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1,107 | Water ($\ce{H2O}$) is a dipole. The reason why is simply because it is not symmetrical, there are more electrons on the oxygen side than on the hydrogen side, and the electronegativity of oxygen. But why isn't $\ce{H2O}$ symmetrical like $\ce{CO2}$? Why isn't $\ce{H2O}$ non-polar like $\ce{CO2}$? Does it have anything to do with the orbitals? | Yep, it has to do with the orbitals. $\ce{CO2}$ is linear, so even though the $\ce{C-O}$ bonds have individual dipole moments, the overall dipole moment is zero as these cancel out (they point in opposite directions, as shown in the diagram below). On the other hand, $\ce{H2O}$ is "bent", which means that the individual dipole moments of the bond are at an angle to each other. They add up to give a net dipole moment (shown with grey in the diagram). The colors indicate electron density, red is more dense/blue is less dense. Dipole moment is from low density to high density. OK, so why do these molecules have differing shapes? This is where orbitals come in. I'll try to explain as much as I can without going into orbitals. Carbon has an outer shell electronic configuration as $2s^22p^2$ . Out of these four electrons, two are used in $\pi$ bonds, and two in $\sigma$ bonds. If you don't know what those are, just look at it like this for now: A set of bonds between two atoms will have one and only one $\sigma$ bond, with the rest of them $\pi$ bonds. So, any single bond is made up of just a $\sigma$ bond, a double bond is made up of one $\sigma$ and one $\pi$ , and a triple bond is made up of one $\sigma$ and two $\pi$ bonds. What these types of bonds actually are can be explained if you know what an orbital is. Now, what VSEPR says is that the geometry of the molecule is only decided by the $\sigma$ bonds and lone pairs on the central atom. You count up the $\sigma$ bonds and lone pairs (lets say they add up to $x$ ), and decide the geometry based on that. The geometry is the most stable configuration of $x$ hybrid orbitals. In simple terms, if we took $x$ balloons and tied them together, the directions the balloons point in help us correspond to where the bonds and lone pairs lie: In $\ce{CO2}$ , we have two $\sigma$ bonds and two $\pi$ bonds (as each double bond has one of each type). Each bond takes up one electron from carbon, so we have no leftover electrons for forming any lone pair. Since we have two $\sigma$ bonds and 0 lone pairs, $x=2$ , giving us the structure given by the first set of balloons, which is linear. And $\ce{CO2}$ is indeed linear: Now let's take water. The central atom (Oxygen) has a valence configuration of $2s^22p^4$ , that is, 6 electrons. In water, since we have two single bonds, we have one $\sigma$ bond each (and no $\pi$ bonds). So we have total two $\sigma$ bonds. But this leaves us with $6-2=4$ unpaired valence electrons. These form two "lone pairs" (pairs of electrons which do not bond). With two lone pairs and two $\sigma$ bonds, $x=4$ . This gives us a tetrahedral structure (third in the balloon diagram). Two of the four points in the tetrahedron are occupied by the lone pairs, and two by bonds: (Note that the angle 104.5 is not the same as the angle in perfect tetrahedra, 109.25--this is due to the lone pairs repelling each other) So finally, we have the following "bent" structure for water: From the structure, as shown above, it is very easy to check if the molecule has a dipole moment. | {
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1,289 | The case of polar solvents is clear to me - we get an attraction between opposite charges. However, how do non-polar substances dissolve in non-polar solvents? How could it be explained on a molecular level? | The electron density distribution in molecules (including nonpolar ones) is not static. Therefor, as a function of time, the electron density is not uniform. Occasionally, randomly, the electron density in a molecule will shift to produce a spontaneous dipole: part of the molecule now has more electron density ($\delta^-$) and part of the molecule now has less electron density ($\delta^+$). This spontaneous dipole is transient. The electron density will shift back to negate it and then shift again to create a new spontaneous dipole. In a vacuum, this behavior would be a curiosity. In the presence of other molecules, these spontaneous dipoles have a propagating effect. If molecule A develops a spontaneous dipole, then the electron density in neighboring molecule B will by respond by developing a spontaneous (but opposite) dipole. The $\delta^+$ region of B will be close to the $\delta^-$ region of A. This new dipole in molecule B is an induced dipole. This induced dipole in B will then induce a new induced dipole in another neighboring molecule C. By this time the transient dipole in A is already fading, perhaps to be replaced by a new induced dipole from another molecule. These random, transient, but continuously propagating dipoles have attractive forces associated with them. These forces are named London dispersion forces after the physicist who proposed them. The magnitude of these forces scales with increasing surface area of the molecule. Thus, larger molecules will overall have stronger London dispersion forces than smaller molecules. Linear rod-like molecules will have stronger forces than spherical molecules (spheres having the smallest surface area to volume ratio of the 3-dimensional solids). More complex attractive forces also arise from nonpolar molecules. Benzene (and other aromatic molecules) can pi stack , which relies on the strong permanent electric quadrupole . Benzene also has a (weak but) permanent magnetic dipole due to its ring current, so some component of the attraction between benzene molecules may be magnetic and not coulombic. Or, the two attractions may be one and the same, as benzene's electric quadrupole may result from its having a magnetic dipole. As his part of the discussion is devolving into physics, I will end here. | {
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2,415 | Thought experiment Given two glasses of water, how would one detect which glass contains heavy water, and which contains potable water without using complicated laboratory equipment? Something like the way a piece of litmus paper (yes, it is laboratory equipment and easily available to the layman at that!) may be used to test whether a solution is an acid, or alkali. | There's a very simple test for $\ce{D_{2}O}$ that springs to mind - ice cubes made with heavy water sink in light water. I assume you're talking about differentiating a glass of essentially pure heavy water from, say, tap water, in which case this test should work rather well and requires equipment no more sophisticated than a freezer, ice cube tray and some glasses. | {
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2,469 | As cited in an answer to this question , the ground state electronic configuration of niobium is: $\ce{Nb: [Kr] 5s^1 4d^4}$ Why is that so? What factors stabilize this configuration, compared to the obvious $\ce{5s^2 4d^3}$ (Aufbau principle), or the otherwise possible $\ce{5s^0 4d^5}$ (half-filled shell)? | There is an explanation to this that can be generalized, which dips a little into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, as seen below: d: ↿ ↿ ↿ _ _ ↿ ↿ ↿ ↿ _ ↿ ↿ ↿ ↿ ↿ ^
OR OR |
s: ⥮ ↿ _ Energy gap (E) The top row is for the d-orbitals, which are higher in energy, and the bottom row is for the s-orbital, which is lower in energy. There is a quantifiable energy gap between the two as denoted on the side (unique for every element). As you may know, electrons like to get in the configuration that is lowest in energy. At first glance, that might suggest putting as many electrons in the s-orbital (lower energy) as possible, and then filling the rest in the d-orbital. This is known as the Aufbau principle and is widely taught in chemistry classes. It's not wrong, and works most of the time, but the story doesn't end there. There is a cost to pairing the electrons in the lower orbital, two costs actually, which I will define now: Repulsion energy: Pretty simple, the idea that e- repel, and having two of them in the same orbital will cost some energy. Normally counted as 1 C for every pair of electrons. Exchange energy: This is a little tricky, and probably the main reason this isn't taught until later in your chemistry education. Basically (due to quantum chemistry which I won't bore you with), there is a beneficial energy associated with having pairs of like energy, like spin electrons. Basically, for every pair of electrons at the same energy level (or same orbital shell in this case) and same spin (so, if you had 2 e- in the same orbital, no dice, since they have to be opposite spin), you accrue 1 K exchange energy, which is a stabilizing energy. (This is very simplified, but really "stabilizing energy" is nothing more than negative energy. I hope your thermodynamics is in good shape!) The thing with exchange (or K) energy is that you get one for every pair, so in the case: ↿ ↿ ↿ from say a p-subshell, you would get 3 K, for each pair, while from this example: ⥮ ↿ ↿ ↿ ↿ from a $\ce{d^6}$, you would get 10 K (for each unique pair, and none for the opposite spin e-) This K is quantifiable as well (and like the repulsion energy is unique for each atom). Thus, the combination of these two energies when compared to the band gap determines the state of the electron configuration. Using the example we started with: d: ↿ ↿ ↿ _ _ ↿ ↿ ↿ ↿ _ ↿ ↿ ↿ ↿ ↿ ^
s: ⥮ OR ↿ OR _ |
PE: 3K + 1C 6K + 0C 10K + 0C Energy gap (E) You can see from the example that shoving 1 e- up from the s to the d-subshell results in a loss of 1C (losing positive or "destabilizing" repulsive energy) and gaining 3K (gaining negative or "stabilizing" exchange energy). Therefore, if the sum of these two is greater than the energy gap (i.e. 3K - 1C > E) then the electron will indeed be found in the d shell in $\ce{Nb}$'s ground state. Which is indeed the case for $\ce{Nb}$. Next, lets look at perhaps exciting the second s e- up to the d-subshell. We gain 4 additional K but don't lose any C, and we must again overcome the energy gap for this electron to be found in the d-subshell. It turns out that for $\ce{Nb}$: 4K + 0C < E (remember that C is considered a negative value, which we're not losing any of), so $\ce{Nb}$ is ultimately found in the $\ce{5s^1 4d^4}$ configuration. | {
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2,509 | How do I type a simple chemical equation in Microsoft Word? I can do subscripts, but long arrows are more difficult. I can't get them to align with the text. Also, I can't figure out how to put a delta above the arrow for heat. I have tried the Chemistry add-on from Microsoft, but that does not seem to help with equations. | If you are using MS Word 2007 or newer, use the equation feature. It is designed for math but works okay for chemistry. Go to the insert tab. (For shortcut you can press Alt+= sight together) Click on the equation button on the far right. Type in your equation. Use the buttons in the ribbon to do superscripts and subscripts. Alternatively you can use _ for subscript and ^ for superscript. The default is to have letters italicized (as variables), so you will want to fix that. There are also shortcut commands to render most the common things you want. For example, underscore _ creates a subscript and a caret ^ creates a superscript Shortcut for typing subscript and superscript in MS Word 2007|2010|2013|2016 and office 365 . You have access to a wide range of arrows from a pull-down menu, but -> will give you a simple right arrow (although it is not very long). This feature on Word will also accept some (but not all) tex commands for formatting equations. To get a long arrow, click on the operator but and choose the arrow with the word "yields" written over it under common operator structures. For up arrow and down arrow showing gas liberation and precipitation use \uparrow or \downarrow followed by space Shortcut for typing arrows of chemical equation in Word 2007 and above . Click on the word "yields" and replace it with as many spaces as you need to create an arrow of whatever length you want. Shortcut for other types of arrows is. Finally, finish your equation. If you need to type above or below arrow just type "\above(text above arrow goes here)[space]".Similarly tying below arrow just type \below(test below goes here)[space]". How to type chemical equation and arrows in Word 2007 and above . For older versions of MS Word, go to the insert menu and click on the equation, which launches the Equation Editor Program (you can also find this program on your computer by searching for eqnedt.exe), which gives you the same ability to create equations. | {
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2,511 | I knew while learning about electrolysis that if the ionic compound is molten it becomes free moving ions. If that is the case, what will happen if I continued heating till it reaches the boiling point so that the ionic compound evaporates? Will it still be free moving ions? Also, shouldn't the result be more efficient at electrolysis than in the liquid state due to increased mobility of ions? If not, why? | Usually not. Boiling point rarely exceeds 4-5 thousand kelvin. A typical ionic bound energy is about 5 eV. 1 eV is roughly 11 thousand kelvin, so ions in low-temperature vapors exist as molecules. When the temperature becomes enough to break ionic molecules, it is enough to strip one or two electrons from atoms, so high-temperature vapors will be plasma with electrons and positive ions. | {
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2,547 | I notice that salt solutions of $\ce{NaCl}$ and $\ce{KCl}$ are colourless while those of $\ce{CuSO4}$ and $\ce{FeSO4}$ are coloured. I got as far as figuring that it has to do with the transition metal ions, but I can't explain why the salt solution of $\ce{ZnSO4}$ is colourless even though zinc itself is one of the transition metals. | You're right--it's got to do with them being transition metals (usually). Transition metal ions form coordination complexes . Their empty $d$ orbitals accept lone pairs from other molecules (called "ligands") and form larger molecules (though we don't call them that--we call them "complexes"). When put in water, the ligand is $\ce{H2O}$, and you get complexes like $\ce{Cu[(H2O)6]^{2+}}$ (this is what you get when you add water to Copper Sulphate) . Thus, metals with no empty $d$ orbitals of comparable energy become colourless. OK, but where do coordination compounds get their color from? Well, $d$ orbitals look like this: Even though the last one looks different, these are all identical in many respects, the most important being that they have the same energy. So if an electron jumps from one orbital to another, it on't make any difference. In fact, since this is a quantum mechanical system, it (sort of) doesn't even make sense to talk about electrons jumping from here to there. All this changes when you have some ligands around. In coordination complexes, the $d$ orbitals are hybridised with the $s$ and $p$ orbitals of the next or previous shell. If we're talking about octahedral complexes, all the orbital lobes have to be aligned along the coordinate axes (since the $p$ orbitals are). This can be said to come from the fact that we're "mixing" these orbitals to get orthogonal hybrid orbitals, but that may not be that good an analogy. Now, the ligands approach along the coordinate axes since their final position will be in hybrid orbitals along the axes. So, they approach the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (as well as the $s$ and $p$ orbitals). The repulsion generated makes these orbitals of higher energy than the other three $d$ orbitals. This is called "d-orbital splitting" {*} Now, electrons may "jump" between the $d$ orbitals (the energy gap isn't so high that the jump is too improbable). Due to the energy gap, some energy is absorbed/released in the form of a photon and it just happens to fall within the visible spectrum (at least, for most transition metals, it does). This jump is called a $d-d$ transition. Using this, we can easily show why $\ce{Zn}$ salts are colourless. In $\ce{Zn}$, the $3d$ orbitals are full, so they're useless from the ligand point of view. The $4d$ orbitals are, on the other hand, empty. Ideal for forming a coordination complex with the $4s$ and $4p$, but there won't be any $d-d$ transitions as there are no electrons in $4d$ to jump around. (Remember, this is before the ligand bonds) *For tetrahedral complexes , the $d$ orbitals take no active part in the bonding, and the ligands approach between the axes. This leads to an opposite gap between the top three(known as $t_{2g}$) and bottom two($e_g$) orbitals, with the $t_{2g}$ orbitals at a higher energy. So we still get $d-d$ transitions and color, though IIRC many tetrahedral compounds do not have a visible color(the energy gap is about half that of the corresponding ocahedral gap--and may not give a photon with a visible frequency) | {
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2,564 | What is the IUPAC nomenclature for alkanes with large numbers of $\ce{C}$ atoms (more than 200)? For example, what is the IUPAC accepted name for $\ce{C_{205}H_{412}}$? | The definitive answer is, of course, in the rules established by the IUPAC’s “Commission on nomenclature in organic chemistry”. The reference you are looking for is: “Extension of Rules A-1.1 and A-2.5 Concerning Numerical Terms Used in Organic Chemical Nomenclature”, Pure Appl. Chem. , 1986 , 58 , 1693-1696 which can be found here as a PDF and here as an HTML version . In particular: NT-1.1 - The fundamental numerical terms for use in hydrocarbon names or as multiplying prefixes for simple features are given in the following list: \begin{array}{|c:c|c:c|c:c|c:c|}\hline
1&\rm \text{mono- or hen-} &10 &\rm \text{deca-}&100&\rm \text{hecta-}& 1000&\rm \text{kilia-} \\\hline
2&\rm \text{di- or do-}&20&\rm \text{icosa-}&200&\rm \text{dicta-}&2000&\rm \text{dilia-}\\\hline
3&\rm \text{tri-}& 30&\rm \text{triaconta-}&300&\rm \text{tricta-}&3000&\rm \text{trilia-} \\\hline
4&\rm \text{tetra-}&40&\rm \text{tetraconta-}&400&\rm \text{tetracta-}&4000&\rm \text{tetralia-} \\\hline
5&\rm \text{penta-}&50&\rm \text{pentaconta-}&500&\rm \text{pentacta-}&5000&\rm \text{pentalia-}\\\hline
6&\rm \text{hexa-}&60&\rm \text{hexaconta-}&600&\rm \text{hexacta-}&6000&\rm \text{hexalia-}\\\hline
7&\rm \text{hepta-}&70&\rm \text{heptaconta-}&700&\rm \text{heptacta-}&7000&\rm \text{heptalia-}\\\hline
8&\rm \text{octa-}&80&\rm \text{octaconta-}&800&\rm \text{ octacta-}&8000&\rm \text{octalia-}\\\hline
9&\rm \text{nona-}&90&\rm \text{nonaconta-}&900&\rm \text{nonacta-}&9000&\rm \text{nonalia-}\\\hline
\end{array} So, $\ce{C_205H_412}$ is the molecular formula of pentadictane. $\ce{C_7547H_15096}$ is heptatetracontapentactaheptaliane. | {
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2,606 | There are lots of videos on YouTube showing sodium, potassium, etc. exploding when dropped into water ( this , for example). I understand that when an alkali metal is exposed to water, a violent exothermic reaction occurs where a hydroxide and hydrogen gas are produced, but why and how does the sample of metal end up detonating and fragmenting? Physically speaking, how can a block of metal seemingly blast apart from the inside when the reaction occurs on the surface of the sample? The Wikipedia article on alkali metals explains this, but I still don't seem to understand how this would result in an explosion. Diagrams would be very helpful. The key of the question is how does the release of hydrogen gas and energy result in an explosion? Is there thermal runaway, and is the metal physically destabilized in some way during the reaction? | Until recently the answer was unknown, but a short time ago it was discovered that the reaction is in fact a Coulombic explosion. The rapid exchange of electrons between the sodium and the water causes the surface of the sodium droplet to become positively charged, and the ions repel each other. This behaves very like a negative surface tension, and the surface of the droplet increases in area rapidly forming a spiny, porcupine like shape as fingers of the molten metal are shot into the liquid at astonishing speed. The larger the surface area gets the faster the reaction occurs, leading to a runaway effect. The study was published in Nature Chem. (Ref.1). References: Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, "Coulomb explosion during the early stages of the reaction of alkali metals with water," Nature Chemistry 2015 , 7 , 250–254 ( https://doi.org/10.1038/nchem.2161 ). | {
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2,832 | Why do the names of most chemical elements end with -um or -ium for both primordial and synthetic elements? | To expand on @BelieveInvis's answer -- in the early 19th century, when the Royal Society was really in the swing of things, the dominant language of scholarship was still Latin. Since Latin didn't have words for the new metallic elements, new words were coined from the existing terms for the substances and given Latinate endings. From the OED's entry on -ium : The Latin names of metals were in -um, e.g. aurum, argentum, ferrum; the names of sodium, potassium, and magnesium, derived from soda, potassa or potash, and magnesia, were given by Davy in 1807, with the derivative form -ium; and although some of the later metals have received names in -um, the general form is in -ium, as in cadmium, iridium, lithium, osmium, palladium, rhodium, titanium, uranium; in conformity with which aluminum has been altered to aluminium. So, I think after that, other elements were simply given the suffix to fit the generally useful naming scheme, and then, metal names which were already in common use kept their common language names ( e.g. gold as opposed to aurum) simply by force of usage. | {
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2,881 | What is the difference between thermodynamic and kinetic stability? I'd like a basic explanation, but not too simple. For example, methane does not burn until lit -- why? | To understand the difference between kinetic and thermodynamic stability, you first have to understand potential energy surfaces , and how they are related to the state of a system. A potential energy surface is a representation of the potential energy of a system as a function of one or more of the other dimensions of a system. Most commonly, the other dimensions are spatial. Potential energy surfaces for chemical systems are usually very complex and hard to draw and visualize. Fortunately, we can make life easier by starting with simple 2-d models, and then extend that understanding to the generalized N-d case. So, we will start with the easiest type of potential energy to understand: gravitational potential energy. This is easy for us because we live on Earth and are affected by it every day. We have developed an intuitive sense that things tend to move from higher places to lower places, if given the opportunity. For example, if I show you this picture: You can guess that the rock is eventually going to roll downhill, and eventually come to rest at the bottom of the valley. However, you also intuitively know that it is not going to move unless something moves it. In other words, it needs some kinetic energy to get going. I could make it even harder for the rock to get moving by changing the surface a little bit: Now it is really obvious that the rock isn't going anywhere until it gains enough kinetic energy to overcome the little hill between the valley it is in, and the deeper valley to the right. We call the first valley a local minimum in the potential energy surface. In mathematical terms, this means that the first derivative of potential energy with respect to position is zero: $$\frac{\mathrm dE}{\mathrm dx} = 0$$ and the second derivative is positive: $$\frac{\mathrm d^2E}{\mathrm dx^2} \gt 0$$ In other words, the slope is zero and the shape is concave up (or convex). The deeper valley to the right is the global minimum (at least as far as we can tell). It has the same mathematical properties, but the magnitude of the energy is lower – the valley is deeper. If you put all of this together, (and can tolerate a little anthropomorphization ) you could say that the rock wants to get to the global minimum, but whether or not it can get there is determined by the amount of kinetic energy it has. It needs at least enough kinetic energy to overcome all of the local maxima along the path between its current local minimum and the global minimum. If it doesn't have enough kinetic energy to move out of its current position, we say that it is kinetically stable or kinetically trapped. If it has reached the global minimum, we say it is thermodynamically stable. To apply this concept to chemical systems, we have to change the potential energy that we use to describe the system. Gravitational potential energy is too weak to play much of a role at the molecular level. For large systems of reacting molecules, we instead look at one of several thermodynamic potential energies . The one we choose depends on which state variables are constant. For macroscopic chemical reactions, there is usually a constant number of particles, constant temperature, and either constant pressure or volume (NPT or NVT), and so we use the Gibbs Free Energy ( $G$ for NPT systems) or the Helmholtz Free Energy ( $A$ for NVT systems). Each of these is a thermodynamic potential under the appropriate conditions, which means that it does the same thing that gravitational potential energy does: it allows us to predict where the system will go, if it gets the opportunity to do so. For kinetic energy, we don't have to change much - the main difference between the kinetic energy of a rock on a hill and the kinetic energy of a large collection of molecules is how we measure it. For single particles, we can measure it using the velocity, but for large groups of molecules, we have to measure it using temperature. In other words, increasing the temperature increases the kinetic energy of all molecules in a system. If we can describe the thermodynamic potential energy of a system in different states, we can figure out whether a transition between two states is thermodynamically favorable – we can calculate whether the potential energy would increase, decrease, or stay the same. If we look at all accessible states and decide that the one we are in has the lowest thermodynamic potential energy, then we are in a thermodynamically stable state. In your example using methane gas, we can look at Gibbs free energy for the reactants and products and decide that the products are more thermodynamically stable than the reactants, and therefore methane gas in the presence of oxygen at 1 atm and 298 K is thermodynamically unstable . However, you would have to wait a very long time for methane to react without some outside help. The reason is that the transition states along the lowest-energy reaction path have a much higher thermodynamic potential energy than the average kinetic energy of the reactants. The reactants are kinetically trapped - or stable just because they are stuck in a local minimum. The minimum amount of energy that you would need to provide in the form of heat (a lit match) to overcome that barrier is called the activation energy . We can apply this to lots of other systems as well. One of the most famous and still extensively researched examples is glasses . Glasses are interesting because they are examples of kinetic stability in physical phases. Usually, phase changes are governed by thermodynamic stability. In glassy solids, the molecules would have a lower potential energy if they were arranged in a crystalline structure, but because they don't have the energy needed to get out of the local minimum, they are "stuck" with a liquid-like disordered structure, even though the phase is a solid. | {
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2,889 | In the documentary about Chernobyl, it was mentioned that Uranium should not come in contact with water, otherwise an explosion occurs. What is the reason for that? What kind of reaction makes it explode? | Interesting question. The documentary was almost certainly referring to the extremely hot lava generated by the nuclear meltdown at Chernobyl coming into contact with water. This lava is made of a substance colloquially known as 'corium' and is a combination of molten fuel rods, moderator, reactor walls and whatever else is melted by the incredible temperature of the reactor failure. This is as far as I know why the failure mode is called a 'meltdown', because the corium melts and pools in the bottom of the reactor vessel. In a molten pool, the fissile material may achieve uncontrolled criticality and just get hotter and hotter. If the temperature is high enough the corium may melt through the bottom of the reactor. This is really bad, because at least hypothetically the lava may melt down to the water table and cause a steam explosion. This is quite a bit like what happens if you try to extinguish an oil fire by throwing water on it (something you should never do ) - the water violently boils and expands and sprays burning oil everywhere. However in this case the burning oil is radioactive lava. This is not a chemical reaction, per se, but rather a case of extremly vigorous heating of water and associated violent expansion. Additional If you remember the Fukushima disaster, at least one of the buildings was destroyed in a hydrogen explosion. As I understand it, in the event of a meltdown, the fissile material can get hot enough to effect the thermolysis of water. (The wikipedia article of the event seems to confirm this) Inside a sealed building, the water will develop a headspace filled with hydrogen and oxygen gas. This mixture is very explosive. | {
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4,027 | I've heard that, even though according to Molecular Orbital Theory there is no chance of having nobel gases bonded to each other, it is not totally impossible . For example, under extreme conditions, Ar2 can be synthesised. So I am wondering whether a carbon can make 5 bonds if the required conditions are provided. | Carbon cannot have more then 4 double-electron bonds in reasonable conditions. However, in can form a bond with 5 or 6 atoms, like $\ce{Fe6C}$ fragment, where iron atoms form octahedron around the carbon atom. However, the sum of orders of 6 $\ce{C-Fe}$ bonds will be still 4. The situation is different if we consider exited states. Indeed, it is possible for hight excited state of carbon to be able to form 5 or 6 bonds. The resulting structure, however, will die quickly. Noble gases are able to form molecules in excited states, or, to be precise, their molecules are stable only in excited state. This kind of molecules is known as excimer (excited dimer). | {
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4,305 | Not to be confused with what is the mechanism of acid-catalyzed ring opening of epoxides . What is the correct order of regioselectivity of acid-catalyzed ring-opening of epoxides: $3^\circ$ > $2^\circ$ > $1^\circ$ or $3^\circ$ > $1^\circ$ > $2^\circ$ ? I am getting ready to teach epoxide ring-opening reactions, and I noticed that my textbook has something different to say about the regioselectivity of acid-catalyzed ring-opening than what I learned. My textbook does not agree with 15 other introductory texts I own, but it does agree with one. None of my Advanced Organic Chemistry texts discuss this reaction at all. Thus, I have no ready references to go read. Edit: My textbook is in the minority, and it is a first edition. Is it wrong, or do the other 15 texts (including some venerable ones) oversimplify the matter? What I learned Although the acid-catalyzed ring-opening of epoxides follows a mechanism with S N 2 features (inversion of stereochemistry, no carbocation rearrangements), the mechanism is not strictly a S N 2 mechanism. The transition state has more progress toward the C-LG bond breaking than an S N 2, but more progress toward the C-Nu bond forming than S N 1. There is significantly more $\delta ^+$ character on the carbon than in S N 2, but not as much as in S N 1. The transition states of the three are compared below: In a More O’Ferrall-Jencks diagram , the acid-catalyzed ring-opening of epoxides would follow a pathway between the idealized S N 2 and S N 1 pathways. Because of the significant $\delta ^+$ character on the carbon, the reaction displays regioselectivity inspired by carbocation stability (even though the carbocation does not form): the nucleophile preferentially attacks at the more hindered position (or the position that would produce the more stable carbocation if one formed). If a choice between a primary and a secondary carbon is presented, the nucleophile preferentially attacks at the secondary position. If a choice between a primary and a tertiary carbon is presented, the nucleophile preferentially attacks at the tertiary position. If a choice between a secondary and a tertiary carbon is presented, the nucleophile preferentially attacks at the tertiary position. The overall order of regioselectiveity is $3^\circ$ > $2^\circ$ > $1^\circ$ . What my textbook says My text agrees that the mechanism is somewhere in between the S N 2 and S N 1 mechanisms, but goes on to say that because it is in between, electronic factors (S N 1) do not always dominate. Steric factors (S N 2) are also important. My text says that in the comparison between primary and secondary, primary wins for steric factors. In other words, the difference between the increased stabilization of the $\delta ^+$ on secondary positions over primary positions is not large enough to overcome the decreased steric access at secondary positions. For the comparison of primary and tertiary, tertiary wins. The increased electronic stabilization at the tertiary position is enough to overcome the decreased steric access at the tertiary position. The comparison between secondary and tertiary is not directly made, but since $3^\circ$ > $1^\circ$ and $1^\circ$ > $2^\circ$ , it is implied that $3^\circ$ > $2^\circ$ . If this pattern is true, then other cyclic "onium" ions (like the bromonium ion and the mercurinium ion) should also behave this way. They don't. Typical of introductory texts, no references are provided. A Google search did not yield satisfactory results and the Wikipedia article on epoxides is less than helpful.
Since I 15 other introductory texts on my bookshelf, I consulted all of them on this reaction. The following is a summary of my findings. Only two of the texts (the one I am using and one other) describe the regioselectivity as $3^\circ$ > $1^\circ$ > $2^\circ$ . All of the other books support the other pattern, including Morrison and Boyd (which lends credence to the pattern that I learned). Books that have $3^\circ$ > $2^\circ$ > $1^\circ$ Brown, Foote, Iverson, and Anslyn Hornsback Ege Wade Bruice Smith Fessenden and Fessenden Volhardt and Schore Solomons and Fryhle Jones Baker and Engel Ouellette and Rawn Carey Morrison and Boyd Streightweiser and Heathcock Books that have $3^\circ$ > $1^\circ$ > $2^\circ$ Klein (the text I am using) McMurray I also surveyed my various Advanced Organic texts (March, Smith, Carey and Sundberg, Wyatt and Warren, Lowry and Richardson, etc.). Interestingly, none of them even mention acid-catalyzed ring-opening of epoxides (either by Brønsted or Lewis acids). I suspect that these omissions mean that this reaction 1) has difficult to predict regioselectivity (despite the predominance of introductory books that suggest otherwise), and thus 2) is synthetically useless. If #2 is true, then why is this reaction in introductory organic texts? | First part It won't decide the issue but the Organic Chemistry text by Clayden, Greeves, Warren and Wothers also mentions that the matter might not be as clear-cut as the majority of your textbooks make it seem. This might strengthen the position of the textbook you're using a bit. But again, there are no references given. Here is the relevant passage (especially the last two paragraphs): Second Part I have found the following passage on the formation of halohydrins from epoxides in the book by Smith and March (7th Edition) , chapter 10-50, page 507: Unsymmetrical epoxides are usually opened to give mixtures of regioisomers. In a typical reaction, the halogen is delivered to the less sterically hindered carbon of the epoxide. In the absence of this
structural feature, and in the absence of a directing group, relatively equal mixtures of
regioisomeric halohydrins are expected. The phenyl is such a group, and in 1-phenyl-2-
alkyl epoxides reaction with $\ce{POCl3}/\ce{DMAP}$ ($\ce{DMAP}$ = 4-dimethylaminopyridine) leads to
the chlorohydrin with the chlorine on the carbon bearing the phenyl.${}^{1231}$ When done in an
ionic liquid with $\ce{Me3SiCl}$, styrene epoxide gives 2-chloro-2-phenylethanol.${}^{1232}$ The
reaction of thionyl chloride and poly(vinylpyrrolidinone) converts epoxides to the corresponding
2-chloro-1-carbinol.${}^{1233}$ Bromine with a phenylhydrazine catalyst, however,
converts epoxides to the 1-bromo-2-carbinol.${}^{1234}$ An alkenyl group also leads to a
halohydrin with the halogen on the carbon bearing the $\ce{C=C}$ unit.${}^{1235}$ Epoxy carboxylic
acids are another example. When $\ce{NaI}$ reacts at pH 4, the major regioisomer is the 2-iodo-3-
hydroxy compound, but when $\ce{InCl3}$ is added, the major product is the 3-iodo-2-hydroxy
carboxylic acid.${}^{1236}$ References: ${}^{1231}$ Sartillo-Piscil, F.; Quinero, L.; Villegas, C.; Santacruz-Juarez, E.; de Parrodi, C.A. Tetrahedron Lett. 2002 , 43 , 15. ${}^{1232}$ Xu, L.-W.; Li, L.; Xia, C.-G.; Zhao, P.-Q. Tetrahedron Lett. 2004 , 45 , 2435. ${}^{1233}$ Tamami, B.; Ghazi, I.; Mahdavi, H. Synth. Commun. 2002 , 32 , 3725. ${}^{1234}$ Sharghi, H.; Eskandari, M.M. Synthesis 2002 , 1519. ${}^{1235}$ Ha, J.D.; Kim, S.Y.; Lee, S.J.; Kang, S.K.; Ahn, J.H.; Kim, S.S.; Choi, J.-K. Tetrahedron Lett. 2004 , 45 , 5969. ${}^{1236}$ Fringuelli, F.; Pizzo, F.; Vaccaro, L. J. Org. Chem. 2001 , 66 , 4719. Also see Concellón, J.M.; Bardales, E.;
Concellón, C.; García-Granda, S.; Díaz, M.R. J. Org. Chem. 2004 , 69 , 6923. | {
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4,390 | I was given the first structure, and then drew the other 5 resonance structures: First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same number of electrons, and no valence orbitals exceeding capacity. My reasoning is that the first structure contributes more. There's no electron deficiency on any of them, so we can disregard that. There is charge separation on the last 5 though, and in combination with this, the negative charge is on the least electronegative atoms, thus making the last 5 structures noticably higher in energy, and thus they contribute less to the true electronic image of the molecule. Hope I'm correct so far; here's my question : In the first structure, nitrogen is sp³ hybridised, but on all others it's clearly sp² hybridised. So what does that mean? It's somewhere in between, but closer to sp³-hybridised? | First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity. Yes , these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It interprets that negative formal charge on the carbon atom as implying a lone pair, since carbon can only have the negative charge if it also has a lone pair. When you add the lone pair and the charge, ChemDraw is suddenly stupid and thinks you have exceeded the octet on carbon (3 bonds + 1 explicit LP + 1 implied LP from the charge). In the first structure, nitrogen is $\mathrm{sp^3}$ hybridised, but on all others it's clearly $\mathrm{sp^2}$ hybridised. So what does that mean? It's somewhere in between, but closer to $\mathrm{sp^3}$ -hybridised? If the lone pair on a nitrogen (or on any atom)) participates in resonance, then that nitrogen (or whatever) atom must be $\mathrm{sp^2}$ -hybridized so that the the lone pair is in a $\mathrm{p}$ -like orbital to ensure appropriate symmetry for $\pi$ -overlap. Put another way, if an atom is $\mathrm{sp^2}$ -hybridized in one resonance structure, then it is $\mathrm{sp^2}$ -hybridized in all of them. Atoms that are $\mathrm{sp^2}$ -hybridized and $\mathrm{sp^3}$ -hybridized have differing geometries, which is not permitted in the resonance phenomenon. In truth, hybridization is an approximation we make to make quantum mechanics jive with molecular geometry. Geometry is real (empirically determinable) and hybridization is fictitious (not empirically determinable), but hybridization makes QM behave better conceptually and mathematically. Hybridization is also a useful predictor of chemical reactivity of various bonds and functional groups in organic chemistry. Experimentally, I would guess that the nitrogen atom is trigonal planar (or very close to it). Trigonal planar implies $\mathrm{sp^2}$ -hybridized (not the other way around). In molecular orbital theory, we do away with the need for both resonance and hybridization. This molecule would have 7 $\pi$ orbitals formed from linear combinations of the p orbitals on the 6 participating carbon atoms and the nitrogen atom. The probability density function plots of these 7 orbitals will be more complex than you might be used to, but they will suggest the same electron density and charge density as a resonance hybrid assembled from your six resonance structures. See the Wikipedia article on conjugated systems , which is not very good but will give you the idea. My reasoning is that the first structure contributes more. The five charge-separated resonance structures are more important than you think, since they imply that the five-membered ring is aromatic like the surprisingly stable cyclopentadienyl carbanion . However, you might not yet be this far along in your studies of organic chemistry. At the introductory level of understanding of resonance in organic molecules, the first structure is most important for the reasons you list. | {
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4,399 | I'm learning how to apply the VSEPR theory to Lewis structures and in my homework, I'm being asked to provide the hybridization of the central atom in each Lewis structure I've drawn. I've drawn out the Lewis structure for all the required compounds and figured out the arrangements of the electron regions, and figured out the shape of each molecule. I'm being asked to figure out the hybridization of the central atom of various molecules. I found a sample question with all the answers filled out: $\ce{NH3}$ It is $\mathrm{sp^3}$ hybridized. Where does this come from? I understand how to figure out the standard orbitals for an atom, but I'm lost with hybridization. My textbook uses $\ce{CH4}$ as an example.
Carbon has $\mathrm{2s^2 \,2p^2}$ , but in this molecule, it has four $\mathrm{sp^3}$ . I understand the purpose of four (there are four hydrogens), but where did the "3" in $\mathrm{sp^3}$ come from? How would I figure out something more complicated like $\ce{H2CO}$ ? | If you can assign the total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR, then you can always automatically assign hybridization. Hybridization was invented to make quantum mechanical bonding theories work better with known empirical geometries. If you know one, then you always know the other. Linear - $\ce{sp}$ - the hybridization of one $\ce{s}$ and one $\ce{p}$ orbital produce two hybrid orbitals oriented $180^\circ$ apart. Trigonal planar - $\ce{sp^2}$ - the hybridization of one $\ce{s}$ and two $\ce{p}$ orbitals produce three hybrid orbitals oriented $120^\circ$ from each other all in the same plane. Tetrahedral - $\ce{sp^3}$ - the hybridization of one $\ce{s}$ and three $\ce{p}$ orbitals produce four hybrid orbitals oriented toward the points of a regular tetrahedron, $109.5^\circ$ apart. Trigonal bipyramidal - $\ce{dsp^3}$ or $\ce{sp^3d}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and one $\ce{d}$ orbitals produce five hybrid orbitals oriented in this weird shape: three equatorial hybrid orbitals oriented $120^\circ$ from each other all in the same plane and two axial orbitals oriented $180^\circ$ apart, orthogonal to the equatorial orbitals. Octahedral - $\ce{d^2sp^3}$ or $\ce{sp^3d^2}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and two $\ce{d}$ orbitals produce six hybrid orbitals oriented toward the points of a regular octahedron $90^\circ$ apart. I assume you haven't learned any of the geometries above steric number 6 (since they are rare), but they each correspond to a specific hybridization also. $\ce{NH3}$ For $\ce{NH3}$, which category does it fit in above? Remember to count the lone pair as an electron domain for determining total electron geometry. Since the sample question says $\ce{NH3}$ is $\ce{sp^3}$, then $\ce{NH3}$ must be tetrahedral. Make sure you can figure out how $\ce{NH3}$ has tetrahedral electron geometry. For $\ce{H2CO}$ Start by drawing the Lewis structure. The least electronegative atom that is not a hydrogen goes in the center (unless you have been given structural arrangement). Determine the number of electron domains on the central atom. Determine the electron geometry using VSEPR. Correlate the geometry with the hybridization. Practice until you can do this quickly. | {
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4,517 | In molecular orbital theory, the fact that a bonding and antibonding molecular orbital pair have different energies is accompanied by the fact that the energy by which the bonding is lowered is less than the energy by which antibonding is raised, i.e. the stabilizing energy of each bonding interaction is less than the destabilising energy of antibonding. How is that possible if their sum has to equal the energies of the combining atomic orbitals and conservation of energy has to hold true? "Antibonding is more antibonding than bonding is bonding." For example, the fact that $\ce{He2}$ molecule is not formed can be explained from its MO diagram, which shows that the number of electrons in antibonding and bonding molecular orbitals is the same, and since the destabilizing energy of the antibonding MO is greater than the stabilising energy of bonding MO, the molecule is not formed. This is the common line of reasoning you find at most places. | Mathematical Explanation When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are: $$\begin{align}
E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\
E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}
\end{align} $$ Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$ , $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the resonance integral , and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' \gt k'$ . You'll note that $j'-k' > 0$ . This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$ . Intuitive Explanation The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the energies of the molecular orbitals get shifted upwards , which is what creates the asymmetric image that we have for these energy levels. Basically, you have two positively charged nuclei getting closer to each other. Now you have two options: Stick some electrons between them. Don't stick some electrons between them. If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly. Further Information I highly recommend the following book, from which most of the information above stems: Peter Atkins and Ronald Friedman, In Molecular Quantum Mechanics ; $5^\text{th}$ ed., Oxford University Press: Oxford, United Kingdom, 2011 (ISBN-13: 978-0199541423). | {
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4,526 | If I have an expression for the Helmholtz free energy (from statistical associating fluid theory for a polymer system), can I still find the Gibbs free energy minimum under constant $T,P$ conditions? Seems that since $\mu_i = \left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j} = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_j}$ where $A$ is the Helmholtz free energy and $G$ is the Gibbs free energy and $\mu_i$ is the chemical potential of component $i$. If I can just find the expressions for $\mu_i=\left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j}$ I should be able to use this in the expression for $G = \sum_i \mu_i n_i$ and find its minimum, without having to worry about an equation of state to relate change in volume ($dV$) with respect to constant pressure ($P$)? Edit I should clarify that I wonder about the Gibbs free energy because I'm interested in the constant $T,P$ (isothermal, isobaric case), and in the case especially for liquids. If you needed to introduce an equation of state to parameterize the change in V as a function of $T,P$ this would not be too difficult for an ideal gas, but for liquids it would be a challenge. So in this case, I wonder how to handle the $PV$ term (for constant $P$) in the Helmholtz free energy which does not appear in the Gibbs. | Mathematical Explanation When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are: $$\begin{align}
E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\
E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}
\end{align} $$ Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$ , $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the resonance integral , and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' \gt k'$ . You'll note that $j'-k' > 0$ . This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$ . Intuitive Explanation The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the energies of the molecular orbitals get shifted upwards , which is what creates the asymmetric image that we have for these energy levels. Basically, you have two positively charged nuclei getting closer to each other. Now you have two options: Stick some electrons between them. Don't stick some electrons between them. If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly. Further Information I highly recommend the following book, from which most of the information above stems: Peter Atkins and Ronald Friedman, In Molecular Quantum Mechanics ; $5^\text{th}$ ed., Oxford University Press: Oxford, United Kingdom, 2011 (ISBN-13: 978-0199541423). | {
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4,667 | Why do transition metals element make colored compounds both in solid form and in solution? Is it related to their electrons? | You are absolutely correct, it all about the metal's electrons and also about their d orbitals. Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level. However when the metal starts bonding with other ligands, this changes. Due to the different symmetries of the d orbitals and the inductive effects of the ligands on the electrons, the d orbitals split apart and become non-degenerate (have different energy levels). This forms the basis of Crystal Field Theory . How these d orbitals split depend on the geometry of the compound that is formed. For example if an octahedral metal complex is formed, the energy of the d orbitals will look like this: As you can see, previously the d orbitals were of the same energy, but now 2 of the orbitals are higher in energy. Now what does this have to do with its colour? Well, electrons are able to absorb certain frequencies of electromagnetic radiation to get promoted to higher energy orbitals. These frequencies have a certain energy which correspond to the energy difference between different orbitals. Now most substances are only able to absorb frequencies of radiation which are outside the visible light spectrum, for example they might be able to absorb radiation which has a frequency of $300$GhZ (that is infrared radiation). This means that it reflects all other types of radiation, including the full spectrum of visible light. So our eyes see a mixture of all the colours; red, green, blue, violet, etc. This is seen as white (this is why several organic compounds are white). However transition metals are special in that the energy difference between the non-degenerate d orbitals correspond to the energy of radiation of the visible light spectrum. This means that when we look at the metal complex, we don't see the entire visible light spectrum, but only a part of it. So for example, if the electrons in an octahedral metal complex are able to absorb green light and get promoted from the $d_{yz}$ orbital to the $d_{z^2}$ orbital, the compound will reflect all other colours except for green. Therefore by using the colour wheel, we can find the complementary colour of green which will be the colour of the compound, which is magneta. This explains why not all transition metal complexes are colourful. For example copper sulfate is a bright blue compound, however zinc sulfate on the hand is a white compound despite being a transition metal. The reason behind this is because zinc's d orbitals are completely filled up with electrons, meaning that it is not possible for any electron to make a d-> d transition as they are all filled up. Hence you might sometimes see zinc referred as not being a transition metal. | {
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4,760 | Both $\ce{SF6}$ and $\ce{SH6}$ and $\ce{SF4}$ and $\ce{SH4}$ have the same central atom and the same hybridization, but my teacher specifically mentioned that $\ce{SH6}$ and $\ce{SH4}$ don't exist. I've looked everywhere but I can't figure out why? I'd appreciate some insight into the problem. | TL;DR Fluorine is electronegative and can support the extra negative charge that is dispersed on the six X atoms in $\ce{SX6}$ , whereas hydrogen cannot. First, let's debunk a commonly taught myth, which is that the bonding in $\ce{SF6}$ involves promotion of electrons to the 3d orbitals with a resulting $\mathrm{sp^3d^2}$ hybridisation. This is not true. Here's a recent and arguably more understandable reference: J. Chem. Educ. 2020, 97 (10), 3638–3646 which explains this. Quoting: The natural ionicity, $i_\ce{SF}$ , of each $\ce{S-F}$ bond [in $\ce{SF6}$ ] is 0.86, indicating a rather ionic σ bond. Each fluorine has an average charge of $−0.45$ , resulting in a sulfur center of charge $+2.69$ . [...] In summary, the electronic structure of this system is best described as a sulfur center with a charge somewhere between $2+$ and $3+$ ; the corresponding negative charge is distributed among the equivalent fluorine atoms. Shown in Figure 12 is the orbital occupation of the sulfur center, $\ce{3s^1 3p^{2.1} 3d^{0.19} 5p^{0.03} 4f^{0.01}}$ . The minimal occupation of d-type orbitals eliminates the possibility of $\mathrm{sp^3d^2}$ hybridization. If not via d-orbital bonding, how does one then describe the structure of $\ce{SF6}$ ? I'll present an LCAO-MO answer. Here's a "simple" MO diagram (I won't go through the details of how to construct it). It's actually fairly similar to that of an octahedral transition metal complex, except that here the 3s and 3p orbitals on sulfur are below the 3d orbitals. Just for the sake of counting electrons, I treated the compound as being "fully ionic", i.e. $\ce{S^6+} + 6\ce{F-}$ . So sulfur started off with 0 valence electrons, and each fluorine started off with 2 electrons in its σ orbitals. I've also neglected the π contribution to bonding, so the fluorine lone pairs don't appear in the diagram. You'll see that, for a total of six $\ce{S-F}$ bonds, we only have four pairs of electrons in bonding MOs. The other two pairs of electrons reside in the $\mathrm{e_g}$ MOs, which are nonbonding and localised on fluorine. If we want to assign a formal charge to sulfur based on this diagram, it would be +2, because there are only actually four bonds. We could perhaps use Lewis diagrams to represent it this way: The "hypervalent" resonance form contributes rather little and does not rely on invoking d-orbital participation; see Martin's comment on my answer below for greater detail about the resonance contributions. I am guessing that its existence can be mostly attributed to negative hyperconjugation , although I'm not 100% sure on this. The trans and cis resonance forms are not equal, so their contribution is not the same, but the contribution from each individual trans resonance form has to be the same by symmetry. Overall, the six fluorines in $\ce{SF6}$ have to be equivalent by the octahedral symmetry of the molecule. You could run a $\ce{^19F}$ NMR of the compound and it should only give you one peak. (An alternative way of looking at it is that two of the $\ce{S-F}$ bonds are "true" 2c2e bonds, and that the other four $\ce{S-F}$ "bonds" are in fact just a couple of 3c4e bonds, but I won't go into that. For more information on multi-centre bonds, this article is a nice introduction: J. Chem. Educ. 1998, 75, 910 ; see also refs. 12 and 13 in that article.) Right from the outset, we can see why $\ce{SH6}$ is not favoured as much. If we use the same framework to describe the bonding in $\ce{SH6}$ , then those "correct" resonance forms that we drew would involve $\ce{H-}$ . I'll leave it to the reader to figure out whether $\ce{F-}$ or $\ce{H-}$ is more stable. Alternatively, if you want to stick to the MO description, the idea is that in $\ce{SH6}$ , the relatively high energy of H1s compared to F2p will lead to the nonbonding $\mathrm{e_g}$ orbitals being relatively higher in energy. All things being equal, it's less favourable for a higher-energy orbital to be occupied, and $\ce{SH6}$ would therefore be very prone to losing these electrons, i.e. being oxidised. In fact, if we do remove those four electrons from the $\mathrm{e_g}$ orbitals, then it's possible that these six-coordinate hydrides could form. But obviously we might not want to have a $\ce{SH6^4+}$ molecule on the loose. It'll probably lose all of its protons in a hurry to get back to being $\ce{H2S + 4H+}$ . Is there anything better? Well, there's the species $\ce{CH6^2+}$ , which is methane protonated twice. It's valence isoelectronic with $\ce{SH6^4+}$ , and if you want to read about it, here's an article: J. Am. Chem. Soc. 1983, 105, 5258 . While it's hardly the most stable molecule on the planet, it's certainly more plausible than $\ce{SH6}$ . Now, just to come back to where we started from: d-orbital participation. Yes, there is an $\mathrm{e_g}$ set of d orbitals that can overlap with the apparently "nonbonding" $\mathrm{e_g}$ linear combination of F2p orbitals, thereby stabilising it. It is true that some degree of this does happen. The issue is how much . Considering the fairly large energy gap between the $\mathrm{e_g}$ orbitals, this interaction is bound to be fairly small, and is nowhere near enough to justify a $\mathrm{sp^3d^2}$ description of it; Martin's comments contain more details. | {
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4,877 | In the TV show "Breaking Bad", Walter White frequently gets rid of people who get in his way by submerging them in a plastic container full of hydrofluoric acid. This, at least in the TV show, completely dissolves the body leaving nothing but a red sludge behind at the end. Is it actually possible to dispose of a body with hydrofluoric acid? If hydrofluoric acid wouldn't work, are there any acids corrosive enough to achieve the stated effect from the show? | Hydrofluoric acid is toxic and corrosive, but actually isn't that strong of an acid compared to other hydrohalic acids; the fluorine has a very good orbital overlap with hydrogen and is also not very polarizable, therefore it resists donating its proton, unlike other hydrohalic acids which are good proton donators. It will break down some tissues, but it will take a relatively long time and won't turn the entire body into stuff that can be rinsed down the drain. Hydrochloric acid is a much stronger acid, and as it has several uses from pH-balancing pool water to preparing concrete surfaces, it's available by the gallon from any hardware store. However, it isn't very good at dissolving bodies either; while it will eventually work by breaking down the connective tissues, it will make a huge stink and take several days to dissolve certain types of tissues and bones. The standard body-dissolving chemical is lye aka sodium hydroxide. The main source is drain clog remover because most drain clogs are formed by hair and other bio-gunk that accumulates naturally when humans shower, exfoliate etc. It works, even though the body's overall chemistry is slightly to the basic side of neutral (about 7.35-7.4) because the hydroxide anion is a strong proton acceptor. That means that it strips hydrogen atoms off of organic molecules to form water (alkaline hydrolysis, aka saponification), and as a result, those organic molecules are turned into simpler molecules with lower melting points (triglycerides are turned into fatty acids, saturated fats are dehydrogenated to form unsaturated fats, alkanes become alcohols, etc). Sodium hydroxide is also a ready source of the sodium ion; sodium salts are always water-soluble (at least I can't think of a single one that isn't). The resulting compounds are thus either liquids or water-soluble alcohols and salts, which flush down the drain. What's left is the brittle, insoluble calcium "shell" of the skeleton; if hydrolyzed by sodium hydroxide, the resulting calcium hydroxide ("slaked lime") won't dissolve completely but is relatively easy to clean up. | {
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5,477 | I was learning about voltaic cells and came across salt bridges. If the purpose of the salt bridge is only to move electrons from an electrolyte solution to the other, then why can I not use a wire? Also, will using $\ce{NaCl}$ instead of $\ce{KNO3}$ in making the salt bridge have any effects on voltage/current output of the cell? why? Plus if it matters, I'm using a Zinc-Copper voltaic cell with a tissue paper soaked in $\ce{KNO3}$ as salt bridge | There's another question related to salt bridges on this site. The purpose of a salt bridge is not to move electrons from the electrolyte, rather it's to maintain charge balance because the electrons are moving from one-half cell to the other. The electrons flow from the anode to the cathode. The oxidation reaction that occurs at the anode generates electrons and positively charged ions. The electrons move through the wire (and your device, which I haven't included in the diagram), leaving the unbalanced positive charge in this vessel. In order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half cell. A similar (but reversed) situation is found in the cathodic cell, where $\ce{Cu^{2+}}$ ions are being consumed, and therefore electroneutrality is maintained by the migration of $\ce{K+}$ ions from the salt bridge into this half cell. Regarding the second part of your question, it is important to use a salt with inert ions in your salt bridge. In your case, you probably won't notice a difference between $\ce{NaCl}$ and $\ce{KNO3}$ since the $\ce{Cu^{2+}}$ and $\ce{Zn^{2+}}$ salts of $\ce{Cl-}$ and $\ce{NO3-}$ are soluble. There will be a difference in the liquid junction potential , but that topic is a bit advanced for someone just starting out with voltaic/galvanic cells. | {
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5,503 | I'm not satisfied with the rationale for the intermolecular attraction known as hydrogen bonding. In my book, it states that Hydrogen bonding is a special type of intermolecular attraction between the hydrogen atom in a polar bond (particularly H ¬ F, H ¬ O, and H ¬ N) and non-bonding electron pair on a nearby small electronegative ion or atom usually F, O, or N (in another molecule). It seems that chemists looked at the data and found they needed a 'fudge factor' to fit the higher boiling points of ammonia, water, and hydrogen fluoride. Why doesn't hydrogen bonding apply to other atoms like Sulfur or Chlorine? They seem to be electronegative enough IMO. Is there an explanation that explains why this is, rather than saying that it's there? What makes Hydrogen so special? Note: Chlorine is more electronegative than Nitrogen. | There are two main ways to look at hydrogen bonding. The first is electrostatic, where the electronegativity of the atoms is used to describe the interaction. Your argument about chlorine being more electronegative than nitrogen is a good one suggesting that the electrostatic argument is only part of the story, and there is at least one study that suggests that hydrogen bonding does occur with chlorine in polar molecules such as chloroform. We can increase dramatically the number of atoms considered to undergo hydrogen bonding if we take a molecular orbital approach. My reasoning here is a summary of what can be found in Inorganic Chemistry by Miessler, Fischer and Tarr. They, in turn, rely heavily on a new definition of hydrogen bonding recommended by the IUPAC Physical and Biophysical Chemistry Division. A hydrogen bond is formed when an $\ce{X-H}$ (where X is more electronegative than H) interacts with a donor atom, $\ce{B}$. The attraction $\ce{X-H...B}$ can be described as consisting of the following components: An electrostatic contribution based on the polarity of $\ce{X-H}$ A partial covalent character which arises from the donor-acceptor nature of the interaction Dispersion forces The first bullet is typically the only phenomenon discussed in General Chemistry classes, and this is not unreasonable since the second bullet requires the introduction of Lewis Acid/Base concepts which may not have been covered. It is important to note that the new definition includes what I like to call "the proof is in the pudding" where the existence of hydrogen bonding requires experimental evidence, which can be found using a number of methods: The $\ce{X-H...B}$ bond angle: a bond angle of 180 degrees indicates strong hydrogen bonding and would be accompanied by shord $\ce{H...B}$ bond distances. A red shift in the IR frequency upon formation of $\ce{X-H...B}$ Hydrogen bonding results in deshielding of the H atom, which can be observed in NMR spectroscopy. Thermodynamic studies should indicate that $\Delta G$ for the formation of $\ce{X-H...B}$ should be larger than the thermal energy of the system. While electrostatics is the dominant contributor to hydrogen bonding, an analysis of frontier orbitals can also be insightful. A thorough answer requires a fair amount of MO theory background, but for those interested, looking at the MO diagram of $\ce{FHF-}$ will be helpful. In brief, the base ($\ce{F-}$ in this case) has a filled $p_z$ orbital that gains access to a delocalized orbital with lower energy, thus validating the formation of a strong hydrogen bond. | {
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5,568 | In High School I learned that an exothermic reactions releases energy, while an endothermic reaction needs energy to occur. Now I learned that there is a separate, somewhat similar classification scheme of exergonic and endergonic reactions. What is the difference between these two classification schemes? Are exothermic reactions always exergonic, and if not, can you give me an example? | The classifications endothermic and exothermic refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$. If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you're going to use reaction enthalpy $\Delta{}_{\mathrm{R}}H$. Then there are three cases to distinguish: $\Delta{}_{\mathrm{R}}H < 0$, an exothermic reaction that releases heat to the surroundings (temperature increases) $\Delta{}_{\mathrm{R}}H = 0$, no net exchange of heat $\Delta{}_{\mathrm{R}}H > 0$, an endothermic reaction that absorbs heat from the surroundings (temperature decreases) In 1876, Thomson and Berthelot described this driving force in a principle regarding affinities of reactions. According to them, only exothermic reactions were possible. Yet how would you explain, for example, wet cloths being suspended on a cloth-line -- dry, even during cold winter? Thanks to works by von Helmholtz , van't Hoff , Boltzmann (and others) we may do. Entropy $S$, depending on the number of accessible realisations of the reactants ("describing the degree of order") necessarily is to be taken into account, too. These two contribute to the maximum work a reaction may produce, described by the Gibbs free energy $G$. This is of particular importance considering reactions with gases, because the number of accessible realisations of the reactants ("degree or order") may change ($\Delta_\mathrm{R} S$ may be large). For a given reaction, the change in reaction Gibbs free energy is $\Delta{}_{\mathrm{R}}G = \Delta{}_{\mathrm{R}}H - T\Delta{}_{\mathrm{}R}S$. Then there are three cases to distinguish: $\Delta{}_{\mathrm{R}}G < 0$, an exergonic reaction, "running voluntarily" from the left to the right side of the reaction equation (react is spontaneous as written) $\Delta{}_{\mathrm{R}}G = 0$, the state of thermodynamic equilibrium, i.e. on a macroscopic level, there is no net reaction or $\Delta{}_{\mathrm{R}}G > 0$, an endergonic reaction, which either needs energy input from outside to run from the left to the right side of the reaction equation or otherwise runs backwards, from the right to the left side (reaction is spontaneous in the reverse direction) Reactions may be classified according to reaction enthalpy, reaction entropy, free reaction enthalpy -- even simultaneously -- always favouring an exergonic reaction: Example, combustion of propane with oxygen, $\ce{5 O2 + C3H8 -> 4H2O + 3CO2}$. Since both heat dissipation ($\Delta_{\mathrm{R}}H < 0$, exothermic) and increase of the number of particles ($\Delta_{\mathrm{R}}S > 0$) favour the reaction, it is an exergonic reaction ($\Delta_{\mathrm{R}}G < 0$). Example, reaction of dioxygen to ozone, $\ce{3 O2 -> 2 O3}$. This is an endergonic reaction ($\Delta_{\mathrm{R}}G > 0$), because the number of molecules decreases ($\Delta_{\mathrm{R}}S < 0$) and simultaneously it is endothermic ($\Delta_{\mathrm{R}}H > 0$), too. Water gas reaction , where water vapour is guided over solid carbon $\ce{H2O + C <=> CO + H2}$. Only at temperatures $T$ yielding an entropic contribution $T \cdot \Delta_{\mathrm{R}}S > \Delta_{\mathrm{R}}H$, an endothermic reaction may become exergonic. Reaction of hydrogen and oxygen to yield water vapour, $\ce{2 H2 + O2 -> 2 H2O}$. This is an exothermic reaction ($\Delta_{\mathrm{R}}H < 0$) with decreasing number of particles ($\Delta_{\mathrm{R}}S < 0$). Only at temperatures at or below $T$ with $|T \cdot \Delta_{\mathrm{R}}S| < |\Delta_{\mathrm{R}}H|$ there is a macroscopic reaction. In other words, while the reaction works fine at room temperature, at high temperatures (e.g. 6000 K), this reaction does not run. After all, please keep in mind this is about thermodynamics, and not kinetics. There are also indications of spontaneity of a reaction. | {
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5,690 | In nucleophilic aromatic substitution reactions, why do fluorides react faster than bromides? Ordinarily bromide is a better leaving group than fluoride, e.g. in $\mathrm{S_N2}$ reactions, so why isn't this the case here? The only thing I can think of is that fluorine is more electron-withdrawing (via the inductive effect), which could stabilise the Meisenheimer complex formed as an intermediate. | The key point to understanding why fluorides are so reactive in the nucleophilic aromatic substitution (I will call it S N Ar in the following) is knowing the rate determining step of the reaction mechanism. The mechanism is as shown in the following picture (Nu = Nucleophile, X = leaving group): Now, the first step (= addition) is very slow as aromaticity is lost and thus the energy barrier is very high. The second step (= elimination of the leaving group) is quite fast as aromaticity is restored. So, since the elimination step is fast compared to the addition step, the actual quality of the leaving group is not very important, because even if you use a very good leaving group (e.g. iodine), which speeds up the elimination step, the overall reaction rate will not increase as the addition step is the bottleneck of the reaction. Now, what about fluorine? Fluorine is not a good leaving group, but that doesn't matter as I said before. It is not the leaving group ability of $\ce{F-}$ which makes the reaction go faster than with, say, bromine or chlorine, but its very high negative inductive effect (due to its large electronegativity). This negative inductive effect helps to stabilize the negative charge in the Meisenheimer complex and thus lowers the activation barrier of the (slow) addition step.
Since this step is the bottleneck of the overall reaction, a speedup here, speeds up the whole reaction. Leaving groups with a negative mesomeric effect (but little negative inductive effect) are not good at stabilizing the Meisenheimer complex, because the negative charge can't be delocalized to them. | {
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5,708 | It's there in every high-school textbook: $\ce{O2}$ is the supporter of combustion. Without $\ce{O2}$ combustion cannot take place. Why? And why only $\ce{O2}$? Why not some other element? And, what happens when a combustible gas burns in air? Say $\ce{H2}$? | Oxygen is not strictly needed, it just happens to be very, very good at what it does! Specifically, combustion requires three things: 1) Fuel : The thing that burns. This is is often a hydrocarbon, or other organic molecule. The simplest possible fuel is pure hydrogen gas. 2) Energy : What gets the reaction started, which is true of most chemical reactions. Note that this so-called activation energy is usually much less than the energy ultimately released from combustion. This is like rolling a boulder some distance in order to roll it down a hill, which releases much potential energy. 3) Oxidizer : The molecule that accepts electrons. It turns out that combustion requires the fuel to be oxidized , that is, it donates electrons. So we need something to accept the electrons, and that's the oxidizer, which is then reduced . Oxygen is a great oxidizer because it is so electronegative , which means it really wants to accept electrons. Only fluorine is more electronegative, and fluorine is a superb oxidizer: blow fluorine gas at nearly any substance and it bursts into flames. For example, see this video: Fluorine . Oxygen is in many ways a perfect oxidizer for supporting life: it doesn't set things on fire like fluorine, but otherwise can oxidize very many things. When you eat, oxygen is oxidizing the food (fuel) you ingested, to generate energy you need to live. So there is "combustion" going on inside you! | {
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5,737 | A friend of mine was looking over the definition of pH and was wondering if it is possible to have a negative pH. From the equation below, it certainly seems mathematically possible—if you have a $1.1$ (or something $\gt 1$ ) molar solution of $\ce{H+}$ ions: $$\text{pH} = -\log([\ce{H+}])$$ (Where $[\ce{X}]$ denotes the concentration of $\ce{X}$ in $\frac{\text{mol}}{\text{L}}$ .) If $[\ce{H+}] = 1.1\ \frac{\text{mol}}{\text{L}}$ , then $\mathrm{pH} = -\log(1.1) \approx -0.095 $ So, it is theoretically possible to create a substance with a negative pH. But, is it physically possible (e.g. can we create a 1.1 molar acid in the lab that actually still behaves consistently with that equation)? | One publication for you: “Negative pH Does Exist”, K. F. Lim, J. Chem. Educ. 2006 , 83 , 1465 . Quoting the abstract in full: The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books. The article text provides some counterexamples: For example, commercially available concentrated HCl solution (37% by mass) has $\mathrm{pH} \approx -1.1$, while saturated NaOH solution has $\mathrm{pH} \approx 15.0$. | {
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5,811 | The following equation is standard in thermodynamics: $$
\Delta G^\circ=-RT\log(K)
$$ where $K$ is the equilibrium constant. In dimensional analysis, Bridgman's theorem tells us that the argument of a transcendental function (like $\log$) must always be dimensionless. But $K$ may have dimensions (depending on the particular equilibrium). Why is this OK? Note: working out dimensions explicitly gives: \begin{align}
[R]&=EN^{-1}\Theta^{-1}\\
[T]&=\Theta\\
[\Delta G^\circ]&=EN^{-1}
\end{align} where $\Theta$ is the dimensions of temperature, $N$ is the dimensions of number and $E$ = dimensions of energy = $MLT^{-2}$. From these, we see that the quantity $\log(K)$ ought to be dimensionless, implying that $K$ should be dimensionless as well. But it isn't always. Furthermore, suppose $K$ has dimensions $NL^{-3}$ (for an equilibrium of the form $A+B\leftrightarrow AB$, say). Suppose now that we scale the units for number by a factor $a$. Then we get new values for $K,R,\Delta G^\circ$, given by: \begin{align}
\hat K&=K/a\\
\hat R&=aR\\
\hat{\Delta G^\circ}&=a\Delta G^\circ
\end{align} From our original equation: \begin{align}
\Delta G^\circ&=-RT\log(K)\\
a\Delta G^\circ&=-aRT\log(K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(a\hat K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(\hat K)-\hat RT\log(a)
\end{align} So the new quantities do not satisfy the old equation. Rather, they satisfy the old equation, but with a constant factor of $-\hat RT\log(a)$ added on. What is going on here? | The problem is that people are often sloppy with the definition of quantities. The equilibrium constant $K$ in your first equation is indeed a dimensionless quantity while the equilibrium constant $K_c$ that is usually used to describe an equilibrium in a solution is not. I will take some detour to show where they come from and how they are connected. From thermodynamics it is known that the Gibbs free energy of reation is given by
\begin{equation}
\Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ ,
\end{equation} where $\xi$ is the extent of reaction and $\nu_{i}$ and $\mu_{i}$ are the stochiometric coefficient and the chemical potential of the $i^{\text{th}}$ component in the reaction, respectively.
Now, imagine the situation for a ideal system consisting of two phases, one purely consisting of component $i$ and the other being a mixed phase comprised of components $1, 2, \dots, k$, in equilibrium.
Since the system is in equilibrium and shows ideal behavior we know that the chemical potential of component $i$ in the mixed phase (having the temperature $T$ and the total pressure $p$), $\mu_{i}(p, T)$, must be equal to the chemical potential, $\mu^{*}_{i}(p_{i}, T)$, of the pure phase having the same temperature but a different pressure $p_{i}$, whereby $p_{i}$ is equal to the partial pressure of component $i$ in the mixed phase, namely
\begin{equation}
\mu_{i}(p, T)= \mu^{*}_{i}(p_{i}, T).
\end{equation}
From Maxwell's relations it is known that
\begin{equation}
\left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = \left( \frac{\partial}{\partial p} \biggl(\frac{\partial G^{*}}{\partial n_{i}}\biggr) \right)_{T} = \Biggl( \frac{\partial}{\partial n_{i}} \underbrace{\biggl(\frac{\partial G^{*}_{i}}{\partial p}\biggr)}_{=\, V_{i}} \Biggr)_{T} = \left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}
\end{equation} but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to
\begin{equation}
\left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}
\end{equation} and one gets
\begin{equation}
\left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = v_{i} \ ,
\end{equation} where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase.
Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get
\begin{equation}
\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,
\end{equation}
so that, introducing the mole fraction $x_{i}$,
\begin{equation}
\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .
\end{equation}
Please, note that there is a dimensionless quantity inside the logarithm.
Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$
\begin{equation}
a_{i} = \frac{\varphi_{i} p_{i}}{p^0}
\end{equation}
where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition.
When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises
\begin{equation}
\mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .
\end{equation}
Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets
\begin{equation}
\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}
where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure.
Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that
\begin{equation}
\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}
So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant
\begin{equation}
K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}
which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via
\begin{equation}
K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
it is more or less proportional to $K$ and thus gives qualitatively the same information. Edit: If the solution at hand behaves like an ideal solution then by definition it's activity/fugacity coefficient is equal to one. Furthermore the state of ideality is defined with respect to standard states: for an ideal solution this is $c^{\ominus} = 1 \, \text{mol}/\text{L}$. Using this together with the ideal gas law on the relation between $K$ and $K_{c}$
\begin{equation}
K = \prod_i [\underbrace{\varphi_{i}}_{= \, 1}]^{\nu_{i}} \Bigl(\underbrace{\frac{RT}{p^{0}}}_{\substack{= \, 1/c^{\ominus} \, = \, 1 \, \text{L} / \text{mol} \\ \text{per definition}}}\Bigr)^{\sum_i \nu_{i}} K_{c} \qquad \Rightarrow \qquad K = \left(\frac{L}{\text{mol}} \right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
one sees that for an ideal solution $K_{c}$ is identical to $K$ scaled by a dimensional prefactor. Edit: I forgot to mention there are also "versions" of the equilibrium constants that are defined in terms of partial pressures or mole fractions which provide a more suitable description for gas equlibria but all those "versions" can be traced back to the original equilibrium constant. | {
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5,844 | It seems that every website on sexual health advises against using oil-based lubricants with condoms. It is claimed that "oil breaks down latex". One source claimed that a latex condom completely breaks down in only 60 seconds. It made me curious, so I made an experiment. I took a piece of rubber latex condom and soaked it into regular canola oil I found in the kitchen. I checked the condom after 1 minute, nothing changed. So I let it soak for about 5 minutes more, and then 5 hours more, still nothing. It was able to hold a large amount of water without leaking or breaking. So I am wondering, is it really true that oil degrades latex? What sort of chemical reaction is supposed to happen? What properties of the latex material and the oil influence this reaction? | First off, may I say that I applaud your decision to test this through an experiment. It's rare to see that than I would like. Now, on to the matter at hand. It's fairly well known from industrial chemistry that non-polar solvents degrade latex quite heavily. I work with latex seals a lot, and the hexanes we use routinely break the seals down in under a day. Of course, if you're lubricating your condoms with hexanes, you're a) an idiot or b) absolutely insane. A paper I managed to find suggests that there really isn't too much direct data on condoms , and it muses that the warnings might have arisen from industry, where nonpolar solvents decidedly do degrade latex. To find out, they did a burst experiment with condoms that had been treated with various oils. Glycerol and Vaseline-treated condoms showed a very, very minor decrease in strength, while mineral oil/baby oil-treated ones burst at less than 10% of the volume of an untreated condom. They also found that 10 month-old condoms have half the burst volume of 1-month old ones, so you could argue that using 1-month-old condoms that have been slathered in Vaseline is still much safer than using older ones. As for the actual chemistry of the weakening, I honestly don't know. If I were to hazard a guess, I would note that the latex looks like a bunch of ethylenes glued together, so my guess would be that the solvents get between the chains and force them apart, weakening them. For this to happen, the solvent must be nonpolar, but still small enough to slip between the chains of the polymer. That's probably why vaseline and canola oil don't have much of an effect---they're just too big to fit between the chains. Again though, I don't know for sure, so don't quote me on this last paragraph. | {
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5,925 | Recently the quack medicine folks online have been promoting the research of a certain Dr. Gerald Pollack who claims to have discovered a "forth phase of water", and who has recently published a book on the said topic. Although it's clear that his research is being deliberately misinterpreted, it's not clear to me whether or not his own claims are valid in the first place. Although some papers on the topic were published in peer reviewed journals, it doesn't seem like anyone else in the scientific community has acknowledged or replicated his results. So, is "EZ-water" a real breakthrough or just the brainchild of a deluded scientist? | EZ-water is not a breakthrough. It is not new, nor is it valid. This appears to be one of the many claims about the healthful benefits of drinking " ionized water ". The original "article" linked, contains a number of misleading statements. Water molecules make up 99% of your body. This is true, but misleading. According to various sources , the human body is between 57% and 75% water by mass. Since water has a much lower molar mass than most biomolecules, 99% of the molecules in your body may well be water. The journal listed in which the research is published is Water , which is not a high impact journal, despite its name. Water had an impact factor of 0.973 in 2012. That means that articles published in 2011 in Water were cited less than 1 time each by anyone in 2012. Compare with Cell , which had in impact factor of 31.957 in the same period - each article in Cell is cited an average of nearly 32 times. Water has a "fourth phase" and no single boiling point or melting point. This is true! Any student of chemistry should be able to tell you this. Take a look at the phase diagram of water below. Water apparently has at least 18 phases (15 solid phases, plus liquid, vapor, and supercritical fluid). You can also see that the solid-liquid and liquid-vapor phase boundaries are not single points. They are curves varying in temperature and pressure. Those curves represent the conditions in which the two phases have the same vapor pressure . This is not a new discovery or a startling secret. This behavior has been known for many decades (if not centuries). Later on the "article" tells us that EZ-water is not $\ce{H2O}$ (note that they cannot be bothered to format their chemical formulas) but $\ce{H3O2}$ , which is alkaline with a negative charge. Hmmm... If it has a different formula and different properties, it must be a different compound. Note that this "formula" is essentially $\ce{H3O2 = H2O + OH}$ , and hints at what happens inside these water ionizers. EZ-water has a negative charge. It is possible for objects to build up charge, but it is not possible for them to remain permanently charged. Overall, the universe is balanced in charge. Consider lightning as a consequence of attempting to maintain separation of charge. This "negative charge" comes from a misunderstanding of pH. The water definitely has a higher pH, which means more $\ce{OH-}$ ions than $\ce{H+}$ ions, but something (probably $\ce{Na+}$ ) needs to replace the missing $\ce{H+}$ . I can go on, but you get the point. A water ionizer electrolyzes water, which requires electricity (not light as suggested by Mercola). I had the good fortune to be involved in the testing and dismantling of one of these ionizers recently. The device claimed to take ordinary tap water and convert it to a "super-cleaning solution that contained only negatively charged oxygen particles" or something like that. From experiments: The ionizer increases the pH of the water. The ionizer does not work with deionized or distilled water! The ionizer does not produce measurable oxygen gas (a smoldering splint will not relight in the vicinity) The device contains an electrolysis cell and an ion exchange resin Here is what happens: Water is electrolyzed. Doing so requires ions in the water already to conduct the electricity, which is why the device must use tap water (and not purified water). At the cathode, water is reduced to hydrogen and hydroxide. At the anode, it is oxidized to oxygen and protons. \begin{align}
\ce{2H2O +2e- &-> H2 + 2OH-}\\
\ce{2H2O &-> O2 +4H+ + 4e-}
\end{align} Protons are exchanged for sodium ions To prevent the recombination of hydroxide and protons to form water and neutralize the solution, there is an ion exchange membrane between the anode and the rest of the solution. $$\ce{H+ + OH- -> H2O}$$ The ion exchange resin collects the protons and releases sodium ions into solution to maintain charge neutrality, which must happen or you would have a lightning gun and not a mystic device to produce super water. Ultimately, drinking this stuff would probably have no effect on your health. But we are no doctors, so don’t listen to us either. It might help or it might hurt, depending on your condition, it also might not matter. First, your stomach secretes acid, so any extra base would be neutralized immediately. Let's say an ionizer gets the water's pH to 8 (slightly basic, and more basic than your blood). This means that the concentration of protons is $\ce{[H+]}=10^{-\text{M}pH}=10^{-8} \text{ M}$ (remember that pH is a logarithmic scale . The concentration of hydroxide is $10^{-14} / 10^{-8} = 10^{-6} \text{ M}$ . The pH in your stomach ranges from 1.5 - 3.5, which represent proton concentrations between $10^{-3.5}= 3.16\times 10^{-4}$ and $10^{-1.5}=3.16\times 10^{-2}$ . If we use the higher pH stomach acid to neutralize an equal volume of EZ-water, the final pH would be: \begin{align}
[\ce{H+}]_f &= \frac{3.16\times 10^{-4} - 1\times 10^{-6}}{2}=\frac{3.15\times 10^{-4}}{2}=1.07_5 \times 10^{-4}\\
\text{pH} &=-\log[\ce{H+}] =-\log(1.07_5 \times 10^{-4})=3.97
\end{align} Since your stomach acid would likely have a greater volume than the ionized water, the change in pH would be even smaller. This calculation assumes your stomach acid is not a buffer, which it is. Second, the acid-base homeostasis of your blood prevents the pH of your body from changing. Yes, your blood pH is 7.4 or so (which is slightly basic), but that does not mean your blood is negatively charged. Outside of range of 7.38 - 7.42 all of your blood proteins (like hemoglobin) start to denature, which will kill you. Well, to be more precise, you will suffocate before that happens as hemoglobin/myoglobin oxygen binding capacity is pH sensitive. And if anything, we can tolerate acidic blood more robustly than basic blood pH (down to 6.8 or 7, but only up to about 7.6) before death. You might want to look up acidosis. | {
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6,233 | There are elements like neptunium and plutonium in the periodic table. Did their discovery have anything to do with Neptune and Pluto? Or are they randomly assigned with such names? | Neptunium (93) and plutonium (94) were so named because they followed uranium (92) on the periodic table. Uranium was named after Uranus . German chemist Martin Heinrich Klaproth discovered uranium in 1789, eight years after William Herschel discovered Uranus. At the time, uranium was the densest element known and Uranus was the farthest planet from the sun known. | {
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6,659 | I know that the mole is widely used in chemistry instead of units of mass or volume as a convenient way to express amounts of reactants or of products of chemical reactions. I'm wondering why people in chemistry still excessively use it for their measurement? To be backward-compatible and consistent with traditional textbooks? Why they don't simply express their quantity measurement per atom, per unit volume, per molecule or etc.? | Simply speaking, because it's an appropriate unit to use. Let's imagine I wanted to measure the length of a rope. What would be an appropriate length to use? Inches? Centimeters? Feet, maybe? It would really be awkward to express it as 0.000189393 miles, or as 304,800,000 nanometers. (Note: if you can't see why these units are awkward, take any page discussing things like this (e.g. biomass of certain species) and change all the units so that they're nonsense like this. Then put it away for a week and try to read it later.) Now let's say I've changed my mind and I'd like to measure all the rope created in the world in a year. Now an appropriate unit is almost certainly miles or kilometers, and not inches or centimeters. Let's consider something else: so far, I've been using length to measure ropes. Would it make sense to measure their combined mass instead? Maybe not for small amounts, since I think I would throttle you if you told me to cut off half a pound of rope, but for global-scale things, tons or metric tons may make sense. On the other hand, using measurements like the average density of the rope or the combined diameters of all the ropes really wouldn't be much use at all. What we've seen here is that when we're measuring things, there are measures that make sense (for rope, length, maybe weight) and some measures that really don't (color, average diameter, etc.). In those measures, there are units that are convenient (inches, feet), and some that aren't (nanometers). This is exactly the problem with chemical units, but much more magnified. When I measure the energy released on hydrolysis of a sample, I'm not measuring the energy of one bond, or two bonds, or a thousand bonds, or even a million. I'm measuring a collection of so insanely many molecules that there's really no number in common language for it. What is an appropriate way to measure things in these molecules? If I'm only interested in how much I have and not what's in it, per gram or per volume is often a good way to do it. Since when I'm measuring the temperature change of something, I don't particularly care what it's made of, measures of this (specific heat, for instance) are done in units of something per gram. On the other hand, if I do care what my sample is made of, then I need a better way to measure it. For example, sodium chloride and calcium chloride look similar, but will have very different energies since there's three ions in a $\ce{CaCl2}$ unit and only two in an $\ce{NaCl}$ unit. Since you can't have half a molecule, the simplest way to do it is to count the darn things. Unfortunately, measuring things per atom is a really awkward way to do things because of the problems I described above. The energy it takes to melt ice is 0.000000000000000000001029 J/molecule. What we need then, is some count of molecules that's convenient. We could go by 1000 molecules, or a million molecules, but it's a pain in the butt to convert between that and macroscopic units (how much does 1 million calcium atoms weigh?) Now it just so happens that $6.022 \times 10^{23}$ atoms of carbon-12 weigh 12 grams, and a single carbon-12 weighs 12 amu. A single molecule of water weighs roughly 18 amu, $6.022 \times 10^{23}$ molecules of water weigh 18 grams. Let's take a look at what using $6.022 \times 10^{23}$ molecules offers us over other units: It is appropriate. $6.022 \times 10^{23}$ molecules will typically fall in the gram to kilogram range of substance, which can easily be
measured out on a balance. It's also within the range of what you'd
usually use in a lab. It allows for easy conversion. You can easily tell when you have
$6.022 \times 10^{23}$ molecules because your atomic mass is equal to
your macroscopic mass. It makes sense. Intuitively, we'd like to measure things per atom or
per molecule, but doing so leads to some ugly units. Instead, if we measure per $6.022 \times 10^{23}$ molecules, we still are doing a counting-based measurement, but the numbers themselves are a lot nicer. You probably know this already, but we call $6.022 \times 10^{23}$ molecules a mole. These are the advantages of using the mole as a unit, and not for backwards-compatibility. I am of the opinion that if we shed all our units today and started from scratch, we would start off with grams and liters, but we would probably start using the measure of a mole again within a year. It's just a very powerful, useful way to measure things. P.S. This is much longer than I was initially intending and was not written while I was in the most awake state of mind. Please let me know in the comments if you think this answer is unclear, or stupid, or just plain wrong. | {
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6,834 | Is it possible to melt diamonds into a liquid? I mean if you heat diamond at open air it will start to burn around 700 degrees Celsius, reacting with oxygen to produce carbon dioxide gas. In the absence of oxygen it will transform into graphite, a more stable form of carbon, long before turning into graphite. My question is, is it possible to melt diamonds into liquid? If so, then, how? If not, then why? | Liquid carbon does indeed exist, but perhaps surprisingly, relatively little is known about it. It only exists above around $4000\ \mathrm{K}$ and $100\ \mathrm{atm}$, which are not trivial conditions to sustain and probe. There certainly are many theoretical studies into the properties of liquid carbon, though. You can find a phase diagram for carbon here , though it is likely just a rough guide. The specific bonding properties of carbon atoms in the liquid phase still seems uncertain to some extent because carbon is very versatile and can create very dissimilar structures (easily seen comparing its two most common phases, graphite and diamond). This source suggests that bonding in the liquid varies continuously between linear acetylenic chains, graphite-like and diamond-like structures, depending on the exact conditions. Edit : Now I realize that obtaining liquid carbon and melting diamonds are two slightly different things. Liquid carbon is trivial to obtain compared to melting a diamond. As you mention, heating diamond at relatively low pressures will cause it to convert into graphite before it could melt. The only way to inhibit that would be to exploit the slight difference in density between graphite and diamond; as diamond is denser, the increased pressure will stabilize it slightly, decreasing the tendency for diamond to revert back to graphite. You'll need a lot of pressure though! Rigorously speaking, according to the phase diagram above, to obtain liquid carbon directly from melting a diamond would require a pressure of at least $10\ \mathrm{GPa}$, which is about $10^5\ \mathrm{atm}$. To put that into perspective, it's about $1/30$ of the pressure at the centre of the Earth. Amazingly, we actually can reach such absurd pressures in the lab by compressing two diamonds against each other . However, if the conditions required conditions for melting diamond (and sustaining it in molten form) are reached, it would probably compromise the integrity of the instrument and cause catastrophic (possibly explosive) failure, so I don't think anyone will be purposefully melting diamonds any time soon! Other techniques could be used ( laser inertial confinement , light-gas guns , etc), but I believe most would only transiently produce liquid diamond. | {
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6,924 | As you may know, the reaction quotient $Q_c$ is defined by the equation
$$
Q_c = \frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}
$$
for the chemical reaction
$$
\alpha A + \beta B \rightarrow \gamma C + \delta D.
$$ This is something I've been struggling to understand at an intuitive level. Why is $Q_c$ the ratio of the product of the reaction's product concentrations and the product of the reaction's reactant concentrations (sorry for the confusing terminology; I can't think of a more precise way to phrase that), rather than the ratio of the sums of concentrations, like this:
$$
Q_c' = \frac{\gamma[C]+\delta[D]}{\alpha[A]+\beta[B]}
$$ This second, incorrect equation makes more sense to me because concentrations of solutes add up. Is the ratio defined as such so that it is directly proportional to the concentrations of each product and inversely proportional to the concentrations of each reactant? Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient? | The derivation cited can help you understand it mathematically. But the answer also turns out to be satisfactory intuitively. Let me try to explain how. The term $RT\ln Q_c$ is derived form the entropy part or $\Delta G=\Delta H-T\Delta S$, and in a slightly abstract way, represents entropy of mixing of the products and the reactants. This is why reactions which are spontaneous initially, as concentrations change, reach to the equilibrium state before complete conversion of reactants into products. If there was no entropy of mixing, a reaction which had negative free energy change(under any set of conditions) would continue to have a negative change (hence, spontaneous) until all of the reactants is converted to products, which will be the state of maximum entropy. But since there is entropy of mixing, the state of maximum entropy is reached before all the reactants are converted into products (because somewhere in between the entropy of mixing due to having a mixture of reactants and products outweighs the negative $\Delta G$ for the reactants to convert into the products) , and the concentrations at which this maximum entropy state can be reached is precisely what the equilibrium constant tells us. To calculate the entropy, we need to realize the fact that certain components of the reaction may have a greater contribution in deciding the equilibrium constant than others because, at the maximum entropy state, the components are present according to their stoichiometric relations and hence one might be in a well-defined excess over other. The relative contributions of each species to the entropy is logarithmic, that is, the increase in entropy is a logarithmic function of concentration (related slightly in a abstract manner to $S=k_B\ln \Omega$). Therefore, the contribution to the free energy and consequently in deciding the $K_c$ is logarithmic in the concentrations. The stoichiometric coefficientts are multiplied to the logarithmic contributions of each(which appear as the power to which the concentrations are raised when brought inside the logarithm) and the relative contributions are simply added, but since the addition of logarithms means multiplications of its arguments, we get the familiar $RT\ln Q_c$ term which gives the mentioned definitions of reaction quotient and the equilibrium constant. I have used reaction quotient and equilibrium constant interchangeably but the RQ represents an unfulfilled equilibrium constant and hence represents the same drive towards maximum entropy as $K_c$ (which is the state of maximum entropy itself). | {
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7,439 | Is there a scientific basis for the coffee making equipment which Gale Boetticher describes in Breaking Bad? He talks about maintaining the right conditions for bringing out the coffee flavor without degrading it. | Yes, there is a scientific basis, but I think you can't do anything with the apparatus shown in the figure. Here you can find much information about it. I will try to summarize. It seems that extracting coffee at a lower temperature makes it tastes better. Gale thinks that the amount of quinic acid is the key variable that makes good coffee. In fact, some studies (McCamey, D. A.; Thorpe, T. M.; and McCarthy, J. P. Coffee Bitterness. In "Developments in Food Science." Vol 25. 169-182. 1990.) report that bitterness of coffee is due to quinic acid but is not the only substance involved (see this link ). I don't have any reference about the dependence between temperature and method of extraction and quantitative of quinic acid extracted. By the way, the apparatus should extract the coffee in a more effective way. The vacuum pump seems to suggest the use of the vacuum to make the water boil at a lower temperature. But the connections in my opinion make no sense, so I will try to describe it: All start with an autoclave that is connected with a Gast vacuum pump . This is connected with an Allihn condenser (not the right thing to do because in there you should connect the water used to refrigerate), and then to a heated Erlenmeyer flask that is connected with a strange steel cylinder (I don't know what it is, maybe a filter), then to a Florence flask and then to another steel cylinder (...). The inspiration comes from a Florence Siphon but this is much simpler and is generally something like this: Here you can read and here you can watch how it works. Of course this is not so nice and intriguing as the previous apparatus and is not a geeky thing that makes us think we are dealing with a great chemist, so the authors thought to make the apparatus more appealing but completely useless! | {
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7,449 | When water temperature reaches $100\ ^\circ \mathrm{C}$, the molecules get so excited that the hydrogen atoms lose the bonds to the oxygen atom and therefore the water starts to become gas. I get that, but at room temperature ($23\ ^\circ \mathrm{C}$), is there no excitation in the atoms or is there? | First, I think I should make it clear that when water boils, the bonds in the water molecule linking the hydrogen and oxygen atom are not broken. During boiling, the intermolecular bonds in water are the ones that get broken, that is the bonds that link the water molecules together. At room temperature, there is evaporation (I wouldn't call it excitation). This is because there are a few molecules of water which can manage to muster enough energy to escape from the large body of molecules and escape into air. This can be explained through a graph depicting the distribution of speed among water molecules worked out by Maxwell and Boltzmann. As you can probably see, there are a lot of water molecules with lower kinetic energy than with higher kinetic energy. Those that have the higher kinetic energy are the ones that are able to break through the water surface to become vapour. Even at low temperatures, there are some water molecules are have enough energy to escape and that's why evaporation in water can occur at any temperature (yes, even if the water is in ice). When the temperature increases, there are more molecules with higher kinetic energy and thus, more water can evaporate. | {
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7,467 | On Outdoors StackExchange on a post about the safety of distilled water , someone claimed that pure water is very corrosive. Water distilled for some laboratory uses (like the Z-pinch reactor) is distilled to 0.00000009% purity, and is caustic enough to burn through most things. Water, purified enough, cannot be distributed through stainless steel pipes. Instead, they use glass pipes to transport it within the plants where it's made. For sure, "caustic" is not the correct word, since water is neutral. But, is it that corrosive? Or is it even less than water with salt? | That quote had false statements all over it. Type I ultra pure water (Milli-Q water, or others) are fairly common among all labs. They are not corrosive to stainless steel. As @chipbuster said, purified water is never distributed through stainless steel partly because concern of contamination. Metal reacts not only with oxygen, but also many pollutants in the air. Overtime on metal surface will build up a layer that has all kind of compounds in them. When water flows through, ions get picked up. However, the most major reason would be that glass is better in all respect. The reason glasswares are commonly used in the lab have little to do with pure water. Glasswares are relatively stable across wide range of pH and very hard to be oxidized or reduced. Plus, glasswares are cheap, extremely easy to manufacture, and transparent, so you can see through. You will really only need metal if extreme pressure or temperature is required. As a matter of fact, it is also unwise to use glassware to distribute ultrapure water--you will get ion contaminations from glass as well. Purified water are actually almost always stored in plastic containers and distributed through plastic tubes. | {
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7,489 | In the highly-rated TV series, Breaking Bad , Walter White, a high school chemistry teacher recently diagnosed with cancer, takes to making the illicit drug, crystal meth (methamphetamine), by two main routes. First, along with his RV-driving accomplice, Jessie Pinkman, Mr. White uses the common small-scale route starting with (1S,2S)-pseudoephedrine (the active ingredient in Sudafed ®️). This method features the use of an optically active starting material to make an optically active end product, (S)-methamphetamine. However, making (S)-methamphetamine on a large scale is limited because it is hard to get sufficient quantities of (1S,2S)-pseudoephedrine. In the second route, Mr. White uses his knowledge of chemistry to move to an alternative synthesis starting with phenylacetone (also know as P2P or phenyl-2-propanone): Racemic methamphetamine was obtained by the Winnebago-based chemists by reductive amination of P2P using methylamine and hydrogen over activated aluminum. While his blue-colored product is considered by his customers to be exceptionally pure, Mr. White clearly knows about the issue of producing the correct enantiomer. In fact, he raises this topic more than once in the series. Since the show might not want to tell us the answer, I am wondering what what other possible methods Mr. White could have used to obtain an enantiomerically pure product? | Intriguing question. First, the best yield would be achieved by selectively producing one enantiomer instead of the other. In this case, White wants D-methamphetamine (powerful psychoactive drug), not L-methamphetamine (Vicks Vapor Inhaler). Reaction processes designed to do this are known as "asymmetric synthesis" reactions, because they favor production of one enantiomer over the other. The pseudoephedrine method for methamphetamine employs one of the more common methods of asymmetric synthesis, called "chiral pool resolution". As you state, starting with an enantiomerically-pure sample of a chiral reagent (pseudoephedrine) as the starting point allows you to preserve the chirality of the finished product, provided the chiral point is not part of any "leaving group" during the reaction. However, again as you show, phenylacetone is achiral, and so the P2P process cannot take advantage of this method. There are other methods of asymmetric synthesis, however none of them seem applicable to the chemistry shown or described on TV either; none of the reagents or catalysts mentioned would work as chiral catalysts, nor are they bio- or organocatalysts. Metal complexes with chiral ligands can be used to selectively catalyze production of one enantiomer, however the aluminum-mercury amalgam is again achiral. I don't remember any mention of using organocatalysis or biocatalysis, but these are possible. The remaining route, then, is chiral resolution; let the reaction produce the 50-50 split, then separate the two enantiomers by some means of reactionary and/or physical chemistry. This seems to be the way it works in the real world. The advantage is that most of the methods are pretty cheap and easy; the disadvantage is that your maximum possible yield is 50% (unless you can then run a racemization reaction on the undesireable half to "reshuffle" the chirality of that half; then your yield increases by 50% of the last increase each time you run this step on the undesirable product). In the case of methamphetamine, this resolution is among the easiest, because methamphetamine forms a "racemic conglomerate" when crystallized. This means, for the non-chemists, that each enantiomer molecule prefers to crystallize with others of the same chiral species, so as the solution cools and the solvent is evaporated off, the D-methamphetamine will form one set of homogeneous crystals and the L-methamphetamine will form another set. This means that all White has to do is slow the evaporation of solvent and subsequent cooling of the pan, letting the largest possible crystals form. Then, the only remaining trick is identifying which crystals have which enantiomer (and as these crystals are translucent and "optically active", observing the polarization pattern of light shone through the crystals will identify which are which). | {
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7,541 | This may be a silly question but why is water wet?
All the previous questions on water do not explain the reason why is water wet. They assume that its a natural property. However we need to understand why is it wet in the first place. This could apply to other liquids such as milk, juices but the essential element is water. Similarly another basic property which we take for granted is heat caused by fire. What causes fire to be hot as an inherent property? For the purpose of discussion consider simple fire by burning wood or paper. | Why is water wet? Chemically speaking saying that water is wet has not much sense.We could, however, say that "Wetting process" is caused by the presence of a thin liquid particles layer over a material. In our specific case, this depends on the chemistry of water and the chemistry of skin. So the main causes of the fact that water is wet, is its wetting properties determined by a force balance between adhesive and cohesive forces over the skin. Mercury, although is a liquid, will not "wet" you because it has a strong cohesive force that doesn't permit the molecules to get in touch with the sensory cells. When you touch water or a water wetted material part of the water molecules are transfer to your hands (a hydrophilic material) and so to the receptor cells giving the sensation of wet, this sensation can be caused however in some case of parasthesias even if there are no liquid particles. In fact, there is no "wet receptor" on your skin but the sensation is due to the elaboration of different stimuli from tactile, pressure and temperature receptor present in your skin.
Try to play with plasticine after a while handling it your hands will become hydrophobic you will no more fell the wet sensation when you wash them even if water is still "wet"! What causes fire to be hot as an inherent property? Fire is an exothermic process. So energy stored in a chemical bond is released as kinetic energy and electromagnetic wave. The fire has a high temperature because the molecules have great kinetic energies when you get in touch with fire (of course wormed air could be a medium as well for short distance) this kinetic energy is transfer to your skin and so to your thermoreceptors . And you feel what human call hot. In fact, you can feel the sensation of hot even without touching the fire because it even irradiates energy as electromagnetic waves that can be absorbed by your skin and so be feel for his receptors. | {
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7,683 | When comparing o,m,p -toluidine basicities, the ortho effect is believed to explain why o -toluidine is weaker. But when comparing o,m,p -toluic acid basicities, the ortho effect is stated as a reason why o -toluic acid is stronger acid. I was told that the ortho effect is a phenomenon in which an ortho- group causes steric hindrance, forcing the $\ce{-COOH}$, $\ce{-NH2}$ or some other bulky group to move out of the plane, inhibiting resonance. Then, if the ortho effect inhibits resonance, why is o -toluic acid the strongest and o -toluidine the weakest? Where am I going wrong in my understanding of the ortho effect? | I'd like to throw a tentative explanation for the ortho effect into the ring: In the molecules in question, an interaction between the protons of the methyl group and the lone pair of the amine nitrogen and the negative charge on the carboxylate, respectively, can be assumed. In the first case, the electron density on the N atom is (slightly) reduced and thus the basicity of o -toluidine. In the latter case, a similar interaction provides additional stabilisation of the carboxylate. As a result, o -toluic acid is more acidic than the isomers. | {
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8,057 | Most books refer to a steep rise in pH when a titration reaches the equivalence point. However, I do not understand why … I mean I am adding the same drops of acid to the alkali but just as I near the correct volume (i.e. the volume required to neutralize the alkali), the pH just suddenly increases quickly. | I've decided to tackle this question in a somewhat different manner. Instead of giving the chemical intuition behind it, I wanted to check for myself if the mathematics actually work out. As far as I understand, this isn't done often, so that's why I wanted to try it, even though it may not make the clearest answer. It turns out to be a bit complicated, and I haven't done much math in a while, so I'm kinda rusty. Hopefully, everything is correct. I would love to have someone check my results. My approach here is to explicitly find the equation of a general titration curve and figure out from that why the pH varies quickly near the equivalence point. For simplicity, I shall consider the titration to be between a monoprotic acid and base. Explicitly, we have the following equilibria in solution $$\ce{HA <=> H^+ + A^-} \ \ \ → \ \ \ K_\text{a} = \ce{\frac{[H^+][A^-]}{[HA]}}$$ $$\ce{BOH <=> B^+ + OH^-} \ \ \ → \ \ \ K_\text{b} = \ce{\frac{[OH^-][B^+]}{[BOH]}}$$ $$\ce{H2O <=> H^+ + OH^-} \ \ \ → \ \ \ K_\text{w} = \ce{[H^+][OH^-]}$$ Let us imagine adding two solutions, one of the acid $\ce{HA}$ with volume $V_\text{A}$ and concentration $C_\text{A}$ , and another of the base $\ce{BOH}$ with volume $V_\text{B}$ and concentration $C_\text{B}$ . Notice that after mixing the solutions, the number of moles of species containing $\ce{A}$ ( $\ce{HA}$ or $\ce{A^-}$ ) is simply $n_\text{A} = C_\text{A} V_\text{A}$ , while the number of moles of species containing $\ce{B}$ ( $\ce{BOH}$ or $\ce{B^+}$ ) is $n_\text{B} = C_\text{B} V_\text{B}$ . Notice that at the equivalence point, $n_\text{A} = n_\text{B}$ and therefore $C_\text{A} V_\text{A} = C_\text{B} V_\text{B}$ ; this will be important later. We will assume that volumes are additive (total volume $V_\text{T} = V_\text{A} + V_\text{B}$ ), which is close to true for relatively dilute solutions. In search of an equation To solve the problem of finding the final equilibrium after adding the solutions, we write out the charge balance and matter balance equations: Charge balance: $\ce{[H^+] + [B^+] = [A^-] + [OH^-]}$ Matter balance for $\ce{A}$ : $\displaystyle \ce{[HA] + [A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}$ Matter balance for $\ce{B}$ : $\displaystyle \ce{[BOH] + [B^+]} = \frac{C_\text{B} V_\text{B}} {V_\text{A} + V_\text{B}}$ A titration curve is given by the pH on the $y$ -axis and the volume of added acid/base on the $x$ -axis. So what we need is to find an equation where the only variables are $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$ . By manipulating the dissociation constant equations and the mass balance equations, we can find the following: $$\ce{[HA]} = \frac{\ce{[H^+][A^-]}}{K_\text{a}}$$ $$\ce{[BOH]} = \frac{\ce{[B^+]}K_\text{w}}{K_\text{b}\ce{[H^+]}}$$ $$\ce{[A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{a}}{K_\text{a} + \ce{[H^+]}}\right)$$ $$\ce{[B^+]} = \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{b}\ce{[H^+]}}{K_\text{b}\ce{[H^+]} + K_\text{w}}\right)$$ Replacing those identities in the charge balance equation, after a decent bit of algebra, yields: $$\ce{[H^+]^4} + \left(K_\text{a} + \frac{K_\text{w}}{K_\text{b}} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^3} + \left(\frac{K_\text{a}}{K_\text{b}}K_\text{w} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} K_\text{a} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}K_\text{a} - K_\text{w}\right) \ce{[H^+]^2} - \left(K_\text{a} K_\text{w} + \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\frac{K_\text{a}}{K_\text{b}} K_\text{w} + \frac{K^2_\text{w}}{K_\text{b}}\right) \ce{[H^+]} - \frac{K_\text{a}}{K_\text{b}} K^2_\text{w} = 0$$ Now, this equation sure looks intimidating, but it is very interesting. For one, this single equation will exactly solve any equilibrium problem involving the mixture of any monoprotic acid and any monoprotic base, in any concentration (as long as they're not much higher than about $1~\mathrm{\small M}$ ) and any volume. Though it doesn't seem to be possible to separate the variables $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$ , the graph of this equation represents any titration curve (as long as it obeys the previous considerations). Though in its full form it is quite daunting, we can obtain some simpler versions. For example, consider that the mixture is of a weak acid and a strong base. This means that $K_\text{b} \gg 1$ , and so every term containing $K_\text{b}$ in the denominator is approximately zero and gets cancelled out. The equation then becomes: Weak acid and strong base: $$\ce{[H^+]^3} + \left(K_\text{a} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\ce{A} + V_\ce{B}} K_\ce{a} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}K_\ce{a} - K_\ce{w}\right) \ce{[H^+]} - K_\ce{a} K_\ce{w} = 0$$ For a strong acid and weak base ( $K_\text{a} \gg 1$ ), you can divide both sides of the equation by $K_\text{a}$ , and now all terms with $K_\text{a}$ in the denominator get cancelled out, leaving: Strong acid and weak base: $$\ce{[H^+]^3} + \left(\frac{K_\ce{w}}{K_\ce{b}}+\frac{C_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}\right) \ce{[H^+]^2} - \left(K_\ce{w} + \frac{C_\text{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}} \frac{K_\ce{w}}{K_\ce{b}}\right) \ce{[H^+]} - \frac{K^2_\ce{w}}{K_\ce{b}} = 0$$ The simplest case happens when adding a strong acid to a strong base ( $K_\ce{a} \gg 1$ and $K_\ce{b} \gg 1$ ), in which case all terms containing either in the denominator get cancelled out. The result is simply: Strong acid and strong base: $$\ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]} - K_\ce{w} = 0$$ It would be enlightening to draw some example graphs for each equation, but Wolfram Alpha only seems to be able to handle the last one, as the others require more than the standard computation time to display. Still, considering the titration of $1~\text{L}$ of a $1~\ce{\small M}$ solution of a strong acid with a $1~\ce{\small M}$ solution of a strong base, you get this graph . The $x$ -axis is the volume of base added, in litres, while the $y$ -axis is the pH. Notice that the graph is exactly as what you'll find in a textbook! Now what? With the equations figured out, let's study how they work. We want to know why the pH changes quickly near the equivalence point, so a good idea is to analyze the derivative of the equation and figure out where they have a very positive or very negative value, indicating a region where $\ce{[H^+]}$ changes quickly with a slight addition of an acid/base. Suppose we want to study the titration of an acid with a base. What we need then is the derivative $\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}}$ . We will obtain this by implicit differentiation of both sides of the equations by $\displaystyle \frac{\ce{d}}{\ce{d}V_\ce{B}}$ . Starting with the easiest case, the mixture of a strong acid and strong base, we obtain: $$\frac{\ce{d[H^+]}}{\ce{d} V_\ce{B}}= \frac{K_\ce{w} - C_\ce{B} \ce{[H^+] - [H^+]^2}}{2(V_\ce{A} + V_\ce{B}\left) \ce{[H^+]} + (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}\right)}$$ Once again a complicated looking fraction, but with very interesting properties. The numerator is not too important, it's the denominator where the magic happens. Notice that we have a sum of two terms ( $2(V_\ce{A} + V_\ce{B})\ce{[H^+]}$ and $(C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A})$ ). The lower this sum is, the higher $\displaystyle \frac{\mathrm{d}\ce{[H^+]}}{\mathrm{d} V_\ce{B}}$ is and the quicker the pH will change with a small addition of the base. Notice also that, if the solutions aren't very dilute, then the second term quickly dominates the denominator because while adding base, the value of $[H^+]$ will become quite small compared to $C_\ce{A}$ and $C_\ce{B}$ . Now we have a very interesting situation; a fraction where the major component of the denominator has a subtraction. Here's an example of how this sort of function behaves . When the subtraction ends up giving a result close to zero, the function explodes. This means that the speed at which $\ce{[H^+]}$ changes becomes very sensitive to small variations of $V_\ce{B}$ near the critical region. And where does this critical region happen? Well, close to the region where $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$ is zero. If you remember the start of the answer, this is the equivalence point! . So there, this proves mathematically that the speed at which the pH changes is maximum at the equivalence point. This was only the simplest case though. Let's try something a little harder. Taking the titration equation for a weak acid with strong base, and implicitly differentiating both sides by $\displaystyle \frac{\ce{d}}{\ce{d} V_\ce{B}}$ again, we get the significantly more fearsome: $$\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}} = \frac{ -\frac{V_\ce{A}}{(V_\ce{A} + V_\ce{B})^2} \ce{[H^+]} (C_\ce{B}\ce{[H^+]} - C_\ce{B} K_\ce{a} + C_\ce{A} K_\ce{a})}{3\ce{[H^+]^2 + 2[H^+]}\left(K_\ce{a} + \frac{C_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}}\right) + \frac{K_\ce{a}}{V_\ce{A} + V_\ce{B}} (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}) -K_\ce{w}}$$ Once again, the term that dominates the behaviour of the complicated denominator is the part containing $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$ , and once again the derivative explodes at the equivalence point. | {
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8,068 | Assuming a perfect diamond with no impurities. Would this diamond considered to be a single large molecule?
Browsing the interweb I found several opinions about this but did not find a clear Yes or No. From what if gathered so far I'm pretty sure a diamond is a monocrystalline solid. Since all its parts are connected with covalent bonds I would assume that it should be considered a molecule even though it has no fixed number of atoms in its molecule formula. | Diamond is a covalent network solid , like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff). Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc. - network solids form one of the two main classes of macromolecules , the other being polymers . You'll note that the Wikipedia article on macromolecules seems to imply that 'macromolecule' and 'polymer' are synonymous. They are not, at least not to those chemists working in the field. | {
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8,073 | I was just thinking what can be the last atomic number that can exist within the range of permissible radioactivity limit and considering all other factors in quantum physics and chemical factors. | Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as the innermost electrons would have to be moving above the speed of light . This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the nucleus is not a point particle, then there seems to be no real issue with arbitrarily high atomic numbers, although unusual effects start happening above $Z \approx 173$. These results may be overturned by an even deeper analysis with current quantum electrodynamics theory, or a new theory altogether. As far as we can tell, however, we will never get anywhere close to such atomic numbers. Very heavy elements are extremely unstable with respect to radioactive decay into lighter elements. Our current method of producing superheavy elements is based on accelerating a certain isotope of an relatively light element and hitting a target made of an isotope of a much heavier element. This process is extremely inefficient, and it takes many months to produce significant amounts of material. For the heaviest elements, it takes years to detect even a handful of atoms. The very short lifetime of the heaviest targets and the very low collision efficiency between projectile and target mean that it will be extremely difficult to go much further than the current 118 elements. It is possible that we may find somewhat more stable superheavy isotopes in the islands of stability around $Z=114$ and $Z=126$, but the predicted most stable isotopes (which even then are not expect to last more than a few minutes) have such a huge amount of neutrons in their nuclei that we have no idea how to produce them; we may be condemned to merely skirt the shores of the islands of stability, while never climbing them. EDIT : Note that the best calculation presented above is based on quantum electrodynamics alone, i.e. only electromagnetic forces are taken into account. Obviously, to predict how nuclei will behave (and therefore how many protons you can stuff into a nucleus before it's impossible to go any further), one needs detailed knowledge of the strong and weak nuclear forces. Unfortunately, the mathematical description of nuclear forces is still an incredibly tough problem in physics today , so no one can hope to provide a rigorous answer from that angle. There must be some limit, as the residual nuclear forces are very short-ranged. At some point there will be so many protons and neutrons in the nucleus (and the resulting nucleus will have become so large) that the diametrically opposite parts of the nucleus won't be able to "detect" each other, as they are too far away. Each additional proton or neutron produces a weaker stabilization via the strong nuclear force. Meanwhile, the electrical repulsion between protons has infinite range, so every additional proton will contribute repulsively just the same. This is why heavier elements need higher and higher neutron-to-proton ratios to remain stable. Thus, at some atomic number, possibly not much higher than our current record of $Z=118$, the electrical repulsion of the protons will always win against the strong nuclear attractions of protons and neutrons, no matter the configuration of the nucleus. Hence, all sufficiently heavy atomic nuclei will suffer spontaneously fission almost immediately after coming into existence, or all the valid reaction pathways to reach an element will require events which are so fantastically unlikely that if even all the nucleons in the entire observable Universe were being collided with each other since the Big Bang in an attempt to synthesize the heaviest element possible, we would statistically expect some sufficiently heavy atom to not have been produced even once. | {
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8,099 | My book writes: When white light shines through a solution of a complex ion of a transition metal, photons of a particular frequency are absorbed and their energy promotes an electron from lower energy level to upper energy level. We perceive the complementary color to the light absorbed. However, I have a question about this. Once the electron is promoted to a higher energy level, won't it fall back down and give off a photon? Surely the electron cannot remain at the higher energy level forever, as that would mean that no further electronic transitions could taking place, and the solution would turn colourless. | Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity. $$\ce{S_0 ->[$h\nu_\mathrm{ex}$] S_1}$$ In most cases, the excited state deactivates through internal conversion in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the solvent is actually heated up, although it might not be measurable. There are however cases, in which the initially formed excited singlet state species first relaxes to its lowest vibrational mode and then emits a photon to reach the ground state. $$\ce{S_1 -> S_0 + $h\nu_\mathrm{f}$}$$ This deactivation via radiation is known as fluorescence . Fluorescence lifetimes are typically rather short (in the ns range) and the emitted photon has a lower energy than the absorbed one. This means that the emission spectrum is shifted to the red end of the uv spectrum ($\lambda_\mathrm{f} > \lambda_\mathrm{ex}$). The effect is known as Stokes shift . A classical case for a fluorescent transition metal complex is $\ce{Ru(bpy)3^{2+}}$. In acetonitrile, it absorbs around $\lambda_\mathrm{ex} = 460~\mathrm{nm}$. The fluorescence is observed around $\lambda_\mathrm{em} = 620~\mathrm{nm}$. Other bidentate ligands - more rigid than 2,2'-bipyridine (bpy) - often used are 1,10-phenanthroline (phen), dipyrido-[3,2- a :2',3'-c]phenazine (dppz), and benzo[ i ]dipyrido-[3,2- a :2',3'-c]phenazine (dppn). In other cases, the excited singlet state undergoes a change in the spin multiplicity ( Intersystem crossing ) to a triplet state. $$\ce{S_1 ->[ISC] T_1}$$ These excited triplet states are usually much longer-lived. When they eventually deactivate via emission of a photon, this process is known as phosphorescence . $$\ce{T_1 -> S_0 + $h\nu_\mathrm{p}$}$$ Keeping all these processes in mind, one question arises - particular from younger students or pupils: If our metal complex absorbs a photon and forms an excited state with such a short lifetime, why can we see a nevertheless observe permanent colour? The answer is in the enormous numbers! Continuous irradiation with white light means that we send a huge number of photons. Moreover, the solution of our transition metal complexes in a cuvette (or a test tube) also contains more than one molecule. In fact, much more! As a result, we have a huge number of incidents The complex absorbs a photon of a certain energy, which means that one colour is "swallowed". The complex forms a short-lived "excited state" the solution reflects the remaining light the excited state complex deactivates again Since this happens all the time while we shine the light on our sample and for a lot of molecules, we observe a summary of the events: a coloured solution. | {
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8,357 | The electron configuration of calcium is 2, 8, 8, 2, where up to that point each shell, asides from the first shell counts up to 8 - why then does scandium have an electron configuration of 2, 8, 9, 2? What causes the 3rd shell to start filling up, rather than the fourth? | The relative energies of the electronic subshells have been calculated for atoms in the vicinity of $Z=20$ ( J. Chem. Educ. , 1994 , 71 (6), 469 ), and the result is surprising: Looking at this graph, by all means the electronic configuration of scandium should in fact be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ $\color{blue}{\ce{3d^3}}$ in order to minimize orbital energies! The graph does not show the energy of the $\ce{4p}$ subshell, but it would lie somewhat above both curves shown. However, there is an important effect not being considered here, which is the destabilizing interelectronic repulsions. Electrons in a same subshell tend to repel each other more (intrasubshell repulsion) than electrons in different subshells (intersubshell repulsion). It turns out (and no one can really qualitatively explain why, see below) the repulsion is such that in an atom with $Z=21$ , the configuration $\ce{[Ar] 4s^2 3d^1}$ is lower in energy than other potential candidates, such as $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^1 3d^2}$ . Theoretically, the odd configurations $\ce{[Ar] 4s^2 4p^1}$ or $\ce{[Ar] 4s^1 3d^1 4p^1}$ would have lower interelectron repulsion energies than the observed ground state, but in these cases the lowered interelectron repulsions do not compensate the requirement of populating the higher-energy $\ce{4p}$ subshell. The comparison of several possible electron configurations and the joint minimization of orbital energy and electron repulsion energies does not seem to be something that can be done without resorting to heavy calculations or direct experimental validation. In addition to some material I linked in comments, this article from Eric Scerri's blog (a chemist who focuses on aspects of periodicity, including electronic distribution) states, regarding the electronic configuration of scandium compared to calcium and to the $\ce{Sc^3+}$ ion (emphasis mine): This amounts to saying that all three of the final electrons enter $\ce{3d}$ but two of them are repelled into an energetically less favourable orbital, the $\ce{4s}$ , because the overall result is more advantageous for the atom as a whole. But this is not something that can be predicted. Why is it 2 electrons, rather than one or even none? In cases like chromium and copper just one electron is pushed into the $\ce{4s}$ orbital. In an analogous case from the second transition series, the palladium atom, the competition occurs between the $\ce{5s}$ and $\ce{4d}$ orbitals. In this case none of the electrons are pushed up into the $\ce{5s}$ orbital and the resulting configuration has an outer shell of $\ce{[Kr] 4d^10}$ . None of this can be predicted in simple terms from a rule of thumb and so it seems almost worth masking this fact by claiming that the overall configuration can be predicted, at least as far as the cases in which two electrons are pushed up into the relevant s orbital. To those who like to present a rather triumphal image of science it is too much to admit that we cannot make these predictions. The use of the sloppy Aufbau seems to avoid this problem since it gives the correct overall configuration and hardly anybody smells a rat. So unfortunately, it seems that even though most of the effects which combine to result in the observed electronic configurations are known, there is no qualitative way to predict where the configurations are going to mismatch with the Aufbau principle or the energy levels of the orbitals. I have read the statement that the Aufbau principle is most decidedly wrong for practically every atom with respect to the placement of the orbital energy levels, but incredibly it happens to predict the configuration of the valence shell for most atoms. | {
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8,537 | Is the name "Dihydrogen monoxide" actually what chemists would use to refer to $\ce{H2O}$ (assuming there was no common name, "water")? Of course, this is all over the internet. I'm a little skeptical though because the similar chemical $\ce{H2S}$ is called "hydrogen sulfide", not "dihydrogen monosulfide". | No, it's not. The "dihydrogen monoxide" name is used as part of a hoax. In the scientific community, there are chemical names for water, and which one is used in the literature generally depends on how it interacts with something else (hydroxic acid and hydrogen hydroxide were two I heard most often in acid-base reactions). IUPAC, the standards committee that sets standard names for chemical structures, suggests "oxidane" as a starting point for the construction of other names for chemicals that are derived from water. However, they do not suggest "oxidane" itself be used to refer to plain water. Most chemists would use "water", even when writing scientific papers. | {
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8,598 | In my textbook, it says that the maximum number of electrons that can fit in any given shell is given by 2n². This would mean 2 electrons could fit in the first shell, 8 could fit in the second shell, 18 in the third shell, and 32 in the fourth shell. However, I was previously taught that the maximum number of electrons in the first orbital is 2, 8 in the second orbital, 8 in the third shell, 18 in the fourth orbital, 18 in the fifth orbital, 32 in the sixth orbital. I am fairly sure that orbitals and shells are the same thing. Which of these two methods is correct and should be used to find the number of electrons in an orbital? I am in high school so please try to simplify your answer and use fairly basic terms. | Shells and orbitals are not the same. In terms of quantum numbers, electrons in different shells will have different values of principal quantum number n . To answer your question... In the first shell (n=1), we have: The 1s orbital In the second shell (n=2), we have: The 2s orbital The 2p orbitals In the third shell (n=3), we have: The 3s orbital The 3p orbitals The 3d orbitals In the fourth shell (n=4), we have: The 4s orbital The 4p orbitals The 4d orbitals The 4f orbitals So another kind of orbitals (s, p, d, f) becomes available as we go to a shell with higher n . The number in front of the letter signifies which shell the orbital(s) are in. So the 7s orbital will be in the 7th shell. Now for the different kinds of orbitals Each kind of orbital has a different "shape", as you can see on the picture below. You can also see that: The s-kind has only one orbital The p-kind has three orbitals The d-kind has five orbitals The f-kind has seven orbitals Each orbital can hold two electrons. One spin-up and one spin-down. This means that the 1s, 2s, 3s, 4s, etc., can each hold two electrons because they each have only one orbital. The 2p, 3p, 4p, etc., can each hold six electrons because they each have three orbitals, that can hold two electrons each (3*2=6). The 3d, 4d etc., can each hold ten electrons, because they each have five orbitals, and each orbital can hold two electrons (5*2=10). Thus, to find the number of electrons possible per shell First, we look at the n=1 shell (the first shell) . It has: The 1s orbital An s-orbital holds 2 electrons. Thus n=1 shell can hold two electrons. The n=2 (second) shell has: The 2s orbital The 2p orbitals s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell can have 8 electrons. The n=3 (third) shell has: The 3s orbital The 3p orbitals The 3d orbitals s-orbitals can hold 2 electrons, p-orbitals can hold 6, and d-orbitals can hold 10, for a total of 18 electrons. Therefore, the formula $2n^2$ holds! What is the difference between your two methods? There's an important distinction between "the number of electrons possible in a shell" and "the number of valence electrons possible for a period of elements" . There's space for $18 \text{e}^-$ in the 3rd shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, elements in the 3rd period only have up to 8 valence electrons. This is because the $3d$-orbitals aren't filled until we get to elements from the 4th period - ie. elements from the 3rd period don't fill the 3rd shell. The orbitals are filled so that the ones of lowest energy are filled first. The energy is roughly like this: $$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s$$ An easy way to visualize this is like this: | {
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8,642 | I have read that combining the DC current with a high-frequency AC current, the electrolysis of water speeds up. Is this true? In that case, how is less energy wasted as heat? Or does it simply catalyze the process? | First of all, have a look at the wikipedia page on electrolysis of water . I also like this review: Zoulias et al. : A Review on Water Electrolysis, TCJST, 4 (2) (2004) 41-71 Specifically they list a number of actually existing installations (context: renewable energy) and their actually achieved efficiency. Speed up does not necessarily have anything to do with higher efficiency. In electrolysis it is often the other way round: if you want to squeeze out the maximum free energy, you need to do the reaction infinitely slowly (despite thermodynamics having a dynamic name, it looks at infinitively slow processes). Thus, speeding up usually means that you find a way to put more power through your system. The big issue is to find a way of doing this without loosing (too much) efficiency. Pulsed/modulated DC: looking through a few papers I liked this one: Shimizu et al. : A novel method of hydrogen generation by water electrolysis using
an ultra-short-pulse power supply, Journal of Applied Electrochemistry (2006) 36:419–423, DOI 10.1007/s10800-005-9090- They aim at avoiding the diffusion controlled situation by having the pulses short enough so that no depletion zone occurs.
Look at these diagrams: So they report one setting where the pulsed electrolysis is actually more efficient than DC in their cell. There's more research going on on this, however the papers I found reported increased efficiency compared to pure DC electrolysis, but the absolute efficiencies are around 10%. However, compare their numbers with the 80% efficiency cited by the review for an industrial alkaline electrolysis. Note that one big difference is the voltage that is applied: for the DC it is around 1.85 - 2.05 V, so much less overvoltage. Note also that when they say that higher voltage speeds up ion transport, then this overvoltage is converted to heat (ions face friction in the medium when they travel) and thus basically lost. Another line that looks real is that if you go to higher temperatures, a (small) part of the energy can be supplied by heat. As heat is cheap, this may help. One point one has to be aware though, that the efficiency calculations may be done with respect to the electric energy only (neglecting the heat input) and thus look artificially nice (like efficiencies of condensing boilers calculated against the lower heating value). I found a bunch of nonsense claims in the internet, about the resonance frequency of water helping to split bonds . The first thing to realize here is that there is no one resonance frequency of water. With suitable energy, you can excite rotational, vibrational and electronic states (I left out translation - there transition energies minute). At room temperature you can say as a rule of thumb that most molecules will be in some excited rotational state, but in the vibrational and electronic ground states. Excitation energies for rotation are in the far infrared or microwave energy/frequency region. Widely used e.g. in the microwave oven at 2.45 GHz ( $\approx$ 12 cm). Actually, the whole region is full of bands where water absorbs. Note that microwave heating of water does not cause electrolysis. Vibrational transitions are around 2.9 μm = 105 THz = 3500 cm⁻¹ and 6μm = 50 THz = 1635 cm⁻¹ with lots of combinations and overtones throughout the near infrared region. Quite exceptionally, the visible region is basically free of water absorption. Electronic transitions (breaking of bonds) need energies in the UV , and here we meet bands that lead to photodissociation , e.g. at 166nm (taken from Wikipedia). That corresponds to 1.8 PHz = $1.8 \cdot 10^{15}$ Hz. Compare this to the kHz and MHz where your link claims dissociation. This doesn't mean that the pulsed DC cannot help, nor that impedance spectoscopy won't give important information. But resonance frequencies in the kHz range are electrical LC-circuit resonances depending on cell and electrode geometries and electrical double layers etc. But neiter on vibrations nor breaking of the bonds of the water molecule. To give the "method" you ask about some real world numbers, at the very end of the Wiki page the energy efficiency for industrial water electrolysis is cited as usually between 50 and 80 %. The paper then proposes to burn the gas in an internal combustion machine. As such a stationary process could be adjusted so that the engine is at its maximum efficiency, we may assume 1/3 or 35% efficiency here. we then need a generator to convert the mechanical energy into electric energy. Fortunately, that step is rather efficient. Say, 95 %. A fuel cell would be more efficient than the combustion - generator combination: ca. 40 - 60 % according to Wikipedia. Unfortunately, also battery charging is not 100% efficient. Let's assume 80–90% (taken from Wikipedia on Li-ion batteries) For batteries that are charged with higher current (or current density) efficiency is less. Example would be lead-acid batteries as used in cars. Wiki quotes efficiencies between 50 and 80 %. Taking these numbers together, I conclude that after going once through the cycle of the proposed "perpetuum mobile", 8 - 24 % of the energy are retained in a "useful state" while 76 - 92 % became heat. With fuel cell, we may be able to "boost" the energy efficiency to 43%. Useful general knowledge (in addition to the law of energy conservation) The US patent system is different from e.g. the German patent system in that here in Germany the patent application has to have commercial/industrial usability. This includes a technical argument why it works (according to the physical laws). A perpetuum mobile would be rejected on these grounds (of course the inventor could prove his case with a prototype). US patents do not have this technical check. generators (mechanic -> electric conversion) are sources of current , while batteries (galvanic cells) are sources of voltage . | {
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8,655 | Nodes are the points in space around a nucleus where the probability of finding an electron is zero. However, I heard that there are two kinds of nodes, radial nodes and angular nodes . What are they and what information do they provide of an atom? | 1. How to get the number and type of nodes for an orbital As you said, nodes are points of zero electron density. From the principal quantum number $n$ and the azimuthal quantum number $\ell$, you can derive the number of nodes, and how many of them are radial and angular . $$\text{number of nodes}=n-1$$ $$\text{angular nodes}=\ell$$ $$\text{radial nodes}=(\text{number of nodes})-(\text{angular nodes})$$ So each type of orbital ($s, p, d$ etc) has its own unique, fixed number of angular nodes, and then as $n$ increases, you add radial nodes. Examples: First shell For the first shell, $n=1$, which means the number of nodes will be 0. Examples: Second shell For the second shell, $n=2$, which yields 1 node. For the $2s$ orbital, $\ell = 0$, which means the node will be radial For the $2p$ orbital, $\ell = 1$, which means the node will be angular Examples: Third shell The third shell, $n=3$, yielding $3-1=2$ nodes. The $3s$ orbital still has $\ell = 0$ meaning no angular nodes, and thus the two nodes must be radial The $3p$ orbital still has one angular node, meaning there will be one radial node as well The $3d$ orbital has two angular nodes, and therefore no radial nodes! 2. The difference between radial and angular nodes Radial nodes are nodes inside the orbital lobes as far as I can understand. Its easiest to understand by looking at the $s$-orbitals, which can only have radial nodes. To see what an angular node is, then, let's examine the $2p$-orbital - an orbital that has one node, and that node is angular. We see that angular nodes are not internal countours of 0 electron probability, but rather is a plane that goes through the orbital. For the $2p_\text{z}$-orbital, the angular node is the plane spanned by the x- and y-axis. For the $2p_\text{y}$-orbital, the angular node is the plane spanned by the z- and x-axis. | {
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8,910 | Quoting from this site : As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol. It's also quite easy to see that this is true glancing at various dye absorption tables. But why is this true? I can understand why the optical cross-section would increase, implying a higher quantum efficiency at some maximum absorbance wavelength (is this generally true?), but why would the energy gap between the ground state and the excited state correspondingly shrink? | Klaus Warzecha's answer pretty much answers your question. But I know that this subject is easier to understand if supported by some pictures. That's why I will take the same route as Klaus at explaining the concept behind why the absorption in conjugated systems is shifted to higher wavelengths but I will provide some pictures on the way. In a conjugated carbon chain or ring system you can think of the $\ce{C}$ atoms as $\text{sp}^{2}$-hybridized. So, each carbon has 3 $\text{sp}^{2}$ orbitals which it uses to form $\sigma$ bonds and 1 $\text{p}$ orbital which is used to form $\pi$ bonds.
It is the $\text{p}$ orbitals that are responsible for the conjugation and their combinations according to the LCAO model are the interesting part since the HOMO and LUMO of the system will be among the molecular orbitals formed from the conjugated $\text{p}$ orbitals. For a start take ethene, the simplest $\pi$-system, being comprised of only 2 carbon atoms.
When you combine two atomic orbitals you get two molecular orbitals.
These result from combining the $\text{p}$ orbitals either in-phase or out-of-phase.
The in-phase combination is lower in energy than the original $\text{p}$ orbitals and the out-of-phase combination is higher in energy than the original $\text{p}$ orbitals.
The in-phase combination accounts for the bonding molecular orbital ($\pi$), whilst the out-of-phase combination accounts for the antibonding molecular orbital ($\pi^{*}$). Now, what happens when you lengthen the conjugated system by combining two ethene fragments?
You get to butadiene.
Butadiene has two $\pi$ bonds and so four electrons in the $\pi$ system.
Which molecular orbitals are these electrons in?
Since each molecular orbital can hold two electrons, only the two molecular orbitals
lowest in energy are filled.
Let's have a closer look at these orbitals.
In $\Psi_1$, the lowest-energy bonding orbital, the electrons are spread out over all four carbon atoms (above and below the plane) in one continuous orbital.
There is bonding between all the atoms.
The other two electrons are in $\Psi_2$.
This orbital has bonding interactions between carbon atoms 1 and 2, and also between 3 and 4 but an antibonding interaction between carbons 2 and 3.
Overall, in both the occupied $\pi$ orbitals there are electrons between carbons 1 and 2 and between 3 and 4, but the antibonding interaction between carbons 2 and 3 in $\Psi_2$ partially cancels out the bonding interaction in $\Psi_1$.
This explains why all the bonds in butadiene are not the same and why the middle bond is more like a single bond while the end bonds are double bonds.
If we look closely at the coefficients on each atom in orbitals $\Psi_1$ and $\Psi_2$, it can be seen that the bonding interaction between the central carbon atoms in $\Psi_1$ is greater than the antibonding one in $\Psi_2$.
Thus butadiene does have some double bond character between carbons 2 and 3, which explains why there is the slight barrier to rotation about this bond. You can construct the molecular orbitals of butadiene by combining the molecular orbitals of the two ethene fragments in-phase and out-of-phase. This method of construction also shows why the HOMO-LUMO gap of butadiene is smaller than that of ethene.
The molecular orbital $\Psi_2$, which is the HOMO of butadiene, is the out-of-phase combination of two ethene $\pi$ orbitals, which are the HOMO of ethene.
Thus, the HOMO of butadiene is higher in energy than the HOMO of ethene.
Furthermore, the molecular orbital $\Psi_3$, which is the LUMO of butadiene, is the in-phase combination of two ethene $\pi^{*}$ orbitals, which are the LUMO of ethene.
Thus, the LUMO of butadiene is lower in energy than the LUMO of ethene.
It follows that the HOMO-LUMO energy gap is smaller in butadiene than in ethene and thus butadiene absorbs light with longer wavelenghts than ethene. If you continue to lengthen the $\pi$ system by adding more ethene fragments you will see that the HOMO and LUMO are getting closer and closer together the longer the $\pi$ system becomes. | {
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8,948 | I mean, how can we reverse the neutralization reaction of $\ce{HCl}$ and $\ce{NaOH}$ to get back $\ce{HCl}$ and $\ce{NaOH}$. I want to extract $\ce{HCl}$ and $\ce{NaOH}$ from a mixture of common salt and distilled water in equal proportions at home. Basically, how can I make this reaction happen? $$\ce{H2O + NaCl -> HCl + NaOH}$$ What temperature or pressure conditions are required? Also, would the extracted components be pure? (I know common salt contains iodine and even distilled water is not pure, that's why I am asking this.) | Unless time travel is an option, you could Electrolyze the $\ce{NaCl}$ solution to obtain a solution of $\ce{NaOH}$, and $\ce{H2}$ and $\ce{Cl2}$ as gases. Collect the gases and photolyze them. The dissociation energy of $\ce{Cl2}$ is $243\, \mathrm{kJ \cdot mol^{-1}}$, irradiation at $\lambda$ < 490 nm will cleave $\ce{Cl2}$ to chlorine radicals and initiate the chain reaction to yield $\ce{HCl}$ gas. $$\ce{Cl2 ->[h\nu] 2 Cl*}$$ $$\ce{Cl* +\ H2 -> HCl + H*}$$
$$\ce{H* +\ Cl2 -> HCl + Cl*}$$ EDIT 1 Unless you know exactly about the risk of both processes, don't do it! This video gives a nice impression of how violently hydrogen and chlorine react upon irradiation! Note that it has nothing to do with the laser involved. A flashlight would give the same dramatic effect! EDIT 2 The answer above, giving practical advices on how to get $\ce{HCl}$ and $\ce{NaOH}$ back apparently suggests that you can not simply revert the neutralisation reaction. The neutralisation is irreversible. In order to find out why this is the case, you might want to have a look at thermodynamics of this exothermic reaction, particularly at the entropy changes. | {
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8,960 | My textbook says that for a chemical reaction with two reactants, $\ce{A}$ and $\ce{B}$ with equation $\ce{aA + bB -> Products}$, where $a$ and $b$ are coefficients, the general rate law is: $\text{Rate} = k[\ce{A}]^m[\ce{B}]^n$ where $m$ and $n$ are the reaction orders for $\ce{A}$ and $\ce{B}$ respectively. Then it says that $m=a$ and $n=b$ if the reaction between $\ce{A}$ and $\ce{B}$ occurs in a single step and with a single activated complex. Why is this true? | Unless time travel is an option, you could Electrolyze the $\ce{NaCl}$ solution to obtain a solution of $\ce{NaOH}$, and $\ce{H2}$ and $\ce{Cl2}$ as gases. Collect the gases and photolyze them. The dissociation energy of $\ce{Cl2}$ is $243\, \mathrm{kJ \cdot mol^{-1}}$, irradiation at $\lambda$ < 490 nm will cleave $\ce{Cl2}$ to chlorine radicals and initiate the chain reaction to yield $\ce{HCl}$ gas. $$\ce{Cl2 ->[h\nu] 2 Cl*}$$ $$\ce{Cl* +\ H2 -> HCl + H*}$$
$$\ce{H* +\ Cl2 -> HCl + Cl*}$$ EDIT 1 Unless you know exactly about the risk of both processes, don't do it! This video gives a nice impression of how violently hydrogen and chlorine react upon irradiation! Note that it has nothing to do with the laser involved. A flashlight would give the same dramatic effect! EDIT 2 The answer above, giving practical advices on how to get $\ce{HCl}$ and $\ce{NaOH}$ back apparently suggests that you can not simply revert the neutralisation reaction. The neutralisation is irreversible. In order to find out why this is the case, you might want to have a look at thermodynamics of this exothermic reaction, particularly at the entropy changes. | {
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9,242 | I'm a mathematician who's currently teaching a course on differential equations. Though I don't know much about chemistry, I like to include examples from chemistry in my course, and I prefer for the details to be accurate. Here is a typical exam problem: When a container of gaseous nitrogen dioxide is heated above 150 °C, the gas begins to decompose into oxygen and nitric oxide:
$$\ce{2NO2 -> O2 + 2NO}$$
The rate of this reaction is determined by the equation
$$\frac{\mathrm d\left[\ce{NO2}\right]}{\mathrm dt} = -k\left[\ce{NO}_2\right]^2$$
where $k$ is a constant. (a) Find the general solution to the above equation. (b) A large container holds 50.0 moles of $\ce{NO2}$ at a constant temperature of 600 °C. After one hour, only 34.3 moles remain. How much $\ce{NO2}$ will there be after another hour? So my questions are: Is the science in this problem reasonably accurate? Is there anything you would change? (I looked up a suitable value of the rate constant $k$ to make sure that the time in part b was reasonable.) What are some other examples of reactions that are governed by simple rate laws? Ideally, I'd like to have several examples each of reactions governed by the equations
$$
\frac{\mathrm dy}{\mathrm dt} = -ky,\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^2,\qquad\text{and}\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^3.
$$
(Is $y^3$ really possible? Are non-integer powers of $y$ possible?) | How thoughtful of you to include chemistry in your differential equations course! We appreciate your effort, especially going the extra mile to make it realistic. I hope you've seen the rate equation page on Wikipedia as it contains a good deal of the mathematics in several examples of reactions. You should also be interested in the reaction order page (you can find examples here of reactions with fraction or negative order with respect to some reagents). Let me do a quick recap on basic chemical kinetics theory. Consider the general reaction equation: $$\ce{a\ A + b\ B + c\ C + d\ D + ... \longrightarrow w\ W + x\ X + y\ Y + z\ Z + ...}$$ , where uppercase letters indicate different molecules in a gas or solution, and lowercase letters indicate the stoichiometric coefficients (negative for the reactants, positive for the products). Naively, we can assume the reaction happens because, while the molecules are jostling about, at some point all of the reactants ($a$ molecules of $\ce{A}$, $b$ molecules of $\ce{B}$, $c$ molecules of $\ce{C}$, etc.) will bump into each other all at the same time and undergo reaction. If you think of collisions as independent events with a probability of happening proportional to the concentration of each species, then it isn't too hard to understand that the frequency with which reactions happens is proportional to $[A]^a[B]^b[C]^c[D]^d...$, and so the rate equation for the reaction would be: $$r=k[A]^a[B]^b[C]^c[D]^d...$$ , where r is the reaction rate and k is a proportionality constant. The consumption or production rate for any species $\Gamma$ with stoichiometric coefficient $\gamma$ is trivially related to the reaction rate by: $$\frac{d\Gamma}{dt}=\gamma r$$ However, experimentally things aren't so simple (fortunately for nature, unfortunately for our minds). A reaction that requires the simultaneous collision of many molecules would be highly unlikely, meaning that it would happen very, very slowly. Even so, experimentally we observe many reactions that involve a large amount of molecules. For example, the combustion of cyclohexane in air is formally given by the equation: $$\ce{C6H12 + 9 O2 → 6CO2 + 6 H2O}$$ If the reaction really required ten molecules to bump into eachother at the same time with sufficient energy and in the right geometry, then this reaction probably could not happen in air, as oxygen would be too rarefied to compensate the extremely low proportionality constant k for a single-step reaction. In actuality, a gaseous mix of cyclohexane and air can react so fast as to cause an explosion, converting all the reactants into products in a miniscule fraction of a second. Clearly something is wrong. It turns out the assumption that reactions occur in one step is incorrect in general. Usually, there is more than one step involved, and indeed there generally is more than one path from reactants to products. The speed at which a reaction occurs is really described by considering all paths at once, with all their steps, and adding their contributions (like a start and end point in a complex tree diagram). In general this is pretty difficult and unnecessarily complex. It is often a reasonable approximation to model the reaction by selecting only the fastest route from reactants to products, and the reaction rate is bottlenecked by the slowest step of the fastest route, which becomes responsible for determining the rate equation. Now, since the reaction rate depends on selecting the fastest route, you shouldn't expect any of the steps in it to contain some very huge bottleneck, such as requiring six molecules to react simultaneously; there would likely be a slightly different route, perhaps with more steps, but in which each step doesn't involve so many molecules interacting at once, and hence being a faster route. Because of this, it turns out that most of the fastest reaction pathways involve slowest steps that very rarely depend on more than two molecules at once. Most reactions which can be modeled simply like this therefore have rate equations of the type $r=k[A]$, $r=k[A]^2$, $r=k[A][B]$ or even $r=k$. Few reactions rates are of the third order, of the type $r=k[A]^3$, $r=k[A]^2[B]$, $r=k[A][B][C]$ or similar. I don't know of any reaction that goes as $r=k[A]^3$ specifically. Fourth order reaction rates are so rare that they are the focus of research when found . I don't expect anyone to know a fifth order reaction. At the end of the day, kinetic theory is hard enough that reaction kinetics end up being determined simply by parameter fitting from experiments. Even if the underlying reaction mechanism is not completely understood, if a fit of the type $r=k[A]^{3/2}[B]^{-1}$ happens to be good, then so be it. Reactions which have been studied in depth are significantly more complex, such as the misleading simplicity of the reaction between hydrogen and bromine . Edit: I don't know why I didn't search for a fifth order reaction rate! Turns out some exist, such as this one . The reaction rate is of the type $r=k[A][B]^4$, so you could have an example where $\frac{d[B]}{dt}=k[B]^4$. | {
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9,395 | Among my friends it is a sort of 'common wisdom' that you should throw away water after a couple of days if it was taken from the tap and stored in a bottle outside the fridge, because it has 'gone bad'.
First of all, the couple of days is not very well defined, which already makes me a bit suspicious. Second, I cannot think of anything in tap water that would make the water undrinkable after a couple of days already. Can someone clarify this issue for me? Does tap water really 'go bad' after a couple of days outside the fridge? Why? | First of all, it depends on how the tap water was treated before it was piped to your house. In most cases, the water was chlorinated to remove microorganisms. By the time the water arrives at your house, there is very little (if any) chlorine left in the water. When you fill you container, there is likely to be some microorganisms present (either in the container or in the water).
In a nutrient rich environment, you can see colonies within 3 days. For tap water, it will probably take 2 to 3 weeks. But that doesn't mean that the small amount of growth doesn't produce bad tasting compounds (acetic acid, urea, etc.). BTW Nicolau Saker Neto, cold water dissolves more gas than hot water. Watch when you heat water on your stove. Before it boils, you will see gas bubbles that form on the bottom and go to the surface (dissolved gases) and bubbles that disappear while rising to the surface (water vapor). | {
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9,847 | Although the wikipedia page on Hydronium indicates a $\mathrm{p}K_\text{a}$ of −1.74, I noticed in the discussion of this page that the subject seems debated (cf. http://en.wikipedia.org/wiki/Talk:Hydronium#pKa.3F ) since alternative (apparently more rigorous) reasoning leads to $\mathrm{p}K_\text{a} = 0$ for $\ce{H3O+}$. Also, many demonstrations lead to $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of 15.74 for $\ce{H2O}$ and thus, using $K_\text{w}=[\ce{H+}][\ce{OH-}]=10^{-14}$, give $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of −1.74 for $\ce{H+}$ and $\ce{OH-}$ respectively. But those demonstrations seem to make use of the concentration $[\ce{H2O}]=55.56~\text{mol/L}$ instead of the activity $a_{\ce{H2O}}=1$ (for a solvent) in the equilibrium constant of the reactions. So it seems to me that the $\mathrm{p}K_\text{a}$ of $\ce{H3O+}$ must be equal to $0$. Is this question still debated? PS1: To add details about the question, the debate in the references mentioned in the discussion on the wikipedia page as well as in the link provided by Nicolau Saker Neto bellow seems to boil down to the coexistence of two different definitions for the equilibrium constant: - one based on the activities of the compounds - the other derived from the 'mass action law' and which thereby uses the molar fractions of the compounds Reading through the Callen (Thermodynamics and an Introduction to Thermostatistics) gives me the impression that the derivation from the mass action law of an equilibrium constant involving molar fractions is based on the assumption of an ideal fluid. However, it is precisely the departure from ideality that seems to justify the use of activities instead of molar fractions. Is that correct? PS2: Another point, which I have not looked in details yet, seems to be a debate about the possibility to apply the definition for $K_\text{a}$ to the water molecules $\ce{H2O}$ considering that they are not a highly diluted solute since they constitute the solvent. Is this a valid issue and should then the acidity of $\ce{H3O+}/\ce{H2O}$ be determined in another solvent than $\ce{H2O}$ to be compared with the other acids? | The controversy surrounding the $\mathrm{p}K_\mathrm{a}$ of hydronium mostly arises from the definition of $K_\mathrm{a}$ or lack thereof. There is no IUPAC definition of $\mathrm{p}K_\mathrm{a}$ or $K_\mathrm{a}$. The closest IUPAC defined term is the standard equilibrium constant , which can be denoted as $K^\circ$ or just $K$. Physical chemistry texts such as Levine and respected works such Bates's "Determination of pH — Theory and Practice" define $K^\circ_\mathrm{a}$ of an acid in water as: $$\frac{a(\ce{A-})a(\ce{H3O+})}{a(\ce{HA})a(\ce{H2O})} \tag{1}$$ Where $a$ is activity . Substituting that the acid is $\ce{H3O+}$: $$\frac{a(\ce{H2O})a(\ce{H3O+})}{a(\ce{H3O+})a(\ce{H2O})} = 1 \tag{2}$$ and of course $-\log(1) = 0$. The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from: omitting the activity of water from the denominator of the $K^\circ_\mathrm{a}$ definition (equation $(1)$); and taking the concentration of water (about $55.5~\mathrm{M}$) as the $K_\mathrm{a}$ of $\ce{H3O+}$. With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$. For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition. However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue: New point of view on the meaning and on the values of Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water , which insists that $$\ce{H2O + H3O+ <=> H3O+ + H2O}$$ "does not correspond to an actual chemical process" and therefore "it is not legitimate" to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water. The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids. The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids. | {
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10,044 | I'm making some graphs and I have to label the axes. I want to be extra careful and put the units in even though the meaning of $\text{pH}$ is well known. But I have a problem (though a simple one): $\text{pH}$ is a minus logarithm (base 10) of concentration of hydrogen ions (or rather their activity). What is the unit then, is it $[-\log(\text{mol}/\text{L})]$? What should I write, could you help me? | The real definition of the $\text{pH}$ is not in terms of concentration but in terms of the activity of a proton, \begin{equation}
\text{pH} = - \log a_{\ce{H+}} \ ,
\end{equation} and the activity is a dimensionless quantity. You can think of the activity as a generalization of the mole fraction that takes into account deviations from the ideal behaviour in real solutions.
By introducing the (dimensionless) activity coefficient $\gamma_{\ce{H+}}$, which represents the effect of the deviations from the ideal behaviour on the concentration, you can link the activity to the concentration via \begin{equation}
a_{\ce{H+}} = \frac{\gamma_{\ce{H+}} c_{\ce{H+}}}{c^0} \ ,
\end{equation} where $c^0$ is the standard concentration of $1 \, \text{mol}/\text{L}$.
If you ignore the non-ideal contributions you can approximately express the $\text{pH}$ in terms of the normalized proton concentration \begin{equation}
\text{pH} \approx - \log \frac{c_{\ce{H+}}}{c^0} \ .
\end{equation} In general, there can be no logarithm of a quantity bearing a unit. If however you encounter such a case it is usually due to sloppy notation: either the argument of the logarithm is implicitly understood to be normalized and thus becomes unitless or the units in the logarithm's argument originate from using the mathematical properties of logarithms to divide the logarithm of a product which is by itself unitless into a sum of logarithms: $\log(a \cdot b) = \log(a) + \log(b)$. | {
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10,419 | Over the course of my studies, my lab coat has absorbed a non-negligible amount of different chemicals, some of them nasty organic polymers, some are more insoluble inorganic compounds like $\ce{Cr2O3}$. Even after washing the lab coat some stains remain. Is there a way to get rid of those, or should I rather buy a new lab coat? | A lab coat will eventually get dirty, that's why it is a good idea to wear it. I had a lot of students complaining about the condition of their coat and they asked me if they could just wash it in the washing machine. My answer was always no . As far as I remember - I am a theoretician now - a labcoat costs much less than a pair of trousers and a T-shirt. It should be renewed annually and disposed of according to the norms how you would dispose of an unknown chemical. You never know where the chemicals incorporated in your coat will end up, so dispose of them as safely as you can. In some countries, it is required by the employer to provide equipment like that and taking care of either cleaning or disposing of it eventually. (In Germany that should be the case.) Wear it as a disposable layer of protection and please - for the safety of the environment and yourself - do never ever put it in a normal washing machine again. As an additional note, if you wash it in the same washer like your normal clothes, how can you assure that it will not end up in your everyday clothes? I can imagine, that having chromium whatever in my underwear would not make me feel alright. | {
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10,653 | In my exam, I was asked why cyclopropane could decolourise bromine water (indicating that it reacted with the bromine). All I could guess was that it is related to the high angle strain in cyclopropane, as the C–C–C bond angle is $60^\circ$ instead of the required $109.5^\circ$. No book I have read mentions this reaction. What is the product formed, and why does it occur? | The following ring opening reaction will occour: You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider p-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of s- and p-type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$. In cyclopropane $\ce{C3H6}$ - which you can also describe as trimethylene $\ce{(CH2)3}$ - bonds have to be bent to overlap at all. A possible way of describing the bonding situation is regarding each $\ce{CH2}$ entity as $\mathrm{sp^2}$ hybridised. Two of these orbitals are used for $\ce{C-H}$ bonds (not shown) and one forms an inner two-electron-three-centre σ bond (left). This leaves p-orbitals to form some kind of degenerate π-like orbitals (middle, right). This very general approach can be derived from a Walsh diagram . Schwarz et.al. { @academia.edu } and Hoffmann { @roaldhoffmann.com } described bonding quite similar and it is in quite good agreement with a calculation (BP86/cc-PVTZ, $D_\mathrm{3h}$) I have done. From this I have prepared a chart of all occupied molecular orbitals formed from valence orbitals and the LUMO . Here is a preview. Each orbital is viewed from three different angles: Especially the symmetrical orbital 8 resembles very well the schematics. A quite rigorous approach for this theory can also be found here . It is noteworthy - as mentioned by ron - that there is no notable increase in electron density in the centre of the ring. This may be due to the fact that there are much more orbitals having nodes in the centre than there are without. Now bromine is known to be easily polarised $\ce{{}^{\delta+}Br-Br^{\delta-}}$ and may intercept at any point of the ring causing a bond break and relaxation to a less strained structure. It will most likely attack at the the $\pi$ type orbitals since bromine is an electrophile. The mechanism is analogous to the addition of bromine to ethene, which is nicely described at chemguide.co.uk . The essential part is the attack of the bromine at the HOMO(s). The ring opening reaction can be reversed by adding sodium . However, when there are bromine radicals present (UV light) then substitution will occur:
\begin{aligned}\ce{
Br2 &->[\ce{h\nu}] 2Br.\\
&+(CH2)3 -> (CH2)2(CHBr) + HBr
}\end{aligned} | {
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11,048 | In mineralogy class, I was taught that metallic and ionic bonds are weaker than covalent bonds and that's why quartz and diamond have such a high hardness value. However, in organic chemistry class, I learned that covalent bonds are weaker than metallic and ionic bonds, thus organic substances have a much lower melting point than that of metals and ionic compounds. What am I getting wrong? Are ionic and metallic bonds weaker than covalent bonds or not? | Quartz and diamond are stronger substances because their molecules form network covalent structures. These structures form a lattice-like structure, much the same as ionic compounds. This molecular network is also the reason that diamond and quartz form a crystalline structures, just like you'd see in ionic substances such as NaCl. Some other structures you might want to look into are Graphite and Graphene, which are both allotropes of carbon (allotropes are, simply put, different molecular arrangements of an element). The network structure combines to make the substance stronger than normal covalent bonded substances. So to answer your question, substances with standard covalent bonds seem to be weaker than those with ionic bonds because the ionic bonds tend to form a lattice structure, that makes them much stronger. You can see this in the fact that the boiling points of ionic salts are much higher than that of a covalent substance like water. However, when covalent bonds form network covalent structures, atoms combine to form a singular macromolecule that is much stronger than singular covalent bonds. | {
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11,169 | I have asked my teacher, as she was introducing the concept of mole to us, why that number was chosen, instead of more convenient one. She told me that it came from the definition of the mole, that is the number of atoms in 12g of Carbon 12. When I asked why that definition was chosen, she answered me that my question wasn't really a question, and I understood from her stare that asking why? twice in a row was not appreciated. However, I certainly hope that this definition wasn't chosen randomly and that there is a reason for it being it. I gave it some thought and I came to believe that it may be for convenience after all. There are 12 nucleons in a carbon atom and if we say that the electrons mass can be neglected, then we get that a mole is the number of nucleon it takes to get one gram of them. However, can we really neglect the mass of electrons, when the atoms get bigger? And if so, why do the element's mass per mole do very rarely end up being integers? This makes me think there may be another reason, or some things I do not understand. And if there are no reason, then I guess I might just as well be doing religious studies..! | Why is the definition of the moles as it is? It is a rather arbitary definition that the mole is the number of atoms in 12g of carbon 12. This has not always been the definition. For example, prior to 1960, the definition was based upon oxygen rather than carbon-12. The first standard was based upon 1 gram of hydrogen. Later, the standard was changed to 16 grams of oxygen being a mole, for convenience because oxygen formed compounds with many other elements. Eventually it was realized that elements including oxygen have different isotopes of different mass, and the mass of a particular isotope is a more specific and measurable standard than the mass of natural abundance oxygen. 12 grams of carbon-12 matches the old standard of 16 grams of natural abundance oxygen more closely than choosing 16 grams of oxygen 16. can we really neglect the mass of electrons, when the atoms get bigger? The mass of a proton or neutron is about 1836 times that of an electron. So depending upon the ratio of neutrons to protons, electrons are at most 1/1836 of the mass of a neutral atom. If accuracy of more than 1 part in 1836 is desired, it is important to consider the electrons, regardless of whether atoms are big or small. why do the element's mass per mole do very rarely end up being integers? No element other than carbon-12 will exactly be an interger. This is because masses of atoms depend upon number of protons, neutrons, electrons and binding energy . In other words, considering that carbon-12 has an equal number of protons and neutrons (and electrons), other isotopes with a 1:1 proton/neutron ratio would have essentially integral atomic masses except for binding energy. For example cadmium 112 is 111.90 instead of exactly 112 because binding energy is slightly more than 0.1 amu, partially offest by a higher number of neutrons than protons, the neutron mass being slightly greater than the combined mass of an electron and proton. As pointed out by Matt Black, the natural abundance atomic masses further deviate because they are weighted averages of the masses of isotopes of a given element, weighted by their abundance on Earth. | {
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12,404 | Our teachers made us accept (without any explanation) that diamond conducts heat better than graphite. What is the reason behind this (alleged) fact? | Diamond is one of the best thermal conductors known, in fact diamond is a better thermal conductor than many metals (thermal conductivity (W/m-K): aluminum=237, copper=401, diamond=895). The carbon atoms in diamond are $\ce{sp^3}$ hybridized and every carbon is bonded to 4 other carbon atoms located at the vertices of a tetrahedron. Hence the bonding in diamond is a uniform, continuous 3-dimensional network of $\ce{C-C}$ single (sigma) bonds. Graphite on the other hand is formed from $\ce{sp^2}$ hybridized carbon atoms that form a continuous 2-dimensional sigma and pi bonding network. This 2-dimensional network forms sheets of graphite, but there is little connection between the sheets, in fact, the sheet-sheet separation is a whopping ~3.4 angstroms. This might lead us to suspect that heat conduction in the 2-dimensional sheet of graphite would be superior to diamond, but that heat conduction between graphite sheets would be very low. This is, in fact, an accurate description of thermal conduction in graphite. Thermal conductivity parallel to the graphite sheets=1950, but thermal conduction perpendicular to the sheet=5.7. Therefor, when we consider thermal conduction over all possible directions (anisotropic) diamond would be superior to graphite. | {
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13,907 | Dipole moment is the degree of polarity, i.e. the seperation of positive and negative charges. But I am not getting the intuition why and how lone pairs affect the polarity and dipole moment. I cannot connect the separation of positive and negative charges definition with lone pair. | How does lone pair of a central atom affect the dipole moment? There is no single answer to your question, let me explain. Unlike a typical covalent bond where the electrons are shared between two nuclei and the electron density is spread out over the entire bond, in a lone pair the electrons are not shared and the electron density is more localized around the atom that has the lone pair of electrons. This increased electron density could lead to a more significant contribution from the lone pair electrons to the molecular dipole moment than from electrons spread out more diffusely in a covalent bond. Next we must understand the directionality of the lone pair of electrons. Consider the two molecules pictured below, ammonia and phosphine. The molecules appear to be very similar, they are in the same column in the Periodic Table. However in ammonia the $\ce{H-N-H}$ angle is around 107 degrees and the molecule is roughly $\ce{sp^3}$ hybridized, the lone pair and the 3 $\ce{N-H}$ bonds roughly pointing towards the corners of a tetrahedron. You can see that in this case (as shown by the arrows, the "arrowhead" end representing the negative end of a dipole), the lone pair on nitrogen will make a contribution to the molecular dipole moment. Next, let's examine phosphine. The $\ce{H-P-H}$ angle is around 90 degrees and the molecule can be viewed as being unhybridized, the lone pair is an $\ce{s}$ orbital and the 3 $\ce{P-H}$ bonds are constructed from phosphorous $\ce{p}$ orbitals. You can see that in this case, the lone pair on phosphorous, due to its spherical symmetry will not make a contribution to the overall molecular dipole moment. So in summary, a lone pair of electrons can make a significant contribution to the magnitude of a molecular dipole moment due to the fact that they are more localized than bonding electrons and consequently there is a high electron density. But, the directionality (or lack thereof) of the lone pair must also be assessed, since a lack of directionality may preclude it from making a significant contribution to the overall molecular dipole moment. | {
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14,087 | $\ce{N}$ & $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized. But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference? | Starting point: 2s orbitals are lower in energy than 2p orbitals. The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees. Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron. In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees. This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus. So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds. There are two choices for the central atom: remain unhybridized: [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another. As a result, substituents will be arranged closer together and destabilizing steric interactions will be increased due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital or hybridize: [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another. As a result, steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character An example: Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$. Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine. Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density. Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital. The orbital will wind up with higher p-character and the s-character that has been "saved" can be used to stabilize other electrons (the lone pair!). Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds. As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$. Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia! Back to our problem: Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2). In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$. Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed. The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine. In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the hybridized case the lowest energy. Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable. | {
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14,267 | I was making pasta, and I noticed the pasta boiling over. I thought about it some more, and I realized I had no idea why this was happening. When the lid is on, the foam rises. When the lid is off, the foam dies back. Clearly there's some surfactant at work allowing bubbles to form, and the bubbles are mostly full of steam. If you remove the lid, they cool and condense down to a tiny size. This article is honestly just not specific enough. Why is the starch important?(It also happens with potatoes) How does a polysaccharide, a hydrophilic molecule, become a surfactant(if indeed it does)? Is it phospholipids from the cells that became pasta? Do starch molecules form a polymer and trap steam underneath like a balloon? Is it some kind of hybrid where bubbles are stabilized by viscosity? What's going on here? This question on seasoned advice tries to tackle it but gets surface tension wrong(I think) so that makes me a little wary of it. Surfactants like soap reduce the surface tension, allowing bubbles to form by letting water molecules spread out into thin films. They also keep your alveoli in your lungs from collapsing. If starches increased the surface tension, wouldn't that reduce the likelihood of forming bubbles by increasing the energetic penalty? High surface tension materials act like mercury. This answer is all over the internet but doesn't make any sense to me. You need less surface tension for bubbles, surely? If it's just phospholipids acting as detergent, could I 'boil over' pasta in cold water with a whisk? (I tried this, you can't.) I considered putting this on seasoned advice, but I'm not interested in how to prevent boil-over(or really any practical results), and the supplied answer there lacks scientific rigor. | The starch forms a loosely bonded network that traps water vapor and air into a foamy mass, which expands rapidly as it heats up. Starch is made of glucose polymers (amylopectin is one of them, shown here): Some of the chains are branched, some are linear, but they all have $\ce{-OH}$ groups which can form hydrogen bonds with each other. Let's follow some starch molecules through the process and see what happens. In the beginning, the starch is dehydrated and tightly compacted - the chains are lined up in nice orderly structures with no water or air between them, maximizing the hydrogen bonds between starch polymers: As the water heats up (or as you let the pasta soak), water molecules begin to "invade" the tightly packed polymer chains, forming their own hydrogen bonds with the starch: Soon, the polymer chains are completely surrounded by water, and are free to move in solution (they have dissolved): However, the water/starch solution is not completely uniform. In the middle of the pot of water, the concentration of starch is low compared to water. There are lots and lots of water molecules available to surround the starch chains and to keep them apart. Near the surface, when the water is boiling, the water molecules escape as vapor. This means that near the surface, the local concentration of starch increases. It increases so much as the water continues to boil, that the starch can collapse back in on itself and hydrogen bond to other starch molecules again. However, this time the orderly structure is broken and there is too much thermal motion to line up. Instead, they form a loosely packed network of molecules connected by hydrogen bonds and surrounding little pockets of water and air (bubbles): This network is very weak, but it is strong enough to temporarily trap the air as it expands due to heating - thus, the bubbles puff up and a rapidly growing foam forms. Since they are very weak, it doesn't take much to disrupt them. Some oil in the water will inhibit the bubbles from breaking the surface as easily, and a wooden spoon across the top will break the network mechanically as soon as it touches it. Many biomolecules will form these types of networks under different conditions. For example, gelatin is a protein (amino acid polymer) that will form elastic hydrogen-bonded networks in hot water. As the gelatin-water mixture cools, the gel solidifies, trapping the water inside to form what is called a sol-gel , or more specifically, a hydrogel . Gluten in wheat is another example, although in this case the bonds are disulfide bonds. Gluten networks are stronger than hydrogen-bonded polysaccharide networks, and are responsible for the elasticity of bread (and of pasta). DISCLAIMER: pictures are not remotely to scale, starch is usually several hundred glucose monomers long, and the relative size of the molecules and atoms isn't shown. there aren't nearly enough water molecules - in reality there would be too many to be able to see the polymer (1,000's). the starch molecules aren't "twisty" enough or showing things like branching - the real network structure and conformations in solution would be much more complicated. But, hopefully you get the idea! | {
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14,473 | Consider an acidic solution with Hydrogen ion concentration, $\ce{[H+]}$ of $10^{-5}\:\mathrm{M}$. Since $\:\mathrm{pH} = -\log \ce{[H+]}$ the $\:\mathrm{pH}$ of solution is $5$. Suppose we dilute solution 10 times with water. Now, $\ce{[H+]}$ is $10^{-6}\:\mathrm{M}$ and $\:\mathrm{pH}$ is $6$. Further dilution should increase $\:\mathrm{pH}$ from $6$ to $7$ and then from $7$ to $8$ and so on. Can this go in for ever? Does this not imply that an acidic solution can be made basic/alkaline simply by adding water? But that doesn't happen? What prevents it? Is there anyone already found answer for this problem or it is just an unsolved basic problem of chemistry? | Water undergoes autoionization , i.e., it reacts as follows: $$
\ce{H2O + H2O <=> H3O+ + OH-}
$$ The equilibrium constant for this reaction at standard conditions is $K_w = [\ce{H3O+}][\ce{OH-}] \approx 1.0 \cdot 10^{-14}$. In pure water, $[\ce{H3O+}] = [\ce{OH-}]$, hence $[\ce{H3O+}] = \sqrt{K_w} \approx 1.0 \cdot 10^{-7}\ \textrm{M}$. Suppose we dilute a solution with some initial concentration of $[\ce{H3O+}]_i = \frac{n_i}{V_i}$, where $n_i$ is the initial number of moles of $\ce{H3O+}$ and $V_i$ is the initial volume. If I now add a volume $\Delta V$ of pure water (which would contain $(1.0 \cdot 10^{-7}) \cdot \Delta V$ moles of $\ce{H3O+}$) the resulting final concentration can be crudely approximated as: $$
[\ce{H3O+}]_f \approx \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V}
$$ We can see what value this expression approaches as we make the solution more and more dilute by taking the limit as $\Delta V \to \infty$: $$
\lim_{\Delta V \to \infty} \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} = 1.0 \cdot 10^{-7}
$$ Therefore, $[\ce{H3O+}]$ tends towards $1.0 \cdot 10^{-7}$ with further dilution, so $pH$ will approach a value of $7.0$. | {
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14,981 | I know that bond angle decreases in the order $\ce{H2O}$, $\ce{H2S}$ and $\ce{H2Se}$. I wish to know the reason for this. I think this is because of the lone pair repulsion but how? | Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths:
\begin{array}{lcc}
\text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\
\hline
\ce{H2O} & 104.5 & 96 \\
\ce{H2S} & 92.3 & 134 \\
\ce{H2Se}& 91.0 & 146 \\
\hline
\end{array} The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization. That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively. Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons. If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider. This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case. The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital. | {
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15,007 | Why is the recipe for Coca-Cola still a secret?
I think that given the current state of technology, it should be proficient enough to find any of the secret ingredients in Coca Cola. Any thoughts?
Can we make a 100% identical clone of Coca-Cola next week? | There are different angles this question can be answered: Chemical point of view: A full analysis of a totally unknown mixture is painful and extremely costly. It is always helpful to know how many components you are looking for; what types etc. In this case, it is not enough to analyse the elemental composition or some pure elements, but Coke contains a lot of natural products and mixtures like caramel. Imagine how you identify (and correctly describe the production parameters) caramel in a dilute solution... Also, liquids with extremely high concentrations of sugar and different acids are unpleasant for analysis, as you often have to separate these main component, so you can identify only small components. That being said, it is not an impossible task; there are many methods to identify e.g. natural products based on DNA traces. But consider another factor: Business point of view: The recipe of Coca Cola itself worth nothing.
Most probably any competent soda maker can make a drink that 99% of the consumers cannot distinguish from Coke by the taste (and could do that decades before). However they don't do because no one would buy a drink that tastes like Coca Cola, but made by others.
The value is in the brand. People buy a fake Rolex if they cannot afford a real. But anyone can afford a Coke- why would buy a fake? If you are in beverage business, it is an imperative to make a drink that is somehow DIFFERENT than the other ones! | {
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15,017 | What is the composition of liquid nitrogen fumes? Is direct contact with these fumes harmful? | The visible fog that forms when liquid nitrogen is poured into an open container is almost entirely water fog: the boiling nitrogen chills the air above it, causing the humidity in the air to condense into fog. There may be some microdroplets of liquid nitrogen in the fog too, but the air, even after it's been chilled, is still much warmer than the boiling point of nitrogen, so there won't be much. (You can condense air by running a continuous stream of liquid nitrogen through a heat exchanger and blowing air over it, but mostly you get liquid oxygen when you do that.) Liquid nitrogen has a relatively small heat of vaporization, as these things go: 199.2 kJ/kg according to Air Liquide . (Compare 572.2 kJ/kg for dry ice .) That plus the Leidenfrost effect means it's actually rather difficult to freeze yourself with the stuff. There's a standard demo where you pour liquid nitrogen into an audience member's cupped hands. The Leidenfrost effect prevents it ever coming in contact with their skin, and it only leaves a mild chill behind. Try this with dry ice and your "volunteer" will need emergency treatment for frostbite. Gaseous nitrogen is second only to the noble gases in its inertness, and it makes up 78% (by volume) of the air you're breathing as you read this, so obviously the "fumes" aren't harmful in themselves. However, as was mentioned in an earlier answer (which seems to have been deleted), if you let liquid nitrogen evaporate in an enclosed space, it can "displace" the oxygen that makes up another 21% of the air, at which point anyone in the area may asphyxiate . The sensation of suffocation is triggered by having too much CO 2 in your bloodstream, not by inadequate oxygen supply, so you might not even notice that anything was wrong until you collapsed. That's the second most dangerous thing about evaporating liquid nitrogen. The most dangerous thing is, if you let liquid nitrogen evaporate in a sealed container , it's liable to explode on you . | {
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15,035 | I have noticed that liquid $\ce{O2}$ (I clarify it as $\ce{O2}$ because oxygen exists in several other forms which have different colors in the liquid state than $\ce{O2}$) has a light blue color to it. In the solid state it has a light blue color like it does in the liquid state and then finally a red color if enough pressure is applied. It can also be pink, black, or metallic again depending on the pressure and temperature. Does it also have a light blue color in the gas phase? | Background $\ce{O2}$ exists as a paramagnetic, triplet since the two electrons in its two (degenerate) HOMO orbitals are unpaired. There are 6 known phases of solid oxygen with color ranging from pale blue to red to black. In the liquid phase it has a light blue color. This color is due to light absorption by the ground state triplet according to the following equation
$$\ce{2 O2(^3\Sigma_{g}) ->C[{h\nu}]\ 2 O2(^1\Delta_{g})}$$
This absorption requires light from the red region of the spectrum (~630 nm). If red light is absorbed , then blue light is transmitted or reflected and this gives rise to the blue color associated with liquid oxygen. Answer Note that the light absorption requires 2 molecules of $\ce{O2}$ and a photon, a 3-body process. If we now consider the probability of this 3-body process occurring in the gas phase, we can see that, since a gas is much more dilute than a liquid, the probability for of these 3 items coming together at the same time will be much smaller. Hence, the probability of photon absorption in the gas phase is much reduced. To the human eye oxygen gas will appear colorless (at normal and reduced pressures). However, if you place the gas in a cell and record its visible spectrum you will still be able to detect this absorption. | {
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15,579 | What should be the oxidation state of oxygen in $\ce{HOF}$ (hypofluorous acid)? Sources on the internet have confused me. Most state its oxidation state to be 0, while the others state that it is -2 (considering the oxidation state of fluorine to be +1, which is a bit surprising for me as it's the most electronegative element in the periodic table) Source stating the oxidation state to be -2: J. Chem. Educ. 1972 , 49 (4), 299 : The two possibilities brought forth in the Note are (1) assign hydrogen +1, oxygen -2, and fluorine +1 or (2) assign H +1, oxy- gen 0, and fluorine -1. On the grounds that $\ce{HOF}$ is a powerful oxidizing agent, the conclusion reached in the Note is that (1) should be favored. Other sources stating the oxidation state to be 0: Yahoo Answers! Answers.com | Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to Electronegativity Considerations in Assigning Oxidation States . When you apply the official pre-2016 definition (via the Internet Archive) from the IUPAC gold book, then you have to assign $\text{OS}(\ce{H}) = +1$ as it is not a compound with a metal, $\text{OS}(\ce{O}) = -2$ as it is no peroxide compound, and leaving the disturbing $\text{OS}(\ce{F})= \color\red{+1}$ . Going with the on electronegativity based alternative description, you will still assign $\text{OS}(\ce{H}) = +1$ as it has the lowest electronegativity. Then you assign $\text{OS}(\ce{F})= -1$ , because of the highest electronegativity. This leaves a oxidation state of $\text{OS}(\ce{O}) = 0$ . This assignment is matched by the 2016 definition, the summary version of which is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The definition refers to the use of Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail). Consequently the heteronuclear approximation yields: The true charges are of course something completely different. Based on a DF-BP86/def2-SVP calculation I ran some population analyses with Multiwfn 3.4.1 (a newer version is available) . The NBO charges are taken from a prior version of this answer, but I since have lost access to the program and the files used. \begin{array}{lrrr}\hline
\text{Method} & \ce{H} & \ce{O} & \ce{F} \\\hline
\text{Hirshfeld} & +0.18 & -0.10 & -0.09 \\
\text{ADCH} & +0.38 & -0.29 & -0.09 \\
\text{VDD} & +0.18 & -0.10 & -0.09 \\
\text{Mulliken} & +0.20 & -0.06 & -0.15 \\
\text{Löwdin} & +0.09 & +0.01 & -0.10 \\
\text{Becke} & +0.38 & -0.21 & -0.17 \\
\text{ADC Becke} & +0.38 & -0.29 & -0.09 \\
\text{CM5} & +0.35 & -0.26 & -0.08 \\\hline
\text{QTAIM} & +0.61 & -0.44 & -0.17 \\\hline
\text{NBO} & +0.45 & -0.32 & -0.12 \\\hline
\end{array} | {
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15,586 | I recently answered a question about Breaking Bad's initial methamphetamine production method (i.e. the reduction of (pseudo)ephedrine). The reaction is as follows: ( Source ) It is infamously known as the $\ce{HI/P}$ reduction and was (and still is somewhat) a real plague in the United States until cough medicine containing ephedrine came under tightened control. That for an introduction, now the question: what is actually the reaction mechanism of the $\ce{HI/P}$ reduction? The first step seems like a simple $\ce{S_{N}2}$ substitution of $\ce{I^{-}}$, catalyzed by the protonation of the alcohol group by hydroiodic acid (making it a far better leaving group). But what is the mechanism after that? How is phosphorus involved? Does the $\ce{HI/P}$ reduction have any legal use? | According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $\ce{I-}$ to $\ce{I2}$ is proposed. Red phosphorus is required for the regeneration of $\ce{HI}$ from $\ce{I2}$. $\ce{P_{red}}$ reacts with $\ce{I2}$ to the phosphorus iodides $\ce{PI3}$ and $\ce{PI5}$, which are subsequently hydrolyzed to $\ce{H3PO3}$/$\ce{H3PO4}$ and $\ce{HI}$, which is reused in further reduction steps. Catalytic amounts of $\ce{HI}$ are therefore sufficient to run the reaction in the presence of phosphorus. However, a too low concentration of $\ce{I-}$ will lead to elimination instead of substitution in the first reaction step. ( Image source ) The SET mechanism of the reduction is not described in detail in the article, so the following is a bit speculative. As a resonance-stabilized radical is involved, homolytic cleavage of the $\ce{C-I}$ bond is one of the first steps, which gives the benzylic radical $\ce{R.}$ and an iodine radical $\ce{I.}$. A single electron is transferred from $\ce{I-}$ to $\ce{R.}$, forming the carbanion $\ce{R-}$ and $\ce{I.}$, which recombines with another iodine radical to $\ce{I2}$. Protonation of $\ce{R-}$ yields the final product. | {
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15,620 | I'm all bent out of shape trying to figure out what Bent's rule means. I have several formulations of it, and the most common formulation is also the hardest to understand. Atomic s character concentrates in orbitals directed toward electropositive substituents Why would this be true? Consider $\ce{H3CF}$. Both the carbon and the fluorine are roughly $\ce{sp^3}$ hybridized. Given that carbon is more electropositive than fluorine, am I supposed to make the conclusion that because carbon is more electropositive than fluorine, there is a great deal of s-character in the $\ce{C-F}$ bond and most of this s-character is around the carbon? Or is this a misunderstanding of "orbitals directed toward electropositive substituents"? The fluorine is $\ce{sp^3}$ hybridized and these orbitals are "directed" toward the carbon in that the big lobe of the hybrid orbital is pointing toward the carbon. So does electron density concentrate near the fluorine? Because that would make more sense. And this s-character concentrated toward the fluorine has the effect of what on the bond angle? I understand that the more s-character a bond has, the bigger the bond angle - consider $\ce{sp}$ vs $\ce{sp^2}$. But since the $\ce{C-F}$ bond now has less s-character around the carbon, the $\ce{H-C-F}$ bond angle can shrink, correct? | That's a good, concise statement of Bent's rule. Of course we could have just as correctly said that p character tends to concentrate in orbitals directed at electronegative elements. We'll use this latter phrasing when we examine methyl fluoride below. But first, let's expand on the definition a bit so that it is clear to all. Bent's rule speaks to the hybridization of the central atom ($\ce{A}$) in the molecule $\ce{X-A-Y}$. $\ce{A}$ provides hybridized atomic orbitals that form $\ce{A}$'s part of its bond to $\ce{X}$ and to $\ce{Y}$. Bent's rule says that as we change the electronegativity of $\ce{X}$ and \ or $\ce{Y}$, $\ce{A}$ will tend to rehybridize its orbitals such that more s character will placed in those orbitals directed towards the more electropositive substituent. Let's examine how Bent's rule might be applied to your example of methyl fluoride. In the $\ce{C-F}$ bond, the carbon hybrid orbital is directed towards the electronegative fluorine. Bent's rule suggests that this carbon hybrid orbital will be richer in p character than we might otherwise have suspected. Instead of the carbon hybrid orbital used in this bond being $\ce{sp^3}$ hybridized it will tend to have more p character and therefore move towards $\ce{sp^4}$ hybridization. Why is this? s orbitals are lower in energy than p orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more s character. The two electrons in the $\ce{C-F}$ bond will spend more time around the electronegative fluorine and less time around carbon. If that's the case (and it is), why "waste" precious, low-energy, s orbital character in a carbon hybrid orbital that doesn't have much electron density to stabilize. Instead, save that s character for use in carbon hybrid orbitals that do have more electron density around carbon (like the $\ce{C-H}$ bonds). So as a consequence of Bent's rule, we would expect more p character in the carbon hybrid orbital used to form the $\ce{C-F}$ bond, and more s-character in the carbon hybrid orbitals used to form the $\ce{C-H}$ bonds. The physically observable result of all this is that we would expect an $\ce{H-C-H}$ angle larger than the tetrahedral angle of 109.5° (reflective of more s character) and an $\ce{H-C-F}$ angle slightly smaller than 109.5° (reflective of more p character). In terms of bond lengths, we would expect a shortening of the $\ce{C-H}$ bond (more s character) and a lengthening of the $\ce{C-F}$ bond (more p character). | {
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15,662 | I am told that because the methyl group is electron donating in the conjugate base of acetic acid, this destabilizes the conjugate base by exacerbating the existing negative formal charge on the deprotonated oxygen, while in formic acid the electron donating methyl is absent in lieu of a hydrogen, which is neither withdrawing nor donating. Is this reasoning correct, and are there any other reasons why formic acid might be stronger than acetic acid? | We are discussing the following equilibrium We can make the acid a stronger acid by pushing the equilibrium to the right. To push the equilibrium to the right we can destabilize the starting acid pictured on the left side of the
equation, and \ or stabilize the carboxylate anion pictured on the right side of the
equation. Comparing acetic acid ($\ce{R~ =~ CH3}$) to formic acid ($\ce{R~ =~ H}$), the methyl group is electron releasing compared to hydrogen. Therefore the methyl group will stabilize the dipolar resonance form of the starting acid where there is a partial positive charge on the carbonyl carbon. This should stabilize the starting acid. Further, this electron releasing ability of the methyl group will tend to destabilize the resultant carboxylate anion which already has a full unit of negative charge. Therefore, because the methyl group 1) stabilizes the starting acid and 2) destabilizes the carboxylate anion product, the methyl group will push the equilibrium to the left, compared to the case where the methyl group is replaced by a hydrogen. Consequently, acetic acid is a weaker acid than formic acid. | {
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15,690 | In the book Organic Chemistry by J. Clayden, N. Greeves, S. Warren, and P. Wothers I found the following reasoning: You may have wondered why it is that, while methyl chloride (chloromethane) is a reactive electrophile that takes part readily in substitution reactions, dichloromethane (DCM) is so unreactive that it can be used as a solvent in which substitution reactions of,other alkyl halides take place. You may think that this is a steric effect: Indeed, $\ce{Cl}$ is bigger than $\ce{H}$ .
But $\ce{CH2Cl2}$ is much less reactive as an electrophile than ethyl chloride or propyl chloride: there must be more to its unreactivity. And there is: Dichloromethane benefits from a sort of 'permanent anomeric effect.' One lone pair of each chlorine is always anti-periplanar to the other $\ce{C–Cl}$ bond so
that there is always stabilization from this effect. So, in MO-terms the situation would look something like this. The reasoning looks plausible to me. The interaction between the free electron pair on $\ce{Cl}$ and the $\sigma^{*}$ orbital of the neighboring $\ce{C-Cl}$ bond, which would be the LUMO of DCM, lowers the energy of the free-electron-pair-orbital, thus stabilizing the compound and it raises the energy of the LUMO, thus making DCM less reactive towards nucleophiles. But how important is this anomeric effect actually for explaining the unreactiveness of DCM toward nucleophiles, especially compared to the steric effect? And how important is the steric effect actually: Is the steric hindrance exerted by the second $\ce{Cl}$ in $\ce{CH2Cl2}$ really that much larger than the steric hindrance exerted by the methyl group in $\ce{CH3CH2Cl}$ (which is a good electrophile for $\mathrm{S_N2}$ reactions)? I'm a little skeptical because if this anomeric effect was very important I would have expected that the $\ce{C-Cl}$ bond length in DCM would be slightly higher than in methyl chloride because there should be some transfer of electron density from the free electron pair into the antibonding $\sigma^{*}$ orbital, which should weaken the $\ce{C-Cl}$ bond. But the actual bond lengths don't show this. They show rather the contrary: $$\begin{array}{c|c}
\hline
\text{Species} & \text{Average }\ce{C-Cl}\text{ bond length / Å} \\
\hline
\ce{CH3Cl} & 1.783 \\
\ce{CH2Cl2} & 1.772 \\
\ce{CHCl3} & 1.767 \\
\ce{CCl4} & 1.766 \\
\hline
\end{array}
$$ (source: Wikipedia ) Now, I know that the stronger polarization of the $\ce{C}$ atomic orbitals in di-, tri-, and tetrachlorinated methane as compared to methyl chloride (due to the electronegativity of $\ce{Cl}$ ) should lead to an overall strengthening of the $\ce{C-Cl}$ bonds in those compounds. But I would have expected a trend that would show only a slight decrease (or maybe even an increase) of bond length when going from methyl chloride to DCM and then a more pronounced decrease when going from DCM to chloroform followed by a similar decrease when going from chloroform to tetrachloromethane. But instead the polarization effect seems to only slightly increase on adding more chlorine atoms. | Introduction (and Abstract, TLDR) In very short words you can say, that the anomeric effect is responsible for the lack of reactiveness. The electronic effect may very well be compensating for the the steric effect, that could come from the methyl moiety. In any way, most of the steric effects can often been seen as electronic effects in disguise. Analysis of Molecular Orbitals I will analyse the bonding picture based on calculations at the density fitted density functional level of theory, with a fairly large basis set: DF-BP86/def2-TZVPP. As model compounds I have chosen chloromethane, dichloromethane, chloroform and chloroethane. First of all let me state, that the bond lengths are a little larger at this level, however, the general trend for shortening can also be observed. In this sense, chloroethane behaves like chloromethane. An attempt to explain this will be given at the end of this article. \begin{array}{lr}\hline
\text{Compound} & \mathbf{d}(\ce{C-Cl})\\\hline
\ce{ClCH3} & 1.797\\
\ce{Cl2CH2} & 1.786\\
\ce{Cl3CH} & 1.783\\\hline
\ce{ClCH2CH3} & 1.797\\\hline
\end{array} In the canonical bonding picture, it is fairly obvious, that the electronic effects dominate and are responsible for the lack of reactivity. In other words, the lowest unoccupied molecular orbital is very well delocalised in the dichloromethane and chloroform case. This is effectively leaving no angle to attack the antibonding orbitals. In the mono substituted cases there is a large coefficient at the carbon, where a nucleophile can readily attack. One can also analyse the bonding situation in terms of localised orbitals. Here I make use of a Natural Bond Orbital (NBO) analysis, that transforms the canonical orbitals into hybrid orbitals, which all have an occupation of about two electrons. Due to the nature of the approach, it is no longer possible to speak of HOMO or LUMO, when analysing the orbitals. Due to the nature of the calculations, i.e. there are polarisation functions, the values do not necessarily add up to 100%. The deviation is so small, that it can be omitted. The following table shows the composition (in $\%$)of the carbon chloro bond and anti bond.
\begin{array}{lrr}\hline
\text{Compound} &\sigma-\ce{C-Cl} & \sigma^*-\ce{C-Cl}\\\hline
\ce{ClCH3} & 45\ce{C}(21s79p) 55\ce{Cl}(14s85p)
& 55\ce{C}(21s79p) 45\ce{Cl}(14s85p)\\
\ce{Cl2CH2} & 46\ce{C}(22s77p) 54\ce{Cl}(14s85p)
& 54\ce{C}(22s77p) 46\ce{Cl}(14s85p)\\
\ce{Cl3CH} & 48\ce{C}(24s76p) 52\ce{Cl}(14s86p)
& 52\ce{C}(24s76p) 48\ce{Cl}(14s86p)\\\hline
\ce{ClCH2CH3} & 44\ce{C}(19s81p) 56\ce{Cl}(14s85p)
& 56\ce{C}(19s81p) 44\ce{Cl}(14s85p)\\\hline
\end{array}
As we go from mono- to di- to trisubstituted methane, the carbon contribution increases slightly, along with the percentage of $s$ character. More $s$ character usually means also a stronger bond, which often results in a shorter bond distance. Of course, delocalization will have a similar effect on its own. The reason, why dichloromethane and chloroform are fairly unreactive versus nucleophiles, has already been pointed out in terms of localised bonding. But we can have a look at these orbitals as well. In the case of chloromethane, the LUMO has more or less the same scope of the canonical orbital, with the highest contribution from the carbon. If we compare this antibonding orbital to an analogous orbital in dichloromethane or chloroform, we can expect the same form. We soon run into trouble, because of the localised $p$ lone pairs of chlorine. Not necessarily overlapping, but certainly in the way of the "backside" of the bonding orbital. In the case of chloroethane we can observe hyperconjugation. However, this effect is probably less strong, and from the canonical bonding picture we could also assume, that this increases the polarisation of the antibonding orbital in favour of carbon. In the following pictures, occupied orbitals are coloured red and yellow, while virtual orbitals are coloured purple and orange. (Note that in chloroform two lone pair orbitals are shown.) Conclusion Even though this article does not use the Valence Bond Approach, one can clearly see the qualitative manifestation of Bent's Rule (compare also: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot? ). A higher $s$ character means a shorter bond. The lack of reactivity towards nucleophiles can be explained electronically with a delocalised LUMO. In terms of localised bonding, the lone pairs of any additional chlorine atom would provide sufficient electron density, to shield the backside attack on the carbon. | {
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15,728 | What is the basic difference between aqueous and alcoholic $\ce{KOH}$? Why does alcoholic $\ce{KOH}$ prefer elimination whereas aqueous $\ce{KOH}$ prefers substitution? | $$\ce{R-OH + OH- <=> RO- + H2O }$$ In alcoholic solution, the $\ce{KOH}$ is basic enough ( $\mathrm{p}K_{\mathrm{a}} =15.74$ ) to deprotonate a small amount of the alcohol molecules ( $\mathrm{p}K_{\mathrm{a}}= 16–17$ ), thus forming alkoxide salts ( $\ce{ROK}$ ). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the alkyl group). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at E2 eliminations. | {
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15,907 | A hybrid sp3 orbital is drawn with one lobe smaller than its other half, the latter which is of equal size when drawing the p orbital. Why is it so? | When combined at a given atomic center, any two atomic orbitals add in a vectorial
manner. For example, consider the orbital $\phi$ defined by $\ce{p_{x}}$ and $\ce{p_{y}}$ atomic orbitals as \begin{align}
\phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}}
\end{align} The orbital addition can be pictured like this for the two cases $c_1 = c_2 > 0$ and $c_l = -c_2 > 0$, respectively. The ratio $c_1 / c_2$ controls how much the orbital $\phi$ is tilted away from the $x$ (or $y$) axis. The mixing of atomic orbitals with different angular momentum quantum number is
also controlled by a vectorial addition and leads to various types of hybrid orbitals
some examples of which are shown here: The example a) would be an $\ce{sp}$ hybrid orbital. It gets its sense of direction from the $\ce{p}$ orbital used to construct it because the $\ce{s}$ orbital is isotropic. Its shape comes about because the parts of the orbitals that have an opposite sign (this is hinted at by the shading in the pictures) cancel each other out to some extent. So, if the left lobe of the $\ce{p}$ orbital is negativ and the right one is positive and the $\ce{s}$ orbital is also positive then the left half of the $\ce{s}$ orbital overlaps with the left lobe of the $\ce{p}$ orbital in a destructive manner and they cancel each other out, such that only a small portion of the initial $\ce{p}$ orbital lobe remains. But the right half of the $\ce{s}$ orbital overlaps with the right lobe of the $\ce{p}$ orbital in a constructive manner and so the right lobe of the initial $\ce{p}$ orbital becomes larger. | {
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15,995 | Let's say a room is filled with butane, I then throw a cigarette into the room. What happens to the atoms/molecules of the butane when they are in contact with the heat from the cigarette? | The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF). A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some way to get the reaction going. Typically the reaction is started by the generation of free radicals . These will react with stable molecules to split them up and generate more free radicals so in effect we have a chain reaction. So the ignition process is basically the generation of free radicals, and that's what the lit cigarette does. The heat of the burning tobacco generates free radicals that then start the butane oxygen reaction. You can also generate free radicals in lots of other ways e.g. using shock, from an electric spark, or by using a suitable catalyst such as the catalytic convertor on your car exhaust. | {
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16,010 | For a solution such as: Luria-Bertani (LB) Broth: 10 g tryptone 5 g yeast extract 10 g of NaCl ($M = 58.44\ \mathrm{g/mol}$) q.s. to 1 l, pH to 7.2, autoclave. I would like to make a bulk amount of premixed powder, i.e. combine 1 kg tryptone, 500 g yeast extract and 1 kg NaCl in a big bottle, shake it and keep it on shelf. Whenever I need to make LB now I just weight 25 g of this premix and add to 1 l of water. I’m concerned about a couple things thought. First how thoroughly it gets mixed, and second any settling of different chemicals over time. Either would cause many different batches to be messed up. I’m wondering if anyone has experience with this and ways to avoid those two problems. I’m interested in doing this for other solutions as well. Do manufacturers who make these premixed powders re pulverize them or process them in a certain way to make it work? | The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF). A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some way to get the reaction going. Typically the reaction is started by the generation of free radicals . These will react with stable molecules to split them up and generate more free radicals so in effect we have a chain reaction. So the ignition process is basically the generation of free radicals, and that's what the lit cigarette does. The heat of the burning tobacco generates free radicals that then start the butane oxygen reaction. You can also generate free radicals in lots of other ways e.g. using shock, from an electric spark, or by using a suitable catalyst such as the catalytic convertor on your car exhaust. | {
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16,055 | While I was looking at the periodic table today, I realised that there were gases that were much lighter than helium such as hydrogen. If hydrogen is lighter than helium, why do we insist on using helium in balloons? | As other answers have noted, the only gas lighter than helium is hydrogen, which has some flammability issues that make it more difficult to handle safely than helium. Also, in practice, hydrogen is not significantly "lighter" than helium. While the molecular mass (and thus, per the ideal gas law , the density) of hydrogen gas is about half that of helium, what determines the buoyancy of a balloon is the difference between the density of the gas inside the balloon and the air outside. The density of air at STP is about $\rho_{\ce{air}}=\pu{1.2754 kg m-3}$ , while the densities of hydrogen and helium gas are $\rho_{\ce{H2}}=\pu{0.08988 kg m-3}$ and $\rho_{\ce{He}}=\pu{0.1786 kg m-3}$ respectively. The buoyant forces of a hydrogen balloon and a helium balloon in air (neglecting the weight of the skin and the pressure difference between the inside and the outside, which both decrease the buoyancy somewhat) are proportional to the density differences $\rho_{\ce{air}} -\rho_{\ce{H2}}=\pu{1.1855 kg m-3}$ and $\rho_{\ce{air}} -\rho_{\ce{He}}=\pu{1.0968 kg m-3}$ . Thus, helium is only about $7.5\%$ less buoyant in air than hydrogen. Of course, if the surrounding air were replaced with a lighter gas, the density difference between hydrogen and helium would become more significant. For example, if you wished to go ballooning on Jupiter , which has an atmosphere consisting mostly of hydrogen and some helium, a helium balloon would simply sink , and even a pure hydrogen balloon (at ambient temperature) would not lift much weight. Of course, you could always just fill the balloon with ambient Jovian air and heat it up to produce a hot hydrogen balloon (not to be confused with a Rozière balloon , which are used on Earth and have separate chambers for hot air and hydrogen / helium). Ps. A quick way to approximately obtain this result is to note that a hydrogen molecule consists of two protons (and some electrons, which have negligible mass), and thus has a molecular mass of about $\pu{2 Da}$ , while a helium atom has two protons and two neutrons, for a total mass of about $\pu{4 Da}$ . Air, meanwhile, is mostly oxygen and nitrogen: oxygen has a molecular mass of about $\pu{32 Da}$ (8 protons + 8 neutrons per atom, two atoms per molecule), while nitrogen is close to $\pu{28 Da}$ (one proton and one neutron per atom less than oxygen). Thus, the average molecular mass of air should be between $28$ and $\pu{32 Da}$ ; in fact, since air is about three quarters nitrogen, it's about $\pu{29 Da}$ , and so the buoyancies of hydrogen and helium in air are proportional to $29 - 2 = 27$ and $29 - 4 = 25$ respectively. Thus, hydrogen should be about $\frac{(27 - 25)}{25} = \frac{2}{25} = \frac{8}{100} = 8\%$ more buoyant than helium, or, in other words, helium should be about $\frac{2}{27} \approx 7.5\%$ less buoyant than hydrogen. Pps. To summarize some of the comments below, there are other possible lifting gases as well, but none of them appear to be particularly viable competitors for helium, at least not at today's helium prices. For example, methane (molecular mass $\approx \pu{16 Da}$ ) has about half the buoyancy of hydrogen or helium in the Earth's atmosphere, and is cheap and easily available from natural gas. However, like hydrogen, it's also flammable, and while it's somewhat less dangerous by some measures (burn speed and flammability range), it's more dangerous by others (total energy content per volume). In any case, the reduced buoyancy, together with the flammability, is probably enough to sink (pun not intended) methane as a viable alternative to helium. A much less flammable choice would be water vapor which, with a molecular mass of $\approx \pu{18 Da}$ , is only slightly less buoyant than methane at the same temperature and pressure. The obvious problem with water is that it's a liquid at ambient temperatures, which means it has to be heated to make it lift anything at all. This wouldn't be so bad (after all, you get extra lift from the expansion due to heat), except for the fact that it makes any failure in the heating system a potential disaster — whereas a hot air balloon will just gently drift down if the burner fails, a hot steam balloon can experience catastrophic buoyancy loss if the vapor condenses. Despite these drawbacks, hot steam balloons have certainly been suggested , studied and tried in the past — alas, not always particularly successfully (although, apparently, there have been much more successful attempts as well). There are various ways in which the condensation issue could potentially be reduced, such as adding extra insulation layers to the balloon envelope, or even surrounding the steam balloon with a more conventional hot air envelope. So far, however, it seems that steam balloons remain firmly in the realm of nifty but impractical ideas. Other potential lifting gases, with molecular mass similar to methane and water, include ammonia and neon . Neon, being a noble gas like helium, would certainly work and be safe, but alas, it's both less buoyant and more expensive than helium. Ammonia, on the other hand, while much less flammable than methane, is rather toxic and corrosive (not to mention really stinky, which, given its other properties, is probably a good thing). I don't think I'd like to fly in an ammonia balloon , but apparently , some people do! It seems that its main advantage (besides being much cheaper than helium) is its relatively low vapor pressure, which makes it easier to store and handle in compressed form. Thus, at least for some niche applications (mainly hobbyists and some weather balloons, AFAICT), ammonia might actually be the most viable alternative to helium (and hot air) today, with methane / natural gas perhaps coming second. If helium were to become more scarce and expensive, these low-cost lifting gases (and possibly other alternatives, like helium recovery or even steam balloons) might become more practical. Then again, so would hydrogen — its safety issues, though well known, are not insurmountable, especially not for things like unmanned weather balloons where the risks are much less.⠀⠀⠀⠀⠀⠀ | {
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16,068 | I have been learning about zwitterions and the isoelectric point, and have been trying to apply this to hydrolysis of peptides. From the diagram in my textbook, I can see that when acid hydrolysis occurs, the amino acids formed from the hydrolysis act as bases because the pH is below their isoelectric point, so they accept the $\ce{H+}$ ions forming positively charged ions. Now when it comes to alkaline hydrolysis, I don't understand why the peptide would not just be hydrolysed, and then act as an acid and donate the extra $\ce{H+}$ ion on its amine group to the $\ce{OH-}$ ions to form negative ions and a $\ce{H2O}$ molecule. Instead, in my text book it shows something like this: $$\ce{-NH-CH(CH2SH)-C(=O)-NH-CH(CH3)-C(=O)- + NaOH \\-> -NH-CH(CH2SH)-C(=O)-O^{-}Na^{+} + H2N-CH(CH3)-C(=O)-}$$ I'm really sorry if this makes no sense whatsoever... | As other answers have noted, the only gas lighter than helium is hydrogen, which has some flammability issues that make it more difficult to handle safely than helium. Also, in practice, hydrogen is not significantly "lighter" than helium. While the molecular mass (and thus, per the ideal gas law , the density) of hydrogen gas is about half that of helium, what determines the buoyancy of a balloon is the difference between the density of the gas inside the balloon and the air outside. The density of air at STP is about $\rho_{\ce{air}}=\pu{1.2754 kg m-3}$ , while the densities of hydrogen and helium gas are $\rho_{\ce{H2}}=\pu{0.08988 kg m-3}$ and $\rho_{\ce{He}}=\pu{0.1786 kg m-3}$ respectively. The buoyant forces of a hydrogen balloon and a helium balloon in air (neglecting the weight of the skin and the pressure difference between the inside and the outside, which both decrease the buoyancy somewhat) are proportional to the density differences $\rho_{\ce{air}} -\rho_{\ce{H2}}=\pu{1.1855 kg m-3}$ and $\rho_{\ce{air}} -\rho_{\ce{He}}=\pu{1.0968 kg m-3}$ . Thus, helium is only about $7.5\%$ less buoyant in air than hydrogen. Of course, if the surrounding air were replaced with a lighter gas, the density difference between hydrogen and helium would become more significant. For example, if you wished to go ballooning on Jupiter , which has an atmosphere consisting mostly of hydrogen and some helium, a helium balloon would simply sink , and even a pure hydrogen balloon (at ambient temperature) would not lift much weight. Of course, you could always just fill the balloon with ambient Jovian air and heat it up to produce a hot hydrogen balloon (not to be confused with a Rozière balloon , which are used on Earth and have separate chambers for hot air and hydrogen / helium). Ps. A quick way to approximately obtain this result is to note that a hydrogen molecule consists of two protons (and some electrons, which have negligible mass), and thus has a molecular mass of about $\pu{2 Da}$ , while a helium atom has two protons and two neutrons, for a total mass of about $\pu{4 Da}$ . Air, meanwhile, is mostly oxygen and nitrogen: oxygen has a molecular mass of about $\pu{32 Da}$ (8 protons + 8 neutrons per atom, two atoms per molecule), while nitrogen is close to $\pu{28 Da}$ (one proton and one neutron per atom less than oxygen). Thus, the average molecular mass of air should be between $28$ and $\pu{32 Da}$ ; in fact, since air is about three quarters nitrogen, it's about $\pu{29 Da}$ , and so the buoyancies of hydrogen and helium in air are proportional to $29 - 2 = 27$ and $29 - 4 = 25$ respectively. Thus, hydrogen should be about $\frac{(27 - 25)}{25} = \frac{2}{25} = \frac{8}{100} = 8\%$ more buoyant than helium, or, in other words, helium should be about $\frac{2}{27} \approx 7.5\%$ less buoyant than hydrogen. Pps. To summarize some of the comments below, there are other possible lifting gases as well, but none of them appear to be particularly viable competitors for helium, at least not at today's helium prices. For example, methane (molecular mass $\approx \pu{16 Da}$ ) has about half the buoyancy of hydrogen or helium in the Earth's atmosphere, and is cheap and easily available from natural gas. However, like hydrogen, it's also flammable, and while it's somewhat less dangerous by some measures (burn speed and flammability range), it's more dangerous by others (total energy content per volume). In any case, the reduced buoyancy, together with the flammability, is probably enough to sink (pun not intended) methane as a viable alternative to helium. A much less flammable choice would be water vapor which, with a molecular mass of $\approx \pu{18 Da}$ , is only slightly less buoyant than methane at the same temperature and pressure. The obvious problem with water is that it's a liquid at ambient temperatures, which means it has to be heated to make it lift anything at all. This wouldn't be so bad (after all, you get extra lift from the expansion due to heat), except for the fact that it makes any failure in the heating system a potential disaster — whereas a hot air balloon will just gently drift down if the burner fails, a hot steam balloon can experience catastrophic buoyancy loss if the vapor condenses. Despite these drawbacks, hot steam balloons have certainly been suggested , studied and tried in the past — alas, not always particularly successfully (although, apparently, there have been much more successful attempts as well). There are various ways in which the condensation issue could potentially be reduced, such as adding extra insulation layers to the balloon envelope, or even surrounding the steam balloon with a more conventional hot air envelope. So far, however, it seems that steam balloons remain firmly in the realm of nifty but impractical ideas. Other potential lifting gases, with molecular mass similar to methane and water, include ammonia and neon . Neon, being a noble gas like helium, would certainly work and be safe, but alas, it's both less buoyant and more expensive than helium. Ammonia, on the other hand, while much less flammable than methane, is rather toxic and corrosive (not to mention really stinky, which, given its other properties, is probably a good thing). I don't think I'd like to fly in an ammonia balloon , but apparently , some people do! It seems that its main advantage (besides being much cheaper than helium) is its relatively low vapor pressure, which makes it easier to store and handle in compressed form. Thus, at least for some niche applications (mainly hobbyists and some weather balloons, AFAICT), ammonia might actually be the most viable alternative to helium (and hot air) today, with methane / natural gas perhaps coming second. If helium were to become more scarce and expensive, these low-cost lifting gases (and possibly other alternatives, like helium recovery or even steam balloons) might become more practical. Then again, so would hydrogen — its safety issues, though well known, are not insurmountable, especially not for things like unmanned weather balloons where the risks are much less.⠀⠀⠀⠀⠀⠀ | {
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16,135 | I have come across many questions where I'm asked to give the number of possible structural isomers. For example number, structural isomers of hexane is 5, while the number structural isomers of decane is 75. How can I determine the possible number of structural isomers of a given organic compound? | It isn't easy but it is an interesting research topic Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. Wikipedia , for example, lists the numbers of isomers and stereoisomers for molecules with up to 120 carbons. But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers. It has been an interesting research topic in computational chemistry and mathematics. This old paper (pdf), for example, list some formulae for simple hydrocarbons among other simple series. Part of the interest arises because of the relationship to the mathematics of graph theory (it seems that chemistry has inspired some new ideas in this field of mathematics partially because enumerating possible isomers of hydrocarbons is strongly related to drawing certain simple trees which is intuitively obvious if you use the standard chemical convention of drawing just the carbon backbone and ignoring hydrogens). You can look up the answers on the fascinating mathematics site OEIS (the online encyclopaedia of integer sequences). The sequence for simple hydrocarbons is here . But the mathematical approach oversimplifies things from the point of view of real-world chemistry. Mathematical trees are idealised abstract objects that ignore real-world chemical constraints like the fact that atoms take up space in three dimensions. This means that some structures that can be drawn cannot exist in the real world because the atoms are too crowded and cannot physically exist without enough strain to cause them to fall apart. Luckily, computation chemists have also studied this. There is, unfortunately, no obvious shortcut other than trying to create models of the possible structures and testing them to see if they are too strained to exist. The first two isomers that are too crowded are for 16 and 17 carbons and have these structures: If you have any intuition of the space filling view of these, you should be able to see why they are problematic. A research group at Cambridge University has produced an applet to enumerate the physically possible isomers for a given number of carbons which is available here if your Java settings allow it. The results are discussed in a paper available in the Journal of Chemical Information and Modelling . | {
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16,215 | In my textbook it is written that the order of basic strength of pnictogen hydrides is $$\ce{NH3 > PH3 > AsH3 > SbH3 > BiH3}$$ I tried but could not find any explanation as to why this happens. What is the explanation? | Each of these molecules has a pair of electrons in an orbital - this is termed a "lone pair" of electrons. It is the lone pair of electrons that makes these molecules nucleophilic or basic. As you move down the column from nitrogen to bismuth, you are placing your outermost shell of electrons, including the lone pair, in a larger and more diffuse orbital (the nitrogen lone pair is contained in the n=2 shell, while the bismuth lone pair is in the n=6 shell). As the electron density of the lone pair is spread over a greater volume and is consequently more diffuse , the lone pair of electrons becomes less nucleophilic, less basic . Commment: But greater the EN of the central atom , lesser would be its tendency to donate electrons right? Response to Comment: The ENs of the central atoms are N (3.04), P (2.19), As (2.18), Sb (2.05), Bi (2.02). All of these atoms, except for nitrogen have similar ENs and I think the electron density argument is valid for them. However, in the case of nitrogen and phosphorous there is a significant EN difference that would tend to argue in the direction opposite to my answer. However there is another important difference between N and P. The lone pair in ammonia is in an sp3 orbital. In all of the other cases the central atom is essentially unhybridized (~90° H-X-H angles) and the lone pair exists in an s orbital. Therefore the lone pair electron density in ammonia (being in an sp3 orbital) is effectively increased compared to phosphine where the lone pair is in an s orbital. So even though the EN difference between N and P is significant, when hybridization differences of the central atom are taken into account the electron density argument still explains the trend in basicity. | {
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16,267 | We've learnt that the electropositive element is written first. Then why is ammonia written as $\ce{NH3}$ ? | According to current nomenclature rules, $\ce{H3N}$ would be correct and acceptable. However some chemical formulas, like $\ce{NH3}$ for ammonia, that were in use long before the rules came out, are still accepted today. | {
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16,342 | I need someone to back me up on this before I go confront my teacher:
I was doing some analysis of the dipole moment of cis-2-butene.
Let's say that the alkyl groups are both on top.
Would the dipole moment be a vector that has its tail on the top and its head on the bottom? How would one calculate that?
I made some calculations assuming all C-H bonds have the same dipole moment vector and came to the conclusion that the three H's attached to the alkyl groups actually end up overcoming the dipole moments of the C-H bonds of the two unsubstituted H's on the double bond. Am I right? | Mini Research Project Time Updates Added CCSD(T) $n_i$ and dipole moments and tweaked discussion (the delay was caused by a system-wide storage upgrade on the machines which took nearly a week to complete). Preamble This response is in no way meant to be contrary to what Geoff has already posted. I happen to enjoy these types of questions and I like to tackle them as little research projects for fun. Therefore, my answer will be brutally detailed and in-depth. That said, this should be a good introduction to how I would approach this problem if I were going to do this at the 'production-level' of research but this definitely goes far beyond the purpose of any reasonable response for an SE answer. Its All About the Dipoles Baby Okay so I wouldn't use THAT heading in a paper but maybe a talk depending on who my audience was... We can easily determine the dipole moment of cis-2-butene via electronic structure theory. I have modeled two conformers of cis-2-butene as shown below. I will refer to the geometry on the left as StructA and the geometry on the right as StructB . A couple of things to note since I don't include any captions to tables. Energies are always reported in $\mathrm{kcal\ mol}^{-1}$ and dipole moments ($\mu$) are given in Debye. Computational Methods Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with second-order Moller-Plesset perturbation (MP2) theory and a variety of density functional theory (DFT) methods using the Gaussian 09 software package for the conformers of cis-2-butene. Both conformers were characterized in $C_{2v}$ symmetry. The DFT methods implemented include B3LYP, B3LYP-GD3(BJ), M06-2X, MN12SX, N12SX, and APFD. The B3LYP-GD3(BJ) method employs Grimme's 3rd generation dispersion correction as well as the Becke-Johnson damping function. All DFT computations employed a pruned numerical integration grid having 90 radial shells and 590 angular points per shell. The heavy-aug-cc-pVTZ basis set was employed for these computations where the heavy (non-hydrogen) atoms were augmented with diffuse functions (e.g. cc-pVTZ for H and aug-cc-pVTZ for carbon). This basis set is abbreviated as haTZ. The CCSD(T) (i.e. coupled-cluster method that includes all single and double substitutions as well as a perturbative treatment of the connected triple excitations) was similarly employed using the CFOUR software package. The magnitudes of the components of the residual Cartesian gradients of the optimized geometries were less than $6.8\times 10^{-6} E_h\ a_0^{-1}$. Single point energies were computed with the explicitly correlated MP2-F12 [specifically MP2-F12 3C(FIX)] and CCSD(T)-F12 [specifically CCSD(T)-F12b with unscaled triples contributions] methods in conjunction with the haTZ. These computations were performed with the Molpro 2010.1 software package using the default density fitting (DF) and resolution of the identity (RI) basis sets. Natural bond orbital (NBO) analyses were performed for the MP2 optimized structures using the haTZ basis set and the SCF density. All computations employed the frozen core approximation (i.e. 1s$^2$ electrons frozen in carbon). Computational Methods in English Two different conformations of cis-2-butene were characterized with a variety of cheap (but usually okay) approximations (i.e. DFT methods) as well as reliable (but generally more expensive) wave function methods (i.e. MP2 and CCSD(T)). The wave function methods are necessary to validate the DFT results. CCSD(T) is the gold-standard and gives very good results for single-reference closed-shell well-behaved systems so we will use this as our 'best estimate'. We use a variety of methods in order to look for agreement in the results. If we see good agreement across the board, we can be confident in our results. If we see massive discrepancies then we will have to be careful when we analyze the results. Geometries have been converged to a tight threshold (i.e. we have good molecules) and our DFT computations use a relatively dense integration grid (which leads to more accurate results). Notice that I employ a heavy-aug-cc-pVTZ (haTZ) basis set. Why leave the diffuse functions off hydrogen? The purpose of diffuse functions is to describe electron density far away from the nucleus of an atom. Therefore we slap these functions onto carbon which are relatively large atoms compared to hydrogen. Hydrogen, on the other hand, has only one electron and therefore has a small electron density when isolated. In cis-2-butene, hydrogen is bonded to carbon via a rather small bond distance. The electron density around the hydrogen is even more reduced than an isolated hydrogen atom in the gas phase. Therefore, it would be impractical to include diffuse functions on hydrogen. Doing so may even lead to erroneous results since we will be trying to describe electron density far away from the hydrogen nucleus when in reality there is virtually none to be found. Finally, we perform single point energies using explicitly correlated methods. Because the CCSD(T) opts and freqs will likely not be done in time for this posting, we can gauge how the resulting geometry from each optimization procedure will vary from another geometry given by a different method. If all of the energies are similar (within a few tenths of a $\mathrm{kcal\ mol}^{-1}$), then we can be confident that our geometries are not only very similar but that small deviations in the geometry will have little effect on the corresponding energies at least in this region of the potential energy surface (PES). Large deviations will usually mean that the method which produced the 'outlier' is not a good approximation for the system (I do not expect cis-2-butene to be problematic at all). Explicitly correlated methods accelerate convergence to the CBS limit. These methods have been shown to give results that a large basis set and a canonical method would provide but with a much smaller basis set. For example, the result that I get with a regular CCSD(T)/aug-cc-pV5Z basis set could be obtained using CCSD(T)-F12/aug-cc-pVTZ. This makes the computations less intensive and much more feasible. Results The number of imaginary frequencies ($n_i$), relative MP2-F12 and CCSD(T)-F12 energetic ($\Delta E^{\mathrm{MP2-F12}}$ and $\Delta E^{\mathrm{CC-F12}}$, respectively, in $\mathrm{kcal\ mol^{-1}}$) and dipole moment ($\mu_z$ in Debye) are given in the following table for both conformers of cis-2-butene for a variety of a methods.
The relative energies were determined by taking the difference of the respective geometry and the reference CCSD(T) geometry [e.g. E(CCSD(T))-E(MP2)]. StructB is a second-order saddle point ($n_i = 2$) on every single PES considered and therefore is not a minimum energy structure. StructA is, however, a minimum ($n_i = 0$) on every PES considered. The characterization of the nature of the stationary point is consistent between CCSD(T), our best estimate, MP2, and DFT methods. Single point energies reveal negligible differences in the optimized geometries. The energies associated with the MP2 optimized structures is the reference point for all other relative energies. Deviations grow no larger than 0.27 $\mathrm{kcal\ mol}^{-1}$ for the MP2-F12 and CCSD(T)-F12 relative energies. In addition, there is good agreement between the MP2-F12 and CCSD(T)-F12 relative energies, suggesting that higher-order correlation effects are small. The dipoles for StructA and StructB are very similar with very small magnitudes, on the order of a couple tenths of a Debye. To put these quantities into perspective, the dipole moment of water is 1.85 D. We all know that water has a pretty large dipole moment so by comparison, cis-2-butene has a very WEAK dipole moment. You can compare to the dipole moments of other molecules by referring to this NIST reference . Clearly, the rotation of the methyl groups have a very small effect on the dipole moments of each conformer. The MP2 and DFT dipole moments deviate from the best estimate by no more than 0.03 D, but remains in qualitative agreement for both StructA and StructB. The figure below shows the directionality of the dipole. The head of the (unscaled) arrow points toward the negative pole while the tail of the (unscaled) arrow is oriented to the positive pole. The numbers on the atoms represent ' natural charges ' from an Natural Bond Orbital (NBO) analysis of the MP2 optimized Struct A . Clearly the carbon atoms have a small negative charge to them as these atoms are sucking (great scientific term here) electron density away from the neighboring hydrogen atoms. This is because the nuclear charge (i.e. the number of protons) on carbon is much greater than that of hydrogen (6 vs. 1). It was suggested that I add some data highlighting the energy difference between the two different conformers. The following table presents the energy difference (where $\Delta E_{\mathrm{A-B}}$ is equivalent to E(A)-E(B)) of each optimized geometry using the MP2-F12 and CCSD(T)-F12 single point energies (in $\mathrm{kcal\ mol}^{-1}$). We can see that StructA is about 1.5 $\mathrm{kcal\ mol}^{-1}$ lower in energy than StructB . This makes sense because StructB is a higher order saddle point and StructA is a minimum on the PES. (I'm actually quite surprised that the energy difference is this large for a couple of methyl rotations...) Conclusions The dipole moment for two conformers of cis-2-butene has been examined using seven different computational approaches and a triple-$\zeta$ quality basis set. The performance of these methods have been tested by evaluating the energies of each geometry with the 'gold standard' CCSD(T) method (the explicitly correlated variant). Good agreement is seen across the board with respect to the Hessian indices, energies, and dipole moments. Only StructA was a minimum on each PES. Both conformers of cis-2-butene have a very weak dipole moment on the order of 0.2 -- 0.3 D. The positive pole is in the vicinity of the methyl groups whereas the negative pole is centered around the sp2 hybridized carbons. FAQ So you may be asking yourself (or rather, should be asking yourself) questions such as these listed below. I will tackle them one at a time. 1.) Why did we look at two conformers of cis-2-butene and why does it matter? StructA is a minimum which means that if you were to characterize this guy in the gas phase, you'd find StructA rather than StructB . This is important because if we were to report these results to other scientists, they will want to know what they can expect to find without wondering. Therefore, StructA is going to be our conformer of interest rather than StructB . The latter still will provide insightful results but that is about it for the purposes of this examination. 2.) Why did we use a variety of methods to characterize these molecular systems? Computational methods are simply approximations. They are not guaranteed to give the 'correct' answer. So we try to address this by using a variety of methods (i.e. approximations) and we analyze the results accordingly. If the results are in agreement, then we can feel confident that the results are correct since we tested them against a list of methods. You can never rely on just one method unless it is rigorous, well-tested and has been shown repeatedly to perform well in the literature. When we say 'perform well', this usually means that the computational results are in reasonable agreement with experimental results. This is important because experimental results is the LAW (for all intents and purposes). If the computations disagree with experiment, 99.9% of the time this means that your computational approach sucked and that your approximation was either flawed or misapplied. By using a host of methods, we can put a little more faith into the results we observe because the chances of massive disagreement between the results of the methods is very unlikely in a normal situation. 3.) What is the point of doing a bunch of energy points? Again, because we used a slew of methods to characterize the geometries of cis-2-butene, we end up with non-identical geometries each time we use a new approximation. For instance, the methyl C-H bond lengths from B3LYP are going to be a little bit different from those obtained with MP2. So then that begs the question, "How do these small differences effect the property of the system that we are interested in?" Generally, minute differences will have little effect on the resulting energies of each molecule under consideration. Energy is a very important property that chemists love to look at. So, if we take each geometry (and each one is unique from the other) and we evaluate the energy of the molecule at the same level of theory (in this case, MP2-F12 and CCSD(T)-F12), then we can quickly see how 'resolved' each geometry was. There should be very good agreement between the relative energies of each geometry (probably to within a few tenths of a $\mathrm{kcal\ mol}^{-1}$). 4.) Okay, so why MP2-F12 AND CCSD(T)-F12? We use MP2-F12 AND CCSD(T)-F12 to test for 'higher-order correlation effects'. MP2 methods are much cheaper than CCSD(T) methods but MP2 is not as rigorous and can be error prone in a host of molecular systems. Therefore, we test the performance of MP2 by busting out our 'gold-standard' which is the CCSD(T) method. If MP2 agrees well with CCSD(T), then we can feel confident in our MP2 results and never have to revisit the difficult, time-consuming CCSD(T) computations ever again. ALso, CCSD(T) will also tell us how DFT performed as well. DFT methods must always be calibrated against something more rigorous since DFT is known for 'getting the right answer for the wrong reasons' and it isn't always right. The 'F12' bit just means that these are 'explicitly-correlated' methods. Rather than give an introduction to what this means, you should understand why we use it instead. You may have noticed that whenever we do a computational job, we specify a method AND a basis set (e.g. heavy-aug-cc-pVTZ). These basis sets can be measured by how many atomic orbitals (or functions) are given in the set. The more that are given, the better the 'basis-set approximation'. Think of this in terms of Riemann sums where you try to approximate the area under a curve using a set of rectangles. Each rectangle is a basis function and the number of rectangles you use forms a basis set. The more rectangles you use, the better your approximation of the area under that curve will be. Basis sets in computational chemistry behave the same way. When you approach an infinite set of rectangles, you approach the exact answer. When you approach an infinite number of basis functions, you approach what is called the CBS (complete basis set) limit. At the CBS limit, you have an exact answer. We cannot implement an infinite basis set in chemistry (for obvious reasons), and very large basis sets are cost prohibitive. Therefore, people have devised these F12 approximations that are constructed in such a way to give results that are comparable to those that you'd get with a large basis set, but you can get them by using a relatively small basis set instead! This is a powerful approach to convergent quantum chemistry that saves you a lot of time while maintaining a set of very good results. 5.) Why didn't you provide more pretty pictures? That's just the nature of the beast. Computational chemistry is usually short on graphics but very dense on spreadsheets. Plus... I'm no artist. I actually spent a good couple hours trying to get some electrostatic potentials posted but the new version of G09 hates the molecular viewer programs I currently use so I ditched that idea. | {
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16,439 | Which is the densest gas known to exist under normal "shirt sleeve" conditions: room temperature, one atmosphere, won't explode on contact with air, won't kill everyone in the building in trace amounts, etc. | Perfluorobutane is inert and has almost twice the density of sulfur hexafluoride. It is non-toxic enough that it is used in fire extinguishers and injected as a contrast agent for ultrasound. Boiling point: $-1.7\ \mathrm{^\circ C}$ . Perfluoropentane is similar and rarer but somewhat higher density $(\sim13\ \mathrm{kg/m^3})$ in proportion to its higher molecular mass. Its boiling point is $28\ \mathrm{^\circ C}$ (uncomfortably warm, but your thermostat can go that high). This is the densest gas that strictly meets all the criteria in the OP. If we relax the criteria a bit: Perfluorohexane is just over the boiling point limit at $56\ \mathrm{^\circ C}$ , but it has a molar mass of $338\ \mathrm{g/mol}$ which makes it slightly denser in gas form than tungsten hexafluoride ( $\ce{WF6}$ ). It's also inert and non-toxic, unlike $\ce{WF6}$ . The ultimate gas density would be uranium hexachloride using depleted uranium and $\ce{Cl-37}$ , with a molar mass of $460\ \mathrm{g/mol}$ , which makes it over 50 % denser in gas form than tungsten hexafluoride, and 3 times denser than sulfur hexafluoride, but it has a boiling point of $75\ \mathrm{^\circ C}$ , decomposes on contact with air, is toxic and slightly radioactive. | {
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16,562 | Pyrene doesn't seem to be aromatic. However, sources claim that it is aromatic. Considerations: Pyrene is cyclic. ✓ Pyrene is flat (planar). ✓ Pyrene has 16 π electrons. Every atom in the ring structure of pyrene is $\ce{sp^2}$ hybridized. ✓ The problem is that pyrene fails the $4n+2$ rule. $4n + 2 \neq 16$ where $n$ is an integer. Something leads me to suspect that Hückel's rule is an oversimplification here. Or is pyrene indeed non-aromatic? | Pyrene is aromatic. The Hückel $4n+2$ rule works best with monocyclic ring systems. If you look at the following resonance structure for pyrene with a central double bond, the monocyclic periphery has 14 π electrons (ignoring the greyed-out central double bond), but that is a rationalization. Nonetheless, pyrene undergoes reactions characteristic of aromatic systems and has ring currents expected from aromatic systems. | {
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16,633 | Bulk gold has a very characteristic warm yellow shine to it, whereas almost all other metals have a grey or silvery color. Where does this come from? I have heard that this property arises from relativistic effects, and I assume that it has to do with some distinct electronic transition energies in the gold atoms. But what changes with the "introduction" of relativistic effects, that then changes the energy of the frontier orbitals in such a drastic manner? | Yes, this is a beautiful question. As you said, in lower rows of the periodic table, there are relativistic effects for the electrons. That is, for core electrons in gold, the electrons are traveling at a significant fraction of the speed of light ( e.g. , ~58% for $\ce{Au}$ $\mathrm{1s}$ electrons). This contracts the Bohr radius of the $\mathrm{1s}$ electrons by ~22%. Source: Wikipedia This also contracts the size of other orbitals, including the $\mathrm{6s}$ . The absorption you see is a $\mathrm{5d \rightarrow 6s}$ transition. For the silver $\mathrm{4d \rightarrow 5s}$ transition, the absorption is in the UV region, but the contraction gives gold a blue absorption ( i.e. less blue is reflected). Our eyes thus see a yellow color reflected. There's a very readable article by Pekka Pyykkö and Jean Paul Desclaux that goes into more detail (if you subscribe to ACS Acc. Chem.Res. ) | {
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16,640 | At the Renaissance fair a few years back I was watching a smith forge metal into shapes. During this time a very odd question came to me. I was wondering what the furnace was made of. My logic stated that whatever the furnace was made of must have a higher melting point than the materials he was melting. This quickly turned into an elemental arms race resulting in an odd question of how do we melt stuff like refractory metals (more specifically the one with the highest melting point) so we can melt other things inside of it. Now I know that (for some odd reason I don't understand) rapid cooling can manipulate the strength of an item. Is there a similar property to manipulate the melting point? Note: My current best guess (like can be done to make weapons harder) is that we take two elements, melt them, and the resulting compound has a higher melting point. | Tungsten's melting point of 3422 °C is the highest of all metals and second only to carbon's, for which melting occurs only at high pressure (there's no standard melting point). This is why tungsten is used in rocket nozzles and reactor linings. There are refractory ceramics and alloys that have higher melting points, notably $\ce{Ta4HfC5}$ with a melting point of 4215 °C, hafnium carbide at 3900 °C and tantalum carbide at 3800 °C. Carbon cannot be used to hold molten tungsten because they will react to form tungsten carbide. Sometimes ladles and crucibles used to prepare or transport high melting point materials like tungsten are lined with the various higher melting ceramics or alloys. More typically tungsten and other refractory materials are fabricated in a non-molten state. A process known as powder metallurgy is used. This process uses 4 basic steps: powder manufacture - a variety of techniques are available to generate small particles of the material being worked powder blending - routine procedures are used to blend the constituent particles into a uniform mixture compacting - the blended powder is placed in a mold and subjected to high pressure sintering - the compacted material is subjected to high temperature and some level of bonding occurs between particles. | {
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16,785 | In a galvanic (voltaic) cell, the anode is considered negative and the cathode is considered positive. This seems reasonable as the anode is the source of electrons and cathode is where the electrons flow. However, in an electrolytic cell, the anode is taken to be positive while the cathode is now negative. However, the reaction is still similar, whereby electrons from the anode flow to the positive terminal of the battery, and electrons from the battery flow to the cathode. So why does the sign of the cathode and anode switch when considering an electrolytic cell? | The anode is the electrode where the oxidation reaction \begin{align}
\ce{Red -> Ox + e-}
\end{align} takes place while the cathode is the electrode where the reduction reaction \begin{align}
\ce{Ox + e- -> Red}
\end{align} takes place. That's how cathode and anode are defined. Galvanic cell Now, in a galvanic cell the reaction proceeds without an external potential helping it along. Since at the anode you have the oxidation reaction which produces electrons you get a build-up of negative charge in the course of the reaction until electrochemical equilibrium is reached. Thus the anode is negative. At the cathode, on the other hand, you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus the cathode is positive. Electrolytic cell In an electrolytic cell, you apply an external potential to enforce the reaction to go in the opposite direction. Now the reasoning is reversed. At the negative electrode where you have produced a high electron potential via an external voltage source electrons are "pushed out" of the electrode, thereby reducing the oxidized species $\ce{Ox}$, because the electron energy level inside the electrode (Fermi Level) is higher than the energy level of the LUMO of $\ce{Ox}$ and the electrons can lower their energy by occupying this orbital - you have very reactive electrons so to speak. So the negative electrode will be the one where the reduction reaction will take place and thus it's the cathode. At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode leaving behind the the reduced species $\ce{Red}$ because the electron energy level inside the electrode (Fermi Level) is lower than the energy level of the HOMO of $\ce{Red}$. So the positive electrode will be the one where the oxidation reaction will take place and thus it's the anode. A tale of electrons and waterfalls Since there is some confusion concerning the principles on which an electrolysis works, I'll try a metaphor to explain it. Electrons flow from a region of high potential to a region of low potential much like water falls down a waterfall or flows down an inclined plane. The reason is the same: water and electrons can lower their energy this way. Now the external voltage source acts like two big rivers connected to waterfalls: one at a high altitude that leads towards a waterfall - that would be the minus pole - and one at a low altitude that leads away from a waterfall - that would be the plus pole. The electrodes would be like the points of the river shortly before or after the waterfalls in this picture: the cathode is like the edge of a waterfall where the water drops down and the anode is like the point where the water drops into. Ok, what happens at the electrolysis reaction? At the cathode, you have the high altitude situation. So the electrons flow to the "edge of their waterfall". They want to "fall down" because behind them the river is pushing towards the edge exerting some kind of "pressure". But where can they fall down to? The other electrode is separated from them by the solution and usually a diaphragm. But there are $\ce{Ox}$ molecules that have empty states that lie energetically below that of the electrode. Those empty states are like small ponds lying at a lower altitude where a little bit of the water from the river can fall into. So every time such an $\ce{Ox}$ molecule comes near the electrode an electron takes the opportunity to jump to it and reduce it to $\ce{Red}$. But that does not mean that the electrode is suddenly missing an electron because the river is replacing the "pushed out" electron immediately. And the voltage source (the source of the river) can't run dry of electrons because it gets its electrons from the power socket. Now the anode: At the anode, you have the low altitude situation. So here the river lies lower than everything else. Now you can imagine the HOMO-states of the $\ce{Red}$ molecules as small barrier lakes lying at a higher altitude than our river. When a $\ce{Red}$ molecule comes close to the electrode it is like someone opening the floodgates of the barrier lake's dam. The electrons flow from the HOMO into the electrode thus creating an $\ce{Ox}$ molecule. But the electrons don't stay in the electrode, so to speak, they are carried away by the river. And since the river is such a vast entity (lots of water) and usually flows into an ocean, the little "water" that is added to it doesn't change the river much. It stays the same, unaltered so that everytime a floodgate gets opened the water from the barrier lake will drop the same distance. | {
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17,038 | When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$. It stays in this form until the water evaporates and then the ions go back to normal $\ce{NaCl}$. So why does water with salt in it still taste like salt? I am asking because if the molecule $\ce{NaCl}$ is broken down into $\ce{Na+}$ and $\ce{Cl-}$ then how does it have the characteristics of $\ce{NaCl}$? | When you taste salt, you're not pushing crystalline $\ce{NaCl}$ into your taste buds. It dissolves in your saliva and dissociates. When one tastes salt, the saltiness taste receptors respond specifically to the sodium cation. That type of taste receptor is a cation channel. This is why lithium and potassium cations also taste salty (though they also stimulate other receptors which make them taste somewhat different). There seem to be at least two types of receptors that respond to saltiness. One responds almost specifically to sodium at low concentrations, but at higher concentrations, the other type responds to many cations. See here | {
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17,124 | According to Wikipedia, $\ce{HeH+}$ and fluoroantimonic acid are the strongest. According to Nature , Carborane acid is the strongest, but Wikipedia says fluoroantimonic acid is stronger. Links: $\ce{HeH+}$ / Wikipedia - stating HeH+ the second most acidic $\ce{HeH+}$ / Wikipedia 2nd source - stating $\ce{HeH+}$ the most acidic (see in 'Notes' section) Fluoroantimonic Acid / Wikipedia - stating it is the strongest acid (super_acid) Carborane Acid / nature.com - According to Nature and even Google itself, it is the strongest acid created but according to wikipedia Flouroantimonic acid is stronger. So, which is the strongest ? | The problem with this question is that the exact answer is " it depends... " First off, it depends on your definition of acidity and how you measure it. Everyone seems to be using Brønstead acids (i.e. $\ce{H+}$ donors). I see two different measures in other answers: Proton affinity : This is a gas-phase measurement of $\ce{A^{−} + H^{+} -> HA}$ Hammett acidity ($H_0$) : This is a solution measurement, given by $\mathrm pK_{\ce{BH^+}} - \log\left(\frac{[\ce{BH^+}]}{[\ce{B}]}\right)$ Secondly, it depends on medium as mentioned by LDC3. Leveling effect : The solvent leveling effect reflects the lowest possible $\mathrm pK_\mathrm a$ in a particular solvent, based on the basicity of the conjugate base. So you need $\ce{HF}$ or fluorosulfuric acid to reach low $\mathrm pK_\mathrm a$. So the problem in my mind is that carboranes and fluoroantimonic acid are solution measurements, but $\ce{HeH+}$ is a gas-phase measurement. It does have the highest gas-phase proton affinity. But I'd put my money on things I can use in lab. Incidentally, the Reed group prepared the fluorinated carborane acid this year: Angew. Chem. Int. Ed. 2014, 53 (4), 1131–1134 . So that compound, $\ce{H(CHB11F11)}$ wins the crown for strongest solution-phase Brønstead acid (for the moment, at least). There's also a nice review article "Myths about the Proton. The Nature of H + in Condensed Media" by the same group in Acc. Chem. Res. 2013, 46 (11), 2567–2575 . | {
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18,466 | What are the definitions of these three things and how are they related? I've tried looking online but there is no concrete answer online for this question. | Here's a graphic I use to explain the difference in my general chemistry courses: All electrons that have the same value for $n$ (the principle quantum number) are in the same shell Within a shell (same $n$ ), all electrons that share the same $l$ (the angular momentum quantum number, or orbital shape) are in the same sub-shell When electrons share the same $n$ , $l$ , and $m_l$ , we say they are in the same orbital (they have the same energy level, shape, and orientation) So to summarize: same $n$ - shell same $n$ and $l$ - sub-shell same $n$ , $l$ , and $m_l$ - orbital Now, in the other answer, there is some discussion about spin-orbitals, meaning that each electron would exist in its own orbital. For practical purposes, you don't need to worry about that - by the time those sorts of distinctions matter to you, there won't be any confusion about what people mean by "shells" and "sub-shells." For you, for now, orbital means "place where up to two electrons can exist," and they will both share the same $n$ , $l$ , and $m_l$ values, but have opposite spins ( $m_s$ ). | {
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18,513 | I have just put together on my mind these two facts: caffeine is an alkaloid and brewed coffee is slightly acidic (pH = 5). My Biology teacher and my Chemistry teacher could not elaborate satisfying answers, but my Physics teacher said it could be because of substances to the industrial coffee powder (hillbillies in Brazil, for instance, do not display gastritis attained from too much coffee). | Coffee contains hundreds, if not thousands, of other compounds in addition to caffeine. Included among these other compounds are many acids. Many small, organic acids such as citric, malic, lactic, pyruvic and acetic acid are present, but both quinic acid and chlorogenic acid (and their derivatives) are usually present in even higher concentration. Phosphoric acid, an inorganic acid, is also present. The exact concentration of these various acids depends upon processing variables such as roast conditions and grind size. Here is a link to an interesting, one-page discussion of the subject. quinic acid chlorogenic acid Edit: In Season 3, episode 6 of "Breaking Bad", we meet Walt's new lab assistant Gale, who is brewing coffee using a lab glassware set-up. Walt says that it is the best coffee he has ever tasted and Gale responds by mentioning " quinic acid " as something to be considered in brewing great coffee. | {
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18,567 | Why does a dissociation reaction shift to the right with the addition of an inert gas? | Dissociation obviously increases the number of moles. The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change. There are two cases on which equilibrium depends. These are: Addition of an inert gas at constant volume : When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change. Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium. Addition of an inert gas at constant pressure : When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases. Consider the following reaction in equilibrium: $$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$ The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants. Please read about the Effect of adding an Inert Gas | {
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18,572 | (a) 0.008 M $\ce{Ba(OH)2}$ (b) 0.010 M $\ce{KI}$ The answer is "a" but I do not understand why. The book says that "a" is a strong electrolyte with a total ion concentration of 0.024. How did they calculate the concentration from the information that they gave? And why is it a strong electrolyte based on that calculation. I thought strong electrolytes were (solutes) ions that dissolved completely in a solvent/solution? | Dissociation obviously increases the number of moles. The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change. There are two cases on which equilibrium depends. These are: Addition of an inert gas at constant volume : When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change. Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium. Addition of an inert gas at constant pressure : When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases. Consider the following reaction in equilibrium: $$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$ The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants. Please read about the Effect of adding an Inert Gas | {
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18,904 | If I understand correctly, when we go from a keto to an enol form, we simply abstract an alpha proton and thereby create a double bond to a carbon-bearing the hydroxyl functional group. When evaluating the stabilities of the two enol forms, it appears that the enol derivative of 2,4-pentadione is more stable because of intramolecular hydrogen bonding. Both have long-range conjugation, so that cannot be a factor. The only difference really is the hydrogen bonding ... and of course ring strain (a 5-membered ring with two double bonds can't be very stable). | Keto-enol equilibria reflect a delicate thermodynamic balance between the two forms. The bonding differences between the keto and enol structures shown above are: keto: C=O double bond, C-C single bond, C-H bond enol: C=C double bond, C-O single bond, O-H bond If we look up the bond energies for these bonds we find keto: 745 + 347 + 413 = 1505 kJ/mol enol: 614 + 358 + 467 = 1439 kJ/mol This rough calculation tells us that, generally, the keto form will predominate. However, the energy difference between the two forms is only 66 kJ/mol, a relatively small number; so it is likely that small differences can cause a dramatic shift in the relative concentrations of the two species. Some general guidelines for predicting shifts in keto-enol equilibria Aromaticity and conjugation - The keto-enol equilibrium for phenol (I) lies entirely on the enol side due to thermodynamic stabilization provided by aromaticity. Compound III would be expected to have a higher enol content than compound II due to the extended conjugation present in the enol of III Substitution - Replacing hydrogen with alkyl groups on a double bond stabilizes the double bond. Therefore, we would expect compound V to have a higher enol content than compound IV Dipolar repulsion - The carbonyl dipoles in butane-2,3-dione (VI) can reduce their electrostatic repulsion by adopting the geometry pictured with the carbonyl groups oriented away from one another. Due to the constraints of the 5-membered ring, cyclopentane-1,2-dione (VII) cannot adopt a similar geometry. In this case, the dipolar repulsion is lessened by increasing the enol content in the equilibrium to a point where the enol predominates. Hydrogen Bonding - Hydrogen bonding can stabilize the enol form. If the hydrogen bond is strong enough, and particularly if other factors also stabilize the enol form, the enol form can predominate. Your molecule, 2,4-pentanedione (IX), is a good example. There are two possible enol forms, VIII and X. X is by far the predominant enol as it has 1) extended conjugation and 2) substitution on the carbon-carbon double bond. In addition, X can form a very stable hydrogen bond involving a 6-membered structure between the two oxygen atoms. Solvent Effects - Especially in cases where hydrogen bonding is involved in stabilizing the enol, the solvent can have a dramatic effect. In benzene, where intramolecular hydrogen bonding predominates, the IX:X ratio is approximately 5:95. In water, where the intramolecular hydrogen bond is replaced by hydrogen bonding to the solvent, the ratio is roughly reversed. Answer to your question Cyclopent-2-enone can form two different enols, XI and XIII. Neither enol has any features that would be expected to provide significant enol stabilization. There is no more conjugation than what is found in the ketone, there is no special hydrogen bonding, and no substitution of the carbon-carbon double bonds. Indeed, the proton nmr of cyclopentenone is just as expected, no evidence for a significant amount of enol. On the other hand, 2,4-pentanedione exists primarily as the enol form, at least in non-polar solvents, for the 3 reasons (conjugation, hydrogen bond, double bond substitution) discussed above. As an aside, you mentioned that a 5-membered ring with two double bonds (a cyclopentadiene) might not be very stable; actually, they are quite stable and quite common, 2 double bonds in a 5-membered ring do not create a lot of strain. | {
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19,001 | Active mass is defined as the molar concentration ie. number of Gram-moles per litre. My book then wrote Active mass of pure solid/liquid is always 1 . The book reasoned that Molar concentration is directly proportional to density. Since density of solid or liquid always remains constant, the active mass is taken as 1 . Really? Doesn't the density of the solid reactant decrease when the forward reaction is about to reach the equilibrium?? In a word, I am greatly confused. What is the reason for taking the active mass 1 ? Please help. | Does the activity of a solid or liquid change over the course of a reaction? The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved. The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too. It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur. What about gases, why don't they show the same phenomenon and have fixed activities? (Asked in comments.) For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, $\mathrm{100mL}$ $\ce{H2O}$ $\mathrm{= 100g}$ and half as much, $\mathrm{50mL}$ $\ce{H2O}$ $\mathrm{= 50g}$. Both have molar densities of $\mathrm{55.5 mol/L}$.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result. Why is active mass given a value of 1 then? Is this simply a convention? Shouldn't its value depend on the compound? (Asked in comments.) No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions. For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar. The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound. With that in mind, consider the activity of solids and liquids. Their standard state is their density at standard conditions. Their experimental state is their density at experimental conditions. It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1. | {
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19,090 | Can someone explain this to me by drawing resonance structures for the cyclopropylmethyl carbocation please? Also one more question, is the tricyclopropylmethyl carbocation more stable than tropylium ion? | It is commonly said, that a cyclopropane fragment behaves somewhat like a double bond. It can conjugate, and pass mesomeric effect similar to a double bond, but the donor orbital is $\sigma_{\ce{C-C}}$ instead of $\pi_{\ce{C=C}}$. Cyclopropane can be considered as a complex of a carbene and an alkene, where the carbene $\mathrm{p}$ orbital interacts with the $\pi^*_{\ce{C=C}}$ orbital while the carbene $\mathrm{sp^2}$ orbital interacts with the $\pi_{\ce{C=C}}$ orbital, so this 'virtual' double bond behaves somewhat like a normal double bond. On the other hand, the structure of the cyclopropylmethyl cation is downright weird. It is well known that both cyclopropylmethyl and cyclobutyl derivatives give very similar product mixture under $\mathrm{S_N1}$ hydrolysis conditions, resulting in both cyclopropylmethyl and cyclobutyl derivatives (see, for example, J. Am. Chem. Soc. 1951, 73 (6), 2509–2520 ). This is commonly described by the conjugation in following manner (here, the 1-cyclopropylethyl cation is depicted): Here bonding of $\ce{C-4}$ with $\ce{C-1}$ and $\ce{C-2}$ can be roughly described as an interaction of the vacant $\mathrm{p}$-orbital of $\ce{C-4}$ with filled orbital of $\pi$-bond between $\ce{C-1}$ and $\ce{C-2}$. It does not matter much, where the original leaving group was - at $\ce{C-2}$ or $\ce{C-1}$. Since the positive charge is more or less symmetrically distributed between three atoms and the small ring is somewhat relieved of its geometrical strain (both cyclopropane and cyclobutane are very strained molecules, not only due to angle strain, but also considerable steric interactions between hydrogens), the cation has remarkable stability. Similar effects are common in the chemistry of small bicyclic systems, with norbornane derivitives being the chosen test subjects for decades with the 2-norbornyl cation being probably the most well-known example. March's Advanced Organic Chemistry , 7th ed., Section 10.C.i discusses such nonclassical carbocations in great detail, with the cyclopropylmethyl system being described on pp 404–406. With further addition of multiple cyclopropyl groups, however, the full conjugation becomes sterically hindered, so extra groups beyond the first have less of an effect. Of course, the stability of these cations is far below that of the tropylium cation, which has very little strain and also possesses aromatic character, distributing the positive charge over seven(!) carbon atoms. In fact, the stability of the tropylium system is so high, that even the cyclooctatrienyl cation (also known as the homotropylium cation) adopts a similar structure. | {
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19,631 | Recently, I am learning the production of soluble and insoluble salts. My friend and I have done this experiment at the school lab. We wanted to taste them to see whether they are salty are not. The teacher luckily stopped us from doing that. So without tasting them, I would really like to know whether all salts are salty. | No. There are sweet, bitter, and various other salts. (Likely, there are tasteless salts too). Pure salty taste is as far as I know exclusive for table salt, though I wouldn't bet on it. Lead and Beryllium salts are said to be sweet, though toxic. Epsom salt, $\ce{MgSO4}$, is bitter. $\ce{CuSO4}$ has an incomprehensible, persistent metallic taste. (Based on personal experience. Copper salts are slightly toxic, but not extremely, so I survived with no consequences.) Salts with hydrolysing cation (various alums) are acidic in addition to other notes. | {
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20,009 | I've asked a similar question here but the answer given shows the behaviour of water under general conditions. I'd like to know what the behaviour of water is like as pressures increase towards infinity without being able to escape it's confinement.. i.e. a ball of water at the core of a galactic mass.. maybe this question is more for theoretical physics since we can't really measure or experiment? | Ron's answer gives us a good idea of what might happen in terms of "normal" chemistry, but if you really mean "increases to infinity", some very strange stuff happens. Suffice it to say, it doesn't stay water after a certain point. The intense temperatures created by the compression will cause the water to break apart, eventually no longer even having oxygen atoms due to nuclear reactions. Because we're talking about an externally applied pressure, the Chandrasekhar limit doesn't apply, so there is a point at which electrons and protons combine (when the electron degeneracy pressure is overcome) and a mass of neutrons remains. Neutrons themselves also have a degeneracy pressure (though we don't have good models to predict the exact pressure that has to be overcome). From here, we don't know what happens with as much certainty, but the formation of quark matter has been predicted. Eventually, we reach a singularity. We can think of this as all the matter we had before being compressed into an infinitesimal volume with infinite density and our applied pressure ceases to mean anything. If we started with enough water, this would behave much like any other black hole, though micro black holes are hypothesized to have some special properties. | {
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