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20,061 | Obviously it depends on the context whether you would call a particular species a nucleophile or a base but are the two terms largely synonymous or is there a difference? | The two are related, in that most nucleophiles are (Lewis) bases and vice versa. Some good nucleophiles are also strong bases, e.g. $\ce{HO-}$. However, a species can be a good nucleophile and a weak base, e.g. $\ce{I-}$; or a species can be a weak nucleophile and a strong base, e.g. $\ce{t-BuO-}$. How can we separate this behavior? Nucleophilicity is a kinetic phenomenon. Nucleophilicity is most often defined based on the relative rate of the reactions of nucleophiles with a standard substrate in a standard solvent. For example, a standard reaction might look like: $$\ce{CH3I ->[Nu-][H2O] CH3Nu}$$ The nucleophilicity will be related to the relative rate constant of reaction with the nucleophile (relative to the rate constant of the reaction with water $\equiv 1$). Basicity is a thermodynamic phenomenon. Basicity is based on the position of equilibrium: $$\ce{B + HSol <=> BH+ + Sol-}$$ | {
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20,076 | I learned the equation $$\ce{H2O + H+ -> H3O+}$$ And I know $\ce{H+}$ and $\ce{H3O+}$ really mean the same thing. But I am confused as to when I should use $\ce{H+}$ and when I should use $\ce{H3O+}$? Is any one of them a more accurate representation? | In addition to entropid's answer, let's remember why we invoke the hydronium ion $\ce{H3O+}$ in the first place. We use $\ce{H3O+}$ as a shorthand for $\ce{H+(aq)}$, which looks more like protonated water clusters of the generic formula $\ce{H+.(H2O)_{$n$}}\equiv \ce{H_{$2n+1$}O_{$n$}+}$. Almost ten years ago, a very interesting paper appeared in Science that examined the structure of these clusters ($n=2,3,4,5,6,7,8$). The linked article has been made freely available by Science , but you need to register for a free account at sciencemag.org. One of the key points of the article was the determination of the inner structure of these clusters. For $n=1$, you have $\ce{H3O+}$, the hydronium ion, which we are reasonably familiar with. When $n=2$, you have $\ce{H5O2+}$, which has a different structure: the proton is evenly shared between two water molecules: $\ce{[H2O\bond{...}H\bond{...}OH2]+}$. Higher order clusters have these two structures, termed the "Eigen" and "Zundel" ions, respectively, at their cores. The study used a combination of theory (MP2/aug-cc-pDVZ) and experiment (photodissociation vibrational spectroscopy). They found the for $n=2,6,7,8$, the clusters have a Zundel core, while the Eigen core exists for $n=3,4,5$. Whoa! And it gets more complicated as you increase the number of water molecules in the clusters (might be behind a pay wall) - though it turns out the Eigen ion is slightly more common in larger clusters. In aqueous solutions, we can have an extensive dynamic network of clusters always in flux. In which case, we have a mix of solvated Eigen $\ce{H3O+}$ ions representing protons closely associated with water molecules and Zundel $\ce{H5O2+}$ ions representing protons in transit. So, we use $\ce{H3O+}$ to keep things simple, and also to satisfy the needs of Brønsted-Lowry acid-base theory, in which every acid-base reaction needs a proton-donor and a proton-acceptor . Thus, the "dissociation" of acids in water is not really a dissociation, but an ionization reaction - an acid-base reaction in which water is the base and hydronium is the conjugate acid. $$\ce{HA + H2O <=> A- + H3O+}$$ Of course, it is only appropriate to use $\ce{H3O+}$ in aqueous solution. In other solvents, the "proton" has a different structure. In general, the proton is attached to a basic site on the solvent molecule. For example, in methanol: $$\ce{HCl + CH3OH <=> Cl- + CH3OH2+}$$ | {
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20,315 | I am a journal typesetter, and recently I came to a problem that I'm not sure I know to solve. The question is how to properly typeset the "NOx" abbreviation for nitrogen oxides. Which is typographically correct: $\mathrm{NO}_x$ (because $x$ is a numeric variable); or $\mathrm{NO}_{\mathrm{x}}$ (because all names of chemical compounds are strictly upright)? | I quote the Green Book by IUPAC, 2 nd printing (2008), section 1.6, enumeration item 2: The overall rule is that symbols representing physical quantities or variables are italic, but symbols representing units, mathematical constants, or labels, are roman. [...] As such, the correct way to write it is $\ce{NO}_x$, because $x$ is a variable. Please note that it is only the sum formula for the compound and not its name. Since the element names still are in roman (upright) font, everything is fine. The same goes for a formula like $\ce{C_{$n$}H_{$2n+2$}}$, which is typeset exactly like that in the Gold Book (hat tip to @Martin ). Here at StackExchange, you can type it as $\ce{NO_$x$}$ ($\ce{NO_$x$}$) or $\ce{NO_x}$ ($\ce{NO_x}$). | {
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20,405 | Why does the electron-donating inductive effect (+I) of the isotopes of hydrogen decrease in the order $\ce{T} > \ce{D} > \ce{H}$?
(where T is Tritium and D is Deuterium) Google has nothing to offer. Does it have to do anything with mass, as the order implies? | Yes, it has a lot to do with mass. Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus (see "Note" below for more detail on this point). The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length. A shorter bond has less volume to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) over resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond. The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon. Similar arguments can be applied to tritium and it's even shorter $\ce{C-T}$ bond should be even more inductively electron donating towards carbon than deuterium. Note: Bohr Radius Detail Most introductory physics texts show the radius of the $n^\text{th}$ Bohr orbit to be given by $$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 m_\mathrm{e}}$$ where $Z$ is the atom's atomic number, $k_\mathrm{c}$ is Coulomb's constant, $e$ is the electron charge, and $m_\mathrm{e}$ is the mass of the electron. However, in this derivation it is assumed that the electron orbits the nucleus and the nucleus remains stationary. Given the mass difference between the electron and nucleus, this is generally a reasonable assumption. However, in reality the nucleus does move too. It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\mathrm{e}$ with the electron's reduced mass, $\mu_\mathrm{e}$ $$\mu_\mathrm{e} = \frac{m_\mathrm{e}\times m_\text{nucleus}}{m_\mathrm{e} + m_\text{nucleus}}$$ Now the equation for the Bohr radius becomes $$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 \mu_\mathrm{e}}$$ Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus $$r_\text{heavy} \lt r_\text{light}$$ and consequently an electron will orbit closer to a deuterium nucleus than it will orbit a protium nucleus. | {
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20,511 | If two alkali metal atoms join with an oxygen atom, an ionic bond forms. Since hydrogen has the same number of valence electrons as alkali metals, why can't water be ionic? This is what I'm thinking: $$\ce{(H+)2O^2-}$$ | First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are "in between" states like polar covalent , where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond. | {
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21,810 | I've been trying to answer my (high school) daughter's questions about the periodic table, and the reactivity series, but we keep hitting gaps in my knowledge. So I showed that the noble gases have a full outer shell, which is why they don't react with anything. And then over the other side of the periodic table we have potassium and sodium, which have only one electron in their outer shell, which is what makes them so reactive, and at the top of our reactivity list. (And the bigger they get, the more reactive, which is why we were not allowed to play with caesium in class...) But then we looked up gold, which is at the bottom of the reactivity series, and found it also has only one electron in its outermost shell (2-8-18-32-18-1). Is there an easy explanation for why gold doesn't fizz like potassium when you drop it in water? (This question could be rephrased as "What properties of each element decide their ranking in the metal reactivity series?" if you prefer; that was the original question we were trying to answer.) | First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answer The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects). A nice summary of some work by Jens K. Norskov can be found here: http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490 In their experiments, they distinguished between gold atoms' ability to break and form bonds and the ease with which they form new compounds, such as gold oxides. The two qualities are related: To make a compound, gold atoms must bond with other atoms, yet they cannot do so until they have sundered their bonds with neighboring gold atoms. I think this is a nice succinct explanation. You always have this trade-off in reactions, but in gold, you don't get much energy in the new compound formation, and you're losing the gold-gold interactions. You can, of course, react gold with aggressive reagents like aqua regia , a 3:1 mix of $\ce{HCl}$ and $\ce{HNO3}$. If properly done, the product is $\ce{HAuCl4}$ or chloroauric acid . | {
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21,820 | Noble gases have full electron shells, which virtually blocks any other element from bonding with it. However, I've heard about cases where they bond to each other - for example, helium can apparently form a dimer $\ce{He2}$. How is this possible? | Noble gases usually do not form strong bonds between their atoms - it takes a fair amount of energy to dimerise them into excimers , but those are short-lived excited molecules. Thanks to excitation, shells of the atoms aren't closed and they react, but very quickly they lose energy and become separate atoms. On the other hand there are many stable molecules created by heavier noble gases (mainly xenon) with other elements. As mentioned in the comments, you may have heard about detection of so called van der Waals molecules of helium, which aren't "true" molecules, but very weakly bound pairs of atoms. In fact helium vdW dimer may have weakest bond even among them, and it was an achievement to observe it. | {
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21,915 | There are two points of view for the answer of this question: The biological view, the only one that I faced during my research, states that since it can trigger perilous conditions like metabolic acidosis, cause a neurologic sequelae, etc. methanol is toxic. The chemical point of view in an answer, which is the one I seek, is the reactions that occur in the body with methanol as a reactant. Why should it be a reaction? Because I assume something will be toxic for humans when it reacts and disables some of their bodies' vital compounds; e.g.: Hydrogen cyanide reacts with enzymes, renders their active sites useless, and finally causes heavy reduction in bio-chemical reactions that results in quick death. So, the question is: What reactions use methanol in the body that makes it toxic for us? | Methanol isn't particularly toxic in and of itself, although it's no walk in the park. If methanol flowed through the body without being broken down, it would cause roughly the same kind of harm as ethanol, i.e. intoxication. The real culprit is one of its metabolic products, methanoic acid, also known as formic acid. To understand how formic acid, present as the formate ion, is toxic, we look to Wikipedia : Formate is toxic because it inhibits mitochondrial cytochrome c oxidase, causing the symptoms of hypoxia at the cellular level, and also causing metabolic acidosis, among a variety of other metabolic disturbances. Edit: As DavePhD points out, an intermediate product in this process is formaldehyde, or methanal. While formaldehyde is also toxic, it is rapidly metabolized to methanoic acid. Reedit: The deeper, more historical reason that this happens is that methanol isn't readily available in nature, meaning that few species have developed biochemical tools to deal with it. There simply hasn't existed an evolutionary pressure to deal with methanol. | {
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21,951 | I know that there is the famous process of heating material and converting them to glass; but what I don't know is, what is the chemical process of the creation of glass? Is it crystallization? How could it result in so much visual differences? What is the description of such a radical process? | Is it crystallization? You are correct. The main difference is that sand is crystalline and glass is not—it is amorphous. The main component (> 95%) of common yellow sand is quartz (the mineral whose composition is SiO 2 ). Note that not all sand is quartz. There are white sands containing calcite (CaCO 3 ) and black sand (containing various heavy minerals). But the most common sand is indeed quartz sand: SiO 2 . Glass, the type you see in your everyday life, on the other hand, is not composed of pure SiO 2 . It has a bunch of other additives such as Na, K, B, and others. This is done to modify the properties of the glass and make it more suitable for human use. It doesn't matter much though for our discussion. So if they are made of the same thing, why the difference? The answer is cooling rate. If you cool molten SiO 2 slow enough, the atoms have enough time to organize themselves into a crystalline structure. In the case of pure SiO 2 , this is a network of SiO 4 tetrahedra: One silicon atom surrounded by four oxygens. If it cools too fast, then the crystalline structure does not form. It may be completely amorphous, or form into a sub-microscopic array of SiO 2 crystals in various structures (CT-opal for example). What determines the cooling rate? Well, in the case of glass it is a matter of minutes. You've seen glass making: The glass is molten and very quickly it solidifies to a solid. In contrast, most of the quartz sand you're seeing is actually broken fragments of rocks called granite. This type of rock has abundant quartz in it, and it forms deep underground (as in 10s of kilometers) at very slow cooling rates. While a glass maker can take his glass and let it cool in the atmosphere or in water, molten silicate magma ("glass") deep in the Earth is surrounded by rocks that are in the hundreds of degrees. This slow cooling facilitates crystallization of the SiO 2 into quartz rather than glass. How slow is this? At least tens of years, more commonly hundreds or even thousands of years. This is much slower than the seconds and minutes in glass making. Slow crystallization of SiO 2 in the Earth is not the only process. There are actually processes that can take molten silicate magma and cool it very rapidly. This commonly happens in volcanic settings: A volcano throws molten magma (again, made mostly out of SiO 2 ) and throws it into the atmosphere where it can cool very rapidly. It's even faster if the magma erupts into water. The result is volcanic glass: sometimes called obsidian. The difference in the visual appearance is actually not due to the presence or absence of a crystalline structure. It's mostly related to grain size. If you take window glass and shatter it so it's all small grains, it will look just like sand. On the other hand, if you take quartz, and make it big and smooth (just like window glass), it will look like this: (By Parent Géry (Own work) [Public domain], via Wikimedia Commons) | {
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22,148 | We recently learned in school that diamonds sparkle as the are very optically dense, meaning that it takes longer for light to pass through them, thus meaning that the light totally internally reflects when you look at a diamond, whereas in glass it would pass through at that angle. Why have we not been able to create a solid material that is so optically dense? Is it because such materials are rare, expensive or hard to produce, or because no-one has looked into it? | There are four properties that make a diamond look like a diamond. I will compare them with the two best diamond simulants at the moment: cubic zirconia $\ce{ZrO2}$ - a special ("cubic") form of zirconium oxide. Don't confuse it with zircon, a different mineral with the forumula $\ce{ZrSiO4}$. moissanite $\ce{SiC}$ - the name for the naturally occurring form of silicon carbide, also used for the synthetic gemstones varieties of it. So what are the differences and why none of them is an all-round replacement for a diamond? Refractive index : determines the brilliance . This is what you were talking about when you said "optical density". This basically means the amount of light the diamond reflects to the observer. If you change the refractive index, you also have to slightly change the cut of the diamond to achieve optimal brilliance, and then your "diamond" will not look exactly like a diamond. The refractive index of diamond is 2.42. Moissanite is higher: 2.65 and cubic zirconia is lower: 2.17. Dispersion : determines the fire . This is what you wanted to say when you mentioned that diamonds are sparkling. Fire is the property of diamonds to break the light into different wavelengths thus creating the characteristic play of colors that you get in diamonds. Here's the interesting thing: The dispersion of diamond is 0.044 and interestingly the dispersion of both cubic zirconia and moissanite is higher: 0.060 and 0.104, respectively. So theoretically, if you consider only fire (the play of colors), then they both look better than diamonds. Birefringence : (the lack of) determines the clarity. When you look through a diamond at all angles, you can clearly see what's behind. This is also true for cubic zirconia. Both minerals are isotropic . On the other hand, moissanite is birefrigent - it will double the image you see through the mineral, causing it to look blurred and lose the visual clarity. Hardness : while not directly related to optical properties, a mineral with higher hardness will retain it's sharp cut. If the cut is damaged, the gemstone will not look as good and lose its value. Diamond has a hardness of 10, moissanite 9.25 and cubic zirconia 8.25. Note that this scale is not linear: 10 is much harder than 9 compared to 9 and 8, for example. As you can see, none of the alternatives fully satisfies the entire set of properties that make a diamond. Moissanite may be slightly better, but it's also more expensive. And remember - one of the most appealing things about diamonds is their rarity and natural occurrence. No synthetic simulant can replace that. | {
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22,195 | Today in chemistry class we were discussing Organic Chemistry. We discussed what organic compounds basically are and then I asked the teacher whether $\ce{CO_2}$ is organic or not. She told that it is as it contains carbon and oxygen with a covalent bond. I told her it can't be as it is not found in animals (naturally). I am very confused about it. I need some good reasons to agree with either explanation. (I have searched the internet already but found no great insights as of now). | It is entirely arbitrary whether you call it an organic compound or not, though most would not. The distinction you make that organic compounds should be found in living things is not a useful criterion. Moreover you are wrong that carbon dioxide isn't: it is both made and used by living things. Animals make it when they metabolise sugars to release energy; plants consume it when they build more complex organic molecules through photosynthesis. In fact most organic molecules are, ultimately, derived from $\ce{CO2}$. Even more importantly most molecules considered organic are neither made by nor are found in living things. Chemists make new carbon compounds all the time (tens of millions in the history of chemistry) and most have never been made by animals or plants. The organic/inorganic terminology is mostly very simple: covalent compounds containing carbon are organic . The only fuzzy area is around very simple molecules like $\ce{CO2}$ where the distinction doesn't matter much. So we would not normally think of diamond or silicon carbide as organic. But we might (though many would not) call calcium carbide organic because it contains a $\ce{C2}$ unit with a carbon-carbon triple bond. However since the terminology is mostly very obvious and also somewhat arbitrary, it isn't worth much argument to sort out those very simple but awkward edge cases. | {
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23,572 | People say it dissolves, but shouldn't the acid form? $$
\ce{NaCl + H2O ->[?] HCl + ???}
$$ | If you dissolve NaCl in water you will get some HCl molecules but there's definitely not going to be a significant concentration of HCl formed. The reaction that you propose - $\ce{Cl- + H2O -> HCl + HO-}$ is highly thermodynamically unfavorable. We can ascertain this fact through consultation of any pKa/pKb table. In the equation above, the product acid (HCl) is a much (as in almost a trillion trillion times) stronger acid than water. Given that HCl is several trillion times stronger than water as a acid, then naturally, HCl will want to protonate hydroxide ion, a byproduct of HCl formation from chloride ion. This is ignoring the fact that hydroxide ion is also a strong base in water, so it has a high proton affinity in water. So even if the products were formed - again, very unfavorable from a thermodynamic standpoint because the reactant base and reactant acid are both so weak - then the products would certainly react with each other and form the reactants again, resulting in no net change in solution contents and pH. As a result, the reaction that you propose is more like this (except that the bottom/reverse arrow should be a lot bigger). $\ce{Cl- + H2O <<=> HCl + HO-}$ | {
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23,610 | I was thinking is there really bond present at microscopic level or atoms/molecules are just nearby and are connected with force which is not visible(like gravitational force between earth and sun) and we make bonds just for understanding. | All credit to Zhang et al. " Real-Space Identification of Intermolecular Bonding with Atomic Force Microscopy " Science Vol. 342 no. 6158 pp. 611-614. Yes, direct images of bonds, not only covalent bonds but also intermolecular hydrogen bonds have been recorded. It is the electron density that is being observed, covalent and hydrogen bonds involving high electron density between the atoms. Scanning Tunneling Microscopy can also be utilized to directly observe the electron density of bonds. | {
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23,899 | My teacher didn't answer this properly: Is toothpaste solid or liquid? You can't say toothpaste is a solid because solid material have a fixed shape but toothpaste doesn't. However, you can't say it's a liquid because liquids flow easily but toothpaste needs a certain force to push it out of the tube. So is it a solid or liquid? And are there any other example just like toothpaste? | Toothpaste is what is called a non-newtonian fluid , more specifically toothpaste is a Bingham plastic . This means that the viscosity of the fluid is linearly dependent on the shear stress, but with an offset called the yield stress (see figure below). This yield stress is what makes it hard to say whether it is liquid or solid. The fact that toothpaste is viscous alone is not sufficient to explain this, because water is also viscous, but doesn't behave like a solid (unless frozen, but that's another phenomenon). What the yield stress does is the following. Below a certain shear threshold the fluid responds as if it were a solid, as you can see happening when you have put toothpaste on your toothbrush, it just sits there without flowing away. A highly viscous but newtonian fluid would flow away (although slowly as pointed out by @ron in his comment to the answer of @freddy). Now if you put sufficient shear stress on the toothpaste, when you squeeze the tube of paste, it will start flowing and respond as a liquid. Other examples, as mentioned in the Wikipedia link in my first sentence, are e.g. mayonnaise and mustard. Another example is silly putty. | {
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23,905 | Is there any famous experiment in chemistry where digital signal processing played an important part? I don't mean using a machine that relies on such techniques (they all do) but an experiment where signals were first recorded (e.g. temperature/pressure/voltage/etc. as a function of time) and then analyzed with some sophisticated methods (e.g. time-frequency analysis with wavelets). | Toothpaste is what is called a non-newtonian fluid , more specifically toothpaste is a Bingham plastic . This means that the viscosity of the fluid is linearly dependent on the shear stress, but with an offset called the yield stress (see figure below). This yield stress is what makes it hard to say whether it is liquid or solid. The fact that toothpaste is viscous alone is not sufficient to explain this, because water is also viscous, but doesn't behave like a solid (unless frozen, but that's another phenomenon). What the yield stress does is the following. Below a certain shear threshold the fluid responds as if it were a solid, as you can see happening when you have put toothpaste on your toothbrush, it just sits there without flowing away. A highly viscous but newtonian fluid would flow away (although slowly as pointed out by @ron in his comment to the answer of @freddy). Now if you put sufficient shear stress on the toothpaste, when you squeeze the tube of paste, it will start flowing and respond as a liquid. Other examples, as mentioned in the Wikipedia link in my first sentence, are e.g. mayonnaise and mustard. Another example is silly putty. | {
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24,035 | Rotten potatoes have a distinctive unpleasant odour, and I can't find any definitive answer as to what the chemical involved is. I was thinking it might be Butyric acid, from the fermentation of the starch, but according to Wikipedia : Butyric acid is present in, and is the main distinctive smell of, human vomit. And those puppies don't smell like vomit. They're more like rotting flesh – the reason I'm asking is that I've spent half the day looking for a dead animal that the cat brought in or something, only to discover that the horrible smell was coming from the potato box. | My first guess was that it was putrescine and other polyamines---but the smell of putrefying potatoes actually comes from methyl mercaptan, dimethyl sulfide, and dimethyl trisulfide. Check it: Source: A. Kamiya, Y. Ose, Study on offensive odor (Report IV): A Consideration of Putrefaction and Offensive Odor of Solid Waste , Journal of Japan Society of Air Pollution, 18(5), 1983, pp 453-463. Notice that the headspace composition changes quite a bit with time. Most of the headspace chromatography studies I found dealt with early detection of disease organisms in infected potatoes, rather than potatoes in full putrefaction mode. You'll also find many references to solanine poisoning from potatoes; solanine is a toxic glycoalkaloid , is nonvolatile, and has nothing at all to do with the foul smell and toxic gas produced by putrid potatoes. Methyl mercaptan ($\rm CH_3SH$) has an odor described as "rotting cabbage" by ATDSR; it's one of the major contributors to the smell of farts (oh, sorry, "flatus" if we're being polite). It has an odor detection threshold as low as 1 ppb. Dimethyl sulfide ($\rm (CH_3)_2S$) is responsible for the smell of the sea (in low concentrations); it too has a cabbagy smell. Dimethyl trisulfide ($\rm CH_3SSSCH_3$) is present in relatively smaller amounts but it has an even stronger odor. The detection threshold is around 1 part per trillion, and it is apparently a strong insect attractant. The gases produced by rotting potatoes are quite toxic (see Rotting potatoes in basement kill four members of Russian family ). (To keep the notes below in context, in my original answer I said that a plot point in "The Walking Dead" was that zombies couldn't smell delicious humans if they were wearing coats smeared with rotting flesh. I suggested that when the zombie apocalypse arrives, packing your pockets with putrid potatoes might work, too. Now I'm not so sure. Can zombies distinguish between polyamines and sulfur compounds? Perhaps they're stench connoisseurs.) | {
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24,520 | I know for a fact that cis alkenes are less stable than trans alkenes due to steric repulsions. But what bugs me is the fact that why systems such as 2-methylprop-1-ene, where I believe steric repulsions are greater than that in cis -but-2-ene, are more stable than its cis and trans counterparts. I saw this being given in the stability order of alkenes in Solomons and Frhyle.
I looked up Morrison and Boyd and it said the same thing. Paula Bruice said otherwise stating trans to be more stable. Which is the correct order, and why? Paula Bruice Solomons and Frhyle For reference, free energies of formation : $$
\begin{array}{c|c} \text{Compound} & \Delta G^{\circ}_{f}\text{ in kJ mol}^{-1} \\\hline \text{2-methylpropene} & 58.1 \\ E\text{-2-butene} & 63.0 \\ Z\text{-2-butene} & 65.9 \\\hline \end{array}
$$ | This is a very good question and if popular books give conflicting answers, then it must be reasoned out. Unfortunately, Paula Bruice has given the wrong answer while the other two books have given no explanation for this comparison. For the answer I assume that you know about hyperconjugation and the various contributing structures it involves. This gives a basic idea of hyperconjugation: Source: tutorcircle.com Note that on hyperconjugation, the other carbon of the alkene gets a negative charge on it. If the hyperconjugation structures of 2-methylprop-1-ene and trans -but-2-ene are drawn: you will notice that: $ \ce{C_1} $ of the former has a negative charge on it $ \ce{C_2} $ of the latter has a negative charge on it The latter's methyl group makes the negatve charge less stable due to +I (inductive effect) The former has no such destablizing effect as the carbon containing the negative charge is just connected to hydrogens As all the hyperconjugation structure have this effect applying on them, the net effect cant be ignored for the overall stability Hence the overall stability of 2-methylprop-1-ene > trans -but-2-ene | {
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24,528 | I'm working on a crime story about cyanide poisoning from apple seeds. I just would like to have an idea of what processes and extraction techniques might be involved in getting cyanide from the seeds. The character is supposed to have access to a high school chemistry lab; so I was thinking equipment and tools that can be found in a typical HS lab. I prefer a detailed answer, step by step, if possible. If not then just generalized terminologies would work. | Forget about the apple seeds, they contain about 1 to 4 mg amygdalin per gramm seeds ( DOI ). Instead, collect apricot seeds during the right season, the amygdalin content varies though the year and can be as high as 5% of the dry weight of the seed ( DOI ). It is probably advantagenous to break the husk with a nut cracker, a plier, etc. and cut the softer inner material to smaller pieces. Extraction of amygdalin can be performed by immersing the material in methanol and subsequent ultrasonification Soxhlet extraction with methanol reflux extraction in water in the presence of citric acid A comparison of the extraction methods is given here . Removal of the solvent in vacuum will yield a crude material with significant amounts of amygdalin. You might want to have a look at this article from the Western Journal of Medicine on the toxicity. Here, an $\mathrm{LD_{50}}$ of 522 mg amygdalin per kg body weight was estimated for the oral application to rats. The online resource of the U.S. National Library of Medicine gives a value of 405 mg/kg. Further information on the health risk of apricot kernels are provided by of the German Bundesinstitut für Risikobewertung (Federal Institute for Risk Assessment) and the British Committee on Toxicity . A note in the a German medical journal, Deutsche Ärzteblatt, ( PDF ) describes a case where boy of four years (110 cm, 18 kg body weight) was given apricot kernels during an alternative cancer treatment. Upon additional treatment with a single dose of 500 mg amygdalin, the kid showed agitation, spasms and the eyes started to roll. I'll leave it up to your fantasy as a writer on how to apply the poison, but spicing some marzipan with it might help ;) | {
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24,691 | I've been using Ruby to write scripts for research, but I want to get into some heavier stuff that Ruby is just too slow for. I noticed there are a few things written in C and C++, but there is an oddly large proportion of software used in computational chemistry that is written in FORTRAN (in which I have zero experience.) Why is FORTRAN used in computational chemistry? From what I understand, FORTRAN is kind of on the ancient side (“punchcard” old.) I was a bit shocked to find fairly recently written tutorials for FORTRAN. Is it a sort of, "this is how we've always done it," thing or is there an efficiency aspect to FORTRAN I'm overlooking? Note: I may have FORTRAN confused with later programming languages with similar names. | I don't think that's really true anymore. Some Fortran use is historical (i.e., early codes were developed in FORTRAN because that was the best programming language for number crunching in the 70s and 80s). Heck, the name stands for "formula translation." Some Fortran use is because of performance. The language was designed to be: especially suited to numeric computation and scientific computing. Many times, I find chemistry coders sticking to Fortran because they know it and have existing highly optimized numeric code-bases. I think the performance side isn't necessarily true anymore when using modern, highly optimizing C and C++ compilers. I write a lot of code in C and C++ for performance and glue a lot of things with Python. I know some quantum programs are written exclusively or primarily in C++. Here are a few open source examples: Psi4 - Written in C++ and Python MPQC - Written in C++ LibInt - Written in C++ for efficient quantum integrals. LibXC - Written in C with Fortran "bindings" for DFT exchange-correlation functionals This is my opinion, but my recommendation for faster performance in chemistry would be Python with some C or C++ mixed in. I find I'm more efficient coding in Python, partly because of the language, partly because of the many packages, partly since I don't have to compile, and that's all important. Also, you can run Python scripts and functions in parallel, on the GPU, and even compile them, e.g. with Numba . As I said, if I think performance is crucial, I'll write pieces in C or usually C++ and link to Python as needed. | {
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24,708 | I wanted to ask an online group of experts/community about how I could improve in chemistry. I study in 10th grade currently and don't get very good grades. Most of the time its a B or C. I have to select science stream, in which physics, chemistry and maths are compulsory. I am good in physics and maths, but what should I do in chemistry? I have a hard time grasping concepts like Ionisation Energy, structure of benzene etc. and these are only going to get 100 times tougher next year! | Chemistry needs a devoting reader. You should try to love chemistry as much as you want! That'll be the ignition for lots of reading and stuff. My pointers: Love chemistry. Learn in an organized matter. (For example, do not jump from studying about covalent bonds' basics to reaction mechanism. Take notes of the most important points of what you learn. Look for open-source online tutorials and not only the books. Some chemistry concepts are related to each other. Learn the first one's basics and then read the other one's. This will build up your understanding, literally. For example, first learn about electronegativity and then study the covalent bonds' basics. This is the exception for rule 2. Try to differentiate between what's taught in schools to be true and what really is. For example, look for exceptions of the periodic trends that are taught in schools. Be imaginatively practical. Don't just say enough for the lewis structure of water, but do it about $\ce{SF6}$. Always try to go relatively higher than what your textbook has written. We have a proverb in my mother language meaning "If you have 100, you do have 90." Ask what misunderstandings you have in here! Never look to doing a research as a waste of time. When approaching to solve a chemical problem, do it step by step. When approaching to solve a problem, don't forget to write units of measurement. When you reached an answer to the problem, ask yourself: "Is this possible?" If you've calculated 5 grams of water to be $\frac{18}{5}$ mol, and you know that a mole of water is 18 grams, shouldn't something be wrong here? For any definitions, uses, notions, and practical experiments you encounter, ask yourself: "What is this for, either in real life, experimental sciences or theoretical chemistry?" Look for online exercises, free of charge, but usually important. Follow the latest news that are being discussed everywhere in the world of chemistry. For specific concepts in chemistry, e.g.: Periodic table, doing a research about the history is very helpful. This is as far as I can count. You can ask "why"s in comments. | {
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26,254 | How is it possible to heat a tiny amount (30 ml)[1] of water to a high enough temperature to make a coffee, in less than 10 seconds and possibly instantly? Most heaters that I know of heat water in no less than 90 seconds (induction heater) or 3 minutes which is way too much for my purposes, and a microwave oven takes me one minute and a half to heat. [1] A single coffee is between 20 - 30 ml and my coffee machine makes 30 ml for each cup. | Well, let's do some math: Assuming 30 mL of water is 30 g, and we want to heat our water from 20 °C to 90 °C, the energy required is:
$$\begin{align}E&=C_Pm\Delta K \\
&=\left(4.18 \mathrm{\frac{ J}{gK}}\right)(30\mathrm{\ g})(70\mathrm{\ K})\\
&=8.778\mathrm{\ kJ}\end{align}$$ So how much power do we need to do this in a given time? "Instant" doesn't really mean anything, so let's go with 10 seconds: $$\begin{align}P&=\frac{E}{t}\\
&=\frac{8778\mathrm{\ J}}{10\mathrm{\ s}}\\
&=877.8 \mathrm{\ W}\end{align}$$ This is not an enormous amount of power, but the trick is that it all has to go into heating the water. A good microwave outputs a fair bit more power than this, but it generally doesn't all get absorbed by such a such a small amount of water in only 10 seconds. Your best bet is probably an electric heating element directly inserted into the liquid, though I don't know if you can get a ~1000 W one small enough to sit in that much water. As Jon Custer notes, it's not necessary to produce all the heat at once. If you heat some kind of thermal reservoir and flow the liquid past/through it, it reduces the demands on your heat source. Edit: Also, I just tried this with a 1200 W microwave and it only took 15 seconds. How fast do you really need this coffee? | {
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26,699 | For example for 4 given elements A,B,C,D, what's the meaning of $\text{AB} \cdot x \text{CD}$? | By writing $\ce{AB.xCD}$ chemists mean that there are CDs are found in the crystalline framework of AB. The most common example of this is water trapped inside the crystal structure of ionic compounds. (See water of crystallization in wikipedia ) An example that's often taught is $\ce{CuSO4.5H2O}$ . See that 5 that's a representative of $x$ ? It means that water can be bound only with intermolecular interactions, but it exists in the structure in a stoichiometric ratio. Source Also, according to this article In summary, when a dot is used to break a formula into subunits, it may signify ignorance of how the subunits are structurally related, as in our inorganic example; or it may correspond to actual significant structural subunits, as in our organic example; or it may represent the combining ratios of the binary starting materials required for the synthesis of the compound, as in our phase diagram example. As a result, for example in direct addition reactions, where there is ignorance about the structural relation of the compounds, we use a dot. | {
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26,749 | I know that this is a rather ambiguous question; but my question is, whenever we take water and freeze it in the freezer, it still tends to stay clear. Since snow is just frozen water, why is it white? Is it due to contents of the air - i.e. dust - that make snow this color? Or is my freezer just weird? | The difference between snow and ordinary ice cubes is mainly about the size of the particles. Snow is made from small, irregular crystals with many edges at a very small scale. Light is refracted or scattered by the edges (or the interface between air and the edges). Snow is white because the scattering effect of those edges dominates what happens to light shining on the snow. In a large block of ice like an ice cube there is very little refraction or scattering as, for any ray of light, there are only two edges (one when light enters the cube, one when it leaves). So transmission dominates (and there is little colour as ice only weakly absorbs visible light). This is common in many other compounds. Titanium dioxide is a transparent mineral but is used as the primary ingredient in many "white" products like paint. The secret is to use TiO 2 particles that are just the right size to maximise the scattering at the particle edges thereby creating "whiteness" as all light is equally scattered. | {
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27,095 | While naming any compound, the numbering should be done such that position of substituted groups gives smallest sum. I'm struggling with the IUPAC naming for cyclic compounds. How exactly does this rule work? | The guideline that you have mentioned (‘While naming any compound, the numbering should be done such that position of substituted groups gives the smallest sum.’) does not exist in the IUPAC recommendations. The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually reads as follows: P-14.3.5 Lowest set of locants The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’. (…) Therefore, the smallest sum of locants can lead to wrong results. For example, for 2,7,8-trimethyldecane, the smallest sum of locants would yield the wrong name ‘3,4,9-trimethyldecane’, since $2+7+8=17 > 3+4+9=16$ . Likewise, for 2,2,6,6,7-pentamethyloctane, the smallest sum of locants would yield the wrong name ‘2,3,3,7,7-pentamethyloctane’, since $2 + 2 + 6 + 6 + 7 = 23 > 2 + 3 + 3 + 7 + 7 = 22$ . For 1,6,7-trimethylnaphthalene, the smallest sum of locants would yield the wrong name ‘2,3,5-trimethylnaphthalene’, since $1 + 6 + 7 = 14 > 2 + 3 + 5 = 10$ . For 1,6-dimethylcyclohex-1-ene, the smallest sum of locants would yield the wrong name ‘2,3-dimethylcyclohex-1-ene’, since $1 + 6 = 7 > 2 + 3 = 5$ . Anyway, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules. | {
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27,302 | I have a very very general question: In DFT functional selection , mostly people speak about the most recent ones. For example my professor always asks: " which DFT Functional did you select ? " and if I say B3LYP, he says : " No ! that's too old ! " but if I answer: M06 , he says : " hmm ... sounds promising, that's a modern functional ". I think it is too naive to select functionals just based on their chronologic sequence. I want to ask if there is any good and reliable criteria for functional selection. For example a criteria that says for an alkene with some specific characteristics, go for M06-L and for an alkane with other characteristics, use B97xxx family and so on.. Is there such a criteria ? I hope this topic can become a good guideline for future reference. ! | What's up with all that magic? (A chapter formerly known as Introduction) The hunt for for the holy grail of density functional theory (DFT) has come a long way. [1] Becke states in the introduction of the cited paper: Density-functional theory (DFT) is a subtle, seductive, provocative business. Its basic premise, that all the intricate motions and pair correlations in a many-electron system are somehow contained in the total electron density alone, is so compelling it can drive one mad. I really like this description, it points out why we use and need DFT, and as it also points out the flaws, that every computational chemist has to deal with: How can something with such a simple approach be correct? Something that is often forgotten about DFT is, that in principle it is correct. It's the implementations and approximations, that make it incorrect, but usable. Becke states this in the following quote: Let us introduce the acronym DFA at this point for “ density-functional approximation .” If you attend DFT meetings, you will know that Mel Levy often needs to remind us that DFT is exact . The failures we report at meetings and in papers are not failures of DFT, but failures of DFAs. I sometimes here that the abbreviation DFT is often used in the wrong context, since we are not talking about the theory itself any more, but about the implementations and approximations of it. One suggestion I heard was, that it should rather be used as density functional technique . With that in mind I would like to state, that I absolutely agree with the previous answer by user1420303 and it's subsequent comment by Geoff Hutchison. Since you asked for a somewhat more practical approach, I like to offer the advice I usually give to students new in the field. Old is bad, isn't it? Some of the functionals are now around for about thirty years. That does not make them bad, maybe even the opposite. It shows, that they are still applicable today, giving reasonable results. One of my personal favourites is the conjunction of Becke 1988 and Perdew 1986, often abbreviated as BP86. [2] It's a pure functional which is available most modern quantum chemical packages. [3] It performs usually well enough for geometries and reasonable well for energies for simple systems, i.e. small organic molecules and reactions. The magical functional B3LYP was one of the first hybrid functionals, and it was introduced by Gaussian's very own developers. [4] A lot of people were surprised how well it worked and it quickly became one of the most popular functionals of all time. It combines Becke's three parameter functional B3 [5] with Lee, Yang and Parr's correlation functional. [6] But why are we surprised it works? The answer is quite simple, it was not fitted to anything. Frish et. al. just reworked the B3PW91 functional to use LYP instead of PW91. As a result, it heavily suffers or benefits from error compensation. Some even go as far as to say: “It is right for the wrong reasons.” [7-9] Is it a bad choice? No. It might not be the best choice, but as long as you know what you are doing and you know it is not failing for your system, it is a reasonable choice. One functional is enough, is it? Now that we established, that old functionals are not out of fashion, we should establish something very, very important: One is never enough. There are a few things, where it is appropriate to do most of the work with one functional, but in these cases the observations have to be validated with other methods. Often it is best to work your way up Jacob's ladder. [10] How do I start? It really depends on your system and what you are looking for. You are trying to elucidate a reaction mechanism? Start with something very simple, to gain structures, many structures. Reaction mechanisms are often about the quantity of the different conformers and later about suitable initial structures for transition states. As this can get complex very fast, it's best to keep it simple. Semi-empirical methods and force fields can often shorten a long voyage. Then use something more robust for a first approach to energy barriers. I rely on BP86 for most of the heavy computing. As a modern alternative, another pure density functional, M06-L is quite a good choice, too. [11] Some of the popular quantum chemistry suites let you use density fitting procedures, which allow you to get even more out of the computer. Just to name a few, without any particular order: Gaussian , MolPro , Turbomole . After you have developed a decent understanding of the various structures you obtained, you would probably want to take it up a notch. Now it really depends on what equipment you have at hand. How much can you afford? Ideally, more is better. At least you should check your results with a pure, a hybrid, and a meta-hybrid functional. But even that can sometimes be a stretch. [12] If you are doing bonding analysis, elucidation of the electronic structure, conformation analysis, or you want to know more about the spectrum, you should try to use at least five different functionals, which you later also check versus ab initio approaches. Most of the times you do not have the hassle to deal with hundreds of structures, so you should focus of getting the most accurate result. As a starting point I would still use a pure functional, the worst thing that could happen is probably, that is reduces the times of subsequent optimisations. Work your way up Jacob's ladder, do what you can, take it to the max. [13] But of course, keep in mind, that some functionals were designed for a specific purpose. You can see that in the Minnesota family of functionals. The basic one is M06-L, as previously stated a pure functional, with the sole purpose of giving fast results. M06 is probably the most robust functional in this family. It was designed for a wide range of applications and is best chosen when dealing with transition metals. M06-2X is designed for main group chemistry. It comes with somewhat built in non-covalent interactions and other features. This functional (like most other though) will fail horribly, if you have multi-reference character in your system. The M06-HF functional incorporates 100% Hartree-Fock exchange and was designed to accurately calculate time dependent DFT properties and spectra. It should be a good choice for charge transfer systems. See the original publication for a more detailed description. [14] Then we have another popular functional: PBE. [15a] In this initial publication an exchange as well as a correlation functional was proposed, both pure density functionals, often used in conjunction. [15b] I don't know much about it's usefulness, since I prefer another quite robust variation of it: PBE0, which is a hybrid functional. [15c,d] Because of its adiabatic connection formula, it is described by the authors as a non-empirical hybrid functional. [15d] Over the years there have been various developments, some of the are called improvement, but it often boils down to personal taste and applicability. For example, Handy and Cohen reintroduced the concept of left-right correlation into their OPTX functional and subsequently used it in combination with LYP, P86 and P91. Aparently, they work well and are now often used also as a reference for other density functionals. They went on and developed a functional analogous to B3LYP but outperforming it. [16] But these were obviously not the only attempts. Xu and Goddard III extended the B3LYP scheme to include long range effects. They claim a good description of dipole moments, polarizabilities and accurate excitation energies. [17] And with the last part in mind, it is also necessary to address long range corrections. Sometimes a system cannot be described accurately without them, sometimes they make the description worse. To name only one, CAM-B3LYP, which uses the coulomb attenuating method. [18] And there are a couple of more, and a couple of more to come, head on over to a similar question: What do short-range and long-range corrections mean in DFT methods? As you can see, there is no universal choice, it depends on your budget and on the properties you are interested in. There are a couple of theoretical/ computational chemists on this platform. I like BP86 as a quick shot and answer questions relating to MO theory with it, shameless self-promotion: Rationalizing the Planarity of Formamide or Rationalising the order of reactivity of carbonyl compounds towards nucleophiles . And sometimes we have overachievers like LordStryker, that use a whole bunch of methods to make a point: Dipole moment of cis-2-butene . So I picked a functional, what else? You still have to pick a basis set. And even here you have to pick one that fits what you need. Since this answer is already way longer than I intended in the first place (Procrastination, yay!), I will keep it short(er). There are a couple of universally applicable basis sets. The most famous is probably 6-31G*. This is a nice ancient basis set that is often used for its elegance and simplicity. Explaining how it was built is easier, than for other basis sets. I personally prefer the Ahlrichs basis set def2-SVP, as it comes with a pre-defined auxiliary basis set suitable for density fitting (even in Gaussian). [19] Worth mentioning is the Dunning basis set family cc-pVDZ, cc-pVTZ, ... . They were specifically designed to be used in correlated molecular calculations. They have been reworked and improved after its initial publication, to fit them to modern computational standards. [20] The range of suitable basis sets is large, most of them are available through the basis set exchange portal for a variety of QC programs. Sometimes an effective core potential can be used to reduce computational cost and is worth considering. *Sigh* What else? When you are done with that, consider dispersion corrections. The easiest way is to pick a functional that has already implemented this, but this is quite dependent on the program of you choice (although the main ones should have this by now, it's not something brand new). However, the standalone DFT-D3 program by Stefan Grimme's group can be obtained from his website. [21] Still reading? Read more! (A chapter formerly known as Notes and References) Axel D. Becke, J. Chem. Phys. , 2014 , 140 , 18A301. (a) A. D. Becke, Phys. Rev. A , 1988 , 38 , 3098-3100. (b) John P. Perdew, Phys. Rev. B , 1986 , 33 , 8822-8824. Unfortunately this functional is not always implemented in the same way, although the differences are pretty small. It basically boils down as to which VWN variation is used in the local spin density approximation term. Also see S. H. Vosko, L. Wilk, and M. Nusair, Can. J. Phys. , 1980 , 58 (8), 1200-1211. P. J. Stephens, F. J. Devlin, C. F. Chabalowski, and M. J. Frisch, J. Phys. Chem. , 1994 , 98 (45), 11623–11627. Axel D. Becke, J. Chem. Phys. , 1993 , 93 , 5648. C. Lee, W. Yang, and R. G. Parr, Phys. Rev. B , 1988 , 37 , 785–789 Unfortunately the B3LYP functional suffers from the same problems that are mentioned in [3]. The failures of B3LYP are known and often well documented. Here are a few recent papers, but there are many, many more. (a) Holger Kruse, Lars Goerigk, and Stefan Grimme, J. Org. Chem. , 2012 , 77 (23), 10824–10834. (b) Joachim Paier, Martijn Marsman and Georg Kresse, J. Chem. Phys. , 2007 , 127 , 024103. (c) Igor Ying Zhang, Jianming Wu and Xin Xu, Chem. Commun. , 2010 , 46 , 3057-3070. ( pdf via researchgate.net ) Just my two cents, that I am hiding in the footnotes: “ Pretty please do not make this your first choice.” John P. Perdew and Karla Schmidt, AIP Conf. Proc. , 2001 , 577 , 1. ( pdf via molphys.org ) Yan Zhao and Donald G. Truhlar, J. Chem. Phys. , 2006 , 125 , 194101. Note, that not always full recomputations of geometries are necessary for all different functionals you apply. Often single point energies can tell you quite much how good your original model performs. Keep the computations to what you can afford. Don't use an overkill of methods, if you already have five functionals agreeing with each other and possibly with an MP2 calculation, you are pretty much done. What can the use of another five functionals tell you more? Y. Zhao, N.E. Schultz, and D.G. Truhlar, Theor. Chem. Account , 2008 , 120 , 215–241. (a) John P. Perdew, Kieron Burke, and Matthias Ernzerhof, Phys. Rev. Lett. , 1996 , 77 , 3865. (b) The exchange functional was revised in Matthias Ernzerhof and John P. Perdew, J. Chem. Phys. , 1998 , 109, 3313. (c) Carlo Adamo and Vincenzo Barone, J. Chem. Phys. , 1999 , 110 , 6158. (d) Kieron Burke, Matthias Ernzerhof, and John P. Perdew, Chem. Phys. Lett. , 1997 , 265 , 115-120. (a) N. C. Handy and A. J. Cohen, Mol. Phys. , 2001 , 99 , 403-12. (b) A. J. Cohen and N. C. Handy, Mol. Phys. , 2001 , 99 607-15. X. Xu and W. A. Goddard III, Proc. Natl. Acad. Sci. USA , 2004 , 101 , 2673-77. T. Yanai, D. P. Tew, and N. C. Handy, Chem. Phys. Lett. , 2004 , 393 , 51-57. (a) Florian Weigend and Reinhart Ahlrichs, Phys. Chem. Chem. Phys. , 2005 , 7 , 3297-3305. (b) Florian Weigend, Phys. Chem. Chem. Phys. , 2006 , 8 , 1057-1065. The point where the use of more basis functions does not effect the calculation. For the correlation consistent basis sets, see a comment by Ernest R. Davidson, Chem. Phys. Rev. , 1996 , 260 , 514-518 and references therein. Also see Thom H. Dunning Jr, J. Chem. Phys. , 1989 , 90 , 1007 as the original source. DFT-D3 Website ; Stefan Grimme, Jens Antony, Stephan Ehrlich, and Helge Krieg, J. Chem. Phys. , 2010 , 132 , 154104. Have fun and good luck! | {
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28,221 | Can someone please explain what exactly the difference between an intermediate and a transition state is? I understand that they are formed as part of the process in converting the reactants to products in a chemical reaction. Other than that, all I know is that usually a transition state exists very briefly (if at all) whereas an intermediate can be isolated. But why? | An intermediate is a short-lived unstable molecule in a reaction which is formed inbetween the reaction when reactants change into products. Whereas, transition state is just the state before formation of new molecule(involves breaking of bonds of reactants and formation of new ones) An intermediate differs from a transition state in that the intermediate has a discrete lifetime (be it a few nanoseconds or many days), whereas a transition state lasts for just one bond vibration cycle. Intermediates may be unstable molecules (in which case they are called reactive intermediates) or highly stable molecules.
The difference between them can be better described through the energy profile diagram. Transition states are local energy maximums and have partial bonds. This might be one of the reasons why they cant be isolated as intermediates. | {
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28,346 | Today, I'm looking for how to find the 2nd perturbation to the base in Rayleigh Schrödinger Perturbation Theory (RSPT). SETUP Starting from the 2nd order perturbation in Dirac's notation: \begin{equation}
\hat{H}^0 |n^2\rangle + \hat{V} |n^1\rangle = E_n^0 |n^2\rangle + E_n^1 |n^1\rangle + E_n^2 |n^0\rangle
\end{equation} Multiply to the left with $\langle k^0 |$ \begin{equation}
\langle k^0| \hat{H}^0 |n^2\rangle + \langle k^0| \hat{V} |n^1\rangle = \langle k^0| E_n^0 |n^2\rangle + \langle k^0| E_n^1 |n^1\rangle + \langle k^0| E_n^2 |n^0\rangle
\end{equation} The hermicity of the Hamilton operator: \begin{equation}
E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V} |n^1\rangle = E_n^0 \langle k^0 |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle + E_n^2 \langle k^0 |n^0\rangle
\end{equation} Cancel the third term on the right \begin{equation}
E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V} |n^1\rangle = E_n^0 \langle k^0 |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle
\end{equation} Gathering same inner products: \begin{equation}
\langle k^0| \hat{V} |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle = ( E_n^0 - E_k^0 ) \langle k^0 |n^2\rangle
\end{equation}
\begin{equation}
( E_n^0 - E_k^0 ) \langle k^0 |n^2\rangle = \langle k^0| \hat{V} |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle
\end{equation}
Replacing $| N^1 \rangle$ and $E_N^1$ with the first order perturbation to the base and to the energy: \begin{equation}
|n^1\rangle =\sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}
\begin{equation}
E_N^{1} = \langle n^0 | \hat{V} | n^0 \rangle
\end{equation} \begin{equation}
( E_n^0 - E_k^0 ) \langle k^0 |n^2\rangle = \langle k^0| \hat{V} \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0} - E_n^1 \langle k^0 \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation} \begin{equation}
( E_n^0 - E_k^0 ) \langle k^0 |n^2\rangle = \sum_{m \neq n} \frac{ \langle k^0| \hat{V} |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0} - \langle n^0 | \hat{V} | n^0 \rangle \sum_{m \neq n} \frac{\langle k^0 |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation} At this point, I'm not sure if the bracket algebra was done properly not even forward. Then ... are another some algebra steps that I cannot figure out the properly path to find the respective coefficient: $$\langle k^0 | n^2 \rangle$$ In order to replace it in: \begin{equation}
|n^2\rangle =\sum_{k \neq n} |k^0 \rangle \frac{\langle k^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_k^0 ???}
\end{equation} QUESTION Please, Would someone help me out to show me how to get the properly 2nd perturbation to the wave function? | Disclaimer This post is some kind of a legacy post. Find the notation used in the question in the other answer. I added this proof as I was not entirely certain I understood the notation correctly. As it turned out, I did not. As a result I posted the complete derivation of RSPT up to second order, trying to guide anyone through it using a different notation.$\require{cancel}$ Obviously, I will come back to the initial problem only at the end, if you are impatient, just skip ahead. Please also note, that I removed the parts that became redundant when I posted the second answer. If you are interested in legacy versions, just look them up in the history. Also, this page is heavily loaded with $\mathcal{M}^{ath}\mathrm{J_{a}X}$, give it time to load completely . In Rayleigh Schrödinger perturbation theory, we set up the Hamiltonian as $$\mathbf{H}=\mathbf{H_0}+\lambda\mathbf{H'}.\tag1$$ The perturbed Schrödinger equation is therefore $$\mathbf{H}\Psi=W\Psi_\mathrm{ST}.\tag2$$
You can develop this into Taylor series:
\begin{align}
W &=
\lambda^0W_0 + \lambda^1W_1 + \lambda^2W_2 + \cdots
&&= \sum_{i=0}\lambda^i W_i\tag3\\
\Psi_\mathrm{ST} &=
\lambda^0\Psi_0 + \lambda^1\Psi_1 + \lambda^2\Psi_2 + \cdots
&&= \sum_{i=0}\lambda^i \Psi_i\tag4\\
\end{align} Your unperturbed reference system is a complete solution denoted as
$$\mathbf{H}_0\Phi_i = E_0\Phi_i,\tag5$$
where $i\in\mathbb{N}$. You chose your system to be intermediate normalised.
$$\langle\Psi_\mathrm{ST}|\Phi_0\rangle = 1\tag6$$
Since all your solutions of the unperturbed system are orthogonal,
$\langle\Phi_i|\Phi_j\rangle=\delta_{ij},$
the same holds for your perturbed system, hence
$$\langle\Psi_{i\neq0}|\Phi_0\rangle = 0.\tag7$$ Now we can combine $(1)$ through $(4)$ and gather them according to their order.
\begin{align}
\lambda^0&:&
\mathbf{H}_0\Psi_0
&= W_0\Psi_0\tag{8a}\\
\lambda^1&:&
\mathbf{H}_0\Psi_1 + \mathbf{H'}\Psi_0
&= W_0\Psi_1 + W_1\Psi_0\tag{8b}\\
\lambda^2&:&
\mathbf{H}_0\Psi_2 + \mathbf{H'}\Psi_1
&= W_0\Psi_2 + W_1\Psi_1 + W_2\Psi_0\tag{8c}\\
\lambda^n&:&
\mathbf{H}_0\Psi_n + \mathbf{H'}\Psi_{n-1}
&= \sum_{i=0}^n W_i\Psi_{n-i}\tag{8d}\\
\end{align} We do have a complete set of solutions, hence we can express perturbations as linear combinations of these.
\begin{aligned}
\Psi_0 &= \Phi_0 &
\Psi_1 &= \sum_i c_i \Phi_i &
\Psi_2 &=\sum_i d_i \Phi_i &
\dots
\end{aligned} Now we project equations $8$ onto $\Psi_0$. For $\mathrm{8a}$ this is trivial and it leads back to the expectation value of the energy for the unperturbed system. (In the last transformation we use the linear combinations.)
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
&= \langle\Phi_0|W_0|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
&= W_0\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9a}\\
W_0 &= \langle\Phi_0|\mathbf{H}|\Phi_0\rangle = E_0\tag{9a'}
\end{align}
The next step is similar, and everything else will follow from there analogously. For $\mathrm{8b}$ we obtain:
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
&= \langle\Phi_0|W_0|\Psi_1\rangle
+ \langle\Phi_0|W_1|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
&= W_0\cancelto{c_0}{\langle\Phi_0|\Psi_1\rangle} +
W_1\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\\
\cancelto{c_o\cdot E_0}{\sum_i c_i \langle\Phi_0|\mathbf{H}_0|\Phi_i\rangle}
+ \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
&= W_0\cdot c_0 + W_1\tag{9b}\\
W_1 &= \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle\tag{9a'}
\end{align} Similarly we can obtain the coefficients by projecting on some other function than $\Phi_0$. I will cut this a little bit shorter.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle
+ \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
&= \langle\Phi_j|W_0|\Psi_1\rangle +
\langle\Phi_j|W_1|\Psi_0\rangle\\
\cancelto{c_j\cdot E_j}{\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle}
+ \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
&= \cancelto{c_j\cdot E_0}{\langle\Phi_j|W_0|\Psi_1\rangle}
+ \cancelto{0}{\langle\Phi_j|W_1|\Psi_0\rangle}\tag{10b}\\
c_j &= \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{E_0-E_j}\tag{10b'}
\end{align} Now this is impossible to solve for $c_0$, but we still have the intermediate normalisation.
\begin{align}
0 &= \langle\Phi_0|\Psi_1\rangle
= \sum_i c_i \langle\Phi_0|\Phi_i\rangle
= c_0\cancelto{1}{\langle\Phi_0|\Phi_i\rangle} +
\sum_{i\neq0} c_i \cancelto{0}{\langle\Phi_0|\Phi_i\rangle} = c_0\\
\end{align} Similarly this applies to all zeroth coefficients, $c_0 = d_0 = \dots = 0$. Let's get back to business and continue with $\mathrm{8c}$.
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle
+ \langle\Phi_0|\mathbf{H'}|\Psi_1\rangle
&= \langle\Phi_0|W_0|\Psi_2\rangle
+ \langle\Phi_0|W_1|\Psi_1\rangle
+ \langle\Phi_0|W_2|\Psi_0\rangle\\
\scriptsize\cancelto{d_0\cdot E_0}{\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle}
+ \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
&= \scriptsize W_0\cancelto{d_0=0}{\langle\Phi_0|\Psi_2\rangle}
+ W_1\cancelto{c_0=0}{\langle\Phi_0|\Psi_1\rangle}
+ W_2\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9c}\\
W_2 &= \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle\tag{9c'}
\end{align} We do now insert $\mathrm{10b'}$ and get the second order correction:
\begin{align}
W_2 &= \sum_{i\neq0}\frac{
\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle
}{E_0-E_i}\tag{9c''}
\end{align} Now we have the expression for the second order energy, let's do the messy stuff for the second order contribution to the wave function. Skip here for the short version, which is still very long, but that's how we roll for the second order correction to the wave function We start by multiplying $\mathrm{8c}$ with $\Phi_j$ from the left and integrating (projection). I will include a few more steps for this problem, hopefully that will help transferring it to your notation.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
\end{align} I will go through this term by term this time, as it is quite a long equation.
Let's start with the most simplest term, the third one on the right hand side. As you noted, it cancels.
\begin{align}
\langle\Phi_j|W_2|\Psi_0\rangle
&= W_2 \cancelto{0}{\langle\Phi_j|\Psi_0\rangle }\\
&= 0
\end{align} Let's continue on the right hand side:
\begin{align}
\langle\Phi_j|W_1|\Psi_1\rangle
&= W_1 \langle\Phi_j|\Psi_1\rangle\\
&= W_1 \sum_i c_i \langle\Phi_j|\Phi_i\rangle\\
&= W_1 \left(c_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle} +
\sum_{i\neq j} c_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
&= c_j W_1
\end{align} And the same for the last contribution:
\begin{align}
\langle\Phi_j|W_0|\Psi_2\rangle
&= W_0 \langle\Phi_j|\Psi_2\rangle\\
&= W_0 \sum_i d_i \langle\Phi_j|\Phi_i\rangle\\
&= W_0 \left(d_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
+ \sum_{i\neq j} d_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
&= d_j E_0 &\mathrm{9a':}&(W_0 = E_0)
\end{align} Let's go to the left side:
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle
&= \sum_i d_i \langle\Phi_j|\mathbf{H}_0|\Phi_i\rangle \\
&= \sum_i d_i E_i \langle\Phi_j|\Phi_i\rangle\\
&= d_j E_j \cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
+ \sum_{i\neq j} d_i E_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\\
&= d_j E_j
\end{align} And the last one can barely be simplified:
\begin{align}
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle
&= \sum_i c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
&= \cancelto{0}{c_0 \langle\Phi_0|\mathbf{H'}|\Phi_0\rangle}
+ \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
\end{align} Now, Avengers reassemble, rearrange, and insert the expression for $W_1$: \begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
d_j E_j + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle
&= c_j W_1 + d_j E_0\\
d_j (E_j - E_0)
&= c_j \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
- \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
d_j &=
c_j \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
- \sum_{i\neq 0} c_i \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}
\end{align} Now introduce the first order base, $c_i$ and $c_j$ from $\mathrm{10b'}$, there will be some minus shuffling as well.
\begin{align}
d_j
&=
\frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)}
\frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
- \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
\frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}\\
&=
\frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)} \cdot
\frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(-1)(E_0 - E_j)}
- \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
\frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(-1)(E_0 - E_j)}\\
d_j &=
\sum_{i\neq 0}
\frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle%
\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle%
}{(E_0 - E_j)(E_0 - E_i)}
-\frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle%
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)^2} \tag{10c'}
\end{align} Well, I am officially exhausted right now. And lastly, a small remark on your notation (I keep this paragraph for nostalgic reasons). It took me quite some time understanding it, and as we saw (in some earlier version of this answer), I was mistaken. While I do not mind the hat notation for operators, indicating order on top of the quantities is problematic in my opinion. To me this place is reserved for powers. I suggest you change it to one of the following: $E_{n,0},\ {}^{(0)}\!E_n,\ E^{(0)}_n$. When you have a look at the other answer, you will notice, I have decided to stick to the last option. But as I do not know which book your are using it is also not easy to recommend. I believe there are as many different notations as there are books out there. I personally preferred the book "Introduction to Computational Chemistry, 2nd Edition"by Frank Jensen . As we settled in the comments, I appreciate that you stick to Dirac's original notation. And also, notation is like Marvel and DC, you pick one and stick with it, you can still appreciate the other, but just not as much. Or a football (Americans pronounce this as "soccer") club, or ... As long as you understand it and can explain it to someone else, you are fine. And now we all deserve some chocolate and some coffee and cake. Thank you for reading this post until the end. | {
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28,399 | Standard Gibbs free energy of formation of liquid water at $\pu{298 K}$ is $\pu{−237.17 kJ mol-1}$ and that of water vapour is $\pu{−228.57 kJ mol-1}$ therefore, $$\ce{H2O (l) -> H2O (g)}\qquad\Delta G =\pu{8.43{kJ mol-1}}$$ Since $\Delta G>0$ , it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference. Why does this happen? | Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, $$\ce{H2O(l)->H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$ Since $\Delta G>0$ , it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference. Your math is correct but you left out a very important symbol from your equations. There is a big difference between $\Delta G$ and $\Delta G^\circ$ . Only $\Delta G^\circ$ means the Gibbs energy change under standard conditions, and as you noted in the question, the free energy values you quoted are the standard gibbs free energy of water and water vapor. Whether or not something is spontaneous under standard conditions is determined by $\Delta G^\circ$ . Whether something is spontaneous under other conditions is determined by $\Delta G$ . To find $\Delta G$ for real conditions, we need to know how they differ from standard conditions. Usually "standard" conditions for gases correspond to one bar of partial pressure for that gas. But the partial pressure of water in our atmosphere is usually much lower than this. Assuming water vapor is an ideal gas, then the free energy change as a function of partial pressure is given by $G = G^\circ + RT \ln{\frac{p}{p^\circ}}$ . If the atmosphere were perfectly 100% dry, then the water vapor partial pressure would be 0, so $\ln{\frac{p}{p^\circ}}$ would be negative infinity. That would translate to an infinitely negative -- i.e. highly spontaneous -- $\Delta G$ for the water evaporation reaction. Small but not-quite zero dryness in the atmosphere would still lead to the $\Delta G$ of water vapor that is more negative than liquid water. So water evaporation is still spontaneous. Extra credit: given the standard formation energies you found, and assuming water is an ideal gas, you could calculate the partial pressure of water vapor at which $\Delta G = 0$ for water evaporation. And the answer had better be the vapor pressure of water, or else there is a thermodynamic inconsistency in your data set! | {
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28,503 | So I came across this diagram,the other day: As a chemistry student, I am well-versed with the dashed and wedged lines, but I was wondering what the wavy and the dotted line represent? | I was wondering what the wavy and the dotted line represent? A dashed line indicates that the bond is extending behind the plane of the drawing surface A bold-wedged line indicates that the bond is protruding out from the plane of the drawing surface A solid line indicates that the bond exists in the plane of the drawing surface. A wavy line indicates that the stereochemistry of the bond is unknown . A dotted line indicates that the bond is not a full bond, it is only a partial bond as in a hydrogen bond or a partially formed or broken bond in a transition state. Also, as Mrigank Pawagi notes in the comments, the dotted line can be used along with the solid line to denote bonds that have partial double bond character due to resonance. An arrow indicates a dative bond in which both electrons in the bond originate from one atom (not common) | {
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28,568 | Caesium has a larger size, and the effective nuclear charge that the valence electron experiences will be far less compared to that of lithium's, right? But lithium is still considered the strongest reducing agent among all the alkali metals, and this is evidenced by its large and negative reduction potential. Why is this so? | The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated. $$\begin{array}{cc}
\hline
\ce{M} & E^\circ(\ce{M+}/\ce{M}) \\
\hline
\ce{Li} & -3.045 \\
\ce{Na} & -2.714 \\
\ce{K} & -2.925 \\
\ce{Rb} & -2.925 \\
\ce{Cs} & -2.923 \\
\hline
\end{array}$$ Source: Chemistry of the Elements 2nd ed., Greenwood & Earnshaw, p 75 However, a full description of the middle three metals is beyond the scope of this question. I just thought it was worth pointing out that the trend is not really straightforward. The $\ce{M+}/\ce{M}$ standard reduction potential is related to $\Delta_\mathrm{r}G^\circ$ for the reaction $$\ce{M(s) -> M+(aq) + e-}$$ by the equation $$E^\circ = \frac{\Delta_\mathrm{r}G^\circ + K}{F}$$ where $K$ is the absolute standard Gibbs free energy for the reaction $$\ce{H+ + e- -> 1/2 H2}$$ and is a constant (which means we do not need to care about its actual value). Assuming that $\Delta_\mathrm{r} S^\circ$ is approximately independent of the identity of the metal $\ce{M}$, then the variations in $\Delta_\mathrm{r}H^\circ$ will determine the variations in $\Delta_\mathrm{r}G^\circ$ and hence $E^\circ$. We can construct an energy cycle to assess how $\Delta_\mathrm{r}H^\circ$ will vary with the identity of $\ce{M}$. The standard state symbol will be dropped from now on. $$\require{AMScd}
\begin{CD}
\ce{M (s)} @>{\large \Delta_\mathrm{r}H}>> \ce{M+(aq) + e-} \\
@V{\large\Delta_\mathrm{atom}H(\ce{M})}VV @AA{\large\Delta_\mathrm{hyd}H(\ce{M+})}A \\
\ce{M (g)} @>>{\large IE_1(\ce{M})}> \ce{M+ (g) + e-}
\end{CD}$$ We can see, as described in Prajjawal's answer, that there are three factors that contribute to $\Delta_\mathrm{r}H$: $$\Delta_\mathrm{r}H = \Delta_\mathrm{atom}H + IE_1 + \Delta_\mathrm{hyd}H$$ (the atomisation enthalpy being the same as the sublimation enthalpy). You are right in saying that there is a decrease in $IE_1$ going from $\ce{Li}$ to $\ce{Cs}$. If taken alone , this would mean that $E(\ce{M+}/\ce{M})$ would decrease going from $\ce{Li}$ to $\ce{Cs}$, which would mean that $\ce{Cs}$ is a better reducing agent than $\ce{Li}$. However, looking at the very first table, this is clearly not true. So, some numbers will be needed. All values are in $\mathrm{kJ~mol^{-1}}$. $$\begin{array}{ccccc}
\hline
\ce{M} & \Delta_\mathrm{atom}H & IE_1 & \Delta_\mathrm{hyd}H & \text{Sum} \\
\hline
\ce{Li} & 161 & 520 & \mathbf{-520} & 161 \\
\ce{Cs} & 79 & 376 & \mathbf{-264} & 211 \\
\hline
\end{array}$$ Source: Inorganic Chemistry 6th ed., Shriver et al., p 160 This is, in fact, an extremely crude analysis. However, it hopefully does serve to show in a more quantitative way why $E(\ce{Cs+}/\ce{Cs}) > E(\ce{Li+}/\ce{Li})$: it's because of the extremely exothermic hydration enthalpy of the small $\ce{Li+}$ ion. Just as a comparison, the ionic radii of $\ce{Li+}$ and $\ce{Cs+}$ ($\mathrm{CN} = 6$) are $76$ and $167~\mathrm{pm}$ respectively (Greenwood & Earnshaw, p 75). | {
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28,702 | We know that there are silicon analogs of alkanes, but according to Wikipedia, silicon analogs of alkenes or alkynes are virtually unknown. How come? | The typical Si–Si single bond length in a silane is around 2.33 Å. This is much longer than a typical C–C single bond (~1.53 Å) and helps explain why silicon-silicon single bonds are so much weaker than carbon-carbon single bonds (bond dissociation energy: ~53 kcal/mol for silicon-silicon vs. ~83 kcal/mol for carbon-carbon single bonds). When we try to form a silicon-silicon double or triple bond, the large separation between the two silicon atoms results in even less effective p-orbital overlap and therefore even weaker π-bonds . The result is that while disilenes and disilynes are known (see the linked SE Chem posts), they are extremely reactive. Only when bulky substituents are placed around the double or triple bonds (to provide some "steric protection" against further reaction) can the molecules be isolated and characterized. | {
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29,196 | Is lead (Pb) radioactive? I have heard about it being radioactive but have also heard it referred to as the heaviest stable element. Radioactivity refers to release of alpha, beta, gamma rays. | Important naturally-occurring radioactive lead nuclides are: $\ce{^{214}Pb}$ $\left(t_{1/2}=26.8\ \mathrm{min}\right)$ from $\ce{^{238}U}$ $\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$ from $\ce{^{238}U}$ $\ce{^{211}Pb}$ $\left(t_{1/2}=36.1\ \mathrm{min}\right)$ from $\ce{^{235}U}$ $\ce{^{212}Pb}$ $\left(t_{1/2}=10.64\ \mathrm{h}\right)$ from $\ce{^{232}Th}$ (Note that 'a' here is the unit symbol for 'year'.) On a geological time scale, their half-lives are very short. Nevertheless, these lead nuclides are members of the $\ce{^{238}U}$ , $\ce{^{235}U}$ , and $\ce{^{232}Th}$ decay series. Thus, they are constantly reproduced by decay of their respective mother nuclides. Therefore, all environmental samples (including soil, water, air, plants, and animals) naturally contain significant amounts of radioactive lead nuclides. However, in various materials (e.g. residues from mining or oil and gas production, fertilizers, building materials), the concentration of natural lead nuclides can be strongly enhanced by technology. When lead is freshly refined (chemically purified), most of the mother nuclides are removed and the short-lived lead nuclides quickly decay. Nevertheless, the lead still contains significant amounts of $\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$ . The presence of $\ce{^{210}Pb}$ is a source of background mainly through the lead X-rays and bremsstrahlung caused by high-energy beta radiation emitted by its decay product $\ce{^{210}Bi}$ . This background is a real problem when lead is used as a shielding material for low-background detectors. Therefore, some shields are made from selected lead with certified low $\ce{^{210}Pb}$ content. Preferably, very old lead is used. Since $\ce{^{210}Pb}$ decays with a half-life of 22.3 years, samples of lead that are many decades old are relatively free of this activity. In one extreme example, 2000-year-old lead was salvaged from a sunken Roman ship. | {
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29,657 | There was a question in one of my exam papers to draw the resonance structures for $\ce{N2O}$. These are the ones I drew, but they were not awarded marks: The ones in the marking scheme are as follows: Can anyone please explain why mine aren't correct? | Firstly, neither of the resonance structures that you drew for your test are possible because they both violate the octet rule. For the structure on the left, the leftmost $\ce{N}$ is in control of only 6 electrons, and will not exist in this form. The structure on the right is not possible because central $\ce{N}$ is participating in 5 bonds, which $\ce{N}$ cannot do. The maximum number of bonds that $\ce{N}$ can participate in is 3 (covalent bonding - like $\ce{NH3}$) or 4 (coordinate bonding - like $\ce{NH4+}$). As long as you satisfy the octet rule (or at least as best as you can - here are some exceptions ) and you wish to check the stability of a molecule that you have drawn, simply use formula for formal charge. This is given by the equation: $$\mathrm{FC} = \mathrm{V} - (\mathrm{N_B} + \frac{\mathrm{B}}{2})$$ Wher $\mathrm{FC}$ is the formal charge, $\mathrm{V}$ is the number of valence electrons that the atom under consideration usually has, $\mathrm{N_B}$ is the number of non-bonded electrons, and $\mathrm{B}$ is the number of electrons shared in a covalent bond on the atom. If a molecule is stable, the sum of the formal charges of each substituent atom should be 0. If you were to test this on each of the resonance structures that the test provides as answers, you would see that this holds true. | {
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30,754 | What would be the effect if someone were to drink ultra-pure water with an electrical resistivity of $18 \, \mathrm{M} \Omega \! \cdot \! \text{cm}$? Would they immediately die? Would they just need to pee more? Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract? What would be the threshold for bad things to happen — a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)? I'm not sure what the science here would be. | What would be the effect if someone were to drink ultra-pure, 18 M-ohm water? Not much, although if they drank many gallons of the water it could be a problem. Would they immediately die? No. Would they just need to pee more? Probably. Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract? It would become orders of magnitude less pure as soon as it mixed with their saliva and then their gastric juices. Same as regular water. What would be the threshold for bad things to happen -- a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)? I don't think the effect would be that different from drinking regular water. All over the internet there is this idea that super pure water is somehow too pure or dangerous, no good evidence or mechanism is ever presented. Water is toxic primarily because it dilutes the sodium and potassium ions in your body, leading to Hyponatremia or possibly the far more dangerous Hypokalemia . The normal level of $\ce{Na+}$ in blood plasma is about 135 millimolar (mM), and the normal level of $\ce{K+}$ is about 5 mM. A very rough approximation to understanding how toxic water can be would then be to compare the concentration difference of sodium (or potassium) between the ingested water and the blood. A 25 micromolar -- i.e. 25 μM or 0.000025 molar solution of sodium hydroxide has a conductivity of about 6.7 microsiemens per cm, more than 100-fold higher than "pure" 18.2 megohm water (0.055 microsiemens per cm). According to this rough difference-based model, the ability of the very dilute sodium ion solution to cause problems is proportional to (135 - 0.025) mM, and the ability of the "pure" water to cause problems is (135 - 0) mM. For nearly biological or medical conditions of interest, 135 mM is not really different from 134.975 mM, and so ultra-pure water is not really more dangerous than everyday drinking water. | {
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30,893 | This may seem silly, but doesn't it seem weird for a compound that's stable (in this context, the tertiary carbocation) to be the most reactive? I mean, wouldn't it be the least, given that it's already stable and wouldn't want to leave that stable configuration. On the same lines, the methyl carbocation sounds like the most reactive due to its high instability, right? wouldn't it want to react in order to form a stable compound? | This is an excellent question. Please correct me if I'm wrong, but I think this is what you are grasping at: First, it is true that tertiary carbocations are generally more stable than primary carbocations (and secondary carbocations) due to having more inductively donating alkyl groups. The hyperconjugative effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations. Second, transition states involving tertiary carbocations as opposed to primary carbocations are more favorable precisely because tertiary carbocations are more stable than primary carbocations. Note that this does not mean that tertiary carbocations are more reactive. We don't generally say that something which exists in a transition state is reactive/unreactive. One reason is that the transition state complex cannot be directly captured or observed. Another reason is that it is the transition state ... of course it's going to be reactive - of course it'll change very rapidly! We may however say that reactants and products are either reactive or unreactive. So if something goes through a transition state involving a tertiary carbocation, it might well be more reactive than something which goes through a transition state involving a primary carbocation. Of course, before labeling anything as reactive or unreactive, you'd do well to note exactly what you mean by reactivity. To put it lightly, there are myriad reactions. An alkyne such as acetylene is reactive with regard to combustion; acetylene torches are commonplace. However, acetylene isn't going to be reactive with regard to, say, backside attack (for a variety of reasons). Ultimately I think your issue lies with semantics rather than the chemistry ... still, a good question! | {
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31,299 | I was holding a piece of plastic earlier, and I ripped it in half. This caused me to start to think about what was happening at the atomic level. If the plastic is being ripped, clearly bonds are being broken and new bonds are being formed, so is this not a chemical reaction? I realize that the plastic retains all of its chemical characteristics despite it being ripped, so I am quite confused about how to classify what is occurring. | When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material. The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align. To some degree this partial alignment makes continued deformation easier. To continue tearing the polymer apart you just need to overcome the van der Waals forces holding the chains together. In addition to the polymer chains being pulled apart, some chemical bonds will be broken. I don't know how many bonds are broken per chain pulled apart, but I suspect it is low. Low or not, some free radicals (or ions) will be generated when the chemical bonds break. If the experiment is performed under ambient conditions, the radicals or ions will quickly react with oxygen or ions in the air. If the experiment is performed in an inert atmosphere, then the radicals will persist ( dangling bonds ) and can be observed in an esr spectrometer . It has been demonstrated that cross-linked polymers, which are less prone to flow deformation and therefore more likely to undergo bond breaking, do produce larger esr signals. So most of what is going on is molecular deformation and pulling apart of the polymer chains - but some bond breaking does occur. It would be an interesting experiment to cut two pieces of plastic and then cut one of them into finer shreds. Place the two samples into separate tubes and then record their esr signals. How much more intense would the signal be in the finely cut sample compared to the bulk sample? Polycarbonate-A, straight out of the reagent bottle, produces an esr signal. | {
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31,375 | Suppose I have two units, Tonks, $T$, and Borks, $B$, as well as the equation $\frac{50T}{20B}=x$. May I manipulate the equation as though $T$ and $B$ were the kinds of symbols that we learned to manipulate in math class? For example: \begin{align}
\frac{50T}{20B}&=x\\
50T&=20Bx\\
\frac{50T}{20}&=Bx\\
\end{align} | Yes, you may. It is quite common to convert units into each other. The simplest conversion might be the prefixing of units, e.g.
$$\mathrm{\frac{km}{m}}=1000 \Longleftrightarrow \mathrm{1~km = 1000~m}.$$ Another example is the interconversion of units of energy. In some parts of chemistry, it is still quite common to use calories, in others Joule, as a SI unit, has taken its place.
$$\mathrm{\frac{cal}{J}}=4.184 \Longleftrightarrow \mathrm{1~cal = 4.184~J}.$$ In principle, all units are defined to be constants, special numbers in a wider sense. You can treat them as such. | {
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31,531 | Does phenol undergo Friedel–Crafts reactions or does it react with Lewis acids like aniline does? | Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction. The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid. In fact most substituents with lone pair would give poor yield. The two pathways involved in the reaction with phenol reduce the overall yeild: Phenols are examples of bidentate nucleophiles, meaning that they can react at two positions: on the aromatic ring giving an aryl ketone via C-acylation, a Friedel-Crafts reaction or, on the phenolic oxygen giving an ester via O-acylation, an esterification | {
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31,618 | Cyclohexenone, an α,β-unsaturated ketone, is electrophilic at both the carbonyl carbon, as well as the β-carbon. So, it can either undergo 1,2-addition to give the allylic alcohol 1 , or 1,4-addition to give the ketone 2 . Where does the Grignard reagent add to? | The correct answer is the 1,2-addition product (i.e. the allylic alcohol 1 ). In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone. This exact transformation was reported by Akai and coworkers recently ( Org. Lett. 2010, 12, 4900 ) and they obtained compound 1 in 95% yield (the supplementary information (PDF) is free). | {
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31,639 | I know how to calculate them and such stuff, but I wanted to know what they actually signify. I have a vague idea that they have something to do with an electron's position in an atom but what do all of them mean? Any help would be greatly appreciated! | Quantum numbers give information about the location of an electron or set of electrons. A full set of quantum numbers describes a unique electron for a particular atom. Think about it as the mailing address to your house. It allows one to pinpoint your exact location out of a set of $n$ locations you could possibly be in. We can narrow the scope of this analogy even further. Consider your daily routine. You may begin your day at your home address but if you have an office job, you can be found at a different address during the work week. Therefore we could say that you can be found in either of these locations depending on the time of day. The same goes for electrons. Electrons reside in atomic orbitals (which are very well defined 'locations'). When an atom is in the ground state, these electrons will reside in the lowest energy orbitals possible (e.g. 1$s^2$ 2$s^2$ and 2$p^2$ for carbon). We can write out the physical 'address' of these electrons in a ground-state configuration using quantum numbers as well as the location(s) of these electrons when in some non-ground (i.e. excited) state. You could describe your home location any number of ways (GPS coordinates, qualitatively describing your surroundings, etc.) but we've adapted to a particular formalism in how we describe it (at least in the case of mailing addresses). The quantum numbers have been laid out in the same way. We could communicate with each other that an electron is "located in the lowest energy, spherical atomic orbital" but it is much easier to say a spin-up electron in the 1$s$ orbital instead. The four quantum numbers allows us to communicate this information numerically without any need for a wordy description. Of course carbon is not always going to be in the ground state. Given a wavelength of light for example, one can excite carbon in any number of ways. Where will the electron(s) go? Regardless of what wavelength of light we use, we know that we can describe the final location(s) using the four quantum numbers. You can do this by writing out all the possible permutations of the four quantum numbers. Of course, with a little more effort, you could predict the exact location where the electron goes but in my example above, you know for a fact you could describe it using the quantum number formalism. The quantum numbers also come with a set of restrictions which inherently gives you useful information about where electrons will NOT be. For instance, you could never have the following possible quantum numbers for an atom: $n$=1; $l$=0; $m_l$=0; $m_s$=1/2 $n$=1; $l$=0; $m_l$=0; $m_s$=-1/2 $n$=1; $l$=0; $m_l$=0; $m_s$=1/2 This set of quantum numbers indicates that three electrons reside in the 1$s$ orbital which is impossible! As Jan stated in his post, these quantum numbers are derived from the solutions to the Schrodinger equation for the hydrogen atom (or a 1-e$^-$ system). There are any number of solutions to this equation that relate to the possible energy levels of they hydrogen atom. Remember, energy is QUANTIZED (as postulated by Max Planck). That means that an energy level may exist (arbitrarily) at 0 and 1 but NEVER in between. There is a discrete 'jump' in energy levels and not some gradient between them. From these solutions a formalism was constructed to communicate the solutions in a very easy, numerical way just as mailing addresses are purposefully formatted in such a way that is easy that anyone can understand with minimal effort. In summary, the quantum numbers not only tell you where electrons will be (ground state) and can be (excited state), but also will tell you where electrons cannot be in an atom (due to the restrictions for each quantum number). Principle quantum number ($n$) - indicates the orbital size. Electrons in atoms reside in atomic orbitals. These are referred to as $s,p,d,f...$ type orbitals. A $1s$ orbital is smaller than a $2s$ orbital. A $2p$ orbital is smaller than a $3p$ orbital. This is because orbitals with a larger $n$ value are getting larger due to the fact that they are further away from the nucleus. The principle quantum number is an integer value where $n$ = 1,2,3... . Angular quantum number ($l$) - indicates the shape of the orbital. Each type of orbital ($s,p,d,f..$) has a characteristic shape associated with it. $s$-type orbitals are spherical while $p$-type orbitals have 'dumbbell' orientations. The orbitals described by $l$=0,1,2,3... are $s,p,d,f...$ orbitals, respectively. The angular quantum number ranges from 0 to $n$-1. Therefore, if $n$ = 3, then the possible values of $l$ are 0, 1, 2. Magnetic quantum number ($m_l$) - indicates the orientation of a particular orbital in space. Consider the $p$ orbitals. This is a set of orbitals consisting of three $p$-orbitals that have a unique orientation in space. In Cartesian space, each orbital would like along an axis (x, y, or z) and would be centered around the origin at 0,0. While each orbital is indeed a $p$-orbital, we can describe each orbital uniquely by assigning this third quantum number to indicate its position in space. Therefore, for a set of $p$-orbitals, there would be three $m_l$, each uniquely describing one of these orbitals. The magnetic quantum number can have values of $-l$ to $l$. Therefore, in our example above (where $l$ = 0,1,2) then $m_l$ would be -2, -1, 0, 1, 2. Spin quantum number ($m_s$) - indicates the 'spin' of the electron residing in some atomic orbital. Thus far we have introduced three quantum numbers that localize a position to an orbital of a particular size, shape and orientation. We now introduce the fourth quantum number that describes the type of electron that can be in that orbital. Recall that two electrons can reside inside one atomic orbital. We can define each one uniquely by indicating the electron's spin. According to the Pauli-exclusion principle, no two electrons can have the exact same four quantum numbers. This means that two electrons in one atomic orbital cannot have the same 'spin'. We generally denote 'spin-up' as $m_s$ =1/2 and spin-down as $m_s$=-1/2. | {
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32,008 | When I start studying electrochemistry, I learn the words “molten” and “aqueous”. I don't have a problem for “aqueous”, but I'm a little bit confused about “molten”. For me, “molten” means melt, which means to become liquid from solid. But why do we say molten lead(II) bromide instead of liquid lead(II) bromide? I mean both are the same, but liquid is arguably easier to understand. | As you said, the meaning is exactly the same. Molten reduces the ambiguity, because you emphasize that you know, that it is solid substance at laboratory conditions and you heat it to become liquid (while staying pure substance). As an example, the "electrolysis of liquid sodium chloride" is in principle enough to tell you all about the process, but as we often like to dissolve the salt in water, you might be slightly unsure, whether the meaning is 100% clear. Therefore "electrolysis of molten sodium chloride" is better choice. | {
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32,158 | The product of $2Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively few protons. Whereas, the product of $2.5Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively many protons. The product of $((2+2.5)/2)Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise a moderate number of protons. What function from the $Z$ values produces a value that maximally approximates the $m_\mathrm{a}$ that corresponds to $Z$? | The short answer is that you can find a power-law fit ($1.61Z^{1.1}$) with low average error. I'd never really thought about it much, but after downloading the IUPAC Atomic Weights , I decided to do some curve fitting. Here's a linear fit between atomic number and atomic mass: As you say, the fit isn't very good for small $Z$, but the overall fit isn't bad - the mean absolute error (MAE) is $2.821 \:\mathrm{u}$, and taken as a whole, the data is surprisingly linear. (Well, surprising to me.) So I thought of a quadratic fit, requiring the intercept to be 0,0 to ensure the best fit for small $Z$: Looks better, right? Certainly the fit is much better for first and second row elements, but the MAE only reduces to $2.749 \:\mathrm{u}$. So I went up to a cubic fit, again requiring 0,0 for the intercepts: Aha, now we're talking! We get the subtle nonlinearities, and the MAE is down to $1.36 \:\mathrm{u}$. Thanks to the comment by Nicolau below, I performed a power-law fit. So that gives a power-law fit with MAE of $0.01 \:\mathrm{u}$ and a fairly easy-to remember function: $m_a \approx 1.61Z^{1.1}$ | {
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32,354 | The wikipedia article on trinitromethane claims that: There is some evidence that the anion (which obeys the $4n+2$ Hückel rule) is aromatic. Unfortunately the citation is behind a paywall so I can't read it but my first obvious question is how can the trinitromethane anion obey Hückel's rule? Two of the requirements for the rule are that the molecule must be planar (or nearly planar) and must be cyclic and I fail to see how either of these conditions are satisfied since the anion is roughly trigonal pyramidal around the carbon and definitely not cyclic. This appears to about Y-aromaticity. All the papers linked are behind paywalls for me so can someone please explain what Y-aromaticity is and how it applies to the example given. I found this previous answer which gives some discussion of Y-aromaticity in nitrate ions. If someone could provide a similar MO analysis of the trinitromethane anion this might be very interesting. | Introduction It is fairly obvious that the statement given by Wikipedia cannot be entirely correct. There is some evidence that the anion (which obeys the 4 n +2 Hückel rule) is aromatic. The trinitromethane anion cannot obey the Hückel (4 n +2) rule because it is neither monocyclic nor planar. And assigning a number of π electrons is fuzzy at best, so the (4 n +2) part is also questionable. Hückel's rule is a very strict set, but an aromatic compound does not necessarily have to fulfil it. The easiest compound to understand this is naphtalene , which is clearly aromatic, but does not fulfil the criteria of Hückel's rule, i.e. it is not monocyclic. In extension to this, helicene is an aromatic compound that fails two of the given conditions. Another case is thiophene, which due to the included sulfur certainly does not entirely consist of trigonally hybridised (sp 2 ) atoms (see Bent's rule). [1] But can the anion still be regarded as aromatic? In order to explain that, we need to look at the concept of Y-aromaticity. What is Y-aromaticity? The term and the concept is not without controversial discussion. It is an attempt to give the "traditional" concept of resonance energy an new, fancy and more catchy name. It is often used to explain the extraordinary stability of the guanidinium cation, which has tree equivalent resonance structures, i.e. the electronic ground state is heavily delocalised. Conventional Lewis structures employ two electron two centre bonds only, which therefore cannot represent the highly delocalised electronic structures, e.g. of aromatic compounds. Hence the concept of resonance had to be introduced to account for such stabilising effects. Obviously the same holds true for any compounds, that exhibit excessively delocalised electronic structures. The term Y-aromaticity has been used for compounds, that are Y-shaped, with a trigonal planar coordinated (carbon) centre. In addition to that the π orbitals hold six electrons. [2] These compounds have a similar orbital picture and could therefore be grouped. The electronic structure of Y-delocalised π systems The π orbitals of a trigonal planar coordinated centre can be generalised by the following orbital picture, which was generated for the nitrate anion at the DF-BP86/def2-SVP level of theory. What we can clearly see is, that there is one totally symmetric bonding orbital. Two degenerate (orthogonal) slightly bonding (or non-bonding) orbitals and another totally symmetric anti-bonding orbital. The remarkable stabilisation comes from the lowest lying π orbital, which is delocalised over all atoms, just like in the more obvious cases like benzene. Additionally, such compounds often retain their planarity by undergoing substitution reactions, rather than addition reactions. Pro and contra of Y-aromaticity The concept itself allows the discussion of certain electronic properties in quite simple terms. It is related enough to the general concept of aromaticity so that it can easily be understood at the same level of teaching. Like aromaticity, it allows predictions - to a certain degree - about the reactivity of such compounds. Peter Grund makes the following assessment, arguing in favour of the concept of acyclic aromaticity. [3] Despite recent trends, it would appear useful to designate systems as "aromatic" if they represent a delocalized system exhibiting peculiar chemical stability, in order to focus attention on such systems even when they are buried in complex structures. Since guanidine and its derivatives appear to possess really exceptional stability, and since their physical and chemical properties appear to be dominated by the tendency to retain the closed-shell, Y-delocalized 6 pi-electron configuration, we suggest that it is profitable to consider such substances as possessing a special type of "aromatic character," despite their acyclic nature. For a general understanding of this feature it certainly does not hurt to keep it in the back of ones head. [4] However, the applicability of this concept is limited to a quite small subset of compound (the same holds true for Hückel's rule, too). It therefore seems to be quite a superfluous concept, since it is in a more general sense already covered by resonance stabilisation. After all, electronic structures are often not well enough described with simple models and in some cases these simple models can lead to wrong assumptions. This is one of these cases, where one can only use this concept as long as its restrictions are well known. Some people might argue, that the naming itself is the problem and it may lead to confusion. I can partly agree with that. That does not change the fact, that the description in resonance, valence bond and molecular orbital theory terms will produce this result. In general I think the term aromaticity is unfortunate right in the beginnings. The concept has changed and extended quite a few times already. I often see problems when the term "Hückel rule" is used synonymously with aromaticity. There is no true theoretical and at the same time all general explanation for aromaticity. It is often determined on a case by case basis experimentally. This limitation also has to apply to any derivations of it. Electronic structure of the trinitromethane anion (TNM ⊖ ) Trinitromethane is a quite acidic substance, which can in part be attributed to the presence of the well delocalised π system. Cioslowski et. al. have investigated the anion and came to this conclusion. In their paper they also use the term Y-aromaticity. [5] Our calculations, however, show that at the HF/6-31G* level, the carbon atom is coplanar with the nitrogens while the oxygens of the nitro groups are rotated out of plane. The result is a structure with $D_3$ symmetry. The $\ce{C(CN)3^-}$ anion is likewise found to be planar, while the $\ce{CF3^-}$ ion exhibits a
pyramidal geometry. In addition, we observe in $\ce{C(CN)3^-}$ a shortening of the carbon-carbon bond and a lengthening of the carbon-nitrogen bond relative to $\ce{CH(CN)3}$. We also find an analogous shortening of the carbon-nitrogen bond and lengthening of the nitrogen-oxygen bond in $\ce{C(NO2)3^-}$ relative to $\ce{CH(NO2)3}$. In contrast, the $\ce{C-F}$ bond length increases upon going from $\ce{CHF3}$ to $\ce{CF3^-}$. The planarity of the $\ce{C(NO2)3^-}$ and $\ce{C(CN)3^-}$ ions and the changes in bond lengths are obvious consequences of Y-aromaticity in these systems. Obviously the bonding situation in the trinitromethane anion is a little more complicated than in guanidine, as the nitro group itself cannot be described by a single Lewis structure. The out of plane rotation of these groups further complicate things, but since the perpendicular $C_2$ axis is retained it is still somewhat possible to compare the resulting orbitals to a planar species. Therefore when you look at the orbitals you still find a remarkable resemblance to the picture above. For this purpose I have included the same schematic drawings next to the orbitals that resemble these best. It is noteworthy, that because of the propeller structure there is no strict separation between σ and π orbitals, since the latter require a horizontal mirror plane, which is absent here. But that is not all. Try counting the number of π electrons. This is far from trivial and it is in no case that makes sense six. [6] Don't let your brain explode about that fact, we are clearly dealing with a borderline case at this point. And we have not even started to talk about the possibility of bonds between the oxygens of the nitro group. [7] Take home message for TNM ⊖ The trinitromethane anion has a strongly delocalised electronic system. It is extraordinarily stable due to this. If you accept the concept of Y-aromaticity as valid, then it is still a borderline case, where additional considerations have to be taken into account. Notes, References and further reading Bent's Rule Bent, H. A. Chem. Rev. 1961, 61 (3), 275–311. DOI: 10.1021/cr60211a005 . What is Bent's rule? Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot? The six π electrons are probably the reason for the statement in the Wikipedia article. Gund, P. J. Chem. Educ. 1972, 49 (2), 100. DOI: 10.1021/ed049p100 . I have used this concept already various times to explain certain electronic structures. From a personal point of view I find it quite helpful. Delocalization of pi electrons in nitrate ion Why is the bond order in the SO₃ molecule 1.33 and not 2? Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻? Cioslowski, J.; Mixon, S. T.; Fleischmann, E. D. J. Am. Chem. Soc. 1991, 113 (13), 4751–4755. DOI: 10.1021/ja00013a007 . There are several possibilities, that can be considered to count electrons towards the number of π electrons. The probably easiest way would be to flatten the molecule for this thought experiment, then count electrons perpendicular to the plane. This gives you two electrons for the central carbon (sp 2 ), no electrons from the nitrogens (sp 2 ) and two electrons each from the oxygens (sp). That gives you a total of 14 π electrons. Maybe as simple as this, but somewhat the easy way out would be to disregard all contributions from the nitro groups, leaving only 2 π electrons. Regard the nitro groups as π-backdonor. Since only one of the intrinsic π orbitals can effectively overlap with the central carbon, each of them would donate two electrons (maybe less, because back donation). Plus two from the central carbon, makes up to eight electrons. This certainly is a very fuzzy approach. Count four electrons per oxygen, which is kind of cheeky, but it could work. This would give 26 π electrons. Write a Lewis structure and count only the double bonds (× 2), which gives a total of eight electrons. I am almost certain, there are still more ways to count the electrons. In [5] they analysed the anion with the quantum theory of atoms in molecules. They found bond critical points between the oxygen atoms. How these can be explained and what they mean is certainly a story for another time. | {
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32,443 | I found different values of Avogadro constant in different places. So what is the correct value? $\pu{6.0221367*10^{23}}$ or $\pu{6.02214129*10^{23}}$ or $\pu{6.0221415*10^{23}}$ or anything else? | Whenever you're looking for accurate fundamental physical constants, CODATA recommended values are the way to go. As of May 20, 2019, Avogadro's constant is now truly an exact value, with infinite significant digits. Behold! $$6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$ It is unlikely this value will change within our lifetimes, which is exciting in its own way! For historical reasons, I'll leave my previous answer below: As of 2015, the latest data for the Avogadro constant is from 2014 . According to CODATA, the most accurate value is: $$6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,074\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2014)}$$ The relative uncertainty in the measurement is thus only 12 parts per billion! Interestingly, the Avogadro constant may be redefined in the near future to be an exact value, that is, a constant with zero uncertainty by definition, much like the speed of light . This would come as a consequence of redefining the SI kilogram as a function of the number of atoms in an ultrapure monoisotopic $\ce{^{28}Si}$ monocrystalline sphere engineered to extreme precision. A great video on this can be found in the Veritasium YouTube channel . All that said, I suspect you don't really have to care which constant should be used. All the suggested values differ by one part in a million, which makes essentially no difference for most chemistry. Edit: As pointed out by Loong in the comments, a few weeks after writing this answer, CODATA released updated values for the physical constants , so I updated this answer for accuracy. The next set of updated values will likely be announced in 2018-2019. For comparison, the previous value was: $$6.022\,141\,29\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,27\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2010)}$$ This represents an uncertainty of 44 parts per billion. This means the uncertainty in the measurement has been cut to almost a fourth of its previous value in four years. Go science! | {
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32,640 | Are there gases that are not transparent at room temperature (i.e. at temperature below the point where the substance starts to radiate visible light due to heating)? | First, a little bit of background. Transparency is not an absolute property of a material. Every substance is opaque, so long as light has to pass through enough of it, and opacity also changes according to ambient conditions. Some substances, such as most metals, are opaque even in $100\ \mathrm{nm}$ thin films, while many gasses will let a small amount of light through mostly unperturbed even after several kilometres. For example, here is a measure of how far photons of different energies can travel on average in the pure elements before interacting once with their atoms (the graph is actually for high energy x-rays rather than visible wavelengths, but they're all photons nonetheless). It makes intuitive sense that a gas should let more light through, as the atoms in it are much more spaced than in a solid or liquid. The two general processes that work to make materials non-transparent are photon absorption and photon scattering (reflection being included as an example of scattering). When we say a material is "transparent" colloquially , what we generally mean is that the material, at the thickness observed, does not scatter much light and thus allows a coherent image to pass through the material (basically, you can see through it to the other side without a lot of distortion, that is, the object is "see-through" or clear ). Interestingly, this means that a clear material need not be colourless , as correctly pointed out by Geoff, because photons of a given colour may be absorbed while letting most other wavelengths pass through without scattering. Gasses in general are highly transparent because they are both highly colourless (absorb little light) and very clear (scatter little light) unless you're looking through several kilometres of gas, as I mentioned previously. However, there are examples of highly clear but coloured gasses, as it is possible for them to absorb a significant amount of photons of a certain energy due to low energy electronic transitions, while allowing the rest of the visible spectrum to pass through unimpeded. While some gasses are coloured and others are not, I believe all pure gasses must be very clear (small amounts of gas will scatter almost no light), and they can only scatter a fair amount of light across a short distance in conditions where they are probably not best described as gasses anymore (plasmas, supercritical fluids, etc). Perhaps the most classic example of a coloured gas is nitrogen dioxide, $\ce{NO2}$, a strongly brown-red noxious gas which is easily formed by decomposition of nitric acid , among several other ways. The unpaired electron in its structure is somewhat uncommon, and it can be easily excited to a higher electronic state upon absorption of photons in the blue region of the visible spectrum. Since blue light is strongly absorbed even in relatively thin samples (a few centimetres of $\ce{NO2}$ gas), the rest of the white light passes through with essentially no scattering, forming a perfect image of the other side but with a strong red tinge. Even more interestingly, $\ce{NO2}$ reacts with itself at low temperatures or high pressures and dimerizes to form dinitrogen tetroxide , $\ce{N2O4}$, a colourless solid/liquid/gas (depending on conditions). In other words, the two substances are in a reversible equilibrium: $$\ce{2 NO2 (g) <=> N2O4 (g) + energy}$$ Coincidentally, this equilibrium is characterized by an equilibrium constant which is close to 1 at ambient conditions, and this constant can easily be changed by varying temperature or pressure to higher or lower values. This means that under relatively easily achieved conditions, it is possible to study the reversible change of a coloured gas into a colourless one! ( Source ) Though I focused on $\ce{NO2}$ because it is a remarkable case, there are several other examples of clear but coloured gasses. As Geoff mentioned, the halogens all form coloured gasses, though bromine is a volatile liquid and iodine is a solid at room temperature, so a small amount of heating is necessary. Ozone, $\ce{O3}$, is somewhat blue ( source ), though its colour is best observed in the liquid phase. The same goes for oxygen gas, $\ce{O2}$, which is slightly blue, though it is not the cause for the blue sky . There are probably a few more examples out there. | {
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33,049 | Which reacts faster in the Cannizzaro Reaction? a) $\ce{OHC-C6H4-NO2}$ b) $\ce{OHC-C6H4-OCH3}$ Obviously, a better hydride releasing group will react faster. Therefore my answer was b , as $\ce{-OCH3}$ shows -I (inductive) effect as well as +R (resonance) effect, as it's in the para position. This increases the electron concentration in the $\ce{C}$ atom of $\ce{CHO}$ and due to the excess electrons, it is a better donor of an electron to the $\ce{H}$ attached to it. Thus, leaving a $\ce{H-}$. But the answer given is a . The answer states that $\ce{C}$ can only give electron to $\ce{H}$ when it accepts an electron from the $\ce{O}$ attached to it. For that the $\ce{C}$ should be electorn deficient. Therefore an electron withdrawing group (in this case $\ce{NO2}$, which shows -R effect at ortho and para positions) better releases hydride. But I don't think that happens. That's because if $\ce{C}$ is electron deficient, then it wouldn't give an electron to $\ce{H}$ (less electronegative) in the first place, even if it takes an electron from oxygen (which is also tough). Can anyone give a proper explanation to the whole problem? | The mechanism of the Cannizzaro reaction is illustrated below. The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate. Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon because it is now $\ce{sp^3}$ hybridized. So we need to look at the step leading to the tetrahedral intermediate - the first step. However, if we examine some of the resonance structures for the starting carbonyl compounds (where the substituents on the aromatic ring can interact with the carbonyl through resonance) we see that the nitro compound places positive charge adjacent to the already positive polarized carbonyl carbon - a destabilizing situation (remember, destabilizing something means making it higher energy, more reactive). On the other hand, the methoxy compound places negative charge adjacent to the already positive polarized carbonyl carbon - a stabilizing situation. The destabilized carbonyl in the nitro compound will be more reactive towards nucleophilic attack. | {
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33,135 | I have seen two different notations for sodium acetate . The first one is: $$\text{NaCH}_3\text{COO}$$ The second one is: $$\text{CH}_3\text{COONa}$$ Now I'm confused, which one is the best to use? | There is nothing wrong with either formula. And you can use even more: $\ce{NaC2H3O2}$ $\ce{C2H3NaO2}$ It really depends on which point you want to bring across. $$\ce{NaCH3COO}$$ This formula, being analogous to formulae like $\ce{NaCl}$ stresses the inorganic salt view more. It shows that there is a cation ($\ce{Na+}$) and an anion ($\ce{CH3COO-}$) which form a salt crystal together much like the anion $\ce{Cl-}$ would do with the same cation. Inorganic nomenclature prefers cations to be written first. $$\ce{NaC2H3O2}$$ Is basically the same except for saying ‘I don’t care how those seven atoms combine to form the anion, all I care for is what’s in there.’ $$\ce{CH3COONa}$$ This one stresses a more organic-chemical point of view where it’s relevant where the cation is actually bound to within the molecule. Oftentimes, organic chemists would even write structural formulae with a bond between oxygen and sodium as if it were hydrogen. The reasoning behind this is ‘I don’t care if the structure ends up being a salt, all I need to know is that the sodium somehow connects to the carboxyl group.’ Note that you can combine the first’s and the third’s idea to give $\ce{NaOOCCH3}$ which works, but is a lot less used than either of them. $$\ce{C2H3NaO2}$$ This is the Beilstein -type lookup formula. You just know the elemental composition of your compound and now want to look it up — you can’t know whether it’s sodium acetate or sodium hydroxyethanalate (if that name is even correct). Or maybe you just don’t care. The reasoning behind this is $\ce{C}$ first, $\ce{H}$ second, the remaining elements following in alphabetic order. Finally, there are also shortened formulae. The most common (the one I actually use most) would be $\ce{NaOAc}$. Here, $\ce{Ac}$ is an abbreviation for $\ce{H3C-C=O}$ where whatever follows or precedes is bound to the carbonyl carbon. Note that this formula is not considered standard, and would need to be included in a list of abbreviations or defined otherwise if you decide to use it. | {
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33,297 | This is quoted from Jim Clark's Chemguide For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series. What is the reason stated above? Why does in only transition series, 4s has more energy than 3d & not in the preceeding elements? | Disclaimer: I now believe this answer to be fully incorrect. Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22. However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in Shriver & Atkins, as well as of the idea that the 3d orbitals are somehow higher in energy than the 4s orbitals. I believe it was mentioned that the energies were calculated with the outdated Thomas–Fermi–Dirac model, but cannot really remember. I will ask another question about the 3d vs 4s issue, but in the meantime I would point the reader in the direction of these articles: Pilar, F. L. 4s is always above 3d! Or, how to tell the orbitals from the wavefunctions. J. Chem. Educ. 1978, 55 (1), 2 DOI: 10.1021/ed055p2 . Melrose, M. P.; Scerri, E. R. Why the 4s Orbital Is Occupied before the 3d. J. Chem. Educ. 1996, 73 (6), 498 DOI: 10.1021/ed073p498 . Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. Transition Metals and the Aufbau Principle. J. Chem. Educ. 1994, 71 (6), 469 DOI: 10.1021/ed071p469 . Scerri, E. R. Transition metal configurations and limitations of the orbital approximation. J. Chem. Educ. 1989, 66 (6), 481 DOI: 10.1021/ed066p481 . Some criticism of Atkins' books by Eric Scerri. While Molly's answer does a good job of explaining why electrons preferentially occupy the 4s subshell over the 3d subshell (due to less inter-electron repulsion), it doesn't directly answer the question of why the order of the 3d/4s energies changes going from Ca to Sc. I stole this figure from Shriver & Atkins 5th ed: The red line represents the energy of the 3d orbital, and the blue line the energy of the 4s orbital. You can see that up to Ca, 3d > 4s but for Sc onwards, 4s < 3d. As chemguide rightly points out, up to Ca, the 4s orbital is lower in energy than the 3d. The energy of an electron in an orbital is given by $$E = -hcR\left(\frac{Z_\text{eff}}{n}\right)^2$$ where $hcR$ is a collection of constants, $Z_\text{eff}$ is the effective nuclear charge experienced by the electron, and $n$ is the principal quantum number. Since $n = 4$ for the 4s orbital and $n = 3$ for the 3d orbital, one would initially expect the 3d orbital to be lower in energy (a more negative energy). However, the 4s orbital is more penetrating than the 3d orbital; this can be seen by comparing the radial distribution functions of the two orbitals, defined as $R(r)^2 r^2$ where $R(r)$ is the radial wavefunction obtained from the Schrodinger equation: The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better. It is therefore shielded less than a 3d electron, which makes $Z_\text{eff}$ larger. Going from the 3d to the 4s orbital, the increase in $Z_\text{eff}$ wins ever so slightly over the increase in $n$ , which makes the energy of the 4s orbital lower. Now, going from Ca to Sc means that you are adding one more proton to the nucleus. This makes the nuclear charge larger and therefore both the 4s and 3d orbitals are stabilised (their energies decrease). The catch is that the energy of the 4s orbital decreases more slowly than that of the 3d orbital, because the 4s orbital is relatively radially diffuse (the maximum in the radial distribution function occurs at a larger value of $r$ ). If you have studied physics, you could think of it as the interaction between two point charges; if the distance between them is large, then increasing the magnitude of one point charge has a smaller effect on the potential energy $U = -\frac{kq_1q_2}{r}$ . The faster decrease of the 3d energy also makes sense because if nuclear charge were to tend to infinity, shielding would become negligible; the orbital energies would then be entirely determined by $n$ , and if this were to be the case, you would expect 3d < 4s in terms of energies, as we said at the very start. However, in Sc, the electrons preferentially occupy the 4s subshell even though it is higher in energy, and this is also because the 4s orbital is radially diffuse - the electrons have more "personal space" and experience less repulsion. One way of putting it is that an empty 4s orbital in Sc has a higher energy than an empty 3d orbital, but a filled 4s orbital has a lower energy than a filled 3d orbital. The fact that 4s > 3d in energy also explains why, for the transition metals, the 4s electrons are removed first upon ionisation ( $\ce{Sc^+}: [\ce{Ar}](3\mathrm{d})^1(4\mathrm{s})^1$ .) I just want to end off with a comment that the factors that determine the electronic configurations of d-block and f-block elements are actually very closely balanced and just a small change in one factor can lead to a completely different electronic configuration. This is why Cr and Cu have an "anomalous" configuration that maximises exchange energy, whereas we don't get carbon adopting a $(1\mathrm{s})^2(2\mathrm{s})^1(2\mathrm{p})^3$ configuration in order to have "stable half-filled shells". | {
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33,313 | I have the formula $\ce{Co(C2O4)*H2O}$ in a reaction that forms cobalt oxide $\ce{Co3O4}$ My data here says that my oxalate hydrate weighed $\pu{0.3283 g}$ and my product oxide weighed $\pu{0.1158 g}$ which I have calculated as: $$0.1158 \times \frac{\pu{176.79 g}\, \mathrm{(cobalt\,oxalate)}} {\pu{240.79 g}\, \mathrm{(cobalt\,oxide)}} =\pu{0.08502 g\,\mathrm{(cobalt)}}$$ Then $\pu{0.3283 g (reactant)} − \pu{0.1158 g (product)}$ gives me $\pu{0.2125 g}$ water removed divided by $\pu{18.016 g/mol}$ to get $\pu{0.01179 mol}$ $\ce{H2O}$ in the sample. How do I use this data to determine the empirical formula? I am not sure how to determine the molar ratios from this point. | Disclaimer: I now believe this answer to be fully incorrect. Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22. However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in Shriver & Atkins, as well as of the idea that the 3d orbitals are somehow higher in energy than the 4s orbitals. I believe it was mentioned that the energies were calculated with the outdated Thomas–Fermi–Dirac model, but cannot really remember. I will ask another question about the 3d vs 4s issue, but in the meantime I would point the reader in the direction of these articles: Pilar, F. L. 4s is always above 3d! Or, how to tell the orbitals from the wavefunctions. J. Chem. Educ. 1978, 55 (1), 2 DOI: 10.1021/ed055p2 . Melrose, M. P.; Scerri, E. R. Why the 4s Orbital Is Occupied before the 3d. J. Chem. Educ. 1996, 73 (6), 498 DOI: 10.1021/ed073p498 . Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. Transition Metals and the Aufbau Principle. J. Chem. Educ. 1994, 71 (6), 469 DOI: 10.1021/ed071p469 . Scerri, E. R. Transition metal configurations and limitations of the orbital approximation. J. Chem. Educ. 1989, 66 (6), 481 DOI: 10.1021/ed066p481 . Some criticism of Atkins' books by Eric Scerri. While Molly's answer does a good job of explaining why electrons preferentially occupy the 4s subshell over the 3d subshell (due to less inter-electron repulsion), it doesn't directly answer the question of why the order of the 3d/4s energies changes going from Ca to Sc. I stole this figure from Shriver & Atkins 5th ed: The red line represents the energy of the 3d orbital, and the blue line the energy of the 4s orbital. You can see that up to Ca, 3d > 4s but for Sc onwards, 4s < 3d. As chemguide rightly points out, up to Ca, the 4s orbital is lower in energy than the 3d. The energy of an electron in an orbital is given by $$E = -hcR\left(\frac{Z_\text{eff}}{n}\right)^2$$ where $hcR$ is a collection of constants, $Z_\text{eff}$ is the effective nuclear charge experienced by the electron, and $n$ is the principal quantum number. Since $n = 4$ for the 4s orbital and $n = 3$ for the 3d orbital, one would initially expect the 3d orbital to be lower in energy (a more negative energy). However, the 4s orbital is more penetrating than the 3d orbital; this can be seen by comparing the radial distribution functions of the two orbitals, defined as $R(r)^2 r^2$ where $R(r)$ is the radial wavefunction obtained from the Schrodinger equation: The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better. It is therefore shielded less than a 3d electron, which makes $Z_\text{eff}$ larger. Going from the 3d to the 4s orbital, the increase in $Z_\text{eff}$ wins ever so slightly over the increase in $n$ , which makes the energy of the 4s orbital lower. Now, going from Ca to Sc means that you are adding one more proton to the nucleus. This makes the nuclear charge larger and therefore both the 4s and 3d orbitals are stabilised (their energies decrease). The catch is that the energy of the 4s orbital decreases more slowly than that of the 3d orbital, because the 4s orbital is relatively radially diffuse (the maximum in the radial distribution function occurs at a larger value of $r$ ). If you have studied physics, you could think of it as the interaction between two point charges; if the distance between them is large, then increasing the magnitude of one point charge has a smaller effect on the potential energy $U = -\frac{kq_1q_2}{r}$ . The faster decrease of the 3d energy also makes sense because if nuclear charge were to tend to infinity, shielding would become negligible; the orbital energies would then be entirely determined by $n$ , and if this were to be the case, you would expect 3d < 4s in terms of energies, as we said at the very start. However, in Sc, the electrons preferentially occupy the 4s subshell even though it is higher in energy, and this is also because the 4s orbital is radially diffuse - the electrons have more "personal space" and experience less repulsion. One way of putting it is that an empty 4s orbital in Sc has a higher energy than an empty 3d orbital, but a filled 4s orbital has a lower energy than a filled 3d orbital. The fact that 4s > 3d in energy also explains why, for the transition metals, the 4s electrons are removed first upon ionisation ( $\ce{Sc^+}: [\ce{Ar}](3\mathrm{d})^1(4\mathrm{s})^1$ .) I just want to end off with a comment that the factors that determine the electronic configurations of d-block and f-block elements are actually very closely balanced and just a small change in one factor can lead to a completely different electronic configuration. This is why Cr and Cu have an "anomalous" configuration that maximises exchange energy, whereas we don't get carbon adopting a $(1\mathrm{s})^2(2\mathrm{s})^1(2\mathrm{p})^3$ configuration in order to have "stable half-filled shells". | {
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33,764 | The following comment by Wildcat made me think about whether density functional theory (DFT) can be considered an ab initio method. @Martin-マーチン, this is sort of nitpicking, but DFT (where the last "T" comes from "Theory") can be considered as an ab-initio method since the theory itself is built from the first principles. The problem with the theory is that the exact functional is unknown, and as a result, in practice we do DFA calculations ("A" from "Approximation") with some approximate functional. It it DFA which is not an ab-initio method then, not DFT. :) I always thought that ab initio refers to wave function based methods only. In principle the wave function is not necessary for the basis of DFT, but it was later introduced by Kohn and Sham for practical reasons. The IUPAC goldbook offers a definition of ab initio quantum mechanical methods : ab initio quantum mechanical methods Synonym : non-empirical quantum mechanical methods Methods of quantum mechanical calculations independent of any experiment other than the determination of fundamental constants. The methods are based on the use of the full Schroedinger equation to treat all the electrons of a chemical system. In practice, approximations are necessary to restrict the complexity of the electronic wavefunction and to make its calculation possible. According to this, most density functional approximations (DFA) cannot be termed ab initio since almost all involve some empirical parameters and/or fitting. DFT on the other hand is independent of any of this. What I have my problems with is the second sentence. It states, that treatment of all electrons is necessary. This is technically not the case for DFT, because here only the electron density is treated. All electrons and the wavefunction are implicitly treated. An earlier definition of ab initio can be found in Leland C. Allen and Arnold M. Karo, Rev. Mod. Phys. , 1960 , 32 , 275. By ab initio we imply: First, consideration of all the electrons simultaneously. Second, use of the exact nonrelativistic Hamiltonian (with fixed nuclei),
$$\mathcal{H} = -\frac12\sum_i{\nabla_i}^2
- \sum_{i,a}\frac{Z_a}{\mathbf{r}_{ia}}
+ \sum_{i>j}\frac{1}{\mathbf{r}_{ij}}
+ \sum_{a,b}\frac{Z_aZ_b}{\mathbf{r}_{ab}}$$
the indices $i$, $j$ and $a$, $b$ refer, respectively, to the electrons and to the nuclei with nuclear charges $Z_a$, and $Z_b$. Third, an effort should have been made to evaluate all integrals rigorously. Thus, calculations are omitted in which the Mulliken integral approximations or electrostatic models have been used exclusively. These approximate schemes are valuable for many purposes, but present experience indicates that they are not sufficiently accurate to give consistent results in ab initio work. This definition obviously does not include DFT, but this is probably due to the fact it was published before the Hohenberg-Kohn theorems. But in general this definition is still largely the same as in the goldbook. Another point which confuses me are titles like: "Potential Energy Surfaces of the Gas-Phase S N 2 Reactions $\ce{X- + CH3X ~$=$~ XCH3 + X-}$ $\ce{(X ~$=$~ F, Cl, Br, I)}$: A Comparative Study by Density Functional Theory and ab Initio Methods" Liqun Deng , Vicenc Branchadell , Tom Ziegler, J. Am. Chem. Soc. , 1994 , 116 (23), 10645–10656. And then again we have titles like: "Ab Initio Density Functional Theory Study of the Structure and Vibrational Spectra of Cyclohexanone and its Isotopomers" F. J. Devlin and P. J. Stephens, J. Phys. Chem. A , 1999 , 103 (4), 527–538. Unfortunately Koch and Holthausen, who wrote the probably most concise book on DFT, A Chemist's Guide to Density Functional Theory , never really refer to DFT as ab initio or clearly draw the line. The closest they come is on page 18: In the context of traditional wave function based ab initio quantum chemistry a large variety of computational schemes to deal with the electron correlation problem has been devised during the years. Since we will meet some of these techniques in our forthcoming discussion on the applicability of density functional theory as compared to these conventional techniques, we now briefly mention (but do not explain) the most popular ones. But that does not really answer my question. Throughout the book they use the term only in the form of conventional ab initio theory or in combination of explicitly stating wave function and variations thereof. In my quite extensive research about DFT selection criteria I never came about the term ' ab initio DFT'. So the question remains: Is density functional theory an ab initio method? | First note that the acronym DFA I used in my comment originates from Axel D. Becke paper on 50 year anniversary of DFT in chemistry: Let us introduce the acronym DFA at this point for “density-functional approximation .” If you attend DFT meetings, you will know that Mel
Levy often needs to remind us that DFT is exact . The failures we
report at meetings and in papers are not failures of DFT, but failures
of DFAs. Axel D. Becke, J. Chem. Phys. , 2014 , 140 , 18A301 . So, there are in fact two questions which must be addressed: "Is DFT ab initio?" and "Is DFA ab initio?" And in both cases the answer depend on the actual way ab initio is defined. If by ab initio one means a wave function based method that do not make any further approximations than HF and do not use any empirically fitted parameters, then clearly neither DFT nor DFA are ab initio methods since there is no wave function out there. But if by ab initio one means a method developed "from first principles" , i.e. on the basis of a physical theory only without any additional input, then DFT is ab initio; DFA might or might not be ab initio (depending on the actual functional used). Note that the usual scientific meaning of ab initio is in fact the second one; it just happened historically that in quantum chemistry the term ab initio was originally attached exclusively to Hartree–Fock based (i.e. wave function based) methods and then stuck with them. But the main point was to distinguish methods that are based solely on theory (termed "ab initio") and those that uses some empirically fitted parameters to simplify the treatment (termed "semi-empirical"). But this distinction was done before DFT even appeared. So, the demarcation line between ab initio and not ab initio was drawn before DFT entered the scene, so that non-wave-function-based methods were not even considered. Consequently, there is no sense to question "Is DFT/DFA ab initio?" with this definition of ab initio historically limited to wave-function-based methods only. Today I think it is better to use the term ab initio in quantum chemistry in its more usual and more general scientific sense rather then continue to give it some special meaning which it happens to have just for historical reasons. And if we stick to the second definition of ab initio then, as I already said, DFT is ab initio since nothing is used to formulate it except for the same physical theory used to formulate HF and post-HF methods (quantum mechanics). DFT is developed from the quantum mechanical description without any additional input: basically, DFT just reformulates the conventional quantum mechanical wave function description of a many-electron system in terms of the electron density. But the situation with DFA is indeed a bit more involved. From the same viewpoint a DFA method with a functional which uses some experimental data in its construction is not ab initio. So, yes, DFA with B3LYP would not qualify as ab initio, since its parameters were fitted to a set of some experimentally measure quantities. However, a DFA method with a functional which does not involve any experimental data (except the values of fundamental constants) can be considered as ab initio method. Say, a DFA using some LDA functional constructed from a homogeneous electron gas model, is ab initio. It is by no means an exact method since it is based on a physically very crude approximation, but so does HF from the family of the wave function based methods. And if the later is considered to be ab initio despite the crudeness of the underlying approximation, why can't the former be also considered ab initio? | {
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33,780 | Hemoglobin is an iron-containing oxygen transport metalloprotein in the red blood cells of most mammals. Simply put, it's a carrier protein. Interestingly it doesn't carry carbon dioxide in the same way it does for oxygen. Oxygen binds to the iron atoms in the protein whereas carbon dioxide is bound to the protein chains of the structure. Carbon dioxide doesn't compete with oxygen in this binding process. However, carbon monoxide is a very aggressive molecule. It's a colourless, odourless, and tasteless gas that is lighter than air and can be fatal to life. It has a greater affinity for hemoglobin than oxygen does. It displaces oxygen and quickly binds, so very little oxygen is transported through the body cells. There are two equilibrium reactions of binding oxygen and becoming oxygenated hemoglobin: $$\ce{Hb (aq) + 4O2 (g) -> Hb(O2)4 (aq)}$$ $$\ce{Hb (aq) + 4O2 (g) <- Hb(O2)4 (aq)}$$ And carbon monoxide binding equation at equilibrium: $$\ce{Hb (aq) + 4CO (g) ⇌ Hb(CO)4 (aq)}$$ It is said the equation is shifted towards right, generating $\ce{Hb(CO)4(aq)}$ , since its bond is much stronger. I have two main related questions: Why is the carboxyhemoglobin bond stronger relative to that of oxygenated hemoglobin? Why is carbon monoxide highly attracted to hemoglobin? Does it have anything to do with the oxidation state of oxygen in each molecule? Please show some reaction equations associating $\ce{Fe}$ . | Excursion into simple coordination chemistry: Bonding, backbonding and simple orbital schemes Please refer to Breaking Bioinformatic’s answer for the MO scheme of carbon monoxide, it is very helpful. You might also look at the orbital pictures in this answer by Martin . Carbon monoxide can bind to metal centres via a σ coordinative bond where the HOMO of $\ce{CO}$ interacts with metal orbitals and also by the π backbonding, Breaking Bioinformatics mentioned. I’ll start by touching the σ bond so we can later better understand the π bond. In figure 1 you can see the molecular orbital scheme of a complex composed of a central metal ion and six ligands that donate in a σ manner exclusively. Figure 1: Molecular orbital scheme of an octahedral complex with six σ donors around a central metal. Copied from this site and first used in this answer of mine . Metal orbitals are 3d, 4s, 4p from bottom to top; ligand orbitals are of s-type. You will notice that figure 1 contains the irreducible representations of the orbitals beneath them. Orbitals can only interact if they have identical irreducible representations; otherwise, their interactions will sum up to zero. We can see that the metal d orbitals are split up into a $\mathrm{t_{2g}}$ set and and $\mathrm{e_g}^*$ set. This is the effect of σ bonding and why it stabilises the entire entity. π bonding can only occur if the ligands have available orbitals of π symmetry. The 2π orbitals of $\ce{CO}$ are a nice example, but one could also simply assume a halide with its p-orbitals for the same effect. Further down in the internet scriptum I originally copied the picture from , you can see a set of two pictures that introduce p-orbitals. Twelve ligand p-type orbitals will transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$ , thus a further stabilising/destabilising interaction is introduced for the $\mathrm{t_{2g}}$ orbitals. Due to the $2\unicode[Times]{x3c0}^*(\ce{CO})$ orbitals being similar in energy to the metal’s $\mathrm{t_{2g}}$ and empty, they can stabilise the former well, creating a possibly strong stabilisation for the overall system. Since this is an interaction of two orbitals creating both bonding and antibonding orbitals, and since the resulting molecular orbital is filled with two electrons that stem from the metal centre this is termed π-backbonding. The binding mode of carbon monoxide to haemoglobin After this extensive background discussion, it may be clear that $\ce{CO}$ is generally a good ligand. For reasons I didn’t go into, a carbonyl complex of metal ions isn’t that stable, but a single carbonyl ligand will almost always be beneficial. Such is the case for the porphyrin-iron(II) system that forms the heart of haemoglobin: The central iron(II) ion is coordinated well from five directions (four from the porphyrin ring and one histidine of haemoglobin) and has a weakly bound water atom in the ground state in its sixth coordination slot, sometimes displaced by a distal histidine. Carbon monoxide can diffuse in and bind very well to this system, displacing the weakly bound water and histidine. In fact, isolated haem can bind carbon monoxide $10^5$ times better that it can oxygen. This is something that nature could not really inhibit, because it is based on fundamental properties, but also something nature didn’t really care for, because the natural concentration of carbon monoxide is very low, and nature rarely had to deal with it competitively inhibiting oxygen binding to haemoglobin. The binding mode of oxygen to haemoglobin Oxygen, the $\ce{O2}$ molecule, is an absolutely lowsy σ donor — especially when compared to $\ce{CO}$ . Its molecular orbital scheme is generally that of carbon monoxide except that it is entirely symmetric and two more electrons are included: They populate the 2π orbitals to give a triplet ground state. These orbitals are now the HOMOs and they hardly extend into space in any meaningful way — plus the still existant lone pair on each oxygen atoms is now at a much lower energy and also does not extend into space and thus cannot bind to a metal centre in a σ manner. What happens here is rather complex, and the last lecture I heard on the topic basically said that final conclusive evidence has not yet been provided. Orthocresol discusses the different viewpoints in detail in this question . The diamagnetic properties of the resulting complex are, however, unquestioned and thus one must assume a singlet ground state or one where antiferromagnetic coupling cancels any spins at molecular levels. Since the ground state of haemoglobin has a high-spin iron(II) centre and the ground state of oxygen is a paramagnetic triplet, it makes sense to assume those two to be the initial competitors. Professor Klüfers now states the steps to be the following: Iron(II) is in a high-spin state with four unpaired electrons; Paramagnetic triplet oxygen approaches and one of its $\unicode[Times]{x3c0}^*$ orbitals coordinates to the iron(II) centre in a σ fashion. This induces a high-spin low-spin transition on iron and slightly reorganises the ligand sphere (pulling oxygen closer to the iron centre). We now have a low-spin singlet iron(II) centre and a coordinative σ half-bond from oxygen. We can attribute that electron to the iron centre. Through linear combination, we can adjust the spin-carrying σ orbital and the populated iron orbitals so that an orbital is generated which can interact with the other $\unicode[Times]{x3c0}^*$ of oxygen in a π fashion. We thus expect an antiferromagnetic coupling and an overall state that can be described best as $\ce{Fe^{III}-^2O2^{.-}}$ — a formal one-electron oxidation of iron to iron(III), reducing oxygen to superoxide ( $\ce{O2-}$ ). If the spin-carrying orbitals are all treated as being metal-centred, we gain a $\ce{Fe^{II}-^1O2}$ state. (Content in the link I quoted from are in German; translation mine and shortened from the original.) Only due to the complexly tuned ligand sphere and also due to the stabilising high-spin low-spin transition (plus reorganising) is oxygen able to bind to iron at all. The distal histidine further stabilises the complex by a hydrogen bond, alleviating the charge slightly. It is to be assumed that nature did a great deal of tuning throughout the evolution since the entire process is rather complex and well-adjusted for collecting oxygen where it is plentiful (in the lungs) and liberating it in tissue where it is scarce. Comparison of the oxygen and carbon monoxide binding modes The simpler picture I drew above for carbon monoxide is not correct. Inside haemoglobin, carbon monoxide also binds in an angular fashion as if it were oxygen — see bonCodigo’s answer for space-filling atom models. This is because the entire binding pocket is made to allow oxygen to bind (as I stated) thus attempting everything to make oxygen a comfortable home. Carbon monoxide is rather strained in there, its binding affinity is reduced by a factor of $1000$ . However, since we started with a binding affinity difference of $10^5$ , carbon monoxide can still bind $100$ times better than oxygen can. Summary Carbon monoxide is generally a good ligand that can bind to metal centres well. Oxygen is generally a poor ligand. Nature did everything to make oxygen a comfortable home in haemoglobin. In doing so, the binding pocket became substantially less comfortable for $\ce{CO}$ . But since carbon monoxide was so good and oxygen so poor to start with, the former still binds better than the latter. Nature didn’t really care, because individuals seldomly come in contact with carbon monoxide so the collateral damage was taken into account. | {
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33,971 | I have searched and searched, oh how I have searched. I am looking for a 3-dimensional visualization of a whole atom, one that that includes all the orbital geometry. A proper "layered" view of the orbitals. What I am not looking for: Individual orbital geometry. A "picture" of an atom (i.e. what looks like television static to me). Little spherical balls. Dot-and-cross diagrams. I think I understand that this cannot be expressed in just one picture alone, so I look forward to discovering software, or a collection of pictures, that could help express this. Maybe something similar to an exploded view, for atoms? Please. The universe is weird and I want to see. | I have searched and searched, oh how I have searched. Do you know what I always tell my mom when she asks me to find something in the Internet she was not able to find herself? I ask her: "Are you sure that the thing you are looking for even exists?" I am looking for a 3 dimensional visualization of a whole (moderately
complex, hydrogen is just a ball) atom that includes 3 dimensional
orbital geometry. Hydrogen atom is not "just a ball". There is no orbitals. In short, except for one-electron systems (such as the hydrogen atom) orbital description is just an approximate model of the reality. Usually when one asks to visualize a physical object, he/she wants to visualize its physical "boundaries" with respect to some medium. The notion of such "boundary" simply looses its meaning in the microscopic world: atoms do not have "boundaries" like macroscopic objects do. Please. The universe is weird and I want to see. Sorry, but you can not do this: you can not visualize some things in the Universe, you can not even imagine them. Our visualization & imagination abilities (unlike an atom) has some boundaries, because they are based ultimately on our senses which in turn has their boundaries. As Paul Dirac said on that matter: […] the main object of physical science is not the provision of
pictures, but is the formulation of laws governing phenomena and the
application of these laws to the discovery of new phenomena. If a
picture exists, so much the better; but whether a picture exists or
not is a matter of only secondary importance. In the case of atomic
phenomena no picture can be expected to exist in the usual sense of
the word 'picture', by which is meant a model functioning essentially
on classical lines. One may, however, extend the meaning of the word
'picture' to include any way of looking at the fundamental laws which
makes their self-consistency obvious . With this extension, one may
gradually acquire a picture of atomic phenomena by becoming familiar
with the laws of the quantum theory. | {
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33,994 | I read that the $\ce{O2}$ molecule is paramagnetic , so I'm wondering: could a strong magnet pull the $\ce{O2}$ to one part of a room – enough to cause breathing problems for the organisms in the room? (I'm not a professional chemist, though I took some college chemistry.) | I'm a physicist, so apologies if the answer below is in a foreign language; but this was too interesting of a problem to pass up. I'm going to focus on a particular question: If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region? The TL;DR is that thermal effects are going to make this idea basically impossible. The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$ , where $\vec{B}$ is the magnetic field. Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point: $\vec{m} = \alpha \vec{B}$ , where $\alpha$ is what we might call the "molecular magnetic susceptibility." Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$ . But potential energy is given by $\vec{F} = - \vec{\nabla} U$ ; which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$ . Now, if we're talking about a sample of gas at a temperature $T$ , then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor: $$
\rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT}
$$ In the limit where $kT \gg \alpha B^2$ , this exponent will be close to zero, and the density will not vary significantly from point to point in the sample. To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$ ; in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field. So how high does this have to be? The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$ ; and so we have 1 $$
\chi_\text{mol} B^2 \gtrsim \mu_0 RT
$$ and so we must have $$
B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}.
$$ If you believe Wikipedia , the molar susceptibility of oxygen gas is $4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$ ; and plugging in the numbers, we get a requirement for a magnetic field of $$
B \gtrsim \pu{258 T}.
$$ This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines. Even at $\pu{91 Kelvin}$ (just above the boiling point of oxygen), you would need a magnetic field of almost $\pu{150 T}$ ; still well out of range. 1 I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules. A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation ; and if the gas gets sufficiently dense, then all bets are off. | {
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34,007 | Is there a compound that absorbs all visible light ? | Substances which absorb almost all the light falling on them appear black. Therefore you are looking for the blackest known compound. The record is currently held by Vantablack [1] , a substance composed of vertically aligned carbon nanotubes, which absorbs up to 99.965% of visible light incident upon it. As you can see it is really black. Previously, the record was held by Superblack [2,3] , a nickel-phosphorus alloy with an absorbance of up to 99.65%. For comparison, regular black paint generally absorbs around 97.5% of visible light. Sources [1] https://en.wikipedia.org/wiki/Vantablack [2] https://www.newscientist.com/article/dn3356-mini-craters-key-to-blackest-ever-black [3] http://pubs.rsc.org/en/Content/ArticleLanding/2002/JM/b204483h#!divAbstract | {
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34,106 | I do not mean at the same time, of course, but I mean it appears from an overview of the common charges formed from ionizing various elements that each element forms one or more of either positive or negative ions, but they never have the opposite charge. For example Fe may have +4 or +3, but never -anything. However, I am aware that this is from a common ion chart. Perhaps an atom can be a cation and an anion under specific circumstances? | Actually, in theory almost all of the elements can be found with both positive and negative oxidation numbers: it's just a matter of finding a system with the proper reagents and conditions to force it. If you isolate chemical species which have a very strong tendency of displaying some specific behaviour (accepting electrons, donating electrons, coordinating ions, releasing a leaving group, bonding to metals, releasing a proton, adopting a specific molecular geometry, or any other myriad properties), then you can often obtain strange results by clashing them with substances which also have the same tendency, but not quite as strong. This often causes the substance with the weaker behaviour to "operate in reverse". Let me give a vivid and related example. As we all know , the alkali metals (group 1 elements) are exclusively present as cations with an oxidation number of +1, except in the pure metals, where it's zero, right ? Well, here's something which might shatter your world: most of the alkali metals (with the exception of lithium, for now) also form alkalides , that is, stable salts containing discrete, clearly observed $\ce{Na^{-}}$, $\ce{K^{-}}$, $\ce{Rb^{-}}$ or $\ce{Cs^{-}}$ anions, with alkali metals displaying oxidation number -1. How is this done? All you need is to find a neutral substance with a much stronger tendency to donate an electron than a neutral alkali metal atom (easier said than done). Since neutral alkali metals atoms form quite stable cations upon loss of an electron, this implies you need to search for a neutral substance capable of donating an electron and forming a cation with exceptional thermodynamic and/or kinetic stability. This can be achieved, for example, by the use of cryptands , which are cyclic molecules capable of coordinating very strongly to cations, strongly enough that they even coordinate alkali metal cations very well. The cryptand-coordinated cation is both thermodynamically and kinetically stable enough that alkalide anions, which would be extremely reactive otherwise, are not reactive enough in this case to immediately cause charge transfer and neutralize the negative charge. Amusingly, it is actually possible to prepare a single compound which contains both alkali metal cations and anions, as exemplified by $\mathrm{[Na(2,2,2- crypt)]^{+} Na^{-}}$, which contains a sodium cation coordinated by a cryptand as a counterion to a natride / sodide ($\ce{Na^{-}}$) anion. One can imagine this compound being made by putting together a neutral sodium atom ($\ce{Na^{0}}$) and the neutral cryptand species $\mathrm{[Na(2,2,2- crypt)]^0}$. As Brian mentions in the comments, this latter species is actually an electride , which can be written as $\mathrm{[Na(2,2,2- crypt)]^{+}e^{-}}$ and thought of as a salt where the anion is a lone electron (!). Both the neutral sodium atom and the electride have a strong tendency to lose an electron in chemical reactions, but this tendency is much stronger for the electride. Thus, the electride ends up having its way, forcing its very loosely bound electron onto the neutral sodium atom, causing the neutral sodium atom to "operate in reverse" and accept an electron rather than donate it, resulting in the $\ce{Na^{-}}$ anion. This table on Wikipedia is far more complete than most "common oxidation number" tables out there, and it lists many negative oxidation numbers for elements, including iron! For many of the transition metals, negative metal oxidation numbers are achievable by using the carbonyl ($\ce{CO}$) ligand, which removes electron density from the metal atom via backbonding. This stabilizes negative charges on the metal atom, once again allowing the resultant species to survive with the appropriate counterion. | {
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34,190 | Polytetrafluoroethylene was discovered by accident. It now is an important material in the industry mainly because of its extremely high bonding energy, which prevents corrosion, halts reaction, and reduces friction (yeah carbon-fluorine bonds! ) And people would have definitely put it to the test, making it contain some of the most vicious and chemically diabolical substances ever created. There is a whole HOST of items it can contain that some chemists have gone so far as to say they were 'evil': Dioxygen Difluoride Known as the gas of Lucifer , there is a whole list of people blown up and killed while just trying to work with one of its components, fluorine. It ignites stuff at temperatures that most of the stuff that we breathe in would be in liquid form . No one really knows about its atomic structure (obviously). Fluoroantimonic Acid With a staggering pH of -25, it chews through stuff you might not even believe could be corroded; like wax or glass. It can even strip hydrogen off of methane ...There are a lot of other chemical demons it can contain, but this is not the point. Let this suffice: Chemical Resistance Comparison (Spoiler: Fluorine is good at this corrosion thing.) With this kind of hyper-resistance to about anything chemically destructive, is there anything that can destroy Teflon through only chemical means? A chemical that reacts exothermically to release heat, which melts the PTFE does not count. You get the drift. Also, I am very curious as to whether there is anything more resilient than Teflon? Polytetrafluoroethylene is made of many carbon-fluorine bonds in series. However, carbon-fluorine is second only to the Si-F bond. Is there an "overclocked" Teflon made of silicon-fluorine bonds that is even stronger? Now I know that some, but very few, solvents can make a mark on Teflon; but my question has not been answered: Is there any more resistant substances? (More Teflon bragging; take that aqua regia) | Just to add a bit to Ben's excellent answer... A number of fluorinating agents also react with PTFE, $\ce{XeF2}$ and $\ce{CoF3}$ being examples Ben mentioned the reaction of magnesium metal. Typically with metals, they must be in intimate contact with the PTFE surface, so molten metals or metals dissolved in anhydrous solvents will react. The magnesium reaction is of special interest because it serves as the basis of the thermite flare. A pyrotechnic device commonly used in the countermeasures aircraft use to evade heat-seeking missiles. The reaction of metals with PTFE is given by the following equation (I think this is the general description for the reaction of metals with PTFE; I'm suspect of the reaction proposed by Ben involving the formation of poly-perfluoroacetylene). $$\ce{2Mg + -(C2F4){-} → 2MgF2 + 2C}$$ The formation of $\ce{MgF2}$ is extremely exothermic. The heat given off along with the carbon soot provides a new, much hotter, target for the attacking missile to lock onto. As to whether there is anything more resistant, I suspect that is unlikely. The $\ce{C-F}$ bond is shorter (135 pm) than the $\ce{Si-F}$ bond (160 pm) and therefore better serves to encase and protect the carbon backbone. While there are some other polymers that have better mechanical or thermal properties, I am not aware of any that have better chemical resistance. In Polymers for Electronic & Photonic Application from 2013, the author states, "PTFE is the most chemically resistant polymer known". | {
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34,193 | Common saying. Diamond possesses: ultra hardness, ( 10 on the Mohs scale; 10000 HV on Vicker's Hard Test (iron merely 30-80)) hyper thermal conductivity, ( $2320~\mathrm{W\, m^{-1}\, K^{-1}}$, or over ten times better than the heatsink in your computer! ) extreme pressure resistance, (withstands a crushing 600 gigapascals; or around 2 times the pressure at the center of the earth, enough to snap carbon nano-tubes and graphene or create metallic oxygen or overcome copper's electron degeneracy pressure, making the maximum chamber pressure of a firing pistol seem literally like popping popcorn... I digress) and excellent luster (what do you expect, it's a diamond ) combine to make the gemstone coveted by all. Diamonds are the stuff of awesome. But do they really exist forever? Wikipedia notes that, Diamond is less stable than graphite, but the conversion rate from diamond
to graphite is negligible at standard conditions. Huh. But Wikipedia doesn't mention how long. So how long would it take for this super-material to convert to the stuff I scribble with? (If you doubt the claims about diamond's seemingly unbelievable properties, check out the link on Wikipedia about diamond and this and this .) Great point Joe made , that $10^{80}$ is just forever to us puny humans. Being a geek I can't resist the urge to compare the time length $10^{80}$: Makes the entire lifespan of a red dwarf star seem like the Planck second. Enough time for you to sift through all the atoms in the entire UNIVERSE at a rate of one atom per second. Getting $67 worth of US quarters and flipping them, one per second, to get all heads-up. Chance of macroscopic quantum tunneling!! (I don't know precisely how much, but quite large) ...and this is $10^{80}$ seconds I'm talking about... | how long would it take for this super-material to convert to the stuff
I scribble with? No, despite the fact that James Bond said " Diamonds are Forever ", that is not exactly the case. Although Bond's statement is a fair approximation of reality it is not a scientifically accurate description of reality. As we will soon see, even though diamond is slightly less stable than graphite (by ~ 2.5 kJ/mol), it is kinetically protected by a large activation energy. Here is a comparative representation of the structures of diamond and graphite. (image source: Satyanarayana T, Rai R. Nanotechnology: The future. J Interdiscip Dentistry 2011;1:93-100 ) ( image source ) Note that diamond is composed of cyclohexane rings and each carbon is bonded to 2 more carbons external to the cyclohexane ring . On the other hand, graphite is comprised of benzene rings and each carbon is bonded to only 1 carbon external to the benzene ring . That means we need to break 6 sigma bonds in diamond and make about 2 pi bonds (remember it's an extended array of rings, don't double count) in graphite per 6-membered ring in order to convert diamond to graphite. A typical aliphatic C–C bond strength is ~340 kJ/mol and a typical pi bond strength is ~260 kJ/mol. So to break 6 sigma bonds and make 2 pi bonds would require ~((6*340)-(2*260)) ~ 1500 kJ/mol. If the transition state were exactly midway between diamond and carbon (with roughly equal bond breaking and bond making), then we might approximate the activation energy as being half that value or ~750 kJ/mol. Since graphite is a bit more stable than diamond, we can refine our model and realize that the transition state will occur a bit before the mid-point. So our refined model would suggest an activation energy something less than 750 kJ/mol. Had we attempted to incorporate the effect of aromaticity in graphite our estimate would be even lower. In any case, this is an extremely large activation energy, so, as we anticipated the reaction would be very slow. An estimate (see p. 171) of the activation energy puts the reverse reaction (graphite to diamond; but since, as noted above, the energy difference between the two is very small the activation energy for the forward reaction is almost the same) at 367 kJ/mol. So at least our rough approximation was in the right ballpark, off by about a factor of 2. However, it appears that the transition state is even further from the midpoint (closer to starting material) than we might have guessed. This activation energy tells us that at 25 °C, it would take well over a billion years to convert one cubic centimeter of diamond to graphite. Note 04/17/20: As mentioned in a comment, the original "estimate" link became defunct and was replaced today with a new estimate link. However the original article and estimate can can still be seen on the Wayback Machine and it estimates the activation energy to be 538.45 kJ/mol, reasonably close to our estimate. | {
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34,929 | Freezing a full bottle of water tends to shatter the glass bottle. What if you used something tougher than glass, like diamond? What would happen if you kept dropping the temperature, but restrained the liquid volume so it couldn't freeze and what sort of force would the liquid exert on the container's walls? | Good question. Let's assume the container is infinitely strong, non-deformable, and constant in volume. Let's also assume that cooling the water is an equilibrium process -- that way, we won't have any supercooling . At equilibrium, the first tiny bit of ice that freezes will take up more volume than the water it froze from. This will raise the pressure on the rest of the water. Eventually the pressure may get so high that additional freezing of more water is not thermodynamically favored. Of course, as the pressure is raised, even the solid ice compresses a bit, freeing up a bit more volume for the liquid water. According to this paper from 2004 , ice is less compressible than water, so as a starting assumption, it may be approximately true to neglect the ice compression effect. Figure 4 from that same paper gives the freezing point depression of water as a function of pressure: To fully answer your question, in addition to that data, an equation that gives pressure as a function of ice volume would also be needed. If we make the assumption I was talking about above -- i.e. that ice is incompressible, then from the data point that water has a constant compressibility of 46.4 ppm per atm we can come up with a very simple version of that equation. $\frac{\Delta V_{water}}{V_{water}}=46.4 \times 10^{-6} \times P$, where P is the pressure in atmospheres. Before freezing of a fraction $X$ of the water: $$V_{ice} = X V_{tot}$$
$$V_{water} = (1-X) V_{tot}$$ After freezing: $$ V_{ice} = X V_{tot} 1.11 $$
$$ V_{water} = (1-X) V_{tot} - \Delta V_{ice} $$ Combining those equations, you can get $$0.11 \frac{X}{k(1-X)}= P$$, where $k$ is the compressibility of water. If even 1% of the water in the container freezes (and all our assumptions are true), then the pressure will be 24 atmospheres! Freezing 10% of the water would mean a pressure of 260 atmospheres. Looking at the chart above, reaching this point would require a temperature of only 271 or 272 K, i.e. only -1 °C or -2°C. Freezing 45% of the water would reach a pressure of 2000 atm, already off the chart above -- but the temperature required to reach that point would only be 253K or -20 °C, the setting of the average home residential freezer! ((Of course, at these extreme pressures, (i) ice is actually compressible, and (ii) the compressibility of liquid water is not constant but also a function of pressure, so the calculations would get quite a bit more complicated.)) The lesson is that for even moderate degrees of cooling, you'd need a very, very strong container. | {
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35,016 | What are the general ways to treat an accident involving highly concentrated or similarly dangerous acids? I have a lab instructor who recently debated with our group about treating an acid spill on the skin. According to him, one should immediately rinse off the acid with running water for about 15-20 minutes. He says that one should never apply bases to the site because an Acid-Base is highly exothermic and would only increase the damage. However, I clearly remember learning once that one should apply baking soda (sodium bicarbonate) to a bee sting or ant bite in order to neutralize the acid and relieve the burning.
I also know that the dilution of an acid in water is done by adding the acid slowly to the water containing vessel because of the heat released in the process. And especially when preparing a dilute solution of sulfuric acid, one wears safety goggles and gently stirs the solution to prevent splashing. So which is it? And does it depend on the severity of the burn or type acid? | My Lab First Aid book$^{*1}$ tells the following: The local therapy mainly consists of the instant, intense rinsing (at least 10 minutes) with tap water (ideally 15 to 20 °C). Don't try to neutralize the chemical burn, because the heat generation could lead to further tissue damages. This means, that your lab instructor is absolutely right. The rinsing water should wash the cold water away faster than the heat from the acid base reaction could harm you. It's not for nothing that there are emergency showers and eye douches in every lab. I have a friend who is a chemical lab assistant and once had to carry concentrated sulfuric acid in a big glass beaker. Suddenly the bottom of the beaker broke off and all the acid poured over her etc. She not even thought of anything else than getting out of her clothes and under the next emergency shower. $^{*1}$ The book is: R. Rossi, Erste hilfe bei akuten Notfällen - Begleitheft zu Jander - Blasius , S. Hirzel Verlag, 2005 , p. 11f.. The text is in german and I translated it. | {
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35,286 | Earlier this year, there was a lot of attention when all-cis 1,2,3,4,5,6-hexafluorocyclohexane was synthesized *: Notably, C&E News quoted the lead author: This compound is remarkable for being the most polar nonionic compound now known to exist Now, the original paper reports a molecular dipole moment of $\pu{6.2 D}$, which is extremely high for an alkane. It also claims the: unusual property of a facially polarized ring in organic chemistry Certainly having one side of the cyclohexane ring completely substituted with fluorines is unusual, so I agree with the facial-polarity property being interesting. But there are many small molecules (e.g., under 20-30 atoms) with large dipole moments, e.g., p-nitroaniline : A quick PM7 calculation gives the predicted dipole moment of $\pu{7.93\pm1 D}$ . So I dispute the claim from C&E News - nitroaniline is clearly more polar and non-ionic. So my question is this: What's the largest dipole moment for a known non-ionic organic molecule under 30 atoms? *: O'Hagan, et. al, "All-cis 1,2,3,4,5,6-hexafluorocyclohexane is a facially polarized cyclohexane" Nature Chemistry 7, 483–488 ( 2015 ) | At first, I thought, that those 6.2 D of O’Hagan et al. have been measured somehow but it as later described, they calculated it (1): A molecular dipole value of 6.2 D for 1 was calculated at the M11/6-311G(2d,p) theory level using natural bond orbital (NBO) analysis with NBO 6.0. I was using your given database for a little search, starting with nitroaniline, and came up with a few small molecules with higher dipole moments. Those seem to be PM7 calculated values and are shown in the following table: \begin{array}{lll}
\hline
\text{Name} & \text{Dipole moment in D} & \text{Source}\\
\hline
\text{4-Nitroacetanilide} & 8.36 \pm 1.08 & (2)\\
\textit N\text,\textit N\text{-Dimethyl-4-nitroaniline} & 9.53 \pm 1.08 & (3)\\
\textit P\text ,\textit P\text{-Dimethyl-}\textit N\text{-(4-nitrophenyl)phosphinic amide} & 10.89 \pm 1.08 & (4)\\
\hline
\end{array} As there were some comments, “it seems that this level of calculation does not provide accurate estimates of dipole moments”, I did some further calculations on levels that might provide better estimates. There is a quite new paper of Hickey and Rowley (5) which says: CCSD, MP2, or hybrid DFT methods using the aug-cc-pVTZ basis set are all able to predict dipole moments with RMSD errors in the 0.12–0.13 D range and polarizabilities with RMSD errors in the 0.30–0.38 ų range. Within the paper, there is later on a comparison between cc-pVTZ, Sadlej-pVTZ and aug-cc-pVTZ: The Sadlej basis set yields results that are improved over the cc-pVTZ basis set and have effectively the same accuracy as the aug-cc-pVTZ basis set. That’s why I chose Sadlej’s basis set at first … but I wanted to look if it could not be easier hence faster. Therefore I also chose the def2-SVPD and looked, what an improvement a def2-TZVPD single point calculation on this geometry would yield. I use ORCA 3.0.3 for calculations – the command line is like the following: ! B3LYP Sadlej-pVTZ TightSCF TightOpt Table 1: "Pure" dipole moments (in Debye) straight from the calculations
\begin{array}{lllll}
\hline
& \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide} & 9.18261 & 9.00380 & 9.07922 & 8.6738\\
\textit N\text ,\textit N\text{-Dimethyl…} & 7.39418 & 7.29188 & 7.30862 & 9.4209\\
\textit P\text ,\textit P\text{-Dimethyl…} & 8.58913 & 8.44430 & 8.51591 & 11.8352\\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.17422 & 5.97013 & 6.04794 & 5.6298\\
\hline
\end{array} What I did then, was to think of how to produce values that could be properly compared to the experimental values from the paper. As they said, with their methods they can yield RMSDs of around 0.12 D. So I took their values for the Sadlej basis set and made my own calculations on their test molecules for the two def2 basis sets (and also for PM6 with Gaussian 09 Rev. A.02). Table 2: Some analytical values to compare with the publication values; the units are in Debye
\begin{array}{lllll}
\hline
& \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{MAE} & 0.08 & 0.08 & 0.08 & 0.40\\
\text{RMSD} & 0.13 & 0.12 & 0.12 & 0.61\\
\hline
\end{array} So for my two ways, the MAE and the RMSD are not too far away from the triple zeta values but are calculated way faster. Now I have made a linear regression between the experimental and the calculated dipole moments from Hickey and Rowley’s test molecules to be able – as said before – to calculate proper error ranges at a given probability. Figure 1: Regression functions for the four tested methods (blue), prediction intervals at 95% (gray); from top left to bottom right: def2-SVPD, def2-TZVPD//def2-SVPD, Sadlej-pVTZ, PM6 Now let’s turn the “pure” values from Table 1 into somehow more real values: \begin{align}
\text{def2-SVPD} \to \text{exp}&: -0.0097439128 + 0.9763275\ x\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: -0.0045050428 + 0.99095691\ x\\
\text{Sadlej-pVTZ} \to \text{exp}&: -0.012922547 + 0.98812009\ x\\
\text{PM6} \to \text{exp}&: -0.16057241 + 0.89545049\ x\\
\end{align} Prediction Intervals (for the error): $$\small\begin{align}
\text{def2-SVPD} \to \text{exp}&: 0.24473369 \sqrt{1.0217391 + 0.010348758 (x-1.4590217)^2}\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: 0.23719452 \sqrt{1.0217391 + 0.010656662 (x-1.4321957)^2}\\
\text{Sadlej-pVTZ} \to \text{exp}&: 0.23333979 \sqrt{1.0217391 + 0.010593477 (x-1.4448261)^2}\\
\text{PM6} \to \text{exp}&: 0.98446665 \sqrt{1.0217391 + 0.0097476423 (x-1.7592391)^2}
\end{align}$$ Table 3: Converted dipole moments (in Debye) from Table 1 including error ranges at 95% probability \begin{array}{llll}
\hline
& \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide} & 8.96 \pm 0.31 & 8.92 \pm 0.30 & 8.96 \pm 0.30 & 7.61 \pm 1.20 \\
\textit N\text ,\textit N\text{-Dimethyl…} & 7.21 \pm 0.29 & 7.22 \pm 0.28 & 7.21 \pm 0.27 & 8.28 \pm 1.24 \\
\textit P\text ,\textit P\text{-Dimethyl…} & 8.38 \pm 0.30 & 8.36 \pm 0.29 & 8.40 \pm 0.29 & 10.44 \pm 1.40 \\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.02 \pm 0.27 & 5.91 \pm 0.26 & 5.96 \pm 0.26 & 4.88 \pm 1.06 \\
\hline
\end{array} What can be seen from this now? The smaller B3LYP/def2-SVPD method yields nearly the same values as the more expensive Sadlej version. Putting the def2-TZVPD single point on top of them, yields slightly deeper results with actually no accuracy improvements. PM6 is no real good option as the error ranges are way higher, with values of about $1-1.4$ D. In the end, the dipole moment of all- cis 1,2,3,4,5,6-hexafluorocyclohexane is probably somewhere around $6$ D but there are probably many other molecules with higher dipole moments … “my” three molecules shown above go at least up to somewhere around $9$ D. As long as no experimental values are available, everything here is only a good guess – but you will know this too good. Sources: O’Hagan et al., Nature Chemistry , 2015 , 7 , 483–488 http://pqr.pitt.edu/mol/NQRLPDFELNCFHW-UHFFFAOYSA-N http://pqr.pitt.edu/mol/QJAIOCKFIORVFU-UHFFFAOYSA-N http://pqr.pitt.edu/mol/FICBIFYYTQXSEY-UHFFFAOYSA-N A. L. Hickey, C. N. Rowley, J. Phys. Chem. A , 2014 , 118 (20), 3678–3687 | {
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35,391 | Some gases are lighter than others and rise. Why don't they continue going up, leave the atmosphere, and then enter outer space? | The atmosphere actually loses gases to outer space. The average velocity $\bar v$ of gas molecules is determined by temperature $T$.
However, not all the molecules travel with the same velocity.
The probability of finding a molecule with a velocity near $v$ is described by the Maxwell distribution of speeds
$$\begin{align}
f{\left(v\right)}&=4\pi\sqrt{{\left(\frac m{2\pi kT}\right)}^3}v^2\exp\left(-\frac{m{v^2}}{2kT}\right)\\[6pt]
&=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right)
\end{align}$$
where $m$ is the mass of the molecule, $k$ is the Boltzmann constant, $M$ is the molar mass of the gas, and $R$ is the molar gas constant. Individual molecules may reach escape velocity $v_\mathrm e$ and thus be able to leave the atmosphere. Escape velocity is the minimum velocity that is sufficient for an object to escape from the gravitational attraction of a massive body. For a planet, the escape velocity may be estimated by using the formula
$${v_\mathrm e}=\sqrt{\frac{2Gm_\text{planet}}r}$$
where $G$ is the gravitational constant, $m_\text{planet}$ is the mass of the planet, and $r$ is the distance from the centre of mass of the planet. Therefore, atmospheric escape depends on the mass of the planet, the temperature of the atmosphere, and the molar mass of the gas. | {
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35,462 | Why does no other element but hydrogen get special names for its isotopes? | Harold Urey and George Murphy used spectroscopy to identify deuterium late in 1931, announcing it at the 1931 Christmas meeting of the American Physical Society. Picking up out of 'From Nuclear Transmutation to Nuclear Fission, 1932-1939" by Per F. Dahl: If anything, the naming of the new isotope proved more problematic than its isolation. At a special session on heavy hydrogen at the general June meeting in 1933 of the APS in Chicago, organized in conjunction with the Century of Progress Exposition, the ensuing discussion on its naming 'threatened to become acrimonious,' according to Francis Aston of the Cavendish Laboratory - the great authority on atomic weight measurements and a guest speaker at the discussion. The argumentation had to do with whether to retain the name 'hydrogen' for the isotope, as Niels Bohr preferred; after all, it was not a new element and had the atomic number 1. Both Gilbert Lewis and Ernest Lawrence opted for 'dygen' for the H$^{2}$ isotope and 'dyon' for its nucleus, wheras Rutherford preferred 'diplogen' and 'diplon' instead. In the end, Urey had the last word, as he was entitled to, settling on 'deuterium' for the isotope and 'deuteron' for its nucleus. Given the heavyweights in the field wanting a separate name for the isotope, even Neils Bohr could not hold back the tide. Again, though, you have to remember that this was in the very early days of nuclear physics. While the existence of the nucleus dated back to Rutherford's experiments pre-WWI, they were still very unclear just what the nucleus was constructed out of. This was all occurring just as Cockroft and Walton were using their new ion accelerator to perform the first human-induced nuclear fission ($^{7}$Li + p -> 2 $^{4}$He), published in the Proceedings of the Royal Society on July 1st, 1932. Alpha, beta, and gamma particles had been identified since the late 1800s, and natural sources were used for nuclear physics experiments. It was readily recognized that a source of energetic protons would be highly desired. Also, remember that 1932 was the year that Chadwick discovered the neutron as a separate entity. Thus, also from Dahl: The discovery of the neutron would soon revise Urey's view of his deuteron as consisting of two protons and an electron. However, as late as October 1932, it was still not obvious among physicists that the deuteron consists of 'one proton and one Chadwick neutron' instead of two protons and one electron. Against this rapid-fire series of experiments on nuclear physics, it really isn't unusual that deuterium got a separate name - it was still not universally appreciated that it wasn't something weird. Than, once it was accepted, it was too useful to readily distinguish experiments with protons vs deuterons vs tritons to revert their names back to normal isotope nomenclature. | {
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37,486 | If protons have a positive charge and electrons have a negative charge,
can we add up several protons and electrons together to create a new element, without adding neutrons to hold the nuclei together ? | Yes and no. Elements are defined by the number of protons only. It does not matter if (say) a carbon nucleus has six or seven (or eight) neutrons, they will all react the same.* With that, to create new elements, you would need to get up to some 115 or so protons fused together. But there is a reason for neutrons: all the positively charged protons in the nucleus repel each other electrostatically so neutrons are there to stabilise the nucleus — you can think of it as cushioning protons apart. For each element there is a set number of neutrons that will create a stable nucleus. E.g. for hydrogen (one proton) it is zero or one, for oxygen it is eight, for carbon six or seven, for tin it can be $62, 64, 65, 66, 67, 68, 69, 70, 72$ or $74$. Nuclei with other neutron numbers decay radioactively on half-life scales from femtoseconds to millions of years. I would expect no-neutron nuclei (non-hydrogen) to be on the very short edge of that spectrum (although I’m no radiochemist). *: There are reactions in certain systems that depend a lot on hydrogen being $\ce{^1H}$ rather than $\ce{^2H}$ or $\ce{D}$ because they rely on the tunneling effect whose probability decreases with mass (I think by the factor $m^2$). That’s also why one shouldn’t drink $\ce{D2O}$ | {
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37,641 | According to Wikipedia , The $\ce{C60}$ molecule is extremely stable, [ 26 ] withstanding high temperatures and high pressures. The exposed surface of the structure can selectively react with other species while maintaining the spherical geometry. [ 27 ] Atoms and small molecules can be trapped within the molecule without reacting . Smaller fullerenes than $\ce{C60}$ have been distorted so heavily they're not stable, even though $\ce{M@C28}$ is stable where $\ce{M\,=\,Ti, Zr, U}$. Some of us have heard and learned about the "rules" of aromaticity: The molecule needs to be cyclic , conjugated , planar and obey Huckel's rule (i.e. the number of the electrons in $\pi$-system must be $4n+2$ where $n$ is an integer). However, I'm now very skeptical to these so-called rules: The cyclic rule is violated due to a proposed expansion of aromaticity. (See what is Y-aromaticity? ) The must-obey-Huckel rule is known to fail in polycyclic compounds. Coronene figure 1 and pyrene figure 2 are good examples with 24 and 16 $\pi$ electrons, respectively. Again, Huckel fails in sydnone . The rule tells you that it's aromatic, while it's not. The planar rule is not a rule at all. We're talking about "2D" aromaticity when we're trying to figure out the $n$ in $4n+2$. The "3D" rule is as following: In 2011, Jordi Poater and Miquel Solà, expended the rule to determine when a fullerene species would be aromatic. They found that if there were $2n^2+2n+1$ π-electrons, then the fullerene would display aromatic properties. - Wikipedia This would mean $\ce{C60}$ is not aromatic, since there is no integer $n$ for which $2n^2+2n+1 = 60$. On the other hand, $\ce{C60-}$ is ($n = 5$). But then this rule strikes me as peculiar because then no neutral or evenly-charged fullerene would be aromatic. Furthermore, outside the page for the rule, Wikipedia never explicitly states that fullerene is not aromatic , just that fullerene is not super aromatic . And any info on superaromaticity is unavailable or unhelpful to me; including the Wikipedia "article" on that topic. So, is $\ce{C60}$ aromatic? Why, or why not? | Aromaticity is not binary, but rather there are degrees of aromaticity. The degree of aromaticity in benzene is large, whereas the spiro-aromaticity in [4.4]nonatetraene is relatively small. The aromaticity in naphthalene is not twice that of benzene. Aromaticity has come to mean a stabilization resulting from p-orbital (although other orbitals can also be involved) overlap in a pi-type system. As the examples above indicate, the stabilization can be large or small. Let's consider $\ce{C_{60}}$: Bond alternation is often taken as a sign of non-aromatic systems. In $\ce{C_{60}}$ there are different bond lengths, ~1.4 and 1.45 angstroms. However, this variation is on the same order as that found in polycyclic aromatic hydrocarbons, and less than that observed in linear polyenes. Conclusion: aromatic, but less so than benzene. Magnetic properties are related to electron delocalization and are often used to assess aromaticity. Both experiment and calculations suggest the existence of ring currents (diamagnetic and paramagnetic) in $\ce{C_{60}}$. Conclusion: Although analysis is complex, analysis is consistent with at least some degree of aromaticity. Reactivity - Substitution reactions are not possible as no hydrogens are present in $\ce{C_{60}}$. When an anion or radical is added to $\ce{C_{60}}$ the electron(s) are not delocalized over the entire fullerene structure. However, most addition reactions are reversible suggesting that there is some extra stability or aromaticity associated with $\ce{C_{60}}$. Conclusion: Not as aromatic as benzene Resonance energy calculations have been performed and give conflicting results, although most suggest a small stabilization. Theoretical analysis of the following isodesmic reaction $$\ce{C_{60} + 120 CH4 -> 30 C2H4 + 60 C2H6}$$ suggested that it only took half as much energy to break all of the bonds in $\ce{C60}$ compared to the same bond-breaking reaction with the appropriate number of benzenes. Conclusion: Some aromatic stabilization, but significantly less than benzene. This brief overview suggests that $\ce{C_{60}}$ does display properties that are consistent with some degree of aromatic stabilization, albeit less than that found with benzene. | {
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37,663 | Why is a burn to the skin caused by steam more serious than a burn caused by the same amount of boiling water at the same temperature? The temperature is the same, which implies that the kinetic energy of the particles in both the steam and the boiling water are the same; and yet the steam is more dangerous. Why so? | Let’s consider the following cases: getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water on one’s skin getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ air on one’s skin getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water vapour on one’s skin With the slightly irrealistic assumption that all of these liberate all their thermal energy to the skin while cooling down to $40\,\mathrm{°C}$. 1 st case Assuming isobaric conditions (constant pressure), the heat energy liberated from $100\,\mathrm{°C}$ water that is cooled to $40\,\mathrm{°C}$ can be calculated as follows: $$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 4.5\,\mathrm{kJ}$$ 2 nd case A similar formula holds true. However, now we are considering a gas all the way, not a liquid. (And ‘air’ is technically not a single substance, but we can work with it.) $$\Delta Q = \Delta T \cdot C_p(\mathrm{air}) \cdot n(\mathrm{air})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 29.19 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 1.7\,\mathrm{kJ}$$ 3 rd case Unlike the previous cases, we start off with steam, which will condense to water. Therefore, we need to add the heat of vapourisation to the equation. $$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O}) + \Delta H_{\mathrm{vap}} \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol} + 40.66 \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 45.2\,\mathrm{kJ}$$ You will note that the third case delivers a lot more heat energy than the first two cases. (Part of the simplifications I used means that I grossly overestimated the heat transferred by hot air. Which is also why one can put one’s hands under a hair dryer and be fine with it.) | {
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39,109 | According to this link , benzene is able to insert itself into the human DNA. It isn't an authoritative source and appears to be quite biased, so I'm wondering if there's any truth to this. The resonance hybrid structure of the benzene molecule causes it to have electrostatic potential, be very stable, and insert itself into human DNA. Taken together, these molecular and electrical characteristics of benzene make it carcinogenic to humans. | I will start my answer with a preface that the website linked to in the question is a pseudoscience website (and I am glad that it has vanished from the face of this earth, only accessible via the Wayback Machine). These people use scientific-sounding jargon that sounds impressive to the lay reader, but any actual scientist will know that it is simply rubbish. The objective is usually to fear monger, or to promote some product of theirs that has no special properties. Unfortunately, these seem to be becoming more and more prevalent recently. However, the premise of the question is sound: benzene itself is known to cause cancer in humans. It is listed as a Class 1 carcinogen , meaning that there is evidence for carcinogenicity in humans . As with many things in molecular biology, we do not actually know the full mechanism by which benzene causes cancer. However, there have been a number of papers written on the topic: some of them are linked at the bottom of this post. Exactly which metabolite of benzene causes cancer is not known. It is likely to be a combination of them. The first step is oxidation of benzene by the enzyme cytochrome P450 2E1 (CYP2E1) to benzene oxide, which exists in equilibrium with oxepin (via a 6π electrocyclic reaction). The cytochrome P450 enzymes are found in the liver, and their role is generally to insert oxygen atoms into molecules, making them more polar and water-soluble, so that they can be excreted. For more information about this process, see Drug metabolism on Wikipedia . These undergo further reactions, catalysed by various other enzymes, to give a large range of metabolites: In particular, The benzoquinones have been shown to inhibit DNA topoisomerase II , an enzyme that makes temporary cuts in double-stranded DNA in order to "unwind" DNA that has been entangled. It plays a key role in many cellular processes such as DNA replication, DNA repair, and chromosome segregation (during cell division); therefore, inhibition may lead to chromosome breakage or failure to segregate. The same quinones can undergo a process called redox cycling, where they undergo an enzymatic reaction in which a single electron is added to them to form a radical anion. These species are then released, and react with molecular oxygen $\ce{O2}$ to give the superoxide anion, $\ce{O2^-}$... and then the process repeats itself. The buildup of $\ce{O2^-}$ (and other reactive oxygen species) leads to oxidative stress and DNA damage. ( E , E )-muconaldehyde has been recently shown to form an adduct with two molecules of deoxyguanosine, i.e. the guanine bases in DNA. Here, R represents the rest of the deoxyribose sugar. Intra- or inter-chain links can be formed via this method, which then lead to inaccurate replication or chromosomal aberrations. A mechanism was proposed in reference 3. It is not reproduced here but it is not difficult to imagine how such a reaction might happen: nitrogen atoms in guanine are nucleophilic, and literally every carbon in muconaldehyde is electrophilic. The literature on the topic contains much more information than I can write in here. Reference 4 is a relatively recent review on the topic, which would be a decent starting point to find further information. Regardless of what mechanism it is, one thing is certain: benzene itself and its molecular properties are not likely to be the cause of its carcinogenicity. It is almost certain that multiple metabolites of benzene are the culprits. References Chen, H.; Eastmond, D. A. Topoisomerase inhibition by phenolic metabolites: a potential mechanism for benzene's clastogenic effects. Carcinogenesis (Oxford) 1995, 16 (10), 2301–2307. DOI:10.1093/carcin/16.10.2301 Rappaport, S. M.; Kim, S.; Lan, Q.; Li, G.; Vermeulen, R.; Waidyanatha, S.; Zhang, L.; Yin, S.; Smith, M. T.; Rothman, N. Human benzene metabolism following occupational and environmental exposures. Chem.-Biol. Interact. 2010, 184 (1–2), 189–195. DOI:10.1016/j.cbi.2009.12.017. Harris, C. M.; Stec, D. F.; Christov, P. P.; Kozekov, I. D.; Rizzo, C. J.; Harris, T. M. Deoxyguanosine Forms a Bis-Adduct with E , E -Muconaldehyde, an Oxidative Metabolite of Benzene: Implications for the Carcinogenicity of Benzene. Chem. Res. Toxicol. 2011, 24 (11), 1944–1956. DOI:10.1021/tx2002838. Hartwig, A. The role of DNA repair in benzene-induced carcinogenesis. Chem.-Biol. Interact. 2010, 184 (1–2), 269–272. DOI:10.1016/j.cbi.2009.12.029. | {
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39,315 | Why do pyrrole and furan have dipoles oriented in different directions? | Both pyrrole and furan have a lone pair of electrons in a p-orbital, this lone pair is extensively delocalized into the conjugated pi framework to create an aromatic 6 pi electron system. Where pyrrole and furan significantly differ is that, in pyrrole there is an $\ce{N-H}$ bond lying in the plane of the ring and directed away from the ring whereas in furan, there is a full lone pair of electrons in roughly the same position. The localized lone pair of electrons pointing away from the ring has a very significant effect on the dipole vector and is enough to cause the observed reversal in dipole moment direction between furan and pyrrole. | {
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39,645 | I have learnt that internal energy, $U$, is a state function and it only depends on temperature... So if $\Delta T = 0$ then $\Delta U = 0$. However when I was studying exothermic and endothermic reactions it was written in my textbook that at a constant temperature and pressure, $\Delta U$ is negative for exothermic reactions. How is that possible? Shouldn't $\Delta U$ be zero since the temperature is constant? | TL;DR: Do not just memorise thermodynamics equations! And if you have an issue with the equations $\Delta U = 0$ or $\Delta H = 0$ for an isothermal process, read on... The first problem You said that an exothermic reaction corresponds to $\Delta U < 0$ . This is not true. It is defined by $\Delta H < 0$ . However, we should still discuss this issue because $H$ is also a state function and in some cases $H$ is dependent only on temperature. How to study thermodynamics This kind of issue in thermodynamics frequently crops up here, and it is a very common mistake amongst students to indiscriminately use equations that they have learnt. So, you have an equation that says $\Delta U = 0$ for an isothermal process, i.e. one at constant $T$ . The important question here is not "what is the equation" or "what is the answer"! Instead, you should be asking yourself "how do I derive this result", and the answer will naturally follow. This should really apply to everything you do - how can you expect to apply a formula that you do not actually understand? Furthermore, if you are able to actually understand how an equation comes about, then you don't need to memorise it - you can derive it. For example, what's the point of memorising $w = -nRT\ln(V_2/V_1)$ , along with the conditions, if you can simply obtain the equation from the very definition of work $đw = -\int p \,\mathrm{d}V$ by substituting in $p = nRT/V$ and integrating? What's more, the fact that you actually substituted in the ideal gas law should tell you something about the conditions that accompany the equations - for one, it's only applicable to ideal gases. What about when you reach the equation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$ and happily memorise it, only to find out later that there are actually three more equations that look exactly like it? Are you going to memorise all four? No - you need to know where they come from. Internal energy In general, the internal energy $U$ of a substance is a function of at least two variables, for example $T$ and $p$ . However, it can be shown using statistical mechanics that for an ideal gas, $$U = n\left(\frac{3NRT}{2} + U_0\right)$$ Here, $N$ is the number of atoms in one molecule of the gas (for example, $N = 2$ for $\ce{H2}$ , and $N = 3$ for $\ce{H2O}$ ). $U_0$ is the molar internal energy at absolute zero (this contains terms such as electronic energy), and crucially, this depends on the exact identity of the gas. This result is one example of the equipartition theorem in action (although the theorem itself is more powerful). We do not need to concern ourselves with its derivation now. The point is that: this equation can only be applied to a fixed amount of a fixed ideal gas at a time . For example: if I have one mole of $\ce{H2}$ at $298\mathrm{~K}$ , it is going to have a different internal energy from one mole of $\ce{Ar}$ at $298\mathrm{~K}$ , even though the amount of substance and the temperature are the same. What does this mean? Firstly, it means that for anything that is not an ideal gas, you cannot assume that $U = U(T)$ and therefore you cannot assume that $\Delta T = 0$ implies $\Delta U = 0$ . Let's say I compress $10\mathrm{~g}$ of water by reversibly increasing the pressure from $1\mathrm{~atm}$ to $10\mathrm{~atm}$ , at a constant temperature of $323\mathrm{~K}$ . Is the change in internal energy equal to zero? No , because water is not an ideal gas. Secondly, it means that if the amount of ideal gas changes in any way, $U$ is going to change. Let's say I have a balloon filled with $1\mathrm{~mol}$ of argon gas, and I add a further $2\mathrm{~mol}$ argon gas into the balloon, all at a fixed temperature of $298\mathrm{~K}$ . Is the change in internal energy equal to zero? No , in fact it will triple because you are changing $n$ from $1\mathrm{~mol}$ to $3\mathrm{~mol}$ ! Lastly, it means that if the chemical composition is changing in any way, you are automatically not allowed to say that $\Delta T = 0$ implies $\Delta U = 0$ . And this is the case even if all reactants and products are ideal gases. Why not? Let's look at this reaction, and let's assume that everything there behaves as an ideal gas, and let's say that the reaction vessel is kept at a constant temperature of $300\mathrm{~K}$ . $$\ce{H2 (g) + Cl2 (g) -> 2HCl (g)}$$ Let's write an expression for the internal energy of the reactants. On the left-hand side, we have: $$U_\mathrm{reactants} = \left[\frac{3}{2}RT + U_0(\ce{H2})\right] + \left[\frac{3}{2}RT + U_0(\ce{Cl2})\right]$$ On the right-hand side, we have: $$U_\mathrm{products} = 3RT + 2U_0(\ce{HCl})$$ So, is $\Delta U$ equal to zero? No; even though the multiples of $RT$ cancel out with each other, the values of $U_0$ for the reacting species are different, which makes $$\Delta U = 2U_0(\ce{HCl}) - U_0(\ce{H2}) - U_0(\ce{Cl2}) \neq 0.$$ Enthalpy Let's go back to our (closed) system where there is only one ideal gas and it's not undergoing any chemical reaction. Within this system, we can safely say that if $\Delta T = 0$ , then $\Delta U = 0$ . So: $$\begin{align}
\Delta H &= \Delta (U + pV) \\
&= \Delta U + \Delta (pV) \\
&= \Delta (pV) \qquad \\
&= \Delta (nRT) \\
&= nR \Delta T \\
&= 0
\end{align}$$ We haven't made any assumptions here apart from the ideality of the gas (the fact that $n$ is constant came from the fact that it is a closed system). So, for an ideal gas, $\Delta T = 0$ implies $\Delta H = 0$ . However, we did use the fact that $\Delta U = 0$ . Therefore, what I said about the cases above where $\Delta U \neq 0$ also applies here. Regarding the terms endothermic and exothermic , they are used to describe chemical reactions . And for chemical reactions, there is absolutely no way you can say that constant temperature implies $\Delta U = 0$ or $\Delta H = 0$ . That's why we can actually use the terms endothermic and exothermic. If every single chemical reaction had $\Delta H = 0$ , life would be rather plain! | {
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40,506 | In inorganic chemistry, when is the prefix meta used? (as in metaborate and metasulphite )
What about the terms pyro and ortho (as in orthophosphorous acid )? For example, I recently came to know about the prefixes per and hypo which are used to refer to the highest and lowest oxidation states of an element respectively (like perchlorate and hypochlorite ). Are there any terms and conventions I should be aware of? | In the older literature, the prefixes “hypo-”, “per-”, “ortho-”, “meta-”, and “pyro-” were used in some cases for the distinction between different acids with the same characteristic element. The prefix “hypo-” was used to denote a lower oxidation state, and the prefix “per-” was used to designate a higher oxidation state. (The prefix “per-” is not be confused with the prefix “peroxo-”.) $\ce{HClO}$ hypochlorous acid $\ce{HClO2}$ chlorous acid $\ce{HClO3}$ chloric acid $\ce{HClO4}$ perchloric acid The prefixes “ortho-” and “meta-” were used to distinguish acids differing in the formal content of water. $\ce{H3BO3}$ orthoboric acid $\ce{(HBO2)$_n$}$ metaboric acid $\ce{H4SiO4}$ orthosilicic acid $\ce{(H2SiO3)$_n$}$ metasilicic acid The prefix “pyro-” was used to designate an acid that is formally formed by removing one molecule of water from two molecules of an ortho-acid. $\ce{H4P2O7}$ pyrophosphoric acid $(\ce{2 H3PO4 -> H4P2O7 + H2O})$ . Various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations. The Second Edition (1970 Rules) of the IUPAC Nomenclature of Inorganic Chemistry (published in 1971) still included all of the above-mentioned prefixes; however, the approved use was limited to only some selected retained names. The further development and the current situation of the nomenclature are well explained in @Jan’s thorough answer . | {
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40,567 | I have learnt that the standard free energy change is related to the equilibrium constant of a reaction by,
$$\Delta G^\circ = -RT \ln K$$ Here, does $K$ refer to $K_p$ or $K_c$? Also, please give me the derivation of this formula. On the net, I saw that we can use either $K_p$ or $K_c$. But doesn't that give two different free energy values? Moreover, if the formula was derived for gaseous reactions (using $K_p$), how can we just extend this to other reactions saying that we can use $K_c$ as well? | As noted in this previous question , the correct definition of the equilibrium constant $K$ depends on activities. If you are interested in the derivation of the equation $\Delta G^\circ = -RT \ln K$ (which requires "proper" thermodynamics), read Philipp's answer to that question. For a reaction $$0 \longrightarrow \sum_i \nu_i \ce{J}_i$$ (this is a fancy way of writing the reaction which makes sure that the stoichiometric coefficients $\nu_i$ of the reactants are negative), the equilibrium constant is defined as follows: $$K = \prod_i a_i^{\nu_i}$$ where $a_i$ is the activity of the species $\ce{J}_i$ at equilibrium . (Note that since $\nu_i$ for reactants is negative, the terms for the reactants will appear in the denominator of a fraction.) The question is, why are $K_c$ and $K_p$ taught at an introductory level despite them not actually being the equilibrium constant $K$ ? The answer is likely partly because it is difficult to introduce the idea of an activity, but also because $K_c$ and $K_p$ are good approximations to the actual equilibrium constant, $K$ . To see why, we have to look at the mysterious quantity called the activity and see how it is related to the concentration of a solution, or the partial pressure of a gas. If we assume certain ideality conditions, then we can write the activity of a gas as: $$a_i = \frac{p_i}{p^\circ}$$ where $p_i$ is the partial pressure of the gas and $p^\circ$ is the standard pressure, defined by IUPAC to be equal to $1~\mathrm{bar}$ . Therefore, $a_i$ is nothing but the partial pressure of the gas in bars, but without the units. If we therefore define $K_p$ as follows: $$K_p = \prod_i p_i^{\nu_i}$$ then, as long as the pressures you use are in bars, the numerical value of $K_p$ will be the same as that of $K$ , the true equilibrium constant. As such, if you ignore the fact that you are taking the natural logarithm of a quantity with units, the value of $\Delta G^\circ$ that you obtain will be "correct" . Similarly, for a solution, we can write: $$a_i = \frac{c_i}{c^\circ}$$ where $c_i$ is the concentration of the species and $c^\circ$ is the standard concentration, again defined by IUPAC to be equal to $1~\mathrm{mol~dm^{-3}}$ . So, the same idea applies. If you keep all concentrations in units of $\mathrm{mol~dm^{-3}}$ , then the numerical value of $K_c$ will match that of $K$ . An example Consider the Haber process (all species gaseous): $$\ce{N2 + 3H2 <=> 2NH3}$$ We would define the "true" equilibrium constant as: $$K = \frac{(a_{\ce{NH3}})^2}{(a_{\ce{N2}})(a_{\ce{H2}})^3}$$ and you would define $K_p$ as: $$K = \frac{(p_{\ce{NH3}})^2}{(p_{\ce{N2}})(p_{\ce{H2}})^3}$$ Let's say (just as an example - the numbers are not correct!) that at a certain temperature $T$ , we have the equilibrium partial pressures: $$\begin{array}{c|c|c}
p_{\ce{NH3}} & p_{\ce{N2}} & p_{\ce{H2}} \\ \hline
20~\mathrm{bar} & 50~\mathrm{bar} & 50~\mathrm{bar}
\end{array}$$ Then you would have $K_p = 6.4 \times 10^{-5}~\mathrm{bar^{-2}}$ . Now you can see why it is not correct to use $K_p$ in the equation above: you would have $$\Delta G^\circ = -RT \ln(6.4 \times 10^{-5}~\mathrm{bar^{-2}})$$ which makes absolutely no sense since you can't take the logarithm of a unit. In a simplistic treatment you would just ignore the fact that a unit existed and just take the logarithm of the number. However, you will not run into that problem if you use the "true" equilibrium constant, as you really should . If we assume ideality, i.e. $a_i = p_i/p^\circ$ , then we find that $K = 6.4 \times 10^{-5}$ . Since activities are dimensionless, you will find that $K$ is dimensionless and you can now take the logarithm. As I mentioned earlier, the numerical quantity of $K$ is exactly equivalent to $K_p$ as long as you keep your pressures in units of bars. I have seen some questions where students are requested to determine $K_c$ for a gas-phase reaction using the ideal gas law and hence $\Delta G$ : $$\begin{align}
pV &= nRT \\
c = \frac{n}{V} &= \frac{p}{RT}
\end{align}$$ The value of $c_i$ and $p_i$ will therefore differ by a factor of $RT$ . Therefore, the numerical values of $K_c$ and $K_p$ in gas-phase reactions will, in general, differ. (If you have the same number of moles of gas on both sides, then the factors of $RT$ will cancel out and numerically $K_c = K_p$ , but that is not always the case.) As established earlier, because of the equation for the activity of an ideal gas, only the use of $K_p$ will lead to the correct numerical value for $\Delta G^\circ$ . If a question requires you to do so, then do it. Just bear in mind that it is not only wrong in terms of taking the natural logarithm of a quantity with units; it will also give you a wrong numerical result for $\Delta G^\circ$ . | {
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40,598 | Though it is highly unlikely, has any carbon compound been found to make an ionic bond and to exhibit ionic properties? | Why unlikely? Ionic compounds of carbon have been known for ages. There are ionic carbides ($\ce{Al4C3}$, $\ce{CaC2}$, etc.), graphite intercalation compounds like $\ce{KC8}$, ionic derivatives of fullerenes and more. Come to think of it, common $\ce{CaCO3}$ is certainly ionic and at the same time a compound of carbon, but this is most likely not what you want. | {
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40,968 | As the title implies, what is the molecular basis of cyanide toxicity? I did some searching around at the CDC and it only states that it prevents cells from using oxygen. I also read how it could take as little as $131~\mathrm{ppm}$ to kill. How can a simple $\ce{CN-}$ ion do so much damage? | Cyanide is a pretty good ligand for coordination compounds. The electron pair on carbon (which, incidentally, also carries the Lewis structure’s formal charge) is located in the HOMO — much like as in $\ce{CO}$, whose molecular orbitals can be found in this answer by Martin (replace oxygen with nitrogen to arrive at $\ce{CN-}$) — making it a good Lewis base and a good σ donor. It is remarkable for being a strong-field ligand, i.e. one that often creates low-spin complexes even for 3d-metals (which generally prefer high-spin if there are no competing reasons for low-spin). While one could draw analogies to $\ce{CO}$ or even $\ce{NO+}$ whose π-antibonding orbitals are strongly Lewis acidic and participate in metal to ligand backbonding, it turns out that for cyanide the main reason is a high covalency of the metal–ligand-σ-bond due to the relevant orbitals being very similar in energy. Thus, cyanide is unique in that it both stabilises higher oxidation states by being an anion and low-spin complexes due to its binding modalities. One might think, in analogy to $\ce{K4[Fe(CN)6]}$, that cyanide preferentially binds to haemoglobin’s iron(II) centre and thereby inhibits oxygen transport, but that is in fact not the case as this centre requires a less-oxidised iron(II). Oxygen (due to the resulting redox-mechanism) and carbon monoxide are much better ligands for haemoglobin. What cyanide does strongly inhibit is cytochrome c oxidase, the last enzyme in the respiratory cycle which collects electrons from four cytochrome c ’s and transfers them to an oxygen molecule to create two water molecules: $$\ce{4 Fe^2+-cytochrome~$c$ + 4 H+ + O2 -> 4 Fe^3+-cytochrome~$c$ + 2 H2O}$$ This is done by a plethora of metal centres consisting of two haems, to cytochromes and two copper centres. I can’t tell you exactly which one, but there will be at least one which needs accessable iron(III) at some point in the catalytic cycle to which $\ce{CN-}$ can bind well and strongly inhibiting the catalysis of the enzyme. Since the inhibited process is directly responsible for cellular respiration by transferring electrons to molecular oxygen (and being the only enzyme dealing with molecular oxygen in the respiratory chain), inhibition of this protein will cause the breakdown of the entire respiratory chain and lead to cellular suffocation. Considering that an amount of $200~\mathrm{mg}$ can be toxic according to Wikipedia , and also considering that a human being contains some 10 trillion cells also according to Wikipedia , that’s still some 400 million cyanide ions per cell demonstrating the efficiency of binding and inhibition. Note that popular antidotes also directly address the fact that cyanide is a strong ligand: One antidote, hydroxycobalamin, presents the cyanide ion with a $\ce{Co^{III}}$ centre and a weakly bound, easily displaced hydroxido ligand. Hydroxide is displaced by cyanide creating cyanidocobalamin, a variant of vitamin B 12 that the body has no problems dealing with. It is ejected via the kidneys. Injected into the circulatory system, it scavenges cyanides before they can enter cells. Another antidote, also injected into the circulatory system, is 4-dimethylaminophenol (not to be abbreviated as DMAP to prevent confusion with 4-dimethylamino pyridine ). It is an oxidising agent designed to oxidise haemoglobin’s iron(II) to methaemoglobin (containing iron(III)) thereby creating a much better acceptor for cyanide. Since there is a lot of iron(II) (a quick estimate with $8~\mathrm{mmol/l}$ gives $40~\mathrm{mmol}$ haemoglobin in human blood) in the blood for oxygen transport, small amounts of this can be sacrificed to capture cyanide before that reaches the respiratory cycle. | {
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41,274 | What is more acidic: $\ce{D3O+}$ in $\ce{D2O}$ or $\ce{H3O+}$ in $\ce{H2O}$ and why? I think it's $\ce{D3O+}$ in $\ce{D2O}$ as I saw somewhere that this property is used in mechanistic studies (the inverse isotope effect), but I need a proper explanation. EDIT: I found a statement about this in Clayden: Water $\ce{H2O}$ is a better solvating agent for $\ce{H_3O^+}$ than $\ce{D2O}$ is for $\ce{D_3O^+}$ because O–H bonds are longer than O–D bonds hence $\ce{D_3O^+}$ in $\ce{D2O}$ is stronger than $\ce{H_3O^+}$ in $\ce{H2O}$. That is the origin of the inverse solvent isotope effect. This is in contrast with all the answers given yet, so I am pretty confused. | For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined. The entire point of the above reference is that $\ce{H3O+ + H2O <=> H2O + H3O+}$ (which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same. $\ce{D3O+ + D2O <=> D2O + D3O+}$ would also correspond to an equilibrium constant of 1 So when Clayden and the OP write $\ce{D3O+}$ in $\ce{D2O}$ is stronger than $\ce{H3O+}$ in $\ce{H2O}$ it is wrong for the above reason. Two genuine thermodynamic equilibriums are $\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$ Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater). But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$ , not $\ce{D2O}$ . $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water. As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$ , The ionization constant of heavy water ( $\ce{D2O}$ ) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$ . Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species. See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data. Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$ , meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles. According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$ , is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$ ." My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network. Also the equilibrium constant for $\ce{H2O + H2DO+ <=> HDO + H3O+}$ can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751 Explanation of Normal/Inverse Solvent Isotope Effect For a kinetic normal/inverse solvent isotope effect
there will be a reactant and transition state. If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect". Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur. Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters. | {
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41,280 | Pure liquid can be supercooled below its freezing point. Conversely, impurities in the liquid will cause a freezing point depression. How then is the freezing point accurate? | For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined. The entire point of the above reference is that $\ce{H3O+ + H2O <=> H2O + H3O+}$ (which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same. $\ce{D3O+ + D2O <=> D2O + D3O+}$ would also correspond to an equilibrium constant of 1 So when Clayden and the OP write $\ce{D3O+}$ in $\ce{D2O}$ is stronger than $\ce{H3O+}$ in $\ce{H2O}$ it is wrong for the above reason. Two genuine thermodynamic equilibriums are $\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$ Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater). But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$ , not $\ce{D2O}$ . $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water. As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$ , The ionization constant of heavy water ( $\ce{D2O}$ ) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$ . Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species. See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data. Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$ , meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles. According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$ , is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$ ." My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network. Also the equilibrium constant for $\ce{H2O + H2DO+ <=> HDO + H3O+}$ can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751 Explanation of Normal/Inverse Solvent Isotope Effect For a kinetic normal/inverse solvent isotope effect
there will be a reactant and transition state. If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect". Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur. Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters. | {
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41,862 | In Brady's Molecular Nature of Matter , I read that $\Delta_\mathrm{r} G^{\circ}$ is $\Delta_\mathrm{r} G$ at $25~^\circ\mathrm{C}$. But later, it gives a value for $\Delta_\mathrm{r} G^\circ$ at another temperature. What is the meaning of $\Delta_\mathrm{r} G^\circ$? Specifically: Does it only refer to $25~^\circ\mathrm{C}$? Does the $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T \Delta_\mathrm{r} S^\circ$ equation only refer to $T=298~\mathrm{K}$? If at $25~^\circ\mathrm{C}$, a system that has reached equilibrium ($\Delta_\mathrm{r} G=0$), and thus $Q=K$ $\Delta_\mathrm{r} G^\circ=0$ $\Delta_\mathrm{r} G^\circ=0=-RT\ln K$ $RT\ln K=0$ $\ln K=0$ $K=1$ But doesn't $K$ vary from reaction to reaction? It shouldn't always equal 1, should it? | Short answer Does it need to be at $25~^\circ\mathrm{C}$? No. $\Delta_\mathrm{r} G^\circ$ can be defined at any temperature you wish to define it at, since the standard state does not prescribe a particular temperature. If you change the temperature, $\Delta_\mathrm{r} G^\circ$ will change. Does $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ$ always use $T = 298~\mathrm{K}$? No. You use whatever temperature you are running your reaction at. (...maths...) Yes, at equilibrium, $\Delta_\mathrm{r} G = 0$ and $Q = K$. However, everything after the first bullet point is wrong. You cannot conclude that $\Delta_\mathrm{r} G^\circ = 0$, nor can you conclude that $K = 1$. The equation $\Delta_\mathrm{r} G^\circ = -RT \ln K$ does not analogously translate into $\Delta_\mathrm{r} G = -RT \ln Q$! The accurate relation is: $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$ Setting $Q = K$ and $\Delta_\mathrm{r} G = 0$ in this equation does not tell you anything about the value of $K$. In fact, if you try doing it, all you will find out is that $\Delta_\mathrm{r} G^\circ = -RT \ln K$ - no surprises there! Long answer Any book that writes that $\Delta_\mathrm{r} G^\circ$ is the "special case" of $\Delta_\mathrm{r} G$ at $T = 298~\mathrm{K}$ is wrong . The Gibbs free energy of a system is defined as follows: $$G = H - TS$$ Under constant temperature and pressure (from now on, I will just assume constant $T$ and $p$ without stating it), all systems will seek to minimise their Gibbs free energy. Equilibrium is reached when $G$ is minimised. When $G$ is at a minimum, any infinitesimal change in $G$, i.e. $\mathrm{d}G$, will be $0$. Therefore, this is equivalent to saying that the condition for chemical equilibrium is $\mathrm{d}G = 0$. Clearly, we need a way to relate this quantity $\mathrm{d}G$ to the actual reactants and products that are in the system. This can be done by using the Maxwell relation (see any physical chemistry text for details): $$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + \sum_i \mu_i\,\mathrm{d}n_i$$ Under constant $T$ and $p$, $\mathrm{d}p = \mathrm{d}T = 0$ and therefore $$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i$$ where $\mu_i$ is the chemical potential of species $i$, defined as a partial derivative: $$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{n_{j\neq i}}$$ So, we now have a refined condition for equilibrium: $$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i = 0$$ We can go further by noting that the values of $\mathrm{d}n_i$ for different components $i$, $j$, etc. are not unrelated. For example, if we have a reaction $i + j \longrightarrow k$, then for each mole of $i$ that is consumed, we must also use up one mole of $j$; this means that $\mathrm{d}n_i = \mathrm{d}n_j$. This can be formalised using the idea of a stoichiometric coefficient $\nu_i$, which is defined to be positive for products and negative for reactants. For example, in the reaction $$\ce{3H2 + N2 -> 2NH3}$$ we have $\nu_{\ce{H2}} = -3$, $\nu_{\ce{N2}} = -1$, and $\nu_{\ce{NH3}} = 2$. By stoichiometry, if $1.5~\mathrm{mol}$ of $\ce{H2}$ is consumed, then $1~\mathrm{mol}$ of $\ce{NH3}$ has to be produced. We could write $\Delta n_{\ce{H2}} = -1.5~\mathrm{mol}$ and $\Delta n_{\ce{NH3}} = 1~\mathrm{mol}$. These quantities are proportional to their stoichiometric coefficients: $$\frac{\Delta n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{-1.5~\mathrm{mol}}{-3} = 0.5~\mathrm{mol} = \frac{1~\mathrm{mol}}{2} = \frac{\Delta n_{\ce{NH3}}}{\nu_{\ce{NH3}}}$$ The quantity $0.5~\mathrm{mol}$ is a constant for all chemical species $\ce{J}$ that participate in the reaction, and it is called the "extent of reaction" and denoted $\Delta \xi$ (that is the Greek letter xi ). If the reaction is going forward, then $\Delta \xi$ is positive, and if the reaction is going backwards, then $\Delta \xi$ is negative. If we generalise the above result, we can write $$\Delta \xi = \frac{\Delta n_i}{\nu_i}$$ and if we make $\Delta n_i$ smaller and smaller until it becomes an infinitesimal, then: $$\begin{align}
\mathrm{d}\xi &= \frac{\mathrm{d}n_i}{\nu_i} \\
\mathrm{d}n_i &= \nu_i\,\mathrm{d}\xi
\end{align}$$ If we go back to our condition for equilibrium, we can substitute in the above to get: $$\mathrm{d}G = \sum_i \mu_i\nu_i\,\mathrm{d}\xi = 0$$ Now, $\mathrm{d}\xi$ is no longer dependent on $i$, since we have established already that $\Delta \xi$ (and by extension $\mathrm{d}\xi$) is a constant for all chemical species. So, we can "divide through" by it to get: $$\Delta_\mathrm{r} G \equiv \frac{\mathrm{d}G}{\mathrm{d}\xi} = \sum_i \mu_i\nu_i = 0$$ where $\Delta_\mathrm{r} G$ is defined to be $\mathrm{d}G/\mathrm{d}\xi$. Note that $\Delta_\mathrm{r} G$ is an intensive property and has units of $\mathrm{kJ~mol^{-1}}$, since $\mathrm{d}\xi$ has units of $\mathrm{mol}$. This ensures that the units we use are consistent: since we know that $\Delta_\mathrm{r}G = \Delta_\mathrm{r} G^\circ + RT\ln Q$, $\Delta_\mathrm{r}G$ must have the same units as $RT$. How do we interpret the physical significance of $\Delta_\mathrm{r} G$, or in other words, what does it even mean? There are two ways, each based on a different mathematical expression. We have $\Delta_\mathrm{r}G = \sum \nu_i \mu_i$. This means that $\Delta_\mathrm{r}G$ is simply the difference between the chemical potentials of the products and the reactants, weighted by their stoichiometric coefficients. For the reaction $\ce{3H2 + N2 -> 2NH3}$, we have: $$\Delta_\mathrm{r} G = \sum_i \mu_i\nu_i = 2\mu_{\ce{NH3}} - 3\mu_{\ce{H2}} -\mu_{\ce{N2}}$$ We have $\Delta_\mathrm{r}G = \mathrm{d}G/\mathrm{d}\xi$. This means that it is the slope of a curve of $G$ against $\xi$: Note that up to this point, we have not stipulated any particular temperature, pressure, amounts of species present, or any conditions whatsoever. We have only said that the temperature and pressure must be constant. It is important to realise that $\Delta_\mathrm{r}G$ is a well-defined quantity at all $T$, all $p$, and all possible values of $n_i, n_j, \cdots$! The shape of the curve will change when you vary the conditions. However, no matter what the curve looks like, it is always possible to find its gradient ($= \Delta_\mathrm{r}G$) at a particular point. What exactly, then, is $\Delta G^\circ$? It is just a special case of $\Delta G$, where all the reactants and products are prepared in a standard state . According to IUPAC, the standard state is defined as: For a gas: pure ideal gas when the pressure $p$ is equal to the standard pressure $p^\circ$. For a liquid or solid: pure liquid or solid at $p = p^\circ$ For a solution: ideal solution when the concentration $c$ is equal to the standard concentration $c^\circ$. $p^\circ$ is most commonly taken to be $\pu{1 bar}$, although older texts may use the value $\pu{1 atm} = \pu{1.01325 bar}$. Since 1982, IUPAC has recommended the value $\pu{1 bar}$ for the standard pressure ( Pure Appl. Chem. 1982, 54 (6), 1239–1250; DOI: 10.1351/pac198254061239 ). However, depending on the context, a different value of $p^\circ$ may prove to be more convenient. Likewise, $c^\circ$ is most commonly – but not necessarily – taken to be $\pu{1 mol dm-3}$. Note that in the above definitions, no temperature is specified. Therefore, by defining the standard Gibbs free energy, we are fixing a particular value of $p$, as well as particular values of $n_i, n_j, \cdots$. However, the value of $T$ is not fixed. Therefore, when stating a value of $\Delta_\mathrm rG^\circ$, it is also necessary to state the temperature which that value applies to. When a reaction vessel is prepared with all its substances in the standard state, all the components of the system will have an activity of exactly $1$ by definition. Therefore, the reaction quotient $Q$ (which is a ratio of activities) will also be exactly equal to $1$. So, we could also say that $\Delta_\mathrm{r}G^\circ$ is the value of $\Delta_\mathrm{r}G$ when $Q = 1$. Returning to the graph of $G$ against $\xi$ above, we note that at the left-most point, $Q = 0$ since there are only reactants; at the right-most point, $Q \to \infty$ as there are only products. As we move from left to right, $Q$ increases continuously, so there must be a point where $Q = 1$. (In general, the point where $Q = 1$ will not be the same as the equilibrium point.) Since $\Delta_\mathrm{r}G$ is the gradient of the graph, $\Delta_\mathrm{r}G^\circ$ is simply the gradient of the graph at that particular point where $Q = 1$ : The gradient of the graph, i.e. $\Delta_\mathrm{r}G$, will vary as you traverse the graph from left to right. At equilibrium, the gradient is zero, i.e. $\Delta_\mathrm{r}G = 0$. However, $\Delta_\mathrm{r}G^\circ$ refers to the gradient at that one specific point where $Q = 1$. In the example illustrated above, that specific gradient is negative, i.e. $\Delta_\mathrm{r}G^\circ < 0$. Again, I reiterate that the temperature has nothing to do with this. If you were to change the temperature, you would get an entirely different graph of $G$ versus $\xi$. You can still find the point on that graph where $Q = 1$, and the gradient of that graph at the point where $Q = 1$ is simply $\Delta_\mathrm{r} G^\circ$ at that temperature . We have established the qualitative relationship between $\Delta_\mathrm{r} G$ and $\Delta_\mathrm{r} G^\circ$, but it is often useful to have an exact mathematical relation. $\Delta_\mathrm{r} G^\circ$ is exactly the same as $\Delta_\mathrm{r} G$ except for the imposition of the standard state. It follows that if we take the equation $$\Delta_\mathrm{r} G = \sum_i \mu_i \nu_i$$ and impose the standard state, we get $$\Delta_\mathrm{r} G^\circ = \sum_i \mu_i^\circ \nu_i$$ Thermodynamics tells us that $$\mu_i = \mu_i^\circ + RT\ln{a_i}$$ where $a_i$ is the thermodynamic activity of species $i$. Substituting this into the expressions for $\Delta G$ and $\Delta G^\circ$ above, we obtain the result: $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$ where the reaction quotient $Q$ is defined as $$Q = \prod_i a_i^{\nu_i}$$ When equilibrium is reached, we necessarily have $\Delta_\mathrm{r} G = 0$ (see the discussion above). The equilibrium constant $K$ is defined to be the value of $Q$ at equilibrium. Therefore, at equilibrium, $Q = K$. Plugging this into the equation above gives us the famous equation: $$\Delta_\mathrm{r} G^\circ = -RT\ln K$$ Again, no temperature is specified ! In general, $K$ depends on the temperature as well; the relationship is given by the van 't Hoff equation . | {
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41,917 | I've heard and read (multiple times) two theories have have been around for a while, both of them obvious nonsense. One states that pressure and friction make the ice melt (turn to water) where the skater's blade comes in contact with it. This is patent nonsense since a puck that weighs very little, or a pencil that weighs next to nothing, slip along the ice as well as any skater. I also find it hard to believe that the temperature of the ice (several degrees below freezing) would suddenly jump several degrees in the split second that it takes for the skate to come in and out of contact with it. The other theory is that the ice surface itself has the same qualities as water due to some hydrogen molecules still in search of home. This is also nonsense: pour water over a concrete floor and try to skate on it; see how far you'll get. So my question is: why is ice slippery? | This is a question has been one of interest since 1850, and I found a paper that nicely summarizes some of the hypotheses that you yourself listed, as well as what is considered to be true. I've paraphrased it below. Michael Faraday was the first to propose that there exists a film of water on ice that freezes when in contact with other pieces of ice, but remains liquid by itself. Though he performed several experiments that supported his claims, most of which involved sticking two pieces of ice together, his ideas were largely dismissed until C. Gurney also suggested that this film of water on play a role in ice's slipperiness. Gurney hypothesized that molecules, inherently unstable at the surface due to the lack of other molecules above them, migrate into the bulk of the solid until the surface becomes unstable, which prompts the formation of the liquid phase. This came to be accepted, and Charles Hosler compiled very compelling evidence on the the forces required to pull ice spheres apart. He discovered that at vapor pressures below ice saturation, no adhesion occurred below $-4\:^{\circ} \mathrm{C}$. From this they inferred that the expected roughness of the surfaces was removed by the presence of a liquid film whose thickness was sufficient to provide a smooth surface of contact. More experiments were performed trying to determine the exact thickness of the water film at different temperatures, and the phenomenon was found to occur at temperatures even below $-22\:^{\circ} \mathrm{C}$. Here is a depiction of what the surface of ice looks like at a molecular level. Rosenberg, R. Why Is Ice Slippery? Phys. Today Physics Today. 2005, 58, 50–54. | {
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42,535 | Methanol is slightly more acidic than water. Their $\mathrm{p}K_\mathrm{a}$ values, in water, are $15.5$ and $15.7$, respectively. All other aliphatic alcohols, however, are less acidic than water. Is the following reasoning correct? This is my best rationalization; is there anything better or anything that can be added? In both hydroxide and methoxide, we have an $\ce{O}$ bearing negative charge. In the former, we have a hydrogen attached to the $\ce{O}$. In the latter, we have a carbon attached to the $\ce{O}$. Carbon is more electronegative than hydrogen, therefore, carbon should be able to better withdraw electron density via induction from the $\ce{O}$. This is a stabilizing interaction. We often think of methyl groups as inductively donating, but that's molecular profiling. They can inductively donate to a carbocation, partly because a carbocation is highly electronegative, so the carbocation pulls the electrons toward itself pretty well (so there's both pushing and pulling of electrons going on). When you have a methyl group attached to an $\ce{O}$, however, the methyl group is often inductively withdrawing. NMR data supports this. Now the question is why aren't other aliphatic alcohols more acidic than water? In longer-chain aliphatic alkoxides, you don't just have a methyl group attached to the $\ce{O}$ bearing the negative charge - you have a bunch of methyl groups strung together. These lessen the amount of inductive withdrawal that the alpha carbon can do. Each$\ce{-CH2 -}$ unit attached to the alpha $\ce{-CH2 -}$ is somewhat inductive donating to the alpha $\ce{-CH2 -}$. | The relative acidities of these molecules in the aqueous solution $^{[1][2]}$ is: $$\text{methanol > water > ethanol > isopropanol > t-butanol}$$ The relative acidities of these molecules in DMSO $^{[3]}$ is: $$\text{methanol > ethanol > isopropanol > water > t-butanol}$$ The problem with only considering the acidities of these molecules in an aqueous solution is that it assumes only one explanation for acidity. The inversion of the order of acidity in DMSO indicates that there are multiple factors at play. The acidity trend in DMSO is explained by polarizability and the anionic hyperconjugation effect $^{[4]}$ of the conjugate bases of the molecules. The larger size of the alkyl substituents in the conjugate bases allows for increased distribution of negative charge over a larger volume. This reduces the charge density, and thereby the Coulombic repulsion between the conjugate base and the solvent. In DMSO, water is much less acidic than even methanol, which is also consistent with the relative polarizability of a methyl group and a hydrogen atom, but fails to explain the relative acidity of t-butanol and methanol. The trend in alcohol acidity is explained by anionic hyperconjugation, which is weakened in larger alkyl substituents. For methanol and the methoxide anion, the primary interaction occurs between a filled lone-pair ( $n$ ) orbital on oxygen and an unoccupied $\pi^*_\ce{Me}$ orbital of the methyl group, resulting in a two-electron stabilizing effect. $\hspace{3.8cm}$ Deprotonation of the alcohol leads to an increase in the energy of the n orbitals on oxygen, and a decreased energy separation between the interacting orbitals $n$ and $\pi^*_\ce{Me}$ , and hence a greater stabilizing interaction $^{[5]}$ . Although DMSO does solvate the solute to an extent, because is an aprotic solvent, it is incapable of hydrogen bonding with the solute. This allows DMSO acidities better reflect the intrinsic acidities of the molecules. In aqueous solution, however, the conjugate bases are stabilized by hydrogen bonding. Because smaller ions are better solvated by water, water's acidity is increased more than methanol's, but because methanol is intrinsically more acidic due to its polarizability, it is still slightly more acidic than water, even in an aqueous solution. As the size of the alkyl substituent increases, the alcohol becomes weaker due to the decreasing ability of water to solvate the conjugate base. This has a greater effect on the alcohol's acidity than the dispersion of the negative charge, and overrides the intrinsic acidity, inverting the order of acidity. $^{[1]}$ [Evans $\mathrm{p}K_\mathrm{a}$ Table, Alcohols][2] $^{[2]}$ [Wikipedia, Ethanol][3] $^{[3]}$ [Bordwell $\mathrm{p}K_\mathrm{a}$ Table, Alcohols][4] $^{[4]}$ [ACS, Acidities of Water and Methanol in Aqueous Solution and DMSO][5] $^{[5]}$ [ACS, Does a Methyl Substituent Stabilize or Destabilize Anions?][6] | {
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42,922 | Thorium series: Uranium series: Actinium series: Why do all the radioactive decay series terminate at lead isotope? Why not hydrogen? (All pictures were taken from Wikipedia .) | There are four main decay chains for actinides and superheavy elements. This is a simple consequence of the fact that one of the main processes to increase a heavy nucleus' stability is the emission of alpha particles, which have a mass number of 4 ($\ce{^4_2\alpha}$); notice that if you take the isotope's mass number and divide it by 4, the remainder of this division (0, 1, 2 or 3, corresponding to the $4n$, $4n+1$, $4n+2$ and $4n+3$ mass number decay chains respectively) stays constant under alpha, beta or gamma decay. There are decays which change the remainder (neutron emission, proton emission, spontaneous fission, etc) and therefore allow hopping between the main decay chains, but for simplicity these can be approximately ignored for the isotopes which are not too unstable or too heavy. Only three out of these four main decay chains "stop" at lead, so already the statement in the question is incorrect. The $4n$, $4n+2$ and $4n+3$ chains "stop" at $\ce{^208_82Pb}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$, respectively. The $4n+1$ decay chain reaches $\ce{^209_82Pb}$, but this lead isotope is quite short-lived ($t_{1/2}=3.25\ \mathrm{h})$ and decays further to $\ce{^209_83Bi}$ where it "stops" (or at least seemed to until 2003 ). Note that I said the chains "stopped" at these isotopes. That's because in reality they don't. It just happens that $\ce{^208_82Pb}$, $\ce{^209_83Bi}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$ are all very long-lived isotopes, with half-lives billions of times greater than the current age of the Universe. This means that any continued decay is severely bottlenecked at these points, so for almost all practical purposes, the decay chains stop there. However, rigorously speaking, if you were really patient there would be further decays in the sequence. In fact, theoretically the heaviest isotope which is not susceptible to any known mode of spontaneous radioactive decay (except proton decay ) is an isotope of zirconium , $\ce{^92_40Zr}$, so the decay chains actually go on well past lead or bismuth. It just takes an immense amount of time for the decays to happen. | {
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43,088 | Why is cubane called pentacyclo[$4$.$2$.$0$.$0^{2,5}$.$0^{3,8}$.$0^{4,7}$]octane? I am especially intrigued by the pentacyclo part. | According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) , polycyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system as follows. The main ring of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible (here: 8 atoms). The main bridge is the bridge that includes as many of the atoms as possible that are not included in the main ring (here: 0 atoms). Bridges other than the main bridge are called secondary bridges : P-23.2.6.1 Naming polycyclic alicyclic hydrocarbons Rings not designated by the bicyclic system described above (…) are defined by citing the number of atoms in each secondary bridge as an arabic number. The locants of the two attachment points of each secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower first) separated by a comma. The numbers indicating independent secondary bridges (bridges that connect atoms of the bicyclic system) are cited
in decreasing order. (…) P-23.2.6.1.1 The prefixes ‘tricyclo’, ‘tetracyclo’, etc., in place of ‘bicyclo’, indicate the number of rings in the polyalicyclic system. The number of rings is equal to the number of bond cuts necessary to transform the polycyclic system into an acyclic skeleton, unbranched or branched. Therefore, the correct systematic name is ‘pentacyclo[4.2.0.0 2,5 .0 3,8 .0 4,7 ]octane’. However, the retained name ‘cubane’ is used in general nomenclature and as preferred IUPAC name (PIN). | {
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44,260 | What is the difference between D and L configuration, and + and −? My textbook says they are two different things.
It also says that the correct way to name glucose is D(+)-glucose. Could someone please explain what D, L and +, − represent, and why they are different? | The D-L system corresponds to the configuration of the molecule: spatial arrangement of its atoms around the chirality center. While (+) and (-) notation corresponds to the optical activity of the substance, whether it rotates the plane of polarized light clockwise (+) or counterclockwise (-). D-L system tells us about the relative configuration of the molecule, compared to the enantiomers of glyceraldehyde as the standard compound. Compounds with the same relative configuration as (+)-glyceraldehyde are assigned the D prefix, and those with the relative configuration of (-)-glyceraldehyde are given the L prefix. It's kind of another way to tell the configuration of molecules beside the Cahn–Ingold–Prelog convention (R/S system), with little difference. (D-L system labels the whole molecule, while R/S system labels the absolute configuration of each chirality center.) In short, the D-L system doesn't have direct connection to (+)/(-) notation. It only relates the stereochemistry of the compound with that of glyceraldehyde, but says nothing about its optical activity. We may have compound with same relative configuration as (+)-glyceraldehyde (thus, it's given the D prefix), yet it rotates the polarized light counterclockwise (-), such as D-(-)-ribose. And also, don't confuse the D-L system with d- and l- naming. d- and l- is the exact same with (+) and (-) notation. Additional explanation D-L system (also called Fischer–Rosanoff convention) is mainly used for naming α-amino acids and sugars. It compares the relative configurations of molecules to the enantiomers of glyceraldehyde. This convention is still in common use today. Rosanoff in 1906 selected the enantiomeric glyceraldehydes as the point of reference [1] ; any sugar derivable by chain lengthening from what is now known as (+)-glyceraldehyde (or named D-glyceraldehyde) belongs to the D series. In other words, we used a D to designate the sugars that degrade to (+)-glyceraldehyde and an L for those that degrade to (-)-glyceraldehyde. In assigning the D and L configurations of sugars, we could direcly look for the OH group of the bottom asymmetric carbon in the Fischer projection. If it's located on the right, we designate it with D, and vice versa, since they would have the same relative configurations with glyceraldehyde for the bottom asymmetric carbon. Reference [1]: IUPAC and IUBMB. Joint Commission On Biochemical Nomenclature. Nomenclature of Carbohydrates . 1996 , 7. | {
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44,272 | This is quoted from Computational Chemistry by Errol J. Lewars ' chapter 2's Stationary Points & Normal-Mode
Vibrations: ZPE : \begin{align}\mathbf H &=\begin{bmatrix}
\dfrac{\partial ^2 E}{\partial q_1 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_3} & \dots & \dfrac{\partial ^2 E}{\partial q_1 \partial q_9} \\
\dfrac{\partial ^2 E}{\partial q_2 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_3} & \dots & \dfrac{\partial ^2 E}{\partial q_2 \partial q_9}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\dfrac{\partial ^2 E}{\partial q_9 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_3} & \dots & \dfrac{\partial ^2 E}{\partial q_9 \partial q_9}
\end{bmatrix}\\ \\ &=\underset{\mathbf P}{\begin{bmatrix}
q_{11} & q_{12} & q_{13} & \dots & q_{19} \\
q_{21} & q_{22} & q_{23} & \dots & q_{29} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
q_{91} & q_{92} & q_{93} & \dots & q_{99}
\end{bmatrix}} \underset{\mathbf k}{\begin{bmatrix}
k_{1} & 0 & 0 & \dots & 0 \\
0 & k_{2} & 0 & \dots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \dots & k_{9}
\end{bmatrix}}\mathbf P^{-1}\end{align} The 9 × 9 Hessian for a tri-atomic molecule
(three Cartesian coordinates for each atom) is decomposed by diagonalization into a P
matrix whose columns are “ direction vectors ” for the vibrations whose force constants
are given by the $\bf k$ matrix. Actually, columns 1, 2 and 3 of $\bf P$ and the corresponding $k_1,k_2,k_3$ of $\bf k$ refer to translational motion of the molecule (motion of the whole
molecule from one place to another in space); these three “force constants” are nearly
zero. Columns 4, 5 and 6 of $\bf P$ and the corresponding $k_4,k_5,k_6$ of $\bf k$ refer to rotational motion about the three principal axes of rotation, and are also nearly zero.
Columns 7, 8 and 9 of $\bf P$ and the corresponding $k_7,k_8,k_9$ of $\bf k$ are the direction
vectors and force constants, respectively, for the normal-mode vibrations: and
refer to vibrational modes 1, 2 and 3, while the 7th, 8th, and 9th columns of $\bf P$ are
composed of the x, y and z components of vectors for motion of the three atoms in mode
1 (column 7), mode 2 (column 8), and mode 3 (column 9). I couldn't conceive what actually $\bf P$ implies; it is, of course , the eigenvector matrix of the Hessian $\bf H $ but what does it represent? Lewars wrote that columns of $\bf P$ are actually direction vectors ; now what did he mean by direction vector ? I thought $q$ represents the geometric coordinate; I got $k$ is the force-constant; but what are those series of $q$ s in $\mathbf P\;?$ Are the $q$ s in $\bf P$ coordinates of the atoms in molecules? If so, aren't they always changing since they are vibrating, rotating...? Can anyone please explain this to me what those $q$ s actually represent? What did Lewars want to mean by direction vectors ? | The D-L system corresponds to the configuration of the molecule: spatial arrangement of its atoms around the chirality center. While (+) and (-) notation corresponds to the optical activity of the substance, whether it rotates the plane of polarized light clockwise (+) or counterclockwise (-). D-L system tells us about the relative configuration of the molecule, compared to the enantiomers of glyceraldehyde as the standard compound. Compounds with the same relative configuration as (+)-glyceraldehyde are assigned the D prefix, and those with the relative configuration of (-)-glyceraldehyde are given the L prefix. It's kind of another way to tell the configuration of molecules beside the Cahn–Ingold–Prelog convention (R/S system), with little difference. (D-L system labels the whole molecule, while R/S system labels the absolute configuration of each chirality center.) In short, the D-L system doesn't have direct connection to (+)/(-) notation. It only relates the stereochemistry of the compound with that of glyceraldehyde, but says nothing about its optical activity. We may have compound with same relative configuration as (+)-glyceraldehyde (thus, it's given the D prefix), yet it rotates the polarized light counterclockwise (-), such as D-(-)-ribose. And also, don't confuse the D-L system with d- and l- naming. d- and l- is the exact same with (+) and (-) notation. Additional explanation D-L system (also called Fischer–Rosanoff convention) is mainly used for naming α-amino acids and sugars. It compares the relative configurations of molecules to the enantiomers of glyceraldehyde. This convention is still in common use today. Rosanoff in 1906 selected the enantiomeric glyceraldehydes as the point of reference [1] ; any sugar derivable by chain lengthening from what is now known as (+)-glyceraldehyde (or named D-glyceraldehyde) belongs to the D series. In other words, we used a D to designate the sugars that degrade to (+)-glyceraldehyde and an L for those that degrade to (-)-glyceraldehyde. In assigning the D and L configurations of sugars, we could direcly look for the OH group of the bottom asymmetric carbon in the Fischer projection. If it's located on the right, we designate it with D, and vice versa, since they would have the same relative configurations with glyceraldehyde for the bottom asymmetric carbon. Reference [1]: IUPAC and IUBMB. Joint Commission On Biochemical Nomenclature. Nomenclature of Carbohydrates . 1996 , 7. | {
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44,296 | How can this be explained on basis of Lewis Acid-Base Reaction ? $\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$ References: https://en.wikipedia.org/wiki/Fluorine#Industrial Google Books reference (click me) | The D-L system corresponds to the configuration of the molecule: spatial arrangement of its atoms around the chirality center. While (+) and (-) notation corresponds to the optical activity of the substance, whether it rotates the plane of polarized light clockwise (+) or counterclockwise (-). D-L system tells us about the relative configuration of the molecule, compared to the enantiomers of glyceraldehyde as the standard compound. Compounds with the same relative configuration as (+)-glyceraldehyde are assigned the D prefix, and those with the relative configuration of (-)-glyceraldehyde are given the L prefix. It's kind of another way to tell the configuration of molecules beside the Cahn–Ingold–Prelog convention (R/S system), with little difference. (D-L system labels the whole molecule, while R/S system labels the absolute configuration of each chirality center.) In short, the D-L system doesn't have direct connection to (+)/(-) notation. It only relates the stereochemistry of the compound with that of glyceraldehyde, but says nothing about its optical activity. We may have compound with same relative configuration as (+)-glyceraldehyde (thus, it's given the D prefix), yet it rotates the polarized light counterclockwise (-), such as D-(-)-ribose. And also, don't confuse the D-L system with d- and l- naming. d- and l- is the exact same with (+) and (-) notation. Additional explanation D-L system (also called Fischer–Rosanoff convention) is mainly used for naming α-amino acids and sugars. It compares the relative configurations of molecules to the enantiomers of glyceraldehyde. This convention is still in common use today. Rosanoff in 1906 selected the enantiomeric glyceraldehydes as the point of reference [1] ; any sugar derivable by chain lengthening from what is now known as (+)-glyceraldehyde (or named D-glyceraldehyde) belongs to the D series. In other words, we used a D to designate the sugars that degrade to (+)-glyceraldehyde and an L for those that degrade to (-)-glyceraldehyde. In assigning the D and L configurations of sugars, we could direcly look for the OH group of the bottom asymmetric carbon in the Fischer projection. If it's located on the right, we designate it with D, and vice versa, since they would have the same relative configurations with glyceraldehyde for the bottom asymmetric carbon. Reference [1]: IUPAC and IUBMB. Joint Commission On Biochemical Nomenclature. Nomenclature of Carbohydrates . 1996 , 7. | {
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44,341 | A friend recently sent me the following chemistry joke: As a lover of puns, I'd like to enjoy it, but I think that it is actually false the way it is phrased. A solution is defined as: "a homogeneous mixture composed of two or more substances" Alcohol alone would not constitute a mixture of multiple substances and would therefore not be a solution. But in English the word "alcohol" is also commonly used to refer to many recreational drinks containing ethyl alcohol in addition to other types of molecules. Are there any common examples of alcoholic beverages which are solutions in the scientific sense that would make the pun valid, such as liquor, wine, or beer? | Technically, alcohol is the name of a class of organic compounds containing one or several hydroxyl groups. Colloquially, the term "alcohol" is understood as you have described: A solution (of varying degree of purity) of ethanol and water. Pure ethanol is impossible to create via the traditional method of purification (atmospheric distillation) as the water-ethanol mixture is azeotropic:¹ An azeotrope [...] is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. This happens because, when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture. Source: Wikipedia This is why a chemist would never think "100% pure ethanol" when you mention "alcohol". If you check for example this vendor page for the purest ethanol (>99.5%) I could find, you see various substances listed as impurities: <0.2% Water <0.1% Methanol Other organic substances, such as isopropanol and acetone Concerning your second question, I would argue that many distilled spirits (vodka, whisky, and cognac just to name a few) are solutions of ethanol, water, and various flavour compounds. Some alcoholic beverages, such as unfiltered beer and wine, usually also contain solid particles and would as such be correctly termed "suspensions". ¹ I wonder now whether it would be possible to produce 100% ethanol from another mixture somehow. The problem is probably that every common starting mixture already has some amount of water. But synthesizing ethanol in, say, an ionic liquid under an inert atmosphere and then distilling it away might be an interesting approach. | {
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45,255 | If we are given $\ce{CH3NC}$ and told to convert it into $\ce{CH3CN}$ by doing some chemical reactions.
My attempt
According to my textbook there are no chemical reaction of isocyanide conpounds. I tried reduction of isocyanide giving a secondary amine. Then I don't know how to convert a secondary amine to cyanide compound.
Hints are too appreciated.
Edit:
I have got a solution but I don't know whether the reaction are spontaneous. I have written these reaction on the basis of theoretical knowledge I don't know whether they are spontaneous or not. Correct me if I am wrong somewhere. | Just heat it up: Isonitriles can be thermally rearranged to nitriles. $$\ce{R-NC ->[\Delta] R-CN}$$ Please have a look at some references, such as: Michael J. S. Dewar, M. C. Kohn, Ground states of $\sigma$-bonded molecules. XVI. Rearrangement of methyl isocyanide to acetonitrile, J. Am. Chem. Soc. , 1972 , 94 , 2704-2706 DOI Michael Meier, Barbara Mueller, Christoph Ruechardt, The isonitrile-nitrile rearrangement. A reaction without a structure-reactivity relationship, J. Org. Chem. , 1987 , 52 , 648-652 DOI UPDATE As far as the sequence outlined in the question is concerned: The synthesis of methyl formamide from methyl isonitrile is pretty much the opposite of how things are done in real life . Isonitriles, used in the Passerini and the Ugi reaction, have shown to be available by dehydration of formamides, e.g. using $\ce{POCl3}$. For an example with cyclohexyl isonitrile, see Org. Synth. , 1961 , 41 , 13. N-Methyl formamide is far from being unstable! Release of methyl amine, which, by the way, has a BP around -6 °C might be possible under alkaline conditions. Making methanol from methylamine, again, is a rather pointless mind game. In real life, methylamine is produced by reacting methanol with ammonia at hight temperatures in the presence of acidic catalysts. You definitely want to make methyl chloride in the lab! With a BP around -24 °C it's very easy to handle, and the GHS classification with H351 (Suspected of causing cancer) and H373 (May cause damage to organs through prolonged or repeated exposure), this is a real gem. Yes, I'm kidding! Don't do it! For a nucleophilic substitution with cyanide, the high volatility of methyl chloride again probably doesn't make your life easier, but just shorter. In summary, my suggestion is simple: Forget about the proposed synthesis! It is laborious, difficult to handle in the lab, and an economical nightmare. | {
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45,267 | Combustion, in chemical terms, is a reaction with a certain molecule and oxygen, and it produces, energy, $\ce{CO2}$, and $\ce{H2O}$ (That's how I learned it at least). So what makes a fire'smokey'? I had always thought that smoke was simply $\ce{CO2}$, but read that it was actually the result of incompletely 'combusted' material, yet I don't fully understand what this means. Is smoke made from material that starts to combust but stops before it finishes, or does it never even begin combustion? What would the chemical reaction be of a wood (Or other similar) fire that produces a lot of smoke? And finally, is there a certain element or compound that gives smoke its 'smokey' smell? | Just heat it up: Isonitriles can be thermally rearranged to nitriles. $$\ce{R-NC ->[\Delta] R-CN}$$ Please have a look at some references, such as: Michael J. S. Dewar, M. C. Kohn, Ground states of $\sigma$-bonded molecules. XVI. Rearrangement of methyl isocyanide to acetonitrile, J. Am. Chem. Soc. , 1972 , 94 , 2704-2706 DOI Michael Meier, Barbara Mueller, Christoph Ruechardt, The isonitrile-nitrile rearrangement. A reaction without a structure-reactivity relationship, J. Org. Chem. , 1987 , 52 , 648-652 DOI UPDATE As far as the sequence outlined in the question is concerned: The synthesis of methyl formamide from methyl isonitrile is pretty much the opposite of how things are done in real life . Isonitriles, used in the Passerini and the Ugi reaction, have shown to be available by dehydration of formamides, e.g. using $\ce{POCl3}$. For an example with cyclohexyl isonitrile, see Org. Synth. , 1961 , 41 , 13. N-Methyl formamide is far from being unstable! Release of methyl amine, which, by the way, has a BP around -6 °C might be possible under alkaline conditions. Making methanol from methylamine, again, is a rather pointless mind game. In real life, methylamine is produced by reacting methanol with ammonia at hight temperatures in the presence of acidic catalysts. You definitely want to make methyl chloride in the lab! With a BP around -24 °C it's very easy to handle, and the GHS classification with H351 (Suspected of causing cancer) and H373 (May cause damage to organs through prolonged or repeated exposure), this is a real gem. Yes, I'm kidding! Don't do it! For a nucleophilic substitution with cyanide, the high volatility of methyl chloride again probably doesn't make your life easier, but just shorter. In summary, my suggestion is simple: Forget about the proposed synthesis! It is laborious, difficult to handle in the lab, and an economical nightmare. | {
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46,337 | For context: I got a question asking, "Which of the following alkaline earth metals do not give flame colour?". I quickly marked $\ce{Be}$ and $\ce{Mg}$ and got negative marks. The following is a quote from my book (emphasis in the original): The elements of Group 2 include beryllium, magnesium, calcium, strontium, barium and radium. These elements with the exception of beryllium are commonly known as the alkaline earth metals . These are so called because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth's crust. As you can see, it's stated that beryllium is not an alkaline earth metal! Is beryllium an alkaline earth metal or not? | There is some disagreement in usage among authors, but IUPAC standard nomenclature approves calling beryllium an alkaline earth metal , as explained on page 51 of IUPAC's last Red Book . In fact, all the elements belonging to group 2, $\ce{Be,Mg,Ca,Sr,Ba,Ra}$, are called alkaline earth metals with IUPAC's approval. Other common traditional names approved by IUPAC are alkali metals for the elements of group 1 except hydrogen, i.e. $\ce{Li,Na,K,Rb,Cs,Fr}$; halogens for $\ce{F,Cl,Br,I,At}$ in group 17,whose only member excluded from such a designation is $\ce{Uus}$ (now $\ce{Ts}$ ), and noble gases for all the elements of group 18 except $\ce{Uuo}$ (now $\ce{Og}$ ), i.e. $\ce{He,Ne,Ar,Kr,Xe,Rn}$. | {
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46,479 | In many organic reactions that I have seen, running the reaction at $\mathrm{-78\ ^\circ C}$ seems to be quite a popular choice, but I've never seen any explanation for this. What makes this temperature such a popular choice? | Dry ice (solid carbon dioxide) sublimes at −78 °C. Dry ice and acetone are a common cold bath for chemical reactions. The melting point of acetone is -95 °C so the bath never gets cold enough to freeze the acetone. The bubbling of the carbon dioxide gas as the dry ice sublimes keeps the cold bath well stirred. Typically, though, the temperature in the flask with an ongoing reaction is at least about 5 °C higher than the one in the cooling bath. (Except if using a thermocouple, working at a scale of 10...25 mL and less often implies that the thermometer is in cooling bath, and not in the reaction mixture. Thus, this temperature difference isn't recorded.) | {
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46,942 | I have seen benzene being depicted in two different ways: For the right-hand structure, does the inscribed circle represent the movement of double bonds or electrons? | The circle, if anything, represents the inability of our everyday physical intuition to cope with the quantum phenomena. See, you would often encounter those two pictures with "double bonds this way" and "double bonds that way", intended to give a vague impression that the molecule switches quickly between the two, but that's not true. It does not switch; it just stays there, in some sense "halfway between" these structures. There is no movement of double bonds, and in fact there are no double bonds. (Also, there is no movement of electrons either. They just sit there on their molecular orbitals, which by itself is an approximation, though a pretty decent one). All $\ce{C-C}$ bonds in benzene are the same; they are neither double nor ordinary. Read the Wikipedia page on aromaticity , or just look at this picture from it: That's what the circle stands for. | {
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47,049 | I have a wonderful reaction of marble chips, $\ce{CaCO3}$, with hydrochloric acid, $\ce{HCl}$, and carbon dioxide was released beautifully (fast, large volume, easy to measure and makes good visual effect too). But there is no reaction between $\ce{CaCO3}$ and $\ce{H2SO4}$. Why not? | Your marble chips react on the surface. In the case of hydrochloric acid, the resulting salt, calcium chloride, is highly soluble in the acid, dissolves and provides further attack to the (new) surface. With sulfuric acid, the highly insoluble calcium sulfate is formed on the surface of the marble chip. With other words: Calcium sulfate acts like a protective layer. | {
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47,056 | I'm wondering why exactly the single bond between two sulfur atoms is stronger than that of two oxygen atoms. According to this page , an $\ce{O-O}$ bond has an enthalpy of $142~\mathrm{kJ~mol^{-1}}$, and a $\ce{S-S}$ bond in $\ce{S8}$ an enthalpy of $226~\mathrm{kJ~mol^{-1}}$. This one reports the $\ce{S-S}$ bond enthalpy to be $268~\mathrm{kJ~mol^{-1}}$, but I'm not sure which molecule they mean, or how they measured it. Anyway, it's still higher than that of $\ce{O-O}$. Searching the Net, the only justification I could find was something similar to concepts they apply in VSEPR, like in this Yahoo Answers thread with such remarkable grammar. Quoting the answer, which might have borrowed some stuff from a high school textbook, due to small size the lone pair of electrons on oxygen atoms repel the bond pair of O-O bond to a greater extent than the lone pair of electrons on the sulfur atoms in S-S bond....as a result S-S bond (bond energy=213 kj/mole)is much more stronger than O-O(bond energy = 138 kj/mole) bond $\ldots$ Other variations of the same argument can be seen here , but it doesn't make sense, since one couldn't apply the same argument to $\ce{O=O}$ and $\ce{S=S}$. The first reference documents the $\ce{S=S}$ and $\ce{O=O}$ bond enthalpies to be $425$ and $494~\mathrm{kJ~mol^{-1}}$, respectively. It's a bit shaky, and I'm looking for a solid explanation using MO or VB, or anything else that actually works. So, why is an $\bf\ce{S-S}$ single bond stronger than $\bf\ce{O-O}$, despite $\bf\ce{O=O}$ being obviously stronger than $\bf\ce{S=S}$? | TL;DR: The $\ce{O-O}$ and $\ce{S-S}$ bonds, such as those in $\ce{O2^2-}$ and $\ce{S2^2-}$, are derived from $\sigma$-type overlap. However, because the $\pi$ and $\pi^*$ MOs are also filled, the $\pi$-type overlap also affects the strength of the bond, although the bond order is unaffected. Bond strengths normally decrease down the group due to poorer $\sigma$ overlap. The first member of each group is an anomaly because for these elements, the $\pi^*$ orbital is strongly antibonding and population of this orbital weakens the bond. Setting the stage The simplest species with an $\ce{O-O}$ bond would be the peroxide anion, $\ce{O2^2-}$, for which we can easily construct an MO diagram. The $\mathrm{1s}$ and $\mathrm{2s}$ orbitals do not contribute to the discussion so they have been neglected. For $\ce{S2^2-}$, the diagram is qualitatively the same, except that $\mathrm{2p}$ needs to be changed to a $\mathrm{3p}$. The main bonding contribution comes from, of course, the $\sigma$ MO. The greater the $\sigma$ MO is lowered in energy from the constituent $\mathrm{2p}$ AOs, the more the electrons are stabilised, and hence the stronger the bond. However, even though the $\pi$ bond order is zero, the population of both $\pi$ and $\pi^*$ orbitals does also affect the bond strength. This is because the $\pi^*$ orbital is more antibonding than the $\pi$ orbital is bonding. (See these questions for more details: 1 , 2 .) So, when both $\pi$ and $\pi^*$ orbitals are fully occupied, there is a net antibonding effect. This doesn't reduce the bond order; the bond order is still 1. The only effect is to just weaken the bond a little. Comparing the $\sigma$-type overlap The two AOs that overlap to form the $\sigma$ bond are the two $\mathrm{p}_z$ orbitals. The extent to which the $\sigma$ MO is stabilised depends on an integral, called the overlap , between the two $n\mathrm{p}_z$ orbitals ($n = 2,3$). Formally, this is defined as $$S^{(\sigma)}_{n\mathrm{p}n\mathrm{p}} = \left\langle n\mathrm{p}_{z,\ce{A}}\middle| n\mathrm{p}_{z,\ce{B}}\right\rangle = \int (\phi_{n\mathrm{p}_{z,\ce{A}}})^*(\phi_{n\mathrm{p}_{z,\ce{B}}})\,\mathrm{d}\tau$$ It turns out that, going down the group, this quantity decreases . This has to do with the $n\mathrm{p}$ orbitals becoming more diffuse down the group, which reduces their overlap. Therefore, going down the group, the stabilisation of the $\sigma$ MO decreases, and one would expect the $\ce{X-X}$ bond to become weaker. That is indeed observed for the Group 14 elements. However, it certainly doesn't seem to work here. That's because we ignored the other two important orbitals. Comparing the $\pi$-type overlap The answer for our question lies in these two orbitals. The larger the splitting of the $\pi$ and $\pi^*$ MOs, the larger the net antibonding effect will be. Conversely, if there is zero splitting, then there will be no net antibonding effect. The magnitude of splitting of the $\pi$ and $\pi^*$ MOs again depends on the overlap integral between the two $n\mathrm{p}$ AOs, but this time they are $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. And as we found out earlier, this quantity decreases down the group; meaning that the net $\pi$-type antibonding effect also weakens going down the group. Putting it all together Actually, to look solely at oxygen and sulfur would be doing ourselves a disservice. So let's look at how the trend continues. $$\begin{array}{|c|c|c|c|}
\hline
\mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} & \mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} \\
\hline
\ce{O} & 144 & \ce{F} & 158 \\
\ce{S} & 266 & \ce{Cl} & 243 \\
\ce{Se} & 192 & \ce{Br} & 193 \\
\ce{Te} & 126 & \ce{I} & 151 \\
\hline
\end{array}$$
(Source: Prof. Dermot O'Hare's web page. ) You can see that the trend goes this way: there is an overall decrease going from the second member of each group downwards. However, the first member has an exceptionally weak single bond. The rationalisation, based on the two factors discussed earlier, is straightforward. The general decrease in bond strength arises due to weakening $\sigma$-type overlap. However, in the first member of each group, the strong $\pi$-type overlap serves to weaken the bond. I also added the Group 17 elements in the table above. That's because the trend is exactly the same, and it's not a fluke! The MO diagram of $\ce{F2}$ is practically the same as that of $\ce{O2^2-}$, so all of the arguments above also apply to the halogens. How about the double bonds? In order to look at the double bond, we want to find a species that has an $\ce{O-O}$ bond order of $2$. That's not hard at all. It's called dioxygen, $\ce{O2}$, and its MO scheme is exactly the same as above except that there are two fewer electrons in the $\pi^*$ orbitals. Since there are only two electrons in the $\pi^*$ MOs as compared to four in the $\pi$ MOs, overall the $\pi$ and $\pi^*$ orbitals generate a net bonding effect. (After all, this is where the second "bond" comes from.) Since the $\pi$-$\pi^*$ splitting is much larger in $\ce{O2}$ than in $\ce{S2}$, the $\pi$ bond in $\ce{O2}$ is much stronger than the $\pi$ bond in $\ce{S2}$. So, in this case, both the $\sigma$ and the $\pi$ bonds in $\ce{O2}$ are stronger than in $\ce{S2}$. There should be absolutely no question now as to which of the $\ce{O=O}$ or the $\ce{S=S}$ bonds is stronger! | {
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47,204 | I was told by my chemistry teacher that $\ce{HCN}$ smells like almonds. She then went on to tell a story about how some of her students tried to play a prank on her by pouring almond extract down the drain to make her think that they had inadvertently created $\ce{HCN}$ gas. She said that she knew that it wasn't $\ce{HCN}$ because if she had smelled the almond scent, then she would have already been dead. I never asked her, but how do people know $\ce{HCN}$ smells like almonds if they would die before they knew what it smells like? | The odour threshold for hydrogen cyanide $(\ce{HCN})$ is in fact quite a bit lower than the lethal toxicity threshold. Data for $\ce{HCN}$ can be found in many places, but here and here are a couple of good references. That subset of the human population that can detect bitter almonds do so at a threshold of $0.58$ to $\pu{5 ppm}$ . The lethal exposure dose is upwards of $\pu{135 ppm}$ . That's a whole $\pu{100 ppm}$ range in which to detect and report the fragrant properties. | {
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48,316 | My textbook mentions that chlorination of alkanes by $\ce{SO2Cl2}$ is preferred over $\ce{SOCl2}$, but I didn't understand why. Could someone explain which of the two reagents I should use in what situations, and briefly tell me why? There is a similar question here: How does SO2Cl2 react with alcohols? However the answers there did not completely clarify my doubts. | The reactivity patterns of $\ce{SO2Cl2}$ and $\ce{SOCl2}$ are quite different. $\ce{SOCl2}$ is a good electrophile, and can be thought of as a source of $\ce{Cl-}$ ions. These ions can go on to react in their typical nucleophilic fashion. $\ce{SO2Cl2}$ however is often a $\ce{Cl2}$ source, as it readily decomposes giving off sulfur dioxide. Usually much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by $\ce{Cl2}$ gas (or something that makes it in solution) happens by a radical mechanism i.e. $\ce{Cl.}$ not $\ce{Cl-}$. | {
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48,971 | $\ce{F-}$ is a hard base since it is small and relatively polarizable. Both $\ce{Cr^6+}$ and $\ce{W^6+}$ are hard acids, but shouldn't $\ce{CrF6}$ be favored since it would be a smaller hard acid? | The answer has to do with two things. Note that HSAB theory is dubious at best and doesn't have very useful predictive power, so I am going to avoid talking about it. (1) The accessibility of the high +6 oxidation state In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove multiple electrons one after another. On the other hand In W, the 5d orbitals have two inner radial nodes . The two pockets of electron density close to the nucleus shield the bulk of the electron density, which is far away from the nucleus. So, when you continuously remove electrons from W, the remaining 5d electrons don't drop in energy so rapidly. This is partly reflected in the ionisation energies for Cr and W (data from NIST Atomic Spectra Database ), all units are in eV). Element IE1 IE2 IE3 IE4 IE5 IE6 Chromium 6.76651 16.486305 30.959 49.16 69.46 90.6349 Tungsten 7.86403 16.37 26.0 38.2 51.6 64.77 For an alternative perspective, here are Latimer diagrams for Cr, Mo and W (from Shriver et al., Inorganic Chemistry 6ed.); you can see that the stability of the higher oxidation states increases going from Cr to Mo to W (i.e. the reduction potentials are lower) This trend is hardly limited to the Group 6 elements. $\ce{OsO4}$ features Os(VIII), whereas the highest oxidation state of iron is Fe(VI), which is already extremely difficult to achieve. Likewise Au(III) is pretty normal but Cu hardly goes above Cu(II). (2) Sterics Cr(VI) can be made, but the only Cr(VI) compounds that we know of are paired with the oxide ion, viz. $\ce{CrO3}$ , $\ce{CrO4^2-}$ , $\ce{Cr2O7^2-}$ . So there is likely also an argument here that chromium, being a smaller metal, cannot quite fit six fluorines around it in the same way that tungsten can. It can fit four oxygens, but that's about it. | {
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49,793 | As we all know that, Any poison is nothing but a chemical compound. And as discussed in the question: Chemicals-do-have-an-expiry-date ! So, my question is: Is poison still poisonous after its 'expiration date'? and Is every poison always as poisonous as just after it was synthesised ? please feel free to update with the suitable tag(s). | It depends on what the poison is.
If we take the colloquial use of the word and include toxins and venoms, many are things like proteins that will certainly denature or otherwise degrade, eventually becoming harmless. e.g. tetrodotoxin, ricin, botulinum, etc.
I would expect that type of poison to have the shortest shelf-life as they are relatively fragile. Many other poisons are small organic molecules.
These can often be degraded by oxidation in air, exposure to UV, hydrolysis etc. and would include things like nicotine and nerve agents like sarin and VX.
Many nerve agents, have shelf lives of a few years and research has actually been done to extend them for use in munitions. Several metals are known to be poisonous (like lead, mercury, and cadmium) and are problematic because they are toxic in not only their elemental forms, but also in inorganic and organic compounds.
There may be a great difference in toxicity of the different forms, (see elemental mercury vs methylmercury), but most forms remain at least somewhat toxic.
These may last a very long time because reactions likely to occur under normal conditions may not render them safe, e.g. a chunk of cinnabar ($\ce{HgS}$ mineral) sitting on your desk will not undergo any significant change to render it safe, even on a geological timescale. | {
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50,106 | If a compound has a carbon atom with four different groups covalently bonded to it, it is called asymmetric and enantiomers of the compound can exist. But imagine if one has a different central atom, such as a nitrogen or a sulfur where one of the four "groups" is an electron pair (see examples below). Would such a thing still behave like a "normal" asymmetric carbon? If not, how else does it behave, i.e. are there enantiomers? Is my assumption that those will still have $\mathrm{sp^3}$ hybrid orbitals wrong? I would also enjoy pointers to literature, I was unable to find any. | Generally, amine nitrogens will not behave like a normal asymmetric carbon. Simple amines are roughly $\mathrm{sp^3}$ hybridiized and the molecules you use as examples do have 4 (we include the lone pair of electrons as a substituent) different substituents around the central nitrogen atom. So in principle me might consider it asymmetric or chiral. But simple amines can undergo a process called nitrogen inversion , which essentially coverts one enantiomer into the other. ( image source ) However, if one can find a way to slow down or eliminate the nitrogen inversion process, then chiral amines can be isolated. One way to achieve this is to incorporate the amine nitrogen into a 3-membered ring (an aziridine). Now to achieve the planar state necessary for nitrogen inversion requires the bond angle in the 3-membered ring to open from 60° to 120°, an impossible task. Aziridines containing a chiral nitrogen atom have been isolated and characterized. | {
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50,216 | We commonly use methane and propane for cooking (and home heating), but not ethane. I would expect ethane to be suitable for this, being in between the two, but I've never heard of anyone using it for this purpose. Why is that? On a related note, why is butane used for cigarette lighters and basically nothing else (in ordinary life, I mean)? | Probably the biggest drivers behind using methane as a fuel is that it is abundant in natural gas and is (currently) mostly useless as a chemical feedstock.
Ethane makes up a few percent of natural gas and can also be obtained as byproducts of petroleum refining, but the big difference from methane is that ethane is extremely useful in chemical synthesis (mostly to make polyethylene).
In fact, household natural gas often contains a bit of ethane, which may vary depending on the current demand for ethane as a feedstock.
In short: it burns just as well, but ethane has other important uses. Propane has some use as a feedstock, but it is an attractive fuel because it is easily stored as a liquid without requiring huge pressures. | {
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50,303 | From Wikipedia's article on pyrimidine : Because of the decreased basicity compared to pyridine, electrophilic substitution of pyrimidine is less facile. But why is pyrimidine less basic than pyridine? Pyrimidine has two $\mathrm{sp^2}$ -hybridised lone pairs available for protonation, compared with pyridine's one. | It is not the number of lone pairs that in any way explains basicity. Take a random sugar and it will have ten times the number of lone pairs (albeit on oxygen, not on nitrogen) without being significantly basic. The problem is electronics. Nitrogen, being a rather electronegative atom, is able to draw the π electrons towards it well — to the extent that pyridine derivatives are termed extremely electron poor and thus very slow in electrophilic substitution reactions. Going from pyridine to pyrimidine we need to double that. Thus, there is a perceived severe shortage of electrons in that ring since both nitrogens are trying to draw them towards themselves. Now assume you protonate that species. We now have two nitrogens of which one is positively charged. Immediately, that positively charged nitrogen multiplies its electron-withdrawing force. Equally immediately, the other nitrogen feels electron density leaving its vicinity. The whole system is much less favourable than the pyridine system with only a single nitrogen. Rephrasing that argument: Basicity is increased when electron- donating neighbours increase the electron density (that equals a partial negative charge) on an atom. In pyrimidine’s case, we have an electron- withdrawing neighbour that reduces the electron density (giving a partial positive charge) and making protonation (i.e. more positive charge) less favourable. | {
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50,304 | I'm having a bit of trouble with the latter questions of this problem. $\pu{15.0 mL}$ of $\pu{1.4 M}$ $\ce{HCl}$ was mixed with $\pu{1.00 g}$ of $\ce{CaCO3}$ until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $\pu{200 mL}$ with water. A $\pu{20.0 mL}$ portion of this solution was then neutralized by $\pu{8.50 mL}$ of a $\pu{0.100 M}$ $\ce{NaOH}$ solution. Calculate: Amount of substance in excess of $\ce{HCl}$ in the $\pu{20.0 mL}$ portion Amount of substance in excess of $\ce{HCl}$ in the $\pu{200.0 mL}$ portion Amount of substance in excess of $\ce{HCl}$ which reacted with $\ce{CaCO3}$ My approach $$\ce{2HCl(aq) + CaCO3(s) -> CaCl2(aq) + H2O(l) + CO2(g)}$$ Finding the amount of substance of each reactant and hence the limiting reactant, For $\ce{HCl}$ : \begin{align}
M[\ce{HCl}] &= \frac{\mathrm{mol}}{\mathrm{L}}\\
\mathrm{mol}[\ce{HCl}] &= \mathrm{M}\times \mathrm{L}\\
\therefore \mathrm{mol}[\ce{HCl}] &= 1.4\times (15.0\times 10^{-3}) = 0.021
\end{align} For $\ce{CaCO_3}$ : \begin{align}
\mathrm{mol}[\ce{CaCO_3}] &= \frac{\mathrm{m}}{\mathrm{mm}}\\
\mathrm{mol}[\ce{CaCO_3}] &= \frac{1.00}{100.0869 }\\
\therefore \mathrm{mol}[\ce{CaCO_3}] &= 9.9913\times10^{-3}
\end{align} Therefore (after dividing by stoichiometric coefficients) $\ce{CaCO_3}$ is limiting. From here I'm not sure if what I did next is correct. I found the amount of substance of $\ce{CaCl_2}$ produced the remaining $\ce{HCl}$ in excess. Which were, \begin{align}
\mathrm{mol}[\ce{CaCl_2}] &= 9.9913\times10^{-3} \text{ and,}\\
\mathrm{mol}[\ce{HCl}] &= 1.1009\times10^{-2}
\end{align} (Meaning that we have $9.9913\times10^{-3}\mathrm{mol}$ of $\ce{CaCl_2}$ and $1.1009\times10^{-2}\mathrm{mol}$ of $\ce{HCl}$ in in $\pu{15 mL}$ of solution?) From here on I'm stumped. | It is not the number of lone pairs that in any way explains basicity. Take a random sugar and it will have ten times the number of lone pairs (albeit on oxygen, not on nitrogen) without being significantly basic. The problem is electronics. Nitrogen, being a rather electronegative atom, is able to draw the π electrons towards it well — to the extent that pyridine derivatives are termed extremely electron poor and thus very slow in electrophilic substitution reactions. Going from pyridine to pyrimidine we need to double that. Thus, there is a perceived severe shortage of electrons in that ring since both nitrogens are trying to draw them towards themselves. Now assume you protonate that species. We now have two nitrogens of which one is positively charged. Immediately, that positively charged nitrogen multiplies its electron-withdrawing force. Equally immediately, the other nitrogen feels electron density leaving its vicinity. The whole system is much less favourable than the pyridine system with only a single nitrogen. Rephrasing that argument: Basicity is increased when electron- donating neighbours increase the electron density (that equals a partial negative charge) on an atom. In pyrimidine’s case, we have an electron- withdrawing neighbour that reduces the electron density (giving a partial positive charge) and making protonation (i.e. more positive charge) less favourable. | {
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50,436 | In every basic coordination chemistry class, at some early point the crystal field theory and LFSE will be taught, explaining that there will be an energy difference between d-orbitals (typically named $\mathrm{t_{2g}}$ and $\mathrm{e_g}$) and that electrons can be excited from one to the other to give colour. The next statement is typically that transitions have to be spin-allowed (the higher-energy orbitals need to be able to accomodate for the lower electron without spin-flipping being required) and Laporte-allowed (in a nutshell: no d→d transition in octahedral complexes). The syllabus then goes on to say that the Laporte rule is not that strict a rule (typically, all near-octahedral complexes of 3d-elements which have spin-allowed transitions have a well visible colour; e.g. iron(II), chromium(III), manganese(III), nickel(II), copper(II) etc.) since small random vibrations can destroy the symmetry making the electron transition allowed and an incoming photon can excite the electron as needed. They also remark that the colour is much less intense than for Laporte-allowed transitions — hence why the colour of the $\mathrm{d^7}$ ion $\ce{[CoCl4]^2-}$ is much more intense than that of $\ce{[Co(H2O)6]^2+}$. A very recent question on the colour of manganese(II) sulphate then had me confused. Manganese(II) salts are typically of a very pale pink. The difference to white can be observed when they are put before a white background. But manganese(II) is a $\mathrm{d^5}$ ion, so the transition is spin-forbidden. I think that being spin-forbidden explains why manganese(II) complexes are so weakly coloured, but what is the mechanism that relieves the spin constraint and how does it work in practice? | Selection rules The intensity of the transition from a state $\mathrm{i}$ to a state $\mathrm{f}$ is governed by the transition dipole moment $\mu_{\mathrm{fi}}$ (strictly, it is proportional to $|\mu_{\mathrm{fi}}|^2$): $$\iint \Psi_\mathrm{f}^*\hat{\mu}\Psi_\mathrm{i}\,\mathrm{d}\tau \,\mathrm{d}\omega \tag{1}$$ where $\mathrm{d}\tau$ is the usual volume element (representing integration over spatial coordinates) and $\omega$ is an additional spin coordinate (since the states $\Psi$ also specify the spin). For an electronic transition, $\hat{\mu}$ is the electric dipole moment operator, and only depends on the spatial coordinates $x$, $y$, and $z$. (I've also ignored the vibrational component i.e. the Franck-Condon factor since it's not really relevant for the purposes of this answer.) Normally in determining selection rules, we are more interested in whether the integral in $(1)$ is zero or not. If the integral is zero, the transition is said to be forbidden. The usual way to determine whether it is zero is by using symmetry: if the integrand is antisymmetric under any symmetry operation, then the integral is zero. (It is exactly analogous to the integral of an odd function $f(x)$ over a symmetric region; for example, $\int_{-a}^a \sin x \,\mathrm{d}x = 0$.) If we assume that both the initial and final states can be separated into spin and spatial components, then we get: $$\iint (\psi_\mathrm{f}^\mathrm{spin}\psi_\mathrm{f}^\mathrm{space})\hat{\mu}(\psi_\mathrm{i}^\mathrm{spin}\psi_\mathrm{i}^\mathrm{space})\,\mathrm{d}\tau\,\mathrm{d}\omega$$ (I also dropped the complex conjugates - they aren't important.) Since $\hat{\mu}$ does not operate on the spin coordinate, we can separate this multiple integral: $$\left(\int \psi_\mathrm{f}^\mathrm{spin} \psi_\mathrm{i}^\mathrm{spin} \,\mathrm{d}\omega\right) \left(\int \psi_\mathrm{f}^\mathrm{space} \hat{\mu} \psi_\mathrm{i}^\mathrm{space} \,\mathrm{d}\tau\right)$$ The first integral is nonzero if $\Delta S = 0$. The quantum number $S$ is related to the eigenvalue of a hermitian operator (the actual eigenvalue is $S(S+1)\hbar^2$), and eigenfunctions of a hermitian operator corresponding to different eigenvalues are necessarily orthogonal, making the integral identically zero if $\Delta S \neq 0$. This is the spin selection rule. The second integral is nonzero if $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ have a different character under the inversion operation, i.e. one is gerade and another is ungerade. This is because $\hat{\mu}$ itself is ungerade, so for the integrand to be gerade, we need $\psi_\mathrm{f}^\mathrm{space}\psi_\mathrm{i}^\mathrm{space}$ to be ungerade ($\mathrm{u \otimes u = g}$). In turn this means that $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ must have different symmetry. This is the Laporte selection rule. It can be relaxed when the geometry is not perfectly octahedral, or tetrahedral for that matter, since inversion is no longer a symmetry operation. However, I digress. Spin-orbit coupling The problems arise because the total states are not separable into spin and spatial components. This occurs because both electron spin, as well as orbital angular momentum, are both angular momenta. They therefore interact with each other, and this is the spin-orbit coupling spoken of. Mathematically, it is treated as a perturbation to the Hamiltonian; the result is that the transition dipole moment cannot be separated into two integrals and this leads to the invalidation of the selection rules. When considering spin-orbit coupling, the term symbols such as $^3\!F$, $^1\!D$, which imply a simultaneous eigenstate of the total spin operator $\hat{S}$ and the total orbital angular momentum operator $\hat{L}$ are no longer meaningful since both angular momenta are coupled to each other (by $\vec{j} = \vec{l} + \vec{s}$) and behave as one "combined" source of angular momentum. The quantities $S$ and $L$ are called "nearly good" quantum numbers: they are no longer strictly conserved, but the "contamination" of different quantum numbers is relatively small. Mathematically we could express this as an admixture of different spin states into the original wavefunction: $$^3\!F_J = c(^3\!F) + c_1(^1\!F) + \cdots$$ where $^3\!F_J$ denotes a state with spin-orbit coupling (not an eigenstate of $\hat{S}$ and $\hat{L}$), and $(^3\!F)$ denotes the original $^3\!F$ state before spin-orbit coupling is considered (which is a true eigenstate of $\hat{S}$ and $\hat{L}$). If $c_i \neq 0$, then one obtains a non-zero transition dipole moment between "states" of nominally different $S$. The fact that the selection rules are still a thing is solely because spin-orbit coupling is a small effect ($c_i$ close to 0), so the compounds behave "almost" as if they obey the selection rules. More properly, the transition dipole moment between states of different $S$ is small, and therefore the d-d bands in $\mathrm{d^5}$ spectra are extremely weak. The magnitude of spin-orbit coupling, however, does vary with $Z^4$ ($Z$ being the atomic number). Therefore, while spin-orbit coupling is rather small in the 3d series, it becomes an extremely important mechanism by which the spin selection rule in the the 4d and 5d congeners, as well as the f-block elements, is relaxed. | {
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50,511 | There have been some claims saying that drinking reboiled water is dangerous because it produces arsenic, nitrates, etc. Is this true? | Reboiling water will do precisely three things: Some volatile contaminants in the water that survived the first boiling might be driven out by the second boiling. Some additional evaporation will occur, causing anything dissolved in the water (salts, contaminants) to be slightly more concentrated. Some additional material from the surface of the vessel you're boiling the water in might dissolve into the water. It's virtually impossible to imagine any scenario in which this would matter, unless you were taking a huge amount of barely drinkable water and boiling away most of it or using a vessel whose lining could produce toxins in your water. Needless to say, you really shouldn't be drinking barely drinkable water at all, nor should you be boiling water even once in a vessel whose lining can produce toxins. If you're doing these kinds of things, the additional risk from reboiling would be miniscule. It will definitely not summon aresenic or lead from nowhere. | {
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