonding molecular orbital. Because each alkali metal (M) has an valence electron configuration, the M molecule has two valence electrons that fill the σ bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li , Na , K , Rb , and Cs ). The general features of these M diagrams are identical to the diagram for the H molecule in Figure \(\Page {2}\). Experimentally, all are found to be stable in the gas phase, and some are even stable in solution. Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be ), in which each metal atom has an valence electron configuration, resemble the diagram for the He molecule in part (c) in Figure \(\Page {3}\) As shown in part (b) in Figure \(\Page {4}\), this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σ bonding orbital and the antibonding orbital and give a bond order of 0. Thus Be , Mg , Ca , Sr , and Ba are all expected to be unstable, in agreement with experimental data. \(Be_2\) is st Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na ion. chemical species molecular orbital energy-level diagram, valence electron configuration, bond order, and stability Because sodium has a [Ne]3 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1 atomic orbitals. The Na ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ molecular orbital, a half-filled σ and a \( \left ( \sigma _{3s} \right)^{2}\left ( \sigma _{3s}^{\star } \right)^{1} \) electron configuration. The bond order is (2-1) 2=1/2 With a fractional bond order, we predict that the Na ion exists but is highly reactive. Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca ion. Ca has a \( \left ( \sigma _{4s} \right)^{2}\left ( \sigma _{4s}^{\star } \right)^{1} \) electron configurations and a bond order of 1/2 and should exist. Atomic orbitals other than orbitals can also interact to form molecular orbitals. Because individual , , and orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each subshell, for example, there are , , and orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations. \[ \sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) \label{Eq6}\] Just as with orbitals, we can form molecular orbitals from orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two atomic orbitals in part (a) in Figure \(\Page {5}\), it is the mathematical of their wave functions that results in interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a \( \sigma _{np_{z}} \) because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the -axis):
\[ \sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) \label{Eq6a}\] The other possible combination of the two orbitals is the mathematical sum: \[ \sigma _{np_{z}}=np_{z}\left ( A \right)+np_{z}\left ( B \right) \label{Eq7}\] In this combination, shown in part (b) in Figure \(\Page {5}\), the positive lobe of one atomic orbital overlaps the negative lobe of the other, leading to interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a \( \sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) \) . Whenever orbitals combine, (more stable) than the atomic orbitals from which it was derived, and (less stable). Overlap of atomic orbital lobes with the produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals. The remaining orbitals on each of the two atoms, and , do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure \(\Page {6}\), we see that we have two pairs of orbitals: the two orbitals lying in the plane of the page, and two orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the orbital on one atom can interact with only the orbital on the other, and the orbital on one atom can interact with only the on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital . The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital . \[ \pi _{np_{x}}=np_{x}\left ( A \right)+np_{x}\left ( B \right) \label{Eq8}\] \[ \pi ^{\star }_{np_{x}}=np_{x}\left ( A \right)-np_{x}\left ( B \right) \label{Eq9}\] The two orbitals can also combine using side-to-side interactions to produce a bonding \( \pi _{np_{y}} \) molecular orbital and an antibonding \( \pi _{np_{y}}^{\star } \) molecular orbital. Because the and atomic orbitals interact in the same way (side-to-side) and have the same energy, the \( \pi _{np_{x}} \) and \( \pi _{np_{y}} \)molecular orbitals are a degenerate pair, as are the \( \pi _{np_{x}}^{\star } \) and \( \pi _{np_{y}}^{\star } \) molecular orbitals. Figure \(\Page {7}\) is an energy-level diagram that can be applied to two identical interacting atoms that have three atomic orbitals each. There are six degenerate atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals. Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an atomic orbital on one atom with an atomic orbital on another. As shown in Figure \(\Page {8}\), the sum of the two atomic wave functions ( + ) produces a σ bonding molecular orbital. Their difference ( − ) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis. If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N , O , and F . When we draw a molecular orbital diagram for a molecule, there are four key points to remember: The number of molecular orbitals is equal to the total number of atomic orbitals we started with. We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F . We use the diagram in part (a) in Figure \(\Page {9}\); the = 1 orbitals (σ and σ ) are located well below those of the = 2 level and are not shown. As illustrated in the diagram, the σ and σ molecular orbitals are much lower in energy than the molecular orbitals derived from the 2 atomic orbitals because of the large difference in energy between the 2 and 2 atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2 orbitals on each F is \( \sigma _{2p_{z}} \) and the next most stable are the two degenerate orbitals, \( \pi _{2p_{x}} \) and \( \pi _{2p_{y}} \) For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the \( \sigma ^{\star }_{2p_{z}} \) orbital is higher in energy than either of the degenerate \( \pi _{2p_{x}}^{\star } \) and \( \pi _{2p_{y}}^{\star } \) orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy. Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ and σ orbitals, 2 fill the \( \sigma _{2p_{z}} \) orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F is predicted to have a stable F–F single bond, in agreement with experimental data. We now turn to a molecular orbital description of the bonding in O . It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O , we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure \(\Page {9}\). We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ and σ orbitals, 2 more to fill the \( \sigma _{2p_{z}} \) orbital, and 4 to fill the degenerate \( \pi _{2p_{x}}^{\star } \) and \( \pi _{2p_{y}}^{\star } \) orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K. None of the other bonding models can predict the presence of two unpaired electrons in O . Chemists had long wondered why, unlike most other substances, liquid O is attracted into a magnetic field. As shown in Figure \(\Page {10}\), it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed. The magnetic properties of O are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H O, CO , and N in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H O, CO , and N , have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O with organic compounds to give H O, CO , and N would require that at least one of the electrons on O change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a . Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke! For period 2 diatomic molecules to the left of N in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the \( \sigma _{2p_{z}} \) molecular orbital is slightly in energy than the degenerate \( \pi ^{\star }_{np_{x}} \) and \( \pi ^{\star }_{np_{y}} \) orbitals. The difference in energy between the 2 and 2 atomic orbitals increases from Li to F due to increasing nuclear charge and poor screening of the 2 electrons by electrons in the 2 subshell. The bonding interaction between the 2 orbital on one atom and the 2 orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ orbital and increases the energy of the \( \sigma _{2p_{z}} \) orbital. Thus for Li , Be , B , C , and N , the \( \sigma _{2p_{z}} \) orbital is higher in energy than the \( \sigma _{3p_{z}} \) orbitals, as shown in Figure \(\Page {11}\) Experimentally, it is found that the energy gap between the and atomic orbitals as the nuclear charge increases (Figure \(\Page {11}\)). Thus for example, the \( \sigma _{2p_{z}} \) molecular orbital is at a lower energy than the \( \pi _{2p_{x,y}} \) pair. Completing the diagram for N in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N versus 141.2 pm in F ), and the bond energy is much greater for N than for F (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N bond is much shorter and stronger than the F bond, consistent with what we would expect when comparing a triple bond with a single bond. Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S , a bright blue gas at high temperatures. chemical species molecular orbital energy-level diagram, bond order, and number of unpaired electrons Sulfur has a [Ne]3 3 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure \(\Page {9}\) and Figure \(\Page {11}\), we need to know how close in energy the 3 and 3 atomic orbitals are because their energy separation will determine whether the \( \pi _{3p_{x,y}} \) or the \( \sigma _{3p_{z}} \) molecular orbital is higher in energy. Because the – energy gap as the nuclear charge increases (Figure \(\Page {11}\)), the \( \sigma _{3p_{z}} \) molecular orbital will be lower in energy than the \( \pi _{3p_{x,y}} \) pair. The molecular orbital energy-level diagram is as follows: Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. Ten valence electrons are used to fill the orbitals through \( \pi _{3p_{x}} \) and \( \pi _{3p_{y}} \), leaving 2 electrons to occupy the degenerate \( \pi ^{\star }_{3p_{x}} \) and \( \pi ^{\star }_{3p_{y}} \) pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S is \( \left ( \sigma _{3s} \right)^{2}\left ( \sigma ^{\star }_{3s} \right)^{2}\left ( \sigma _{3p_{z}} \right)^{2}\left ( \pi _{3p_{x,y}} \right)^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right)^{2} \) with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond. Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O ). \( \left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \sigma _{2p_{z}} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right)^{4} \) bond order of 1; no unpaired electrons Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χ > χ ), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure \(\Page {12}\). The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond. A molecular orbital energy-level diagram is always skewed toward the more electronegative atom. Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O with N at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O to produce NO , which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals. Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure \(\Page {13}\)) shows that the general pattern is similar to that for the O molecule (Figure \(\Page {11}\)). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2 atomic orbitals, the 11th electron must occupy one of the degenerate π orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N and O molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot. Note that electronic structure studies show the ground state configuration of \(\ce{NO}\) to be \( \left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{2} \) in order of increasing energy. Hence, the \( \pi _{2p_{x,y}}\) orbitals are lower in energy than the \(\sigma _{2p_{z}}\) orbital. This is because the \(\ce{NO}\) molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond ( Molecular orbital theory can also tell us something about the of \(NO\). As indicated in the energy-level diagram in Figure \(\Page {13}\), NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, \(NO\) is easily oxidized to the \(NO^+\) cation, which is isoelectronic with \(N_2\) and has a bond order of 3, corresponding to an N≡O triple bond. Molecular Orbital Bonding for Second Row Elements: Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure \(\Page {14}\) that the 1 orbital of atomic hydrogen is closest in energy to the 3 orbitals of chlorine. Consequently, the filled Cl 3 atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1 and Cl 3 orbitals. Of the three orbitals, only one, designated as 3 , can interact with the H 1 orbital. The 3 and 3 atomic orbitals have no net overlap with the 1 orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3 , 3 , and 3 orbitals do not change when HCl forms, they are called . A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3 than to the H 1 atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give \( H^{\delta +} -- Cl^{\delta -} \). Electrons in nonbonding molecular orbitals have on bond order. Use a “skewed” molecular orbital energy-level diagram like the one in Figure \(\Page {12}\) to describe the bonding in the cyanide ion (CN ). What is the bond order? chemical species “skewed” molecular orbital energy-level diagram, bonding description, and bond order The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure \(\Page {12}\) fills the five lowest-energy orbitals, as shown here: Because χ > χ , the atomic orbitals of N (on the right) are lower in energy than those of C. The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N . Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl ). What is the bond order? All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1. Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods. A is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the of bonding is a approach. Molecular orbitals are constructed using , which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals. A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely contains at least one node perpendicular to the internuclear axis. A (bonding) or a (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a (bonding) and a (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane. The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an . The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the , defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist. Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered , molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called , using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry. ( ) | 40,518 | 2,690 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.09%3A_Solutions_of_Electrolytes |
Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of \(NaCl\) in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of \(NaCl\) and KCl are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution. The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor (\(i\)) and is defined as follows:Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions. \[i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.9.1}\] As the solute concentration increases the van’t Hoff factor decreases. The van’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van ’t Hoff factor, the greater the deviation. As the data in Table \(\Page {1}\) show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution. Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure \(\Page {1}\)). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as \(Mg^{2+}\), \(Al^{3+}\), \(\ce{SO4^{2−}}\), and \(\ce{PO4^{3−}}\) have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations. A 0.0500 M aqueous solution of \(FeCl_3\) has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor \(i\) for the solution. : solute concentration, osmotic pressure, and temperature : van’t Hoff factor : : A If \(FeCl_3\) dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be \[\Pi=MRT=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm\] B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of \(FeCl_3\) dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm: \[4.15\; atm=M \left[ 0.0821 \;(L⋅atm)/(K⋅mol)\right] (298 \;K) \] \[0.170 mol/L=M\] The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of \(FeCl_3\) dissolved. From Equation \ref{13.9.1}, the van’t Hoff factor for the solution is \[i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; FeCl_3}=3.40\] Calculate the van’t Hoff factor for a 0.050 m aqueous solution of \(MgCl_2\) that has a measured freezing point of −0.25°C. : 2.7 (versus an ideal value of 3 Ionic compounds may not completely dissociate in solution due to effects, in which case observed colligative effects may be less than predicted. | 4,406 | 2,691 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.02%3A_Charles'_Law |
Quantitative experiments establishing the law were first published in 1802 by Gay-Lussac, who credited Jacques Charles with having discovered the law earlier. Charles’ law relates the volume and temperature of a gas when measurements are made at constant pressure. We can imagine rediscovering Charles’ law by trapping a sample of gas in a tube and measuring its volume as we change the temperature, while keeping the pressure constant. This presumes that we have a way to measure temperature, perhaps by defining it in terms of the volume of a fixed quantity of some other fluid—like liquid mercury. At a fixed pressure, \(P_1\), we observe a linear relationship between the volume of a sample of gas and its temperature, like that in Figure 2. If we repeat this experiment with the same gas sample at a higher pressure, \(P_2\), we observe a second linear relationship between the volume and the temperature of the gas. If we extend these lines to their intersection with the temperature axis at zero volume, we make a further important discovery: Both lines intersect the temperature axis at the same point. We can represent this behavior mathematically as \[V={\beta }^*\left(n,P\right)T^*+{\gamma }^*(n,P)\] where we recognize that both the slope and the -axis intercept of the graph depend on the pressure of the gas and on the number of moles of gas in the sample. A little reflection shows that here too the slope and intercept must be directly proportional to the number of moles of gas, so that we can rewrite our equation as \[V=n\beta \left(P\right)T^*+n\gamma (P)\] When we repeat these experiments with different gaseous substances, we discover an additional important fact: \(\beta (P)\) and \(\gamma (P)\) are the same for any gas. This means that the temperature at which the volume extrapolates to zero is the same for any gas and is independent of the constant pressure we maintain as we vary the temperature (Figure 2). | 1,952 | 2,693 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.05%3A_Hybrid_Orbitals |
The localized uses a process called , in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining ( ) two or more atomic orbitals from the same atom. In BeH , we can generate two equivalent orbitals by combining the 2 orbital of beryllium and any one of the three degenerate 2 orbitals. By taking the sum and the difference of Be 2 and 2 atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the -axes, as shown in Figure \(\Page {1}\). This gives us Equation \ref{9.5.1b}, where the value \(\frac{1}{\sqrt{2}}\) is needed mathematically to indicate that the 2 and 2 orbitals contribute equally to each hybrid orbital. \[sp = \dfrac{1}{\sqrt{2}} (2s + 2p_z) \label{9.5.1a} \] and \[sp = \dfrac{1}{\sqrt{2}} (2s - 2p_z) \label{9.5.1b} \] The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called because they are formed from one and one orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure and orbitals, as illustrated in this diagram: Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals (Figure \(\Page {4}\)). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds. The concept of hybridization also explains why boron, with a 2 2 valence electron configuration, forms three bonds with fluorine to produce BF , as predicted by the Lewis and VSEPR approaches. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2 electrons to an unoccupied 2 orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2 and two 2 orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here: Looking at the 2 2 valence electron configuration of carbon, we might expect carbon to use its two unpaired 2 electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2 orbital and the three 2 orbitals on carbon to give a set of four degenerate (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron: In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH or CF ), but these species are highly reactive, unstable intermediates that only form in certain chemical reactions. Valence bond theory explains the number of bonds formed in a compound and the relative bond strengths. The bonding in molecules such as NH or H O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH , for example, N, with a 2 2 valence electron configuration, can hybridize its 2 and 2 orbitals to produce four hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons: The three singly occupied lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H O has an hybridized oxygen atom that uses two singly occupied lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH and H O. Unfortunately, however, recent experimental evidence indicates that in NH and H O, the hybridized orbitals are entirely equivalent in energy, making this bonding model an active area of research. Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. two chemical compounds number of electron pairs and molecular geometry, hybridization, and bonding Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. B is hybridized; F is also hybridized so it can accommodate one B–F bond and three lone pairs. The molecular geometry is tetrahedral. Each N atom is hybridized and uses one hybrid orbital to form the N–N bond, two to form N–H bonds, and one to accommodate a lone pair. The molecular geometry about each N is trigonal pyramidal. The number of hybrid orbitals used by the central atom is the same as the number of electron pairs around the central atom. Hybridization is not restricted to the and atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence ( − 1) orbitals as well as its and orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF and SF ). Using the orbital, all three orbitals, and one ( − 1) orbital gives a set of five hybrid orbitals that point toward the vertices of a trigonal bipyramid (part (a) in Figure \(\Page {7}\)). In this case, the five hybrid orbitals are all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other. Similarly, the combination of the orbital, all three orbitals, and orbitals gives a set of six equivalent hybrid orbitals oriented toward the vertices of an octahedron (part (b) in Figure 9.5.6). In the VSEPR model, PF and SF are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which or hybrid orbitals are used for bonding. What is the hybridization of the central atom in each species? Describe the bonding in each species. three chemical species hybridization of the central atom To accommodate five electron pairs, the sulfur atom must be hybridized. Filling these orbitals with 10 electrons gives four hybrid orbitals forming S–F bonds and one with a lone pair of electrons. What is the hybridization of the central atom in each species? Describe the bonding. with four P–Cl bonds with three Br–F bonds and two lone pairs with six Si–F bonds Hybridization using orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S). Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF and SiF ), only SiF reacts with F to give a stable hexafluoro dianion, SiF . Because there are no 2 atomic orbitals, the formation of octahedral CF would require hybrid orbitals created from 2 , 2 , and 3 atomic orbitals. The 3 orbitals of carbon are so high in energy that the amount of energy needed to form a set of hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF have never been prepared. What is the hybridization of the oxygen atom in OF ? Is OF likely to exist? chemical compound hybridization and stability The VSEPR model predicts that OF will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. To accommodate five electron pairs, the O atom would have to be hybridized. The only orbital available for forming a set of hybrid orbitals is a 3 orbital, which is higher in energy than the 2 and 2 valence orbitals of oxygen. As a result, the OF molecule is unlikely to exist. In fact, it has not been detected. What is the hybridization of the boron atom in \(BF_6^{3−}\)? Is this ion likely to exist? hybridization; no Hybridization increases the overlap of bonding orbitals and explains the molecular geometries of many species whose geometry cannot be explained using a VSEPR approach. The model (called ) assumes that covalent bonds are formed when atomic orbitals overlap and that the strength of a covalent bond is proportional to the amount of overlap. It also assumes that atoms use combinations of atomic orbitals ( ) to maximize the overlap with adjacent atoms. The formation of can be viewed as occurring via of an electron from a filled subshell to an empty or ( − 1) valence orbital, followed by , the combination of the orbitals to give a new set of (usually) equivalent orbitals that are oriented properly to form bonds. The combination of an and an orbital gives rise to two equivalent oriented at 180°, whereas the combination of an and two or three orbitals produces three equivalent or four equivalent , respectively. The bonding in molecules with more than an octet of electrons around a central atom can be explained by invoking the participation of one or two ( − 1) orbitals to give sets of five or six orbitals, capable of forming five or six bonds, respectively. The spatial orientation of the hybrid atomic orbitals is consistent with the geometries predicted using the VSEPR model. | 10,943 | 2,694 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.02%3A_Glassware_and_Equipment/1.2D%3A_Greasing_Joints |
Ground glass joints are manufactured to fit quite well with one another, and yet they are not perfectly airtight. In some situations (e.g. when using reduced pressure inside an apparatus), grease must be applied to each joint to ensure a good seal. Grease is also used whenever the joint may be in contact with a highly basic solution, as basic solutions can form sodium silicates and etch glass. Grease can be applied with a syringe full of grease (Figure 1.8a), wood splint, or toothpick. Grease should be lightly applied in portions around the male joint, closer to the glass end than the end which will be in contact with reagents (Figure 1.8a). If grease is allowed near the end which will contact the reagents, there is a possibility the reagent will dissolve the grease and become contaminated. The female joint should then be connected, and the joints twisted to spread the grease in a thin layer. The joint should become transparent all the way around the joint, but to a depth of only one-third to one-half of the joint (Figure 1.8b). If the entire joint becomes transparent or if grease is seen spilling out of the joint, too much grease has been used (Figure 1.8c). Excess grease should be wiped off with a KimWipe (one is used in Figure 1.9). To clean grease from a joint after a process is complete, wipe of the majority of the grease using a paper towel or KimWipe. Then wet a KimWipe with some hydrocarbon solvent and rub the moistened KimWipe onto the joint to dissolve the grease (Figure 1.9). Hydrocarbon solvents (e.g. hexanes) work much better than acetone to dissolve residual grease. | 1,618 | 2,695 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/26%3A_Chemical_Equilibrium/26.07%3A_The_Van_'t_Hoff_Equation |
We can use Gibbs-Helmholtz to get the temperature dependence of \(K\) \[ \left( \dfrac{∂[ΔG^o/T]}{∂T} \right)_P = \dfrac{-ΔH^o}{T^2} \nonumber \] At equilibrium, we can equate \(ΔG^o\) to \(-RT\ln K\) so we get: \[ \left( \dfrac{∂[\ln K]}{∂T} \right)_P = \dfrac{ΔH^o}{RT^2} \nonumber \] We see that whether \(K\) increases or decreases with temperature is linked to whether the reaction enthalpy is positive or negative. If temperature is changed little enough that \(ΔH^o\) can be considered constant, we can translate a \(K\) value at one temperature into another by integrating the above expression, we get a similar derivation as with melting point depression: \[\ln \dfrac{K(T_2)}{K(T_1)} = \dfrac{-ΔH^o}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \nonumber \] If more precision is required we could correct for the temperature changes of \(ΔH^o\) by using heat capacity data. How \(K\) increases or decreases with temperature is linked to whether the reaction enthalpy is positive or negative. The expression for \(K\) is a rather sensitive function of temperature given its exponential dependence on the difference of stoichiometric coefficients One way to see the sensitive temperature dependence of equilibrium constants is to recall that \[K=e^{−\Delta_r{G^o}/RT}\label{18} \] However, since under constant pressure and temperature \[\Delta{G^o}= \Delta{H^o}−T\Delta{S^o} \nonumber \] Equation \(\ref{18}\) becomes \[ K=e^{-\Delta{H^o}/RT} e^{\Delta {S^o}/R}\label{19} \] Taking the natural log of both sides, we obtain a linear relation between \(\ln K \)and the standard enthalpies and entropies: \[\ln K = - \dfrac{\Delta {H^o}}{R} \dfrac{1}{T} + \dfrac{\Delta{S^o}}{R}\label{20} \] which is known as the van ’t Hoff equation. It shows that a plot of \(\ln K\) vs. \(1/T\) should be a line with slope \(-\Delta_r{H^o}/R\) and intercept \(\Delta_r{S^o}/R\). Hence, these quantities can be determined from the \(\ln K\) vs. \(1/T\) data without doing calorimetry. Of course, the main assumption here is that \(\Delta_r{H^o}\) and \(\Delta_r{S^o}\) are only very weakly dependent on \(T\), which is usually valid. | 2,158 | 2,696 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/04%3A_Evaluating_Analytical_Data/4.02%3A_Characterizing_Experimental_Errors |
Characterizing a penny’s mass using the data in suggests two questions. First, does our measure of central tendency agree with the penny’s expected mass? Second, why is there so much variability in the individual results? The first of these questions addresses the accuracy of our measurements and the second addresses the precision of our measurements. In this section we consider the types of experimental errors that affect accuracy and precision. Accuracy is how close a measure of central tendency is to its expected value, \(\mu\). We express accuracy either as an absolute error, \[e = \overline{X} - \mu \label{4.1}\] or as a percent relative error, % \[\% e = \frac {\overline{X} - \mu} {\mu} \times 100 \label{4.2}\] Although Equation \ref{4.1} and Equation \ref{4.2} use the mean as the measure of central tendency, we also can use the median. The convention for representing a statistical parameter is to use a Roman letter for a value calculated from experimental data, and a Greek letter for its corresponding expected value. For example, the experimentally determined mean is \(\overline{X}\) and its underlying expected value is \(\mu\). Likewise, the experimental standard deviation is and the underlying expected value is \(\sigma\). We identify as determinate an error that affects the accuracy of an analysis. Each source of a has a specific magnitude and sign. Some sources of determinate error are positive and others are negative, and some are larger in magnitude and others are smaller in magnitude. The cumulative effect of these determinate errors is a net positive or negative error in accuracy. It is possible, although unlikely, that the positive and negative determinate errors will offset each other, producing a result with no net error in accuracy. We assign determinate errors into four categories—sampling errors, method errors, measurement errors, and personal errors—each of which we consider in this section. A determinate sampling error occurs when our sampling strategy does not provide a us with a representative sample. For example, if we monitor the environmental quality of a lake by sampling from a single site near a point source of pollution, such as an outlet for industrial effluent, then our results will be misleading. To determine the mass of a U. S. penny, our strategy for selecting pennies must ensure that we do not include pennies from other countries. An awareness of potential sampling errors especially is important when we work with heterogeneous materials. Strategies for obtaining representative samples are covered in . In any analysis the relationship between the signal, , and the absolute amount of analyte, , or the analyte’s concentration, , is \[S_{total} = k_A n_A + S_{mb} \label{4.3}\] \[S_{total} = k_A C_A + S_{mb} \label{4.4}\] where is the method’s sensitivity for the analyte and is the signal from the method blank. A exists when our value for or for is in error. For example, a method in which is the mass of a precipitate assumes that is defined by a pure precipitate of known stoichiometry. If this assumption is not true, then the resulting determination of or is inaccurate. We can minimize a determinate error in by calibrating the method. A method error due to an interferent in the reagents is minimized by using a proper method blank. The manufacturers of analytical instruments and equipment, such as glassware and balances, usually provide a statement of the item’s maximum , or . For example, a 10-mL volumetric pipet (Figure 4.2.1
) has a tolerance of ±0.02 mL, which means the pipet delivers an actual volume within the range 9.98–10.02 mL at a temperature of 20 C. Although we express this tolerance as a range, the error is determinate; that is, the pipet’s expected volume, \(\mu\), is a fixed value within this stated range. Volumetric glassware is categorized into classes based on its relative accuracy. Class A glassware is manufactured to comply with tolerances specified by an agency, such as the National Institute of Standards and Technology or the American Society for Testing and Materials. The tolerance level for Class A glassware is small enough that normally we can use it without calibration. The tolerance levels for Class B glassware usually are twice that for Class A glassware. Other types of volumetric glassware, such as beakers and graduated cylinders, are not used to measure volume accurately. Table 4.2.1
provides a summary of typical measurement errors for Class A volumetric glassware. Tolerances for digital pipets and for balances are provided in Table 4.2.2
and Table 4.2.3
. The tolerance values for the volumetric glassware in Table 4.2.1
are from the ASTM E288, E542, and E694 standards. The measurement errors for the digital pipets in Table 4.2.2
are from www.eppendorf.com. We can minimize a determinate measurement error by calibrating our equipment. Balances are calibrated using a reference weight whose mass we can trace back to the SI standard kilogram. Volumetric glassware and digital pipets are calibrated by determining the mass of water delivered or contained and using the density of water to calculate the actual volume. It is never safe to assume that a calibration does not change during an analysis or over time. One study, for example, found that repeatedly exposing volumetric glassware to higher temperatures during machine washing and oven drying, led to small, but significant changes in the glassware’s calibration [Castanheira, I.; Batista, E.; Valente, A.; Dias, G.; Mora, M.; Pinto, L.; Costa, H. S. , , 719–726]. Many instruments drift out of calibration over time and may require frequent recalibration during an analysis. Finally, analytical work is always subject to , examples of which include the ability to see a change in the color of an indicator that signals the endpoint of a titration, biases, such as consistently overestimating or underestimating the value on an instrument’s readout scale, failing to calibrate instrumentation, and misinterpreting procedural directions. You can minimize personal errors by taking proper care. Determinate errors often are difficult to detect. Without knowing the expected value for an analysis, the usual situation in any analysis that matters, we often have nothing to which we can compare our experimental result. Nevertheless, there are strategies we can use to detect determinate errors. The magnitude of a constant is the same for all samples and is more significant when we analyze smaller samples. Analyzing samples of different sizes, therefore, allows us to detect a constant determinate error. For example, consider a quantitative analysis in which we separate the analyte from its matrix and determine its mass. Let’s assume the sample is 50.0% w/w analyte. As we see in Table 4.2.4
, the expected amount of analyte in a 0.100 g sample is 0.050 g. If the analysis has a positive constant determinate error of 0.010 g, then analyzing the sample gives 0.060 g of analyte, or an apparent concentration of 60.0% w/w. As we increase the size of the sample the experimental results become closer to the expected result. An upward or downward trend in a graph of the analyte’s experimental concentration versus the sample’s mass (Figure 4.2.2
) is evidence of a constant determinate error. A , in which the error’s magnitude depends on the amount of sample, is more difficult to detect because the result of the analysis is independent of the amount of sample. Table 4.2.5
outlines an example that shows the effect of a positive proportional error of 1.0% on the analysis of a sample that is 50.0% w/w in analyte. Regardless of the sample’s size, each analysis gives the same result of 50.5% w/w analyte. One approach for detecting a proportional determinate error is to analyze a standard that contains a known amount of analyte in a matrix similar to our samples. Standards are available from a variety of sources, such as the National Institute of Standards and Technology (where they are called Standard Reference Materials) or the American Society for Testing and Materials. Table 4.2.6
, for example, lists certified values for several analytes in a standard sample of leaves. Another approach is to compare our analysis to an analysis carried out using an independent analytical method that is known to give accurate results. If the two methods give significantly different results, then a determinate error is the likely cause. The primary purpose of this Standard Reference Material is to validate analytical methods for determining flavonoids, terpene lactones, and toxic elements in or other materials with a similar matrix. Values are from the official Certificate of Analysis available at www.nist.gov. Constant and proportional determinate errors have distinctly different sources, which we can define in terms of the relationship between the signal and the moles or concentration of analyte (Equation \ref{4.3} and Equation \ref{4.4}). An invalid method blank, , is a constant determinate error as it adds or subtracts the same value to the signal. A poorly calibrated method, which yields an invalid sensitivity for the analyte, , results in a proportional determinate error. As we saw in , precision is a measure of the spread of individual measurements or results about a central value, which we express as a range, a standard deviation, or a variance. Here we draw a distinction between two types of precision: repeatability and reproducibility. is the precision when a single analyst completes an analysis in a single session using the same solutions, equipment, and instrumentation. , on the other hand, is the precision under any other set of conditions, including between analysts or between laboratory sessions for a single analyst. Since reproducibility includes additional sources of variability, the reproducibility of an analysis cannot be better than its repeatability. The ratio of the standard deviation associated with reproducibility to the standard deviation associated with repeatability is called the Horowitz ratio. For a wide variety of analytes in foods, for example, the median Horowtiz ratio is 2.0 with larger values for fatty acids and for trace elements; see Thompson, M.; Wood, R. “The ‘Horowitz Ratio’–A Study of the Ratio Between Reproducibility and Repeatability in the Analysis of Foodstuffs,” , , , 375–379. Errors that affect precision are indeterminate and are characterized by random variations in their magnitude and their direction. Because they are random, positive and negative tend to cancel, provided that we make a sufficient number of measurements. In such situations the mean and the median largely are unaffected by the precision of the analysis. We can assign indeterminate errors to several sources, including collecting samples, manipulating samples during the analysis, and making measurements. When we collect a sample, for instance, only a small portion of the available material is taken, which increases the chance that small-scale inhomogeneities in the sample will affect repeatability. Individual pennies, for example, may show variations in mass from several sources, including the manufacturing process and the loss of small amounts of metal or the addition of dirt during circulation. These variations are sources of indeterminate sampling errors. During an analysis there are many opportunities to introduce indeterminate method errors. If our method for determining the mass of a penny includes directions for cleaning them of dirt, then we must be careful to treat each penny in the same way. Cleaning some pennies more vigorously than others might introduce an indeterminate method error. Finally, all measuring devices are subject to indeterminate measurement errors due to limitations in our ability to read its scale. For example, a buret with scale divisions every 0.1 mL has an inherent indeterminate error of ±0.01–0.03 mL when we estimate the volume to the hundredth of a milliliter (Figure 4.2.3
). Indeterminate errors associated with our analytical equipment or instrumentation generally are easy to estimate if we measure the standard deviation for several replicate measurements, or if we monitor the signal’s fluctuations over time in the absence of analyte (Figure 4.2.4
) and calculate the standard deviation. Other sources of indeterminate error, such as treating samples inconsistently, are more difficult to estimate. To evaluate the effect of an indeterminate measurement error on our analysis of the mass of a circulating United States penny, we might make several determinations of the mass for a single penny (Table 4.2.7
). The standard deviation for our original experiment (see ) is 0.051 g, and it is 0.0024 g for the data in Table 4.2.7
. The significantly better precision when we determine the mass of a single penny suggests that the precision of our analysis is not limited by the balance. A more likely source of indeterminate error is a variability in the masses of individual pennies. In Section 4.5 we will discuss a statistical method—the -test—that you can use to show that this difference is significant. Analytical chemists make a distinction between error and uncertainty [Ellison, S.; Wegscheider, W.; Williams, A. , , 607A–613A]. is the difference between a single measurement or result and its expected value. In other words, error is a measure of . As discussed earlier, we divide errors into determinate and indeterminate sources. Although we can find and correct a source of determinate error, the indeterminate portion of the error remains. expresses the range of possible values for a measurement or result. Note that this definition of uncertainty is not the same as our definition of precision. We calculate precision from our experimental data and use it to estimate the magnitude of indeterminate errors. Uncertainty accounts for all errors—both determinate and indeterminate—that reasonably might affect a measurement or a result. Although we always try to correct determinate errors before we begin an analysis, the correction itself is subject to uncertainty. Here is an example to help illustrate the difference between precision and uncertainty. Suppose you purchase a 10-mL Class A pipet from a laboratory supply company and use it without any additional calibration. The pipet’s tolerance of ±0.02 mL is its uncertainty because your best estimate of its expected volume is 10.00 mL ± 0.02 mL. This uncertainty primarily is determinate. If you use the pipet to dispense several replicate samples of a solution and determine the volume of each sample, the resulting standard deviation is the pipet’s precision. Table 4.2.8
shows results for ten such trials, with a mean of 9.992 mL and a standard deviation of ±0.006 mL. This standard deviation is the precision with which we expect to deliver a solution using a Class A 10-mL pipet. In this case the pipet’s published uncertainty of ±0.02 mL is worse than its experimentally determined precision of ±0.006 ml. Interestingly, the data in Table 4.2.8
allows us to calibrate this specific pipet’s delivery volume as 9.992 mL. If we use this volume as a better estimate of the pipet’s expected volume, then its uncertainty is ±0.006 mL. As expected, calibrating the pipet allows us to decrease its uncertainty [Kadis, R. , , 167–173]. | 15,401 | 2,697 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.01%3A_Basics_of_metabolism |
A large number of chemical reactions take place in living things almost all the time. Some of these reactions synthesize the substances needed by living things from the raw materials available. Usually, these reactions convert simple molecules or compounds to more complex molecules or compounds at the expense of energy. For example, photosynthesis converts carbon dioxide (\(\ce{CO2}\)) and water (\(\ce{H2O}\)) into glucose (\(\ce{C6H12O6}\)) by utilizing energy from sunlight. \(\ce{6CO2 + 6H2O + Energy (2808 kJ/mol) -> C6H12O2}\) Other reactions break down the complex molecules to release the energy needed as heat, to do work, to provide energy for synthetic reactions, to dispose of waster byproducts, etc. For example, reverse of photosynthesis happens in the digestion of glucose. \(\ce{6CO2 + 6H2O + Energy (2808 kJ/mol)C6H12O2 -> 6CO2 + 6H2O + Energy (2808 kJ/mol)}\) Metabolism is chemical reactions taking place in living things needed for the sustenance of life. Metabolism can be subdivided into two categories of reactions. The figure on the right summarizes metabolism and its sub-categories: catabolism and anabolism (Copyright: Linares-Pastén, J. A. (2018), CCBYSA 4, via Wikimedia commons) Metabolism usually happens through a series of interconnected chemical reactions, called . Metabolic pathways have a lot of similarities in different species. It indicates their common origin in the early stages of the evolution of species and their retention due to their efficiency. Some diseases, such as type II diabetes and cancer, disrupt normal metabolism, and the difference allows scientists to find therapeutic interventions. Some of those metabolic pathways, particularly the catabolic pathways related to the digestion of food and its conversions to obtain energy will be described in this chapter. The catabolism of food starts from the digestion of food. It can be divided into four major stages, as illustrated in the figure on the right (Copyright: modified from Tim Vickers, vectorized by Fvasconcellos, Public domain, via Wikimedia Commons). is mainly the hydrolysis of food molecules that happens in the digestive system shown in the figure on the left (Copyright: National Cancer Institute, Public domain, via Wikimedia Commons). Polysaccharides are hydrolyzed to monosaccharides, fats are hydrolyzed to glycerol and fatty acids, and proteins are hydrolyzed to amino acids. The products of the stage 1 reactions diffuse into the bloodstream and are transported to cells where the next stages of food catabolism take place. happens in the cells which is the degradation of monosaccharides, fatty acids, and amino acids. It yields smaller groups which are usually a two-carbon acetyl group or, in the case of amino acids, four-carbon carboxylate groups, which enter the next stage. usually happens in mitochondria in the cells. It oxidizes the two-carbon acetyl or four-carbon carboxylate groups in the citric acid cycle to carbon dioxide (\(\ce{CO2}\)) at the expense of the reduction of coenzymes \(\ce{NAD^+}\) and \(\ce{FAD}\) to \(\ce{NADH}\) and \(\ce{FADH2}\). also happens in mitochondria. In this stage the reduces coenzymes \(\ce{NADH}\) and \(\ce{FADH2}\) are oxidized back to their oxidized forms \(\ce{NAD^+}\) and \(\ce{FAD}\) at the expense of reduction of oxygen (\(\ce{O2}\)) into water (\(\ce{H2O}\)) via electron transport process with release of energy. The energy released is used to synthesize adenosine triphosphate (ATP) which is a high-energy molecule from adenosine diphosphate (ADP) and phosphate (Pi) which are lower-energy molecules via the oxidative phosphorylation process. The energy is temporarily stored in the form of ATP and released wherever it may be needed by the reverse reaction, i.e., conversion of ATP into ADP and Pi. Stages 2, 3, and 4 of food catabolism happen in the cells. A typical is surrounded by a cell membrane. are organized or specialized structures within the cells. A typical animal cell is illustrated in the figure on the right with some of the organelles labeled (Copyright; , Public domain). The in the cell contains hereditary material, i.e., DNA. The space between the cell membrane and the nucleus is called the . The is the fluid part of the cytoplasm containing an aqueous solution of electrolytes and enzymes that catalyze many of the cell's chemical reactions. Within the cytoplasm are organelles that perform specialized functions. For example, are the sites for protein synthesis, and are the energy factory of the cell where stage 3 and stage 4 of food catabolism take place. The mitochondrion (plural mitochondria) has an and and an between the two, as shown in the figure on the left (Copyright; LadyofHats, Public domain, via Wikimedia Commons). The fluid section surrounded by the inner membrane is called the . The enzymes that catalyze the chemical reactions of stage 3 of food catabolism are located in the matrix and along the inner membrane. These reactions ultimately convert the food molecules into \(\ce{CO2}\), \(\ce{H2O}\), and energy. The enzymes that catalyze stage 4 reactions that use this energy to produce high-energy ATP also take place in the matrix along the inner membrane. The principal compounds involved in the common metabolic pathways are adenosine triphosphate (\(\ce{ATP}\)) which is the agent for the temporary storage of energy and transfer of phosphate (\(\ce{PO4^{3-}}\) or \(\ce{Pi}\)) group; nicotinamide adenine dinucleotide (\(\ce{NAD^+}\)) and flavin adenine dinucleotide (\(\ce{FAD}\)) which are agents for the transfer of electrons during the biological oxidation-reduction sections; and coenzyme A (\(\ce{CoA}\)) which is the agent for the transfer of acetyl (\(\ce{CH3CO{-}}\)) group. These compounds are described here. Adenosine triphosphate (\(\ce{ATP}\)) is a nucleotide composed of adenine base bonded to ribose sugar by an N-glycosidic bond and triphosphate bonded to ribose by an ester bond, as shown in Figure \(\Page {1}\). The N-glycoside of adenine with ribose is a nucleoside, i.e., adenosine. When adenosine is bonded to diphosphate it makes adenosine diphosphate (\(\ce{ADP}\)). Hydrolysis of \(\ce{ATP}\) splits one phosphate ( \(\ce{PO4^{3-}}\) or \(\ce{Pi}\)) and convert it to \(\ce{ADP}\) with the release of 30.5 kJ/mol energy. \(\ce{ATP + H2O -> ADP + Pi}\) \(\delta\)H = -30.5 kJ/mol This energy is used in the processes that require energy, such as muscle contraction, nerve signal conduction, and biosynthesis. Catabolism of food releases energy that is temporarily stored in the form of \(\ce{ATP}\) by reversing the above reaction. \(\ce{ADP + Pi -> ATP + H2O}\) \(\delta\)H = +30.5 kJ/mol The cycle of reactions between \(\ce{ATP}\) and \(\ce{ADP}\) shown in Figure \(\Page {1}\) happen rapidly, producing 1 to 2 million \(\ce{ATP}\) molecules per second. An average human body produces \(\ce{ATP}\) about equal to the human body mass per day but it contains about 1 g of \(\ce{ATP}\) at a time. \(\ce{ADP}\) can also hydrolyze with water releasing energy and converting to adenosine monophosphate (\(\ce{AMP}\) \(\ce{ADP + H2O -> AMP + Pi}\) \(\delta\)H = -30.5 kJ/mol \(\ce{ATP}\) is also a phosphorylation agent in metabolic reactions. For example, D-glucose is phosphorylated by the reaction shown below. Oxidation is: Reduction is: In metabolic reactions \(\ce{H}\) is represented as a proton \(\ce{H^+}\) and an electron \(\ce{e^-}\). Usually \(\ce{2H^{+}}\) + \(\ce{2e^{-}}\) are transferred from or to a coenzyme in metabolic reactions. Generally, oxidation reactions release energy and reduction reactions gain energy. Nicotinamide adenine dinucleotide is a coenzyme composed of two nucleotides, an adenosine diphosphate (ADP) and a second nucleotide in which the nitrogen base is nicotinamide provided by vitamin niacin. The two nucleotides are linked by diphosphate linkage, as shown in Figure \(\Page {2}\). The oxidation oxidized form is represented as \(\ce{NAD^{+}}\) and its reduced form is represented as \(\ce{NADH}\). The \(\ce{NAD^{+}}\) is reduced to \(\ce{NADH}\) by reacting with two hydrogen (\(\ce{2H^{+}}\) + \(\ce{2e^{-}}\)) leaving one \(\ce{H^{+}}\) in the products, as shown in Figure \(\Page {2}\). Oxidation of \(\ce{NADH}\) to \(\ce{NAD^{+}}\) is the reverse reaction. An example of an oxidation reaction process in metabolism in which \(\ce{NAD^{+}}\) is required as an oxidizing agent is the oxidation of alcohols (\(\ce{-OH}\)) group to carbonyl (\(\ce{C=O}\)) group. For example, ethanol is oxidized to ethanol in the liver at the expense of reduction of \(\ce{NAD^{+}}\) to \(\ce{NADH}\) as shown below. Flavin adenine dinucleotide is a coenzyme composed of two nucleotides, an adenosine diphosphate (ADP) and a second nucleotide which is riboflavin (vitamin B ). Riboflavin is composed of flavin and ribitol which is a sugar alcohol. The two nucleotides are linked by an ester linkage between ribitol and phosphate, as shown in Figure \(\Page {3}\). The oxidation oxidized form is represented as \(\ce{FAD}\) and its reduced form is represented as \(\ce{FADH }\). The oxidation-reduction happens in the \(\ce{N}\) containing rings of the flavin part. The \(\ce{FAD}\) is reduced to \(\ce{FADH }\) by reacting with two hydrogen (\(\ce{2H^{+}}\) + \(\ce{2e^{-}}\)), as shown in Figure \(\Page {2}\). Oxidation of \(\ce{FADH }\) to \(\ce{FAD}\) is the reverse reaction. An example of a metabolic process in which \(\ce{FAD}\) is required as an oxidizing agent is the conversion of carbon-carbon single bond (\(\ce{C-C}\)) into a double bond (\(\ce{C=C}\)). For example, a \(\ce{C-C}\) bond is oxidized to \(\ce{C=C}\) bond of fumarate at the expense of reduction of \(\ce{FAD}\) to \(\ce{FADH }\) as shown below. Coenzyme A (\(\ce{CoA}\)) is composed of several components as illustrated in Figure \(\Page {4}\). The reactive part of \(\ce{CoA}\) is the thiol (\(\ce{-SH}\) group which forms a high-energy thioester bond with acyl groups. Free coenzyme is usually represented as \(\ce{HS-CoA}\) and when it is bonded with an acyl group, it is represented as \(\ce{Ac-S-CoA}\), where \(\ce{Ac-}\) is an acyl group. For example, pyruvate transfers its acetyl groups to \(\ce{HS-CoA}\) in the following reaction which is a step in the catabolism of carbohydrates. The \(\ce{-S-CoA}\) group in \(\ce{Ac-S-CoA}\) is a good leaving group that easily transfers the acyl group to other compounds during biosynthesis. For example, the acetyl group is transferred from \(\ce{Ac-S-CoA}\) to an acyl carrier protein (ACP) which is a step in the synthesis of fatty acids. Nearly all metabolic reactions in living things are catalyzed by enzymes. Unlike chemical reactions in chemical laboratories where different solvents, extreme conditions of temperature and pressure, can be applied and strong acids or bases can be used to catalyze the reactions, the biomedical reactions in living things have to take place under physiological conditions medium at physiological pH and body temperature. Enzymes are the specialized catalysts that catalyze the reactions under physiological conditions. The major factors responsible for the catalytic activity of the enzymes are the following. The following two examples from the catabolism of glucose illustrate the factors. Oxidation of malate to oxaloacetate in the citric acid cycle is catalyzed by the enzyme malate dehydrogenase. The substrate (oxaloacetate) is bound by the enzyme in such a way that i) the charge of its two carboxylates (\(\ce(-COO^{-}}\)) groups are neutralized by argentine side chains of the enzyme, ii) histidine side acts as a base catalyst and removes the proton from the \(\ce{-OH}\) group of the substrate, and at the same time, iii) the \(\ce{H}\) on the \(\ce{C}\) carrying the \(\ce{-OH}\) group is exposed to the nicotine amide ring of \(\ce{NAD^{+}}\) that removes the proton, converting malate into oxaloacetate that leaves the enzyme, as illustrated in Figure \(\Page {1}\), Another example is the formation of citrate from oxaloacetate and acetyl-\(\ce{CoA}\) in the citric acid cycle, catalyzed by the enzyme citrate synthase as illustrated in Figure \(\Page {2}\). Nearly every metabolic reaction is catalyzed by an enzyme in similar ways as described above. | 12,277 | 2,699 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Wolff-Kishner_Reduction |
Aldehydes and ketones can be converted to a hydrazine derivative by reaction with hydrazine. These "hydrazones" can be further converted to the corresponding alkane by reaction with base and heat. These two steps can be combined into one reaction called the Wolff-Kishner Reduction which represents a general method for converting aldehydes and ketones into alkanes. Typically a high boiling point solvent, such as ethylene glycol, is used to provide the high temperatures needed for this reaction to occur. Note! Nitrogen gas is produced as part of this reaction. Example 1) Deprotonation of Nitrogen 2) Protonation of the Carbon 3) Deprotonation of Nitrogen 4) Protonation of Carbon 1) Please draw the products of the following reactions. 1) ) ), | 771 | 2,700 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/02%3A_Structural_Organic_Chemistry/2.01%3A_Structural_Formulas |
The building block of structural organic chemistry is the tetravalent carbon atom. With few exceptions, carbon compounds can be formulated with four covalent bonds to each carbon, regardless of whether the combination is with carbon or some other element. The two-electron bond, which is illustrated by the carbon-hydrogen bonds in methane or ethane and the carbon-carbon bond in ethane, is called a . In these and many related substances, each carbon is attached to four other atoms: There exist, however, compounds such as ethene (ethylene), \(C_2H_4\), in which two electrons from each of the carbon atoms are mutually shared, thereby producing two-electron bonds, an arrangement which is called a . Each carbon in ethene is attached to only three other atoms: Similarly, in ethyne (acetylene), \(C_2H_2\), three electrons from each carbon atom are mutually shared, producing two-electron bonds, called a , in which each carbon is attached to only two other atoms: Of course, in all cases each carbon has a full octet of electrons. Carbon also forms double and triple bonds with several other elements that can exhibit a covalence of two or three. The carbon-oxygen (or carbonyl) double bond appears in carbon dioxide and many important organic compounds such as methanal (formaldehyde) and ethanoic acid (acetic acid). Similarly, a carbon-nitrogen triple bond appears in methanenitrile (hydrogen cyanide) and ethanenitrile (acetonitrile). By convention, a single straight line connecting the atomic symbols is used to represent a single (two-electron) bond, two such lines to represent a double (four-electron) bond, and three lines a triple (six-electron) bond. Representations of compounds by these symbols are called ; some examples are A point worth noting is that structural formulas usually do not indicate the electron pairs. This is perhaps unfortunate because they play as much a part in the chemistry of organic molecules as do the bonding electrons and their omission may lead the unwary reader to overlook them. However, when it is important to represent them, this can be done best with pairs of dots, although a few authors use lines: To save space and time in the representation of organic structures, it is common practice to use "condensed formulas" in which the bonds are not shown explicitly. In using condensed formulas, normal atomic valences are understood throughout. Examples of condensed formulas are Another type of abbreviation that often is used, particularly for ring compounds, dispenses with the symbols for carbon and hydrogen atoms and leaves only the lines in a structural formula. For instance, cyclopentane, \(C_5H_{10}\), often is represented as a regular pentagon in which it is understood that each apex represents a carbon atom with the requisite number of hydrogens to satisfy the tetravalence of carbon: Likewise, cyclopropane, \(C_3H_6\); cyclobutane, \(C_4H_8\); and cyclohexane, \(C_6H_{12}\), are drawn as regular polygons: Although this type of line drawing is employed most commonly for cyclic structures, its use for open chain (acyclic) structures is becoming increasingly widespread. There is no special merit to this abbreviation for simple structures such as butane, \(C_4H_{10}\); 1-butene, \(C_4H_8\); or 1,3-butadiene, \(C_4H_6\), but it is of value in representing more complex molecules such as \(\beta\)-carotene, \(C_{40}H_{56}\): Line structures also can be modified to represent the three-dimensional of molecules, and the way that this is done will be discussed in detail in Chapter 5. At the onset of you study of organic chemistry, you should write out the formulas rather completely until you are thoroughly familiar with what these abbreviations stand for. and (1977) | 3,768 | 2,701 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Kc |
This page defines the equilibrium constant and introduces the equilibrium constant expressed in terms of concentrations, K . It assumes familiarity with the concept of dynamic equilibrium, as well as the terms "homogeneous" and "heterogeneous" as applied to chemical reactions. The two types of dynamic equilibria (homogeneous and heterogeneous) are discussed separately below, because the equilibrium constants are defined differently. This is the more straightforward case. It applies where everything in the equilibrium mixture is in the same phase. An example of a gaseous homogeneous equilibrium is the conversion of sulfur dioxide to sulfur trioxide at the heart of the Contact Process: \[2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \tag{1}\] A commonly used liquid example is the esterification reaction between an organic acid and an alcohol, as in the example below: \[ CH_3COOH_{(l)} + CH_3CH_2OH_{(l)} \rightleftharpoons CH_3COOCH_2CH_{3(l)} + H_2O_{(l)}\tag{2}\] Consider the general equilibrium reaction: \[ aA + bB \rightleftharpoons cC + dD\tag{3}\] No state symbols are specified, but the reaction is assumed to be homogeneous here. The measured equilibrium concentrations of the reactants and products are used to calculate the equilibrium constant, as shown in the figure below. The equilibrium constant always has the same value (provided the temperature does not change), irrespective of the initial concentrations of A, B, C and D. It is also unaffected by a change in pressure or the presence (or absence) of a catalyst. Compare this with the chemical equation for the equilibrium. The convention is that the substances on the right side of the equation are written in the numerator of the K expression, and those on the left side in the denominator. The exponents are the coefficients of the reactants and products from the equation. A typical reaction is given below: \[ CH_3COOCH_2CH_{3(l)} + H_2O_{(l)} \rightleftharpoons CH_3COOH_{(l)} + CH_3CH_2OH_{(l)}\] There is only one molecule of each reactant or product involved in the reaction, so all the exponents in the equilibrium constant expression are "1," and need not be included in the \(K_c\) ex pression, shown below: As long as the temperature is held constant, \(K_c\) has a constant value regardless of the proportions of acid and alcohol used. At room temperature, this value is approximately 4 for this reaction. This is the reverse of the previous esterification reaction: \[ CH_3COOCH_2CH_{3(l)} + H_2O_{(l)} \rightleftharpoons CH_3COOH_{(l)} + CH_3CH_2OH_{(l)}\] The \(K_c\) ex pression is written as follows: \[ K_c = \dfrac{[CH_3COOH,CH_3CH_2OH]}{[CH_3COOCH_2CH_3,H_2O]}\] Compared with the previous \(K_c\), this expressio n is simply inverted. Its value at room temperature is approximately 1/4 (0.25). It is helpful to write down the chemical equation for a equilibrium reaction whenever discussing an equilibrium constant, to ensure the expression is written correctly (products over reactants). Recall that the equation for this process is as follows: \[ 2SO_{2(g)} + O_2 {(g)} \rightleftharpoons 2SO_{3(g)} \tag{4}\] In this case, the \(K_c\) expressio n includes some nontrivial exponents: \[ K_c = \dfrac{[SO_3]^2}{[SO_2]^2[O_2]} \tag{5}\] Although everything is present as a gas, because \(K_c\) is used he re, the activities are measured in mol dm . There is another equilibrium constant, \(K_p\), that i s more frequently used for gases. The chemical equation for the Haber process is given below: \[ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \tag{6}\] The \(K_c\) expression is as follows: \[ K_c = \dfrac{[NH_3]^2}{[N_2,H_2]^3} \tag{7}\] An examples of a heterogeneous equilibrium is the equilibrium established if steam is in contact with red hot carbon. In this reaction, a gas reacts with a solid: \[ H_2O_{(g)} + C_{(s)} \rightleftharpoons H_{2(g)} + CO_{(g)} \tag{8}\] Another heterogeneous equilibrium involves shaking copper with silver nitrate solution; this reaction involves solids and aqueous ions: \[ Cu_{(s)} + 2Ag^+_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2Ag_{(s)} \tag{9}\] The important consideration for a heterogeneous equilibrium is that there are no solid terms in an equilibrium constant expression; therefore, not every reactant or product will be accounted in \(K_c\). \[ H_2O_{(g)} + C_{(s)} \rightleftharpoons H_{2(g)} + CO_{(g)}\] The equilibrium constant expression is written the same way as in previous examples, omitting the solid carbon term: \[ K_c = \dfrac{[H_2,CO]}{[H_2O]}\] Consider the redox reaction between solid copper and silvir ions in solution: \[ Cu_(s)+ + 2Ag^+_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2Ag_{(s)}\] Both the copper on the left-hand side and the silver on the right are solids. Both are left out of the equilibrium constant expression: \[ K_c = \dfrac{[Cu_{2+}]}{[Ag^+]^2}\] This equilibrium is only established if calcium carbonate is heated in a closed system, preventing carbon dioxide from escaping: \[CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)} \] The only nonsolid species in this system is carbon dioxide, so it is the only term in the equilibrium constant expression: \[ K_c = [CO_2]\] Jim Clark ( ) | 5,218 | 2,702 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Kp |
This page explains equilibrium constants expressed in terms of partial pressures of gases, K . It covers an explanation of the terms mole fraction and partial pressure, and looks at \(K_p\) for both homogeneous and heterogeneous reactions involving gases. The page assumes that you are already familiar with the concept of an equilibrium constant, and that you know about \(K_c\) - an equilibrium constant expressed in terms of concentrations Before we can go any further, there are two terms relating to mixtures of gases that you need to be familiar with. If you have a mixture of gases (A, B, C, etc), then the mole fraction of gas A is worked out by dividing the number of moles of A by the total number of moles of gas. The mole fraction of gas A is often given the symbol x . The mole fraction of gas B would be x - and so on. Pretty obvious really! For example, in a mixture of 1 mole of nitrogen and 3 moles of hydrogen, there are a total of 4 moles of gas. The mole fraction of nitrogen is 1/4 (0.25) and of hydrogen is 3/4 (0.75). The partial pressure of one of the gases in a mixture is the pressure which it would exert if it alone occupied the whole container. The partial pressure of gas A is often given the symbol P . The partial pressure of gas B would be P - and so on. There are two important relationships involving partial pressures. The first is again fairly obvious. The total pressure of a mixture of gases is equal to the sum of the partial pressures. It is easy to see this visually: Gas A is creating a pressure (its partial pressure) when its molecules hit the walls of its container. Gas B does the same. When you mix them up, they just go on doing what they were doing before. The total pressure is due to both molecules hitting the walls - in other words, the sum of the partial pressures. The more important relationship is the second one: Learn it! That means that if you had a mixture made up of 20 moles of nitrogen, 60 moles of hydrogen and 20 moles of ammonia (a total of 100 moles of gases) at 200 atmospheres pressure, the partial pressures would be calculated like this: 0.2 x 200 = 40 atm Partial pressures can be quoted in any normal pressure units. The common ones are atmospheres or pascals (Pa). Pascals are exactly the same as N m (newtons per square meter). A homogeneous equilibrium is one in which everything in the equilibrium mixture is present in the same phase. In this case, to use K , everything must be a gas. A good example of a gaseous homogeneous equilibrium is the conversion of sulfur dioxide to sulfur trioxide at the heart of the Contact Process: We are going to start by looking at a general case with the equation: If you allow this reaction to reach equilibrium and then measure (or work out) the equilibrium partial pressures of everything, you can combine these into the equilibrium constant, K . Just like K , \(K_p\) always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with. K has exactly the same format as K , except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom. You will remember that the equation for this is: K is given by: The equation for this is: . . . and the \(K_p\) expression is: A typical example of a heterogeneous equilibrium will involve gases in contact with solids. Exactly as happens with K , you don't include any term for a solid in the equilibrium expression. The next two examples have already appeared on the \(K_c\) page. Everything is exactly the same as before in the expression for K , except that you leave out the solid carbon. This equilibrium is only established if the calcium carbonate is heated in a closed system, preventing the carbon dioxide from escaping. The only thing in this equilibrium which isn't a solid is the carbon dioxide. That is all that is left in the equilibrium constant expression. Jim Clark ( ) | 4,056 | 2,703 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reaction_Fundamentals/Reaction_Coordinates_in_Potential_Energy_Diagrams |
Reaction are graphs that show the energy of a process as a function of the extent to which that process has occurred. As these are graphs showing mathematical functions, there must be a numerical coordinate axis that shows the independent variable. This coordinate is called the , and it reflects the geometry of the system. Very often, the reaction coordinate reflects extent to which a reaction has progressed from reactants to products, starting with reactants near the y-axis (the energy coordinate) and progressing toward products. The simplest reaction coordinate is found for a diatomic system. In this case, the geometry in the molecular frame is completely described by the intermolecular distance, r. Therefore, the potential energy diagram for the dissociation of a simple diatomic molecule consists of a plot of the energy of the system as a function of separation. For systems of 3 or more atoms, the geometry coordinates get more complicated. A non-linear molecule consisting of N atoms will have 3N-6 different geometry coordinates, and there are 3N-5 for linear molecules (this also applies to N = 2). That means that a depiction of the potential energy for a non-linear molecule isa 3N - 6 dimensional surface (3N - 5 dimensions for a linear molecule). Formally, the coordinate axes consist of the mathematical that describe motion of the atoms in the molecule, although more physical meaningful coordinates can also be utilized. A non-linear triatomic molecule can be described with three coordinates. The best example is water, HOH. The three coordinates in water are nominally the two O-H bond lengths and the H-O-H bond angle. Mathematically, these are described as symmetric and asymmetric OH stretching motion, and the HOH bend. Considering the challenge in representing a 4-D system (3 geometry coordinates and an energy coordinate), there are simplifications and/or approximations that can be made. For example, it is to create a plot where one or two of the variables are fixed. For example, the energy can be plotted against the HOH bond angle, holding the OH bonds at their equilibrium bond lengths. Because the optimal OH bond lengths do not vary significantly as a function of bond angle, a plot of energy vs bond angle in water will be very close to the lowest energy part of the potential energy surface. Alternatively, it is possible to fix the bond angle and one of the OH bond lengths, and look at the energy as a function of the length of the other OH bond. Although this is likely a reasonable approximation of the lowest energy path for breaking the O-H bond, it does not account for changes in the preferred bond angle that might occur, or, to a lesser extent, the change in the OH bond length between water and hydroxyl radical. Alternatively, it is possible to plot a for O-H breaking where the other variables (HOH bond angle, other O-H bond length) are always at the optimal values as a function of the breaking O-H bond length. In this plot, the energy does account for the changes in the preferred bond angle and the change to the hydroxyl bond length. If there is a readily available numerical coordinate to use, then a relaxed surface scan is very convenient and is an accurate reflection of the surface, since it refers to the lowest energy pathway in that coordinate. The potential energy surface for bond rotation is an example of a relaxed surface scan. The reaction coordinate is the dihedral angle between groups on the two atoms, which can be easily observed in a . Potential energy surfaces for bond rotation are commonly used for conformational analysis of molecules like and . Most organic chemical reactions cannot be described by a single simple coordinate. Even when they can, it may be more convenient to use a more complicated coordinate. For example, chloride undergoes S 2 reaction with methyl bromide through a 5-centered transition state. In principle, the energy of the process can be plotted vs the Cl-C bond length. Therefore, the energy increases gradually as chloride approaches from long distance until the bond length reaches that in the transition state, which is approximately 200 pm. From there, the energy will decrease precipitously as the Cl-C bond length approaches the value in CH Cl, 175 pm. A more convenient coordinate might utilize the Cl-C bond length until the formation of the transition state, and switch to the C-Br bond length after the transition state. Alternatively, if using the Cl-C bond length, it might be better to not to use a linear scale, or even consistent scaling. Consequently, most potential energy surfaces in organic chemistry are not drawn with any single, numerical coordinate, but with a generalized reaction coordinate that reflects the geometry changes in the reaction. Instead of numerical labels on the axis, the positions are labeled with their structures, and the regions in-between are the paths that connect the indicated geometries. Because of this, it is possible to reduce the highly-dimensional surfaces of complicated reactions to 1-dimensional curves. Energy is on the y-axis, and the x-axis indicates geometry. Energy is a function of geometry. . Each point on the diagram has the same molecular formula (same atoms and electrons). The bonding can change, however. For example, in the S 2 process shown above, the reactant is Cl + CH Br, and the product is CH Cl + Br . Even if this is not explicitly shown, it is implicit in the diagram. It is not possible to compare energies of different atoms. A lot of important details about the reaction can be lost in the projection. It can be helpful to think of different steps in the reaction as happening in orthogonal directions - for example, if one occurs in the plane of the drawing, the next step might come out toward you, or away from you. | 5,848 | 2,705 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Michael_Additions_and_Robinson_Annulation |
Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942). Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred. ) | 738 | 2,706 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO8._Semi-Anionic_Nucleophiles |
Some nucleophiles are added to carbonyls in the form of salts, such as sodium cyanide. In a salt, there is an anion and a cation. The anion can act as a nucleophile, donating a lone pair to the carbonyl. The cation is just a counterion; it is there to balance the charge but doesn't usually play an active role. Some anions are too unstable and reactive to be used as salts. This is especially true with a number of carbon nucleophiles. C-H bonds are not usually acidic enough to deprotonate with a strong base. That makes it hard to make a simple salt containing such nucleophiles. There are exceptions, such as acetylide or alkynyl protons like CH CCH. In that case, the resulting anion is relatively stable because the lone pair is in a lower-energy orbital with more s character, so it is held more tightly to the nucleus. Less stable carbon anions can be be stabilized through a covalent bond. If the carbon is covalently attached to a less electronegative atom, the carbon has a partially negative charge. It can still act as though it were an anion. However, the covalent bond stabilizes the would-be "lone pair." Compounds like this can be considered to be "semi-anionic." Frequently, they are described as polar covalent compounds, although that is really a much more general term. These polar covalent bonds can be found any time a carbon atom is bound to a metal. Remember, the metals are the aroms in the coloured boxes in the periodic table below. One of the most common classes of this type of compounds is the family of organomagnesium halides or Grignard reagents (Green-yard reagents). was awarded the Nobel Prize in Chemistry for his development of these reagents. These compounds are made by reacting an alkyl halide (such as CH CH CH Cl) with magnesium metal. The metal undergoes an insertion into the C-Cl bond, forming CH CH CH MgCl. (You don't need to worry about how this happens.) Because magnesium is less electronegative than carbon, the C-Mg bond acts as though it were a lone pair on the carbon and the magnesium acts as though it were a cation. What's striking about Grignard formation is that polarity is reversed in this reaction. In the alkyl halide, the carbon attached to the halogen has a partially positive charge, because carbon is further to the left than halogens in the periodic table. After magnesium insertion, this same carbon has a partial negative charge, because carbon is farther to the right in the periodic table than magnesium. This sort of reversal in reactivity is sometimes called "umpolung chemistry." Propylmagnesium chloride and other Grignard reagents can deliver alkyl nucleophiles to carbonyls. Just like with simple anionic nucleophiles, an alkoxide ion results. Subsequent treatment of the alkoxide with acid provides a proton, resulting in an alcohol. Remember, the order of these two steps is very important. Adding the acid before the Grignard reagent would not work, because the Grignard reagent would become protonated at the carbon. Although the Mg-C bond is covalent, it is still polar enough so that the carbon can act as a nucleophile or as a base. Once propylmagnesium has become protonated, it forms propane, which isn't likely to act as a nucleophile. Show the products of the following reactions. Assume workup with aqueous acid after each reaction. The situation with Grignard reagents is even more delicate than that. Solvents must be chosen very carefully for Grignard reactions. Grignard reagents are basic enough that they can't tolerate protic solvents. Protic solvents are solvents that are capable of hydrogen-bonding. Although they don't seem very acidic, they can still give up a proton to a strong enough base. A Grignard reagent is a strong enough base to take that proton. Show why Grignard reagents cannot be used with ethanol as a solvent. In fact, Grignard reagents are even fussier than that. Not only do they not get along well with acidic or even semi-acidic protons, but they tend to need coordinating solvents to help support the magnesium atom and keep the complex stable. The most common solvents for this use are (diethyl) ether and tetrahydrofuran (THF). Because coordination of magnesium by these weakly donating solvents is crucial, Grignard reagents can't generally be isolated. Instead, they are sold and used as solutions in ethereal solvents. There are plenty of other semi-anionic nucleophiles. For example, alkyl lithium reagents are also very common, and they are prepared by treatment of alkyl halides with finely divided lithium metal. The reaction produces lithium chloride as a side product. Another class of semi-anionic nucleophiles is the family of complex metal hydrides. Examples include sodium borohydride, NaBH , and lithium aluminum hydride (LAH), LiAlH . There are many other variations. Barbier reactions are a general class of reactions involving metal alkyls and carbonyls. Treatment of a halide such as propargyl bromide (HCCCH Br) with zinc metal in the presence of an aldehyde such as benzaldehyde (C H CHO) results in nucleophilic addition of the propargyl group to the aldehyde. Semi-anionic nucleophiles do not just react with carbonyls. They are also frequently used to prepare organometallic compounds via "transmetallation." For example, treatment of tantalum pentachloride, TaCl , with dimethylzinc, (CH ) Zn, affords trimethyltantalum dichloride, (CH ) TaCl . , | 5,412 | 2,707 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkenes_from_Aldehydes_and_Ketones_-_Wittig_Reaction |
The or Wittig olefination is a chemical reaction of an aldehyde or ketone with a triphenyl phosphonium ylide (often called a ) to give an alkene and triphenylphosphine oxide. The Wittig reaction was discovered in 1954 by Georg Wittig, for which he was awarded the Nobel Prize in Chemistry in 1979. It is widely used in organic synthesis for the preparation of alkenes. It should not be confused with the Wittig rearrangement. Wittig reactions are most commonly used to couple aldehydes and ketones to singly substituted phosphine ylides. With unstabilised ylides this results in almost exclusively the Z-alkene product. In order to obtain the E-alkene, stabilised ylides are used or unstabilised ylides using the Schlosser modification of the Wittig reaction can be performed. The steric bulk of the ylide influences the stereochemical outcome of nucleophilic addition to give a predominance of the betaine (cf. Bürgi–Dunitz angle). Note that for betaine both R and R as well as PPh3+ and O− are positioned anti to one another. Carbon-carbon bond rotation gives the betaine , which then forms the oxaphosphetane . Elimination gives the desired Z-alkene and triphenylphosphine oxide . With simple Wittig reagents, the first step occurs easily with both aldehydes and ketones, and the decomposition of the betaine (to form ) is the rate-determining step. However, with stabilised ylides (where R stabilises the negative charge) the first step is the slowest step, so the overall rate of alkene formation decreases and a bigger proportion of the alkene product is the E-isomer. This also explains why stabilised reagents fail to react well with sterically hindered ketones. Mechanistic studies have focused on unstabilized ylides, because the intermediates can be followed by NMR spectroscopy. The existence and interconversion of the betaine ( and ) is subject of ongoing research. Phosphonium ylides react with carbonyl compounds via a π²s/π²a [2+2] cycloaddition to directly form the oxaphosphetanes and . The stereochemistry of the product is due to the addition of the ylide to the carbonyl and to the equilibration of the intermediates. Maryanoff and Reitz identified the issue about equilibration of Wittig intermediates and termed the process "stereochemical drift". For many years, the stereochemistry of the Wittig reaction, in terms of carbon-carbon bond formation, had been assumed to correspond directly with the Z/E stereochemistry of the alkene products. However, certain reactants do not follow this simple pattern. Lithium salts can also exert a profound effect on the stereochemical outcome. Mechanisms differ for aliphatic and aromatic aldehydes and for aromatic and aliphatic phosphonium ylides. Evidence suggests that the Wittig reaction of unbranched aldehydes under lithium-salt-free conditions do not equilibrate and are therefore under kinetic reaction control. Vedejs has put forth a theory to explain the stereoselectivity of stabilized and unstabilized Wittig reactions. Wittig reagents are usually prepared from a phosphonium salt, which is in turn prepared by the quaternization of triphenylphosphine with an alkyl halide. The alkylphosphonium salt is deprotonated with a strong base such as -butyllithium: One of the simplest ylides is methylenetriphenylphosphorane (Ph P=CH ). It is also a precursor to more elaborate Wittig reagents. Alkylation of Ph P=CH with a primary alkyl halide R−CH −X, produces substituted phosphonium salts: These salts can be deprotonated in the usual way to give Ph P=CH−CH R. The Wittig reagent may be described in the form (the more familiar representation) or the form: The ylide form is a significant contributor, and the carbon is nucleophilic. Simple phosphoranes are reactive. Most hydrolyze and oxidize readily. They are therefore prepared using air-free techniques. Phosphoranes are more air-stable when they contain an electron withdrawing group. Some examples are Ph P=CHCO R and Ph P=CHPh. These ylides are sufficiently stable to be sold commercially The Wittig reaction is a popular method for the synthesis of alkene from ketones and aldehydes. The Wittig reagent can generally tolerate carbonyl compounds containing several kinds of functional groups such as OH, OR, aromatic nitro and even ester groups. There can be a problem with sterically hindered ketones, where the reaction may be slow and give poor yields, particularly with stabilized ylides, and in such cases the Horner–Wadsworth–Emmons (HWE) reaction (using phosphonate esters) is preferred. Another reported limitation is the often labile nature of aldehydes which can oxidize, polymerize or decompose. In a so-called Tandem Oxidation-Wittig Process the aldehyde is formed n situ by oxidation of the corresponding alcohol. As mentioned above, the Wittig reagent itself is usually derived from a primary alkyl halide. Quaternization of triphenylphosphine with most secondary halides is inefficient. For this reason, Wittig reagents are rarely used to prepare tetrasubstituted alkenes. However the Wittig reagent can tolerate many other variants. It may contain alkenes and aromatic rings, and it is compatible with ethers and even ester groups. Even C=O and nitrile groups can be present if conjugated with the ylide- these are the stabilised ylides mentioned above. Bis-ylides (containing two P=C bonds) have also been made and used successfully. One limitation relates to the stereochemistry of the product. With simple ylides, the product is usually mainly the Z-isomer, although a lesser amount of the E-isomer is often formed also – this is particularly true when ketones are used. If the reaction is performed in DMF in the presence of LiI or NaI, the product is almost exclusively the Z-isomer. If the E-isomer is the desired product, the Schlosser modification may be used. With stabilised ylides the product is mainly the E-isomer, and this same isomer is also usual with the HWE reaction. The major limitation of the traditional Wittig reaction is that the reaction proceeds mainly via the erythro betaine intermediate, which leads to the Z-alkene. The erythro betaine can be converted to the threo betaine using phenyllithium at low temperature. This modification affords the E-alkene. Allylic alcohols can be prepared by reaction of the betaine ylid with a second aldehyde. For example: Because of its reliability and wide applicability, the Wittig reaction has become a standard tool for synthetic organic chemists. The most popular use of the Wittig reaction is for the introduction of a methylene group using methylenetriphenylphosphorane (Ph P=CH ). Using this reagent even a sterically hindered ketone such as camphor can be converted to its methylene derivative. In this case, the Wittig reagent is prepared by deprotonation of methyltriphenylphosphonium bromide with potassium tert-butoxide. In another example, the phosphorane is produced using sodium amide as a base, and this reagent converts the aldehyde shown into alkene in 62% yield. The reaction is performed in cold THF, and the sensitive nitro, azo and phenoxide groups are tolerated. The product can be used to incorporate a photostabiliser into a polymer, to protect the polymer from damage by UV radiation. Another example of its use is in the synthesis of leukotriene A methyl ester. The first step uses a stabilised ylide, where the carbonyl group is conjugated with the ylide preventing self condensation, although unexpectedly this gives mainly the product. The second Wittig reaction uses a non-stabilised Wittig reagent, and as expected this gives mainly the product. Note that the epoxide and ester functional groups survive intact. Methoxymethylenetriphenylphosphine is a Wittig reagent for the homologation of aldehydes. | 7,813 | 2,708 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/11%3A_Chemical_Kinetics_I/11.07%3A_The_Method_of_Initial_Rates |
The is a commonly used technique for deriving rate laws. As the name implies, the method involves measuring the initial rate of a reaction. The measurement is repeated for several sets of initial concentration conditions to see how the reaction rate varies. This might be accomplished by determining the time needed to exhaust a particular amount of a reactant (preferably one on which the reaction rate does not depend!) A typical set of data for a reaction \[A + B \rightarrow products \nonumber \] might appear as follows: The analysis of this data involves taking the ratios of rates measured where one of the concentrations does not change. For example, assuming a rate law of the form \[ \text{rate} = k [A]^{\alpha}[B]^{\beta} \label{orRL} \] The ratio of runs \(i\) and \(j\) generate the following relationship. \[ \dfrac{\text{rate}_i}{\text{rate}_j} = \dfrac{k [A]_i^{\alpha}[B]_i^{\beta}}{k [A]_j^{\alpha}[B]_j^{\beta}} \nonumber \] So using runs \(1\) and \(2\), \[ \dfrac{0.0347\, M/s}{0.0694\, M/s} = \dfrac{\cancel{k} (0.01\,M/s)^{\alpha} \cancel{(0.01\,M/s)^{\beta}}}{\cancel{k} (0.02\,M/s)^{\alpha} \cancel{(0.01\,M/s)^{\beta}}} \nonumber \] this simplifies to \[ \dfrac{1}{2} = \left( \dfrac{1}{2} \right)^{\alpha} \nonumber \] So clearly, \(\ = 1\) and the reaction is 1 order in \(A\). Taking the ratio using runs 2 and 3 yields \[ \dfrac{0.0694\, M/s}{0.2776\, M/s} = \dfrac{\cancel{k} (0.02\,M)^{\alpha} \cancel{(0.01\,M)^{\beta}}}{\cancel{k} (0.02\,M)^{\alpha} \cancel{(0.02\,M)^{\beta}}} \nonumber \] This simplifies to \[ \dfrac{1}{4} = \left( \dfrac{1}{2} \right)^{\beta} \label{Me1} \] By inspection, one can conclude that \(\ = 2\), and that the reaction is second order in B. But if it is not so clear (as it might not be if the concentration is not incremented by a factor of 2), the value of \(\ can be determined by taking the natural logarithm of both sides of the Equation \(\ref{Me1}\). \[ \ln \dfrac{1}{4} = \ln \left( \dfrac{1}{2} \right)^{\beta} \nonumber \] \[ = \beta \ln \left( \dfrac{1}{2} \right) \nonumber \] dividing both sides by \(\ln(1/2)\) \[ \dfrac{ \ln\left( \dfrac{1}{4} \right)}{\ln\left( \dfrac{1}{2} \right)} =\beta \dfrac{ \ln \left( \dfrac{1}{2} \right)}{\ln \left( \dfrac{1}{2} \right)} \nonumber \] or \[ \beta = \dfrac{-1.3863}{-0.69315} = 2 \nonumber \] And so the rate law (Equation \ref{orRL}) can be expressed as \[ \text{rate} = k [A,B]^{2} \nonumber \] And is 1 order in A, 2 order in B, and 3 order overall. The rate constant can then be evaluated by substituting one of the runs into the rate law (or using all of the data and taking an average). Arbitrarily selecting the first run for this, \[ 0.0347 \,M/s = k (0.01 \, M/s)(0.01 \, M/s)^{2} \nonumber \] This results in a value of \(k\) \[ k = \dfrac{0.0347 \,M/s} {(0.01 \, M/s)(0.01 \, M/s)^2} = 3.47 \times 10^{5} \, M^{-2} s^{-1} \nonumber \] It is useful to note that the units on \(k\) are consistent with a 3 order rate law. | 2,985 | 2,709 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/05%3A_Cooperativity/18%3A_Cooperativity/18.02%3A_Two-State_Thermodynamics |
Here we describe the basic thermodynamics of two-state systems, which are commonly used for processes such as protein folding, binding, and DNA hybridization. Working with the example of protein folding analyzed through the temperature-dependent folded protein content. \[K=\dfrac{k_f}{k_u} = \dfrac{[F]}{[U]} = \dfrac{\phi_F}{1-\phi_F} \nonumber \] where φ is the fraction of protein that is folded, and the fraction that is unfolded is (1 ‒ φ ). \[ \begin{aligned} &\phi_F =\dfrac{K}{K+1} \\ &K= e^{-\Delta G^0/RT} \\ &\phi_F = \dfrac{1}{1+e^{-\Delta G^0/RT}} = \dfrac{1}{1+e^{\Delta H^0/RT} e^{-\Delta S^0/R}} \end{aligned} \] Define the melting temperature T as the temperature at which φ = 0.5. Then at T , \(\Delta G^0=0 \mathrm{or} T_m = \Delta H^0/\Delta S^0 \). Characteristic melting curves for T = 300 K are below: We can analyze the slope of curve at T using a van’t Hoff analysis: \[ \begin{aligned} \dfrac{d\phi_F}{dT}&=\dfrac{d\phi_F}{dK} \cdot \dfrac{dK}{dT}=\dfrac{d\phi_F}{dK}\cdot K \dfrac{d\ln{K}}{dT}\\ \dfrac{d\ln{K}}{dT} &= \dfrac{\Delta H^0}{RT^2} \\ \dfrac{d\phi_F}{dK} &=K^{-2}(1+K)^{-2} \\ \left( \dfrac{d\phi_F}{dT} \right)_{T=T_m} &= \dfrac{\Delta H^0}{4RT^2_m} \quad \mathrm{since} \; K=1 \; \mathrm{at} \; T_m \end{aligned} \] This analysis assumes that there is no temperature dependence to ΔH, although we know well that it does from our earlier discussion of hydrophobicity. A more realistic two-state model will allow for a change in heat capacity between the U and F states that describes the temperature dependence of the enthalpy and entropy. \[ \Delta G^0(T) = \Delta H^0(T_m)-T\Delta S^0(T_m)+ \Delta C_p \left[ T-T_m-T \ln{\dfrac{T}{T_m}} \right] \nonumber \] | 1,725 | 2,710 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/The_Mass_Spectra_of_Elements |
This page looks at the information you can get from the mass spectrum of an element. It shows how you can find out the masses and relative abundances of the various isotopes of the element and use that information to calculate the relative atomic mass of the element. It also looks at the problems thrown up by elements with diatomic molecules - like chlorine, Cl . Monatomic elements include all those except for things like chlorine, Cl , with molecules containing more than one atom. : The two peaks in the mass spectrum shows that there are 2 isotopes of boron - with relative isotopic masses of 10 and 11 on the C scale. : The relative sizes of the peaks gives you a direct measure of the relative abundances of the isotopes. The tallest peak is often given an arbitrary height of 100 - but you may find all sorts of other scales used. It doesn't matter in the least. You can find the relative abundances by measuring the lines on the stick diagram. In this case, the two isotopes (with their relative abundances) are: The relative atomic mass (RAM) of an element is given the symbol and is defined as: The relative atomic mass of an element is the weighted average of the masses of the isotopes on a scale on which a carbon-12 atom has a mass of exactly 12 units A "weighted average" allows for the fact that there won't be equal amounts of the various isotopes. The example coming up should make that clear. Suppose you had 123 typical atoms of boron. 23 of these would be B and 100 would be B. The total mass of these would be (23 x 10) + (100 x 11) = 1330 The average mass of these 123 atoms would be 1330 / 123 = 10.8 (to 3 significant figures). 10.8 is the relative atomic mass of boron. Notice the effect of the "weighted" average. A simple average of 10 and 11 is, of course, 10.5. Our answer of 10.8 allows for the fact that there are a lot more of the heavier isotope of boron - and so the "weighted" average ought to be closer to that. The 5 peaks in the mass spectrum shows that there are 5 isotopes of zirconium - with relative isotopic masses of 90, 91, 92, 94 and 96 on the C scale. : This time, the relative abundances are given as percentages. Again you can find these relative abundances by measuring the lines on the stick diagram. In this case, the 5 isotopes (with their relative percentage abundances) are: Suppose you had 100 typical atoms of zirconium. 51.5 of these would be Zr, 11.2 would be Zr and so on. The total mass of these 100 typical atoms would be (51.5 x 90) + (11.2 x 91) + (17.1 x 92) + (17.4 x 94) + (2.8 x 96) = 9131.8 The average mass of these 100 atoms would be 9131.8 / 100 = 91.3 (to 3 significant figures). 91.3 is the relative atomic mass of zirconium. Chlorine is taken as typical of elements with more than one atom per molecule. We'll look at its mass spectrum to show the sort of problems involved. Chlorine has two isotopes, Cl and Cl, in the approximate ratio of 3 atoms of Cl to 1 atom of Cl. You might suppose that the mass spectrum would look like this: You would be wrong! The problem is that chlorine consists of molecules, not individual atoms. When chlorine is passed into the ionization chamber, an electron is knocked off the molecule to give a molecular ion, Cl . These ions won't be particularly stable, and some will fall apart to give a chlorine atom and a Cl ion. The term for this is fragmentation. \[\ce{ Cl_2^{+} \rightarrow Cl + Cl^{+}}\] If the Cl atom formed is not then ionized in the ionization chamber, it simply gets lost in the machine - neither accelerated nor deflected. The Cl ions will pass through the machine and will give lines at 35 and 37, depending on the isotope and you would get exactly the pattern in the last diagram. The problem is that you will also record lines for the unfragmented Cl ions. Think about the possible combinations of chlorine-35 and chlorine-37 atoms in a Cl ion. Both atoms could be Cl, both atoms could be Cl, or you could have one of each sort. That would give you total masses of the Cl ion of: 35 + 35 = 70 35 + 37 = 72 37 + 37 = 74 That means that you would get a set of lines in the m/z = 70 region looking like this: These lines would be in addition to the lines at 35 and 37. The relative heights of the 70, 72 and 74 lines are in the ratio 9:6:1. If you know the right bit of math, it's very easy to show this. If not, don't worry. Just remember that the ratio is 9:6:1. What you can't do is make any predictions about the relative heights of the lines at 35/37 compared with those at 70/72/74. That depends on what proportion of the molecular ions break up into fragments. That's why you've got the chlorine mass spectrum in two separate bits so far. You must realise that the vertical scale in the diagrams of the two parts of the spectrum isn't the same. The overall mass spectrum looks like this: Jim Clark ( ) | 4,871 | 2,711 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/16%3A_Targeted_Diffusion/16.03%3A_Mean_First_Passage_Time |
Another way of describing diffusion-to-target rates is in terms of first passage times. The mean first passage time (MFPT), ⟨τ⟩, is the average time it takes for a diffusing particle to reach a target position for the first time. The inverse of ⟨τ⟩ gives the rate of the corresponding diffusion-limited reaction. A first passage time approach is particularly relevant to problems in which a description the time-dependent averages hide intrinsically important behavior of outliers and rare events, particularly in the analysis of single molecule kinetics. To describe first passage times, we begin by defining the reaction probability R and the survival probability S. R is a conditional probability function that describes the probability that a molecule starting at a point \(x_0=0\) at time t will reach a reaction boundary at x = x after time t: R(x ,t|x ,t ). S is just the conditional probability that the molecule has reached x = b during that time interval: S(x ,t|x ,t ). Therefore \[ R+S=1 \nonumber \] Next, we define F(τ,x |x ), the first passage probability density. F(τ)dτ is the probability that a molecule passes through x = x for the first time between times τand τ+dτ. R, S, and F are only a function of time for a fixed position of the reaction boundary, i.e. they integrate over any spatial variations. To connect F with the survival probability, we recognize that the reaction probability can be obtained by integrating over all possible first passage times for time intervals τ < t. Dropping space variables, recognizing that (t‒t ) = τ, and setting x = 0, \[R(t) = \int^t_0 F(\tau )d\tau \nonumber \] This relation implies that the first passage time distribution can be obtained by differentiating S \[ F(t) = \dfrac{\partial }{\partial t}R(t) = - \dfrac{\partial }{\partial t} S(t) \] Then the MFPT is obtained by averaging over F(t) \[ \langle \tau \rangle = \int^{\infty }_0 \tau F(\tau ) d\tau \] To evaluate these quantities for a particular problem, we seek to relate them to the time-dependent probability density, P(x,t|x ,t ), which is an explicit function of time and space. The connection between P and F is not immediately obvious because evaluating P at x = x without the proper boundary conditions includes trajectories that have passed through x = x before returning there again later. The key to relating these is to recognize that the survival probability can be obtained by calculating a diffusion problem with an absorbing boundary condition at x = x that does not allow the particle to escape: P(x ,t|x ) = 0. The resulting probability distribution P (x,t|x ,t ) is not conserved but gradually loses probability density with time. Hence, we can see that the survival probability is an integral over the remaining probability density that describes particles that have not yet reached the boundary: \[ S(t) = \int^{x_f}_{-\infty }dxP_a(x,t) \] The mean free passage time can be written as \[ \langle \tau \rangle = \int^{x_f}_{-\infty }dx \int^{\infty}_{0}dt \: P_a(x,t) \nonumber \] The next important realization is that the first passage time distribution is related to the flux of diffusing particles through x . Combining eq. (16.3.1) and (16.3.3) shows us \[ F(t) = -\int^{x_f}_{-\infty }dx \dfrac{\partial }{\partial t} P_a(x,t) \] Next we make use of the continuity expression for the probability density \[ \dfrac{\partial P}{\partial t} = -\dfrac{\partial j}{\partial x} \nonumber \] j is a flux, or probability current, with units of s , not the flux density we used for continuum diffusion J (m s ). Then eq. (16.3.4) becomes \[ \begin{aligned} F(t) &= \int^{x_f}_{-\infty}dx \dfrac{\partial }{\partial x} j_a(x,t) \\ &=j_a(x_f,t) \end{aligned} \] \[\] So the first passage time distribution is equal to the flux distribution for particles crossing the boundary at time t. Furthermore, from eq. (16.3.2), we see that the MFPT is just the inverse of the average flux of particles crossing the absorbing boundary: \[\langle \tau \rangle = \dfrac{1}{\langle j_a(x_f) \rangle } \] In chemical kinetics, \( \langle j_a (x_f) \rangle \) is the rate constant from transition state theory. To calculate F one needs to solve a Fokker–Planck equation for the equivalent diffusion problem with an absorbing boundary condition. As an example, we can write these expressions explicitly for diffusion from a point source. This problem is solved using the Fourier transform method, applying absorbing boundary conditions at x to give \[P_a (x,t) = P(x,t)-P(2x_f-x,t) \qquad \qquad (x \leq x_f) \nonumber \] which is expressed in terms of the probability distribution in the absence of absorbing boundary conditions: \[ P(x,t) = (4\pi Dt)^{1/2}\mathrm{exp}\left[ \dfrac{-(x-x_0)^2}{4Dt} \right] \nonumber \] The corresponding first passage time distribution is: \[F(t) = \dfrac{x_f-x_0}{(4\pi Dt^3)^{1/2}} \mathrm{exp}\left[ -\dfrac{(x-x_0)^2}{4Dt} \right] \nonumber \] F(t) decays in time as t , leading to a long tail in the distribution. The mean of this distribution gives the MFPT \[ \langle \tau \rangle = x^2_f/2D \nonumber \] and the most probable passage time is x /6D. Also, we can use eq. (16.3.3) to obtain the survival probability \[ S(t) = \mathrm{erf}\left( \dfrac{x_f}{\sqrt{4Dt}} \right) = \mathrm{erf}\left( \sqrt{\dfrac{\langle \tau \rangle }{2t}} \right) \nonumber \] S(t) depends on the distance of the target and the rms diffusion length over time t. At long times S(t) decays as t . It is interesting to calculate the probability that the diffusing particle will reach xf at any time. From eq. (16.3.4), we can see that this probability can be calculated from \(\int^{\infty}_0 F(\tau )d\tau \). For the current example, this integral over F gives unity, saying that a random walker in 1D will eventually reach every point on a line. Equivalently, it is guaranteed to return to the origin at some point in time. This observation holds in 1D and 2D, but not 3D. From eq. (16.3.6) we see that it is also possible to calculate the MFPT by solving for the flux at an absorbing boundary in a steady state calculation. As a simple example, consider the problem of releasing a particle on the left side of a box, \(P(x, 0) = \delta (x,x_0)\), and placing the reaction boundary at the other side of the box x = b. We solve the steady-state diffusion equation \(\partial^2P_a/\partial x^2=0\) with an absorbing boundary at x = b, i.e., \(P(b,t)=0\). This problem is equivalent to absorbing every diffusing particle that reaches the right side and immediately releasing it again on the left side. The steady-state solution is \[ P_a(x) = \dfrac{2}{b}\left( 1-\dfrac{x}{b} \right) \nonumber \] Then, we can calculate the flux of diffusing particles at x=b: \[ j(b) = \left. -D\dfrac{\partial P}{\partial x} \right|_{x=b} = \dfrac{2D}{b^2}\nonumber \] and from the inverse we obtain the MFPT: \[\langle \tau \rangle = \dfrac{1}{j(b)} = \left( \dfrac{b^2}{2D} \right) \nonumber \] To extend this further, let’s examine a similar 1D problem in which a particle is released at x = 0, and diffuses in x toward a reaction boundary at x = b, but this time under the influence of a potential U(x). We will calculate the MFPT for arrival at the boundary. Such a problem could be used to calculate the diffusion of an ion through an ion channel under the influence of the transmembrane electrochemical potential. From our earlier discussion of diffusion in a potential, the steady state flux is \(j=\dfrac{-D\left[ P(b) e^{U(b)/k_BT}-P(x)e^{U(x)/k_BT} \right]}{\int^b_x e^{U(x')/k_BT}dx'} \) Applying the absorbing boundary condition, P(b) = 0, the steady state probability density is \[P_a(x) = \dfrac{j}{D}e^{-U(x)/k_BT} \int^b_x e^{U(x')/k_BT} dx' \] Now integrating both sides over the entire box, the left side is unity, so we obtain an expression for the flux \[ \dfrac{1}{j} = \dfrac{1}{D} \int^b_0 e^{-U(x)/k_BT}\left[ \int^b_x e^{U(x')/k_BT}/k_BT dx' \right] dx \] But j is just the MFPT, so this expression gives us ⟨τ⟩. Note that if we set U to be a constant in eq. (16.3.8), that we recover the expressions for ⟨τ⟩, j, and P in the preceding example. For the case of a linear external potential, we can write the potential in terms of a constant external force \(U=-fx\). Solving this with the steady state solution, we substitute U into eq. (16.3.8) and obtain \[ \langle \tau \rangle = \dfrac{1}{j} = \dfrac{1}{D\underset{\sim}{f }^2} \left[ e^{-\underset{\sim}{f }b}-1+\underset{\sim}{f }b \right] \] where \(\underset{\sim}{f}=f/k_BT\) is the force expressed in units of thermal energy. Substituting into eq. (16.3.7) gives the steady state probability density \[ P(x) = \dfrac{\underset{\sim}{f }\left( 1-e^{\underset{\sim}{f }(b-x)} \right) }{e^{-\underset{\sim}{f }b}-1+\underset{\sim}{f }b} \nonumber\] Now let’s compare these results from calculations using the first passage time distribution. This requires solving the diffusion equation in the presence of the external potential. In the case of a linear potential, we can solve this by expressing the constant force as a drift velocity \[ v_x = \dfrac{f}{\zeta} = \dfrac{fD}{k_BT} = \underset{\sim}{f }D \nonumber \] Then the solution is obtained from our earlier example of diffusion with drift: \[P(x,t) = -\dfrac{1}{\sqrt{4\pi Dt}}\mathrm{exp} \left[ -\dfrac{(x-\underset{\sim}{f }Dt)^2}{4Dt} \right] \nonumber \] The corresponding first passage time distribution is \[F(t) = \dfrac{b}{\sqrt{4\pi Dt^3}}\mathrm{exp}\left[ -\dfrac{(b-\underset{\sim}{f }Dt)^2}{4Dt} \right] \nonumber \] and the MFPT is given by eq. (16.3.9). __________________________________ | 9,638 | 2,712 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.02%3A_Heat_Engines_and_the_Carnot_Cycle |
Sadi Carnot (1796 – 1832) (Mendoza, 2016), a French physicist and engineer was very interested in the improvement of steam engines to perform the tasks needed by modern society. To simplify his analysis of the inner workings of an engine, Carnot devised a useful construct for examining what affect engine efficiency. His construct is the . The idea behind a heat engine is that it will take energy in the form of heat, and transform it into an equivalent amount of work. Unfortunately, such a device is impractical. As it turns out, nature prevents the complete conversion of energy into work with perfect efficiency. This leads to an important statement of the . It is impossible to convert heat into an equivalent amount of work without some other changes occurring in the universe. As such, a more reasonable picture of the heat engine is one which will allow for losses of energy to the surroundings. The fraction of energy supplied to the engine that can be converted to work defines the efficiency of the engine. The Carnot cycle is a theoretical cyclic heat engine that can used to examine what is possible for an engine for which the job is convert heat into work. For simplicity, all energy provided to the engine occurs isothermally (and reversibly) at a temperature \(T_h\) and all of the energy lost to the surroundings also occurs isothermally and reversibly at temperature \(T_l\). In order to insure this, the system must change between the two temperatures adiabatically. Thus, the cycle consists of four reversible legs, two of which are isothermal, and two of which are adiabatic. Plotted on a pressure-volume diagram, the Carnot cycle looks as follows: Because this is a closed cycle (the ending state is identical initial state) any state function must have a net change of zero as the system moves around the cycle. Furthermore, the efficiency of the engine can be expressed by the net amount of work the engine produces per unit of heat supplied to power the engine. \[\epsilon = \dfrac{w_{net}}{q_h} \nonumber \] In order to examine this expression, it is useful to write down expressions fo the heat and work flow in each of the four legs of the engine cycle. The total amount of work done is given by the sum of terms in the thirst column. Clearly the terms for the two adiabatic legs cancel (as they have the same magnitude, but opposite signs.) So the total work done is given by \[ w_{tot} = nRT_h \ln \left( \dfrac{V_2}{V_1} \right) + nRT_l \ln \left( \dfrac{V_4}{V_3} \right) \nonumber \] The efficiency of the engine can be defined as the total work produced per unit of energy provided by the high temperature reservoir. \[\epsilon = \dfrac{w_{tot}}{q_h} \nonumber \] or \[\epsilon = \dfrac{ nRT_h \ln \left( \dfrac{V_2}{V_1} \right) + nRT_l \ln \left( \dfrac{V_4}{V_3} \right)}{nRT_h \ln \left( \dfrac{V_2}{V_1} \right) } \label{eff1} \] That expression has a lot of variables, but it turns out that it can be simplified dramatically. It turns out that by the choice of pathways connecting the states places a very important restriction on the relative values of V , V , V and V . To understand this, we must consider how the work of adiabatic expansion is related to the initial and final temperatures and volumes. In Chapter 3, it was shown that the initial and final temperatures and volumes of an adiabatic expansion are related by \[V_iT_i^{C_V/R} = V_fT_f^{C_V/R} \nonumber \] or \[ \dfrac{V_i}{V_f} = \left( \dfrac{T_f}{T_i} \right)^{C_V/R} \nonumber \] Using the adiabatic expansion and compression legs (II and IV), this requires that \[\dfrac{V_2}{V_2} = \left( \dfrac{T_h}{T_l} \right)^{C_V/R} \nonumber \] and \[\dfrac{V_4}{V_1} = \left( \dfrac{T_l}{T_h} \right)^{C_V/R} \nonumber \] Since the second terms are reciprocals of one another, the first terms must be as well! \[ \dfrac{V_2}{V_2}=\dfrac{V_1}{V_4} \nonumber \] A simple rearrangement shows that \[ \dfrac{V_2}{V_1}=\dfrac{V_3}{V_4} \nonumber \] This is very convenient! It is what allows for the simplification of the efficiency expression (Equation \ref{eff1}) becomes \[\epsilon = \dfrac{ \cancel{nR}T_h \cancel{\ln \left( \dfrac{V_2}{V_1} \right)} + \cancel{nR}T_l \cancel{ \ln \left( \dfrac{V_2}{V_1} \right)}}{\cancel{nR}T_h \cancel{\ln \left( \dfrac{V_2}{V_1} \right)} } \nonumber \] Canceling terms in the numerator and denominator yields \[\epsilon = \dfrac{T_g-T_l}{T_h} \label{eff2} \] This expression gives the maximum efficiency and depends only on the high and low temperatures! Also, it should be noted that the heat engine can be run backwards. By providing work to the engine, it can be forces to draw heat from the low temperature reservoir and dissipate it into the high temperature reservoir. This is how a refrigerator or heat pump works. The limiting efficiency of such a device can also be calculated using the temperatures of the hot can cold reservoirs. What is the maximum efficiency of a freezer set to keep ice cream at a cool -10 C, which it is operating in a room that is 25 C? What is the minimum amount of energy needed to remove 1.0 J from the freezer and dissipate it into the room? The efficiency is given by Equation \ref{eff2} and converting the temperatures to an absolute scale, the efficiency can be calculated as \[\epsilon = \dfrac{298\,K - 263\,K}{298\,K} \nonumber \] This value can be used in the following manner \[energy_{transferred} = \epsilon (work_{required}) \nonumber \] So \[ 1.0 \,J = 0.1174(w) \nonumber \] or \[w = 8.5\, J \nonumber \] It is interesting to note that any arbitrary closed cyclical process can be described as a sum of infinitesimally small Carnot cycles, and so all of the conclusions reached for the Carnot cycle apply to any cyclical process. | 5,747 | 2,713 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/15%3A_Passive_Transport/15.01%3A_Dimensionality_Reduction |
One approach that does not require energy input works by recognizing that displacement is faster in systems with reduced dimensionality. Let’s think about the time it takes to diffusively encounter a small fixed target in a large volume, and how this depends on the dimensionality of the search. We will look at the mean first passage time to find a small target with radius b centered in a spherical volume with radius R, where R≫b. If the molecules are initially uniformly distributed within the volume the average time it takes for them to encounter the target (i.e., MFPT) is \( \begin{aligned} &\langle \tau_{3D} \rangle \simeq \dfrac{R^2}{3D_3} \left( \dfrac{R}{b} \right) \qquad \qquad R \gg b \\ &\langle \tau_{2D} \rangle \simeq \dfrac{R^2}{2D_2} \ln \left( \dfrac{R}{b} \right) \qquad \: \: \: \: \: \: R \gg b \\ &\langle \tau_{1D} \rangle \simeq \dfrac{R^2}{3D_1} \end{aligned} \) Here D is the diffusion constants in n dimensions (cm /sec). If we assume that the magnitude of D does not vary much with n, the leading terms in these expressions are about equal, and the big differences are in the last factor \( \left( \dfrac{R}{b} \right) > \ln \left( \dfrac{R}{b} \right) \gg 1 \) \( \langle \tau_{3D} \rangle > \langle \tau_{2D} \rangle \gg \langle \tau_{1D} \rangle\) Based on the volume that needs searching, there can be a tremendous advantage to lowering the dimensionality. _______________________________________________ | 1,457 | 2,714 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/27%3A_More_about_Spectroscopy/27.04%3A_Why_Spin-Spin_Splitting |
In , we outlined the structural features that lead to observation of spin-spin splitting in the NMR spectra of organic compounds. Rules for predicting the multiplicities and intensities of spin-spin splitting patterns also were discussed. However, we did not discuss the underlying basis for spin-spin splitting, which involves perturbation of the nuclear magnetic energy levels shown in Figure 9-21 by magnetic interaction between the nuclei. You may wish to understand more about the origin of spin-spin splitting than is provided by the rules for correlating and predicting spin-spin splitting given previously, but having a command of what follows is not necessary to the qualitative use of spin-spin splitting in structural analysis. However, it will provide you with an understanding of the origins of the line spacings and line multiplicities. We will confine our attention to protons, but the same considerations apply to other nuclei (\(\ce{^{13}C}\), \(\ce{^{15}N}\), \(\ce{^{19}F}\), and \(\ce{^{31}P}\)) that have the spin \(I = \frac{1}{2}\). The main differences between proton-proton splittings and those of other nuclei are in the magnitudes of the splitting constants (\(J\) values) and their variation with structure. Why does splitting occur? Let us start by comparing the two-proton systems of \(7\) and \(8\): The protons in each compound will have the shift differences typical of \(\ce{Cl_2CH}-\) and \(\ce{-CHO}\) and, at \(60 \: \text{MHz}\), can be expected from the data in Table 9-4 and Equation 9-4 to be observed at about \(350 \: \text{Hz}\) and \(580 \: \text{Hz}\), respectively, from TMS. Now consider a frequency-sweep experiment\(^4\) arranged so that the \(\ce{-CH=O}\) proton will come into resonance first. For \(7\) the two protons are separated by seven bonds in all (five carbon-carbon and two carbon-hydrogen bonds), thus we expect spin-spin splitting to be negligible. We can construct an energy diagram (Figure 27-5) for the magnetic energies of the possible states of protons at \(60 \: \text{MHz}\) with the aid of Figure 9-24. (If this diagram is not clear to you, we suggest you review before proceeding.) Now consider \(\ce{Cl_2CH-CH=O}\), in which the protons are in close proximity to one another, three bonds apart. Each of these protons has a magnetic field and two possible magnetic states that correspond to a compass needle pointing either north or south (see Figure 9-21). The interactions between a north-north set of orientations (\(+\frac{1}{2}\), \(+\frac{1}{2}\) or \(\rightrightarrows\)) of the two protons or a south-south set (\(-\frac{1}{2}\), \(-\frac{1}{2}\) or \(\leftleftarrows\)) will make these states stable, whereas the interactions between either a north-south (\(+\frac{1}{2}\), \(-\frac{1}{2}\) or \(\rightleftarrows\)) or a south-north (\(-\frac{1}{2}\), \(+\frac{1}{2}\) or \(\leftrightarrows\)) orientation will make these states more stable. Why? Because north-south or south-north orientations of magnets attract each other, whereas north-north or south-south repel each other.\(^5\) Let us suppose the \(+\frac{1}{2}\), \(+\frac{1}{2}\) or \(-\frac{1}{2}\), \(-\frac{1}{2}\) orientations are by \(1.25 \: \text{Hz}\). The \(+\frac{1}{2}\), \(-\frac{1}{2}\) or \(-\frac{1}{2}\), \(+\frac{1}{2}\) states must then be \(1.25 \: \text{Hz}\). Correction of the energy levels and the transition energies for these is shown in Figure 27-6. There are four possible combinations of the magnetic quantum numbers of the two protons of \(\ce{CHCl_2CHO}\), as shown in Figure 27-6. Because the differences in energy between the magnetic states corresponding to these four combinations is very small (see ), there will be almost equal numbers of \(\ce{CHCl_2CHO}\) molecules with the \(\left( +\frac{1}{2}, \: +\frac{1}{2} \right)\), \(\left( -\frac{1}{2}, \: +\frac{1}{2} \right)\), \(\left( +\frac{1}{2}, \: -\frac{1}{2} \right)\), and \(\left( -\frac{1}{2}, \: -\frac{1}{2} \right)\) spin combinations. The transitions shown in Figure 27-6 will be observed for those molecules with the two protons in the \(\left( +\frac{1}{2}, \: +\frac{1}{2} \right)\) state going to the \(\left( -\frac{1}{2}, \: +\frac{1}{2} \right)\) state or for the molecules with \(\left( +\frac{1}{2}, \: +\frac{1}{2} \right)\) to \(\left( +\frac{1}{2}, \: -\frac{1}{2} \right)\) as well as from \(\left( -\frac{1}{2}, \: +\frac{1}{2} \right) \rightarrow \left( -\frac{1}{2}, \: -\frac{1}{2} \right)\) and \(\left( +\frac{1}{2}, \: -\frac{1}{2} \right) \rightarrow \left( -\frac{1}{2}, \: -\frac{1}{2} \right)\).\(^6\) It is very important to remember that the transitions shown in Figure 27-6 involve molecules that have protons in spin states, and by the uncertainty principle ( ) the lifetimes of these spin states must be long if resonance lines are to be observed. Now if we plot the energies of the transitions shown in Figures 27-5 and 27-6, we get the predicted line positions and intensities of Figure 27-7. Four lines in two equally spaced pairs appear for \(\ce{Cl_CH-CH=O}\), as expected from the naive rules for spin-spin splitting. A harder matter to explain, and what indeed is beyond the scope of this book to explain, is why, as the chemical shift is decreased at constant spin-spin interactions, the outside lines arising from a system of two nuclei of the type shown in Figure 27-7 become progressively weaker in intensity, as shown in Figure 9-44. Furthermore, the inside lines move closer together, become more intense, and finally coalesce into a single line as the chemical-shift difference, \(\delta\), approaches zero. All we can give you here is the proposition that the outside lines become "forbidden" by spectroscopic selection rules as the chemical shift approaches zero. At the same time, the transitions leading to the inside lines become more favorable so that the integrated peak intensity of the overall system remains constant.\(^7\) In a similar category of difficult explanations is the problem of why second-order splittings are observed, as in Figure 9-32. The roots of the explanation again lie in quantum mechanics which we cannot cover here, but which do permit very precise quantitative prediction and also qualitative understanding.\(^7\) The important point to remember is that whenever the chemical shifts and couplings begin to be of similar magnitude, you can expect to encounter NMR spectra that will have more lines and lines in different positions than you would expect from the simple treatment we developed in this chapter and in Chapter 9. In extreme cases, such as with the protons of 4-deuterio-1-buten-3-yne, shown in Figure 27-8, none of the line positions or spacings correspond to any one chemical shift or spin-spin coupling. However, it is important to recognize that such spectra by no means defy analysis and, as also is seen in Figure 27-8, excellent correspondence can be obtained between calculated and observed line positions and intensities by using appropriate chemical shift and coupling parameters. However, such calculations are numerically laborious and are best made with the aid of a high-speed digital computer. \(^4\)We will use frequency sweep simply because it is easier to talk about energy changes in frequency units. However, the same arguments will hold for a field-sweep experiment. \(^5\)Such interactions with simple magnets will average to if the magnets are free to move around each other at a fixed distance. However, when electrons are between the magnets, as they are for magnetic nuclei in molecules, a small residual stabilization (or destabilization) is possible. Because these magnetic interactions are "transmitted" through the bonding electrons, we can understand in principle why it is that the number of bonds between the nuclei, the bond angles, conjugation, and so on, is more important than just the average distance between the nuclei in determining the size of the splittings. \(^6\)The transitions \(\left( +\frac{1}{2}, \: +\frac{1}{2} \right) \rightarrow \left( -\frac{1}{2}, \: -\frac{1}{2} \right)\) and \(\left( -\frac{1}{2}, \: +\frac{1}{2} \right) \rightarrow \left( +\frac{1}{2}, \: -\frac{1}{2} \right)\) are in quantum-mechanical terms known as "forbidden" transitions and are not normally observed. Notice that the changes by \(\pm\)1 for "allowed" transitions. \(^7\)A relatively elementary exposition of these matters is available in J. D. Roberts, , W. A. Benjamin, Inc., Menlo Park, Calif., 1961. and (1977) | 8,519 | 2,715 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Benzene/Substitution_Reactions_of_Benzene_Derivatives |
When substituted benzene compounds undergo electrophilic substitution reactions, two related features must be considered. The first is the relative reactivity of the compound compared with benzene itself. Experiments have shown that substituents on a benzene ring can influence reactivity in a profound manner. For example, a hydroxy or methoxy substituent increases the rate of electrophilic substitution about ten thousand fold, as illustrated by the case of anisole in the virtual demonstration (above). In contrast, a nitro substituent decreases the ring's reactivity by roughly a million. This or of the benzene ring toward electrophilic substitution may be correlated with the electron donating or electron withdrawing influence of the substituents, as measured by molecular dipole moments. In the following diagram we see that electron donating substituents (blue dipoles) activate the benzene ring toward electrophilic attack, and electron withdrawing substituents (red dipoles) deactivate the ring (make it less reactive to electrophilic attack). The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects: is the of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring (left-hand diagram below). is the result of of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance), as in the middle diagram. Finally, polar double and triple bonds conjugated with the benzene ring may withdraw electrons, as in the right-hand diagram. Note that in the resonance examples all the contributors are not shown. In both cases the charge distribution in the benzene ring is greatest at sites ortho and para to the substituent. In the case of the nitrogen and oxygen activating groups displayed in the top row of the previous diagram, electron donation by resonance dominates the inductive effect and these compounds show exceptional reactivity in electrophilic substitution reactions. Although halogen atoms have non-bonding valence electron pairs that participate in p-π conjugation, their strong inductive effect predominates, and compounds such as chlorobenzene are less reactive than benzene. The three examples on the left of the bottom row (in the same diagram) are examples of electron withdrawal by conjugation to polar double or triple bonds, and in these cases the inductive effect further enhances the deactivation of the benzene ring. Alkyl substituents such as methyl increase the nucleophilicity of aromatic rings in the same fashion as they act on double bonds. The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs. Since a mono-substituted benzene ring has two equivalent ortho-sites, two equivalent meta-sites and a unique para-site, three possible constitutional isomers may be formed in such a substitution. If reaction occurs equally well at all available sites, the expected statistical mixture of isomeric products would be 40% ortho, 40% meta and 20% para. Again we find that the nature of the substituent influences this product ratio in a dramatic fashion. Bromination of methoxybenzene (anisole) is very fast and gives mainly the para-bromo isomer, accompanied by 10% of the ortho-isomer and only a trace of the meta-isomer. Bromination of nitrobenzene requires strong heating and produces the meta-bromo isomer as the chief product. Some additional examples of product isomer distribution in other electrophilic substitutions are given in the table below. It is important to note here that the reaction conditions for these substitution reactions are not the same, and must be adjusted to fit the reactivity of the reactant C H -Y. The high reactivity of anisole, for example, requires that the first two reactions be conducted under very mild conditions (low temperature and little or no catalyst). The nitrobenzene reactant in the third example is very unreactive, so rather harsh reaction conditions must be used to accomplish that reaction. Y in C H –Y Reaction % Ortho-Product % Meta-Product % Para-Product These observations, and many others like them, have led chemists to formulate an empirical classification of the various substituent groups commonly encountered in aromatic substitution reactions. Thus, substituents that activate the benzene ring toward electrophilic attack generally direct substitution to the ortho and para locations. With some exceptions, such as the halogens, deactivating substituents direct substitution to the meta location. The following table summarizes this classification.
The information summarized in the above table is very useful for rationalizing and predicting the course of aromatic substitution reactions, but in practice most chemists find it desirable to understand the underlying physical principles that contribute to this empirical classification. We have already analyzed the activating or deactivating properties of substituents in terms of , and these same factors may be used to rationalize their influence on substitution orientation. The first thing to recognize is that the proportions of ortho, meta and para substitution in a given case reflect the relative rates of substitution at each of these sites. If we use the nitration of benzene as a reference, we can assign the rate of reaction at one of the carbons to be 1.0. Since there are six equivalent carbons in benzene, the total rate would be 6.0. If we examine the nitration of toluene, tert-butylbenzene, chlorobenzene and ethyl benzoate in the same manner, we can assign relative rates to the ortho, meta and para sites in each of these compounds. These relative rates are shown (colored red) in the following illustration, and the total rate given below each structure reflects the 2 to 1 ratio of ortho and meta sites to the para position. The overall relative rates of reaction, referenced to benzene as 1.0, are calculated by dividing by six. Clearly, the alkyl substituents activate the benzene ring in the nitration reaction, and the chlorine and ester substituents deactivate the ring. From rate data of this kind, it is a simple matter to calculate the proportions of the three substitution isomers. Toluene gives 58.5% ortho-nitrotoluene, 37% para-nitrotoluene and only 4.5% of the meta isomer. The increased bulk of the tert-butyl group hinders attack at the ortho-sites, the overall product mixture being 16% ortho, 8% meta and 75% para-nitro product. Although chlorobenzene is much less reactive than benzene, the rate of ortho and para-substitution greatly exceeds that of meta-substitution, giving a product mixture of 30% ortho and 70% para-nitrochlorobenzene. Finally, the benzoic ester gave predominantly the meta-nitro product (73%) accompanied by the ortho (22%) and para (5%) isomers, as shown by the relative rates. Equivalent rate and product studies for other substitution reactions lead to similar conclusions. For example, electrophilic chlorination of toluene occurs hundreds of times faster than chlorination of benzene, but the relative rates are such that the products are 60% ortho-chlorotoluene, 39% para and 1% meta-isomers, a ratio similar to that observed for nitration. The manner in which specific substituents influence the orientation of electrophilic substitution of a benzene ring is shown in the following interactive diagram. As noted on the opening illustration, the product-determining step in the substitution mechanism is the first step, which is also the slow or rate determining step. It is not surprising, therefore, that there is a rough correlation between the rate-enhancing effect of a substituent and its . The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites. This can be seen in seven representative substituents beneath the following diagram. In the case of alkyl substituents, charge stabilization is greatest when the alkyl group is bonded to one of the positively charged carbons of the benzenonium intermediate. This happens only for ortho and para electrophilic attack, so such substituents favor formation of those products. Interestingly, primary alkyl substituents, especially methyl, provide greater stabilization of an adjacent charge than do more substituted groups (note the greater reactivity of toluene compared with tert-butylbenzene). Nitro (NO ), sulfonic acid (SO H) and carbonyl (C=O) substituents have a full or partial positive charge on the atom bonded to the aromatic ring. Structures in which like-charges are close to each other are destabilized by charge repulsion, so these substituents inhibit ortho and para substitution more than meta substitution. Consequently, meta-products predominate when electrophilic substitution is forced to occur. Halogen (X), OR and NR substituents all exert a destabilizing inductive effect on an adjacent positive charge, due to the high electronegativity of the substituent atoms. By itself, this would favor meta-substitution; however, these substituent atoms all have non-bonding valence electron pairs which serve to stabilize an adjacent positive charge by pi-bonding, with resulting delocalization of charge. Consequently, all these substituents direct substitution to ortho and para sites. The balance between inductive electron withdrawal and p-π conjugation is such that the nitrogen and oxygen substituents have an overall stabilizing influence on the benzenonium intermediate and increase the rate of substitution markedly; whereas halogen substituents have an overall destabilizing influence. | 10,331 | 2,716 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.08%3A_Orbitals/5.8.02%3A_Lecture_Demonstrations |
Observe aquarium lights (long, thin, vertically mounted tungsten bulbs) and similar shaped socket mounted fluorescent lamps through "Rainbow Glasses" or "Spectrum Glasses" to see the difference between and . Typical "Spectrum Tubes" available from scientific suppliers, and a variety of colored lamps may also be used. Sources: American Paper Optics, Inc. Bartlett, TN 901.381.1515; 800.767.8427 ; www.holidayspecs.com Rainbow Symphony, Inc.,6860 Canby Ave. Suite 120 • Reseda · CA · 91335; (818) 708-8400 (818) 708-8400 • (818) 708-8470 (818) 708-8470 Fax Show that atoms are the transducers of energy from all forms to light energy; 1. heat energy in, light out: Flame Tests 2. UV in, light out: phosphorescence 3. chemical energy in, light out: chemiluminescence (glow sticks, luminol, etc.) 4. electrical energy in: neon signs, lights (as above) | 866 | 2,720 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX8._Condensation_Polymers |
Carboxyloids, such as esters, can interconvert with each other in the presence of the appropriate nucleophile. In the case of esters, an equilibrium will result. If an alcohol is added to an ester, under the right conditions we might get a new ester. A new alcohol would also appear, originating from the old OR group of the original ester. Two different esters and two different alcohols would be in equilibrium. This equilibrium might be perturbed in one direction or another, for example, by the addition of an excess of one nucleophile. In the jargon of synthetic organic chemistry, an ester is a functional group. It is a site on the molecule at which reactions take place. It is also a site on the molecule that is easily subject to synthetic transformations. In other words, one functional group might easily be converted into another. In the case of a trans-esterification reaction, one ester simply gets converted into another. Some compounds have more than one functional group. A compound could be both an ester and an alcohol, for instance. We might call this compound a hydroxyester. But wait -- if an alcohol can react with an ester, to make a new ester, doesn't this compound have both components of a reaction built in? Could it react with itself? There are a couple of ways that could happen. If the chain between the carbonyl and the hydroxyl group is long enough (remember, a six-atom interaction between the hydroxyl oxygen and the carbonyl carbon may be optimal), the hydroxyl could wrap around and form a cyclic ester. That's an intramolecular reaction -- a reaction within one molecule. Alternatively, if there is another one of these molecules around someplace, an intermolecular reaction might occur. That's a reaction between two different molecules. The hydroxyl group on one molecule can reach out and react with the carbonyl on another molecule. Now there are two molecules of the same kind bonded to each other. This double molecule is called a dimer. The individual molecules that have been linked togethr to make the dimer are called monomers. That dimer has two esters in it, not just one. Of course, it still has a hydroxyl group on one end. That hydroxyl group can still react with another carbonyl on another molecule. Now there are three molecules bonded together. This molecule is called a trimer. This process could keep going on indefinitely, of course. We might end up with a very large molecule, composed of many individual (former) molecules that have bonded together. This very large molecule made up of repeating units is called a polymer. A polymer is built up from many monomers linked together. This particular kind of polymer is called a polyester. This polymer is frequently drawn in a way that emphasizes the repeating pattern of monomers that have been incorporated into the chain. Polyesters can also be made by co-polymerizing two different monomers together. One monomer could be a diester, for example. The other monomer could be a diol. These two molecules are ready to react together, with one molecule acting as an electrophile and the other molecule acting as the nucleophile. Together, the diester and the diol could be polymerized. The result would be an alternating copolymer, in which diester and diol monomer units alternate all along the polymer chain. Polymers make up an important class of materials with many uses. Many polymers are lightweight, strong materials used to make parts for automobiles and other products. Polymers can also be very flexible or elastic. The physical properties of polymers are very different from the properties of other molecular compounds. These differences are a direct result of the very large size of polymer molecules. A polymer molecule might be thousands of monomers long, with a molecular weight in the millions. Express the following polymer structures in abbreviated structures showing n repeating units in parentheses. Show the structures of the polymers that would result from the following monomers. In each case, show a drawing with several enchained monomers. Ring-opening polymerization involves a multi-step reaction in which a cyclic compound, such as a lactone (below) is opened into a chain through the addition of a nucleophile (called the "initiator"). The resulting chain is able to act as a nucleophile and open the next lactone, and so on, until a polymer has formed. Show a mechanism for formation of the oligomer in which n = 3. Ring-opening polymerizations are frequently accelerated through the addition of small amounts of metal compounds, such as diethylzinc (Et Zn) or tin octoate (Sn(O CCH(CH CH )CH CH CH CH ) . Explain the role of these compounds in the reaction mechanism. Karen Wooley at Texas A&M recently reported the following synthesis of a polyphosphoramidate for use as a pharmaceutical delivery agent. The goal is to use a benign delivery agent that is easily broken down and excreted by the body, resulting in low toxicity and minimal side effects. , | 5,006 | 2,721 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.3%3A_Acid-Base_Reactions |
Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases. We defined as substances that dissolve in water to produce H ions, whereas were defined as substances that dissolve in water to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of and have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO ) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive. The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the , an acid is a substance like hydrochloric acid that dissolves in water to produce H ions (protons; Equation 5.3.1), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH ) ions (Equation 5.3.2): \[ \underset{an\: Arrhenius\: acid}{HCl_{(g)}} \xrightarrow {H_2 O_{(l)}} H^+_{(aq)} + Cl^-_{(aq)} \label{5.3.1}\] \[ \underset{an\: Arrhenius\: base}{NaOH_{(s)}} \xrightarrow {H_2O_{(l)}} Na^+_{(aq)} + OH^-_{(aq)} \label{5.3.2}\] According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H and OH ions, respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations: \[NH_{3\;(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \label{5.3.3}\] Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), who defined acid–base reactions in terms of the transfer of a proton (H ion) from one substance to another. According to Brønsted and Lowry, an acid ( is any substance that can donate a proton, and a base is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form \(NH_4^+\), so in Equation \(\ref{5.3.3}\), \(NH_3\) is a Brønsted–Lowry base and \(HCl\) is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified. Definitions of Acids and Bases: Acids differ in the number of protons they can donate. For example, monoprotic acids are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO , and HNO . All carboxylic acids that contain a single −CO H group, such as acetic acid (CH CO H), are monoprotic acids, dissociating to form RCO and H . can donate more than one proton per molecule. For example, H SO can donate two H ions in separate steps, so it is a diprotic acid (a and H PO , which is capable of donating three protons in successive steps, is a triprotic acid (Equations \(\ref{5.3.4}\), \(\ref{5.3.5}\), and \(\ref{5.3.6}\)): \[ H_3 PO_4 (l) \overset{H_2 O(l)}{\rightleftharpoons} H ^+ ( a q ) + H_2 PO_4 ^- (aq) \label{5.3.4}\] \[ H_2 PO_4 ^- (aq) \rightleftharpoons H ^+ (aq) + HPO_4^{2-} (aq) \label{5.3.5}\] \[ HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq) \label{5.3.6}\] In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in equilibrium We will not discuss the strengths of acids and bases quantitatively until next semester. Qualitatively, however, we can state that strong acids ( react essentially completely with water to give \(H^+\) and the corresponding anion. Similarly, strong bases ( dissociate essentially completely in water to give \(OH^−\) and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids ( \(H^+\) and weak bases \(OH^-\) react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form. In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO , HClO , and H SO (H PO is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH) . Common weak acids include HCN, H S, HF, oxoacids such as HNO and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows: \[ CH_3 CO_2 H(l) \overset{H_2 O(l)}{\rightleftharpoons} H^+ (aq) + CH_3 CO_2^- (aq) \label{5.3.7}\] Although acetic acid is soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates), as we stated in section 8.1. Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation 5.3.8) but a weak acid when it donates its second proton (Equation 5.3.9) as indicated by the single and double arrows, respectively: \[ \underset{strong\: acid}{H_2 SO_4 (l)} \xrightarrow {H_2 O(l)} H ^+ (aq) + HSO_4 ^- (aq) \label{5.3.8}\] \[ \underset{weak\: acid}{HSO_4^- (aq)} \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq) \label{5.3.9}\] Consequently, an aqueous solution of sulfuric acid contains \(H^+_{(aq)}\) ions and a mixture of \(HSO^-_{4\;(aq)}\) and \(SO^{2−}_{4\;(aq)}\) ions, but no \(H_2SO_4\) molecules. The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion: \[ NH_3 (g) + H_2 O(l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \label{5.3.10}\] Most of the ammonia (>99%) is present in the form of NH (g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S ). There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Table \(\Page {1}\) lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that Many weak acids and bases are extremely soluble in water. Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these. compound acid or base strength Determine whether the compound is organic or inorganic. If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H or OH ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H SO are weak acids. Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these. Definition of Strong/Weak Acids & Bases: Because isolated protons are unstable and hence very reactive, an acid never simply “loses” an H ion. Instead, the proton is always transferred to another substance, which acts as a base in the Brønsted–Lowry definition. Thus in every acid–base reaction, one species acts as an acid and one species acts as a base. Occasionally, the same substance performs both roles, as you will see later. When a strong acid dissolves in water, the proton that is released is transferred to a water molecule that acts as a proton acceptor or base, as shown for the dissociation of sulfuric acid: \[ \underset{acid\: (proton\: donor)}{H_2 SO_4 (l)} + \underset{base\: (proton\: acceptor)} {H_2 O(l)} \rightarrow \underset{acid}{H _3 O^+ (aq)} + \underset{base}{HSO_4^- (aq)} \label{5.3.11}\] Technically, therefore, it is imprecise to describe the dissociation of a strong acid as producing \(H^+_{(aq)}\) ions, as we have been doing. The resulting \(H_3O^+\) ion, called the hydronium ion and is a more accurate representation of \(H^+_{(aq)}\). For the sake of brevity, however, in discussing acid dissociation reactions, we often show the product as \(H^+_{(aq)}\) (as in Equation 5.3.7) with the understanding that the product is actually the\(H_3O^+ _{(aq)}\) ion. Conversely, bases that do not contain the hydroxide ion accept a proton from water, so small amounts of OH are produced, as in the following: \[ \underset{base}{NH_3 (g)} + \underset{acid}{H_2 O(l)} \rightleftharpoons \underset{acid}{NH_4^+ (aq)} + \underset{base}{OH^- (aq)} \label{5.3.12}\] Again, the double arrow indicates that the reaction does not go to completion but rather reaches a state of equilibrium. In this reaction, water acts as an acid by donating a proton to ammonia, and ammonia acts as a base by accepting a proton from water. Thus water can act as either an acid or a base by donating a proton to a base or by accepting a proton from an acid. Substances that can behave as both an acid and a base are said to be amphoteric . The products of an acid–base reaction are also an acid and a base. In Equation \(\ref{5.3.11}\), for example, the products of the reaction are the hydronium ion, here an acid, and the hydrogen sulfate ion, here a weak base. In Equation \(\ref{5.3.12}\), the products are NH , an acid, and OH , a base. The product NH is called the conjugate acid of the base NH , and the product OH is called the conjugate base of the acid H O. Thus all acid–base reactions actually involve two conjugate acid–base pairs ; in Equation \(\ref{5.3.12}\), they are NH /NH and H O/OH . A neutralization reaction is one in which an acid and a base react in stoichiometric amounts to produce water and a salt , the general term for any ionic substance that does not have OH as the anion or H as the cation. If the base is a metal hydroxide, then the general formula for the reaction of an acid with a base is described as follows: . For example, the reaction of equimolar amounts of HBr and NaOH to give water and a salt (NaBr) is a neutralization reaction: \[ \underset{acid}{HBr(aq)} + \underset{base}{NaOH(aq)} \rightarrow \underset{water}{H_2 O(l)} + \underset{salt}{NaBr(aq)} \label{5.3.13}\] Acid plus base yields water plus salt. If we write the complete ionic equation for the reaction in Equation \(\ref{5.3.13}\), we see that \(Na^+_{(aq)}\) and \(Br^−_{(aq)}\) are spectator ions and are not involved in the reaction: \[ H^+ (aq) + \cancel{Br^- (aq)} + \cancel{Na^+ (aq)} + OH^- (aq) \rightarrow H_2 O(l) + \cancel{Na^+ (aq)} + \cancel{Br^- (aq)} \label{5.3.14}\] The overall reaction is therefore simply the combination of H (aq) and OH (aq) to produce H O, as shown in the net ionic equation: \[H^+(aq) + OH^-(aq) \rightarrow H_2O(l)\label{5.3.15}\] The net ionic equation for the reaction of any strong acid with any strong base is identical to Equation \(\ref{5.3.15}\). The strengths of the acid and the base generally determine whether the reaction goes to completion. The reaction of strong acid with strong base goes essentially to completion, as does the reaction of a strong acid with a weak base, and a weak acid with a strong base. Examples of the last two are as follows: \[ \underset{strong\: acid}{HCl(aq)} + \underset{weak\: base}{NH_3 (aq)} \rightarrow \underset{salt}{NH_4 Cl(aq)} \label{5.3.16}\] \[ \underset{weak\: acid} {CH_3 CO _2 H(aq)} + \underset{strong\: base}{NaOH(aq)} \rightarrow \underset{salt}{CH _3 CO _2 Na(aq)} + H_2 O(l) \label{5.3.17}\] Sodium acetate is written with the organic component first followed by the cation, as is usual for organic salts. Most reactions of a weak acid with a weak base also go essentially to completion. One example is the reaction of acetic acid with ammonia: \[ \underset{weak\: acid}{CH _3 CO _2 H(aq)} + \underset{weak\: base}{NH_3 (aq)} \rightarrow \underset{salt}{CH_3 CO_2 NH_4 (aq)} \label{5.3.18}\] An example of an acid–base reaction that does not go to completion is the reaction of a weak acid or a weak base with water, which is both an extremely weak acid and an extremely weak base. Except for the reaction of a weak acid or a weak base with water, acid–base reactions essentially go to completion. In some cases, the reaction of an acid with an anion derived from a weak acid (such as HS ) produces a gas (in this case, H S). Because the gaseous product escapes from solution in the form of bubbles, the reverse reaction cannot occur. Therefore, these reactions tend to be forced, or driven, to completion. Examples include reactions in which an acid is added to ionic compounds that contain the HCO , CN , or S anions, all of which are driven to completion: \[ HCO_3^- (aq) + H^+ (aq) \rightarrow H_2 CO_3 (aq) \label{5.3.19}\] \[ H_2 CO_3 (aq) \rightarrow CO_2 (g) + H_2 O(l) \] \[ CN^- (aq) + H^+ (aq) \rightarrow HCN(g) \label{5.3.20}\] \[ S ^{2-} (aq) + H^+ (aq) \rightarrow HS^- (aq) \label{5.3.21}\] \[ HS^- (aq) + H^+ (aq) \rightarrow H_2 S(g) \] The reactions in Equations \(\ref{5.3.19}\)-\(\ref{5.3.21}\) are responsible for the rotten egg smell that is produced when metal sulfides come in contact with acids. Acid/Base Neutralization Reactions & Net Ionic Equations: Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH CH CO H) with aqueous calcium hydroxide [Ca(OH) ] to give calcium propionate. Do you expect this reaction to go to completion, making it a feasible method for the preparation of calcium propionate? reactants and product balanced chemical equation and whether the reaction will go to completion Write the balanced chemical equation for the reaction of propionic acid with calcium hydroxide. Based on their acid and base strengths, predict whether the reaction will go to completion. Propionic acid is an organic compound that is a weak acid, and calcium hydroxide is an inorganic compound that is a strong base. The balanced chemical equation is as follows: \[2CH_3CH_2CO_2H(aq) + Ca(OH)_2(aq) \rightarrow (CH_3CH_2CO_2)_2Ca(aq) + 2H_2O(l) \nonumber\] The reaction of a weak acid and a strong base will go to completion, so it is reasonable to prepare calcium propionate by mixing solutions of propionic acid and calcium hydroxide in a 2:1 mole ratio. Write a balanced chemical equation for the reaction of solid sodium acetate with dilute sulfuric acid to give sodium sulfate. \[2CH_3CO_2Na(s) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) + 2CH_3CO_2H(aq) \nonumber\] One of the most familiar and most heavily advertised applications of acid–base chemistry is , which are bases that neutralize stomach acid. The human stomach contains an approximately 0.1 M solution of hydrochloric acid that helps digest foods. If the protective lining of the stomach breaks down, this acid can attack the stomach tissue, resulting in the formation of an . Because one factor that is believed to contribute to the formation of stomach ulcers is the production of excess acid in the stomach, many individuals routinely consume large quantities of antacids. The active ingredients in antacids include sodium bicarbonate and potassium bicarbonate (NaHCO and KHCO ; Alka-Seltzer); a mixture of magnesium hydroxide and aluminum hydroxide [Mg(OH) and Al(OH) ; Maalox, Mylanta]; calcium carbonate (CaCO ; Tums); and a complex salt, dihydroxyaluminum sodium carbonate [NaAl(OH) CO ; original Rolaids]. Each has certain advantages and disadvantages. For example, Mg(OH) is a powerful laxative (it is the active ingredient in milk of magnesia), whereas Al(OH) causes constipation. When mixed, each tends to counteract the unwanted effects of the other. Although all antacids contain both an anionic base (OH , CO , or HCO ) and an appropriate cation, they differ substantially in the amount of active ingredient in a given mass of product. Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets are required to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO ? (Neutralizing all of the stomach acid is not desirable because that would completely shut down digestion.) volume and molarity of acid and mass of base in an antacid tablet number of tablets required for 90% neutralization We first write the balanced chemical equation for the reaction: \(2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2CO_3(aq)\) Each carbonate ion can react with 2 mol of H to produce H CO , which rapidly decomposes to H O and CO . Because HCl is a strong acid and CO is a weak base, the reaction will go to completion. Next we need to determine the number of moles of HCl present: \( 75\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .20\: mol\: HCl} {\cancel{L}} \right) = 0. 015\: mol\: HCl \) Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90: \((0.015\: mol\: HCl)(0.90) = 0.014\: mol\: HCl\) We know from the stoichiometry of the reaction that each mole of CaCO reacts with 2 mol of HCl, so we need \( moles\: CaCO_3 = 0 .014\: \cancel{mol\: HCl} \left( \dfrac{1\: mol\: CaCO_3}{2\: \cancel{mol\: HCl}} \right) = 0 .0070\: mol\: CaCO_3 \) Each Tums tablet contains \( \left( \dfrac{500\: \cancel{mg\: CaCO_3}} {1\: Tums\: tablet} \right) \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg\: CaCO_3}} \right) \left( \dfrac{1\: mol\: CaCO_3} {100 .1\: \cancel{g}} \right) = 0 .00500\: mol\: CaCO_ 3 \) Thus we need \(\dfrac{0.0070\: \cancel{mol\: CaCO_3}}{0.00500\: \cancel{mol\: CaCO_3}}= 1.4\) Tums tablets. Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must be consumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH) CO ? The neutralization reaction can be written as follows: \( NaAl(OH)_2CO_3(s) + 4HCl(aq) \rightarrow AlCl_3(aq) + NaCl(aq) + CO_2(g) + 3H_2O(l) \) 6.4 tablets One of the key factors affecting reactions that occur in dilute solutions of acids and bases is the concentration of H and OH ions. The pH scale provides a convenient way of expressing the hydrogen ion (H ) concentration of a solution and enables us to describe acidity or basicity in quantitative terms. Pure liquid water contains extremely low, but measurable concentrations of H O (aq) and OH (aq) ions produced via an , in which water acts simultaneously as an acid and as a base: \[H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)\label{5.3.22}\] The concentration of hydrogen ions in pure water is only 1.0 × 10 M at 25°C. Because the autoionization reaction produces both a proton and a hydroxide ion, the OH concentration in pure water is also 1.0 × 10 M. Pure water is a neutral solution [H ] = [OH ] = 1.0 × 10 M. The pH scale describes the hydrogen ion concentration of a solution in a way that avoids the use of exponential notation. pH is often defined as the negative base-10 logarithm of the hydrogen ion concentration; \[pH = -log[H^+]\label{5.3.23}\] Conversely, \[[H^+] = 10^{-pH}\label{5.3.24}\] (If you are not familiar with logarithms or using a calculator to obtain logarithms and antilogarithms) Because the hydrogen ion concentration is 1.0 × 10 M in pure water at 25°C, the pH of pure liquid water (and, by extension, of any neutral solution) is \[pH = -\log[1.0 \times 10^{-7}] = 7.00\label{5.3.25}\] Adding an acid to pure water increases the hydrogen ion concentration and decreases the hydroxide ion concentration because a neutralization reaction occurs, such as that shown in Equation 5.3.15. Because the negative exponent of [H ] becomes smaller as [H ] increases, the pH with increasing [H ]. For example, a 1.0 M solution of a strong monoprotic acid such as HCl or HNO has a pH of 0.00: \[pH = -\log[1.0] = 0.00\label{5.3.26}\] pH decreases with increasing [H ]. Conversely, adding a base to pure water increases the hydroxide ion concentration and decreases the hydrogen ion concentration. Because the autoionization reaction of water does not go to completion, neither does the neutralization reaction. Even a strongly basic solution contains a detectable amount of H ions. For example, a 1.0 M OH solution has [H ] = 1.0 × 10 M. The pH of a 1.0 M NaOH solution is therefore \[pH = -log[1.0 \times 10^{-14}] = 14.00\label{5.3.27}\] For practical purposes, the pH scale runs from pH = 0 (corresponding to 1 M H ) to pH 14 (corresponding to 1 M OH ), although pH values less than 0 or greater than 14 are possible. We can summarize the relationships between acidity, basicity, and pH as follows: Keep in mind that the pH scale is logarithmic, so a change of 1.0 in the pH of a solution corresponds to a change in the hydrogen ion concentration. The foods and consumer products we encounter daily represent a wide range of pH values, as shown in Figure \(\Page {2}\). molarity of acid or pH pH or [H ] Using the balanced chemical equation for the acid dissociation reaction and Equation 5.3.25 or 5.3.24, determine [H ] and convert it to pH or vice versa. \(HClO_4(l) \rightarrow H^+(aq) + ClO_4^-(aq)\) The H ion concentration is therefore the same as the perchloric acid concentration. The pH of the perchloric acid solution is thus \(pH = -log[H^+] = -log(2.1 \times 10^{-2}) = 1.68\) The result makes sense: the H ion concentration is between 10 M and 10 M, so the pH must be between 1 and 2. : The assumption that [H ] is the same as the concentration of the acid is valid for strong acids. Because weak acids do dissociate completely in aqueous solution, a more complex procedure is needed to calculate the pH of their solutions. \(10^{-pH} = [H^+]\) Thus \([H^+] = 10^{-3.80} = 1.6 \times 10^{-4}\: M\). Tools have been developed that make the measurement of pH simple and convenient (Figures 5.3.3 and 5.3.4). For example, pH paper consists of strips of paper impregnated with one or more acid–base indicators which are intensely colored organic molecules whose colors change dramatically depending on the pH of the solution. Placing a drop of a solution on a strip of pH paper and comparing its color with standards give the solution’s approximate pH. A more accurate tool, the pH meter, uses a glass electrode, a device whose voltage depends on the H ion concentration (Figure \(\Page {4}\)). Introduction to pH: An acidic solution and a basic solution react together in a neutralization reaction that also forms a salt. Acid–base reactions require both an acid and a base. In Brønsted–Lowry terms, an is a substance that can donate a proton (H ), and a is a substance that can accept a proton. All acid–base reactions contain two acid–base pairs: the reactants and the products. Acids can donate one proton ( ), two protons ( ), or three protons ( ). Compounds that are capable of donating more than one proton are generally called . Acids also differ in their tendency to donate a proton, a measure of their acid strength. react completely with water to produce H O (aq) (the ), whereas dissociate only partially in water. Conversely, react completely with water to produce the hydroxide ion, whereas react only partially with water to form hydroxide ions. The reaction of a strong acid with a strong base is a , which produces water plus a . The acidity or basicity of an aqueous solution is described quantitatively using the . The of a solution is the negative logarithm of the H ion concentration and typically ranges from 0 for strongly acidic solutions to 14 for strongly basic ones. Because of the of water, which produces small amounts of hydronium ions and hydroxide ions, a of water contains 1 × 10 M H ions and has a pH of 7.0. An is an intensely colored organic substance whose color is pH dependent; it is used to determine the pH of a solution. ( ) | 25,632 | 2,723 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.9%3A_Qualitative_Cation_Analysis |
The composition of relatively complex mixtures of metal ions can be determined using , a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in \(\Page {1}\). Most metal chloride salts are soluble in water; only \(\ce{Ag^{+}}\), \(\ce{Pb^{2+}}\), and \(\ce{Hg2^{2+}}\) form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M \(\ce{HCl}\), thereby causing \(\ce{AgCl}\), \(\ce{PbCl2}\), and/or \(\ce{Hg2Cl2}\) to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation. Next, the acidic solution is saturated with \(\ce{H2S}\) gas. Only those metal ions that form very insoluble sulfides, such as \(\ce{As^{3+}}\), \(\ce{Bi^{3+}}\), \(\ce{Cd^{2+}}\), \(\ce{Cu^{2+}}\), \(\ce{Hg^{2+}}\), \(\ce{Sb^{3+}}\), and \(\ce{Sn^{2+}}\), precipitate as their sulfide salts under these acidic conditions. All others, such as \(\ce{Fe^{2+}}\) and \(\ce{Zn^{2+}}\), remain in solution. Once again, the precipitates are collected by filtration or centrifugation. Ammonia or \(\ce{NaOH}\) is now added to the solution until it is basic, and then \(\ce{(NH4)2S}\) is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions \(\ce{Co^{2+}}\), \(\ce{Fe^{2+}}\), \(\ce{Mn^{2+}}\), \(\ce{Ni^{2+}}\), and \(\ce{Zn^{2+}}\) precipitate as their sulfides, and the trivalent metal ions \(\ce{Al^{3+}}\) and \(\ce{Cr^{3+}}\) precipitate as their hydroxides: \(\ce{Al(OH)3}\) and \(\ce{Cr(OH)3}\). If the mixture contains \(\ce{Fe^{3+}}\), sulfide reduces the cation to \(\ce{Fe^{2+}}\), which precipitates as \(\ce{FeS}\). The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When \(\ce{Na2CO3}\) is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding \(\ce{(NH4)2HPO4}\) causes the same metal ions to precipitate as insoluble phosphates. At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals (\(\ce{Li^{+}}\), \(\ce{Na^{+}}\), \(\ce{K^{+}}\), \(\ce{Rb^{+}}\), and \(\ce{Cs^{+}}\)) and ammonium (\(\ce{NH4^{+}}\)). We now take a second sample from the original solution and add a small amount of \(\ce{NaOH}\) to neutralize the ammonium ion and produce \(\ce{NH3}\). (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. The other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present. Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing \(\ce{Ag^{+}}\), \(\ce{Pb^{2+}}\), and \(\ce{Hg2^{2+}}\), are all quite insoluble in water. Because \(\ce{PbCl2}\) is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any \(\ce{PbCl2}\) present. Isolating the solution and adding a small amount of \(\ce{Na2CrO4}\) solution to it will produce a bright yellow precipitate of \(\ce{PbCrO4}\) if \(\ce{Pb^{2+}}\) were in the original sample ( \(\Page {2}\)). As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any \(\ce{AgCl}\) because \(\ce{Ag^{+}}\) forms a stable complex with ammonia: \(\ce{[Ag(NH3)2]^{+}}\). In addition, \(\ce{Hg2Cl2}\) in ammonia \[\ce{2Hg2^{2+} \rightarrow Hg + Hg^{2+}} \nonumber \] to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution: \[\ce{Hg2Cl2(s) + 2NH3(aq) \rightarrow Hg(l) + Hg(NH_2)Cl(s) + NH^{+}4(aq) + Cl^{−}(aq)} \nonumber \] Any silver ion in the solution is then detected by adding \(\ce{HCl}\), which reverses the reaction and gives a precipitate of white \(\ce{AgCl}\) that slowly darkens when exposed to light: \[\ce{[Ag(NH3)2]^{+} (aq) + 2H^{+}(aq) + Cl^{−}(aq) \rightarrow AgCl(s) + 2NH^{+}4(aq)} \nonumber \] Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups. In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together. | 5,673 | 2,724 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/09%3A_Mixtures/9.07%3A_Activity_of_an_Uncharged_Species |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The \(a_i\) of uncharged species \(i\) (i.e., a substance) is defined by the relation \begin{gather} \s{ a_i \defn \exp\left(\frac{\mu_i-\mu_i\st}{RT}\right) } \tag{9.7.1} \cond{(uncharged species)} \end{gather} or \begin{gather} \s{ \mu_i = \mu_i\st + RT\ln a_i } \tag{9.7.2} \cond{(uncharged species)} \end{gather} where \(\mu_i\st\) is the standard chemical potential of the species. The activity of a species in a given phase is a dimensionless quantity whose value depends on the choice of the standard state and on the intensive properties of the phase: temperature, pressure, and composition. Some chemists define the activity by \(\mu_i = \mu_i\rf + RT\ln a_i\). The activity defined this way is not the same as the activity used in this e-book unless the phase is at the standard pressure. The quantity \(a_i\) is sometimes called the of \(i\), because it depends on the chemical potential relative to a standard chemical potential. An important application of the activity concept is the definition of equilibrium constants (Sec. 11.8.1). For convenience in later applications, we specify that the value of \(a_i\) is the same in phases that have the same temperature, pressure, and composition but are at different elevations in a gravitational field, or are at different electric potentials. Section 9.8 10.1 will describe a modification of the defining equation \(\mu_i = \mu_i\st + RT\ln a_i\) for a system with phases of different elevations, and Sec. 10.1 will describe the modification needed for a charged species. The standard states of different kinds of mixture components have the same definitions as those for reference states (Table 9.3), with the additional stipulation in each case that the pressure is equal to the standard pressure \(p\st\). When component \(i\) is in its standard state, its chemical potential is the standard chemical potential \(\mu\st_i\). It is important to note from Eq. 9.7.2 that when \(\mu_i\) equals \(\mu_i\st\), the logarithm of \(a_i\) is zero and the activity in the standard state is therefore unity. The following equations in the form of Eq. 9.7.2 show the notation used in this e-book for the standard chemical potentials and activities of various kinds of uncharged mixture components: \begin{equation} \tx{Substance \(i\) in a gas mixture} \qquad \mu_i = \mu_i\st\gas + RT\ln a_i\gas \tag{9.7.3} \end{equation} \begin{equation} \tx{Substance \(i\) in a liquid or solid mixture} \qquad \mu_i = \mu_i\st + RT\ln a_i \tag{9.7.4} \end{equation} \begin{equation} \tx{Solvent A of a solution} \qquad \mu\A = \mu\A\st + RT\ln a\A \tag{9.7.5} \end{equation} \begin{equation} \tx{Solute B, mole fraction basis} \qquad \mu\B = \mu\xbB\st + RT\ln a\xbB \tag{9.7.6} \end{equation} \begin{equation} \tx{Solute B, concentration basis} \qquad \mu\B = \mu\cbB\st + RT\ln a\cbB \tag{9.7.7} \end{equation} \begin{equation} \tx{Solute B, molality basis} \qquad \mu\B = \mu\mbB\st + RT\ln a\mbB \tag{9.7.8} \end{equation} We need to be able to relate the activity of component \(i\) to the mixture composition. We can do this by finding the relation between the chemical potential of component \(i\) in its reference state and in its standard state, both at the same temperature. These two chemical potentials, \(\mu_i\rf\) and \(\mu_i\st\), are equal only if the mixture is at the standard pressure \(p\st\). It will be useful to define the following dimensionless quantity: \begin{equation} \G_i \defn \exp\left(\frac{\mu_i\rf-\mu_i\st}{RT}\right) \tag{9.7.9} \end{equation} The symbol \(\G_i\) for this quantity was introduced by Pitzer and Brewer ( , 2nd edition, McGraw-Hill, New York, 1961, p. 249). They called it . To see why, compare the definition of activity given by \(\mu_i = \mu_i\st + RT\ln a_i\) with a rearrangement of Eq. 9.7.9: \(\mu_i\rf = \mu_i\st + RT\ln\G_i\). At a given temperature, the difference \(\mu_i\rf-\mu_i\st\) depends only on the pressure \(p\) of the mixture, and is zero when \(p\) is equal to \(p\st\). Thus \(\G_i\) is a function of \(p\) with a value of 1 when \(p\) is equal to \(p\st\). This e-book will call \(\G_i\) the of species \(i\). To understand how activity is related to composition, let us take as an example the activity \(a\mbB\) of solute B based on molality. From Eqs. 9.5.18 and 9.7.8, we have \begin{equation} \begin{split} \mu\B & = \mu\mbB\rf + RT\ln\left(\g\mbB\frac{m\B}{m\st}\right) \cr & = \mu\mbB\st + RT\ln a\mbB \end{split} \tag{9.7.10} \end{equation} The activity is then given by \begin{equation} \begin{split} \ln a\mbB & = \frac{\mu\mbB\rf-\mu\mbB\st}{RT} + \ln\left(\g\mbB\frac{m\B}{m\st}\right)\cr & = \ln\G\mbB + \ln\left(\g\mbB\frac{m\B}{m\st}\right) \end{split} \tag{9.7.11} \end{equation} \begin{equation} a\mbB = \G\mbB \g\mbB\frac{m\B}{m\st}\hspace{2.28cm} \tag{9.7.12} \end{equation} The activity of a constituent of a condensed-phase mixture is in general equal to the product of the pressure factor, the activity coefficient, and the composition variable divided by the standard composition. We are now able to write explicit formulas for \(\G_i\) for each kind of mixture component. They are collected in Table 9.6. Considering a constituent of a condensed-phase mixture, by how much is the pressure factor likely to differ from unity? If we use the values \(p\st=1\br\) and \(T=300\K\), and assume the molar volume of pure \(i\) is \(V_i^*=100\units{cm\(^3\) mol\(^{-1}\)}\) at all pressures, we find that \(\G_i\) is \(0.996\) in the limit of zero pressure, unity at \(1\br\), \(1.004\) at \(2\br\), \(1.04\) at \(10\br\), and \(1.49\) at \(100\br\). For a solution with \(V\B^{\infty}=100\units{cm\(^3\) mol\(^{-1}\)}\), we obtain the same values as these for \(\G\xbB\), \(\G\mbB\), and \(\G\cbB\). These values demonstrate that it is only at high pressures that the pressure factor differs appreciably from unity. For this reason, it is common to see expressions for activity in which this factor is omitted: \(a_i=\g_i x_i\), \(a\mbB=\g\mbB m\B/m\st\), and so on. In principle, we can specify any convenient value for the standard pressure \(p\st\). For a chemist making measurements at high pressures, it would be convenient to specify a value of \(p\st\) within the range of the experimental pressures, for example \(p\st=1\units{kbar}\), in order that the value of each pressure factor be close to unity. | 13,742 | 2,725 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.6%3A_Exceptions_to_the_Octet_Rule |
Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions: The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide (NO refer to figure one). Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the possible. The formal charge is . No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitrogen monoxide. Nitrogen monoxide has 11 valence electrons. If you need more information about formal charges, see . If we were to imagine nitrogen monoxide had ten valence electrons we would come up with the Lewis Structure (Figure \(\Page {1}\)): at the formal charges of Figure \(\Page {2}\) based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure \(\Page {1}\), it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure \(\Page {1}\), oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitrogen monoxide (Figure \(\Page {2}\)): There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with \(\cdot OH\), the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted \(\cdot Cl\). Interestingly, odd Number of Valence Electrons will result in the molecule being paramagnetic. The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH (Borane). If one was to make a Lewis structure for BH following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure \(\Page {3}\)): The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH with Lewis theory. One of the things that may account for BH 's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps. Let's take a look at another incomplete octet situation dealing with boron, BF (Boron trifluorine). Like with BH , the initial drawing of a Lewis structure of BF will form a structure where boron has only six electrons around it (Figure \(\Page {4}\)). If you look Figure \(\Page {4}\), you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure \(\Page {5}\)): Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1. This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF is less than what would be typical for a single bond (see However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure \(\Page {5}\), a formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table (\(\chi=4.0\)). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible. However the large electronegativity difference here, as opposed to in BH , signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure \(\Page {6}\): None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a of all three structures: the one with the incomplete octet (Figure \(\Page {4}\)), the one with the double bond (Figure \(\Page {5}\)), and the one with the ionic bond (Figure \(\Page {6}\)). The most contributing structure is probably the incomplete octet structure (due to Figure \(\Page {5}\) being basically impossible and Figure \(\Page {6}\) not matching up with the behavior and properties of BF ). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure. As a side note, it is important to note that BF frequently bonds with a F ion in order to form BF rather than staying as BF . This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF . Draw the Lewis structure for boron trifluoride (BF ). 1. Add electrons (3*7) + 3 = 2. Draw connectivities: 3. Add octets to outer atoms: 4. Add extra electrons (24-24=0) to central atom: 5. Does central electron have octet? 6. The central Boron now has an octet (there would be three resonance Lewis structures) BF reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron: More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. , which those terminal atoms bond to. For example, \(PCl_5\) is a legitimate compound (whereas \(NCl_5\)) is not: Expanded valence shells are observed for elements in period 3 (i.e. n=3) and beyond The 'octet' rule is based upon available n and n orbitals for valence electrons (2 electrons in the orbitals, and 6 in the orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available ( =2). The orbital diagram for the valence shell of phosphorous is: Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration: There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule. One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule. Such is the case for the sulfate ion, SO . A strict adherence to the octet rule forms the following Lewis structure: If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2. If instead we made a structure for the sulfate ion with an expanded octet, it would look like this: Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure \(\Page {12}\), as opposed to +2 and -1 (difference of 3) in Figure \(\Page {12}\)) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case. Draw the Lewis structure for \(ICl_4^-\) ion. 1. Count up the valence electrons: 7+(4*7)+1 = electrons 2. Draw the connectivities: 3. Add octet of electrons to outer atoms: 4. Add extra electrons (36-32= ) to central atom: 5. The ICl ion thus has 12 valence electrons around the central Iodine (in the 5 orbitals) Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies. Expanded Valence Shell Bonding: ( ) | 14,125 | 2,726 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/11%3A_Chemical_Bonding_II%3A_Additional_Aspects/11.4%3A_Multiple_Covalent_Bonds |
So far in our molecular orbital descriptions we have not dealt with polyatomic systems with multiple bonds. To do so, we can use an approach in which we describe \(\sigma\) bonding using localized electron-pair bonds formed by hybrid atomic orbitals, and \(\pi\) bonding using molecular orbitals formed by unhybridized atomic orbitals. We begin our discussion by considering the bonding in ethylene (C H ). Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are hybridized, which means that a singly occupied orbital on one carbon overlaps with a singly occupied orbital on each H and a singly occupied lobe on the other C. Thus each carbon forms a set of three \(\sigma\) bonds: two C–H ( + ) and one C–C ( + ) (part (a) in Figure \(\Page {1}\)). The hybridization can be represented as follows: After hybridization, each carbon still has one unhybridized 2 orbital that is perpendicular to the hybridized lobes and contains a single electron (part (b) in Figure \(\Page {1}\)). The two singly occupied 2 orbitals can overlap to form a \(\pi\) bonding orbital and a \(\pi\)* antibonding orbital, which produces the energy-level diagram shown in Figure \(\Page {2}\). With the formation of a \(\pi\) bonding orbital, electron density increases in the plane between the carbon nuclei. The \(\pi\)* orbital lies outside the internuclear region and has a nodal plane perpendicular to the internuclear axis. Because each 2 orbital has a single electron, there are only two electrons, enough to fill only the bonding (\(\pi\)) level, leaving the \(\pi\)* orbital empty. Consequently, the C–C bond in ethylene consists of a \(\sigma\) bond and a \(\pi\) bond, which together give a C=C double bond. Our model is supported by the facts that the measured carbon–carbon bond is shorter than that in ethane (133.9 pm versus 153.5 pm) and the bond is stronger (728 kJ/mol versus 376 kJ/mol in ethane). The two CH fragments are coplanar, which maximizes the overlap of the two singly occupied 2 orbitals. sp2 Hybridization: Triple bonds, as in acetylene (C H ), can also be explained using a combination of hybrid atomic orbitals and molecular orbitals. The four atoms of acetylene are collinear, which suggests that each carbon is hybridized. If one lobe on each carbon atom is used to form a C–C \(\sigma\) bond and one is used to form the C–H \(\sigma\) bond, then each carbon will still have two unhybridized 2 orbitals (a 2 pair), each with one electron (part (a) in Figure \(\Page {3}\)). In complex molecules, and y can be used to describe \(\sigma\) bonding, and unhybridized orbitals and can be used to describe \(\pi\) bonding. Describe the bonding in HCN using a combination of hybrid atomic orbitals and molecular orbitals. The HCN molecule is linear. chemical compound and molecular geometry bonding description using hybrid atomic orbitals and molecular orbitals Because HCN is a linear molecule, it is likely that the bonding can be described in terms of hybridization at carbon. Because the nitrogen atom can also be described as hybridized, we can use one hybrid on each atom to form a C–N \(\sigma\) bond. This leaves one hybrid on each atom to either bond to hydrogen (C) or hold a lone pair of electrons (N). Of 10 valence electrons (5 from N, 4 from C, and 1 from H), 4 are used for \(\sigma\) bonding: We are now left with 2 electrons on N (5 valence electrons minus 1 bonding electron minus 2 electrons in the lone pair) and 2 electrons on C (4 valence electrons minus 2 bonding electrons). We have two unhybridized 2 atomic orbitals left on carbon and two on nitrogen, each occupied by a single electron. These four 2 atomic orbitals can be combined to give four molecular orbitals: two \(\pi\) (bonding) orbitals and two \(\pi\)* (antibonding) orbitals. With 4 electrons available, only the \(\pi\) orbitals are filled. The overall result is a triple bond (1 \(\sigma\) and 2 \(\pi\)) between C and N. Describe the bonding in formaldehyde (H C=O), a trigonal planar molecule, using a combination of hybrid atomic orbitals and molecular orbitals. sp Hybridization: To describe the bonding in more complex molecules with multiple bonds, we can use an approach that uses hybrid atomic orbitals to describe the \(\sigma\) bonding and molecular orbitals to describe the \(\pi\) bonding. In this approach, unhybridized orbitals on atoms bonded to one another are allowed to interact to produce bonding, antibonding, or nonbonding combinations. For \(\pi\) bonds between two atoms (as in ethylene or acetylene), the resulting molecular orbitals are virtually identical to the \(\pi\) molecular orbitals in diatomic molecules such as O and N . | 4,802 | 2,727 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/12%3A_Intermolecular_Forces%3A_Liquids_And_Solids/12.1%3A_Intermolecular_Forces |
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to molecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, molecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances . The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid. Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. In this section, we explicitly consider three kinds of intermolecular interactions. The first two are often described collectively as van der Waals forces. Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a ). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\Page {1a}\). These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\Page {1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\Page {1b}\), are , whereas those in Figure \(\Page {1d}\) are . Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(\Page {2}\). On average, however, the attractive interactions dominate. Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/ , where is the distance between the ions. Doubling the distance ( → 2 ) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/ , so doubling the distance between the dipoles decreases the strength of the interaction by 2 , or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure. Conversely, \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(\Page {1}\). The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6. Discussing Dipole Intermolecular Forces. Arrange ethyl methyl ether (CH OCH CH ), 2-methylpropane [isobutane, (CH ) CHCH ], and acetone (CH COCH ) in order of increasing boiling points. Their structures are as follows: compounds. order of increasing boiling points. Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points. The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D. Arrange carbon tetrafluoride (CF ), ethyl methyl sulfide (CH SC H ), dimethyl sulfoxide [(CH ) S=O], and 2-methylbutane [isopentane, (CH ) CHCH CH ] in order of decreasing boiling points. dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C) Thus far, we have considered only interactions between polar molecules. Other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature; why others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(\Page {2}\)). What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances. Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(\Page {3}\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/ . Doubling the distance therefore decreases the attractive energy by 2 , or 64-fold. Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H molecules in part (b) in Figure \(\Page {3}\), tends to become more pronounced as atomic and molecular masses increase (Table \(\Page {2}\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1 electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more than lighter ones. For similar substances, London dispersion forces get stronger with increasing molecular size. The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus, London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(\Page {4}\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(\Page {4}\) shows 2,2-dimethylpropane (neopentane) and -pentane, both of which have the empirical formula C H . Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas -pentane has an extended conformation that enables it to come into close contact with other -pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of -pentane (36.1°C). All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate. Discussing London/Dispersion Intermolecular Forces. Arrange -butane, propane, 2-methylpropane [isobutene, (CH ) CHCH ], and -pentane in order of increasing boiling points. compounds order of increasing boiling points Determine the intermolecular forces in the compounds, and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and -pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and -butane has the more extended shape. Consequently, we expect intermolecular interactions for -butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < -butane (−0.5°C) < -pentane (36.1°C). Arrange GeH , SiCl , SiH , CH , and GeCl in order of decreasing boiling points. GeCl (87°C) > SiCl (57.6°C) > GeH (−88.5°C) > SiH (−111.8°C) > CH (−161°C) Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent, Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(\Page {5}\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H Te and H Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C. Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(\Page {6}\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the ) and the atom that has the lone pair of electrons (the ). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water, rather than sinks. Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion. Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor. Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze. Discussing Hydrogen Bonding Intermolecular Forces. Considering CH , C H , Xe, and (CH ) N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. compounds formation of hydrogen bonds and structure Of the species listed, xenon (Xe), ethane (C H ), and trimethylamine [(CH ) N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors. The one compound that can act as a hydrogen bond donor, methanol (CH OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows: Considering CH CO H, (CH ) N, NH , and CH F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. CH CO H and NH ; Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as can form only two hydrogen bonds at a time as can, on average, pure liquid NH . Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms hydrogen bonds at a time. Arrange C (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N O in order of increasing boiling points. compounds. order of increasing boiling points. Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N O have very similar molar masses (40 and 44 g/mol, respectively), but N O is polar while Ar is not. Consequently, N O should have a higher boiling point. A C molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N O (−88.5°C) < C (>280°C) < NaCl (1465°C). Arrange 2,4-dimethylheptane, Ne, CS , Cl , and KBr in order of decreasing boiling points. KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS (46.6°C) > Cl (−34.6°C) > Ne (−246°C) Identify the most significant intermolecular force in each substance. Identify the most significant intermolecular force in each substance. hydrogen bonding dipole-dipole interactions Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid, or the liquid and gas phases, are due to changes in intermolecular interactions, but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as ), and hydrogen bonds. arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ , where is the distance between dipoles. are due to the formation of in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an in adjacent molecules; their energy falls off as 1/ . Larger atoms tend to be more than smaller ones, because their outer electrons are less tightly bound and are therefore more easily perturbed. are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the ) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the ). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cagelike structure that is less dense than liquid water. | 22,288 | 2,729 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Carbonates |
Carbonate is a polyatomic anion with the formula \(CO_3^{2-}\) and has a trigonal planar molecular structure which consists of a carbon atom surrounded by three oxygen atoms. The carbonate ion is a moderately strong base, so by definition of a , it attracts protons in aqueous solutions. It carries a formal charge of -2. Carbonate bonds to metal cations, generally forming insoluble compounds. The term "carbonate" is usually used to refer to one of its salts or carbonate minerals. The more commonly known carbonates are calcium carbonate (\(CaCO_3\)) and sodium carbonate (\(Na_2CO_3\)). \(CO_3^{2−}\) All of the react with carbonate ions and create thermally stable compounds. The exception to that rule is \(Li_2CO_3\). Lithium and magnesium have very similar properties. Their similarities are referred to as a diagonal relationship, possibly due to their comparable size. Therefore, lithium and its compounds do not react the same as other group 1 elements. Some of the examples of alkali metal carbonates are shown below: The group 2 carbonates are the most important minerals of the . Their insolubility in water and their solubility in acidic solution makes them ideal reservoirs for petroleum. One of the most significant group 2 carbonates is calcium carbonate, which is the chief constituent of . Limestones are used primarily for building stones including the manufacturing of glasses, Portland cement, and the formation of limestone caves. Here is the reaction of carbonate calcium: \(Ca^{2+} + CO_3^{2-} \longrightarrow CaCO_3\) To obtain pure CaCO from limestone, three steps must be taken: Permanent hard water contains HCO . By adding Na CO (washing soda), the water is softened and hard water precipitates calcium and magnesium. Ammonium sulfide group filtrate, when treated with CO 2-, yields precipitate from the fourth group (Mg, Ca, Sr, Ba). Aqueous carbonate anion is the key reagent, earning the name carbonate group. After the series of precipitations, the solution will contain Na, K, NH (common water soluble salts). Bicarbonates are used in the lab to prevent injury or damage from use of strong acids; for instance, by laying out bicarbonate powder in areas of potential acid leakage, accidental spills get neutralized. | 2,278 | 2,731 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.01%3A_Boyle's_Law |
Robert Boyle discovered Boyle’s law in 1662. Boyle’s discovery was that the pressure, P, and volume, V, of a gas are inversely proportional to one another if the temperature, T, is held constant. We can imagine rediscovering Boyle’s law by trapping a sample of gas in a tube and then measuring its volume as we change the pressure. We would observe behavior like that in Figure 1. We can represent this behavior mathematically as \[PV={\alpha }^*(n,T)\] where we recognize that the “constant”, \({\alpha }^*\), is actually a function of the temperature and of the number of moles, \(n\), of gas in the sample. That is, the product of pressure and volume is constant for a fixed quantity of gas at a fixed temperature. A little thought convinces us that we can be more specific about the dependence on the quantity of gas. Suppose that we have a volume of gas at a fixed pressure and temperature, and imagine that we introduce a very thin barrier that divides the volume into exactly equal halves, without changing anything else. In this case, the pressure and temperature of the gas in each of the new containers will be the same as they were originally. But the volume is half as great, and the number of moles in each of the half-size containers must also be half of the original number. That is, the pressure–volume product must be directly proportional to the number of moles of gas in the sample: \[PV=n\alpha (T)\] where \(\alpha (T)\) is now a function only of temperature. When we repeat this experiment using different gaseous substances, we discover a further remarkable fact: Not only do they all obey Boyle’s law, but also the value of \(\alpha (T)\) is the same for any gas. | 1,699 | 2,732 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/09%3A_Molecular_Geometry_and_Bonding_Theories/9.06%3A_Multiple_Bonds |
So far in our valence bond orbital descriptions we have not dealt with polyatomic systems with multiple bonds. To do so, we can use an approach in which we describe \(\sigma\) bonding using localized electron-pair bonds formed by hybrid atomic orbitals, and \(\pi\) bonding using formed by unhybridized atomic orbitals. We begin our discussion by considering the bonding in ethylene (C H ). Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are hybridized, which means that a singly occupied orbital on one carbon overlaps with a singly occupied orbital on each H and a singly occupied lobe on the other C. The hybridization can be represented as follows: When considering multiple bonds, we can introduce the more sophisticated to better understand how orbital overlap creates bonding orbitals. Recall that atomic orbitals represent ; this implies that "orbital overlap" actually involves the combination of those wavefunctions, which can occur via both constructive and destructive interference. Thus, when the two singly occupied 2 orbitals in ethylene overlap, they actually create both a \(\pi\) bonding orbital (constructive combination) and a \(\pi\)* antibonding orbital (destructive combination), which produces the energy-level diagram shown in Figure \(\Page {2}\). With the formation of a \(\pi\) bonding orbital, electron density increases in the plane between the carbon nuclei. Electrons occupying this orbital lower the potential energy of the combination and tend to hold the two nuclei together (i.e., they form a bond.) The \(\pi\)* orbital lies outside the internuclear region and has a nodal plane perpendicular to the internuclear axis; electrons in this orbital would tend to push the nuclei apart, so it is called an antibonding orbital. Because each 2 orbital has a single electron, there are only two electrons, enough to fill only the bonding (\(\pi\)) level, leaving the \(\pi\)* orbital empty. Consequently, the C–C bond in ethylene consists of a \(\sigma\) bond and a \(\pi\) bond, which together give a C=C double bond. Our model is supported by the facts that the measured carbon–carbon bond is shorter than that in ethane (133.9 pm versus 153.5 pm) and the bond is stronger (728 kJ/mol versus 376 kJ/mol in ethane). The two CH fragments are coplanar, which maximizes the overlap of the two singly occupied 2 orbitals. Triple bonds, as in acetylene (C H ), can also be explained using a combination of hybrid atomic orbitals and molecular orbitals. The four atoms of acetylene are collinear, which suggests that each carbon is hybridized. If one lobe on each carbon atom is used to form a C–C \(\sigma\) bond and one is used to form the C–H \(\sigma\) bond, then each carbon will still have two unhybridized 2 orbitals (a 2 pair), each with one electron (part (a) in Figure \(\Page {3}\)). In complex molecules, and y can be used to describe \(\sigma\) bonding, and unhybridized orbitals and can be used to describe \(\pi\) bonding. Describe the bonding in using a combination of hybrid atomic orbitals and molecular orbitals. The HCN molecule is linear. chemical compound and molecular geometry bonding description using hybrid atomic orbitals and molecular orbitals Because HCN is a linear molecule, it is likely that the bonding can be described in terms of hybridization at carbon. Because the nitrogen atom can also be described as hybridized, we can use one hybrid on each atom to form a C–N \(\sigma\) bond. This leaves one hybrid on each atom to either bond to hydrogen (C) or hold a lone pair of electrons (N). Of 10 valence electrons (5 from N, 4 from C, and 1 from H), 4 are used for \(\sigma\) bonding: We are now left with 2 electrons on N (5 valence electrons minus 1 bonding electron minus 2 electrons in the lone pair) and 2 electrons on C (4 valence electrons minus 2 bonding electrons). We have two unhybridized 2 atomic orbitals left on carbon and two on nitrogen, each occupied by a single electron. These four 2 atomic orbitals can be combined to give four molecular orbitals: two \(\pi\) (bonding) orbitals and two \(\pi\)* (antibonding) orbitals. With 4 electrons available, only the \(\pi\) orbitals are filled. The overall result is a triple bond (1 \(\sigma\) and 2 \(\pi\)) between C and N. Describe the bonding in formaldehyde (H C=O), a trigonal planar molecule, using a combination of hybrid atomic orbitals and molecular orbitals. Resonance structures can be used to describe the bonding in molecules such as ozone (O ) and the nitrite ion (NO ). Ozone can be represented by either of these Lewis electron structures: Although the model correctly predicts that both species are bent, it gives no information about their bond orders. Experimental evidence indicates that ozone has a bond angle of 117.5°. Because this angle is close to 120°, it is likely that the central oxygen atom in ozone is trigonal planar and hybridized. If we assume that the terminal oxygen atoms are also hybridized, then we obtain the \(\sigma\)-bonded framework shown in Figure \(\Page {4}\). Two of the three lobes on the central O are used to form O–O sigma bonds, and the third has a lone pair of electrons. Each terminal oxygen atom has two lone pairs of electrons that are also in lobes. In addition, each oxygen atom has one unhybridized 2 orbital perpendicular to the molecular plane. The \(\sigma\) bonds and lone pairs account for a total of 14 electrons (five lone pairs and two \(\sigma\) bonds, each containing 2 electrons). Each oxygen atom in ozone has 6 valence electrons, so O has a total of 18 valence electrons. Subtracting 14 electrons from the total gives us 4 electrons that must occupy the three unhybridized 2 orbitals. With a molecular orbital approach to describe the \(\pi\) bonding, three 2 atomic orbitals give us three molecular orbitals, as shown in Figure \(\Page {5}\). One of the molecular orbitals is a \(\pi\) bonding molecular orbital, which is shown as a banana-shaped region of electron density above and below the molecular plane. This region has nodes perpendicular to the O plane. The molecular orbital with the highest energy has two nodes that bisect the O–O \(\sigma\) bonds; it is a \(\pi\)* antibonding orbital. The third molecular orbital contains a single node that is perpendicular to the O plane and passes through the central O atom; because the orbital nodes do not directly touch, this is a molecular orbital. Because electrons in nonbonding orbitals are neither bonding nor antibonding, they are ignored in calculating bond orders. We can now place the remaining four electrons in the three energy levels shown in Figure \(\Page {5}\), thereby filling the \(\pi\) bonding and the nonbonding levels. The result is a single \(\pi\) bond holding three oxygen atoms together, or \(½ \pi\) bond per O–O. We therefore predict the overall O–O bond order to be \(½ \pi\) bond plus 1 \(\sigma\) bond), just as predicted using resonance structures. The molecular orbital description, however, makes it more clear that resonance really means that electrons are over all three atoms at once. The molecular orbital approach also shows that the \(\pi\) nonbonding orbital is localized on the terminal O atoms, which suggests that they are more electron rich than the central O atom (corresponding to the "extra" lone pair seen on one of the terminal O atoms in the Lewis resonance structures). The reactivity of ozone is consistent with the predicted charge localization. Resonance structures are a crude way of describing molecular orbitals that extend over more than two atoms. Describe the bonding in the nitrite ion in terms of a combination of hybrid atomic orbitals and molecular orbitals. Lewis dot structures and the VSEPR model predict that the NO ion is bent. chemical species and molecular geometry bonding description using hybrid atomic orbitals and molecular orbitals The lone pair of electrons on nitrogen and a bent structure suggest that the bonding in NO is similar to the bonding in ozone. This conclusion is supported by the fact that nitrite also contains 18 valence electrons (5 from N and 6 from each O, plus 1 for the −1 charge). The bent structure implies that the nitrogen is hybridized. If we assume that the oxygen atoms are hybridized as well, then we can use two hybrid orbitals on each oxygen and one hybrid orbital on nitrogen to accommodate the five lone pairs of electrons. Two hybrid orbitals on nitrogen form \(\sigma\) bonds with the remaining hybrid orbital on each oxygen. The \(\sigma\) bonds and lone pairs account for 14 electrons. We are left with three unhybridized 2 orbitals, one on each atom, perpendicular to the plane of the molecule, and 4 electrons. Just as with ozone, these three 2 orbitals interact to form bonding, nonbonding, and antibonding \(\pi\) molecular orbitals. The bonding molecular orbital is spread over the nitrogen and both oxygen atoms. Placing 4 electrons in the energy-level diagram fills both the bonding and nonbonding molecular orbitals and gives a \(\pi\) bond order of 1/2 per N–O bond. The overall N–O bond order is \(1\;\frac{1}{2}\), consistent with a resonance structure. Describe the bonding in the formate ion (HCO ), in terms of a combination of hybrid atomic orbitals and molecular orbitals. Like nitrite, formate is a planar polyatomic ion with 18 valence electrons. The \(\sigma\) bonding framework can be described in terms of hybridized carbon and oxygen, which account for 14 electrons. The three unhybridized 2 orbitals (on C and both O atoms) form three \(\pi\) molecular orbitals, and the remaining 4 electrons occupy both the bonding and nonbonding \(\pi\) molecular orbitals. The overall C–O bond order is therefore \(frac{3}{2}\) Hydrocarbons in which two or more carbon–carbon double bonds are directly linked by carbon–carbon single bonds (called structures) are generally more stable than expected because of resonance. Because the double bonds are close enough to interact electronically with one another, the \(\pi\) electrons are shared over all the carbon atoms, as illustrated for 1,3-butadiene in Figure \(\Page {6}\). As the number of interacting atomic orbitals increases, the number of molecular orbitals increases, the energy spacing between molecular orbitals decreases, and the systems become more stable (Figure \(\Page {7}\)). Thus as a chain of alternating double and single bonds becomes longer, the energy required to excite an electron from the highest-energy occupied (bonding) orbital to the lowest-energy unoccupied (antibonding) orbital decreases. If the chain is long enough, the amount of energy required to excite an electron corresponds to the energy of visible light. For example, vitamin A is yellow because its chain of five alternating double bonds is able to absorb violet light. Many of the colors we associate with dyes result from this same phenomenon; most dyes are organic compounds with alternating double bonds. As the number of atomic orbitals increases, the difference in energy between the resulting molecular orbital energy levels decreases, which allows light of lower energy to be absorbed. As a result, organic compounds with long chains of carbon atoms and alternating single and double bonds tend to become more deeply colored as the number of double bonds increases. As the number of interacting atomic orbitals increases, the energy separation between the resulting molecular orbitals steadily decreases. A derivative of vitamin A called is used by the human eye to detect light and has a structure with alternating C=C double bonds. When visible light strikes retinal, the energy separation between the molecular orbitals is sufficiently close that the energy absorbed corresponds to the energy required to change one double bond in the molecule from , where like groups are on the same side of the double bond, to , where they are on opposite sides, initiating a process that causes a signal to be sent to the brain. If this mechanism is defective, we lose our vision in dim light. Once again, a molecular orbital approach to bonding explains a process that cannot be explained using any of the other approaches we have described. Polyatomic systems with multiple bonds can be described using hybrid atomic orbitals for \(\sigma\) bonding and molecular orbitals to describe \(\pi\) bonding. To describe the bonding in more complex molecules with multiple bonds, we can use an approach that uses hybrid atomic orbitals to describe the \(\sigma\) bonding and molecular orbitals to describe the \(\pi\) bonding. In this approach, unhybridized orbitals on atoms bonded to one another are allowed to interact to produce bonding, antibonding, or nonbonding combinations. For \(\pi\) bonds between two atoms (as in ethylene or acetylene), the resulting molecular orbitals are virtually identical to the \(\pi\) molecular orbitals in diatomic molecules such as O and N . Applying the same approach to \(\pi\) bonding between three or four atoms requires combining three or four unhybridized orbitals on adjacent atoms to generate \(\pi\) bonding, antibonding, and nonbonding molecular orbitals extending over all of the atoms. Filling the resulting energy-level diagram with the appropriate number of electrons explains the bonding in molecules or ions that previously required the use of resonance structures in the Lewis electron-pair approach. | 13,572 | 2,733 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.07%3A_Structure_of_Solids |
Crystalline solids have regular ordered arrays of components held together by uniform intermolecular forces, whereas the components of amorphous solids are not arranged in regular arrays. With few exceptions, the particles that compose a solid material, whether ionic, molecular, covalent, or metallic, are held in place by strong attractive forces between them. When we discuss solids, therefore, we consider the positions of the atoms, molecules, or ions, which are essentially fixed in space, rather than their motions (which are more important in liquids and gases). The constituents of a solid can be arranged in two general ways: they can form a regular repeating three-dimensional structure called a crystal lattice, thus producing a crystalline solid, or they can aggregate with no particular order, in which case they form an amorphous solid (from the Greek ámorphos, meaning “shapeless”). Crystalline solids, or crystals, have distinctive internal structures that in turn lead to distinctive flat surfaces, or faces. The faces intersect at angles that are characteristic of the substance. When exposed to x-rays, each structure also produces a distinctive pattern that can be used to identify the material. The characteristic angles do not depend on the size of the crystal; they reflect the regular repeating arrangement of the component atoms, molecules, or ions in space. When an ionic crystal is cleaved (Figure \(\Page {2}\), for example, repulsive interactions cause it to break along fixed planes to produce new faces that intersect at the same angles as those in the original crystal. In a covalent solid such as a cut diamond, the angles at which the faces meet are also not arbitrary but are determined by the arrangement of the carbon atoms in the crystal. Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbors; that is, the regularity of the crystalline lattice creates local environments that are the same. Thus the intermolecular forces holding the solid together are uniform, and the same amount of thermal energy is needed to break every interaction simultaneously. Amorphous solids have two characteristic properties. When cleaved or broken, they produce fragments with irregular, often curved surfaces; and they have poorly defined patterns when exposed to x-rays because their components are not arranged in a regular array. An amorphous, translucent solid is called a glass. Almost any substance can solidify in amorphous form if the liquid phase is cooled rapidly enough. Some solids, however, are intrinsically amorphous, because either their components cannot fit together well enough to form a stable crystalline lattice or they contain impurities that disrupt the lattice. For example, although the chemical composition and the basic structural units of a quartz crystal and quartz glass are the same—both are SiO and both consist of linked SiO tetrahedra—the arrangements of the atoms in space are not. Crystalline quartz contains a highly ordered arrangement of silicon and oxygen atoms, but in quartz glass the atoms are arranged almost randomly. When molten SiO is cooled rapidly (4 K/min), it forms quartz glass, whereas the large, perfect quartz crystals sold in mineral shops have had cooling times of thousands of years. In contrast, aluminum crystallizes much more rapidly. Amorphous aluminum forms only when the liquid is cooled at the extraordinary rate of 4 × 10 K/s, which prevents the atoms from arranging themselves into a regular array. In an amorphous solid, the local environment, including both the distances to neighboring units and the numbers of neighbors, varies throughout the material. Different amounts of thermal energy are needed to overcome these different interactions. Consequently, amorphous solids tend to soften slowly over a wide temperature range rather than having a well-defined melting point like a crystalline solid. If an amorphous solid is maintained at a temperature just below its melting point for long periods of time, the component molecules, atoms, or ions can gradually rearrange into a more highly ordered crystalline form. Crystals have sharp, well-defined melting points; amorphous solids do not. Because a crystalline solid consists of repeating patterns of its components in three dimensions (a crystal lattice), we can represent the entire crystal by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a unit cell. For example, the unit cell of a sheet of identical postage stamps is a single stamp, and the unit cell of a stack of bricks is a single brick. In this section, we describe the arrangements of atoms in various unit cells. Unit cells are easiest to visualize in two dimensions. In many cases, more than one unit cell can be used to represent a given structure, as shown for the Escher drawing in the chapter opener and for a two-dimensional crystal lattice in Figure \(\Page {4}\). Usually the smallest unit cell that completely describes the order is chosen. The only requirement for a valid unit cell is that repeating it in space must produce the regular lattice. Thus the unit cell in Figure \(\Page {4d}\) is not a valid choice because repeating it in space does not produce the desired lattice (there are triangular holes). The concept of unit cells is extended to a three-dimensional lattice in the schematic drawing in Figure \(\Page {5}\). There are seven fundamentally different kinds of unit cells, which differ in the relative lengths of the edges and the angles between them (Figure \(\Page {6}\)). Each unit cell has six sides, and each side is a parallelogram. We focus primarily on the cubic unit cells, in which all sides have the same length and all angles are 90°, but the concepts that we introduce also apply to substances whose unit cells are not cubic. If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called simple cubic (Figure \(\Page {7a}\)). If the unit cell also contains an identical component in the center of the cube, then it is body-centered cubic (bcc) (\(\Page {7b}\)). If there are components in the center of each face in addition to those at the corners of the cube, then the unit cell is face-centered cubic (fcc) (Figure \(\Page {7c}\)). As indicated in Figure \(\Page {7}\), a solid consists of a large number of unit cells arrayed in three dimensions. Any intensive property of the bulk material, such as its density, must therefore also be related to its unit cell. Because density is the mass of substance per unit volume, we can calculate the density of the bulk material from the density of a single unit cell. To do this, we need to know the size of the unit cell (to obtain its volume), the molar mass of its components, and the number of components per unit cell. When we count atoms or ions in a unit cell, however, those lying on a face, an edge, or a corner contribute to more than one unit cell, as shown in Figure \(\Page {7}\). For example, an atom that lies on a face of a unit cell is shared by two adjacent unit cells and is therefore counted as \({1\over 2}\) atom per unit cell. Similarly, an atom that lies on the edge of a unit cell is shared by four adjacent unit cells, so it contributes \({1 \over 4}\) atom to each. An atom at a corner of a unit cell is shared by all eight adjacent unit cells and therefore contributes \({1 \over 8}\) atom to each. The statement that atoms lying on an edge or a corner of a unit cell count as \({1 \over 4}\) or \({1 \over 8}\) atom per unit cell, respectively, is true for all unit cells except the hexagonal one, in which three unit cells share each vertical edge and six share each corner (Figure \(\Page {7}\):), leading to values of \({1 \over 3}\) and \({1 \over 6}\) atom per unit cell, respectively, for atoms in these positions. In contrast, atoms that lie entirely within a unit cell, such as the atom in the center of a body-centered cubic unit cell, belong to only that one unit cell. For all unit cells except hexagonal, atoms on the faces contribute \({1\over 2}\) atom to each unit cell, atoms on the edges contribute \({1 \over 4}\) atom to each unit cell, and atoms on the corners contribute \({1 \over 8}\) atom to each unit cell. Metallic gold has a face-centered cubic unit cell (\(\Page {7c}\)). How many Au atoms are in each unit cell? : unit cell : number of atoms per unit cell Using Figure \(\Page {7}\), identify the positions of the Au atoms in a face-centered cubic unit cell and then determine how much each Au atom contributes to the unit cell. Add the contributions of all the Au atoms to obtain the total number of Au atoms in a unit cell. As shown in Figure \(\Page {7}\), a face-centered cubic unit cell has eight atoms at the corners of the cube and six atoms on the faces. Because atoms on a face are shared by two unit cells, each counts as \({1 \over 2}\) atom per unit cell, giving \(6\times{1 \over 2}=3\) Au atoms per unit cell. Atoms on a corner are shared by eight unit cells and hence contribute only \({1\over 8}\) atom per unit cell, giving \(8\times{1\over 8}=1\) Au atom per unit cell. The total number of Au atoms in each unit cell is thus \(3 + 1 = 4\). Metallic iron has a body-centered cubic unit cell (Figure \(\Page {7b}\)). How many Fe atoms are in each unit cell? two Now that we know how to count atoms in unit cells, we can use unit cells to calculate the densities of simple compounds. Note, however, that we are assuming a solid consists of a perfect regular array of unit cells, whereas real substances contain impurities and defects that affect many of their bulk properties, including density. Consequently, the results of our calculations will be close but not necessarily identical to the experimentally obtained values. Calculate the density of metallic iron, which has a body-centered cubic unit cell (Figure \(\Page {7b}\)) with an edge length of 286.6 pm. : unit cell and edge length : density : : We know from Example \(\Page {1}\) that each unit cell of metallic iron contains two Fe atoms. The molar mass of iron is 55.85 g/mol. Because density is mass per unit volume, we need to calculate the mass of the iron atoms in the unit cell from the molar mass and Avogadro’s number and then divide the mass by the volume of the cell (making sure to use suitable units to get density in g/cm ): \[ \textit{mass of Fe} =\left ( 2 \; \cancel{atoms} \; Fe \right )\left ( \dfrac{ 1 \; \cancel{mol}}{6.022\times 10^{23} \; \cancel{atoms}} \right )\left ( \dfrac{55.85 \; g}{\cancel{mol}} \right ) =1.855\times 10^{-22} \; g \nonumber \] \[ volume=\left [ \left ( 286.6 \; pm \right )\left ( \dfrac{10^{-12 }\; \cancel{m}}{\cancel{pm}} \right )\left ( \dfrac{10^{2} \; cm}{\cancel{m}} \right ) \right ] =2.345\times 10^{-23} \; cm^{3} \nonumber \] \[ density = \dfrac{1.855\times 10^{-22} \; g}{2.345\times 10^{-23} \; cm^{3}} = 7.880 g/cm^{3} \nonumber \] This result compares well with the tabulated experimental value of 7.874 g/cm . Calculate the density of gold, which has a face-centered cubic unit cell (Figure \(\Page {7c}\)) with an edge length of 407.8 pm. 19.29 g/cm Our discussion of the three-dimensional structures of solids has considered only substances in which all the components are identical. As we shall see, such substances can be viewed as consisting of identical spheres packed together in space; the way the components are packed together produces the different unit cells. Most of the substances with structures of this type are metals. The arrangement of the atoms in a solid that has a simple cubic unit cell was shown in Figure \(\Page {5a}\). Each atom in the lattice has only six nearest neighbors in an octahedral arrangement. Consequently, the simple cubic lattice is an inefficient way to pack atoms together in space: only 52% of the total space is filled by the atoms. The only element that crystallizes in a simple cubic unit cell is polonium. Simple cubic unit cells are, however, common among binary ionic compounds, where each cation is surrounded by six anions and vice versa (Figure \(\Page {8}\)). The body-centered cubic unit cell is a more efficient way to pack spheres together and is much more common among pure elements. Each atom has eight nearest neighbors in the unit cell, and 68% of the volume is occupied by the atoms. As shown in Figure \(\Page {8}\), the body-centered cubic structure consists of a single layer of spheres in contact with each other and aligned so that their centers are at the corners of a square; a second layer of spheres occupies the square-shaped “holes” above the spheres in the first layer. The third layer of spheres occupies the square holes formed by the second layer, so that each lies directly above a sphere in the first layer, and so forth. All the alkali metals, barium, radium, and several of the transition metals have body-centered cubic structures. The most efficient way to pack spheres is the close-packed arrangement, which has two variants. A single layer of close-packed spheres is shown in Figure \(\Page {6a}\). Each sphere is surrounded by six others in the same plane to produce a hexagonal arrangement. Above any set of seven spheres are six depressions arranged in a hexagon. In principle, all six sites are the same, and any one of them could be occupied by an atom in the next layer. Actually, however, these six sites can be divided into two sets, labeled B and C in Figure \(\Page {9a}\). Sites B and C differ because as soon as we place a sphere at a B position, we can no longer place a sphere in any of the three C positions adjacent to A and vice versa. If we place the second layer of spheres at the B positions in Figure \(\Page {9a}\), we obtain the two-layered structure shown in Figure \(\Page {9b}\). There are now two alternatives for placing the first atom of the third layer: we can place it directly over one of the atoms in the first layer (an A position) or at one of the C positions, corresponding to the positions that we did not use for the atoms in the first or second layers (Figure \(\Page {9c}\)). If we choose the first arrangement and repeat the pattern in succeeding layers, the positions of the atoms alternate from layer to layer in the pattern ABABAB…, resulting in a hexagonal close-packed (hcp) structure (Figure \(\Page {9a}\)). If we choose the second arrangement and repeat the pattern indefinitely, the positions of the atoms alternate as ABCABC…, giving a cubic close-packed (ccp) structure (Figure \(\Page {9b}\)). Because the ccp structure contains hexagonally packed layers, it does not look particularly cubic. As shown in Figure \(\Page {9b}\), however, simply rotating the structure reveals its cubic nature, which is identical to a fcc structure. The hcp and ccp structures differ only in the way their layers are stacked. Both structures have an overall packing efficiency of 74%, and in both each atom has 12 nearest neighbors (6 in the same plane plus 3 in each of the planes immediately above and below). Table \(\Page {1}\) compares the packing efficiency and the number of nearest neighbors for the different cubic and close-packed structures; the number of nearest neighbors is called the coordination number. Most metals have hcp, ccp, or bcc structures, although several metals exhibit both hcp and ccp structures, depending on temperature and pressure. A crystalline solid can be represented by its unit cell, which is the smallest identical unit that when stacked together produces the characteristic three-dimensional structure. Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array (a crystal lattice), which results in a crystalline solid, or more or less randomly to produce an amorphous solid. Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures. The smallest repeating unit of a crystal lattice is the unit cell. The simple cubic unit cell contains only eight atoms, molecules, or ions at the corners of a cube. A body-centered cubic (bcc) unit cell contains one additional component in the center of the cube. A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. Simple cubic and bcc arrangements fill only 52% and 68% of the available space with atoms, respectively. The hexagonal close-packed (hcp) structure has an ABABAB… repeating arrangement, and the cubic close-packed (ccp) structure has an ABCABC… repeating pattern; the latter is identical to an fcc lattice. The hcp and ccp arrangements fill 74% of the available space and have a coordination number of 12 for each atom in the lattice, the number of nearest neighbors. The simple cubic and bcc lattices have coordination numbers of 6 and 8, respectively. | 17,359 | 2,734 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 2,735 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Kinetics_of_Nucleophilic_Substitution_Reactions |
We will be contrasting about two types of nucleophilic substitution reactions. One type is referred to as , whereby the rate determining step is unimolecular and , whereby the rate determining step is bimolecular. We will begin our discussion with S 2 reactions, and discuss S 1 reactions . In the term S 2, the S stands for substitution, the N stands for nucleophilic, and the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane. If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate. If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate. The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics: Bimolecular nucleophilic substitution (SN ) reactions are , meaning they are a . This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen. The potential energy diagram for an SN reaction is shown below. Upon nucleophilic attack, a single transition state is formed. A transition state, unlike a reaction intermediate, is a very short-lived species that cannot be isolated or directly observed. Again, this is a single-step, concerted process with the occurrence of a single transition state. | 2,365 | 2,737 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/04%3A_Evaluating_Analytical_Data/4.06%3A_Statistical_Methods_for_Normal_Distributions |
The most common distribution for our results is a normal distribution. Because the area between any two limits of a normal distribution curve is well defined, constructing and evaluating significance tests is straightforward. One way to validate a new analytical method is to analyze a sample that contains a known amount of analyte, \(\mu\). To judge the method’s accuracy we analyze several portions of the sample, determine the average amount of analyte in the sample, \(\overline{X}\), and use a significance test to compare \(\overline{X}\) to \(\mu\). Our null hypothesis is that the difference between \(\overline{X}\) and \(\mu\) is explained by indeterminate errors that affect the determination of \(\overline{X}\). The alternative hypothesis is that the difference between \(\overline{X}\) and \(\mu\) is too large to be explained by indeterminate error. \[H_0 \text{: } \overline{X} = \mu \nonumber\] \[H_A \text{: } \overline{X} \neq \mu \nonumber\] The test statistic is , which we substitute into the confidence interval for \(\mu\) given by \[\mu = \overline{X} \pm \frac {t_\text{exp} s} {\sqrt{n}} \label{4.1}\] Rearranging this equation and solving for \(t_\text{exp}\) \[t_\text{exp} = \frac {|\mu - \overline{X}| \sqrt{n}} {s} \label{4.2}\] gives the value for \(t_\text{exp}\) when \(\mu\) is at either the right edge or the left edge of the sample's confidence interval (Figure 4.6.1
a) To determine if we should retain or reject the null hypothesis, we compare the value of to a critical value, \(t(\alpha, \nu)\), where \(\alpha\) is the confidence level and \(\nu\) is the degrees of freedom for the sample. The critical value \(t(\alpha, \nu)\) defines the largest confidence interval explained by indeterminate error. If \(t_\text{exp} > t(\alpha, \nu)\), then our sample’s confidence interval is greater than that explained by indeterminate errors (Figure 4.6.1
b). In this case, we reject the null hypothesis and accept the alternative hypothesis. If \(t_\text{exp} \leq t(\alpha, \nu)\), then our sample’s confidence interval is smaller than that explained by indeterminate error, and we retain the null hypothesis (Figure 4.6.1
c). Example 4.6.1
provides a typical application of this significance test, which is known as a of \(\overline{X}\) to \(\mu\). You will find values for \(t(\alpha, \nu)\) in . Another name for the -test is Student’s -test. Student was the pen name for William Gossett (1876-1927) who developed the -test while working as a statistician for the Guiness Brewery in Dublin, Ireland. He published under the name Student because the brewery did not want its competitors to know they were using statistics to help improve the quality of their products. Before determining the amount of Na CO in a sample, you decide to check your procedure by analyzing a standard sample that is 98.76% w/w Na CO . Five replicate determinations of the %w/w Na CO in the standard gave the following results \(98.71 \% \quad 98.59 \% \quad 98.62 \% \quad 98.44 \% \quad 98.58 \%\) Using \(\alpha = 0.05\), is there any evidence that the analysis is giving inaccurate results? The mean and standard deviation for the five trials are \[\overline{X} = 98.59 \quad \quad \quad s = 0.0973 \nonumber\] Because there is no reason to believe that the results for the standard must be larger or smaller than \(\mu\), a two-tailed -test is appropriate. The null hypothesis and alternative hypothesis are \[H_0 \text{: } \overline{X} = \mu \quad \quad \quad H_\text{A} \text{: } \overline{X} \neq \mu \nonumber\] The test statistic, , is \[t_\text{exp} = \frac {|\mu - \overline{X}|\sqrt{n}} {s} = \frac {|98.76 - 98.59| \sqrt{5}} {0.0973} = 3.91 \nonumber\] The critical value for (0.05, 4) from is 2.78. Since is greater than (0.05, 4), we reject the null hypothesis and accept the alternative hypothesis. At the 95% confidence level the difference between \(\overline{X}\) and \(\mu\) is too large to be explained by indeterminate sources of error, which suggests there is a determinate source of error that affects the analysis. There is another way to interpret the result of this -test. Knowing that is 3.91 and that there are 4 degrees of freedom, we use to estimate the \(\alpha\) value corresponding to a (\(\alpha\), 4) of 3.91. From , (0.02, 4) is 3.75 and (0.01, 4) is 4.60. Although we can reject the null hypothesis at the 98% confidence level, we cannot reject it at the 99% confidence level. For a discussion of the advantages of this approach, see J. A. C. Sterne and G. D. Smith “Sifting the evidence—what’s wrong with significance tests?” , , 226–231. To evaluate the accuracy of a new analytical method, an analyst determines the purity of a standard for which \(\mu\) is 100.0%, obtaining the following results. \(99.28 \% \quad 103.93 \% \quad 99.43 \% \quad 99.84 \% \quad 97.60 \% \quad 96.70 \% \quad 98.02 \%\) Is there any evidence at \(\alpha = 0.05\) that there is a determinate error affecting the results? The null hypothesis is \(H_0 \text{: } \overline{X} = \mu\) and the alternative hypothesis is \(H_\text{A} \text{: } \overline{X} \neq \mu\). The mean and the standard deviation for the data are 99.26% and 2.35%, respectively. The value for is \[t_\text{exp} = \frac {|100.0 - 99.26| \sqrt{7}} {2.35} = 0.833 \nonumber\] and the critical value for (0.05, 6) is 2.477. Because is less than (0.05, 6) we retain the null hypothesis and have no evidence for a significant difference between \(\overline{X}\) and \(\mu\). Earlier we made the point that we must exercise caution when we interpret the result of a statistical analysis. We will keep returning to this point because it is an important one. Having determined that a result is inaccurate, as we did in Example 4.6.1
, the next step is to identify and to correct the error. Before we expend time and money on this, however, we first should examine critically our data. For example, the smaller the value of , the larger the value of . If the standard deviation for our analysis is unrealistically small, then the probability of a type 2 error increases. Including a few additional replicate analyses of the standard and reevaluating the -test may strengthen our evidence for a determinate error, or it may show us that there is no evidence for a determinate error. If we analyze regularly a particular sample, we may be able to establish an expected variance, \(\sigma^2\), for the analysis. This often is the case, for example, in a clinical lab that analyze hundreds of blood samples each day. A few replicate analyses of a single sample gives a sample variance, , whose value may or may not differ significantly from \(\sigma^2\). We can use an -test to evaluate whether a difference between and \(\sigma^2\) is significant. The null hypothesis is \(H_0 \text{: } s^2 = \sigma^2\) and the alternative hypothesis is \(H_\text{A} \text{: } s^2 \neq \sigma^2\). The test statistic for evaluating the null hypothesis is , which is given as either \[F_\text{exp} = \frac {s^2} {\sigma^2} \text{ if } s^2 > \sigma^2 \text{ or } F_\text{exp} = \frac {\sigma^2} {s^2} \text{ if } \sigma^2 > s^2 \label{4.3}\] depending on whether is larger or smaller than \(\sigma^2\). This way of defining ensures that its value is always greater than or equal to one. If the null hypothesis is true, then should equal one; however, because of indeterminate errors usually is greater than one. A critical value, \(F(\alpha, \nu_\text{num}, \nu_\text{den})\), is the largest value of that we can attribute to indeterminate error given the specified significance level, \(\alpha\), and the degrees of freedom for the variance in the numerator, \(\nu_\text{num}\), and the variance in the denominator, \(\nu_\text{den}\). The degrees of freedom for is – 1, where is the number of replicates used to determine the sample’s variance, and the degrees of freedom for \(\sigma^2\) is defined as infinity, \(\infty\). Critical values of for \(\alpha = 0.05\) are listed in for both one-tailed and two-tailed -tests. A manufacturer’s process for analyzing aspirin tablets has a known variance of 25. A sample of 10 aspirin tablets is selected and analyzed for the amount of aspirin, yielding the following results in mg aspirin/tablet. \(254 \quad 249 \quad 252 \quad 252 \quad 249 \quad 249 \quad 250 \quad 247 \quad 251 \quad 252\) Determine whether there is evidence of a significant difference between the sample’s variance and the expected variance at \(\alpha = 0.05\). The variance for the sample of 10 tablets is 4.3. The null hypothesis and alternative hypotheses are \[H_0 \text{: } s^2 = \sigma^2 \quad \quad \quad H_\text{A} \text{: } s^2 \neq \sigma^2 \nonumber\] and the value for is \[F_\text{exp} = \frac {\sigma^2} {s^2} = \frac {25} {4.3} = 5.8 \nonumber\] The critical value for (0.05, \(\infty\), 9) from is 3.333. Since is greater than (0.05, \(\infty\), 9), we reject the null hypothesis and accept the alternative hypothesis that there is a significant difference between the sample’s variance and the expected variance. One explanation for the difference might be that the aspirin tablets were not selected randomly. We can extend the -test to compare the variances for two samples, and , by rewriting Equation \ref{4.3} as \[F_\text{exp} = \frac {s_A^2} {s_B^2} \nonumber\] defining and so that the value of exp is greater than or equal to 1. shows results for two experiments to determine the mass of a circulating U.S. penny. Determine whether there is a difference in the variances of these analyses at \(\alpha = 0.05\). The standard deviations for the two experiments are 0.051 for the first experiment ( ) and 0.037 for the second experiment ( ). The null and alternative hypotheses are \[H_0 \text{: } s_A^2 = s_B^2 \quad \quad \quad H_\text{A} \text{: } s_A^2 \neq s_B^2 \nonumber\] and the value of is \[F_\text{exp} = \frac {s_A^2} {s_B^2} = \frac {(0.051)^2} {(0.037)^2} = \frac {0.00260} {0.00137} = 1.90 \nonumber\] From , the critical value for (0.05, 6, 4) is 9.197. Because < (0.05, 6, 4), we retain the null hypothesis. There is no evidence at \(\alpha = 0.05\) to suggest that the difference in variances is significant. To compare two production lots of aspirin tablets, we collect an analyze samples from each, obtaining the following results (in mg aspirin/tablet). Lot 1: \(256 \quad 248 \quad 245 \quad 245 \quad 244 \quad 248 \quad 261\) Lot 2: \(241 \quad 258 \quad 241 \quad 244 \quad 256 \quad 254\) Is there any evidence at \(\alpha = 0.05\) that there is a significant difference in the variances for these two samples? The standard deviations are 6.451 mg for Lot 1 and 7.849 mg for Lot 2. The null and alternative hypotheses are \[H_0 \text{: } s_\text{Lot 1}^2 = s_\text{Lot 2}^2 \quad \quad \quad H_\text{A} \text{: } s_\text{Lot 1}^2 \neq s_\text{Lot 2}^2 \nonumber\] and the value of is \[F_\text{exp} = \frac {(7.849)^2} {(6.451)^2} = 1.480 \nonumber\] The critical value for (0.05, 5, 6) is 5.988. Because < (0.05, 5, 6), we retain the null hypothesis. There is no evidence at \(\alpha = 0.05\) to suggest that the difference in the variances is significant. Three factors influence the result of an analysis: the method, the sample, and the analyst. We can study the influence of these factors by conducting experiments in which we change one factor while holding constant the other factors. For example, to compare two analytical methods we can have the same analyst apply each method to the same sample and then examine the resulting means. In a similar fashion, we can design experiments to compare two analysts or to compare two samples. It also is possible to design experiments in which we vary more than one of these factors. We will return to this point in . Before we consider the significance tests for comparing the means of two samples, we need to make a distinction between and . This is a critical distinction and learning to distinguish between these two types of data is important. Here are two simple examples that highlight the difference between unpaired data and paired data. In each example the goal is to compare two balances by weighing pennies. In both examples the samples of 10 pennies were drawn from the same population; the difference is how we sampled that population. We will learn why this distinction is important when we review the significance test for paired data; first, however, we present the significance test for unpaired data. One simple test for determining whether data are paired or unpaired is to look at the size of each sample. If the samples are of different size, then the data must be unpaired. The converse is not true. If two samples are of equal size, they may be paired or unpaired. Consider two analyses, and with means of \(\overline{X}_A\) and \(\overline{X}_B\), and standard deviations of and . The confidence intervals for \(\mu_A\) and for \(\mu_B\) are \[\mu_A = \overline{X}_A \pm \frac {t s_A} {\sqrt{n_A}} \label{4.4}\] \[\mu_B = \overline{X}_B \pm \frac {t s_B} {\sqrt{n_B}} \label{4.5}\] where and are the sample sizes for and for . Our null hypothesis, \(H_0 \text{: } \mu_A = \mu_B\), is that and any difference between \(\mu_A\) and \(\mu_B\) is the result of indeterminate errors that affect the analyses. The alternative hypothesis, \(H_A \text{: } \mu_A \neq \mu_B\), is that the difference between \(\mu_A\)and \(\mu_B\) is too large to be explained by indeterminate error. To derive an equation for , we assume that \(\mu_A\) equals \(\mu_B\), and combine Equation \ref{4.4} and Equation \ref{4.5} \[\overline{X}_A \pm \frac {t_\text{exp} s_A} {\sqrt{n_A}} = \overline{X}_B \pm \frac {t_\text{exp} s_B} {\sqrt{n_B}} \nonumber\] Solving for \(|\overline{X}_A - \overline{X}_B|\) and using a propagation of uncertainty, gives \[|\overline{X}_A - \overline{X}_B| = t_\text{exp} \times \sqrt{\frac {s_A^2} {n_A} + \frac {s_B^2} {n_B}} \label{4.6}\] Finally, we solve for \[t_\text{exp} = \frac {|\overline{X}_A - \overline{X}_B|} {\sqrt{\frac {s_A^2} {n_A} + \frac {s_B^2} {n_B}}} \label{4.7}\] and compare it to a critical value, \(t(\alpha, \nu)\), where \(\alpha\) is the probability of a type 1 error, and \(\nu\) is the degrees of freedom. Problem 9 asks you to use a propagation of uncertainty to show that Equation \ref{4.6} is correct. Thus far our development of this -test is similar to that for comparing \(\overline{X}\) to \(\mu\), and yet we do not have enough information to evaluate the -test. Do you see the problem? With two independent sets of data it is unclear how many degrees of freedom we have. Suppose that the variances \(s_A^2\) and \(s_B^2\) provide estimates of the same \(\sigma^2\). In this case we can replace \(s_A^2\) and \(s_B^2\) with a pooled variance, \(s_\text{pool}^2\), that is a better estimate for the variance. Thus, Equation \ref{4.7} becomes \[t_\text{exp} = \frac {|\overline{X}_A - \overline{X}_B|} {s_\text{pool} \times \sqrt{\frac {1} {n_A} + \frac {1} {n_B}}} = \frac {|\overline{X}_A - \overline{X}_B|} {s_\text{pool}} \times \sqrt{\frac {n_A n_B} {n_A + n_B}} \label{4.8}\] where , the pooled standard deviation, is \[s_\text{pool} = \sqrt{\frac {(n_A - 1) s_A^2 + (n_B - 1)s_B^2} {n_A + n_B - 2}} \label{4.9}\] The denominator of Equation \ref{4.9} shows us that the degrees of freedom for a pooled standard deviation is \(n_A + n_B - 2\), which also is the degrees of freedom for the -test. Note that we lose two degrees of freedom because the calculations for \(s_A^2\) and \(s_B^2\) require the prior calculation of \(\overline{X}_A\) amd \(\overline{X}_B\). So how do you determine if it is okay to pool the variances? Use an -test. If \(s_A^2\) and \(s_B^2\) are significantly different, then we calculate using Equation \ref{4.7}. In this case, we find the degrees of freedom using the following imposing equation. \[\nu = \frac {\left( \frac {s_A^2} {n_A} + \frac {s_B^2} {n_B} \right)^2} {\frac {\left( \frac {s_A^2} {n_A} \right)^2} {n_A + 1} + \frac {\left( \frac {s_B^2} {n_B} \right)^2} {n_B + 1}} - 2 \label{4.10}\] Because the degrees of freedom must be an integer, we round to the nearest integer the value of \(\nu\) obtained using Equation \ref{4.10}. Equation \ref{4.10}, which is from ; Miller, J.N. , 2nd Ed., Ellis-Horward: Chichester, UK, 1988. In the 6th Edition, the authors note that several different equations have been suggested for the number of degrees of freedom for when and differ, reflecting the fact that the determination of degrees of freedom an approximation. An alternative equation—which is used by statistical software packages, such as R, Minitab, Excel—is \[\nu = \frac {\left( \frac {s_A^2} {n_A} + \frac {s_B^2} {n_B} \right)^2} {\frac {\left( \frac {s_A^2} {n_A} \right)^2} {n_A - 1} + \frac {\left( \frac {s_B^2} {n_B} \right)^2} {n_B - 1}} = \frac {\left( \frac {s_A^2} {n_A} + \frac {s_B^2} {n_B} \right)^2} {\frac {s_A^4} {n_A^2(n_A - 1)} + \frac {s_B^4} {n_B^2(n_B - 1)}} \nonumber\] For typical problems in analytical chemistry, the calculated degrees of freedom is reasonably insensitive to the choice of equation. Regardless of whether we calculate using Equation \ref{4.7} or Equation \ref{4.8}, we reject the null hypothesis if is greater than \(t(\alpha, \nu)\) and retain the null hypothesis if is less than or equal to \(t(\alpha, \nu)\). provides results for two experiments to determine the mass of a circulating U.S. penny. Determine whether there is a difference in the means of these analyses at \(\alpha = 0.05\). First we use an -test to determine whether we can pool the variances. We completed this analysis in , finding no evidence of a significant difference, which means we can pool the standard deviations, obtaining \[s_\text{pool} = \sqrt{\frac {(7 - 1)(0.051)^2 + (5 - 1)(0.037)^2} {7 + 5 - 2}} = 0.0459 \nonumber\] with 10 degrees of freedom. To compare the means we use the following null hypothesis and alternative hypotheses \[H_0 \text{: } \mu_A = \mu_B \quad \quad \quad H_A \text{: } \mu_A \neq \mu_B \nonumber\] Because we are using the pooled standard deviation, we calculate using Equation \ref{4.8}. \[t_\text{exp} = \frac {|3.117 - 3.081|} {0.0459} \times \sqrt{\frac {7 \times 5} {7 + 5}} = 1.34 \nonumber\] The critical value for (0.05, 10), from , is 2.23. Because is less than (0.05, 10) we retain the null hypothesis. For \(\alpha = 0.05\) we do not have evidence that the two sets of pennies are significantly different. One method for determining the %w/w Na CO in soda ash is to use an acid–base titration. When two analysts analyze the same sample of soda ash they obtain the results shown here. Analyst A: \(86.82 \% \quad 87.04 \% \quad 86.93 \% \quad 87.01 \% \quad 86.20 \% \quad 87.00 \%\) Analyst B: \(81.01 \% \quad 86.15 \% \quad 81.73 \% \quad 83.19 \% \quad 80.27 \% \quad 83.93 \% \quad\) Determine whether the difference in the mean values is significant at \(\alpha = 0.05\). We begin by reporting the mean and standard deviation for each analyst. \[\overline{X}_A = 86.83\% \quad \quad s_A = 0.32\% \nonumber\] \[\overline{X}_B = 82.71\% \quad \quad s_B = 2.16\% \nonumber\] To determine whether we can use a pooled standard deviation, we first complete an -test using the following null and alternative hypotheses. \[H_0 \text{: } s_A^2 = s_B^2 \quad \quad \quad H_A \text{: } s_A^2 \neq s_B^2 \nonumber\] Calculating , we obtain a value of \[F_\text{exp} = \frac {(2.16)^2} {(0.32)^2} = 45.6 \nonumber\] Because is larger than the critical value of 7.15 for (0.05, 5, 5) from , we reject the null hypothesis and accept the alternative hypothesis that there is a significant difference between the variances; thus, we cannot calculate a pooled standard deviation. To compare the means for the two analysts we use the following null and alternative hypotheses. \[H_0 \text{: } \overline{X}_A = \overline{X}_B \quad \quad \quad H_A \text{: } \overline{X}_A \neq \overline{X}_B \nonumber\] Because we cannot pool the standard deviations, we calculate using Equation \ref{4.7} instead of Equation \ref{4.8} \[t_\text{exp} = \frac {|86.83 - 82.71|} {\sqrt{\frac {(0.32)^2} {6} + \frac {(2.16)^2} {6}}} = 4.62 \nonumber\] and calculate the degrees of freedom using Equation \ref{4.10}. \[\nu = \frac {\left( \frac {(0.32)^2} {6} + \frac {(2.16)^2} {6} \right)^2} {\frac {\left( \frac {(0.32)^2} {6} \right)^2} {6 + 1} + \frac {\left( \frac {(2.16)^2} {6} \right)^2} {6 + 1}} - 2 = 5.3 \approx 5 \nonumber\] From , the critical value for (0.05, 5) is 2.57. Because is greater than (0.05, 5) we reject the null hypothesis and accept the alternative hypothesis that the means for the two analysts are significantly different at \(\alpha = 0.05\). To compare two production lots of aspirin tablets, you collect samples from each and analyze them, obtaining the following results (in mg aspirin/tablet). Lot 1: \(256 \quad 248 \quad 245 \quad 245 \quad 244 \quad 248 \quad 261\) Lot 2: \(241 \quad 258 \quad 241 \quad 244 \quad 256 \quad 254\) Is there any evidence at \(\alpha = 0.05\) that there is a significant difference in the variance between the results for these two samples? This is the same data from . To compare the means for the two lots, we use an unpaired -test of the null hypothesis \(H_0 \text{: } \overline{X}_\text{Lot 1} = \overline{X}_\text{Lot 2}\) and the alternative hypothesis \(H_A \text{: } \overline{X}_\text{Lot 1} \neq \overline{X}_\text{Lot 2}\). Because there is no evidence to suggest a difference in the variances (see Exercise 4.6.2
) we pool the standard deviations, obtaining an of \[s_\text{pool} = \sqrt{\frac {(7 - 1) (6.451)^2 + (6 - 1) (7.849)^2} {7 + 6 - 2}} = 7.121 \nonumber\] The means for the two samples are 249.57 mg for Lot 1 and 249.00 mg for Lot 2. The value for is \[t_\text{exp} = \frac {|249.57 - 249.00|} {7.121} \times \sqrt{\frac {7 \times 6} {7 + 6}} = 0.1439 \nonumber\] The critical value for (0.05, 11) is 2.204. Because is less than (0.05, 11), we retain the null hypothesis and find no evidence at \(\alpha = 0.05\) that there is a significant difference between the means for the two lots of aspirin tablets. Suppose we are evaluating a new method for monitoring blood glucose concentrations in patients. An important part of evaluating a new method is to compare it to an established method. What is the best way to gather data for this study? Because the variation in the blood glucose levels amongst patients is large we may be unable to detect a small, but significant difference between the methods if we use different patients to gather data for each method. Using paired data, in which the we analyze each patient’s blood using both methods, prevents a large variance within a population from adversely affecting a -test of means. Typical blood glucose levels for most non-diabetic individuals ranges between 80–120 mg/dL (4.4–6.7 mM), rising to as high as 140 mg/dL (7.8 mM) shortly after eating. Higher levels are common for individuals who are pre-diabetic or diabetic. When we use paired data we first calculate the difference, , between the paired values for each sample. Using these difference values, we then calculate the average difference, \(\overline{d}\), and the standard deviation of the differences, . The null hypothesis, \(H_0 \text{: } d = 0\), is that there is no difference between the two samples, and the alternative hypothesis, \(H_A \text{: } d \neq 0\), is that the difference between the two samples is significant. The test statistic, , is derived from a confidence interval around \(\overline{d}\) \[t_\text{exp} = \frac {|\overline{d}| \sqrt{n}} {s_d} \nonumber\] where is the number of paired samples. As is true for other forms of the -test, we compare to \(t(\alpha, \nu)\), where the degrees of freedom, \(\nu\), is – 1. If is greater than \(t(\alpha, \nu)\), then we reject the null hypothesis and accept the alternative hypothesis. We retain the null hypothesis if is less than or equal to (a, o). This is known as a Marecek et. al. developed a new electrochemical method for the rapid determination of the concentration of the antibiotic monensin in fermentation vats [Marecek, V.; Janchenova, H.; Brezina, M.; Betti, M. , , 15–19]. The standard method for the analysis is a test for microbiological activity, which is both difficult to complete and time-consuming. Samples were collected from the fermentation vats at various times during production and analyzed for the concentration of monensin using both methods. The results, in parts per thousand (ppt), are reported in the following table. Is there a significant difference between the methods at \(\alpha = 0.05\)? Acquiring samples over an extended period of time introduces a substantial time-dependent change in the concentration of monensin. Because the variation in concentration between samples is so large, we use a paired -test with the following null and alternative hypotheses. \[H_0 \text{: } \overline{d} = 0 \quad \quad \quad H_A \text{: } \overline{d} \neq 0 \nonumber\] Defining the difference between the methods as \[d_i = (X_\text{elect})_i - (X_\text{micro})_i \nonumber\] we calculate the difference for each sample. The mean and the standard deviation for the differences are, respectively, 2.25 ppt and 5.63 ppt. The value of is \[t_\text{exp} = \frac {|2.25| \sqrt{11}} {5.63} = 1.33 \nonumber\] which is smaller than the critical value of 2.23 for (0.05, 10) from . We retain the null hypothesis and find no evidence for a significant difference in the methods at \(\alpha = 0.05\). Suppose you are studying the distribution of zinc in a lake and want to know if there is a significant difference between the concentration of Zn at the sediment-water interface and its concentration at the air-water interface. You collect samples from six locations—near the lake’s center, near its drainage outlet, etc.—obtaining the results (in mg/L) shown in the table. Using this data, determine if there is a significant difference between the concentration of Zn at the two interfaces at \(\alpha = 0.05\). Complete this analysis treating the data as (a) unpaired and as (b) paired. Briefly comment on your results. Complete this analysis treating the data as (a) unpaired and as (b) paired. Briefly comment on your results. : The mean and the standard deviation for the concentration of Zn at the air-water interface are 0.5178 mg/L and 0.1732 mg/L, respectively, and the values for the sediment-water interface are 0.4445 mg/L and 0.1418 mg/L, respectively. An -test of the variances gives an of 1.493 and an (0.05, 5, 5) of 7.146. Because is smaller than (0.05, 5, 5), we have no evidence at \(\alpha = 0.05\) to suggest that the difference in variances is significant. Pooling the standard deviations gives an of 0.1582 mg/L. An unpaired -test gives as 0.8025. Because is smaller than (0.05, 11), which is 2.204, we have no evidence that there is a difference in the concentration of Zn between the two interfaces. : To treat as paired data we need to calculate the difference, , between the concentration of Zn at the air-water interface and at the sediment-water interface for each location, where \[d_i = \left( \text{[Zn}^{2+} \text{]}_\text{air-water} \right)_i - \left( \text{[Zn}^{2+} \text{]}_\text{sed-water} \right)_i \nonumber\] The mean difference is 0.07333 mg/L with a standard deviation of 0.0441 mg/L. The null hypothesis and the alternative hypothesis are \[H_0 \text{: } \overline{d} = 0 \quad \quad \quad H_A \text{: } \overline{d} \neq 0 \nonumber\] and the value of is \[t_\text{exp} = \frac {|0.07333| \sqrt{6}} {0.0441} = 4.073 \nonumber\] Because is greater than (0.05, 5), which is 2.571, we reject the null hypothesis and accept the alternative hypothesis that there is a significant difference in the concentration of Zn between the air-water interface and the sediment-water interface. The difference in the concentration of Zn between locations is much larger than the difference in the concentration of Zn between the interfaces. Because out interest is in studying the difference between the interfaces, the larger standard deviation when treating the data as unpaired increases the probability of incorrectly retaining the null hypothesis, a type 2 error. One important requirement for a paired -test is that the determinate and the indeterminate errors that affect the analysis must be independent of the analyte’s concentration. If this is not the case, then a sample with an unusually high concentration of analyte will have an unusually large . Including this sample in the calculation of \(\overline{d}\) and gives a biased estimate for the expected mean and standard deviation. This rarely is a problem for samples that span a limited range of analyte concentrations, such as those in Example 4.6.6
or Exercise 4.6.4
. When paired data span a wide range of concentrations, however, the magnitude of the determinate and indeterminate sources of error may not be independent of the analyte’s concentration; when true, a paired -test may give misleading results because the paired data with the largest absolute determinate and indeterminate errors will dominate \(\overline{d}\). In this situation a regression analysis, which is the subject of the next chapter, is more appropriate method for comparing the data. Earlier in the chapter we examined several data sets consisting of the mass of a circulating United States penny. Table 4.6.1
provides one more data set. Do you notice anything unusual in this data? Of the 112 pennies included in and , no penny weighed less than 3 g. In Table 4.6.1, however, the mass of one penny is less than 3 g. We might ask whether this penny’s mass is so different from the other pennies that it is in error. A measurement that is not consistent with other measurements is called outlier. An might exist for many reasons: the outlier might belong to a different population (Is this a Canadian penny?); the outlier might be a contaminated or otherwise altered sample (Is the penny damaged or unusually dirty?); or the outlier may result from an error in the analysis (Did we forget to tare the balance?). Regardless of its source, the presence of an outlier compromises any meaningful analysis of our data. There are many significance tests that we can use to identify a potential outlier, three of which we present here. One of the most common significance tests for identifying an outlier is . The null hypothesis is that there are no outliers, and the alternative hypothesis is that there is an outlier. The -test compares the gap between the suspected outlier and its nearest numerical neighbor to the range of the entire data set (Figure 4.6.2
). The test statistic, , is \[Q_\text{exp} = \frac {\text{gap}} {\text{range}} = \frac {|\text{outlier's value} - \text{nearest value}|} {\text{largest value} - \text{smallest value}} \nonumber\] This equation is appropriate for evaluating a single outlier. Other forms of Dixon’s -test allow its extension to detecting multiple outliers [Rorabacher, D. B. , , 139–146]. The value of is compared to a critical value, \(Q(\alpha, n)\), where \(\alpha\) is the probability that we will reject a valid data point (a type 1 error) and is the total number of data points. To protect against rejecting a valid data point, usually we apply the more conservative two-tailed -test, even though the possible outlier is the smallest or the largest value in the data set. If is greater than \(Q(\alpha, n)\), then we reject the null hypothesis and may exclude the outlier. We retain the possible outlier when is less than or equal to \(Q(\alpha, n)\). Table 4.6.2
provides values for \(Q(\alpha, n)\) for a data set that has 3–10 values. A more extensive table is in . Values for \(Q(\alpha, n)\) assume an underlying normal distribution. Although Dixon’s -test is a common method for evaluating outliers, it is no longer favored by the International Standards Organization (ISO), which recommends the . There are several versions of Grubb’s test depending on the number of potential outliers. Here we will consider the case where there is a single suspected outlier. For details on this recommendation, see International Standards ISO Guide 5752-2 “Accuracy (trueness and precision) of measurement methods and results–Part 2: basic methods for the determination of repeatability and reproducibility of a standard measurement method,” 1994. The test statistic for Grubb’s test, , is the distance between the sample’s mean, \(\overline{X}\), and the potential outlier, \(X_\text{out}\), in terms of the sample’s standard deviation, . \[G_\text{exp} = \frac {|X_\text{out} - \overline{X}|} {s} \nonumber\] We compare the value of to a critical value \(G(\alpha, n)\), where \(\alpha\) is the probability that we will reject a valid data point and is the number of data points in the sample. If is greater than \(G(\alpha, n)\), then we may reject the data point as an outlier, otherwise we retain the data point as part of the sample. Table 4.6.3
provides values for (0.05, ) for a sample containing 3–10 values. A more extensive table is in . Values for \(G(\alpha, n)\) assume an underlying normal distribution. Our final method for identifying an outlier is . Unlike Dixon’s -Test and Grubb’s test, you can apply this method to any distribution as long as you know how to calculate the probability for a particular outcome. Chauvenet’s criterion states that we can reject a data point if the probability of obtaining the data point’s value is less than (2 ) , where is the size of the sample. For example, if = 10, a result with a probability of less than \((2 \times 10)^{-1}\), or 0.05, is considered an outlier. To calculate a potential outlier’s probability we first calculate its standardized deviation, \[z = \frac {|X_\text{out} - \overline{X}|} {s} \nonumber\] where \(X_\text{out}\) is the potential outlier, \(\overline{X}\) is the sample’s mean and is the sample’s standard deviation. Note that this equation is identical to the equation for in the Grubb’s test. For a normal distribution, we can find the probability of obtaining a value of using the probability table in . contains the masses for nine circulating United States pennies. One entry, 2.514 g, appears to be an outlier. Determine if this penny is an outlier using a -test, Grubb’s test, and Chauvenet’s criterion. For the -test and Grubb’s test, let \(\alpha = 0.05\). For the -test the value for exp is \[Q_\text{exp} = \frac {|2.514 - 3.039|} {3.109 - 2.514} = 0.882 \nonumber\] From Table 4.6.2
, the critical value for (0.05, 9) is 0.493. Because is greater than (0.05, 9), we can assume the penny with a mass of 2.514 g likely is an outlier. For Grubb’s test we first need the mean and the standard deviation, which are 3.011 g and 0.188 g, respectively. The value for is \[G_\text{exp} = \frac {|2.514 - 3.011} {0.188} = 2.64 \nonumber\] Using Table 4.6.3
, we find that the critical value for (0.05, 9) is 2.215. Because is greater than (0.05, 9), we can assume that the penny with a mass of 2.514 g likely is an outlier. For Chauvenet’s criterion, the critical probability is \((2 \times 9)^{-1}\), or 0.0556. The value of is the same as , or 2.64. Using , the probability for = 2.64 is 0.00415. Because the probability of obtaining a mass of 0.2514 g is less than the critical probability, we can assume the penny with a mass of 2.514 g likely is an outlier. You should exercise caution when using a significance test for outliers because there is a chance you will reject a valid result. In addition, you should avoid rejecting an outlier if it leads to a precision that is much better than expected based on a propagation of uncertainty. Given these concerns it is not surprising that some statisticians caution against the removal of outliers [Deming, W. E. ; Wiley: New York, 1943 (republished by Dover: New York, 1961); p. 171]. You also can adopt a more stringent requirement for rejecting data. When using the Grubb’s test, for example, the ISO 5752 guidelines suggests retaining a value if the probability for rejecting it is greater than \(\alpha = 0.05\), and flagging a value as a “straggler” if the probability for rejecting it is between \(\alpha = 0.05\) and \(\alpha = 0.01\). A “straggler” is retained unless there is compelling reason for its rejection. The guidelines recommend using \(\alpha = 0.01\) as the minimum criterion for rejecting a possible outlier. On the other hand, testing for outliers can provide useful information if we try to understand the source of the suspected outlier. For example, the outlier in represents a significant change in the mass of a penny (an approximately 17% decrease in mass), which is the result of a change in the composition of the U.S. penny. In 1982 the composition of a U.S. penny changed from a brass alloy that was 95% w/w Cu and 5% w/w Zn (with a nominal mass of 3.1 g), to a pure zinc core covered with copper (with a nominal mass of 2.5 g) [Richardson, T. H. , , 310–311]. The pennies in , therefore, were drawn from different populations. | 37,263 | 2,738 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Carbohydrates_Fundamentals/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 2,739 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Alpha-carbon_Reactions |
Many aldehydes and ketones undergo substitution reactions at alpha carbons. Many aldehydes and ketones undergo substitution reactions at an alpha carbon, as shown in the following diagram (alpha-carbon atoms are colored blue). These reactions are acid or base catalyzed, but in the case of halogenation the reaction generates an acid as one of the products, and is therefore autocatalytic. If the alpha-carbon is a chiral center, as in the second example, the products of halogenation and isotopic exchange are racemic. Indeed, treatment of this ketone reactant with acid or base alone serves to racemize it. Not all carbonyl compounds exhibit these characteristics, the third ketone being an example. Reactions at the \(\alpha\) carbon Two important conclusions may be drawn from these examples. , these substitutions are limited to carbon atoms alpha to the carbonyl group. Cyclohexanone (the first ketone) has two alpha-carbons and four potential substitutions (the alpha-hydrogens). Depending on the reaction conditions, one or all four of these hydrogens may be substituted, but none of the remaining six hydrogens on the ring react. The second ketone confirms this fact, only the alpha-carbon undergoing substitution, despite the presence of many other sites. , the substitutions are limited to hydrogen atoms. This is demonstrated convincingly by the third ketone, which is structurally similar to the second but has no alpha-hydrogen. Kinetic studies of these reactions provide additional information. The rates of halogenation and isotope exchange are essentially the same (assuming similar catalysts and concentrations), and are identical to the rate of racemization for those reactants having chiral alpha-carbon units. At low to moderate halogen concentrations, the rate of halogen substitution is proportional (i.e. ) to aldehyde or ketone concentration, but independent of halogen concentration. This suggests the existence of a common reaction intermediate, formed in a slow (rate-determining step) prior to the final substitution. Acid and base catalysts act to increase the rate at which the common intermediate is formed, and their concentration also influences the overall rate of substitution.
From previous knowledge and experience, we surmise that the common intermediate is an of the carbonyl reactant. Several facts support this proposal: The reactions shown above, and others to be described, may be characterized as an electrophilic attack on the electron rich double bond of an enol tautomer. This resembles closely the first step in the . Therefore, if electrophilic substitution reactions of this kind are to take place it is necessary that nucleophilic character be established at the alpha-carbon. A full description of the acid and base-catalyzed keto-enol tautomerization process (shown below) discloses that only two intermediate species satisfy this requirement. These are the itself and its conjugate base (common with that of the keto tautomer), usually referred to as an . Clearly, the proportion of enol tautomer present at equilibrium is a critical factor in alpha substitution reactions. In the case of simple aldehydes and ketones this is very small, as noted above. A complementary property, the acidity of carbonyl compounds is also important, since this influences the concentration of the more nucleophilic enolate anion in a reaction system. Ketones such as cyclohexanone are much more acidic than their parent hydrocarbons (by at least 25 powers of ten); nevertheless they are still very weak acids (pK = 17 to 21) compared with water. Together with some related acidities, this is listed in the following table. Even though enol tautomers are about a million times more acidic than their keto isomers, their low concentration makes this feature relatively unimportant for many simple aldehydes and ketones. In cases where more than one activating function influences a given set of alpha-hydrogens, the enol concentration and acidity is increased. Examples of such doubly (and higher) activated carbon acids are . In view of these facts it may seem surprising that alpha-substitution reactions occur at all. However, we often fail to appreciate the way in which a rapid equilibrium involving a minor reactive component may spread the consequences of its behavior throughout a much larger population. Consider, for example, a large group of hungry, active hamsters running about in a big cage. Opening onto the cage there is a small annex that can hold a maximum of three hamsters. Out of two hundred hamsters in the cage, there are an average of two hamsters in the annex at any given time. The hamsters are free to enter and exit the annex, but any hamster that does so is marked by a bright red dye. Although the hamster concentration in the annex is small relative to the whole population, it will not be long before all the hamsters are dyed red. If we substitute molecules for hamsters, their numbers will be extraordinarily large (recall the size of Avogadro's number), but the equilibrium between keto tautomers (hamsters in the cage) and enol tautomers (hamsters in the annex) is so rapid that complete turnover of all the molecules in a sample may occur in fractions of a second rather than minutes or hours. The principle is the same in both cases. Racemization and isotope exchange are due to the rapid equilibrium between chiral keto tautomers and achiral enol tautomers, as well as statistical competition between hydrogen and its deuterium isotope. For halogenation there is also a thermodynamic driving force, resulting from increased bond energy in the products. For example, the alpha-chlorination of cyclohexane, shown above, is exothermic by over 10 kcal/mole. A useful carbon-carbon bond-forming reaction known as the or the Aldol Condensation is yet another example of electrophilic substitution at the alpha carbon in enols or enolate anions. Three examples of the base-catalyzed aldol reaction are shown in the following diagram, and equivalent acid-catalyzed reactions also occur. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. By clicking the "Structural Analysis" button below the diagram, a display showing the nucleophilic enolic molecule and the electrophilic molecule together with the newly formed carbon-carbon bond will be displayed. Stepwise mechanisms for the base-catalyzed and acid-catalyzed reactions may be seen by clicking the appropriate buttons. In the presence of acid or base catalysts the aldol reaction is reversible, and the beta-hydroxy carbonyl products may revert to the initial aldehyde or ketone reactants. In the absence of such catalysts these aldol products are perfectly stable and isolable compounds. Because of this reversibility, the yield of aldol products is related to their relative thermodynamic stability. In the case of aldehyde reactants (as in reactions #1 & 2 above), the aldol reaction is modestly exothermic and the yields are good. However, aldol reactions of ketones are less favorable (e.g. #3 above), and the equilibrium product concentration is small. A clever way of overcoming this disadvantage has been found. A comparatively insoluble base, Ba(OH) , is used to catalyze the aldol reaction of acetone, and the product is removed from contact with this base by filtration and recirculation of the acetone. The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this reaction is an α,β-unsaturated aldehyde or ketone, as shown in the following diagram. Acid-catalyzed conditions are more commonly used to effect this elimination (examples #1, 2 & 5), but base-catalyzed elimination also occurs, especially on heating (examples #3, 4 & 5). The additional stability provided by the conjugated carbonyl system of the product makes some ketone aldol reactions thermodynamically favorable (#4 & 5), and mixtures of stereoisomers (E & Z) are obtained from reaction #4. Reaction #5 is an interesting example of an intramolecular aldol reaction; such reactions create a new ring. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water are termed . Hence the following examples are properly referred to as . The dehydration step of an aldol condensation is also reversible in the presence of acid and base catalysts. Consequently, on heating with aqueous solutions of strong acids or bases, many α, β-unsaturated carbonyl compounds fragment into smaller aldehyde or ketones, a process known as the . The acid-catalyzed elimination of water is not exceptional, since this was noted as a . Nevertheless, the conditions required for the beta-elimination are found to be milder than those used for simple alcohols. The most surprising aspect of beta-elimination, however, is that it can be base-catalyzed. In earlier discussions we have noted that . Why then should the base-catalyzed elimination of water occur in aldol products? To understand this puzzle we need to examine plausible mechanisms for beta-elimination, and these will be displayed by clicking the "Beta-Elimination Mechanism" button under the diagram. As shown by the equations, these eliminations might proceed from either the keto or enol tautomers of the beta-hydroxy aldol product. Although the keto tautomer route is not unreasonable (recall the enhanced acidity of the alpha-hydrogens in carbonyl compounds), the enol tautomer provides a more favorable pathway for both acid and base-catalyzed elimination of the beta oxygen. Indeed, the base-catalyzed loss of hydroxide anion from the enol is a conjugated analog of the base-catalyzed decomposition of a hemiacetal. The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called or mixed reactions, and under certain conditions such crossed aldol condensations can be effective. Some examples are shown below, and in most cases beta-elimination of water occurs under the conditions used. The exception, reaction #3, is conducted under mild conditions with an excess of the reactive aldehyde formaldehyde serving in the role of electrophilic acceptor. The first reaction demonstrates that ketones having two sets of alpha-hydrogens may react at both sites if sufficient acceptor co-reactant is supplied. The interesting difference in regioselectivity shown in the second reaction (the reactants are in the central shaded region) illustrates some subtle differences between acid and base-catalyzed aldol reactions. The base-catalyzed reaction proceeds via an enolate anion donor species, and the kinetically favored proton removal is from the less substituted alpha-carbon. The acid-catalyzed aldol proceeds via the enol tautomer, and the more stable of the two enol tautomers is that with the . Finally, reaction #4 has two reactive alpha-carbons and a reversible aldol reaction may occur at both. Only one of the two aldol products can undergo a beta-elimination of water, so the eventual isolated product comes from that reaction sequence. The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the reaction. The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult. CH CHO + CH CHO + NaOH – + – + – + – The effectiveness of the aldol reaction as a synthetic tool has been enhanced by controlling the enolization of donor compounds, and subsequent reactions with acceptor carbonyls. To see how this is done . In its simplest form the aldol reaction is reversible, and normally forms the thermodynamically favored product. To fully appreciate the complex interplay of factors that underlie this important synthesis tool, we must evaluate the significance of several possible competing reaction paths. Since the negative charge of an enolate anion is delocalized over the alpha-carbon and the oxygen, as , electrophiles may bond to either atom. Reactants having two or more reactive sites are called , so this term is properly applied to enolate anions. Modestly electrophilic reactants such as are not sufficiently reactive to combine with neutral enol tautomers, but the increased nucleophilicity of the enolate anion conjugate base permits such reactions to take place. Because alkylations are usually irreversible, their products should reflect the inherent (kinetic) reactivity of the different nucleophilic sites. If an alkyl halide undergoes an S 2 reaction at the carbon atom of an enolate anion the product is an alkylated aldehyde or ketone. On the other hand, if the S 2 reaction occurs at oxygen the product is an ether derivative of the enol tautomer; such compounds are stable in the absence of acid and may be isolated and characterized. These alkylations (shown above) are irreversible under the conditions normally used for S 2 reactions, so the product composition should provide a measure of the relative rates of substitution at carbon versus oxygen. It has been found that this competition is sensitive to a number of factors, including negative charge density, solvation, cation coordination and product stability.
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in , and hydroxide or alkoxide bases induce competing S 2 and E2 reactions of alkyl halides. It is necessary, therefore, to by treatment with a very strong base (pK > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases having pK 's greater than 30 were , and some others that have been used for enolate anion formation are: NaH (sodium hydride, pK > 45), NaNH (sodium amide, pK = 34), and (C H ) CNa (trityl sodium, pK = 32). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly . Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane. Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced. The following equation provides examples of electrophilic substitution at both carbon and oxygen for the enolate anion derived from cyclohexanone. A full analysis of the factors that direct substitution of enolate anions to carbon or oxygen is beyond the scope of this text. However, an outline of some significant characteristics that influence the two reactions shown above is illustrative. Reactant Important Factors CH –I The negative charge density is greatest at the oxygen atom (greater electronegativity), and coordination with the sodium cation is stronger there. Because methyl iodide is only a modest electrophile, the S 2 transition state resembles the products more than the reactants. Since the C-alkylation product is thermodynamically more stable than the O-alkylated enol ether, this is reflected in the transition state energies. (CH ) Si–Cl Trimethylsilyl chloride is a stronger electrophile than methyl iodide (note the electronegativity difference between silicon and chlorine). Relative to the methylation reaction, the S 2 transition state will resemble the reactants more than the products. Consequently, reaction at the site of greatest negative charge (oxygen) will be favored. Also, the high Si–O bond energy (over 25 kcal/mole greater than Si–C) thermodynamically favors the silyl enol ether product. The reaction of alkyl halides with enolate anions presents the same problem of competing S 2 and E2 reaction paths that was encountered earlier in the . Since enolate anions are very strong bases, they will usually cause elimination when reacted with 2º and 3º-halides. Halides that are incapable of elimination and/or have enhanced S 2 reactivity are the best electrophilic reactants for this purpose. Four examples of the C-alkylation of enolate anions in synthesis are displayed in the following diagram. The first two employ the versatile strong base , which is the reagent of choice for most intermolecular alkylations of simple carbonyl compounds. The dichloro alkylating agent used in reaction #1 nicely illustrates the high reactivity of allylic halides and the unreactive nature of vinylic halides in . The additive effect of carbonyl groups on alpha-hydrogen acidity is demonstrated by reaction #3. Here the two hydrogen atoms activated by both carbonyl groups are over 10 times more acidic than the methyl hydrogens on the ends of the carbon chain. Indeed, they are sufficiently acidic (pK = 9) to allow in aqueous or alcoholic solutions. As shown (in blue), the negative charge of the enolate anion is delocalized over both oxygen atoms and the central carbon. The oxygens are hydrogen bonded to solvent molecules, so the kinetically favored S 2 reaction occurs at the carbon. The monoalkylated product shown in the equation still has an acidic hydrogen on the central carbon, and another alkyl group may be attached there by repeating this sequence.
The last example (reaction #4) is an interesting case of intramolecular alkylation of an enolate anion. Since alkylation reactions are irreversible, it is possible to form small highly strained rings if the reactive sites are in close proximity. Reversible bond-forming reactions, such as the aldol reaction, cannot be used for this purpose. The use of aqueous base in this reaction is also remarkable, in view of the very low enolate anion concentration for such systems. It is the rapid intramolecular nature of the alkylation that allows these unfavorable conditions to be used. The five-carbon chain of the dichloroketone can adopt many conformations, two of which are approximated in the preceding diagram. Although conformer II of the enolate anion could generate a stable five-membered ring by an intramolecular S 2 reaction, assuming proper orientation of the α and γ' carbon atoms, the concentration of this ideally coiled structure will be very low. In this case O-alkylation of the enolate anion, rather than C-alkylation, is preferred from stereoelectronic arguments (see ). On the other hand, conformations in which the α and γ-carbons are properly aligned for three-membered ring formation are much more numerous, the result being that as fast as the enolate base is formed it undergoes rapid and irreversible cyclization. Ring closures to four, five, six and seven-membered are also possible by intramolecular enolate alkylation, as illustrated by the following example. In general, five and six-membered rings are thermodynamically most stable, whereas three-membered ring formation is favored kinetically. Several enolate-like compounds and ions have been studied as alternative intermediates for synthesis. To learn more about these possibilities . ), | 20,858 | 2,741 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.02%3A_Simple_Distillation/5.2C%3A_Step-by-Step_Procedures |
The condenser is an intricate piece of glassware, and allows for cold water to circulate through the distillation apparatus. The circulating water does not mix with the sample to be purified, but instead passes through another jacket surrounding the hollow tube where the gaseous sample travels. It is important that the water jacket be of cold water, to maximize the efficiency of condensing the gaseous sample. It is for this reason that the water hoses must be attached to the condenser in a certain way. A hose should connect from the water spigot to the arm of the condenser, forcing water to travel against gravity through the condenser (this is shown correctly in Figure 5.17b). The hose connecting the arm of the condenser should then drain to the sink. By forcing the water uphill, it will completely fill the condenser. The hoses are connected incorrectly in Figure 5.17a, with water flowing into the upper arm of the condenser. With a downhill flow of water, only the lower portion of the condenser fills with the circulating water, leading to inefficient cooling of gas traveling through the inner tube. Whether the condenser hoses point up or down seems to be a personal choice, as there are pros and cons to both orientations. When the hoses point downward, there is a small portion of the condenser that does not fill with cooling water (Figure 5.18a), but the effect is likely minimal. When the hoses point upwards, there is sometimes a tendency for the hoses to bend and pop off, spraying water all over the lab (this can sometimes be avoided by securing wire around the hose joints). Personal judgment may be applied to whether the condenser hoses point up or down. An assembled simple distillation apparatus is shown in Figure 5.19. Assembly of this complicated apparatus is shown in this section piece by piece. The glassware used for this apparatus is quite expensive, and undoubtedly your instructor would appreciate care being taken when using this experiment. A single condenser costs $72, and a complete kit containing all the glassware needed for distillation costs $550!\(^8\) Fill the distilling flask with sample 1/3-1/2 full. Always use an extension clamp on the distilling flask. Add a few boiling stones or stir bar to flask. Position the thermometer bulb just below the arm of the three-way adapter, where vapors turn toward the condenser. Wet condenser hoses with water before attaching. Connect the condenser hoses such that water flow uphill: bring water from the faucet into the lower arm, and drain out the upper arm. Be sure all of the connections are secure (especially between the distilling flask and 3-way adapter: potential of fire!) Turn on the condenser water. Apply the heat source to the distilling flask. Collect distillate at a rate of . Record the temperature where liquid is actively distilling and thermometer bulb is immersed. Record the pressure. Stop the distillation when the temperature changes dramatically or if the distilling flask is nearly empty (never distill to dryness!) Lower and remove the heat source, but keep water circulating until the flask is just warm to the touch. A few variations of the simple distillation setup are commonly encountered: It is not uncommon for new organic chemistry students to expect the temperature to rise on the thermometer the very moment boiling is seen in the distilling flask. Students should remember that the thermometer measures the temperature at the location of the bulb, which is a distance away from the boiling liquid. Therefore, when a solution begins to boil, it may still take a while for the thermometer to register anything other than room temperature. It is important to record a temperature only when the thermometer bulb is fully engulfed by vapors. Figure 5.31a shows a solution that is boiling but not yet distilling - the temperature at this point would not represent the boiling point of the liquid. Figure 5.31b shows an actively distilling solution, and notice that the thermometer bulb is completely enveloped in vapors and condensation; the temperature at this point would correspond to the boiling point of the solution. For high-boiling liquids, it may be difficult for vapors to reach the condenser as they too quickly are cooled by the glassware which is in contact with the air in the room. It may be helpful to insulate the distilling flask and three-way adapter to better retain heat and allow the sample to remain in the gas phase longer. To insulate a portion of the distillation, wrap the parts prior to the condenser with (Figure 5.32c), then secure with an outer wrapping of aluminum foil (Figure 5.32d). A small gap can be left in the insulation in order to "peek in" on activity inside the apparatus. Glass wool has an appearance similar to cotton, but unlike cotton is not flammable so is useful as an insulating material when an apparatus is to be heated. Glass wool comes in two forms: a fibrous form and a cottony form. The fibrous form (Figure 5.32a) has tiny fibers that can become embedded in skin in the most painful way, similar to several simultaneous paper cuts. This type of glass wool should not be manipulated with bare hands, but only when wearing thick gloves. The more cotton-like glass wool (Figure 5.32b), however, does not hurt when handling with bare hands. If glass wool is unavailable, aluminum foil can be used alone to insulate a portion of the distillation. It will not insulate well if the foil is wrapped too tightly to the glass, but works well if a small pocket of air is allowed between the foil and glass. If the expected boiling point of the compound to be distilled is quite high \(\left( > 150^\text{o} \text{C} \right)\) and distillation continues to be difficult, vacuum distillation might be attempted. If vapor is noticed escaping out of the vacuum adapter like a tea kettle, the condenser is not doing a good enough job of trapping the gas (Figure 5.33). Reasons for this may be that you have forgotten to turn on the water in the condenser, the water stream is too weak, or the heating is too vigorous. \(^8\)Prices were found in the Chemglass catalog in March 2016. \(^9\)Certain compounds (e.g. ethers, dioxanes, tetrahydrofuran) are well known to form peroxides when left in contact with air over a period of time. However, many compounds are somewhat susceptible to peroxide formation, including compounds with allylic hydrogen atoms (e.g. cyclohexene). A more complete listing of compounds that form peroxides can be found in: , CRC Press, 84\(^\text{th}\) edition, , 16-6. | 6,584 | 2,742 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/06%3A_Bonding_in_Organic_Molecules/6.01%3A_Prelude_to_Bonding |
In previous chapters, we have shown how you can use ball-and-stick models to predict the general arrangements in space of organic molecules. The sticks correspond to chemical bonds, which we represent in structural formulas as lines, or in Lewis structures as pairs of dots denoting shared pairs of electrons. Remembering that electrons and nuclei are charged particles, and that it is electrical forces of attraction and repulsion between the electrons and nuclei that determine the bonding, perhaps we should be surprised that such simple mechanical models provide so much useful information. What we will try to do in this chapter is to show you how the modern electronic theory of chemical bonding provides strong support for the use of ball-and-stick models for many organic molecules, and also where it indicates that the models need to be modified or cannot properly represent the structural arrangements. There are several qualitative approaches to bonding in polyatomic molecules, but we shall discuss here the most widely used and currently popular approach. This approach involves setting up appropriate atomic orbitals for the atoms and considering that each bond arises from the attractive electrical forces of two or more nuclei for a pair of electrons in overlapping atomic orbitals, with each orbital on a different atom. The geometry of the bonds is assumed to be determined by the geometry of the orbitals and by the repulsive forces between the electrons. In the course of showing how this approach can be applied, we shall discuss ways of formulating bonding and geometries for several important kinds of organic compounds. Finally, we will show you some of the results currently being obtained by sophisticated quantum-mechanical calculations, which provide strong support for our qualitative formulations. and (1977) | 1,857 | 2,743 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/The_Molecular_Ion_(M_)_Peak |
This page explains how to find the relative formula mass (relative molecular mass) of an organic compound from its mass spectrum. It also shows how high resolution mass spectra can be used to find the molecular formula for a compound. When the vaporised organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion. The molecular ion is often given the symbol \(\ce{M^{+}}\) or\(\ce{M^{\cdot +}}\) - the dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process. The molecular ions tend to be unstable and some of them break into smaller fragments. These fragments produce the familiar stick diagram. Fragmentation is irrelevant to what we are talking about on this page - all we're interested in is the molecular ion. In the mass spectrum, the heaviest ion (the one with the greatest m/z value) is likely to be the molecular ion. A few compounds have mass spectra which don't contain a molecular ion peak, because all the molecular ions break into fragments. For example, in the mass spectrum of pentane, the heaviest ion has an m/z value of 72. Because the largest m/z value is 72, that represents the largest ion going through the mass spectrometer - and you can reasonably assume that this is the molecular ion. The relative formula mass of the compound is therefore 72. Finding the relative formula mass (relative molecular mass) from a mass spectrum is therefore trivial. Look for the peak with the highest value for m/z, and that value is the relative formula mass of the compound. There are, however, complications which arise because of the possibility of different isotopes (either of carbon or of chlorine or bromine) in the molecular ion. These cases are dealt with on separate pages. So far we've been looking at m/z values in a mass spectrum as whole numbers, but it's possible to get far more accurate results using a high resolution mass spectrometer. You can use that more accurate information about the mass of the molecular ion to work out the molecular formula of the compound. For normal calculation purposes, you tend to use rounded-off relative isotopic masses. For example, you are familiar with the atomic mass numbers (\(Z\)). To 4 decimal places, however, these are the relative isotopic masses. The carbon value is 12.0000, of course, because all the other masses are measured on the carbon-12 scale which is based on the carbon-12 isotope having a mass of exactly 12. Two simple organic compounds have a relative formula mass of 44 - propane, C H , and ethanal, CH CHO. Using a high resolution mass spectrometer, you could easily decide which of these you had. On a high resolution mass spectrometer, the molecular ion peaks for the two compounds give the following m/z values: You can easily check that by adding up numbers from the table of accurate relative isotopic masses above. A gas was known to contain only elements from the following list: The gas had a molecular ion peak at m/z = 28.0312 in a high resolution mass spectrometer. What was the gas? After a bit of playing around, you might reasonably come up with 3 gases which had relative formula masses of approximately 28 and which contained the elements from the list. They are N , CO and C H . Working out their accurate relative formula masses gives: The gas is obviously C H . Jim Clark ( ) | 3,694 | 2,744 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Diffraction_Scattering_Techniques/Bragg's_Law |
The structures of crystals and molecules are often being identified using x-ray diffraction studies, which are explained by Bragg’s Law. The law explains the relationship between an x-ray light shooting into and its reflection off from crystal surface. Bragg’s Law was introduced by Sir W.H. Bragg and his son Sir W.L. Bragg. The law states that when the x-ray is incident onto a surface, its angle of incidence, \(\theta\), will reflect back with a same angle of scattering, \(\theta\). And, when the path difference, \(d\) is equal to a whole number, \(n\), of wavelength, a constructive interference will occur. Consider a single crystal with aligned planes of lattice points separated by a distance . Monochromatic X-rays A, B, and C are incident upon the crystal at an angle . They reflect off atoms X, Y, or Z. The path difference between the ray reflected at atom X and the ray reflected at atom Y can be seen to be 2YX. From the Law of Sines we can express this distance YX in terms of the lattice distance and the \[n\lambda = 2d\sin\theta\] where: The principle of Bragg’s law is applied in the construction of instruments such as Bragg spectrometer, which is often used to study the structure of crystals and molecules. | 1,253 | 2,745 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO10._Activation_of_Carbonyls |
A secondary theme in carbonyl chemistry centers on the role played by the oxygen lone pairs. A compound with lone pairs can act as a Lewis base. Can carbonyl compounds also act as Lewis bases? The answer is yes, although it is most important to think about carbonyls primarily as Lewis acids. One of the reasons the basicity of the lone pair matters is because of carbonyl activation. If a carbonyl donates a lone pair to a Lewis acid, forming a bond, the carbonyl gets a formal positive charge. If the carbonyl has a formal positive charge, it attracts electrons more strongly. In that case, nucleophiles react more easily with the carbonyl. The carbonyl is said to be activated. Figure CO10.1. Activation of a carbonyl via donation to a proton. A carbonyl can be activated by the addition of a proton donor, such as HCl or other common acids. Most common mineral acids are used as aqueous solutions (the familiar HCl, H SO , HNO , H PO and so on). Sometimes, in a laboratory reaction, it isn't helpful to have all that water around (the reasons will become clear later). Other, organic acids are sometimes used instead, such as camphorsulfonic acid or toluenesulfonic acid; these are both solids that are easy to weight out and add to a reaction, and they don't add a bunch of water to the reaction. Figure CO10.2. Some protic acids useful in activating carbonyls. Carbonyls are also activated by more general Lewis acids. Often, metal chloride salts are used. These may include main group metals, such as aluminum, bismuth or indium, or transition metals such as scandium, titanium or iron. Figure CO10.3. Activation of a carbonyl by a metal ion. Once the carbonyl is activated, nucleophiles are more strongly attracted to the carbon. The carbon was already partially positive, but with a full positive charge on the molecule, electrons are attracted much more strongly. It is tempting to donate electrons from a nucleophile to the positive oxygen. However, the oxygen already has three bonds and an octet. Remember, donating a lone pair from a nucleophile means the lone pair is becoming a bond between the nucleophile and the electrophile. Giving a pair of electrons directly to the oxygen would give it four bonds and more than an octet-- it would have 10 electrons. Instead, donation to the neighbouring carbon allows the C=O pi bond to move to the oxygen and become a lone pair. The positive charge on the oxygen disappears. Figure CO10.4. Donation of nucleophile to an activated carbonyl. Show, with arrows, the activation of the following carbonyls. Show, with arrows, the activation of the following carbonyls, followed by donation from the nucleophile. , | 2,680 | 2,746 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.02%3A_Simple_Distillation/5.2D%3A_Microscale_Distillation |
There are a variety of methods used to distill small amounts of material \(\left( < 10 \: \text{mL} \right)\). The goal of all of these variations is to minimize the loss of material. A certain amount of sample is always necessary to fill the apparatus before distillation occurs, and this quantity (called the " ") normally condenses after cooling, often back into the original distilling flask. Therefore, minimizing the path length between distilling and receiving flasks can increase the recovery with distillation. Another approach to increase recovery on the microscale is to minimize the number of joints in the apparatus, which may not be perfectly airtight, and can contribute to leaking of material. Most microscale versions are simple distillations, as use of a fractional column adsorbs too much material. Therefore, these techniques are generally used to remove non-volatile or very high-boiling or low-boiling impurities. A semi-microscale apparatus is essentially a macroscale apparatus but lacks the condenser (Figure 5.34). This shortens the path length between distilling and receiving flask. It can be used for \(5\)-\(10 \: \text{mL}\) of material. The distillation needs to be done with care, as too high of temperatures may cause vapor to be lost out of the vacuum adapter. A Hickman head is provided in many microscale kits, and can be used to distill \(1\)-\(3 \: \text{mL}\) of material. It is often connected to a conical vial, using an O-ring and screw-capped connector to secure the joint (Figure 5.35a). The solution boils, condenses on the sides of the flask and collects in the lip of the Hickman head (indicated with an arrow in Figure 5.35b). The bulb of the thermometer should be positioned below the lip (Figure 5.35a), although it may be difficult to accurately measure the vapor temperature as so little material is used. After cooling, the flask must be tilted to retrieve the distillate by pipette (Figure 5.35c), and the lip holds at the most \(1 \: \text{mL}\) of distillate so may need to be emptied several times. Some Hickman heads have a side port so distillate can be removed without stopping the distillation (Figure 5.35d). An all-in-one distillation apparatus, called a " ", is sometimes used to distill small quantities of material (Figure 5.36). These pieces of glassware are quite expensive ($200 each),\(^{10}\) and so are normally used in research settings, not teaching labs. When assembling the glassware, it is important that the joint connecting the distilling flask to the apparatus is held securely: verify it is not loose by grasping the pieces with your fingers (Figure 5.36b). The apparatus requires a thermometer that has a ground-glass fitting (Figure 5.36d). \(^{10}\)Prices were found in the Chemglass catalog in March 2016. | 2,803 | 2,747 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/07%3A_Mixtures_and_Solutions/7.04%3A_The_Gibbs-Duhem_Equation |
For a system at , the Gibbs-Duhem equation must hold: \[\sum_i n_i d\mu_i = 0 \label{eq1} \] This relationship places a compositional constraint upon any changes in the chemical potential in a mixture at constant temperature and pressure for a given composition. This result is easily derived when one considers that \(\mu_i\) represents the partial molar Gibbs function for component \(i\). And as with other partial molar quantities, \[ G_{tot} = \sum_i n_i \mu_i \nonumber \] Taking the derivative of both sides yields \[ dG_{tot} = \sum_i n_i d \mu_i + \sum_i \mu_i d n_i \nonumber \] But \(dG\) can also be expressed as \[dG = Vdp - sdT + \sum_i \mu_i d n_i \nonumber \] Setting these two expressions equal to one another \[ \sum_i n_i d \mu_i + \cancel{ \sum_i \mu_i d n_i } = Vdp - sdT + \cancel{ \sum_i \mu_i d n_i} \nonumber \] And after canceling terms, one gets \[ \sum_i n_i d \mu_i = Vdp - sdT \label{eq41} \] For a system at constant temperature and pressure \[Vdp - sdT = 0 \label{eq42} \] Substituting Equation \ref{eq42} into \ref{eq41} results in the (Equation \ref{eq1}). This expression relates how the chemical potential can change for a given composition while the system maintains equilibrium. So for a binary system, consisting of components \(A\) and \(B\) (the two most often studied compounds in all of chemistry) \[ d\mu_B = -\dfrac{n_A}{n_B} d\mu_A \nonumber \] | 1,411 | 2,748 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Units_of_Measure/Metric%2F%2FImperial_Conversion_Errors |
To learn more about systems of measurement, visit the SI Unit page and the Non-SI Unit page. As the four examples below can attest, small errors in these unit systems can harbor massive ramifications. Although NASA declared the metric system as its official unit system in the 1980s, conversion factors remain an issue. The Mars Climate Orbiter, meant to help relay information back to Earth, is one notable example of the unit system struggle. The orbiter was part of the Mars Surveyor ’98 program, which aimed to better understand the climate of Mars. As the spacecraft journeyed into space on September 1998, it should have entered orbit at an altitude of 140-150 km above Mars, but instead went as close as 57km. This navigation error occurred because the software that controlled the rotation of the craft’s thrusters was not calibrated in SI units. The spacecraft expected newtons, while the computer, which was inadequately tested, worked in pound forces; one pound force is equal to about 4.45 newtons. Unfortunately, friction and other atmospheric forces destroyed the Mars Climate Orbiter. The project cost $327.6 million in total. Tom Gavin, an administrator for NASA's Jet Propulsion Laboratory in Pasadena, stated, "This is an end-to-end process problem. A single error like this should not have caused the loss of Climate Orbiter. Something went wrong in our system processes in checks and balances that we have that should have caught this and fixed it." Another NASA related conversion concern involves the Constellation project, which is focused mainly on manned spaceflight. Established in 2005, it includes plans for another moon landing. The Constellation project is partially based upon decades-old projects such as the rocket and the crew capsule. These figures and plans are entirely in British Imperial units; converting this work into metric units would cost approximately $370 million. Work on the Con NASA/ Tokyo Disneyland’s Space Mountain roller coaster came to a sudden halt just before the end of a ride on December 5, 2003. This startling incident was due a broken axle. The axle in question fractured because it was smaller than the design’s requirement; because of the incorrect size, the gap between the bearing and the axle was over 1 mm – when it should have been a mere 0.2 mm (the thickness of a dime vs. the thickness of two sheets of common printer paper.) The accumulation of excess vibration and stress eventually caused it to break. Though the coaster derailed, there were no injuries. Once again, unit systems caused the accident. In September 1995, the specifications for the coaster’s axles and bearings were changed to metric units. In August 2002, however, the British-Imperial-unit plans prior to 1995 were used to order 44.14 mm axels instead of the needed 45 mm axels. A Boeing 767 airplane flying for Air Canada on July 23, 1983 diminished its fuel supply only an hour into its flight. It was headed to Edmonton from Montreal, but it received low fuel pressure warnings in both fuel pumps at an altitude of 41,000 feet; engine failures followed soon after. Fortunately, the captain was an experienced glider pilot and the first officer knew of an unused air force base about 20 kilometers away. Together, they landed the plan on the runway, and only a few passengers sustained minor injuries. This incident was due partially to the airplane’s fuel indication system, which had been malfunctioning. Maintenance workers resorted to manual calculations in order to fuel the craft. They knew that 22,300 kg of fuel was needed, and they wanted to know how much in liters should be pumped. They used 1.77 as their density ratio in performing their calculations. However, 1.77 was given in pounds per liter, not kilograms per liter. The correct number should have been 0.80 kilograms/liter; thus, their final figure accounted for less than half of the necessary fuel. If Jimmy walks 5 miles, how many kilometers did he travel? \[5 \;\cancel{miles} \times \left (\dfrac{1.6\; kilometers }{1\; \cancel{mile}}\right) = 8\; kilometers \nonumber\] A solid rocket booster is ordered with the specification that it is to produce a total of 10 million pounds of thrust. If this number is mistaken for the thrust in Newtons, by how much, in pounds, will the thrust be in error? (1 pound = 4.5 Newtons) 10,000,000 Newtons x (1 pound / 4.448 Newtons) = 2,200,000 pounds. 10,000,000 pounds - 2,200,000 pounds = 7.800,000 pounds. The error is a missing 7,800,000 pounds of thrust. \[3,027,400\; \cancel{liters} \times \left(\dfrac{0.264 \;gallons}{1\; \cancel{liter}}\right) = 800,000\;\text{gallons remaining in tank} \nonumber\] The volume of the space fish is 1,300,000 - 800,000 = 500,000 gallons, which converts to 1,892,100 liters worth of fish swimming around the solar system. A bolt is ordered with a thread diameter of 1.25 inches. What is this diameter in millimeters? If the order was mistaken for 1.25 centimeters, by how many millimeters would the bolt be in error? \[1.25\; \cancel{\rm{ inches}} \times \dfrac{25.4\; \rm{millimeters}}{1\; \cancel{ \rm{inch}}} = 31.75 \; \rm{millimeters} \nonumber\] Since 1.25 centimeters x (10 millimeters / 1 centimeter) = 12.5 millimeters, the bolt would delivered 31.75 - 12.5 = 19.25 millimeters too small. The Mars Climate Orbiter was meant to stop about 160 km away from the surface of Mars, but it ended up within 36 miles of the surface. How far off was it from its target distance (in km)? If the Orbiter is able to function as long as it stays at least 85 km away from the surface, will it still be functional despite the mistake? \[36 \; \cancel{\text{miles}} \times \dfrac {1.6 \; \text{kilometers} }{1\; \cancel{\text{mile}}} = 57.6 \;\text{km kilometers from surface} \nonumber\] The difference then is (in kilometers): 160 - 57.6 kilometers = 102.4 kilometers away from targeted distance. Hence, the Orbiter is unable to function due to this mistake since it is beyond the 85 km error designed into its function. | 6,032 | 2,749 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/03%3A_First_Law_of_Thermodynamics/3.06%3A_Reaction_Enthalpies |
Reaction enthalpies are important, but difficult to tabulate. However, because enthalpy is a state function, it is possible to use to simplify the tabulation of reaction enthalpies. Hess’ Law is based on the addition of reactions. By knowing the reaction enthalpy for constituent reactions, the enthalpy of a reaction that can be expressed as the sum of the constituent reactions can be calculated. The key lies in the canceling of reactants and products that °Ccur in the “data” reactions but not in the “target reaction. Find \(\Delta H_{rxn}\) for the reaction \[2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \nonumber \] Given \[C(gr) + ½ O_2(g) \rightarrow CO(g) \nonumber \] with \(\Delta H_1 = -110.53 \,kJ\) \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with \(\Delta H_2 = -393.51\, kJ\) The target reaction can be generated from the data reactions. \[ {\color{red} 2 \times} \left[ CO(g) \rightarrow C(gr) + O_2(g) \right] \nonumber \] plus \[ { \color{red} 2 \times} \left[ C(gr) + 2 O_2(g) \rightarrow 2 CO_2(g) \right] \nonumber \] equals \[2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \nonumber \] so \[{ \color{red} 2 \times} \Delta H_1 = -787.02 \, kJ \nonumber \] \[{ \color{red} 2 \times} \Delta H_2 = 221.06\, kJ \nonumber \] \[ { \color{red} 2 \times} \Delta H_1 + { \color{red} 2 \times} \Delta H_2 = -565.96 \,kJ \nonumber \] One of the difficulties with many thermodynamic state variables (such as enthalpy) is that while it is possible to measure changes, it is impossible to measure an absolute value of the variable itself. In these cases, it is necessary to define a zero to the scale defining the variable. For enthalpy, the definition of a zero is that the standard enthalpy of formation of a pure element in its standard state is zero. All other enthalpy changes are defined relative to this standard. Thus it is essential to very carefully define a standard state. The standard state of a substance is the most stable form of that substance at 1 atmosphere pressure and the specified temperature. Using this definition, a convenient reaction for which enthalpies can be measured and tabulated is the . This is a reaction which forms one mole of the substance of interest in its standard state from elements in their standard states. The enthalpy of a standard formation reaction is the (\(\Delta H_{f^o}\)). Some examples are It is important to note that the standard state of a substance is . For example, the standard state of water at -10 °C is solid, whereas the standard state at room temperature is liquid. Once these values are tabulated, calculating reaction enthalpies becomes a snap. Consider the heat combustion (\(\Delta H_c\)) of methane (at 25 °C) as an example. \[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \nonumber \] The reaction can expressed as a sum of a combination of the following standard formation reactions. \[C(gr) + 2 H_2(g) \rightarrow CH_4(g) \nonumber \] with \(\Delta H_f^o = -74.6\, kJ/mol\) \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with \(\Delta H_f^o = -393.5\, kJ/mol\) \[H_2(g) + ½ O_2(g) \rightarrow H_2O(l) \nonumber \] with \(\Delta H_f^o = -285.8 \,kJ/mol\) The target reaction can be generated from the following combination of reactions \[{ \color{red} -1 \times} \left[ C(gr) + 2 H_2(g) \rightarrow CH_4(g)\right] \nonumber \] \[CH_4(g) \rightarrow C(gr) + 2 H_2(g) \nonumber \] with \(\Delta H_f^o ={ \color{red} -1 \times} \left[ -74.6\, kJ/mol \right]= 74.6\, kJ/mol\) \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with \(\Delta H_f^o = -393.5\, kJ/mol\) \[{ \color{red} 2 \times} \left[ H_2(g) + ½ O_2(g) \rightarrow H_2O(l) \right] \nonumber \] \[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \nonumber \] with \(\Delta H_f^o = {\color{red} 2 \times} \left[ -285.8 \,kJ/mol \right] = -571.6\, kJ/mol\). \[CH_4(g) + 2 O_2(g) \rightarrow CO2_(g) + 2 H_2O(l) \nonumber \] with \(\Delta H_c^o = -890.5\, kJ/mol\) Alternately, the reaction enthalpy could be calculated from the following relationship \[\Delta H_{rxn} = \sum_{products} \nu \cdot \Delta H_f^o - \sum_{reactants} \nu \cdot \Delta H_f^o \nonumber \] where \(\nu\) is the stoichiometric coefficient of a species in the balanced chemical reaction. For the combustion of methane, this calculation is \[ \begin{align} \Delta _{rxn} & = (1\,mol) \left(\Delta H_f^o(CO_2)\right) + (2\,mol) \left(\Delta H_f^o(H_2O)\right) - (1\,mol) \left(\Delta H_f^o(CH_4)\right) \\ & = (1\,mol) (-393.5 \, kJ/mol) + (2\,mol) \left(-285.8 \, kJ/mol \right) - (1\,mol) \left(-74.6 \, kJ/mol \right) \\ & = -890.5 \, kJ/mol \end{align} \nonumber \] A note about units is in order. Note that reaction enthalpies have units of kJ, whereas enthalpies of formation have units of kJ/mol. The reason for the difference is that enthalpies of formation (or for that matter enthalpies of combustion, sublimation, vaporization, fusion, etc.) refer to specific substances and/or specific processes involving those substances. As such, the total enthalpy change is scaled by the amount of substance used. General reactions, on the other hand, have to be interpreted in a very specific way. When examining a reaction like the combustion of methane \[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \nonumber \] with \(\Delta H_{rxn} = -890.5\, kJ\). The correct interpretation is that the reaction of one mole of CH (g) with two moles of O (g) to form one mole of CO (g) and two moles of H O(l) releases 890.5 kJ at 25 °C. Ionized species appear throughout chemistry. The energy changes involved in the formation of ions can be measured and tabulated for several substances. In the case of the formation of positive ions, the enthalpy change to remove a single electron at 0 K is defined as the . \[ M(g) \rightarrow M^+(g) + e^- \nonumber \] with \(\Delta H (0 K) \equiv 1^{st} \text{ ionization potential (IP)}\) The removal of subsequent electrons requires energies called the 2 Ionization potential, 3 ionization potential, and so on. \[M^+(g) \rightarrow M^{2+}(g) + e^- \nonumber \] with \(\Delta H(0 K) ≡ 2^{nd} IP\) \[M^{2+}(g) \rightarrow M^{3+}(g) + e^- \nonumber \] with \(\Delta H(0 K) ≡ 3^{rd} IP\) An atom can have as many ionization potentials as it has electrons, although since very highly charged ions are rare, only the first few are important for most atoms. Similarly, the can be defined for the formation of negative ions. In this case, the first electron affinity is defined by \[X(g) + e^- \rightarrow X^-(g) \nonumber \] with \(-\Delta H(0 K) \equiv 1^{st} \text{ electron affinity (EA)}\) The minus sign is included in the definition in order to make electron affinities mostly positive. Some atoms (such as noble gases) will have negative electron affinities since the formation of a negative ion is very unfavorable for these species. Just as in the case of ionization potentials, an atom can have several electron affinities. \[X^-(g) + e^- \rightarrow X^{2-}(g) \nonumber \] with \(-\Delta H(0 K) ≡ 2^{nd} EA\). \[X^{2-}(g) + e^- \rightarrow X^{3-}(g) \nonumber \] with \(-\Delta H(0 K) ≡ 3^{rd} EA\). In the absence of standard formation enthalpies, reaction enthalpies can be estimated using average bond enthalpies. This method is not perfect, but it can be used to get ball-park estimates when more detailed data is not available. A \(D\) is defined by \[XY(g) \rightarrow X(g) + Y(g) \nonumber \] \(\Delta H \equiv D(X-Y)\) In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed. \[\Delta H_{rxn} = \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \nonumber \] As an example, consider the combustion of ethanol: In this reaction, five C-H bonds, one C-C bond, and one C-O bond, and one O=O bond must be broken. Also, four C=O bonds, and one O-H bond are formed. The reaction enthalpy is then given by \[ \begin{align} \Delta H_c = \, &5(413 \,kJ/mol) + 1(348\, kJ/mol) + 1(358 \,kJ/mol) \nonumber \\ & + 1(495\, kJ/mol) - 4(799 \,kJ/mol) – 2(463\, kJ/mol) \nonumber \\ =\,& -856\, kJ/mol \end{align} \nonumber \] Because the bond energies are defined for gas-phase reactants and products, this method does not account for the enthalpy change of condensation to form liquids or solids, and so the result may be off systematically due to these differences. Also, since the bond enthalpies are averaged over a large number of molecules containing the particular type of bond, the results may deviate due to the variance in the actual bond enthalpy in the specific molecule under consideration. Typically, reaction enthalpies derived by this method are only reliable to within ± 5-10%. | 8,639 | 2,750 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/07%3A_Mixtures_and_Solutions/7.01%3A_Thermodynamics_of_Mixing |
A natural place to begin a discussion of mixtures is to consider a mixture of two gases. Consider samples of the two gases filling two partitions in a single container, both at the same pressure, temperature, having volumes \(V_A\) and \(V_B\). After being allowed to mix isothermally, the partial pressures of the two gases will drop by a factor of 2 (although the total pressure will still be the original value) and the volumes occupied by the two gases will double. Assuming ideal behavior, so that interactions between individual gas molecules are unimportant, it is fairly easy to calculate \( H) for each gas, as it is simply an isothermal expansion. The total enthalpy of mixing is then given by \[\Delta H_{mix} = \Delta H_A + \Delta H_B \nonumber \] And since the enthalpy change for an , \[\Delta H_{mix} =0 \nonumber \] is a straight-forward conclusion. This will be the criterion for an . In general, real mixtures will deviate from this limiting ideal behavior due to interactions between molecules and other concerns. Also, many substances undergo chemical changes when they mix with other substances. But for now, we will limit ourselves to discussing mixtures in which no chemical reactions take place. The entropy change induced due to isothermal mixing (assuming again no interactions between the molecules in the gas mixture) is again going to be the sum of the contributions from isothermal expansions of the two gases. Fortunately, entropy changes for isothermal expansions are easy to calculate for ideal gases. \[\Delta S = nR \ln \left( \dfrac{V_2}{V_1}\right) \nonumber \] If we use the initial volumes V and V for the initial volumes of gases A and B, the total volume after mixing is \(V_A + V_B\), and the total entropy change is \[\Delta S_{mix} = n_AR \ln \left( \dfrac{V_A + V_B}{V_A}\right) + n_AR \ln \left( \dfrac{V_A + V_B}{V_B}\right) \nonumber \] Noting that the term \[ \dfrac{V_A + V_B}{V_A} = \dfrac{1}{\chi_A} \nonumber \] where \(\chi_A\) is the mole fraction of \(A\) after mixing, and that \(n_A\) can be expresses as the product of \(\chi_A\) and the total number of moles, the expression can be rewritten \[\Delta S_{mix} = n_{tot} R \left[ -\chi_A \ln (\chi_A) - \chi_B \ln (\chi_B) \right] \nonumber \] It should be noted that because the mole fraction is always between 0 and 1, that \(\ln (\chi_B) < 0\). As such, the entropy change for a system undergoing isothermal mixing is always positive, as one might expect (since mixing will make the system less ordered). The entropy change for a system undergoing isothermal mixing is always positive. Calculating \(\Delta G_{mix}\) should be no more difficult than calculating \(\Delta S_{mix}\). For and constant total pressure \[ \Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix} \nonumber \] and so it follows from above that for the isothermal mixing of two gases at constant total pressure \[\Delta G_{mix} = n_{tot} RT \left[ -\chi_A \ln (\chi_A) + \chi_B \ln (\chi_B) \right] \nonumber \] The relationships describing the isothermal mixing of two ideal gases \(A\) and \(B\) is summarized in the graph below. Again, because \(\ln (\chi_i) < 0\), then \(\Delta G_{mix} < 0\) implying that mixing is a spontaneous process for an . This is true for gases. But for many combinations of liquids or solids, the strong intermolecular forces may make mixing unfavorable (for example in the case of vegetable oil and water). Also, these interactions may make the volume non-additive as well (as in the case of ethanol and water). Mixing is always a spontaneous process for an ideal solution. | 3,611 | 2,753 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.01%3A_Introduction |
Until now, you have studied chemical processes in which atoms share or transfer electrons to form new compounds, leaving the atomic nuclei largely unaffected. In this chapter, we examine some properties of the atomic nucleus and the changes that can occur in atomic nuclei. Nuclear reactions differ from other chemical processes in one critical way: in a nuclear reaction, the identities of the elements change. In addition, nuclear reactions are often accompanied by the release of enormous amounts of energy, as much as a times more than the energy released by chemical reactions. Moreover, the yields and rates of a nuclear reaction are generally unaffected by changes in temperature, pressure, or the presence of a catalyst. We begin by examining the structure of the atomic nucleus and the factors that determine whether a particular nucleus is stable or decays spontaneously to another element. We then discuss the major kinds of nuclear decay reactions, as well as the properties and uses of the radiation emitted when nuclei decay. You will learn how radioactive emissions can be used to study the mechanisms of chemical reactions and biological processes and how to calculate the amount of energy released during a nuclear reaction. You will also discover why houses are tested for radon gas, how radiation is used to probe organs such as the brain, and how the energy from nuclear reactions can be harnessed to produce electricity. Last, we explore the nuclear chemistry that takes place in stars, and we describe the role that stars play in producing most of the elements in the universe. The high-energy particles ejected into the surrounding water or air by an intense radioactive source such as this nuclear reactor core produce a ghostly bluish glow. | 1,782 | 2,754 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.10%3A_Ions_and_Electrolytes/8.10.9A%3A_8.10.9A%3A_Electrolytes_and_Electrolytic_Solutions |
Make sure you thoroughly understand the following essential ideas: are those that are capable of conducting an electric current. A substance that, when added to water, renders it conductive, is known as an . A common example of an electrolyte is ordinary salt, sodium chloride. Solid NaCl and pure water are both non-conductive, but a solution of salt in water is readily conductive. A solution of sugar in water, by contrast, is incapable of conducting a current; sugar is therefore a . These facts have been known since 1800 when it was discovered that an electric current can decompose the water in an electrolytic solution into its elements (a process known as ). By mid-century, Michael Faraday had made the first systematic study of electrolytic solutions. Faraday recognized that for a sample of matter to conduct electricity, two requirements must be met: In metallic solids, the charge carriers are electrons rather than ions; their mobility is a consequence of the quantum-mechanical uncertainty principle which promotes the escape of the electrons from the confines of their local atomic environment. In the case of electrolytic solutions, Faraday called the charge carrier (after the Greek word for "wanderer"). His most important finding was that each kind of ion (which he regarded as an electrically-charged atom) carries a definite amount of charge, most commonly in the range of ±1-3 units. The fact that the smallest charges observed had magnitudes of ±1 unit suggested an "atomic" nature for electricity itself, and led in 1891 to the concept of the "electron" as the unit of electric charge — although the identification of this unit charge with the particle we now know as the electron was not made until 1897. An ionic solid such as NaCl is composed of charged particles, but these are held so tightly in the crystal lattice that they are unable to move about, so the second requirement mentioned above is not met and solid salt is not a conductor. If the salt is melted or dissolved in water, the ions can move freely and the molten liquid or the solution becomes a conductor. Since positively-charged ions are attracted to a negative electrode that is traditionally known as the , these are often referred to as . Similarly, negatively-charged ions, being attracted to the positive electrode, or , are called . (These terms were all coined by Faraday.) Although we tend to think of the solvent (usually water) as a purely passive medium within which ions drift around, it is important to understand that electrolytic solutions would not exist without the in reducing the strong attractive forces that hold solid salts and molecules such as HCl together. Once the ions are released, they are stabilized by interactions with the solvent molecules. Water is not the only liquid capable of forming electrolytic solutions, but it is by far the most important. It is therefore essential to understand those properties of water that influence the stability of ions in aqueous solution. According to Coulomb's law, the force between two charged particles is directly proportional to the product of the two charges, and inversely proportional to the square of the distance between them: The proportionality constant is the dimensionless . Its value in empty space is unity, but in other media it will be larger. Since appears in the denominator, this means that the force between two charged particles within a gas or liquid will be less than if the particles were in a vacuum. Water has one of the highest dielectric constants of any known liquid; the exact value varies with the temperature, but 80 is a good round number to remember. When two oppositely-charged ions are immersed in water, the force acting between them is only 1/80 as great as it would be between the two gaseous ions at the same distance. It can be shown that in order to separate one mole of Na and Cl ions at their normal distance of 23.6 pm in solid sodium chloride, the work required will be 586 J in a vacuum, but only 7.3 J in water The dielectric constant is a of matter, rather than being a property of the molecule itself, as is the dipole moment. It is a cooperative effect of all the molecules in the liquid, and is a measure of the extent to which an applied electric field will cause the molecules to line up with the negative ends of their dipoles pointing toward the positive direction of the electric field. The high dielectric constant of water is a consequence of the small size of the H O molecule in relation to its large dipole moment. When one molecule is reoriented by the action of an external electric field, local hydrogen bonding tends to pull neighboring molecules into the same alignment, thus producing an additive effect that would be absent if the molecules were all acting independently. When an ion is introduced into a solvent, the attractive interactions between the solvent molecules must be disrupted to create space for the ion. This costs energy and would by itself tend to inhibit dissolution. However, if the solvent has a high permanent dipole moment, the energy cost is more than recouped by the between the ion and the surrounding solvent molecules. Water, as you know, has a sizeable dipole moment that is the source of the hydrogen bonding that holds the liquid together. The greater strength of ion-dipole attractions compared to hydrogen-bonding (dipole-dipole) attractions stabilizes the dissolved ion. Water is not the only electrolytic solvent, but it is by far the best. For some purposes, chemists occasionally need to employ non-aqueous solvents when studying electrolytes. Here are a few examples: | 5,674 | 2,755 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.01%3A_Prelude_to_Descriptive_Chemistry |
So far we have been devoted to explaining fundamental concepts and principles such as the atomic theory, electronic structure and chemical bonding, intermolecular forces and their effects on solids, liquids, or gases, and classes of reactions such as redox or acid-base. It is well to remember, though, that all these concepts and principles have been developed and used by chemists in order to better understand, recall, and systematize macroscopic laboratory observations. In other words, although the concepts we have described so far have their own inherent beauty as great ideas, they are primarily important because they reduce the quantity of memorization which is necessary to master descriptive chemistry, allowing people to recall facts that otherwise might be forgotten. Such a reduction in memory work is only possible if you know how to apply principles to specific elements, their compounds, and the reactions they undergo. This is not as easy as it may seem, but neither is it impossibly difficult. We will describe some of the chemistry of the representative elements, showing as we do so how their properties may be rationalized on the basis of concepts and principles. Explanation of these properties is organized according to the concept of periodicity, with each subsequent section corresponding to one of the eight groups of representative elements. As you read, you should try to see why a certain reaction occurs, as well as what actually happens, and instead of memorizing each specific equation, you should try to organize the chemistry of these elements according to the generalizations you have already learned. At this point it is useful to look and consolidate some of the general trends observed for the representative elements. First, metals on the far left of the periodic table are good reducing agents, while nonmetals on the far right (excluding noble gases) are strong oxidizing agents. Thus these elements are quite reactive, especially when one from the left combines with one from the right. Hydrogen compounds (hydrides) of the alkali and alkaline-earth metals contain strongly basic H ions and produce basic solutions. Toward the middle of the periodic table acid-base properties of hydrogen compounds are harder to predict. Some, like CH in Group IVA, are neither acids nor bases, but others, like NH , have lone pairs of electrons and can accept protons. Protons can be easily donated and are acidic only when they are bound to halogens or oxygen. The acidic behavior of oxides also increases from left to right across the periodic table and decreases from top to bottom. The situation is complicated by the fact that the higher the oxidation state of an atom, the more covalent its oxide and the more acidic it will be. Thus SO dissolves in water to give a strong acid, while SO gives a weak one. Taking account of both of these trends, one can fairly well predict which oxides are likely to be basic, which amphoteric, and which acidic. General rules can also be used to predict which oxidation states will be most common. On the left of the periodic table the group number gives the most common oxidation state. From group IIIA on, the group number minus 2 (for the electrons) is also common, especially for elements near the bottom of the table. .The group number is a good choice when an element combines with a highly electronegative element, but the group number minus 2 is more common when one element is bonded to another element of intermediate electron-withdrawing power. For example from the Chalcogens, SF and SF are both stable, but SF is the most stable sulfur chloride. From group VA on to the right of the table, the group number minus 8 is an important oxidation state, especially for the first member of a group. Oxidation numbers other than those already mentioned usually differ by increments of 2. For example, chlorine exhibits –1, +1, +3, +5 and +7 oxidation states in stable compounds, and sulfur is found in –2, +2, +4, and +6 states. In conclusion, do not overlook the forest by concentrating too much on individual trees. Look for and try to understand and use the generalizations and correlations that have been developed in this chapter. If you do this, you will retain the facts presented here much more efficiently. | 4,311 | 2,756 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/05%3A_Surface_Analytical_Techniques/5.04%3A_Vibrational_Spectroscopy |
Vibrational spectroscopy provides the most definitive means of identifying the surface species generated upon molecular adsorption and the species generated by surface reactions. In principle, any technique that can be used to obtain vibrational data from solid state or gas phase samples (IR, Raman etc.) can be applied to the study of surfaces - in addition there are a number of techniques which have been specifically developed to study the vibrations of molecules at interfaces (EELS, SFG etc.). There are, however, only two techniques that are routinely used for vibrational studies of molecules on surfaces - these are: There are a number of ways in which the IR technique may be implemented for the study of adsorbates on surfaces. For solid samples possessing a high surface area:
It can be shown theoretically that the best sensitivity for IR measurements on metallic surfaces is obtained using a grazing-incidence reflection of the IR light. Furthermore, since it is an optical (photon in/photon out) technique it is not necessary for such studies to be carried out in vacuum. The technique is not inherently surface-specific, but One major problem, is that of sensitivity (i.e. the signal is usually very weak owing to the small number of adsorbing molecules). Typically, the sampled area is ca. 1 cm with less than 10 adsorbed molecules (i.e. about 1 nanomole). With modern FTIR spectrometers, however, such small signals (0.01% - 2% absorption) can still be recorded at relatively high resolution (ca. 1 cm ). For a number of practical reasons, low frequency modes ( < 600 cm ) are not generally observable - this means that it is not usually possible to see the vibration of the metal-adsorbate bond and attention is instead concentrated on the intrinsic vibrations of the adsorbate species in the range 600 - 3600 cm . The observation of vibrational modes of adsorbates on metallic substrates is subject to the . This states that only those vibrational modes which give rise to an oscillating dipole perpendicular (normal) to the surface are IR active and give rise to an observable absorption band. Further information on the selection rules for surface IR spectroscopy can be found in the review by Sheppard & Erkelens [Appl. Spec. 38, 471 (1984)]. It also needs to be remembered that even if a transition is allowed it may still be very weak if the transition moment is small. The sequence of spectra shown below demonstrate how IR spectroscopy can clearly reveal changes in the adsorption geometry of chemisorbed molecules. In this particular instance, all the bands are due to the stretching mode of the N-O bond in NO molecules adsorbed on a Pt surface, but the vibrational frequency is sensitive to changes in the coordination and molecular orientation of the adsorbed NO molecules.
Note - the surface coverage of adsorbed NO molecules decreases as the temperature is raised and little NO remains adsorbed at temperatures of 450 K and above. The RAIRS spectra shown below were observed during HCN adsorption on Pt at sub-ambient temperatures; the surface species which are generated give rise to much weaker absorptions than NO, and signal:noise considerations become much more important. These spectra also illustrate the effect of the surface normal selection rule for metallic surfaces. This is a technique utilising the inelastic scattering of low energy electrons in order to measure vibrational spectra of surface species: superficially, it can be considered as the electron-analogue of Raman spectroscopy. To avoid confusion with other electron energy loss techniques it is sometimes referred to as HREELS - high resolution EELS VELS - vibrational ELS. Since the technique employs low energy electrons, it is necessarily restricted to use in high vacuum (HV) and UHV environments - however, the use of such low energy electrons ensures that it is a surface specific technique and, arguably, it is the vibrational technique of choice for the study of most adsorbates on single crystal substrates. The basic experimental geometry is fairly simple as illustrated schematically below - it involves using an electron monochromator to give a well-defined beam of electrons of a fixed incident energy, and then analysing the scattered electrons using an appropriate electron energy analyser. A substantial number of electrons are elastically scattered (\( E = E_o\)) - this gives rise to a strong in the spectrum. On the low kinetic energy side of this main peak (\(E < E_o\)), additional weak peaks are superimposed on a mildly sloping background. These peaks correspond to electrons which have undergone discrete energy losses during the scattering from the surface. The magnitude of the energy loss, \(ΔE = (E_o - E)\), is equal to the vibrational quantum (i.e. the energy) of the vibrational mode of the adsorbate excited in the inelastic scattering process. In practice, the incident energy ( ) is usually in the range 5-10 eV (although occasionally up to 200 eV) and the data is normally plotted against the energy loss (frequently measured in meV). The selection rules that determine whether a vibrational band may be observed depend upon the nature of the substrate and also the experimental geometry: specifically the angles of the incident and (analysed) scattered beams with respect to the surface. and a geometry, scattering is principally by a long-range dipole mechanism. In this case the loss features are relatively intense, but only those vibrations giving rise to a dipole change normal to the surface can be observed. By contrast, in an geometry, electrons lose energy to surface species by a short-range impact scattering mechanism. In this case the loss features are relatively weak but all vibrations are allowed and may be observed. If spectra can be recorded in specular and off-specular modes the selection rules for metallic substrates can be put to good use - helping the investigator to obtain more definitive identification of the nature and geometry of the adsorbate species. The resolution of the technique (despite the HREELS acronym !) is generally rather poor; 40-80 cm is not untypical. A measure of the instrumental resolution is given by looking at the FWHM (full-width at half maximum) of the elastic peak. This poor resolution can cause problems in distinguishing between closely similar surface species - however, recent improvements in instrumentation have opened up the possibility of much better spectral resolution ( < 10 cm ) and will undoubtedly enhance the utility of the technique. In summary, there are both advantages and disadvantages in utilising EELS, as opposed to IR techniques, for the study of surface species It offers the advantages of ... but suffers from the limitations of ... One of the classic examples of an area in which vibrational spectroscopy has contributed significantly to the understanding of the surface chemistry of an adsorbate is that of: Adsorbed carbon monoxide usually gives rise to strong absorptions in both the IR and EELS spectra at the (C-O) stretching frequency. The metal-carbon stretching mode (ca. 400 cm ) is usually also accessible to EELS. The interpretation of spectra of CO as an adsorbed surface species draws heavily upon IR spectra from related inorganic cluster and coordination complexes - the structure of such complexes usually being available from x-ray single crystal diffraction measurements. This comparison suggests that the CO stretching frequency can provide a good indication of the surface coordination of the molecule: as a rough guideline, (C-O) CO ( gas phase ) 2143 cm Terminal CO 2100 - 1920 cm Bridging ( 2f site ) 1920 - 1800 cm Bridging ( 3f / 4f site ) < 1800 cm The RAIRS spectrum shown below was obtained for a saturation coverage of CO on a Pt surface at 300 K. The data indicate that the majority of the CO molecules are bound in a terminal fashion to a single Pt surface atom (ν ∼ 2090 cm ), whilst a smaller number of molecules are co-ordinated in a bridging site between two Pt atoms (ν ∼ 1870 cm ). The reduction in the stretching frequency of terminally-bound CO from the value observed for the gas phase molecule ( 2143 cm ) can be explained in terms of the Dewar-Chatt or for the bonding of CO to metals. This simple model considers the metal-CO bonding to consist of two main components: : this is a -bonding interaction due to overlap of a filled -"lone pair" orbital on the carbon atom with empty metal orbitals of the correct symmetry - this leads to electron density transfer from the CO molecule to the metal centre. : this is a bonding interaction due to overlap of filled metal d (and p ?) orbitals with the * antibonding molecular orbital of the CO molecule. Since this interaction leads to the introduction of electron density into the CO antibonding orbital there is a consequent reduction in the CO bond strength and its intrinsic vibrational frequency (relative to the isolated molecule). For a more detailed discussions on the bonding of CO to metals, you are recommended to refer to one of the following: | 9,124 | 2,758 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.01%3A_Factors_that_Affect_Reaction_Rates |
There are many factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, the solvent, and the presence of a catalyst. Although a balanced chemical equation for a reaction describes the quantitative relationships between the amounts of reactants present and the amounts of products that can be formed, it gives us no information about whether or how fast a given reaction will occur. This information is obtained by studying the chemical kinetics of a reaction, which depend on various factors: reactant concentrations, temperature, physical states and surface areas of reactants, and solvent and catalyst properties if either are present. By studying the kinetics of a reaction, chemists gain insights into how to control reaction conditions to achieve a desired outcome. Two substances cannot possibly react with each other unless their constituent particles (molecules, atoms, or ions) come into contact. If there is no contact, the reaction rate will be zero. Conversely, the more reactant particles that collide per unit time, the more often a reaction between them can occur. Consequently, the reaction rate usually increases as the concentration of the reactants increases. Increasing the temperature of a system increases the average kinetic energy of its constituent particles. As the average kinetic energy increases, the particles move faster and collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature. Conversely, the reaction rate of virtually all reactions decreases with decreasing temperature. For example, refrigeration retards the rate of growth of bacteria in foods by decreasing the reaction rates of biochemical reactions that enable bacteria to reproduce. In systems where more than one reaction is possible, the same reactants can produce different products under different reaction conditions. For example, in the presence of dilute sulfuric acid and at temperatures around 100°C, ethanol is converted to diethyl ether: \[\mathrm{2CH_3CH_2OH}\xrightarrow{\mathrm{H_2SO_4}}\mathrm{CH_3CH_2OCH_2CH_3}+\mathrm{H_2O} \label{14.1.1} \] At 180°C, however, a completely different reaction occurs, which produces ethylene as the major product: \[\mathrm{CH_3CH_2OH}\xrightarrow{\mathrm{H_2SO_4}}\mathrm{C_2H_4}+\mathrm{H_2O} \label{14.1.2} \] When two reactants are in the same fluid phase, their particles collide more frequently than when one or both reactants are solids (or when they are in different fluids that do not mix). If the reactants are uniformly dispersed in a single homogeneous solution, then the number of collisions per unit time depends on concentration and temperature, as we have just seen. If the reaction is heterogeneous, however, the reactants are in two different phases, and collisions between the reactants can occur only at interfaces between phases. The number of collisions between reactants per unit time is substantially reduced relative to the homogeneous case, and, hence, so is the reaction rate. The reaction rate of a heterogeneous reaction depends on the surface area of the more condensed phase. Automobile engines use surface area effects to increase reaction rates. Gasoline is injected into each cylinder, where it combusts on ignition by a spark from the spark plug. The gasoline is injected in the form of microscopic droplets because in that form it has a much larger surface area and can burn much more rapidly than if it were fed into the cylinder as a stream. Similarly, a pile of finely divided flour burns slowly (or not at all), but spraying finely divided flour into a flame produces a vigorous reaction The nature of the solvent can also affect the reaction rates of solute particles. For example, a sodium acetate solution reacts with methyl iodide in an exchange reaction to give methyl acetate and sodium iodide. \[CH_3CO_2Na_{(soln)} + CH_3I_{(l)} \rightarrow CH_3CO_2CH_{3\; (soln)} + NaI_{(soln)} \label{14.1.3} \] Solvent viscosity is also important in determining reaction rates. In highly viscous solvents, dissolved particles diffuse much more slowly than in less viscous solvents and can collide less frequently per unit time. Thus the reaction rates of most reactions decrease rapidly with increasing solvent viscosity. A catalyst is a substance that participates in a chemical reaction and increases the reaction rate without undergoing a net chemical change itself. Consider, for example, the decomposition of hydrogen peroxide in the presence and absence of different catalysts. Because most catalysts are highly selective, they often determine the product of a reaction by accelerating only one of several possible reactions that could occur. Most of the bulk chemicals produced in industry are formed with catalyzed reactions. Recent estimates indicate that about 30% of the gross national product of the United States and other industrialized nations relies either directly or indirectly on the use of catalysts. | 5,199 | 2,759 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.05%3A_Thermodynamic_Probability_W_and_Entropy |
The has shown that if we want to predict whether a chemical change is spontaneous or not, we must find some general way of determining whether the final state is more probable than the initial. This can be done using a number , called the . is defined as the number of alternative microscopic arrangements which correspond to the same macroscopic state. The significance of this definition becomes more apparent once we have considered a few examples. Figure \(\Page {1}\) illustrates a crystal consisting of only eight atoms at the absolute zero of temperature. Suppose that the temperature is raised slightly by supplying just enough energy to set one of the atoms in the crystal vibrating. There are eight possible ways of doing this, since we could supply the energy to any one of the eight atoms. All eight possibilities are shown in Figure \(\Page {1}\) . Since all eight possibilities correspond to the crystal having the same temperature, we say that = 8 for the crystal at this temperature. Also, we must realize that the crystal will not stay perpetually in any of these eight arrangements. Energy will constantly be transferred from one atom to the other, so that all the eight arrangements are . Let us now supply a second quantity of energy exactly equal to the first, so that there is just enough to start two molecules vibrating. There are 36 different ways in which this energy can be assigned to the eight atoms (Figure \(\Page {1}\) ). We say that = 36 for the crystal at this second temperature. Because energy continually exchanges from one atom to another, there is an equal probability of finding the crystal in any of the 36 possible arrangements. A third example of is our eight-atom crystal at the absolute zero of temperature. Since there is no energy to be exchanged from atom to atom, only one arrangement is possible, and = 1. This is true not only for this hypothetical crystal, but also presumably for a real crystal containing a large number of atoms, perfectly arranged, at absolute zero. The thermodynamic probability enables us to decide how much more probable certain situations are than others. Consider the flow of heat from crystal to crystal , as shown in Figure \(\Page {2}\). We shall assume that each crystal contains 100 atoms. Initially crystal is at absolute zero. Crystal is at a higher temperature and contains 64 units of energy-enough to set 64 of the atoms vibrating. If the two crystals are brought together, the molecules of lose energy while those of gain energy until the 64 units of energy are evenly distributed between both crystals. In the initial state the 64 units of energy are distributed among 100 atoms. Calculations show that there are 1.0 × 10 alternative ways of making this distribution. Thus W , initial thermodynamic probability, is 1.0× 10 . The 100 atoms of crystal continually exchange energy among themselves and transfer from one of these 1.0 × 10 arrangements to another in rapid succession. At any instant there is an equal probability of finding the crystal in any of the 1.0 × 10 arrangements. When the two crystals are brought into contact, the energy can distribute itself over twice as many atoms. The number of possible arrangements rises enormously, and , the thermodynamic probability for this new situation, is 3.6 × 10 . In the constant reshuffle of energy among the 200 atoms, each of these 3.6 × 10 arrangements will occur with equal probability. However, only 1.0 × 10 of them correspond to all the energy being in crystal . Therefore the probability of the heat flow reversing itself and all the energy returning to crystal A is This example shows how we can use as a general criterion for deciding whether a reaction is spontaneous or not. Movement from a less probable to a more probable molecular situation corresponds to movement from a state in which is smaller to a state where is larger. In other words . If we can find some way of calculating or measuring the initial and final values of , the problem of deciding in advance whether a reaction will be spontaneous or not is solved. If is greater than , then the reaction will occur of its own accord. Although there is nothing wrong in principle with this approach to spontaneous processes, in practice it turns out to be very cumbersome. For real samples of matter (as opposed to 200 atoms in the example of Figure 2) the values of are on the order of 10 —so large that they are difficult to manipulate. The logarithm of , however, is only on the order of 10 , since log 10 = . This is more manageable, and chemists and physicists use a quantity called the which is proportional to the logarithm of . This way of handling the extremely large thermodynamic probabilities encountered in real systems was first suggested in 1877 by the Austrian physicist Ludwig Boltzmann (1844 to 1906). The equation \[\text{ln } x = 2.303 \text{ log } x \nonumber \] The thermodynamic probability for 1 mol propane gas at 500 K and 101.3 kPa has the value 10 . Calculate the entropy of the gas under these conditions. Since \(W = 10 ^ {10^{25}}\) \( \text{log } W = 10^{25} \) Thus \( S = 2.303k \text{ log } W = 1.3805 \times 10^{-23} \text {J K}^{-1} \times 2.303 \times 10^{25} = 318 \text{J K}^{-1} \) Note also that the dimensions of entropy are energy/temperature. The statement that the entropy increases when a spontaneous change occurs is called the . (The first law is the law of conservation of energy.) The second law, as it is usually called, is one of the most fundamental and most widely used of scientific laws. In this book we shall only be able to explore some of its chemical implications, but it is of importance also in the fields of physics, engineering, astronomy, and biology. Almost all environmental problems involve the second law. Whenever pollution increases, for instance, we can be sure that the entropy is increasing along with it. The second law is often stated in terms of an entropy difference Δ . If the entropy increases from an initial value of to a final value of as the result of a spontaneous change, then \[\Delta S = S_{2} - S_{1} \label{4} \] \[ \begin{align} \Delta S=S_{2}-S_{1} & =2.303\times k\times \text{ log }W_{2}-2.303\times k\times \text{ log }W_{1} \\ & = 2.303 \times k \times \text{ log } \frac{W_{\text{2}}}{W_{\text{1}}} \\ & = 2.303 \times 1.3805 \times 10^{-23} \text{ J K}^{-1} \times 1.813 \times 10^{23} \end{align} \nonumber \] | 6,529 | 2,760 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/19%3A_Spontaneous_Change%3A_Entropy_and_Gibbs_Energy/19.8%3A_Coupled_Reactions |
Endergonic reactions can also be pushed by coupling them to another reaction, which is strongly exergonic, often through shared intermediates. Many chemicals' reactions are (i.e., not spontaneous (\(\Delta G > 0\))) and require energy to be externally applied to occur. However, these reaction can be to a separate, (thermodynamically favorable \(\Delta G <0\)) reactions that 'drive' the thermodynamically unfavorable one by coupling or 'mechanistically joining' the two reactions often via a share intermediate. Since Gibbs Energy is a , the \(\Delta G\) values for each half-reaction may be summed, to yield the combined \(\Delta G\) of the coupled reaction. One simple example of the coupling of reaction is the decomposition of calcium carbonate: \[CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)} \;\;\;\;\;\;\; \Delta G^o = 130.40 \;kJ/mol \label{1}\] The strongly positive \(\Delta G\) for this reaction is reactant-favored. If the temperature is raised above 837 ºC, this reaction becomes spontaneous and favors the products. Now, let's consider a second and completely different reaction that can be coupled ot this reaction. The combustion of coal released by burning the coal \(\Delta G^o = -394.36 \;kJ/mol\) is greater than the energy required to decompose calcium carbonate (\(\Delta G^o = 130.40 \;kJ/mo\)). \[C_{(s)} + O_2 \rightleftharpoons CO_{2(g)} \;\;\;\;\;\;\; \Delta G^o = -394.36 \;kJ/mol \label{2}\] If reactions \(\ref{1}\) and \(\ref{2}\) were added \[CaCO_{3(s)} + C_{(s)} + O_2 \rightleftharpoons CaO_{(s)} + 2CO_[{2(g)} \;\;\;\; \Delta G^o = -263.96 \;kJ/mol \label{3}\] and then were applied, the combined reaction (Equation \(\ref{3}\)) is product-favored with \(\Delta G^o = -263.96 \;kJ/mol\). This is because the reactant-favored reaction (Equation \(\ref{2}\)) is linked to a strong spontaneous reaction so that both reactions yield products. Notice that the \(\Delta G\) for the coupled reaction is the sum of the constituent reactions; this is a consequence of Gibbs energy being a state function: \[\Delta G^o = 130.40 \;kJ/mol+ -394.36 \;kJ/mol = -263.96 \;kJ/mol\] This is a common feature in biological systems where some enzyme-catalyzed reactions are interpretable as two coupled half-reactions, one spontaneous and the other non-spontaneous. Organisms often the hydrolysis of (adenosine triphosphate) to generate (adenosine diphosphate) as the spontaneous coupling reaction (Figure \(\Page {1}\)). \[ATP + H_2O \rightleftharpoons ADP + P_i \label{4}\] The phosphoanhydride bonds formed by ejecting water between two phosphate group of ATP exhibit a large negative \(-\Delta G\) of hydrolysis and are thus often termed "high energy" bonds. However, as with all bonds, energy is requires to break these bonds, but the thermodynamic Gibbs energy difference is strongly "energy releasing" when including the solvation thermodynamics of the phosphate ions; \(\Delta G \) for this reaction is - 31 kJ/mol. ATP is the major 'energy' molecule produced by metabolism, and it serves as a sort of 'energy source' in cell: ATP is dispatched to wherever a non-spontaneous reaction needs to occurs so that the two reactions are coupled so that the overall reaction is thermodynamically favored. \(RCHO\) and nicotinamide adenine dinucleotide (NAD) is a coenzyme found in all living cells and in the reduced form, \(NAD^+\), it acts as an oxidizing agent that can accept electrons from other molecules. The NAD -linked oxidation of an aldehyde is practically irreversible with an equilibrium that strongly favors the products (\(\Delta G >> 0\): \[RCHO + NAD^+ + H_2O \rightleftharpoons RCOOH + NADH + H^+ \label{Spontaneous}\] The position of equilibrium for phosphorylating carboxylic acids lies very much to the left: \[RCOOH + P_i \rightleftharpoons RC(=O)(O-P_i) + H_2O \label{Nonspontaneous}\] The non-spontaneous formation of a phosphorylated carboxylic acid can be driven by coupling it to the (spontaneous) NAD -linked oxidation of an aldehyde? Similarly, ATP hydrolysis can be used to combine amino acids together to generate polypeptides (and proteins) as graphically illustrated by Figure \(\Page {2}\). In this case, the reverse of Equation \(\ref{4}\) is initially coupled to the oxidizing glucose by oxygen \[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \label{5}\] Reaction \(\ref{5}\) is strongly spontaneous with \(\Delta G = −2880 \;kJ/mol \) o Two (or more) reactions may be combined such that a spontaneous reaction may be made 'drive' an nonspontaneous one. Such reactions may be considered . Changes in Gibbs energy of the coupled reactions are additive. | 4,643 | 2,761 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4E%3A_Hotplates |
Hotplates are perhaps the most versatile heat source in the laboratory (Figure 1.48) and can be used to heat beakers, Erlenmeyer flasks, and various hot baths (water, sand, and oil baths). They can also be used to develop stained TLC plates. Hotplates work by passing electricity through a heating element covered by a ceramic top. the hotplate surface can reach temperatures up to \(350^\text{o} \text{C}\),\(^6\) which is hot enough to ignite many low-boiling solvents. Diethyl ether, pentane, hexane, low-boiling petroleum ether, and acetone should therefore be heated in an open vessel with a hotplate. Caution should be used when heating flammable organic liquid in an open vessel on a hotplate, as organic vapors may spill out of containers and ignite upon contact with the heating element, which may be hotter than the ceramic surface. This can be especially true if the hotplate is turned to "high". Therefore, a low setting must be used when cautiously heating certain flammable organic liquids (e.g. ethanol) with a hotplate. \(^6\)As reported in the Fischer Scientific catalog. | 1,106 | 2,762 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.07%3A_Orbital_Descriptions_of_Multiple_Bonds |
The association of four with an octet can be applied to multiple bonds as well as single bonds. A simple example is ethene (ethylene), C H . The for this molecule is Since the orbitals which overlap are not pointing directly at each other, each of these bonds is referred to as a or (more frivolously) as a banana bond. In ball-and-stick models of molecules, a double bond is usually represented by two springs or by curved sticks (shown in Figure \(\Page {2}\) ) joining the two atoms together. In making such a model, it is necessary to bend the springs a fair amount in order to fit them into the appropriate holes in the balls. The ability to bend or stretch is characteristic of all chemical bonds—not just those between doubly bonded atoms. Thus each atom can vibrate about its most stable position. Perhaps ball-and-spring models would be more appropriate than ball-and-stick models in all cases. The bent-bond picture makes it easy to explain several characteristics of double bonds. As noted in , the distance between two atomic nuclei connected by a double bond is shorter than if they were connected by a single bond. In the case of carbon-carbon bonds, for example, the distance is 133 pm, while the C—C distance is 156 pm. This makes sense when we realize that each bent bond extends along a curved path. The distance between the ends of such a path (the C nuclei) is necessarily shorter than the path itself. Another characteristic of double bonds is that they make it difficult to twist one end of a molecule relative to the other. This phenomenon usually is called a . Such a barrier accounts for the fact that it is possible to prepare different compounds with the formula C H F . Their structures are shown in Figure 2. Structure ( ) is unique because both F atoms are attached to the same C atom, but ( ) and ( ) differ only by a 180° flip of the right-hand group. If there were no barrier to rotation around the double bond, structures ( ) and ( ) could interconvert very rapidly whenever they collided with other molecules. It would then be impossible to prepare a sample containing only type ( ) molecules or only type ( ) molecules. Since they have the same molecular formula, ( ), ( ) and ( ) are isomers. Structure ( ) in which the two F atoms are on sides of the double bond is called the isomer, while structure ( ) in which two like atoms are on the side is called the isomer. It is easy to explain why there is a barrier to rotation preventing the interconversion of these cis and trans isomers in terms of our bent-bond model. Rotation of one part of the molecule about the line through the C atoms will cause one of the bent-bond electron clouds to twist around the other. Unless one-half of the double bond breaks, it is impossible to twist the molecule through a very large angle. The jmol above allows you to view all of the molecular orbitals for ethene. By using the scroll bar, you can choose any of the associated with the molecule and view them. Molecular Orbitals are discussed further in the section on . By choosing orbital , you can view a sigma bond orbital for ethene. Choosing the orbital will display a pi orbital for the molecule. | 3,220 | 2,763 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.1%3A_Types_of_Chemical_Compounds_and_their_Formulas |
The atoms in all substances that contain multiple atoms are held together by electrostatic interactions—interactions between electrically charged particles such as protons and electrons. Electrostatic attraction between oppositely charged species (positive and negative) results in a force that causes them to move toward each other, like the attraction between opposite poles of two magnets. In contrast, electrostatic repulsion between two species with the same charge (either both positive or both negative) results in a force that causes them to repel each other, as do the same poles of two magnets. Atoms form chemical compounds when the attractive electrostatic interactions between them are stronger than the repulsive interactions. Collectively, the attractive interactions between atoms are called chemical bonds. Chemical bonds are generally divided into two fundamentally different types: ionic and covalent. In reality, however, the bonds in most substances are neither purely ionic nor purely covalent, but lie on a spectrum between these extremes. Although purely ionic and purely covalent bonds represent extreme cases that are seldom encountered in any but very simple substances, a brief discussion of these two extremes helps explain why substances with different kinds of chemical bonds have very different properties. Ionic compounds consist of positively and negatively charged ions held together by strong electrostatic forces, whereas covalent compounds generally consist of molecules, which are groups of atoms in which one or more pairs of electrons are shared between bonded atoms. In a covalent bond, atoms are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons they share. This discussion of structures and formulas begins by describing covalent compounds. The energetic factors involved in bond formation are described in more quantitative detail in later. Ionic compounds consist of ions of opposite charges held together by strong electrostatic forces, whereas pairs of electrons are shared between bonded atoms in covalent compounds. Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the diatomic (“two atoms”) molecules H , N , O , F , Cl , Br , and I (part (a) in Figure \(\Page {1}\)). Similarly, a few pure elements exist as polyatomic (“many atoms”) molecules, such as elemental phosphorus and sulfur, which occur as P and S (part (b) in Figure \(\Page {1}\)). Each covalent compound is represented by a molecular formula, which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as \(H_2O\). Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as \(CO_2\). Covalent compounds that predominantly organic compounds CH O inorganic compounds SF ClF NH silane SiH H O OH HCl H SO For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up. Write the molecular formula of each compound. : identity of elements present and number of atoms of each : molecular formula : Identify the symbol for each element in the molecule. Then identify the substance as either an organic compound or an inorganic compound. If the substance is an organic compound, arrange the elements in order beginning with carbon and hydrogen and then list the other elements alphabetically. If it is an inorganic compound, list the elements beginning with the one farthest left in the periodic table. List elements in the same group starting with the lower element and working up. From the information given, add a subscript for each kind of atom to write the molecular formula. : a. b. c. Write the molecular formula for each compound. : Molecular formulas give only the elemental composition of molecules. In contrast, structural formulas show which atoms are bonded to one another and, in some cases, the approximate arrangement of the atoms in space. Knowing the structural formula of a compound enables chemists to create a three-dimensional model, which provides information about how that compound will behave physically and chemically. The structural formula for H can be drawn as H–H and that for I as I–I, where the line indicates a single pair of shared electrons, a single bond. Two pairs of electrons are shared in a double bond, which is indicated by two lines—for example, O is O=O. Three electron pairs are shared in a triple bond, which is indicated by three lines—for example, N is N≡N ( ). Carbon is unique in the extent to which it forms single, double, and triple bonds to itself and other elements. The number of bonds formed by an atom in its covalent compounds is not arbitrary. Hydrogen, oxygen, nitrogen, and carbon have very strong tendencies to form substances in which they have one, two, three, and four bonds to other atoms, respectively ( ). The structural formula for water can be drawn as follows: Because the latter approximates the experimentally determined shape of the water molecule, it is more informative. Similarly, ammonia (NH ) and methane (CH ) are often written as planar molecules: As shown in however, the actual three-dimensional structure of NH looks like a pyramid with a triangular base of three hydrogen atoms. The structure of CH , with four hydrogen atoms arranged around a central carbon atom as shown in is tetrahedral: the hydrogen atoms are positioned at every other vertex of a cube. Many compounds—carbon compounds, in particular—have four bonded atoms arranged around a central atom to form a tetrahedron. res \(\Page {3}\)-\(\Page {3}\) illustrate different ways to represent the structures of molecules. It should be clear that there is no single “best” way to draw the structure of a molecule; the method used depends on which aspect of the structure should be emphasized and how much time and effort is required. shows some of the different ways to portray the structure of a slightly more complex molecule: methanol. These representations differ greatly in their information content. For example, the molecular formula for methanol (part (a) in ) gives only the number of each kind of atom; writing methanol as CH O tells nothing about its structure. In contrast, the structural formula (part (b) in ) indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in ) and the perspective drawing (part (d) in ) show the three-dimensional structure of the molecule. The latter (also called a wedge-and-dash representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. In addition, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.” Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a condensed structural formula (part (f) in ), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH OH, which indicates that the molecule contains a CH unit that looks like a fragment of methane (CH ). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point. Write the molecular formula for each compound. The condensed structural formula is given. : condensed structural formula : molecular formula : : The molecular formula lists the elements in the molecule and the number of atoms of each. Write the molecular formula for each molecule. CHCl N H C H N The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively-charged protons in the nucleus is equal to the number of negatively-charged electrons. In contrast, ions are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cations. Conversely, ions that contain more electrons than protons have a net negative charge and are called anions. Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge. Ionic compounds contain both cations and anions in a ratio that results in electrical charge. In covalent compounds, electrons are shared between bonded atoms and are simultaneously attracted to more than one nucleus. In contrast, ionic compounds contain cations and anions rather than discrete neutral molecules. Ionic compounds are held together by the attractive electrostatic interactions between cations and anions. In an ionic compound, the cations and anions are arranged in space to form an extended three-dimensional array that maximizes the number of attractive electrostatic interactions and minimizes the number of repulsive electrostatic interactions ( ). As shown in , the electrostatic energy of the interaction between two charged particles is proportional to the product of the charges on the particles and inversely proportional to the distance between them: \[ \text {electrostatic energy} \propto {Q_1Q_2 \over r} \label{3.1.1}\] where When \(Q_1\) and \(Q_2\) are both positive, corresponding to the charges on cations, the cations repel each other and the electrostatic energy is positive. When \(Q_1\) and \(Q_2\) are both negative, corresponding to the charges on anions, the anions repel each other and the electrostatic energy is again positive. The electrostatic energy is negative only when the charges have opposite signs; that is, positively charged species are attracted to negatively charged species and vice versa. As shown in , the strength of the interaction is proportional to the magnitude of the charges and decreases as the distance between the particles increases. These energetic factors are discussed in greater quantitative detail later. If the electrostatic energy is positive, the particles repel each other; if the electrostatic energy is negative, the particles are attracted to each other. One example of an ionic compound is sodium chloride (NaCl), formed from sodium and chlorine. In forming chemical compounds, many elements have a tendency to gain or lose enough electrons to attain the same number of electrons as the noble gas closest to them in the periodic table. When sodium and chlorine come into contact, each sodium atom gives up an electron to become a Na ion, with 11 protons in its nucleus but only 10 electrons (like neon), and each chlorine atom gains an electron to become a Cl ion, with 17 protons in its nucleus and 18 electrons (like argon), as shown in part (b) in . Solid sodium chloride contains equal numbers of cations (Na ) and anions (Cl ), thus maintaining electrical neutrality. Each Na ion is surrounded by 6 Cl ions, and each Cl ion is surrounded by 6 Na ions. Because of the large number of attractive Na Cl interactions, the total attractive electrostatic energy in NaCl is great. Consistent with a tendency to have the same number of electrons as the nearest noble gas, when forming ions, elements in groups 1, 2, and 3 tend to lose one, two, and three electrons, respectively, to form cations, such as Na and Mg . They then have the same number of electrons as the nearest noble gas: neon. Similarly, K , Ca , and Sc have 18 electrons each, like the nearest noble gas: argon. In addition, the elements in group 13 lose three electrons to form cations, such as Al , again attaining the same number of electrons as the noble gas closest to them in the periodic table. Because the lanthanides and actinides formally belong to group 3, the most common ion formed by these elements is M , where M represents the metal. Conversely, elements in groups 17, 16, and 15 often react to gain one, two, and three electrons, respectively, to form ions such as Cl , S , and P . Ions such as these, which contain only a single atom, are called monatomic ions. The charges of most monatomic ions derived from the main group elements can be predicted by simply looking at the periodic table and counting how many columns an element lies from the extreme left or right. For example, barium (in Group 2) forms Ba to have the same number of electrons as its nearest noble gas, xenon; oxygen (in Group 16) forms O to have the same number of electrons as neon; and cesium (in Group 1) forms Cs , which has the same number of electrons as xenon. Note that this method is ineffective for most of the transition metals, as discussed in Some common monatomic ions are listed in . Elements in Groups 1, 2, and 3 tend to form 1+, 2+, and 3+ ions, respectively; elements in Groups 15, 16, and 17 tend to form 3−, 2−, and 1− ions, respectively. Li lithium Be beryllium N nitride (azide) O oxide F fluoride Na sodium Mg magnesium Al aluminum P phosphide S sulfide Cl chloride K potassium Ca calcium Sc scandium Ga gallium As arsenide Se selenide Br bromide Rb rubidium Sr strontium Y yttrium In indium Te telluride I iodide Cs cesium Ba barium La lanthanum Predict the charge on the most common monatomic ion formed by each element. : element : ionic charge : Identify the group in the periodic table to which the element belongs. Based on its location in the periodic table, decide whether the element is a metal, which tends to lose electrons; a nonmetal, which tends to gain electrons; or a semimetal, which can do either. After locating the noble gas that is closest to the element, determine the number of electrons the element must gain or lose to have the same number of electrons as the nearest noble gas. : Predict the charge on the most common monatomic ion formed by each element. : Molecular and Ionic Compounds: In general, ionic and covalent compounds have different physical properties. Ionic compounds form hard crystalline solids that melt at high temperatures and are resistant to evaporation. These properties stem from the characteristic internal structure of an ionic solid, illustrated schematically in part (a) in which shows the three-dimensional array of alternating positive and negative ions held together by strong electrostatic attractions. In contrast, as shown in part (b) in most covalent compounds consist of discrete molecules held together by comparatively weak intermolecular forces (the forces between molecules), even though the atoms within each molecule are held together by strong intramolecular covalent bonds (the forces within the molecule). Covalent substances can be gases, liquids, or solids at room temperature and pressure, depending on the strength of the intermolecular interactions. Covalent molecular solids tend to form soft crystals that melt at low temperatures and evaporate easily.Some covalent substances, however, are not molecular but consist of infinite three-dimensional arrays of covalently bonded atoms and include some of the hardest materials known, such as diamond. This topic will be addressed elsewhere. The covalent bonds that hold the atoms together in the molecules are unaffected when covalent substances melt or evaporate, so a liquid or vapor of independent molecules is formed. For example, at room temperature, methane, the major constituent of natural gas, is a gas that is composed of discrete CH molecules. A comparison of the different physical properties of ionic compounds and covalent molecular substances is given in When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formula, the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is C H . The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore C H . The formula unit is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane has the empirical formula C H , but it contains two C H formula units, giving a molecular formula of C H . Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as Na and Cl , then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be NaCl. If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of Mg and Cl , for example, two Cl ions are needed to balance the two positive charges on each Mg ion, giving an empirical formula of MgCl . Similarly, the formula for the ionic compound that contains Na and O ions is Na O. Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions. An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl , a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows: . When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg and O . Using the absolute values of the charges on the ions as subscripts gives the formula Mg O : This simplifies to its correct empirical formula MgO. The empirical formula has one Mg ion and one O ion. Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair. : ions or elements : empirical formula for binary ionic compound : If not given, determine the ionic charges based on the location of the elements in the periodic table. Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the lowest numbers to write the empirical formula. Check to make sure the empirical formula is electrically neutral. a. Using the absolute values of the charges on the ions as the subscripts gives Ga3As3: Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga and As have charges of equal magnitude but opposite signs. One Ga ion balances the charge on one As ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is gallium arsenide, which is widely used in the electronics industry in transistors and other devices. b. Because Eu has a charge of +3 and O has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion: The subscript for Eu is 2 (from O ), and the subscript for O is 3 (from Eu ), giving Eu O ; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu O is neutral. Europium oxide is responsible for the red color in television and computer screens. c. Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca . Chlorine lies in group 17, so it should gain one electron to form Cl . Two Cl ions are needed to balance the charge on one Ca ion, which leads to the empirical formula CaCl . We could also cross charges, using the absolute value of the charge on Ca as the subscript for Cl and the absolute value of the charge on Cl as the subscript for Ca: The subscripts in CaCl cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter. Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair. : Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH ) and the methylammonium (CH NH ) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in . The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K and NO ions combine in a 1:1 ratio to form KNO (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca and SO form CaSO (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH and NO form NH NO (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca and PO ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca (PO ) ; the parentheses around PO in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca P O gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO ions. Write the empirical formula for the compound formed from each ion pair. : ions : empirical formula for ionic compound : If it is not given, determine the charge on a monatomic ion from its location in the periodic table. Use to find the charge on a polyatomic ion. Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the smallest whole numbers when writing the empirical formula. : a. Because HPO has a charge of −2 and Na has a charge of +1, the empirical formula requires two Na ions to balance the charge of the polyatomic ion, giving Na HPO . The subscripts are reduced to the lowest numbers, so the empirical formula is Na HPO . This compound is sodium hydrogen phosphate, which is used to provide texture in processed cheese, puddings, and instant breakfasts. b. The potassium cation is K , and the cyanide anion is CN . Because the magnitude of the charge on each ion is the same, the empirical formula is KCN. Potassium cyanide is highly toxic, and at one time it was used as rat poison. This use has been discontinued, however, because too many people were being poisoned accidentally. c. The calcium cation is Ca , and the hypochlorite anion is ClO . Two ClO ions are needed to balance the charge on one Ca ion, giving Ca(ClO) . The subscripts cannot be reduced further, so the empirical formula is Ca(ClO) . This is calcium hypochlorite, the “chlorine” used to purify water in swimming pools. Write the empirical formula for the compound formed from each ion pair. : The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most covalent compounds consist of molecules, groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The molecular formula of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called organic compounds, whereas compounds that consist primarily of elements other than carbon and hydrogen are inorganic compounds. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two. A structural formula indicates the composition and approximate structure and shape of a molecule. Single bonds, double bonds, and triple bonds are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions). Ions can consist of one atom (monatomic ions) or several (polyatomic ions). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure. An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration. | 28,594 | 2,764 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.5%3A_The_Iron_Triad%3A_Iron_Cobalt_and_Nickel |
The Iron Triad is composed of three elements: iron (Fe), cobalt (Co), and nickel (Ni), which share similar chemical and physical characteristics. They are found adjacent to each other in period 4 of the . The Iron Triad is known for possessing ferromagnetic elements similar to gadolinium (Gd), and neodymium (Nd). These types of ferromagnetic elements have the ability to create a large magnetic pole due to their unpaired electrons. When one of these elements is inside an environment where the temperature is at its individual (T ), however, the specific paramagnetic lining of atoms is broken down by the energy within the element and ferromagnetism is lost. The (T )'s for iron, cobalt, and nickel are 768°C, 1121°C, and 354°C respectively and are taken advantage of to make use of these elements in industry. In addition, elements in the iron triad are commonly combined with carbon and each other to create various types of alloys. It is because of these magnetic properties and use in alloys that the three elements are typically grouped together and labeled as the "Iron Triad". While very similar in magnetic properties and reaction, these elements are also very unique and used differently in both nature and industry. Iron (Fe) is a transition metal with an atomic weight of 55.845 and an atomic number of 26. Its most common and comfortable oxidation state is +3 and is usually a shiny silver color. While it can be found in the sun and the stars, it is extremely common on planet earth and its abundance in the earth’s crust is 4.7%. The pure metal form of iron is rarely found due to its ability to easily react with other metals and environments. Iron's most common use in worldwide industry is to be used in alloys and steels for everyday use, some which even contain other elements in the Triad. By combining metals like iron with carbon to make alloys, very strong materials can be made that are used for steel housings, supports in television tubes, and other important products. Pure iron is available commercially in the United States and production of about 500 million tons of Iron is created every year throughout the world. From a biological perspective, iron is essential to organisms since it’s needed for the creation of hemoglobin. Humans need hemoglobin since these red pigments supply oxygen to our body through red blood cells. To increase intake of iron, foods like eggs, groundnuts and green leafy vegetables should be consumed. Without iron, our bodies would not be able to form hemoglobin and an organism would fail to survive. Cobalt (Co) is a transition metal with an atomic weight of 58.93 and an atomic number of 27, right in between iron and nickel. Cobalt, however, is not as abundant as iron and only makes up of about 0.0020% of the Earth’s crust. This makes cobalt a little more rare and valuable than the other members of the triad because it is still used heavily in international industry. It is for this reason that this hard gray element is mined for and traded around the world. Like iron, this element is commonly combined with other metals to create alloys. Superalloys that contain cobalt are used for parts of gas turbine engines that benefit both commercial and military devices. In current times, however, the rise of rechargeable batteries has made demand for the element rise to new heights since it is commonly used for electrodes in this relatively new technology. While these are its most popular applications, cobalt is also utilized in the production of petroleum, dyes, magnets, and electronics due to its ferromagnetic properties. Cobalt and its alloys have recently made strides in the biomedical sphere by playing a large role in neurological, dental, orthopedic, and cardiovascular implant devices. This is due to the fact that the alloys function well with their corrosion resistance, mechanical properties, and biocompatibility. Nickel (Ni), a transition metal with an atomic weight of 58.69 and an atomic number of 28, is usually recognized by its silvery shine with a hint of gold color. It was first discovered by Axel Fredrik Cronstedt in 1751 and named nickel from the ore "kupfernickel". Although nickel is not extremely common, it still ranks as #24 on the list of the most abundant elements on earth. This element is most abundant in Canada and about 300 million pounds are used in America every year. Nickel is most commonly used as a key ingredient in low alloy steels, stainless steels, and cast irons. In fact, four out of five of these 300 million pounds of Nickel go to this type of alloy making. Like cobalt, it shares the unique characteristic of corrosion resistance and is ideal for corrosion and heat resistant coatings, magnetic alloys, and controlled-expansion alloys. New nickel-cadmium rechargeable batteries have also brought the use of nickel into a new light. Like cobalt, the turning of the millennium and the increased use of rechargeable batteries has given nickel a new major use that will likely surpass others in the future. Nickel-cadmium batteries have improved the performance of cordless power tools, portable computers, and other portable electronic devices. Atomic Number 26 27 28 Mass 55.85 58.93 58.69 Electron Configuration 3d 4s 3d 4s 3d 4s Metallic radius, pm 124 125 125 Ionization Energy, kJ mol 762.5 760.4 737.1 759 758 737 1561 1646 1753 2957 3232 3393 E°, V -0.440 -0.277 -0.257 Common (+) Oxidation States 2, 3, 6 2, 3 2, 3 Melting Point (°C) 1530 1495 1455 Boiling Point (°C) 2862 2927 2732 Density, g cm 7.87 8.90 8.91 Electrical Conductivity 16 25 23 | 5,611 | 2,765 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.03%3A_The_Simple_Gas_Laws-_Boyles_Law_Charless_Law_and_Avogadros_Law |
Early scientists explored the relationships among the pressure of a gas ( ) and its temperature ( ), volume ( ), and amount ( ) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the . As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal. The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure \(\Page {1}\). In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure \(\Page {2}\)). A simple plot of \(V\) versus \(P\) gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows: \[PV = \rm constant \label{10.3.1} \] Dividing both sides by \(P\) gives an equation illustrating the inverse relationship between \(P\) and \(V\): or \[V \propto \dfrac{1}{P} \label{10.3.3} \] where the ∝ symbol is read “is proportional to.” A plot of versus 1/ is thus a straight line whose slope is equal to the constant in Equations \(\ref{10.3.1}\) and \(\ref{10.3.3}\). Dividing both sides of Equation \(\ref{10.3.1}\) by instead of gives a similar relationship between and 1/ . The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: This law in practice is shown in Figure \(\Page {2}\). At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air. The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure \(\Page {3}\)).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure \(\Page {3}\) that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure \(\Page {3}\), plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis. The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K). We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure \(\Page {3}\) is often referred to as Charles’s law and is stated mathematically as \[V ={\rm const.}\; T \label{10.3.4} \] or \[V \propto T \label{10.3.5} \] with not Charles’s law is valid for virtually all gases at temperatures well above their boiling points. We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure \(\Page {4}\)). This is the historic “Avogadro’s hypothesis.” A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: Stated mathematically, This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures. For a sample of gas, The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure \(\Page {5}\). Volume with increasing temperature or amount, but with increasing pressure. The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure ( ), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure ( ), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present ( ). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is , the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | 7,633 | 2,766 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4I%3A_Heat_Guns |
Heat guns are inexpensive tools for delivering strong heat in a more flexible manner than other heating methods. Heat can be directed from every direction, and the gun can be manually waved about in order to dissipate the heating intensity. Heat guns are commonly used to quickly develop stained plates (Figures 1.56a+b), and result in more even heating and less charring than when using a hotplate. They are also ideal for sublimations (Figure 1.56c), as the heat can be directed to the sides of the flask to coax off crystals deposited on the sides. A disadvantage of using heat guns is that they must be continually held, which makes them most ideal for short processes. A heat gun is not simply a hair dryer, and the nozzle gets (temperatures can be between \(150\)-\(450^\text{o} \text{C}\))!\(^8\) Care should be taken to not touch the nozzle after use, and the gun should be set down carefully, as it may mark the benchtop or cord. \(^8\)As reported in the Fischer Scientific catalog. | 1,007 | 2,767 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.04%3A_Ionization_Energy |
We have seen that when elements react, they often gain or lose enough electrons to achieve the valence electron configuration of the nearest noble gas. Why is this so? In this section, we develop a more quantitative approach to predicting such reactions by examining periodic trends in the energy changes that accompany ion formation. Because atoms do not spontaneously lose electrons, energy is required to remove an electron from an atom to form a cation. Chemists define the ionization energy (\(I\)) of an element as the amount of energy needed to remove an electron from the gaseous atom \(E\) in its ground state. \(I\) is therefore the energy required for the reaction \[ E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ \text{energy required=I } \label{7.4.1} \] Because an input of energy is required, the ionization energy is always positive (\(I > 0\)) for the reaction as written in Equation \(\Page {1}\). Larger values of mean that the electron is more tightly bound to the atom and harder to remove. Typical units for ionization energies are kilojoules/mole (kJ/mol) or electron volts (eV): If an atom possesses more than one electron, the amount of energy needed to remove successive electrons increases steadily. We can define a first ionization energy (\(I_1\)), a second ionization energy (\(I_2\)), and in general an nth ionization energy (\(I_n\)) according to the following reactions: \[ \ce{E(g) \rightarrow E^+(g) +e^-} \;\;\ I_1=\text{1st ionization energy} \label{7.4.2} \] \[ \ce{E^{+}(g) \rightarrow E^{2+}(g) +e^-} \;\;\ I_2=\text{2nd ionization energy} \label{7.4.3} \] \[ \ce{E^{2+}(g) \rightarrow E^{3+}(g) +e^-} \;\;\ I_3=\text{3rd ionization energy} \label{7.4.4} \] Values for the ionization energies of \(Li\) and \(Be\) listed in Table \(\Page {1}\) show that successive ionization energies for an element increase as they go; that is, it takes more energy to remove the second electron from an atom than the first, and so forth. There are two reasons for this trend. First, the second electron is being removed from a positively charged species rather than a neutral one, so in accordance with Coulomb’s law, more energy is required. Second, removing the first electron reduces the repulsive forces among the remaining electrons, so the attraction of the remaining electrons to the nucleus is stronger. Successive ionization energies for an element . The increase in successive ionization energies, however, is not linear, but increases drastically when removing electrons in lower \(n\) orbitals closer to the nucleus. The most important consequence of the values listed in Table \(\Page {1}\) is that the chemistry of \(\ce{Li}\) is dominated by the \(\ce{Li^+}\) ion, while the chemistry of \(\ce{Be}\) is dominated by the +2 oxidation state. The energy required to remove the electron from \(\ce{Li}\): \[\ce{Li^+(g) \rightarrow Li^{2+}(g) + e^-} \label{7.4.5} \] is more than 10 times greater than the energy needed to remove the first electron. Similarly, the energy required to remove the electron from \(\ce{Be}\): \[\ce{Be^{2+}(g) \rightarrow Be^{3+}(g) + e^-} \label{7.4.6} \] is about 15 times greater than the energy needed to remove the first electron and around 8 times greater than the energy required to remove the second electron. Both \(\ce{Li^+}\) and \(\ce{Be^{2+}}\) have 1 closed-shell configurations, and much more energy is required to remove an electron from the 1 core than from the 2 valence orbital of the same element. The chemical consequences are enormous: lithium (and all the alkali metals) forms compounds with the 1+ ion but not the 2+ or 3+ ions. Similarly, beryllium (and all the alkaline earth metals) forms compounds with the 2+ ion but not the 3+ or 4+ ions. The energy required to remove electrons from a filled core is prohibitively large under normal reaction conditions. Ionization Energy: Ionization energies of the elements in the third row of the periodic table exhibit the same pattern as those of \(Li\) and \(Be\) (Table \(\Page {2}\)): successive ionization energies increase steadily as electrons are removed from the valence orbitals (3 or 3 , in this case), followed by an especially large increase in ionization energy when electrons are removed from filled core levels as indicated by the bold diagonal line in Table \(\Page {2}\). Thus in the third row of the periodic table, the largest increase in ionization energy corresponds to removing the fourth electron from \(Al\), the fifth electron from Si, and so forth—that is, removing an electron from an ion that has the valence electron configuration of the preceding noble gas. This pattern explains why the chemistry of the elements normally involves only valence electrons. Too much energy is required to either remove or share the inner electrons. From their locations in the periodic table, predict which of these elements has the highest fourth ionization energy: B, C, or N. three elements element with highest fourth ionization energy These elements all lie in the second row of the periodic table and have the following electron configurations: The fourth ionization energy of an element (\(I_4\)) is defined as the energy required to remove the fourth electron: \[E^{3+}_{(g)} \rightarrow E^{4+}_{(g)} + e^- \nonumber \] Because carbon and nitrogen have four and five valence electrons, respectively, their fourth ionization energies correspond to removing an electron from a partially filled valence shell. The fourth ionization energy for boron, however, corresponds to removing an electron from the filled 1 subshell. This should require much more energy. The actual values are as follows: B, 25,026 kJ/mol; C, 6223 kJ/mol; and N, 7475 kJ/mol. From their locations in the periodic table, predict which of these elements has the lowest second ionization energy: Sr, Rb, or Ar. \(\ce{Sr}\) The first column of data in Table \(\Page {2}\) shows that first ionization energies tend to increase across the third row of the periodic table. This is because the valence electrons do not screen each other very well, allowing the effective nuclear charge to increase steadily across the row. The valence electrons are therefore attracted more strongly to the nucleus, so atomic sizes decrease and ionization energies increase. These effects represent two sides of the same coin: stronger electrostatic interactions between the electrons and the nucleus further increase the energy required to remove the electrons. However, the first ionization energy decreases at Al ([Ne]3 3 ) and at S ([Ne]3 3 ). The electron configurations of these "exceptions" provide the answer why. The electrons in aluminum’s filled 3 subshell are better at screening the 3 electron than they are at screening each other from the nuclear charge, so the electrons penetrate closer to the nucleus than the electron does and the electron is more easily removed. The decrease at S occurs because the two electrons in the same orbital repel each other. This makes the S atom slightly less stable than would otherwise be expected, as is true of all the group 16 elements. The first ionization energies of the elements in the first six rows of the periodic table are plotted in Figure \(\Page {1}\) and are presented numerically and graphically in Figure \(\Page {2}\). These figures illustrate three important trends: Generally, \(I_1\) increases diagonally from the lower left of the periodic table to the upper right. Gallium (Ga), which is the first element following the first row of transition metals, has the following electron configuration: [Ar]4 3 4 . Its first ionization energy is significantly lower than that of the immediately preceding element, zinc, because the filled 3 subshell of gallium lies inside the 4 subshell, shielding the single 4 electron from the nucleus. Experiments have revealed something of even greater interest: the second and third electrons that are removed when gallium is ionized come from the 4 orbital, the 3 subshell. The chemistry of gallium is dominated by the resulting Ga ion, with its [Ar]3 electron configuration. This and similar electron configurations are particularly stable and are often encountered in the heavier -block elements. They are sometimes referred to as . In fact, for elements that exhibit these configurations, . As we noted, the first ionization energies of the transition metals and the lanthanides change very little across each row. Differences in their second and third ionization energies are also rather small, in sharp contrast to the pattern seen with the - and -block elements. The reason for these similarities is that the transition metals and the lanthanides form cations by losing the electrons before the ( − 1) or ( − 2) electrons, respectively. This means that transition metal cations have ( − 1) valence electron configurations, and lanthanide cations have ( − 2) valence electron configurations. Because the ( − 1) and ( − 2) shells are closer to the nucleus than the shell, the ( − 1) and ( − 2) electrons screen the electrons quite effectively, reducing the effective nuclear charge felt by the electrons. As increases, the increasing positive charge is largely canceled by the electrons added to the ( − 1) or ( − 2) orbitals. That the electrons are removed before the ( − 1) or ( − 2) electrons may surprise you because the orbitals were filled in the reverse order. In fact, the , the ( − 1) , and the ( − 2) orbitals are so close to one another in energy, and interpenetrate one another so extensively, that very small changes in the effective nuclear charge can change the order of their energy levels. As the orbitals are filled, the effective nuclear charge causes the 3 orbitals to be slightly lower in energy than the 4 orbitals. The [Ar]3 electron configuration of Ti tells us that the 4 electrons of titanium are lost before the 3 electrons; this is confirmed by experiment. A similar pattern is seen with the lanthanides, producing cations with an ( − 2) valence electron configuration. Because their first, second, and third ionization energies change so little across a row, these elements have important similarities in chemical properties in addition to the expected vertical similarities. For example, all the first-row transition metals except scandium form stable compounds as M ions, whereas the lanthanides primarily form compounds in which they exist as M ions. Use their locations in the periodic table to predict which element has the lowest first ionization energy: Ca, K, Mg, Na, Rb, or Sr. six elements element with lowest first ionization energy Locate the elements in the periodic table. Based on trends in ionization energies across a row and down a column, identify the element with the lowest first ionization energy. These six elements form a rectangle in the two far-left columns of the periodic table. Because we know that ionization energies increase from left to right in a row and from bottom to top of a column, we can predict that the element at the bottom left of the rectangle will have the lowest first ionization energy: Rb. Use their locations in the periodic table to predict which element has the highest first ionization energy: As, Bi, Ge, Pb, Sb, or Sn. \(\ce{As}\) The tendency of an element to lose electrons is one of the most important factors in determining the kind of compounds it forms. Periodic behavior is most evident for , the energy required to remove an electron from a gaseous atom. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell. Consequently, only valence electrons can be removed in chemical reactions, leaving the filled inner shell intact. Ionization energies explain the common oxidation states observed for the elements. Ionization energies increase diagonally from the lower left of the periodic table to the upper right. Minor deviations from this trend can be explained in terms of particularly stable electronic configurations, called , in either the parent atom or the resulting ion. | 12,210 | 2,768 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/22%3A_Chemistry_of_The_Main-Group_Elements_II/22.3%3A_Group_17%3A_The_Halogens |
Because the halogens are highly reactive, none is found in nature as the free element. Hydrochloric acid, which is a component of aqua regia (a mixture of HCl and HNO that dissolves gold), and the mineral fluorspar (CaF ) were well known to alchemists, who used them in their quest for gold. Despite their presence in familiar substances, none of the halogens was even recognized as an element until the 19th century. Because the halogens are highly reactive, none is found in nature as the free element. Chlorine was the first halogen to be obtained in pure form. In 1774, Carl Wilhelm Scheele (the codiscoverer of oxygen) produced chlorine by reacting hydrochloric acid with manganese dioxide. Scheele was convinced, however, that the pale green gas he collected over water was a compound of oxygen and hydrochloric acid. In 1811, Scheele’s “compound” was identified as a new element, named from the Greek chloros, meaning “yellowish green” (the same stem as in chlorophyll, the green pigment in plants). That same year, a French industrial chemist, Bernard Courtois, accidentally added too much sulfuric acid to the residue obtained from burned seaweed. A deep purple vapor was released, which had a biting aroma similar to that of Scheele’s “compound.” The purple substance was identified as a new element, named iodine from the Greek iodes, meaning “violet.” Bromine was discovered soon after by a young French chemist, Antoine Jérôme Balard, who isolated a deep red liquid with a strong chlorine-like odor from brine from the salt marshes near Montpellier in southern France. Because many of its properties were intermediate between those of chlorine and iodine, Balard initially thought he had isolated a compound of the two (perhaps ICl). He soon realized, however, that he had discovered a new element, which he named bromine from the Greek bromos, meaning “stench.” Currently, organic chlorine compounds, such as PVC (polyvinylchloride), consume about 70% of the Cl produced annually; organobromine compounds are used in much smaller quantities, primarily as fire retardants. Because of the unique properties of its compounds, fluorine was believed to exist long before it was actually isolated. The mineral fluorspar (now called fluorite [CaF ]) had been used since the 16th century as a “flux,” a low-melting-point substance that could dissolve other minerals and ores. In 1670, a German glass cutter discovered that heating fluorspar with strong acid produced a solution that could etch glass. The solution was later recognized to contain the acid of a new element, which was named fluorine in 1812. Elemental fluorine proved to be very difficult to isolate, however, because both HF and F are extraordinarily reactive and toxic. After being poisoned three times while trying to isolate the element, the French chemist Henri Moissan succeeded in 1886 in electrolyzing a sample of KF in anhydrous HF to produce a pale green gas (Figure \(\Page {1}\)). For this achievement, among others, Moissan narrowly defeated Mendeleev for the Nobel Prize in Chemistry in 1906. Large amounts of fluorine are now consumed in the production of cryolite (Na AlF ), a key intermediate in the production of aluminum metal. Fluorine is also found in teeth as fluoroapatite [Ca (PO ) F], which is formed by reacting hydroxyapatite [Ca (PO ) OH] in tooth enamel with fluoride ions in toothpastes, rinses, and drinking water. The heaviest halogen is astatine (At), which is continuously produced by natural radioactive decay. All its isotopes are highly radioactive, and the most stable has a half-life of only about 8 h. Consequently, astatine is the least abundant naturally occurring element on Earth, with less than 30 g estimated to be present in Earth’s crust at any one time. All the halogens except iodine are found in nature as salts of the halide ions (X ), so the methods used for preparing F , Cl , and Br all involve oxidizing the halide. Reacting CaF with concentrated sulfuric acid produces gaseous hydrogen fluoride: \[CaF_{2(s)} + H_2SO_{4(l)} \rightarrow CaSO_{4(s)} + 2HF_{(g)} \label{1}\] Fluorine is produced by the electrolysis of a 1:1 mixture of HF and K HF at 60–300°C in an apparatus made of Monel, a highly corrosion-resistant nickel–copper alloy: Fluorine is one of the most powerful oxidants known, and both F and HF are highly corrosive. Consequently, the production, storage, shipping, and handling of these gases pose major technical challenges. Although chlorine is significantly less abundant than fluorine, elemental chlorine is produced on an enormous scale. Fortunately, large subterranean deposits of rock salt (NaCl) are found around the world (Figure \(\Page {2}\)), and seawater consists of about 2% NaCl by mass, providing an almost inexhaustible reserve. Inland salt lakes such as the Dead Sea and the Great Salt Lake are even richer sources, containing about 23% and 8% NaCl by mass, respectively. Chlorine is prepared industrially by the chloralkali process, which uses the following reaction: \[2NaCl_{(aq)} +2H_2O_{(l)} \xrightarrow{electrolysis} 2NaOH(aq) + Cl_{2(g)} + H_{2(g)} \label{3}\] Bromine is much less abundant than fluorine or chlorine, but it is easily recovered from seawater, which contains about 65 mg of Br per liter. Salt lakes and underground brines are even richer sources; for example, the Dead Sea contains 4 g of Br per liter. Iodine is the least abundant of the nonradioactive halogens, and it is a relatively rare element. Because of its low electronegativity, iodine tends to occur in nature in an oxidized form. Hence most commercially important deposits of iodine, such as those in the Chilean desert, are iodate salts such as Ca(IO ) . The production of iodine from such deposits therefore requires reduction rather than oxidation. The process is typically carried out in two steps: reduction of iodate to iodide with sodium hydrogen sulfite, followed by reaction of iodide with additional iodate: \[2IO^−_{3(aq)} + 6HSO^−_{3(aq)} \rightarrow 2I^−_{(aq)} + 6SO^2−_{4(aq)} + 6H^+_{(aq)} \label{4}\] \[5I^−_{(aq)} + IO^−_{3(aq)} + 6H^+_{(aq)} \rightarrow 3I_{2(s)} + 3H_2O_{(l)} \label{5}\] Because the halogens all have ns np electron configurations, their chemistry is dominated by a tendency to accept an additional electron to form the closed-shell ion (X ). Only the electron affinity and the bond dissociation energy of fluorine differ significantly from the expected periodic trends shown in Table \(\Page {1}\). Electron–electron repulsion is important in fluorine because of its small atomic volume, making the electron affinity of fluorine less than that of chlorine. Similarly, repulsions between electron pairs on adjacent atoms are responsible for the unexpectedly low F–F bond dissociation energy. (As discussed earlier, this effect is also responsible for the weakness of O–O, N–N, and N–O bonds.) Electrostatic repulsions between lone pairs of electrons on adjacent atoms cause single bonds between N, O, and F to be weaker than expected. Because it is the most electronegative element in the periodic table, fluorine forms compounds in only the −1 oxidation state. Notice, however, that all the halogens except astatine have electronegativities greater than 2.5, making their chemistry exclusively that of nonmetals. The halogens all have relatively high ionization energies, but the energy required to remove electrons decreases substantially as we go down the column. Hence the heavier halogens also form compounds in positive oxidation states (+1, +3, +5, and +7), derived by the formal loss of ns and np electrons. Because ionization energies decrease down the group, the heavier halogens form compounds in positive oxidation states (+1, +3, +5, and +7). Fluorine is the most reactive element in the periodic table, forming compounds with every other element except helium, neon, and argon. The reactions of fluorine with most other elements range from vigorous to explosive; only O , N , and Kr react slowly. There are three reasons for the high reactivity of fluorine: With highly electropositive elements, fluorine forms ionic compounds that contain the closed-shell F ion. In contrast, with less electropositive elements (or with metals in very high oxidation states), fluorine forms covalent compounds that contain terminal F atoms, such as SF . Because of its high electronegativity and 2s 2p valence electron configuration, fluorine normally participates in only one electron-pair bond. Only a very strong Lewis acid, such as AlF , can share a lone pair of electrons with a fluoride ion, forming AlF . Oxidative strength decreases down group 17. The halogens (X ) react with metals (M) according to the general equation \[M_{(s,l)} + nX_{2(s,l,g)} \rightarrow MX_{n(s,l)} \label{6}\] For elements that exhibit multiple oxidation states fluorine tends to produce the highest possible oxidation state and iodine the lowest. For example, vanadium reacts with the halogens to give VF , VCl , VBr , and VI . Metal halides in the +1 or +2 oxidation state, such as CaF , are typically ionic halides, which have high melting points and are often soluble in water. As the oxidation state of the metal increases, so does the covalent character of the halide due to polarization of the M–X bond. With its high electronegativity, fluoride is the least polarizable, and iodide, with the lowest electronegativity, is the most polarizable of the halogens. Halides of small trivalent metal ions such as Al tend to be relatively covalent. For example, AlBr is a volatile solid that contains bromide-bridged Al Br molecules. In contrast, the halides of larger trivalent metals, such as the lanthanides, are essentially ionic. For example, indium tribromide (InBr ) and lanthanide tribromide (LnBr ) are all high-melting-point solids that are quite soluble in water. As the oxidation state of the metal increases, the covalent character of the corresponding metal halides also increases due to polarization of the M–X bond. All halogens react vigorously with hydrogen to give the hydrogen halides (HX). Because the H–F bond in HF is highly polarized (H –F ), liquid HF has extensive hydrogen bonds, giving it an unusually high boiling point and a high dielectric constant. As a result, liquid HF is a polar solvent that is similar in some ways to water and liquid ammonia; after a reaction, the products can be recovered simply by evaporating the HF solvent. (Hydrogen fluoride must be handled with extreme caution, however, because contact of HF with skin causes extraordinarily painful burns that are slow to heal.) Because fluoride has a high affinity for silicon, aqueous hydrofluoric acid is used to etch glass, dissolving SiO to give solutions of the stable SiF ion. Except for fluorine, all the halogens react with water in a disproportionation reaction, where X is Cl, Br, or I: \[X_{2(g,l,s)} + H_2O_{(l)} \rightarrow H^+_{(aq)} + X^−_{(aq)} + HOX_{(aq)} \label{7}\] The most stable oxoacids are the perhalic acids, which contain the halogens in their highest oxidation state (+7). The acid strengths of the oxoacids of the halogens increase with increasing oxidation state, whereas their stability and acid strength decrease down the group. Thus perchloric acid (HOClO , usually written as HClO ) is a more potent acid and stronger oxidant than perbromic acid. Although all the oxoacids are strong oxidants, some, such as HClO , react rather slowly at low temperatures. Consequently, mixtures of the halogen oxoacids or oxoanions with organic compounds are potentially explosive if they are heated or even agitated mechanically to initiate the reaction. Because of the danger of explosions, oxoacids and oxoanions of the halogens should never be allowed to come into contact with organic compounds. Both the acid strength and the oxidizing power of the halogen oxoacids decrease down the group. The halogens react with one another to produce interhalogen compounds, such as ICl , BrF , and IF . In all cases, the heavier halogen, which has the lower electronegativity, is the central atom. The maximum oxidation state and the number of terminal halogens increase smoothly as the ionization energy of the central halogen decreases and the electronegativity of the terminal halogen increases. Thus depending on conditions, iodine reacts with the other halogens to form IF (n = 1–7), ICl or ICl , or IBr, whereas bromine reacts with fluorine to form only BrF, BrF , and BrF but not BrF . The interhalogen compounds are among the most powerful Lewis acids known, with a strong tendency to react with halide ions to give complexes with higher coordination numbers, such as the IF ion: \[IF_{7(l)} + KF_{(s)} \rightarrow KIF_{8(s)} \label{8}\] All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7). The interhalogen compounds are also potent oxidants and strong fluorinating agents; contact with organic materials or water can result in an explosion. All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7), but the importance of the higher oxidation states generally decreases down the group. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form. Predict the products of each reaction and write a balanced chemical equation for each reaction. The halogens are highly reactive. All halogens have relatively high ionization energies, and the acid strength and oxidizing power of their oxoacids decreases down the group. The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X ion. Their chemistry is exclusively that of nonmetals. Consistent with periodic trends, ionization energies decrease down the group. Fluorine, the most reactive element in the periodic table, has a low F–F bond dissociation energy due to repulsions between lone pairs of electrons on adjacent atoms. Fluorine forms ionic compounds with electropositive elements and covalent compounds with less electropositive elements and metals in high oxidation states. All the halogens react with hydrogen to produce hydrogen halides. Except for F , all react with water to form oxoacids, including the perhalic acids, which contain the halogens in their highest oxidation state. Halogens also form interhalogen compounds; the heavier halogen, with the lower electronegativity, is the central atom. | 14,629 | 2,770 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Theoretical_and_Actual_Yields |
Reactants not completely used up are called excess reagents, and the reactant that completely reacts is called the limiting reagent. This concept has been illustrated for the reaction: \[\mathrm{2 Na + Cl_2 \rightarrow 2 NaCl}\] Amounts of products calculated from the complete reaction of the are called theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of actual yield to theoretical yield expressed in percentage is called the percentage yield. \(\mathrm{percent\: yield = \dfrac{actual\: yield}{theoretical\: yield}\times100}\) Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the process or inefficiency of the chemical reaction. Methyl alcohol can be produced in a high-pressure reaction \(\mathrm{CO_{\large{(g)}} + 2 H_{2\large{(g)}} \rightarrow CH_3OH_{\large{(l)}}}\) If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess amount of \(\ce{CO}\), estimate the theoretical and the percentage yield? To calculate the theoretical yield, consider the reaction \(\begin{alignat}{2}
\ce{&CO_{\large{(g)}} +\, &&2 H_{2\large{(g)}} \rightarrow \, &&CH_3OH_{\large{(l)}}}\\
&\:28.0 &&\:4.0 &&\:\:\:32.0 \hspace{45px}\ce{(stoichiometric\: masses\: in\: g,\: kg,\: or\: tons)}
\end{alignat}\) \(\mathrm{1.2\: tons\: H_2 \times\dfrac{32.0\: CH_3OH}{4.0\: H_2}= 9.6\: tons\: CH_3OH}\) Thus, the theoretical yield from 1.2 metric tons (1.2x10 g) of hydrogen gas is 9.6 tons. The actual yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is \(\mathrm{\%\: yield =\dfrac{6.1\: tons}{9.6\: tons}\times 100 = 64 \%}\) Due to or the , the limiting reagent may not be completely consumed. Thus, a lower yield is expected in some cases. Losses during the recovery process of the product will cause an even lower actual yield. A solution containing silver ion, \(\ce{Ag+}\), has been treated with excess of chloride ions \(\ce{Cl-}\). When dried, 0.1234 g of \(\ce{AgCl}\) was recovered. Assuming the percentage yield to be 98.7%, how many grams of silver ions were present in the solution? HINT The reaction and relative masses of reagents and product are: \(\begin{alignat}{2}\ce{
Ag^+_{\large{(aq)}}} &+ \mathrm{Cl^-_{\large{(aq)}}} &&\rightarrow \ce{AgCl_{\large{(s)}}} \\
107.868 &+ 35.453 &&= 143.321
\end{alignat}\) The calculation, \(\mathrm{0.1234\: g\: AgCl \times \dfrac{107.868\: g\: Ag^+}{143.321\: g\: AgCl}= 0.09287\: g\: Ag^+}\) shows that 0.1234 g dry \(\ce{AgCl}\) comes from 0.09287 g \(\ce{Ag+}\) ions. Since the actual yield is only 98.7%, the actual amount of \(\ce{Ag+}\) ions present is therefore \(\mathrm{\dfrac{0.09287\: g\: Ag^+}{0.987}= 0.09409\: g\: Ag^+}\) DISCUSSION One can also calculate the theoretical yield of \(\ce{AgCl}\) from the percentage yield of 98.7% to be \(\mathrm{\dfrac{0.1234\: g\: AgCl}{0.987}= 0.1250\: g\: AgCl}\) From 0.1250 g \(\ce{AgCl}\), the amount of \(\ce{Ag+}\) present is also 0.09409 g. Hint: \(\ce{AgNO3}\) is the excess reagent Skill -
Apply the concept of excess and limiting reagents for work. You can add \(\ce{AgNO3}\) slowly until the clear portion of the solution gives no precipitate when a drop of \(\ce{AgNO3}\) solution is added. This indicates that all the \(\ce{I-}\) ions are consumed. Hint: Test for excess or limiting reagent. Skill -
Excess reagent can be tested for its presence, and limiting reagent can be tested for its absence. Hint: 0.05670 g \(\ce{Ag}\) and 0.06670 g \(\ce{I}\) Skill -
Calculate the amount of limiting reagent from the amount of products. Hint: 0.05256 mol/L Skill -
Calculate the concentration when the amount of solute is known. The concept of concentration will be covered in the unit dealing with solution, but you should be able to convert; see the following relationship. 0.1234 g \(\ce{AgI}\) = 0.0005256 mol = 0.5256 mili-mol \(\ce{AgI}\) or \(\ce{Ag}\) or \(\ce{I}\). | 4,178 | 2,771 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/09%3A_The_Periodic_Table_and_Some_Atomic_Properties/9.6%3A_Magnetic_Properties |
The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic forms: including paramagnetism, and diamagnetism, ferromagnetism, and anti-ferromagnetism. Only paramagnetism, and diamagnetism are discussed here. Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can orient in either direction, they exhibit magnetic moments that can align with a magnet. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, \(O_2\) is a good example of paramagnetism (that is best understood with molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet: As shown in \(\Page {1}\), since molecular oxygen (\(O_2\) is paramagnetic, it is attracted to the magnet. In contrast, molecular nitrogen, \(N_2\), has no unpaired electrons and it is diamagnetic (discussed below); it is therefore unaffected by the magnet. Paramagnetism is a form of magnetism whereby materials are by an externally applied magnetic field. There are some exceptions to the paramagnetism rule; these concern some transition metals, in which the unpaired electron is not in a d-orbital. Examples of these metals include \(Sc^{3+}\), \(Ti^{4+}\), \(Zn^{2+}\), and \(Cu^+\). These metals are the not defined as paramagnetic: they are considered diamagnetic because all d-electrons are paired. Paramagnetic compounds sometimes display bulk magnetic properties due to the clustering of the metal atoms. This phenomenon is known as ferromagnetism, but this property is not discussed here. Diamagnetic substances are characterized by paired electrons—except in the previously-discussed case of transition metals, there are no unpaired electrons. According to the which states that no two identical electrons may take up the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly by a magnetic field as demonstrated with the pyrolytic carbon sheet in Figure \(\Page {1}\). Diamagnetic materials are by the applied magnetic field. Diamagnetism, to a greater or lesser degree, is a property of all materials and always makes a weak contribution to the material's response to a magnetic field. For materials that show some other form of magnetism (such paramagntism), the diamagnetic contribution becomes negligible. The magnetic form of a substance can be determined by examining its electron configuration: if it shows unpaired electrons, then the substance is paramagnetic; if all electrons are paired, the substance is diamagnetic. This process can be broken into four steps: Determining Magnetic Properties from Orbital Diagrams: Are chlorine atoms paramagnetic or diamagnetic? Follow the four steps outlines above. For Cl atoms, the electron configuration is 3s 3p Ignore the core electrons and focus on the valence electrons only. There is one unpaired electron. Since there is an unpaired electron, \(Cl\) atoms are paramagnetic (but weakly since only one electron is unpaired). Indicate whether boron atoms are paramagnetic or diamagnetic. : The B atom has 2s 2p as the electron configuration. Because it has one unpaired electron, it is paramagnetic. Are zinc atoms paramagnetic or diamagnetic? For Zn atoms, the electron configuration is 4s 3d There are no unpaired electrons. Because there are no unpaired electrons, Zn atoms are diamagnetic. Indicate whether F ions are paramagnetic or diamagnetic. The F ion has 2s 2p has the electron configuration. Because it has no unpaired electrons, it is diamagnetic. | 4,508 | 2,772 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX5._Getting_Towed_Uphill |
The uphill carboxyloids are useful materials in terms of being able to make other compounds. For example, a thioester such as acetyl coenzyme A may be able to make a variety of acyl esters. In the laboratory, acid chlorides are very common starting materials to make other carboxyloids. However, if they are so far uphill, how are they formed in the first place? The two most common methods of making acid chlorides are treatment with thionyl chloride or with oxalyl chloride. Figure CX5.1. Possible syntheses of an acid chloride. The key part of making the uphill acid chloride out of the downhill carboxylic acid is the reagent used. Structurally, the reagents can be compared to acid chlorides themselves. They can be though of as being a little bit like uphill carboxyloids themselves. Thus, as one compound gives its chloride and gets oxygenated on its way downhill, it provides the energy needed to drive the carboxylic acid uphill. Figure CX5.2. Conversion of thionyl chloride to sulfur dioxide and hydrochloric acid. Figure CX5.3. Conversion of oxalyl chloride to carbon dioxide, carbon monoxide and hydrochloric acid. Reaction of a carboxylic acid with oxalyl chloride starts with a carboxyloid substitution using the oxalyl chloride as electrophile and carboxylic acid as nucleophile. The chloride ion that is liberated then acts as a nucleophile in a cascade reaction that releases CO and CO as an acid chloride forms. Draw the mechanism. Reaction of a carboxylic acid with thionyl chloride is very similar to reaction with oxalyl chloride, described above. Draw the mechanism of the reaction. Provide additional factors (such as energetics, equilibrium concepts) that explain why oxalyl chloride and thionyl chloride can drive the conversion of a carboxylic acid to an acid chloride. Acid anhydrides are usually made from the corresponding carboxylic acids. One molecule of carboxylic acid acts as a nucleophile and a second acts as an electrophile. Because the same kind of molecule acts as nucleophile and electrophile, acid anhydrides are typically symmetric: they havae a (C=O)O(C=O) unit in the middle, with the same alkyl groups on either side of it. To aid formation of acid anhydrides, carboxylic acids are often heated strongly (well above 100 C). Otherwise, they are sometimes heated in the presence of a strong drying agent, such as (empirically, P O ). In the presence of water, phosphorus pentoxide is converted to phosphoric acid, H PO . Draw a mechanism for the conversion of ethanoic acid to ethanoic anhydride. Show the reverse reaction to the conversion of ethanoic acid to ethanoic anhydride (the same reaction, in the other direction). Which of the two directions do you think is favourable, based on the ski hill? Draw a mechanism for the conversion of phosphorus pentoxide to phosphoric acid. Explain why the conditions outlined above lead to acid anhydride formation. , | 2,921 | 2,773 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.1%3A_Lewis_Theory%3A_An_Overview |
Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond wil be discussed and the general properties found in typical substances in which the bond type occurs Each bond classification is discussed in detail in subsequent sections of the chapter. Let's look at the preferred arrangements of electrons in atoms when they form chemical compounds. At the beginning of the 20 century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called (often shortened to ) that can be used for predicting the number of bonds formed by most elements in their compounds. Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons. Lewis Dot symbols: To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. For example, the electron configuration for atomic sulfur is [Ne]3s 3p , thus there are valence electrons. Its Lewis symbol would therefore be: Fluorine, for example, with the electron configuration [He]2 2 , has seven valence electrons, so its Lewis dot symbol is constructed as follows: Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure \(\Page {2}\). The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds. In 1904, Richard Abegg formulated what is now known as , which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. The refers to the tendency of atoms to prefer to have eight electrons in the . When atoms have fewer than eight electrons, they tend to react and form more stable compounds. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light ;reactions that decrease stability must absorb energy, getting colder. When discussing the octet rule, we do not consider or electrons. Only the s and p electrons are involved in the octet rule, making it a useful rule for the (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with s p . A stable arrangement is attended when the atom is surrounded by eight electrons. This octet can be made up by own electrons and some electrons which are shared. Thus, an atom continues to form bonds until an octet of electrons is made. This is known as octet rule by Lewis. The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration. The formula for table salt is NaCl. It is the result of Na ions and Cl ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron. The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na and Cl , which is much less stable than Na and Cl . Atoms are more stable when they have no charge, or a small charge. Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows: No dots are shown on Cs in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs ion, which has the valence electron configuration of Xe, and the F ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements. Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table. As you might expect for such a qualitative approach to bonding, there are exceptions to the octet rule, which we describe elsewhere. These include molecules in which one or more atoms contain fewer or more than eight electrons. Lewis Theory of Bonding: One convenient way to predict the number and basic arrangement of bonds in compounds is by using , which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of eight valence electrons in their compounds, the so-called . Hydrogen, with only two valence electrons, does not obey the octet rule. ( ) | 7,354 | 2,774 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.03%3A_Factors_Affecting_Solubility |
Experimentally it is found that the solubility of most compounds depends strongly on temperature and, if a gas, on pressure as well. As we shall see, the ability to manipulate the solubility by changing the temperature and pressure has several important consequences. Figure \(\Page {1}\) shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and \(\ce{CH_3CO_2Na}\)) exhibit a dramatic increase in solubility with increasing temperature. Others (such as \(\ce{NaCl}\) and \(\ce{K_2SO_4}\)) exhibit little variation, and still others (such as \(\ce{Li_2SO_4}\)) become less soluble with increasing temperature. Notice in particular the curves for \(\ce{NH4NO3}\) and \(\ce{CaCl2}\). The dissolution of ammonium nitrate in water is endothermic (\(ΔH_{soln} = +25.7\; kJ/mol\)), whereas the dissolution of calcium chloride is exothermic (\(ΔH_{soln} = −68.2 \;kJ/mol\)), yet Figure \(\Page {1}\) shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature. The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate (\(\ce{CH_3CO_2Na}\)) and 50 g of \(\ce{KBr}\), we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure \(\Page {1}\), both compounds dissolve in water at 80°C, and all 50 g of \(\ce{KBr}\) remains in solution at 0°C. Only about 36 g of \(\ce{CH3CO2Na}\) are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of \(\ce{CH_3CO_2Na}\) crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original \(\ce{CH_3CO_2Na}\) in essentially pure form in only one step. Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure \(\Page {1}\) and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was \(\ce{KBr}\) in this example) and preferably present in relatively small amounts. The solubility of gases in liquids decreases with increasing temperature, as shown in Figure \(\Page {2}\). Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process (\(ΔH_{soln} < 0\)). Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas. The phenomenon is similar to that involved in the increase in the vapor pressure of a pure liquid with increasing temperature. In the case of vapor pressure, however, it is attractive forces between solvent molecules that are being overcome by the added thermal energy when the temperature is increased. The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (Figure \(\Page {3}\)). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex and will be described elsewhere, but the driving force is the loss of dissolved \(\ce{CO2}\) from solution. Hard water contains dissolved \(\ce{Ca^{2+}}\) and \(\ce{HCO3^{-}}\) (bicarbonate) ions. Calcium bicarbonate (\(\ce{Ca(HCO3)2}\) is rather soluble in water, but calcium carbonate (\(\ce{CaCO3}\)) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water: \[\ce{2HCO3^{-}(aq) -> CO3^{2-}(aq) + H2O(l) + CO2(aq)} \label{13.9} \] Heating the solution decreases the solubility of \(\ce{CO2}\), which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale. In , lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of \(\ce{O2}\) at higher temperatures (Figure \(\Page {2}\)), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water. A similar effect is seen in the rising temperatures of bodies of water such as the k0oi89Chesapeake Bay, the largest estuary in North America, where \lobal warming has been implicated as the cause. For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality. External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure \(\Page {4}\), which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures. The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836): \[C = kP \label{13.3.1} \] where Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table \(\Page {1}\). As the data in Table \(\Page {1}\) demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the elements, the Henry’s law constants increase smoothly from \(\ce{He}\) to \(\ce{Ne}\) to \(\ce{Ar}\). Nitrogen and oxygen are the two most prominent gases in the Earth’s atmosphere and they share many similar physical properties. However, as Table \(\Page {1}\) shows, \(\ce{O2}\) is twice as soluble in water as \(\ce{N2}\). Many factors contribute to solubility including the nature of the intermolecular forces at play. For a details discussion, see "The O /N Ratio Gas Solubility Mystery" by Rubin Battino and Paul G. Seybold ( ), Gases that react chemically with water, such as \(\ce{HCl}\) and the other hydrogen halides, \(\ce{H2S}\), and \(\ce{NH3}\), do obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, \(\ce{HCl}\) reacts with water to give \(\ce{H^{+}(aq)}\) and \(\ce{Cl^{-}(aq)}\), not dissolved \(\ce{HCl}\) molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule. Gases that with water do not obey Henry’s law. Henry’s law has important applications. For example, bubbles of \(\ce{CO2}\) form as soon as a carbonated beverage is opened because the drink was bottled under \(\ce{CO2}\) at a pressure greater than 1 atm. When the bottle is opened, the pressure of \(\ce{CO2}\) above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of \(\ce{N2}\) to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases. Due to the low Henry’s law constant for \(\ce{O2}\) in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the \(\ce{O2}\) concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind \(\ce{O2}\) reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds \(\ce{O2}\) and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four \(\ce{O2}\) molecules. Although the concentration of dissolved \(\ce{O2}\) in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved \(\ce{O2}\) concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for \(\ce{O2}\) result in dissolved oxygen concentrations comparable to those in normal blood. A Discussing Henry's Law. Link: The Henry’s law constant for \(\ce{O2}\) in water at 25°C is \(1.27 \times 10^{-3} M/atm\), and the mole fraction of \(\ce{O2}\) in the atmosphere is 0.21. Calculate the solubility of \(\ce{O2}\) in water at 25°C at an atmospheric pressure of 1.00 atm. : Henry’s law constant, mole fraction of \(\ce{O2}\), and pressure : : A According to , the partial pressure of \(\ce{O2}\) is proportional to the mole fraction of \(\ce{O2}\): \[\begin{align*} P_A &= \chi_A P_t \\[4pt] &= (0.21)(1.00\; atm) \\[4pt] &= 0.21\; atm \end{align*} \nonumber \] B From Henry’s law, the concentration of dissolved oxygen under these conditions is \[\begin{align*} [\ce{CO2}] &= k P_{\ce{O2}} \\[4pt] &=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) \\[4pt] &=2.7 \times 10^{-4}\; M \end{align*} \nonumber \] To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved \(\ce{CO2}\) in a soft drink: \(0.17 M\) \(1 \times 10^{-5} M\) The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. | 14,105 | 2,775 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/RR._Radical_Reactions/RR3._Initiation%3A_Radical_Stability |
Bond strength isn't just about the interaction of the two fragments bonded together. It is also influenced by the stability of those two species on their own. When the bond is broken, what pieces are left over? The formation of radicals may be driven by the weakness of a particular bond. In terms of radical formation via bond homolysis, the reaction is more product-favoured if the bond being broken is weak. In other words, the bond is not very low in energy, so the overall reaction may become more downhill (or at least less uphill). In that case, forward reaction is favoured because of reactant destabilization. However, a downhill reaction could also occur through product stabilization. We have seen that the stability of anions and cations is strongly influenced by delocalization. Factors that spread the excess charge onto multiple atoms, rather than allowing charge to concentrate on one atom, make charged species much more stable. For example, carbon-based anions are relatively unstable, but a delocalized carbanion is within the realm of possibility. Enolate ions are particularly easy to obtain because negative charge is partially delocalized onto a more electronegative oxygen atom. Delocalization also strongly stabilizes radicals. It is one of the most important factors in the stability of carbon-based radicals. Illustrate the resonance stabilization in the following radicals a) allyl, CH CHCH b) benzyl, CH C H c) cyclopentadienyl, C H , | 1,478 | 2,777 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/The_Atom |
The is the smallest unit of matter that is composed of three sub-atomic particles: the proton, the neutron, and the electron. Protons and neutrons make up the nucleus of the atom, a dense and positively charged core, whereas the negatively charged electrons can be found around the nucleus in an electron cloud. An atom consists of a positively charged nucleus surrounded by one or more negatively charged particles called electrons. The number of protons found in the nucleus equals the number of electrons that surround it, giving the atom a neutral charge (neutrons have zero charge). Most of an atom’s mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. Although the nucleus is heavy, it is small compared with the overall size of an atom. The radius of a typical atom is around 1 to 2.5 angstroms (Å), whereas the radius of a nucleus is about 10 Å. If an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily fit inside a small football stadium.The nucleus of an atom contains protons and neutrons. Protons and neutrons have nearly equal masses, but they differ in charge. A neutron has no charge, whereas a proton has a positive charge that exactly balances the negative charge on an electron. The atomic mass unit (amu) is defined as exactly one-twelfth the mass of a carbon atom that has six protons and six neutrons in its nucleus. On this scale, protons and neutrons have masses that are close to, but not precisely, 1 u each. In fact, there are 6.022 x 10 u in 1 gram. This number is known as Avogadro’s number, N. The number of protons in the nucleus of an atom is known as the atomic number, Z. It is the same as the number of electrons around the nucleus of the electrically-neutral atom. The mass number of an atom is equal to the total number of protons and neutrons. When two atoms are close enough to combine chemically and form chemical bonds with one another, each atom “sees” the outermost electrons of the other atom. These outer electrons, called , are the most important factors in the chemical behavior of atoms. Both the neutron and the proton have key roles in the characteristics of an atom. The number of neutrons determine the type of isotope the element could be and the proton determines the number of electrons that surround its nucleus. All atoms with the same atomic number exhibit the same chemical behaviors and are classified as the same chemical element. Each element has its own name and a one- or two-letter symbol (usually derived from the element’s English or Latin name). For example, the symbol for carbon is C, and the symbol for calcium is Ca. It is important to note that for elements with two letters, the first letter is capital and the second letter is lowercase. The symbol for sodium is Na—the first two letters of its Latin (and German) name, natrium to distinguish it from nitrogen, N, and sulfur, S. What is the atomic symbol for bromine, and what is its atomic number? Why is it incorrect to use the letter B? What other element preempts the symbol B? (Refer to the periodic table) Bromine’s atomic number is 35, and its symbol is Br; B is the symbol for boron The of an atom is comprised of protons and neutrons; it is therefore positively charged. The number of protons within the nucleus of a given atom is equal to the of the corresponding element, which can be found on the periodic table. For example, the atomic number of helium is two. Therefore, the number of protons is also two. The number of neutrons within the nucleus of a given atom can be found by subtracting the atomic number from the atomic mass. The is the sum of protons and neutrons. Atomic Mass Number = Number of Protons + Number of Neutrons To find the number of neutrons, subtract the atomic number, the number of protons, from the mass number. Notation of a specific element follows this format: \(^A_Z {\rm E} ^c\) where E is a specific element, A is mass number, Z is the atomic number, and C is the charge. For helium, the notation is as follows: \(^4_2 {\rm He}\) Helium has 2 protons, 2 neutrons and a charge of zero. Atoms of the same element that have a different number of neutrons are known as . Most elements have several naturally occurring isotopes. The atomic mass of a particular element is equal to the average of the relative abundance of all its isotopes found in nature. For example, there are three naturally occurring isotopes of carbon: carbon-12, carbon-13, and carbon-14. Carbon-12 is the most common of these three, making up about 98.89% of all carbon, whereas carbon-13 has 1.11% natural abundance. Carbon-14 occurs rarely in nature. Atomic masses for other elements uses the carbon-12 scale as a reference. Early physicists assigned the atomic mass of 12 to the carbon-12 isotope (which is the most common carbon isotope) so that it would be easier to determine the atomic masses of other atoms. Using this information, we can determine the average atomic mass of carbon. (Use 13 for the approximate mass of carbon-13.) average atomic mass = mass of carbon-12 x (% natural abundance/100) + mass of carbon-13 x (% natural abundance/100) = 12 x .9889 + 13 x .0111 = 12.0111 Table 2 gives examples of common isotopes and their percent abundances in nature: How many protons, neutrons, and electrons are there in an atom of uranium-238? Write the symbol for this isotope. The atomic number of uranium (see ) is 92, and the mass number of the isotope is 238. Therefore, it has 92 protons, 92 electrons, 146 neutrons (238 amu - 92 protons). Its symbol is \(^{238}_{92}{\rm U}\) (or U). The total mass of an atom is called its atomic weight, and this is approximately the sum of the masses of its constituent protons, neutrons, and electrons. When protons, neutrons, and electrons combine to form an atom, some of their mass is converted to energy and is given off (this is the source of energy in nuclear fusion reactions; because the atom cannot be broken down into its fundamental particles unless the energy for the missing mass is supplied from outside it, this energy is called the of the nucleus). Each isotope of an element is characterized by an atomic number (total number of protons), a mass number (total number of protons and neutrons), and an atomic weight (mass of atom in atomic mass units). Because the reduction in mass upon the formation of an atom is small, the mass number is usually the same as the atomic weight rounded to the nearest integer (for example, the atomic weight of chlorine-37 is 36.966, which is rounded to 37) If there are several isotopes of an element in nature, then the experimentally observed atomic weight (the natural atomic weight) is the weighted average of the isotope weights. The average is weighted according to the percent abundance of the isotopes. In general, all isotopes of an element have similar chemical properties. Their behaviors differ with regard to mass-sensitive properties such as diffusion rates. Magnesium (Mg) has three significant natural isotopes: 78.70% of all magnesium atoms have an atomic weight of 23.985 u, 10.13% have an atomic weight of 24.986 u, and 11.17% have an atomic weight of 25.983 u. How many protons and neutrons are present in each of these three isotopes? What are the symbols for each isotope? What is the weighted average of the atomic weights? There are 12 protons in each magnesium isotope. The isotope whose atomic weight is 23.985 u has a mass number of 24 (protons and neutrons), so 24 - 12 protons gives 12 neutrons. The symbol for this isotope is Mg. Similarly, the isotope whose atomic weight is 24.986 amu has a mass number of 25, 13 neutrons, and Mg as a symbol. The third isotope (25.983 amu) has a mass number of 26, 14 neutrons, and Mg as a symbol. We calculate the average atomic weight as follows: (0.7870 x 23.985) + (0.1013 x 24.986) + (0.1117 x 25.983) = 24.31 u Boron has two naturally occurring isotopes, B and B. We know that 80.22% of its atoms are B, atomic weight 11.009 u. From the natural atomic weight given on the inside front cover, calculate the atomic weight of the B isotope. If 80.22% of all boron atoms are B, then 100.00 - 80.22 or 19.78% are the unknown isotope, because the percentage abundance of isotopes must add up to 100%. In the the atomic weight of boron is found to be 10.81 u. represents the unknown atomic weight: \[\begin{eqnarray} (0.8022 \times 11.009) + (0.1978 \times W) &=& 10.81 {\rm u} \quad {\rm (natural~atomic~weight)} \\ W &=& \frac{10.81-8.831}{0.1978} \\ &=& 10.01 {\rm u} \end{eqnarray} \] What appears to be a cloud of electrons surrounds the nucleus of an atom. The electron cloud serves as a model to help visualize the location of electrons in an atom. These electrons are held in place by their attraction to protons in the nucleus. This attraction is known as an . Electrons move around the nucleus very quickly, creating the image of a cloud; however, they actually form . Each electron shell can hold a certain amount of electrons. The first shell holds two electrons, and subsequent shells hold eight electrons (the movement of electrons is discussed in Unit IV). The number of electrons in a given atom is equal to the number of protons in that atom. However, units of the same element can have different numbers of electrons. Such units are known as . Atoms that lose one or more electrons become positively charged and are called . Cations are smaller than the original atom due to the loss of an electron. Examples of cations include Ca and Al (note: An H cation is simply a proton). Atoms can also gain electrons, forming a negative ion known as an . Anions are bigger that the original atoms due to electron gain. Examples of anions include Cl and O . Although some elements may only gain one electron, some atoms can gain up to four electrons. 1. Complete the chart below. 2. Boron has two naturally occurring isotopes, boron-10 and boron-11, with the masses of 10.013 g and 11.009 g, respectively. What is the average atomic mass of boron? 3. What is the mass of chlorine-37, if the mass of chlorine-35 is 34.969 g and the atomic mass of chlorine is 35.453 g? (Hint: Refer to table.1.) 4. For the atom Pt , indicate the number of protons, neutrons and electrons. 5. Write the notation for an atom containing 24 protons, 28 neutrons and 21 electrons. 1. 2. Referring to table.1, the percent natural abundance of boron-10 is 19.9% and the percent natural abundance of boron-11 is 80.1%. M (10.0129 x .199) + (11.00931 x .801) = 10.811 The average atomic mass of boron is . 3. If we refer to table.1, we find the percent natural abundance of chlorine-35 is 75.77% and the percent natural abundance for chlorine-37 is 24.23. Using the same general formula as in the previous problem, solve for the mass of chlorine-37 (represented by n)/; (34.969 x .7577) + (n x .2423) = 35.453 26.496 + .2423n = 35.453 .2423n= 8.957 n = 36.967 The mass of chlorine-37 is . 4. Number of protons = 78. Number of neutrons = 118. Number of electrons = 74. 5. Cr | 11,139 | 2,778 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Affinity_Chromatography/03_Practice_Problems |
Several classroom practice problems and laboratory experiments have been consolidated into PDFs for use as teaching aids. They may be accessed and downloaded via the links below. to view the classroom practice problems PDF file. to view the laboratory experiments PDF file. Serum contains proteins, antibodies, peptides, hormones, cytokines, glucose, fats, and inorganic substances. You are required to isolate the protein albumin from a horse serum sample. What methods would be useful to isolate albumin from the serum sample? Some points you should consider before you select a method are the following: The following material can be useful in the purification of horse serum albumin: Author: DiResta Dan, Deaprtment of Biology, University of Miami Copyright University of Miami | 801 | 2,779 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/Indicators |
are substances whose solutions change color due to changes in pH. These are called acid-base indicators. They are usually weak acids or bases, but their conjugate base or acid forms have different colors due to differences in their absorption spectra. Did you know that the color of hydrangea flowers depends on the pH of the soil in which they are grown? Indicators are organic weak acids or bases with complicated structures. For simplicity, we represent a general indicator by the formula \(\mathrm{\color{Blue} HIn}\), and its ionization in a solution by the equilibrium, \[\mathrm{ {\color{Blue} HIn} \rightleftharpoons H^+ + {\color{Red} In^-}}\] and define the equilibrium constant as , \[K_{\large\textrm{ai}} = \mathrm{\dfrac{[H^+,{\color{Red} In^-}]}{[{\color{Blue} HIn}]}}\] which can be rearranged to give \[\mathrm{\dfrac{[{\color{Red} In^-}]}{[{\color{Blue} HIn}]}} = \dfrac{K_{\large\textrm{ai}}}{\ce{[H+]}}\] When \(\ce{[H+]}\) is greater than 10 , \(\mathrm{\color{Red} In^-}\) color dominates, whereas color due to \(\mathrm{\color{Blue} HIn}\) dominates if \(\ce{[H+]} < \dfrac{K_{\large\textrm{ai}}}{10}\). The above equation indicates that the color change is the most sensitive when \(\ce{[H+]} = K_{\large\textrm{ai}}\) in numerical value. We define p = - log( ), and the p value is also the pH value at which the color of the indicator is most sensitive to pH changes. Taking the negative log of gives, \[-\log K_{\large\textrm{ai}} = -\log\ce{[H+]} - \log\mathrm{\dfrac{[{\color{Red} In^- }]}{[{\color{Blue} HIn}]}}\] or \[\mathrm{pH = p\mathit K_{\large{ai}}} + \log\mathrm{\dfrac{[{\color{Red} In^-}]}{[{\color{Blue} HIn}]}}\] This is a very important formula, and its derivation is very simple. Start from the definition of the equilibrium constant ; you can easily derive it. Note that pH = p when \([\mathrm{\color{Red} In^-}] = [\mathrm{\color{Blue} HIn}]\). In other words, when the pH is the same as p , there are equal amounts of acid and base forms. When the two forms have equal concentration, the color change is most noticeable. Colors of substances make the world a wonderful place. Because of the colors and structures, flowers, plants, animals, and minerals show their unique characters. Many indicators are extracted from plants. For example, red cabbage juice and tea pigments show different colors when the pH is different. The color of tea darkens in a basic solution, but the color becomes lighter when lemon juice is added. Red cabbage juice turns blue in a basic solution, but it shows a distinct red color in an acidic solution. There is a separate file for this, and it can also be accessed from the Chemical Handbook menu. Find an indicator for the titration of a 0.100 M solution of a weak acid \(\ce{HA}\) (\(K_a = 6.2 \times 10^{-6}\)) with 0.100 M \(\ce{NaOH}\) solution. First, you should estimate the pH at the equivalence point, at which the solution is 0.0500 M \(\ce{NaA}\). This is a hydrolysis problem, but the following method employs the general principle of equilibrium. \[\begin{array}{ccccccc}
\ce{A- &+ &H2O &\rightleftharpoons &HA &+ &OH-}\\
0.0500-y &&&&y &&y
\end{array} \nonumber\] If we multiply the numerator and the denominator by \(\ce{[H+]}\), rearrange the terms, note that \(\ce{[H+,OH- ]} = K_{\large\textrm w}\), and by the definition of of the acid, we have the following relationship: \[\dfrac{y^2}{0.0500-y} = \ce{\dfrac{[HA,OH- ]}{[A- ]} \dfrac{[H+]}{[H+]}} = \dfrac{K_{\large\textrm w}}{K_{\large\textrm a}}\] \[\begin{align*}
y &= \left(0.0500\left(\dfrac{K_{\large\textrm w}}{K_{\large\textrm a}}\right)\right)^{1/2}\\
&= 9.0 \times 10^{-6}
\end{align*}\] \[\begin{align*}
\ce{pOH} &= -\log \ce{[OH- ]} = -\log 9.0 \times 10^{-6}\\
&= 5.05\\ \\
\ce{pH} &= 14 - 5.05\\
&= 8.95
\end{align*}\] Phenolphthalein in the table above has a p value of 9.7, which is the closest to the pH of equivalence point in this titration. This indicator is colorless in acidic solution, but a light appears when the pH is > 8. The color becomes more as the pH rises. A parade of the color intensities is shown below: The equivalence point is when the color changes most rapidly, not when the solution has changed color. Improper use of indicators will introduce inaccuracy to titration results. Indicators change color gradually at various pH. Let us assume that the acid form has a blue color and the basic form has red color. The variation of colors at different pH is shown below. The background color affects their appearance and our perception of them. The long stretched color in the middle of the last line has equal intensity of and . If a solution has a color matching this, the pH would be the same as the p of the indicator, provided that the conjugate forms of the indicator have the and colors. | 4,831 | 2,780 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Aquatic_Chemistry/1._Acid-Base_Chemistry_of_Natural_Aquatic_Systems |
Natural waters contain a wide variety of solutes that act together to determine the pH, which typically ranges from 6 to 9. Some of the major processes that affect the acid-base balance of natural systems are: In this chapter and also in the next one which deals specifically with the carbonate system, we will consider acid-base equilibria as they apply to natural waters. We will assume that you are already familiar with such fundamentals as the Arrhenius and Brfinsted concepts of acids and bases and the pH scale. You should also have some familiarity with the concepts of free energy and activity. The treatment of equilibrium calculations will likely extend somewhat beyond what you encountered in your General Chemistry course, and considerable emphasis will be placed on graphical methods of estimating equilibrium concentrations of various species. ) | 875 | 2,781 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Units_of_Measure/Metric_Prefixes_-_from_yotta_to_yocto |
In introductory chemistry we use only a few of the most common metric prefixes, such as milli, centi, and kilo. Our various textbooks and lab manuals contain longer lists of prefixes, but few if any contain a list. There is no point of memorizing this, but it is nice to have a place to look them up. You will find prefixes from throughout the range as you read the scientific literature. In particular, the smaller prefixes such as nano, pico, femto, etc., are becoming increasingly common as analytical chemistry and biotechnology develop more sensitive methods. To help you visualize the effect of these prefixes, there is a column "a sense of scale", which gives some examples of the magnitudes represented. a sense of scale (for some) Most are approximate. You have probably heard words such as kilobyte, in the context of computers. What does it mean? It might seem to mean 1000 bytes, since kilo means 1000. But in the computer world it often means 1024 bytes. That is 2 - a power of two very close to 1000. Now, in common usage it often does not matter whether the intent was 1000 bytes or 1024 bytes. But they are different numbers and sometimes it does matter. So, a new set of "binary prefixes", distinguished by "bi" in the name or "i" in the abbreviation, was introduced in 1998. By this new system, 1024 bytes would be properly called a kibibyte or KiB. (Sounds like something you would feed the dog.) This new system of binary prefixes has been endorsed by the International Electrotechnical Commission (IEC) for use in electrical technology. See the NIST page at . Whether these will catch on remains to be seen, but at least if you see such an unusual prefix you might want to be aware of this. >Robert Bruner ( ) | 1,754 | 2,782 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Affinity_Chromatography/02_Theory/03_Reading |
Rogue AC, Lowe CR., Affinity chromatography: history, perspectives, limitations and prospects, Methods Mol Biol. 2008;421:1-21. Collins, MO; Choudhary, J, Mapping multiprotein complexes by affinity purification and mass spectrometry, Current Opinion in Biotechnology (2008), 19(4), 324-330. Sousa, F; Prazeres, DMF; Queiroz, JA, Affinity chromatography approaches to overcome the challenges of purifying plasmid DNA, Trends in Biotechnology (2008), 26(9), 518-525. Fang, Xiangming; Zhang, Wei-Wei, Affinity separation and enrichment methods in proteomic analysis, Journal of Proteomics (2008), 71(3), 284-303. Chong, Shaorong, Edited by Cox, Michael H.; Phillips, George N, Proteins: affinity tags, Handbook of Proteins (2007), 2, 1270-1275. Kaur-Atwal, Gushinder; Weston, Daniel J.; Bonner, Philip L. R.; Crosland, Susan; Green, Philip S.; Creaser, Colin S., lmmobilized metal affinity chromatography for the analysis of proteins and peptides, Current Analytical Chemistry (2008), 4(2), 127-135. Hage, David S., Chromatography, affinity, Kirk-Othmer Separation Technology (2nd Edition) (2008), 1, 453-470. Ngo, That T.; Jogie-Brahim, Sherryline; Narinesingh, Dyer, Affinity Chromatographic Purification of Antibodies, Analytical Letters (2007), 40(15), 2799-2820. Rogue, Ana C. A.; Silva Claudia S. O.; Taipa, M. Angela, Affinity-based methodologies and ligands for antibody purification: Advances and perspectives, Journal of Chromatography, A (2007), 1160(1-2), 44-55. Hage, David S., Edited by Seidel, Arza, Chromatography, affinity, Kirk-Othmer Encyclopedia of Chemical Technology (5th Edition) (2004), 6, 390-407. Gutierrez, R; Martin del Valle, EM; Galan, MA, Immobilized metal-ion affinity chromatography: Status and trends, Separation and Purification Reviews (2007), 36(1), 71-111. Chan, Nora; Lewis, Darren; Kelly, Michele; Ng, Ella S. M.; Schriemer, David C., Frontal affinity chromatography - mass spectrometry for ligand discovery and characterization, Methods and Principles in Medicinal Chemistry (2007), 36(Mass Spectrometry in Medicinal Chemistry), 217-246. Okanda, Fred M.; El Rassi, Ziad, Biospecific interaction (affinity) CEC and affinity nano-LC, Electrophoresis (2007), 28(1-2), 89-98. Mallik, Rangan; Hage, David S., Journal of Separation Science (2006), 29(12),1686-1704. Varilova, Tereza; Madera, Milan; Pacakova, Vera; Stulik, Karel, Separation media in affinity chromatography of proteins - a critical review, Current Proteomics (2006), 3(1), 55-79. Pichon, Valerie; Haupt Karsten, Affinity separations on molecularly imprinted polymers with special emphasis on solid-phase extraction, Journal of Liquid Chromatography & Related Technologies, (2006), 29(7&8), 989-1023. Wang, Guangquan; SaIm, Jeffrey R.; Gurgel, Patrick V.; Carbonell, Ruben G., Edited by Galan, Miguel A.; Martin del Valle, Eva, Small peptide ligands for affinity separations of biological molecules, Chemical Engineering (2005), 63-83. Cuatrecasas, Pedro; Wilchek, Meir, Edited by Lennarz, William J.; Lane, M. Daniel, Affinity chromatography, Encyclopedia of Biological Chemistry (2004), 1, 51-56. : Bailon, P; Ehrlich, G.K.; Fung, W.; Berthold, W., Eds.; Methods in Molecular Biology; Humana Press, 2000, Vol. 147. : Bio-Medicine (2009). Hemdan, E.S.; Zhao, Y.; Sulkowski, E.; Porath, J. Surface topography of histidine residues: A facile probe by immobilized metal ion affinity chromatography. Proc. Natl. Acad. Sci. USA 1989, 86, 1811-1815. Jameson, G.W.; Elmore, D.T. . Biochem. J. 1974, 141, 555-565. Large-Scale Affinity Chromatography: Genetic Engineering & Biotechnology News (2007). : Mohr, P.; Pommerening, K.; Chromatographic Science Series; CRC Press, 1985. Vol. 33. | 3,690 | 2,783 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Synthesis_of_Alcohols/Alcohols_from_Hydroboration-Oxidation_of_Alkenes |
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH or BHR ) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the bouble bond. Furthermore, the borane acts as a lewisAnti-Markovnikov acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Anti-MarkovnikovHydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific ( meaning that the hydroboration takes place on the same face of the double bond, this leads stereochemistry. Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An Additional feature of this reaction is that it occurs without rearrangement. First off it is very imporatnt to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B H (diborane). Additionally, the dimer B H ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a lewis acid-base complex, which allows boron to have an electron octet. \[ 2BH_3 \rightarrow B_2H_6\] . EpoxidationEpoxidation If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well. If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the | 1,899 | 2,784 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.03%3A_Cell_potentials_and_Thermodynamics |
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic. It has long been known that some metals are more "active" than others in the sense that a more active metal can "displace" a less active one from a solution of its salt. The classic example is the one we have already mentioned on the preceding page: \[\ce{ Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)} \nonumber\] Here zinc is more active because it can displace (precipitate) copper from solution. If you immerse a piece of metallic zinc in a solution of copper sulfate, the surface of the zinc quickly becomes covered with a black coating of finely-divided copper, and the blue color of the hydrated copper(II) ion diminishes. Similar comparisons of other metals made it possible to arrange them in the order of their increasing electron-donating (reducing) power. This sequence became known as the electromotive or activity series of the metals. The activity series has long been used to predict the direction of oxidation-reduction reactions. Consider, for example, the oxidation of Cu by metallic zinc that we have mentioned previously. The fact that zinc is near the top of the activity series means that this metal has a strong tendency to lose electrons. By the same token, the tendency of Zn to accept electrons is relatively small. Copper, on the other hand, is a poorer electron donor, and thus its oxidized form, Cu, is a fairly good electron acceptor. We would therefore expect the reaction \[Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)\] to proceed in the direction indicated, rather than in the reverse direction. An old-fashioned way of expressing this is to say that "zinc will displace copper from solution". The above table is of limited practical use because it does not take into account the concentrations of the dissolved species. In order to treat these reactions quantitatively, it is convenient to consider the oxidation and reduction steps separately. When a net reaction proceeds in an electrochemical cell, oxidation occurs at one electrode (the anode) and reduction takes place at the other electrode (the cathode.) We can think of the cell as consisting of two half-cells joined together by an external circuit through which electrons flow and an internal pathway that allows ions to migrate between them so as to preserve electroneutrality. Each half-cell has associated with it an electrode-solution potential difference whose magnitude depends on the nature of the particular electrode reaction and on the concentrations of the dissolved electroactive species. The sign of this potential difference depends on the direction (oxidation or reduction) in which the electrode reaction proceeds. To express them in a uniform way, we adopt the convention that half-cell potentials are always defined for the reduction direction. Thus the half-cell potential for the Zn/Zn electrode (or as it is sometimes called) always refers to the reduction reaction \[Zn^{2+} + 2e^– \rightarrow Zn (s)\] In the cell the zinc appears on the left side, indicating that it is being oxidized, not reduced. For this reason, the potential difference contributed by the left half-cell has the opposite sign to its conventional half-cell potential. More generally, we can define the cell potential (or cell EMF) as \[E_{cell} = \Delta V = E_{right} – E_{left} \label{1}\] in which "right" and "left" refer to the cell notation convention (" eduction on the ight") and not, of course, to the physical orientation of a real cell in the laboratory. If we expand the above expression we see that the cell potential \[E_{cell} = V_{Cu} – V_{solution} + V_{solution} – V_{Zn}\] is just the difference between the two half-cell potentials \(E_{right}\) and \(E_{left}\). The fact that individual half-cell potentials are not directly measurable does not prevent us from defining and working with them. Although we cannot determine the absolute value of a half-cell potential, we can still measure its value in relation to the potentials of other half cells. In particular, if we adopt a reference half-cell whose potential is arbitrarily defined as zero, and measure the potentials of various other electrode systems against this reference cell, we are in effect measuring the half-cell potentials on a scale that is relative to the potential of the reference cell. The reference cell that has universally been adopted for this purpose is the \[Pt | ½ H_{2(g)} | H^+_{(aq)} || \;... \] in which hydrogen gas is allowed to bubble over a platinum electrode having a specially treated surface which catalyzes the reaction \[½ H_{2(g)} → H^+ + e^–\] When this electrode is operated under standard conditions of 1 atm H pressure, 25°C, and pH = 0, it becomes the standard hydrogen electrode, sometimes abbreviated SHE. To measure the relative potential of some other electrode couple M /M, we can set up a cell \[Pt | H_{2(g)} | H^+ || M^{2+}_{ (aq)} | M_{(s)}\] whose net reaction is H + M → 2H + M the potential difference between the platinum and M electrodes will be but since the difference is by definition zero for the hydrogen half-cell, the cell potential we measure corresponds to which is just the potential (relative to that of the SHE) of the half-cell whose reaction is \[M^{2+} + 2e^– → M_{(s)}\] The M /M half-cell is on the left, and the standard hydrogen cell is on the right. The two half-cells are joined through the salt bridge in the middle. The more "active" the metal M (the greater its tendency to donate electrons to H ), the more negative will be = Δ = – Standard [reduction] potentials are commonly denoted by the symbol . values for hundreds of electrodes have been determined (mostly in the period 1925-45, during which time they were referred to as "oxidation potentials") and are usually tabulated in order of increasing tendency to accept electrons (increasing oxidizing power.) +.222 +.268 +.337 +.535 +.771 +.799 +1.23 +1.36 Note particularly that Given the values for two half reactions, you can easily predict the potential difference of the corresponding cell: simply add the reduction potential of the reduction half-cell to the of the reduction potential (that is, to the oxidation potential) of the oxidation reaction. Find the standard potential of the cell Cu | Cu || Cl | AgCl | Ag and predict the direction of electron flow when the two electrodes are connected. The net reaction corresponding to this cell will be: Cu 2AgCl → 2 Ag + 2 Cl + Cu = E + E Or Written another way = E - E =(.222 + (– .337)) v = Since E is negative, the reaction will run in the direction. The correct net reaction will be: 2 Ag + 2 Cl + Cu → 2AgCl + Cu is being Reduced = E + E Or Written another way = E - E =(.337 + (– .222)) v = Since this potential is positive, the reaction will proceed to the right; electrons will be withdrawn from the silver electrode and flow through the external circuit into the copper electrode. Note carefully that in combining these half-cell potentials, we did not multiply for the Cu /Cu couple by two. The reason for this will be explained later. From the above, it should be apparent that the potential difference between the electrodes of a cell is a measure of the tendency for the cell reaction to take place: the more positive the cell potential, the greater the tendency for the reaction to proceed to the right. But we already know that the standard free energy change expresses the tendency for any kind of process to occur under the conditions of constant temperature and pressure. Thus Δ ° and measure the same thing, and are related in a simple way: \[\Delta G° = –nFE° \label{2}\] ... or in more detail (see below for explanations of the units given for voltage) A few remarks are in order about this very fundamental and important relation: 1 J = 1 watt-sec = 1 (amp-sec) × volts Thus the has the dimensions of joules/coulomb– the energy produced per quantity of charge passing through the cell. Because voltage is the quotient of two extensive quantities, it is itself . When we multiply the anodic and cathodic half-reactions by the stoichiometric factors required to ensure that each involves the same quantity of charge, the free energy change and the number of coulombs both increase by the same factor, leaving the potential (voltage) unchanged. This explains why we do not have to multiply the s of the anode and cathode reactions by stoichiometric factors when we are finding the potential of a complete cell. If Eq. 2 is solved for , we have \[ E° =\dfrac{\Delta G°}{nF} \label{3}\] This states explicitly that the cell potential is a measure of the free energy change , which is a brief re-statement of the principle explained immediately above. To see this more clearly, consider the cell Cu | Cu || Cl | AgCl | Ag for which we list the standard reduction potentials and °s of the half-reactions: 2 × [AgCl(s) + → Ag(s) + Cl ] Cu(s) → Cu + 2 +.222 v –(+.337) V –42800 J +65000 J 2 Ag + 2 Cl + Cu → AgCl + Cu Cu | Cu || AgCl | Cl | Ag –.115 v +22200 J Here we multiply the cathodic reaction by two in order to balance the charge. Because the anodic reaction is written as an oxidation, we reverse the sign of its and obtain = – = for the cell potential. The negative cell potential tells us that this reaction will not proceed spontaneously. Note, however, that if we are combining two half reactions to obtain a third half reaction, the values are not additive, since this third half-reaction is not accompanied by another half reaction that causes the charges to cancel. Free energies are always additive, so we combine them, and use Δ to find the cell potential. Calculate for the electrode Fe /Fe from the standard potential of the couples Fe /Fe and Fe /Fe Tabulate the values and calculate the Δ s as follows: A table of standard half-cell potentials summarizes a large amount of chemistry, for it expresses the relative powers of various substances to donate and accept electrons by listing reduction half-reactions in order of increasing E° values, and thus of increasing spontaneity. The greater the value of E°, the greater the tendency of the substance on the left to acquire electrons, and thus the stronger this substance is as an oxidizing agent. If you have studied elementary chemical thermodynamics, you will have learned about the role that a quantity called the , usually referred to as simply the " ", plays in determining the direction of any chemical change. The rule is that all spontaneous change (that is, all reactions that proceed to the "right") is associated with a fall in the free energy, and the greater the degree of that fall (Δ ), the greater will be the tendency for the reaction to take place. Since oxidation-reduction processes involve the transfer of an electron from a donor to an acceptor, it makes sense to focus on the electron and to consider that it falls from a higher-free energy environment (the reductant, or "source") to a lower-free energy one (the oxidant, or "sink".) As can be seen from the diagram below, this model makes it far easier to predict what will happen when two or more oxidants and reducants are combined; the electron "falls" as far as it can, filling up oxidizing agents (sinks) from the bottom up, very much in the same way as electrons fill atomic orbitals as we build up larger atoms. This chart is essentially an abbreviated form of a table of standard potentials in which the various couples are displayed on a vertical scale corresponding to
E° = –Δ / . Any available sink on the right side will tend to drain electrons from a source above it. For example, immersion of metallic zinc in a solution of CuSO will result in the reduction of Cu to metallic copper (red arrows.) Similarly, addition of chlorine to water will tend to oxidize the water, producing O and Cl (blue arrows.) Note especially the positions of the H / H and H O/O ,H couples on this chart, as they define the range of E°s for substances that will not decompose water (green region.) A more detailed table with a more complete explanation can be seen on the " ; it is strongly recommended that you take the time to acquire a thorough understanding of this concept. At this point, it might be worth calling your attention to the similar way of depicting acid-base reactions as representing the "fall of the proton" as shown below and described much more thoroughly . Acids are (donors), bases are . Protons "fall" (in free energy) whenever a base is present that presents proton-empty free energy levels. The red arrows show what happens when acetic acid is titrated with a strong base; the results are acetate ion and water. Note here again the crucial role of water, both as a proton acceptor (forming hydronium ion) and as a proton donor (forming hydroxide ion.) Note also that the pH of a solution is a direct measure of the average free energy of protons in the solution (relative to H O .) An important difference between proton transfer and electron transfer reactions is that the latter can vary greatly in speed, from almost instantaneous to so slow as to be unobservable. Acid-base reactions are among the fastest known. Considerable insight into the chemistry of a single element can be had by comparing the standard electrode potentials (and thus the relative free energies) of the various oxidation states of the element. The most convenient means of doing this is the . As examples, diagrams for iron and chlorine are shown below. The formulas of the species that represent each oxidation state of the element are written from left to right in order of decreasing oxidation number, and the standard potential for the reduction of each species to the next on the right is written in between the formulas. Potentials for reactions involving hydrogen ions will be pH dependent, so separate diagrams are usually provided for acidic and alkaline solutions (effective hydrogen ion concentrations of 1M and 10 M, respectively). The more positive the reduction potential, the greater will be the tendency of the species on the left to be reduced to the one on the right. To see how Latimer diagrams are used, look first at the one for iron in acid solution. The line connecting Fe and Fe represents the reaction Fe + → Fe whose positive E° (.440 v) indicates that metallic iron will dissolve in acidic solution to form Fe . Because the oxidation of this species to the +3 state has a negative potential (-0.771v; moving to the left on the diagram reverses the sign), the +2 state will be the stable oxidation state of iron under these conditions. An important condition to recognize in a Latimer diagram is when the potential on the left of a species is less positive than that on the right. This indicates that the species can oxidize and reduce itself, a process known as . As an example, consider Cl in alkaline solution. The potential for its reduction to Cl is sufficiently positive (+1.35 v) to supply the free energy necessary for the oxidation of one atom of chlorine to hypochlorite. Thus elemental chlorine is thermodynamically unstable with respect to disproportionation in alkaline solution, and the same it true of the oxidation product, ClO (hypochlorite ion). Cl can oxidize water (green arrows, top) and also undergo disproportionation (purple arrows, bottom). In the latter process, one Cl molecule donates electrons to another. Bear in mind that many oxidation-reduction reactions, unlike most acid-base reactions, tend to be very slow, so the fact that a species is thermodynamically unstable does not always mean that it will quickly decompose. Thus the two reactions shown in the figure are normally very slow. The free energy change for a process represents the maximum amount of non- work that can be extracted from it. In the case of an electrochemical cell, this work is due to the flow of electrons through the potential difference between the two electrodes. Note, however, that as the rate of electron flow (i.e., the current) increases, the potential difference must decrease; if we short-circuit the cell by connecting the two electrodes with a conductor having negligible resistance, the potential difference is zero and no work will be done. The full amount of work can be realized only if the cell operates at an infinitessimal rate; that is, reversibly. You should recall that this is exactly analogous to the expansion of an ideal gas. The full amount of work is extracted only under the special condition that the external pressure opposing expansion is only infinitessimally smaller than the pressure of the gas itself. If the gas is allowed to expand into a vacuum ( = 0), no work will be done. The total amount of energy a reaction can supply under standard conditions at constant pressure and temperature is given by °. If the reaction takes place by combining the reactants directly (no cell) or in a short-circuited cell, no work is done and the heat released is Δ . If the reaction takes place in a cell that performs electrical work, then the heat released is diminished by the amount of electrical work done. In the limit of reversible operation, the heat released becomes Δ = Δ + Δ Pt | H | H || M | M whose left half consists of a (SHE) and whose net reaction is H + M → 2H + M | 17,542 | 2,785 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/19%3A_Electrochemistry/19.08%3A_Electrolysis |
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis , occurs: an external voltage is applied to drive a nonspontaneous reaction ( ). In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications. In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.Electrolytic Cells If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu solution and the other electrode is cadmium metal immersed in a 1 M Cd solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd ) and is the anode, while metallic copper will be deposited on the copper electrode (Cu is reduced to Cu), which is the cathode (part (a) in ). The overall reaction is as follows:
\( Cd\left ( s \right ) +Cu^{2+}\left ( aq \right )\rightarrow Cu\left ( s \right ) +Cd^{2+}\left ( aq \right ) \tag{19.7.1}\) This reaction is thermodynamically spntaneous as written (
\( \Delta G^{o}=nFE_{cell}^{o}= \left ( 2\; \cancel{mol \; e^{-}} \right )\left [ 96,486 \;J/\left ( \cancel{V}\cdot \cancel{mol} \right ) \right ]\left ( 0.74 \; \cancel{V} \right ) =-140 \; kJ/mol \tag{19.7.2}\) In this direction, the system is acting as a galvanic cell. The reverse reaction, the reduction of Cd by Cu, is thermodynamically nonspontaneous and will occur only with an of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd is reduced) (part (b) in ). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows:
\(cathode:\; Cd^{2+}\left ( aq \right )+ 2e^{-} \rightarrow Cd\left ( s \right ) \;\;\; E_{cathode}^{o} = -0.40 \; V \tag{19.7.3}\) \(anode:\; Cu\left ( s \right )\rightarrow Cu^{2+} \left ( aq \right ) + 2e^{-} \;\;\; E_{anode}^{o}= 0.34 \; V \tag{19.7.4}\) \(overall:\; Cd^{2+}\left ( aq \right ) + Cu\left ( s \right )\rightarrow Cu^{2+} \left ( aq \right )+ Cd \left ( s \right ) \;\; E_{cell}^{o}=-0.74 \; V \tag{19.7.5}\) Because ° < 0, the overall reaction—the reduction of Cd by Cu—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in . At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten NaCl, for example, and an electrical potential is applied, Cl is oxidized at the anode, and Na is reduced at the cathode. The overall reaction is as follows:
\( 2NaCl\left ( l \right ) \rightarrow Na\left ( l \right ) +Cl_{2}\left ( g \right ) \tag{19.7.6}\) This is the reverse of the formation of NaCl from its elements. The product of the reduction reaction is sodium because the melting point of sodium metal is 97.8°C, well below that of NaCl (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten NaCl in a ( ). In this specialized cell, CaCl (melting point = 772°C) is first added to the NaCl to lower the melting point of the mixture to about 600°C, thereby lowering operating costs. Similarly, in the used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al O ; melting point = 2054°C) and 95% cryolite (Na AlF ; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO gas at the carbon anode. The overall reaction is as follows:
\( 2Al_{2}O_{3}\left ( l \right ) + 3C\left ( s \right ) \rightarrow 4Al\left ( l \right ) + 3CO_{2}\left ( g \right ) \tag{19.7.7}\) Oxide ions react with oxidized carbon at the anode, producing CO (g). There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general. In the Hall–Heroult process, C is oxidized instead of O or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O or F . Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na more difficult to reduce. In fact, the reduction of Na to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome. If a molten mixture of MgCl and KBr is electrolyzed, what products will form at the cathode and the anode, respectively? identity of salts electrolysis products List all the possible reduction and oxidation products. Based on the electronegativity values shown in , determine which species will be reduced and which species will be oxidized. Identify the products that will form at each electrode. The possible reduction products are Mg and K, and the possible oxidation products are Cl and Br . Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl is a stronger oxidant than Br . Electrolysis will therefore produce Br at the anode and Mg at the cathode. Exercise Predict the products if a molten mixture of AlBr and LiF is electrolyzed. Br and Al Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H and O . However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H SO or Na SO ) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H and O ( ). The reactions that occur are as follows:
\(cathode:\; 2H^{+}\left ( aq \right )+ 2e^{-} \rightarrow H_{2}\left ( g \right ) \;\;\; E_{cathode}^{o} = 0 \; V \tag{19.7.8}\) \(anode:\; 2H_{2}O\left ( l \right ) \rightarrow O_{2}\left ( g \right ) + 4H^{+} \left ( g \right ) + 4e^{-} \;\;\; E_{anode}^{o}= 1.23 \; V \tag{19.7.9}\) \(overall:\; 2H_{2}O\left ( l \right )\rightarrow O_{2}\left ( g \right )+ 2H_{2}\left ( g \right ) \;\; E_{cell}^{o}=-1.23 \; V \tag{19.7.10}\) For a system that contains an electrolyte such as Na SO , which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H ] = [OH ] = 1.0 × 10 . Assuming that = = 1 atm, we can use the standard potentials and to calculate for the overall reaction: \( E_{cell}= E_{cell}^{o}-\left ( \dfrac{0.0591 \; V}{n} \right )log\;\left ( P_{O_{2}}P_{H_{2}}^{2} \right ) \tag{19.7.11}\)
\( E_{cell}= -1.23 \; V -\left ( \dfrac{0.0591 \; V}{4} \right )log\;\left ( 1 \right ) = -1.23 \; V \) Thus is −1.23 V, which is the value of ° if the reaction is carried out in the presence of 1 M H rather than at pH 7.0. In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltage , represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car. In general, any metal that does react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The -block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO , SO , PO , are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. In a process called electroplating , a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in . The half-reactions in electroplating a fork, for example, with silver are as follows:
\(cathode\left ( fork \right ):\; Ag^{+}\left ( aq \right )+ e^{-} \rightarrow Ag\left ( s \right ) \;\;\; E_{cathode}^{o} = 0.80 \; V \tag{19.7.12}\) \(anode\left ( silver bar \right ):\; Ag\left ( s \right ) \rightarrow Ag^{+} \left ( aq \right ) + e^{-} \;\;\; E_{anode}^{o}= 0.80 \; V \tag{19.7.13}\) The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because ° = 0 V, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating. If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material. The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction Ag (aq) + e → Ag(s), 1 mol of electrons reduces 1 mol of Ag to Ag metal. In contrast, in the reaction Cu (aq) + 2e → Cu(s), 1 mol of electrons reduces only 0.5 mol of Cu to Cu metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 ), which is equal to 96,486 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (C) transferred is the product of the current (A) and the time ( , in seconds):
\( C=A\times t \tag{19.7.14}\) The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process. For example, if a current of 0.60 A passes through an aqueous solution of CuSO for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
\( charge=\left (0.60 \;A \right )\left ( 6.0 \; \cancel{min} \right ) \left ( 60 \; s/\cancel{min} \right ) =220 A \cdot s \tag{19.7.15}\) The number of moles of electrons transferred to Cu is therefore \( moles \; e^{-}= \dfrac{220 \; \cancel{C}}{96,486 \; \cancel{C}/mol}=2.3\times 10^{-3} \; mol \; e^{-} \tag{19.7.16}\) Because two electrons are required to reduce a single Cu ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 /s, and reaction times are on the order of 3–4 weeks. A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%? mass of metal, time, and efficiency current required Calculate the number of moles of metal corresponding to the given mass transferred. Write the reaction and determine the number of moles of electrons required for the electroplating process. Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes. We must first determine the number of moles of Ag corresponding to 2.00 g of Ag: \( moles \; Ag= \dfrac{2.0 \; \cancel{g}}{107.868 \; \cancel{g}/mol}=1.85\times 10^{-2} \; mol \; Ag \) The reduction reaction is Ag (aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver. Using the definition of the faraday, \( Coulombs=\left ( 1.85\times 10^{-2} \; \cancel{mol} \right )\left ( 96,486 \; C/\cancel{mol} \right ) =1.78 \times 10^{3} \; C\) The current in amperes needed to deliver this amount of charge in 12.0 h is therefore \( amperes=\dfrac{1.78 \times 10^{3} \; C}{ \left ( 12.0 \; \cancel{h} \right ) \left ( 60 \; \cancel{min}/ \cancel{h} \right ) \left ( 60 \; s/\cancel{min} \right ) } \) \( \;\;\;\; =4.12\times 10^{-2} \; C/s = 4.12\times 10^{-2} \; A \) Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A. Exercise A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al O Na AlF mixture? 5.8 h In , an external voltage is applied to drive a nonspontaneous reaction. A Downs cell is used to produce sodium metal from a mixture of salts, and the Hall–Heroult process is used to produce aluminum commercially. Electrolysis can also be used to produce H and O from water. In practice, an additional voltage, called an , must be applied to overcome factors such as a large activation energy and a junction potential. is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. : C = A × Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur? How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants? Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis? Two solutions, one containing Fe(NO ) ·6H O and the other containing the same molar concentration of Fe(NO ) ·6H O, were electrolyzed under identical conditions. Which solution produced the most metal? Justify your answer. The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons? Electrolysis is the most direct way of recovering a metal from its ores. However, the Na (aq)/Na(s), Mg (aq)/Mg(s), and Al (aq)/Al(s) couples all have standard electrode potentials ( °) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead? What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl is electrolyzed using a current of 12.4 A for 1.0 h? What mass of copper metal is deposited if a 5.12 A current is passed through a Cu(NO ) solution for 1.5 h. What mass of PbO is reduced when a current of 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery? Electrolysis of Cr (aq) produces Cr (aq). If you had 500 mL of a 0.15 M solution of Cr (aq), how long would it take to reduce the Cr to Cr using a 0.158 A current? Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed. Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed. 5.2 L | 17,641 | 2,786 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/14%3A_Hydrodynamics/14.03%3A_Laminar_and_Turbulent_Flow |
The Reynolds number is a dimensionless number is used to indicate whether flow conditions are in the laminar or turbulent regimes. It indicates whether the motion of a particle in a fluid is dominated by inertial or viscous forces.1 \[ \mathcal{R} = \dfrac{inertial\: forces}{viscous \: forces} \nonumber\] When \(\mathcal{R}>1\), the particle moves freely, experiencing only weak resistance to its motion by the fluid. If \(\mathcal{R}<1\), it is dominated by the resistance and internal forces of the fluid. For the latter case, we can consider the limit m → 0 in eq. , and find that the velocity of the particle is proportional to the random fluctuations: \(v(t)=f_r(t)/\zeta\). We can also express the Reynolds number in other forms: Hydrodynamically, for a sphere of radius r moving through a fluid with dynamic viscosity η and density ρ at velocity v, \[ \mathcal{R} =\dfrac{rv\rho}{\eta} \nonumber \] Consider for an object with radius 1 cm moving at 10 cm/s through water: \(\mathcal{R}=10^3\). Now compare to a protein with radius 1 nm moving at 10 m/s: \(\mathcal{R}=10^{-2}\). The drag force on an object is determined by the force required to displace the fluid against the direction of flow. A sphere, rod, or cube with the same mass and surface area will respond differently to flow. Empirically, the drag force on an object can be expressed as \[ f_d = \left[ \dfrac{1}{2} \rho C_d v^2 \right] a \nonumber \] This expression takes the form of a pressure (term in brackets) exerted on the cross-sectional area of the object along the direction of flow, a. C is the drag coefficient, a dimensionless proportionality constant that depends on the shape of the object. In the case of a sphere of radius r: a = πr2 in the turbulent flow regime (\(\mathcal{R} >1000\)) C = 0.44–0.47. Determination of C is somewhat empirical since it depends on \(\mathcal{R}\) and the type of flow around the sphere. The drag coefficient for a sphere in the viscous/laminar/Stokes flow regimes (\(\mathcal{R}<1\)) is \(C_d=24/\mathcal{R}\). This comes from using the Stokes Law for the drag force on a sphere \(f_d=6\pi \eta v r\) and the Reynolds number \(\mathcal{R}=\rho vd/\eta\). ________________________________ | 2,232 | 2,788 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/20%3A_Carbohydrates/20.08%3A_Polysaccharides |
The fibrous tissue in the cell walls of plants contains the polysaccharide , which consists of long chains of glucose units, each of which is connected by a \(\beta\)-glucoside link to the \(\ce{C_4}\) hydroxyl of another glucose as in the disaccharide (i.e., \(\beta\)-1,4): Indeed, enzymatic hydrolysis of cellulose leads to cellobiose. The molecular weight of cellulose varies with the source but is usually high. Cotton cellulose appears to have about 3000 glucose units per molecule. The natural fibers obtained from cotton, wood, flax, hemp, and jute all are cellulose fibers and serve as raw materials for the textile and paper industries. In addition to its use as a natural fiber and in those industries that depend on wood as a construction material, cellulose is used to make cellulose acetate (for making rayon acetate yarn, photographic film, and cellulose acetate butyrate plastics), nitric acid esters (gun cotton and celluloid\(^7\)), and cellulose xanthate (for making viscose rayon fibers). The process by which viscose rayon is manufactured involves converting wood pulp or cotton linters into cellulose xanthate by reaction with carbon disulfide and sodium hydroxide: The length of the chains of the cellulose decreases about 300 monomer units in the process. At this point, the cellulose is regenerated in the form of fine filaments by forcing the xanthate solution through a spinneret into an acid bath: A few animals (especially ruminants and termites) are able to metabolize cellulose, but even these animals depend on appropriate microorganisms in their intestinal tracts to hydrolyze the \(\beta\)-1,4 links; other animals, including man, cannot utilize cellulose as food because they lack the necessary hydrolytic enzymes. However, such enzymes are distributed widely in nature. In fact, deterioration of cellulose materials - textiles, paper, and wood - by enzymatic degradation (such as by dry rot) is an economic problem that is not yet adequately solved. Efforts to turn this to advantage through enzymatic hydrolysis of cellulose to glucose for practical food production have not been very successful ( ). A second, very widely distributed polysaccharide is , which is stored in the seeds, roots, and fibers of plants as a food reserve - a potential source of glucose. The chemical composition of starch varies with the source, but in any one starch there are two structurally different polysaccharides. Both consist entirely of glucose units, but one is a linear structure ( ) and the other is a branched structure ( ). The amylose form of starch consists of repeating 1,4-glucopyranose links as in cellulose, but unlike cellulose the linkage is \(\alpha\) rather than \(\beta\) (i.e., \(\alpha\)-1,4): Hydrolysis by the enzyme diastase leads to maltose. In amylopectin, amylose chains are joined by \(\alpha\)-1,6 linkages: Animals also store glucose in the form of starchlike substances called . These substances resemble amylopectin more than amylose in that they are branched chains of glucose units with \(\alpha\)-1,4- and \(\alpha\)-1,6-glucoside links. Starch is used in paper manufacture and in the textile and food industries. Fermentation of grain starches is an important source of ethanol. Hydrolysis of starch catalyzed by hydrochloric acid results in a syrupy mixture of glucose, maltose, and higher-molecular-weight saccharides. This mixture is called and is marketed as corn syrup. The hydrolysis does not proceed all the way to glucose because the \(\alpha\)-1,6 glucosidic link at the branch point is not easily hydrolyzed. Enzymes also catalyze hydrolysis of starch, but the enzyme \(\alpha\) is specific for \(\alpha\)-1,4 links and, like acid-catalyzed hydrolysis, gives a mixture of glucose, maltose, and polysaccharides (dextrin). The enzyme \(\alpha\)-1,6- can hydrolyze the \(\alpha\)-1,6 links at the branch points and, when used in conjunction with \(\alpha\) amylase, completes the hydrolysis of starch to glucose. A very interesting group of polysaccharides isolated from cornstarch hydrolysates are known as . One of these compounds, , is a large doughnut-shaped molecule with a central cavity that literally can engulf a small, relatively nonpolar organic molecule and hold it in water solution, similar to a micelle ( ). As with micelles, unusual reactivity is exhibited by the bound molecules. An example is the change in the ortho-para ratio in electrophilic substitution of methoxybenzene by hypochlorous acid, \(\ce{HOCl}\), in the presence and absence of cyclohexaamylose: Apparently the cyclohexaamylose wraps around the methoxybenzene in such a way as to protect the ortho carbons from attack by \(\ce{HOCl}\) but to leave the para carbon exposed. It is this kind of specificity that we need to generate in reactions before we can claim to have synthetic reactions under control. Many polysaccharides besides starch and cellulose are important components of animal tissues, or play a vital role in biochemical processes. One example is , a celluloselike material that is the structural component of the hard shells of insects and crustaceans. The difference between chitin and cellulose is that instead of being a polymer of glucose, chitin is a polymer of 2-deoxy-2-\(\ce{N}\)-ethanamidoglucose (\(\ce{N}\)-acetyl-\(\beta\)-\(D\)-glucosamine): is a very important and complex polysaccharide derivative that occurs in intestinal walls and has a major use as a blood anticoagulant, especially in connection with artificial kidney therapy. Heparin also has shown great promise in the treatment of patients with extensive burns, by promoting blood circulation to burn-damaged tissue. The structure of heparin can not be defined precisely because its composition depends on the source of supply. The major components of the polysaccharide chain are \(D\)-glucuronic acid, \(L\)-iduronic acid, and the same 2-deoxy-2-aminoglucose (\(D\)-glucosamine) that is a constituent of chitin (although in heparin it occurs as the \(\alpha\) anomer): The general construction of heparin involves the linkage of the anomeric carbons of one of the components with the 4-hydroxyl of another. A key feature of the heparin structure is the presence of sulfate groups that occur as hydrogen sulfate esters ( ) and as sulfamido groups, \(\ce{-NHSO_3H}\), on the 2-deoxy-2-amino-\(D\)-glucose units in the chain. Hydrogen sulfate groups also are located on the 2-hydroxyls of the \(L\)-iduronic acid units of the chain. In addition there are \(\ce{N}\)-ethanoyl groups attached to some of the 2-deoxy-2-amino-\(D\)-glucose nitrogens that are not connected to \(\ce{-SO_3H}\). Heparin is clearly an extraordinarily complex substance with many highly polar groups, and its mode of action as an anticoagulant is not clear. At present, because of increases in the use of artificial kidney machines, heparin is in rather short supply. Among the plant polysaccharides are the , which are used as jelling agents in the making of preserves and jellies from fruit. Also important are the from seaweeds and from trees, which are used as stabilizers and emulsifiers in the food, pharmaceutical, cosmetic, and textile industries. The pectins principally are polysaccharides of the methyl ester of \(D\)-galacturonic acid, whereas the alginates are polysaccharides made up of varying proportions of \(D\)-mannuronic acid and \(L\)-guluronic acid. The plant gums are similar materials. There are other polysaccharides besides cellulose in the cell walls of plants. These are called hemicelluloses, but the name is misleading because they are unrelated to cellulose. Those that are made of pentose sugars (mainly xylose) are most abundant. They accumulate as wastes in the processing of agricultural products, and on treatment with acids that yield a compound of considerable commercial importance, oxa-2,4-cyclopentadiene-2-carbaldehyde (furfural): \(^7\)Celluloid, one of the first plastics, is partially nitrated cellulose (known as pyroxylin) plasticized with camphor. and (1977) | 8,056 | 2,789 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods/1.5C%3A_Gravity_Filtration |
When there is a need to separate a solid-liquid mixture, it is common that the particles are so fine that they swirl and disperse when the flask is tilted. These mixtures cannot be , and an alternative method is gravity filtration. Gravity filtration is generally used when the filtrate (liquid that has passed through the filter paper) will be retained, while the solid on the filter paper will be discarded. A common use for gravity filtration is for separating anhydrous magnesium sulfate (\(\ce{MgSO4}\)) from an organic solution that it has dried (Figure 1.68b). Anhydrous magnesium sulfate is powdery, and with swirling in an organic solvent creates a fine dispersal of particles like a snow globe. To gravity filter a mixture, pour the mixture through a quadrant-folded filter paper (Figure 1.69) or fluted filter paper in a funnel and allow the liquid to filter using only the force of gravity (Figure 1.68c). It is best to pour as if attempting to decant, meaning to keep the solid settled in the flask for as long as possible. When solid begins to pour onto the filter paper, it has the possibility of clogging the filter paper pores or slowing filtration. After finished pouring, rinse the solid on the filter paper (and in the flask) with a few portions of fresh solvent to remove residual compound adhering to the solid. | 1,348 | 2,791 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Formulas_of_Inorganic_and_Organic_Compounds |
Chemistry is the experimental and theoretical study of materials on their properties at both the macroscopic and microscopic levels. Understanding the relationship between properties and structures/bonding is also a hot pursuit. Chemistry is traditionally divided into organic and inorganic chemistry. The former is the study of compounds containing at least one carbon-hydrogen bonds. By default, the chemical study of all other substances is called inorganic chemistry, a less well defined subject. However, the boundary between organic and inorganic compounds is not always well defined. For example, oxalic acid, H C O , is a compound formed in plants, and it is generally considered an organic acid, but it does not contain any C-H bond. Inorganic chemistry is also closely related to other disciplines such as materials sciences, physical chemistry, thermodynamics, earth sciences, mineralogy, crystallography, spectroscopy etc. A chemical formula is a format used to express the structure of atoms. The formula tells which elements and how many of each element are present in a compound. Formulas are written using the elemental symbol of each atom and a subscript to denote the number of elements. This notation can be accredited to Swedish chemist Jons Jakob Berzeliu. The most common elements present in organic compounds are carbon, hydrogen, oxygen, and nitrogen. With carbon and hydrogen present, other elements, such as phosphorous, sulfur, silicon, and the halogens, may exist in organic compounds. Compounds that do not pertain to this rule are called inorganic compounds. Understanding how atoms in a molecules are arranged and how they are bonded together is very important in giving the molecule its identity. Isomers are compounds in which two molecules can have the same number of atoms, and thus the same molecular formula, but can have completely different physical and chemical properties because of differences in structural formula. A polymer is formed when small molecules of identical structure, monomers, combine into a large cluster. The monomers are joined together by covalent bonds. When monomers repeat and bind, they form a polymer. While they can be comprised of natural or synthetic molecules, polymers often include plastics and rubber. When a molecule has more than one of these polymers, square parenthesis are used to show that all the elements within the polymer are multiplied by the subscript outside of the parenthesis. The subscript (shown as n in the example below) denotes the number of monomers present in the macromolecule (or polymer). Ethylene becomes the polymer polyethylene. The molecular formula is based on the actual makeup of the compound. Although the molecular formula can sometimes be the same as the empirical formula, molecular compounds tend to be more helpful. However, they do not describe how the atoms are put together. Molecular compounds are also misleading when dealing with isomers, which have the same number and types of atoms (see above in molecular geometry and structural formula). Ex. Molecular Formula for Ethanol: C H O. An empirical formula shows the most basic form of a compound. Empirical formulas show the number of atoms of each element in a compound in the most simplified state using whole numbers. Empirical formulas tend to tell us very little about a compound because one cannot determine the structure, shape, or properties of the compound without knowing the molecular formula. Usefulness of the empirical formula is decreased because many chemical compounds can have the same empirical formula. Ex. Find the empirical formula for C H Answer: C H O (divide all subscripts by 2 to get the smallest, whole number ratio). A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which empirical and molecular formulas cannot always represent. Ex. Structural Formula for Ethanol: Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together. Ex. Condensed Structural Formula for Ethanol: CH CH OH (Molecular Formula for Ethanol C H O). Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are then assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas. Ex. Line-Angle Formula for Ethanol: Inorganic compounds are typically not of biological origin. Inorganic compounds are made up of atoms connected using ionic bonds. These inorganic compounds can be binary compounds, binary acids, or polyatomic ions. Binary compounds are formed between two elements, either a metal paired with a nonmetal or two nonmetals paired together. When a metal is paired with a nonmetal, they form ionic compounds in which one is a negatively charged ion and the other is positvely charged. The net charge of the compound must then become neutral. Transition metals have different charges; therefore, it is important to specify what type of ion it is during the naming of the compound. When two nonmetals are paired together, the compound is a molecular compound. When writing out the formula, the element with a positive oxidation state is placed first. Ex. Ionic Compound: BaBr (Barium Bromide) Ex. Molecular Compound: N O (Dinitrogen Tetroxide) Binary acids are binary compounds in which hydrogen bonds with a nonmetal forming an acid. However, there are exceptions such as NH , which is a base. This is because it shows no tendency to produce a H . Because hydrogen is positively charged, it is placed first when writing out these binary acids. Ex. HBr (Hydrobromic Acid) Polyatomic ions is formed when two or more atoms are connected with covalent bonds. Cations are ions that have are postively charged, while anions are negatively charged ions. The most common polyatomic ions that exists are those of anions. The two main polyatomic cations are Ammonium and Mercury (I). Many polyatomic ions are typically paired with metals using ionic bonds to form chemical compounds. Ex. MnO (Polyatomic ion); NaMnO (Chemical Compound) Many acids have three different elements to form ternary compounds. When one of those three elements is oxygen, the acid is known as a oxoacid. In other words, oxacids are compounds that contain hydrogen, oxgygen, and one other element. Ex. HNO (Nitric Acid) Certain compounds can appear in multiple forms yet mean the same thing. A common example is hydrates: water molecules bond to another compound or element. When this happens, a dot is shown between H O and the other part of the compound. Because the H O molecules are embedded within the compound, the compound is not necessarily "wet". When hydrates are heated, the water in the compound evaporates and the compound becomes anhydrous. These compounds can be used to attract water such as CoCl . When CoCl is dry, CoCl is a blue color wherease the hexahydrate (written below) is pink in color. Ex. CoCl 6 H O Organic compounds contain a combination carbon and hydrogen or carbon and hydrogen with nitrogen and a few other elements, such as phosphorous, sulfur, silicon, and the halogens. Most organic compounds are seen in biological origin, as they are found in nature. Hydrocarbons are compounds that consist of only carbon and hydrogen atoms. Hydrocarbons that are bonded together with only single bonds are alkanes. The simplest example is methane (shown below). When hydrocarbons have one or more double bonds, they are called alkenes. The simplest alkene is Ethene (C H ) which contains a double bond between the two carbon atoms. Ex. Methane on left, Ethene on right Functional groups are atoms connected to carbon chains or rings of organic molecules. Compounds that are within a functional group tend to have similar properties and characteristics. Two common functional groups are hydroxyl groups and carboxyl groups. Hydroxyl groups end in -OH and are alcohols. Carboxyl groups end in -COOH, making compounds containing -COOH carboxylic acids. Functional groups also help with nomenclature by using prefixes to help name the compounds that have similar chemical properties. Ex. Hydroxyl Group on top; Carboxyl Group on bottom 1. b and c. 2. Propane. 3. a. carboxyl group, b. hydroxyl group, c. carboxyl group. 4. a. C H O , b. C H , c. H O. 5. Methylbutane, C H | 9,274 | 2,792 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.13%3A_Organic_Compounds-Some_Additional_Classes |
Although a tremendous number of different hydrocarbons occur as a result of carbon’s ability to bond in long chains, an even greater variety of substances is possible when oxygen, nitrogen, and several other elements with carbon and hydrogen. The presence of highly electronegative atoms like oxygen or nitrogen in combination with hydrogen permits hydrogen bonds to form between molecules of many of these substances. We have deferred discussion of their properties until now so that you can apply your knowledge of hydrogen bonding to them. Elsewhere, we that alkanes were relatively unreactive and that the presence of a double or triple bond made unsaturated molecules more likely to combine chemically. A site which makes an organic molecule more reactive than a simple hydrocarbon chain is called a . Many of the important organic functional groups involve oxygen atoms, nitrogen atoms, or both. We will discuss substances containing some of these functional groups in this section. In addition to having different shapes, organic molecules can combine carbon backbones with other atoms, such as oxygen, nitrogen, or sulfur, to create functional groups. The specific arrangement of these atoms cause molecules to have certain properties. For instance, carboxylic acids always end up having a low pH. Chemists have long studied different compounds in nature, and many of the names for functional groups came about before we even understood what combination of atoms created their function. In organic chemistry, putting a dash before something like “-OH” means the O is bonded to the carbon structure. An R or R’ is used to denote “more of the carbon structure goes here.” See the example pictures. In the following pages of boxes. -OH attached to a carbon with only carbons and hydrogens
otherwise attached to it. -ol -NH , -NHR, or -NR attached to a carbon with only carbons and hydrogens otherwise attached to it. -amine =O attached to a carbon on the . -al =O attached to a carbon in the -one =O attached to a carbon that also has an alcohol: -OH. Denoted: -COOH, -CO H, -C(O)OH, or: -oic acid (when acidic) -ate (when basic) =O attached to a carbon that also has an amine, -NH , -NHR, or -NR attached to it. The central structure: with one H on the N, and a carbon chain continuing on both ends is commonly called a “peptide link” or “peptide bond” -amide or peptide Thiol -S-H attached to a carbon at the end of the chain -thiol Sulfide Disulfide -S- in the middle of the chain -S-S- in the middle of the chain -sulfide -disulfide | 2,573 | 2,793 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.06%3A_Aggregate_Particles |
Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in . Properties of Liquid Solutions, Colloids, and Suspensions Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word (from the Greek , meaning “glue”) to describe these substances, as well as the words sol and gel to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosols , which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible. Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effect , whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in The glass on the left has dilute colloidal silver, the one on the right tap water. The red laser pointer beam is scattered by the Tyndall effect in the glass on the left. Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic. Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called , unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid. In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine, ). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries ( ). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin. The characteristic shape of sickled red blood cells is the result of fibrous aggregation of hemoglobin molecules inside the cell. This satellite photograph shows the Mississippi River delta from New Orleans ( ) to the Gulf of Mexico ( ). Where seawater mixes with freshwater from the Mississippi River, colloidal clay particles in the river water precipitate ( ). Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in . Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels. Emulsions are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”: Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate [CH (CH ) CO Na ], and detergents, such as sodium dodecyl sulfate [CH (CH ) OSO Na ], whose structures are as follows: When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise. A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called into the small intestine. Bile contains a variety of , detergent-like molecules that emulsify the fats. Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micelles , which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (part (a) in ). A large class of biological molecules called phospholipids consists of detergent-like molecules with a hydrophilic head and hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayers , extended sheets consisting of a double layer of molecules. As shown in part (b) in , the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution. A cell membrane is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cell is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks. A is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a are smaller and do not separate on standing. A colloid can be classified as a , a dispersion of solid particles in a liquid or solid; a , a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an , a dispersion of solid or liquid particles in a gas; or an , a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the . Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called . are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. are collections of molecules that are surrounded by a phospholipid bilayer called a and are able to reproduce themselves. How does a colloid differ from a suspension? Which has a greater effect on solvent properties, such as vapor pressure? Is homogenized milk a colloid or a suspension? Is human plasma a colloid or a suspension? Justify your answers. How would you separate the components of an emulsion of fat dispersed in an aqueous solution of sodium chloride? | 11,724 | 2,794 |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.