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Is it possible to have a diatomic molecule of sodium in gaseous state?
Already I know that hydrogen, all the halogens, nitrogen and oxygen forms diatomic molecules. But I am confused about $\ce{Na}$? So I would like to know about that.

I haven't seen any cases approving $\ce{Na2}$'s existence, but $\ce{Li2}$ exists for sure. http://en.wikipedia.org/wiki/Dilithium

The molecule $\ce{Na2}$ is well known in the gas phase (and $\ce{ Li2, K2}$), and has been studied since approx 1929! The ground state has a (long) bond length  of  $0.3078$ nm, vibrational frequency of $159 \pu{cm^{-1}}$ and rotational constant $0.1547 \pu{cm^{-1}}$. At least four excited state are known.

According to molecular orbital theory, disodium should be stable in the gas phase, with a bond order of one. The molecular orbital diagram is the same for all the alkali metals since they all have one valence electron in an $s$ orbital.

(image source)
NIST chemistry webbook has a small page on disodium and quotes an enthalpy change of formation of $142.07~\mathrm{kJ~mol^{-1}}$ which is moderately endothermic.
This page has a brief discussion of alkali metal dimers and claims that they are all stable in the gas phase, in agreement with the prediction from MO theory.

And are there any elements behave like this except what I have mentioned above and Na?

The quoted enthalpy of formation, +142 kJ/mol Na2, is not the whole story. It should be compared with the atomization energy given by Chemistry Libretexts as +107 kJ/mol Na atoms. Thus sodium atoms in the gas phase pair up with a bond enthalpy of -72 kJ/mol Na2.

The disodium molecule $\ce{Na2}$ has first been observed by M. Polanyi and collaborators in the diluted flame of sodium vapor and chlorine $\ce{Cl2}$. When sodium metal is heated in a vacuum, it gets vaporized. If now a tiny amount of gas $\ce{Cl2}$ is sent into this vapor, the famous yellow flame of $\ce{Na}$ is produced at the point where both gases are in contact to one another. This diffusion flame is due to three consecutive reactions. First : $$\ce{Na + Cl2 -> NaCl + Cl}$$ Then the chlorine atom cannot react with atomic sodium $\ce{Na}$ since there is no third body to remove the reaction energy. So the $\ce{Cl}$ atom does react with the dimer $\ce{Na2}$ in the reaction $$\ce{Cl  +  Na2 -> NaCl + Na'}$$ and this reaction is exothermic enough to produce an excited $\ce{Na'}$ atom. Sorry ! this $\ce{Na'}$ should have been printed as $\ce{Na}$ with a star, but my keyboard refuses to print this symbol (*) in index. Now this excitation energy ir reemitted as the yellow D-line of sodium ($\pu{589.89 nm}$) $$\ce{Na' -> Na + h\nu }$$ This was the first proof that the dimer $\ce{Na2}$ exists in the sodium vapor.
Ref,: 1) M. Polanyi, Atomic reactions, William and Norgate, London, 1932.
2) M. G. Evans, M. Polanyi, Transactions Faraday Soc. 35, 178, 1935


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Are more than one correct Lewis structures possible (not counting resonance)?
I am currently learning chemical bonding, specifically Lewis structures. And these nitrogen oxides, nitrogen-oxygen ions and other atoms showing variable oxidation states are confusing me. As many examples I have understood with bonding atoms having no $d$ or $f$ orbitals, the variable oxidation states are a result of dative bonds.
So when drawing the Lewis structure for $\ce{NO3-}$, I drew:

My book however has only one dative bond, and has this structure:

To me, both of them look correct. The octets seem to be satisfied in both cases.
So my questions are:

Are both of them correct? If yes, how?
If my structure is incorrect, what rule am I violating?

nitrogen can make 3 covalent bond and has one lone pair so 1 lone pair will be distributed to 1 oxygen and then nitrogen will form 1 double bond with another oxygen now nitrogen can form only one single bond which will be possible with oxygen having negative sign just like the second structure in your question

@shashanksharma You are explaining the correctness of the structure in the book, which I know is correct. I am asking what's wrong with my structure? It too satisfies the octet.

your questions says to draw a lewis structure of n o 3 - but in your  structure you showed negative sign in nitrogen which is not mentioned in question.

@shashanksharma what do you mean by "not mentioned in the question"? I had to draw the structure for $\ce{NO3-}$, I drew it by assigning negative charge on N, in the book it is drawn by assigning negative charge on one of the O atoms. I am simply asking if my approach is correct or not? And if not, then why?

@swapnil you have to note that nitrate is an ion. So, the negative charge on one of the O atoms is simply demonstrating the existence of an extra negative charge, not the existence of a second dative bond.

@M.A.Ramezani I drew the first structure myself without looking anywhere. I am asking what is wrong WITh assigning the negative charge on N and forming dative bonds with 2 oxygen atoms, """INSTEAD""" of the structure drawn in my book.

@ShashankSharma What question? $\ce{NO3-}$ means negative charge on the whole ion. Writing $\ce{NO3-}$ does not mean that only oxygen can be assigned a negative sign. The net charge on the ion must be -1. That's all. There is NO strict condition for negative charge to be present only on oxygen atom.

I can't help now try this en.wikipedia.org/wiki/Nitrate

Aha! Then you think a little about the electronegativity differences. What does electronegativity mean? Which of the the two $\ce{N}$ and $\ce{O}$ is more electronegative?

@M.A.Ramezani put that as an answer.

The octet is violated in your proposed structure, by the central nitrogen atom which has nine electrons. A dative bond does not mean an electron is removed from the donating atom; that is ionization. There is in fact no discernible difference between a dative covalent and a normal covalent bond. A dative bond will count as two electrons for each connected atom, just like a regular covalent bond.

As $\ce{O}$ is more electronegative than $\ce{N}$ the $\ce{1-}$ charge in the $\ce{O-N}$ will be on the oxygen. Even if it were to be a dative bond, its donation in an electron pair makes it very easy for oxygen to take. Oxygen is, after all, an oxidising agent.

Although I now have an answer to my question (from my teacher), your answer seems to be wrong, based on the fact in $\ce{CN-}$, N is more electronegative than C, yet the -1 charge is still on C.

@Swapnil For $\ce{CN^-}$ it is a different story, note that the -1 on the C allows for both bonding atoms to have a full octet. In the previous example both the O and the N had a full octet already so the extra -1 goes the the electronegative one.

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Reduction of vanadyl ion
I want to reduce vanadyl to vanadium (III), can somebody please suggest some appropriate methods for it? I am trying to make a redox flow battery for a project but I have only vanadyl sulfate.

Have you consulted a table of reduction potentials such as here?

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/d-Block_Elements/Group_05%3A_Transition_Metals/Chemistry_of_Vanadium

http://chemed.chem.purdue.edu/demos/main_pages/12.7.html

my guess is that in acidic environment it should be easily reduced. So, 2VOSO4/H2SO4  + reducer should do the trick. Maybe, oxalic acid in boiling diluted acid?

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State of a Racemic Mixture
Is a racemic mixture always an equi-molar solution of two enantiomers, both in the same state? Is it necessary for both isomers to be in the liquid state for racemisation to occur, or does racemisation also occur in the gaseous and solid states?
Thirdly,can the two enantiomers still form a racemic mixture if they are initially at different temperatures?

Is a Racemic Mixture always an equi-molar solution of two enantiomers, both in the same state?

Yes and yes. A racemic mixture is equimolar in both enantiomers. Otherwise you would call the reaction enantioselective with a so-and-so selectivity.
A racemic mixture contains both enantiomers in the same state, as energy differences between enantiomers are so minuscule that you would never get one solid and one liquid at the same time.

Is it necessary for both the mixture to be liquid in nature, or does racemisation also occur in the gaseous or solid state?

Racemisation is the process where you start out with a pure enantiomer and in the end get a mixture. For complete racemisation you end up with an equimolar mixture as discussed above.
It is thus possible that racemisation occurs in gaseous and solid phases, not only in solution or liquid. However, the reaction times will differ significantly. I could imagine that the solid racemisation reaction will take a long time.

Can the two enantiomers still form a Racemic mixture if they are initially at different temperatures?

Higher temperatures just mean that the racemisation process proceeds faster. In the end, the warmer enantiomer will have racemised further on than the colder one, but for $t\to\infty$ the result will be the same.

Thanks a lot for the answers! As I was reading your answer, another question hit me. Can the two enantiomers react with each other? And would the reaction be exothermic? If so, then how would this affect the mixture in any way?

@tennispro1213 I think that is a question that warrants its own link. I haven't heard of anything like that, but just because it's not on my radar doesn't mean it doesn't happen.


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Is ClH7 possible?
Chlorine can be $sp^3d^3$ hybridized.
If so, it can form $\ce{ClH_7}$ and then chlorine, being more electronegative, will have (-7) as its oxidation number. But we know that the oxidation number of $\ce{Cl}$ lies between (-1) and (+7).
Thoughts?

http://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit

http://chemistry.stackexchange.com/questions/31584/on-the-geometry-of-sp3d-and-sp3d3-hybridization

I would not be so quick to dismiss the existence of other hydrogen chlorides.

Your question is based on a skewed premise, as gsurfer04 already pointed out. Rather than invoking d-orbitals, which are usually rather far away in energy from the corresponding p-orbitals (in fact, usually higher in energy than the next shell’s s-orbital, so for chlorine 3p → 4s → 3d), these hypervalent halogen compounds are considered to derive from multi-centre bonds.
For details in the bonding mechanism, I am going to forward you to this excellent answer.
I haven’t seen any multi-hydrogen chloride yet. In fact, when discussing interhalogens like $\ce{ClF3}$, our professor in the introductory chemistry course went as far as saying even $\ce{H3Cl}$ is impossible. I’ve forgotten the exact reasoning (and unfortunately it’s not explicitly stated in the web material any more) but I believe it to be along the lines of hydrogen’s s-orbitals not being able to form the outside parts of a four-electron three-centre bond.

First of all, $d$ orbitals play no part in hypervalency.
If you look at existing hypervalent molecules, you'll see that the central atom is always less electronegative. As H is less electronegative than $\ce{Cl}$, $\ce{ClH7}$ will be very unstable.

The central atom isn't always less electron negative http://pubs.acs.org/doi/abs/10.1021/ja00293a020 and http://pubs.acs.org/doi/pdf/10.1021/ja961081v

@DavePhD Fascinating find on the triprotonated methane! Thanks for the link.

@DavePhD Although the central atom isn't necessarily less electronegative, in order for the molecule to have 3c-4e bonds, theoretically, the peripheral atoms should be electronegative atoms to withdraw electron density away from the central atom. After all, two of the electrons of the 4 electrons in the 3c-4e bond becomes a lone pair on the peripheral atoms.

authors: عصام الحسيني


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Reactions between enantiomers in a racemic mixture
As a follow up to this question I started thinking about another question. During racemization, enantiomers are mixed in a equal ratios to form an optically inactive mixture.
What type of reactions can take place between the enantiomers and how would this affect the mixture, if at all?

Racemisation isn't reaction between enantiomers but reaction turning part of one enantiomer into other.

If a chiral molecule (an enantiomer) can react with itself, dimerize for example, to form a new compound, then formation of this new compound would contaminate the racemic mixture.

(image source)
Perhaps more importantly is interaction of the racemic mixture with itself in the solid state.  Just as the interaction between a right hand and left hand is different than the interaction between two right hands, the interaction between the R form and S form of the enantiomer will be different than the interaction between two R (or two S) forms.
In liquids and gases, molecular encounters are fleeting and cannot be preserved or "locked in" for significant periods of time as they can be in a solid. In other words, any favorable, lower energy molecular encounters are quickly broken apart by the energetically larger effects of thermal motion.  However, when the racemic mixture is in a solid phase, if the interaction between R and R (or S and S) is lower in energy then the interaction between R and S, then the R and S enantiomers will tend to form separate crystals and a mixture of pure R and pure S crystals will result.  Such mixtures are often termed "conglomerates".  On the other hand if the interaction between R and S is more favorable (lower in energy), then a racemic compound or racemate is formed where each crystal contains R-S pairs.
A conglomerate, being a simple mixture, will show a melting point depression compared to the pure enantiomer.  A racemic compound may have a melting point higher or lower than the pure enantiomer.
Further reading on racemic conglomerates and racemic compounds can be found here and here.

Can you give more links about formation of conglomerates and why they are not formed in liquid and gaseous state?

@NoahJ.Standerson  I modified the answer to address your comment.

This is a difficult question to answer without having specific compounds to work with, but let's say that you have a pair of enantiomers with a single stereocenter, so there's an (R)-enantiomer and an (s)-enantiomer. There are four possible reactions:
$\ce{R + R -> RR}$
$\ce{R + S -> RS}$
$\ce{S + R -> SR}$
$\ce{S + S -> SS}$
The set of products are four stereoisomers of which there are two pairs of enantiomers: (RR)/(SS) and (RS)/(SR). [Note that I'm assuming that the chemical reaction renders the (RS)/(SR) compounds asymmetrical. If they were symmetrical, then these would be achiral, meso compounds.]
Where this could get more interesting is that the rates of those reactions are not necessarily the same, since we have two chiral reactants interacting. Consider the transition states leading to RR and RS. Those transition states diastereomeric, and thus require different activation energies. This shouldn't be surprising since RR and RS are diastereomers, which have different energies. The reaction with the lower activation energy is called the matched case, and the reaction with the higher activation energy is called the mismatched case. I'm not aware of any chemistry where the possible starting materials are all stereoisomers of each other, but this phenomenon can be important for setting new stereocenters in compounds that already have stereocenters.

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After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this?
After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this?
I am looking for a general answer; nothing too detailed. It happened to me in lab and I am wondering what error I could have committed.

Can't help but notice it's still too broad.

Preface: The problem is that you have overstated your percentage yield, and the symptom is that it is above 100%.
Example: You and your friend each do an experiment where the literature states the yield should be ~65%. You both use poor technique and fail to dry your product, causing you to attribute water mass as product mass. Your friend's yield is 101%, and yours is 99%. Both you and your friend are wrong, for the same reason! Don't make the mistake of stopping your analysis when percentage yield drops below 100%. Don't chastise your friend for having an "impossible" yield while patting yourself on the back for your "99%"!
Overstatement of percentage yield happens for the following reasons:
Zero) Your model of the reaction chemistry is just wrong. Unlikely in an undergraduate setting, but worth mentioning.
1) Your math is wrong. Worth double checking and easy to rule out as a source of error.
2) You used more reactant than you thought. Most errors in reactant measurement cause you to use less reactant than recorded (e.g. transfer errors, impure reactants), but you could possibly have used more than you thought. Beyond simple measurement errors, perhaps your limiting reagent was provided to you in solution, and that solution was of higher concentration than reported. Put a bit of time into thinking about this, but [especially in educational experiments] protocols usually minimize the chance of this kind of error.
3) You actually have less product than you think. This is the most likely culprit. You have to be really honest with yourself here. In most experiments, you take the mass of something at the end, and call it the mass of your product. Is it really your product? Or does that mass contain impurities which cause you to misrepresent impurities as product. Likely culprits are solvents (e.g. water from improper drying), side-products, and/or unreacted reagent. Have you really done your best to filter and remove them? If not, your yield will be inflated.

Below is a basic example.
If percent yield is based on a mass, then a >100% yield can result from a product contains materials that should not be there.  For instance, a product that should be anhydrous may not be completely dry and water would add mass to the product increasing your yield.  Other present contaminants can also do this and can even appear from a product that reacted with the surrounding atmosphere.

Are you implying that an additional product besides the dry substance and water could have possibly been produced?

Absolutely.  Perhaps you added too much reactant to your reactant vessel.  Perhaps your reaction mixture oxidized with the air around it.  Maybe the reaction didn't goto completion like you thought.  A lot of things can happen to throw off your % yield.

Imagine a reaction where your starting material is cleaved and the product has a lower molecular weight. If the conversion is incomplete and the product contaminated with some starting material, the resulting weight may suggest a yield higher than 100%.

Yields above 100% are obviously impossible, so either you used more educts than you thought, or you don't have as much product as you think.
An error when measuring the amount of educts for the reaction is one possible reason. A miscalculation at any point of the amounts needed or the final yield is also something you might want to check again.
A really common reason is the presence of other material in your product, often solvents like water are still present if you weren't careful enough when drying. This is really the most likely cause, especially in student labs.
You should think about the analytics you performed, and check if they show the presence of anything aside from your product. And then think about what kinds of substances could be in there (solvents, educts, other reaction products) that wouldn't be visible in the kind of analytics you performed.

Percentage yield exceeds 100% if impurities are present due to inadequate purification. Such impurities could be present for many reasons, such as if the vessels used to collect the product are contaminated. Also, it is possible that the product absorbs components from air like water vapor or carbon dioxide.

Like others said, contaminants may be there or unreacted chemicals can be there. It happened to one or two times while I was doing summer project and left the compound in vaccumm and a trap to get rid of solvents. You should use proper solvents to separate the product and unreacted or side product stuff.

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Which elements can be diatomic?
Which elements can be diatomic and why?

Motivation
Hydrogen, Nitrogen, Oxygen and the Halogens tend to be thermodynamically stable as a diatomic molecule at room temperature, and are usually labelled as diatomic nonmetals. (Although I am unsure about Astatine). These are the ones you learn in school and I am fine with their reasoning. I want to know about other elements in particular.
Looking at the rest of the non-metals: Carbon, Phosphorus and Sulfur have diatomic molecules in gaseous states. All with a common theme of being heated at very high temperatures.
This leaves us with diatomic Selenium which I could not find anything on.

Further notes
Diatomicity doesn't seem to be just a non-metal thing:

Lithium forms diatomic molecules and all diatomic alkali metals are claimed to be stable in gaseous states. Found from this question.
Silicon has been reported to have a diatomic molecule similar in structure to carbon.
Tungsten, Chromium, Uranium and Molybdenum are claimed to be diatomic with a sextuple bond.
Supposedly Noble gases can form dimer molecules but if these are considered diatomic is debatable.

I have a feeling almost every element can be found as a diatomic molecule in the right conditions (or at least present as a measurable mole fraction), though high temperatures and low pressures may be needed for many of them. In the majority of cases, any chemical bond will be more stable than no chemical bond. Plus there are some stranger cases, such as bound diatomic noble gas molecules, stable when encapsulated by a fullerene molecule.

Alkali-earth metals and noble gases definitely can't form stable diatomic molecules, Zn/Cd/Hg likely can't too.  Homoatomic bonds for most heavy non-transition metals are very weak so detecting diatomic molecules is unlikely.

@permeakra Look at the bottom of further notes for noble gases.

@permeakra Diberyllium exists: 'diberyllium molecule exists (and has been observed in the gas phase). It nevertheless still has a low dissociation energy of only $59~\mathrm{kJ~mol^{-1}}$.' Also see here.

@bon Interesting, it shouldn't. Well, Palladium clasters shouldn't exist to, but they do.

Depends what you call diatomic - if you count helium vdV dimer than two atoms of any element which will have occasion and time will get bound.

"Stable" is a somewhat nebulous property. In general a chemist would think of stable to mean that the molecule would survive some large ( how big is large??) number of intermolecular collisions.   In the cold of deep space where collisions are infrequent I'm sure that any two atoms would form a "molecule." If you increased the partial pressure so that such "molecules" started colliding, then the list would be different.

I think sodium in the gas phase exists as a dimer, but I could be wrong.

This leaves us with diatomic Selenium which I could not find anything on.

Selenium forms a dimer with bond energy of 79 kcal/mol, according to table 8.6 (page 451) of Comprehensive Handbook of Chemical Bond Energies, which provides 5 references.
Tellenium dimer is 61 kcal/mol
Polonium dimer 45 kcal/mol


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Why is Sulfur trioxide a neutral electrophile?
My chemistry book says, an electrophile is a reagent which is atrracted to electrons, therefore positive ions and reagents with incomplete octets act as electrophiles, Sulfur trioxide belongs to neither category, so how does it act as an electrophile?

An electrophile is an electron pair acceptor.

In organic chemistry almost all electrophiles are like that - only have partial positive charge. "positive ions and reagents with incomplete octets" you're rather talking about Lewis acids it seems - why SO3 is lewis acid is more interesting.

Sulfur is bonded to three oxygens, twice each. The sulfur in the middle technically has a formal charge of zero, but the oxygens it is bonded to are extremely electronegative (they're electron hogs), and so the sulfur atom in the middle has a partial plus charge on it. Ergo, it will be willing to accept electrons to try to compensate.

Not really, structures with formal charge describe it better.


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What's the bonding in the diborane(6)- hydrogen chloride dimer?
I was researching and Googling and researching again, about this very interesting molecule: Diborane.
Then I faced a very interesting structure, something that I have never seen before. Something that turns up lesser than 50 results in Google, most of which are behind paywalls, and thus leave me in pain of not understanding this species. I'm talking about diborane - hydrogen chloride.

Source
So, I'd like to understand this species better.
The bonding certainly doesn't seem ionic. What is the attraction force between $\ce{HCl}$ and $\ce{B2H6}$? There must be one or they wouldn't have written these two compounds together.

Edit: I'm specifically looking for the hydrogen chloride - diborane(6) dimer you can see in the image. And the Cl is pointing towards the H in diborane, so this isn't about dipole/induced dipole interactions. You can find more info about it in here. Interestingly, in calculating molar mass, both the $M$s of $\ce{HCl}$ and $\ce{B2H6}$ are taken into account.

I am currently running SAPT2+3(CCD) on this to determine the attractive and repulsive contributions to the interaction energy.  I will post the results once completed. It is clear, however, that this would be a non-covalent interaction.

This paper calls it a hydrogen bond, this one provides structural information

I would imagine that it would be a dipole/induced-dipole judging by the second paper that ron posted (with the hydrogen on HCl pointing toward the diborane).  The first paper shows HCl pointing up toward the centroid of  diborane which is completely different than the geometry presented in paper 2.

@Lord The first paper Ron linked to speaks about diborane (4), while I'm looking for the diborane (6) "version". There's some info here. The interesting part is that in computing molecular mass, it takes both HCl and B2H6 into account.

@Ron please see the edit, and my comment above. I'm specifically aiming to understand the interactions in the dimer of diborane (6) - hydrogen chloride, with chloride pointing towards hydrogen.

Any chance an H2O-HX (where X is a halogen) dimer system has been examined?  That would be a very analogous model system to your diborane-HCl dimer assuming an appropriate orientation.

This is a good question. Related to it I would wonder, why it is even stable in that way. There are a couple of ways I could imagine it to decompose... @LordStryker It seems the water hydrogenhalogenide dimer system has a few more electrons, compared to diborane. I would not be too sure to find similarities.

@Martin You don't see a useful analogy to a dimer system containing a halogen pointed toward two hydrogen atoms (sort of like a bifurcated H-bond except in my example with a Cl)?  I'm rather surprised.

@LordStryker Oh, yes, I was not thinking right. That would be useful.

I find this question to be quite interesting as it clearly pertains to non-covalent interactions.
I have analyzed two conformations of the diborane-HCl dimer system, one with the chlorine on HCl orientated toward diborane (HCl-Dib) and one with the chlorine oriented away from the diborane (ClH-Dib).  For comparison, I have similarly analyzed the Water-HCl dimer with the chlorine oriented toward the two hydrogens on water.
Methodology
Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with Gaussian 09 using Grimme's B97D density functional along with the 6-31+G(d',p') basis set.  All geometries were characterized to C$_{2V}$ symmetry.  Symmetry adapted perturbation theory (SAPT) was implemented at the SAPT2+3(CCD) level using the Psi4 software package.  These computations implemented Dunning's heavy-aug-cc-pVTZ (haTZ) basis set (i.e. aug-cc-pVTZ for non-hydrogen atoms and cc-pVTZ for all hydrogen atoms).  Finally, natural charges were quantified using the Natural Bond Orbital (NBO) analysis routine available in Gaussian 09 (see http://nbo6.chem.wisc.edu/ for more details).
Results
Both diborane-containing complexes were minima on the B97D potential energy surface.  The H$_2$O-HCl dimer is a higher order saddle point exhibiting two imaginary modes of vibration.  The interaction energy ($E_{\mathrm{int}}$) can be quantified by attractive and repulsive contributions, the former consisting of electrostatics ($E_{\mathrm{elst}}$), induction ($E_{\mathrm{ind}}$), and dispersion ($E_{\mathrm{disp}}$) and the latter consisting of exchange ($E_{\mathrm{exch}}$). The table below summarizes the SAPT results.

Clearly, the dominant attractive force in each case is dispersion.  The ClH-Dib complex exhibits the strongest interaction energy and contains a dipole that points from the diborane monomer to the chlorine.  The HCl-Dib complex has an interaction energy that is about 1 kcal mol$^{-1}$ smaller in magnitude and exhibits a dipole pointing toward the diborane monomer.  The analogous H$_2$O-HCl system is qualitatively similar to the HCl-Dib complex in terms of the magnitudes of the components to the interaction energy.  Surprisingly, this dipole-dipole interaction is slightly weaker (by about 0.1 kcal mol$^{-1}$) than that seen in the dipole/induced-dipole system of HCl-Dib.
I have included below figures for the HCl-Dib, ClH-Dib, and H$_2$O-HCl complexes, respectively, with corresponding natural atomic charges shown on each atom.

Figure 1 - HCl-Dib

Figure 2 - ClH-Dib

Figure 3 - H$_2$O-HCl

Interestingly, when chlorine is pointed toward diborane (Figure 1), the natural atomic charges for diborane remain rather symmetrical and diborane barely exhibits an individual dipole moment.  The H$\cdots$Cl separation between the two monomers is about 3.4 Angstroms, a long-range interaction.  However, when chlorine is pointed away from diborane (Figure 2), the H$\cdots$H interaction is much smaller, exhibiting a separation of about 2.2 Angstroms.  When considering water, the separation increases once again (Figure 3) to about 3.2 Angstroms.
Conclusions
The dominate attractive force to the interaction energy for these dimer systems is dispersion, though we see a competing contribution from electrostatics in ClH-Dib (Figure 2).  Surprisingly, the orientation of the halogen significantly affects the geometry and energetics of these systems.  When chlorine is oriented toward the two hydrogens, a weaker interaction and a larger inter-monomer separation is seen.  However, when hydrogen is oriented toward the two hydrogens, the interaction energy increases in magnitude and the inter-monomer separation decreases.
Q/A
I find it more convenient to address some of Martin's comments within the answer itself (sometimes responses can become quite long).
Which compound is lower in energy?

Based on absolute energies, H$_2$O-Cl is the lowest in energy of the three dimers (by about 23 Hartrees).

How stable are they wrt decomposition?

The interaction energy will give you this information (defined as the energy difference between the interacting monomers in an optimized dimer geometry and infinitely separated monomers adopting the geometry they have in the dimer complex).  Given that the magnitude of $E_{\mathrm{int}}$ is less than 2 kcal mol$^{-1}$, I would say that these are not very 'stable'.

Have you tried asymmetric states?

No.  I do not believe breaking symmetry in the diborane-containing complexes will change the energy/geometry appreciably.

I find it quite surprising, that the BH--ClH distance is much longer (0.6A) than the sum of the respective vdW radii, while the BH--HCl distance is about the sum of the vdW radii. Still, in both cases there should be (about) no contribution from the orbitals. How do the bond lengths change in the two conformations?

In the diborane-containing complexes, the bond lengths deviate by about 0.01 Angstroms (so... no change).

Can you make a rough comparison between the B--Cl distance in the two cases?
They appear to be quite similar.

In HCl-Dib, the B$\cdots$Cl separation is about 3.79 Angstroms.  In ClH$\cdots$Dib, the B$\cdots$Cl separation is about 3.86 Angstroms.  Similar they are indeed.

I have so many questions about this, but first things first: Congratulations on your choice of functional ;) Secondly but not lesser: This is good work as usual. Thirdly: I am curious, so I am pretty sure I will come up with a terrible lot of questions. (Probably also due to my lack of knowledge in SAPT. I'll do some reading.)  Let's start with the obvious. Which compound is lower in energy? How stable are they wrt decomposition? Have you tried asymmetric states?

I find it quite surprising, that the BH--ClH distance is much longer (0.6A) than the sum of the respective vdW radii, while the BH--HCl distance is about the sum of the vdW radii. Still, in both cases there should be (about) no contribution from the orbitals. How do the bond lengths change in the two conformations? Can you make a rough comparison between the B--Cl distance in the two cases? They appear to be quite similar.

@Martin-マーチン I thank you for your insightful questions and I've responded to them in the body of the answer.   Are you happy with B97D for these complexes or should we look for method effects on these geometries?  I could always implement B3LYP. ;)


text:
Stable natural helium hydride?
Reading the transcript of the Royal Society of Chemistry podcast Helium Hydride, they state that helium hydride is possibly the most ancient compound to form in the Universe.  They make the assertion that despite helium hydride being most stable as an ion, but:

The neutral molecule can briefly exist, but only as an excimer, where the excess electron has enough energy to keep it at a higher level.

Under what natural conditions could a stable neutral helium hydride molecule exist?

As explained in Emission Spectra of Bound Helium Hydride Physical Review Letters vol. 77 pages 2941-2944:

The ground state [of neutral HeH] is repulsive, as shown by
  numerous experiments and calculations.

but excited states of HeH that are bound exist (as first experimentally observed by this paper, but previously calculated).
Given that only excited states are bound, it don't think it proper to say HeH stably exists under any conditions natural or otherwise.
The paper does state:

Because of the abundance of He and $\ce{H_2}$ in stellar atmospheres
  and interstellar space, HeH is expected to
  be formed there. Therefore our data may be useful to
  find the HeH lines there.

However, that it exists does not mean it stably exists.


text:
Equivalents Chemistry
Why equivalents of reactants and products come out to be equal in a reaction? I am seeking logic behind the law of chemical equivalence and i have just asked it for clarifying the concept.


text:
Autocatalysis / Electroless plating of mercury
As a project I am trying to plate mercury onto a powdered metal amalgam of $\ce{Cu}$ & $\ce{Ag}$. I am using solution of $\ce{HgCl2}$ in normal $\ce{HCl}$ (a reducing agent), Any thoughts for increasing the efficiency of plating process on the metal alloy powder? How can we improve the $\ce{Hg}$ plating on the metal powder? Can the concept of autocatalysis / electroless plating be helpful in this case?


text:
Subtracting wave functions?
In molecular orbital theory bonding of two atoms can be seen as addition or subtraction of the wave functions of these atoms. Addition seems pretty logical and straight-forward - atoms overlap their electron clouds and probability to find an electron in it increases. Subtraction is the thing I don't get. How can the probability to find an electron be negative? Am I missing something?

I think you are confusing the wavefunction $\Psi$ with the probability density function $\Psi^2$. The former is a state function describing the behavior of an electron. The latter describes the "probability of finding an electron".
Wavefunctions are usually complex functions, but even the simplest sine wave functions can be negative in the algebraic sense: the value of $\Psi=f(x)<0$. These regions where $Psi=f(x)$ has different algebraic signs are the regions corresponding to the orbital lobes of different phase, like in a p-orbital. However, $\Psi^2$ is always $\Psi^2 =f^2(x)\ge 0$. Information about the "location" of the electron is gained, information about its phase is lost.
Take as an example the two wave functions representing two atomic orbitals $AO_1$ and $AO_2$ on the graph below. $\Psi_{AO_1}=\cos{(\frac{\pi x}{2})}$. $\Psi_{AO_2}=\cos{(\frac{3\pi x}{2})}$. $\Psi_{AO_2}$ has regions where $\Psi_{AO_2}<0$ when $\frac{1}{3}<|x|<1$.

Now let's look at the probability density functions $\Psi^2_{AO_1}=\cos^2 {(\frac{\pi x}{2})}$ and $\Psi^2_{AO_2}=\cos^2 {(\frac{3\pi x}{2})}$. Notice that both functions are always $\ge0$.

Okay... now, when atomic orbitals combine to form molecular orbitals, one of those conservation laws (angular momentum) gets involved, and the number of orbitals has to remain constant. Thus, molecular orbitals are linear combinations of atomic orbitals. For example, if two atomic orbitals combine, two molecular orbitals are produced.
Let's go back to our two sample sine wavefunctions above and pretend they are atomic orbitals overlapping. One of the molecular orbitals could then be: $\Psi_{MO_1}=\frac{1}{2}\Psi_{AO_1}+\frac{1}{2}\Psi_{AO_2}$. This is the orbital representing constructive overlap between these two wavefunctions. To get the other orbital, we can't just add again, or we would get the same orbital. If we subtract, we get a different orbital - the one representing destructive overlap. The graphs of these two "molecular orbitals" and their probability density functions are below.

Again note the difference between the MO wavefunctions and their density functions. If these were atomic nuclei at $x=\frac{1}{3}$ and $x=-\frac{1}{3}$, then $\Psi_{MO_1}$ would be bonding and $\Psi_{MO2}$ would be antibonding (or perhaps nonbonding).

Teaching primarily general and organic chemistry, I spend a lot of time thinking about how to explain the fundamentals concepts of quantum mechanical bonding theories without calculus. The analogy to simpler types of waves useful because it is intuitive. The maths of quantum mechanics are more complex, but at the end of the day, QM waves behave like other types of waves.

Yep, I am confusing those two. Now I am not sure what Ψ actually means. And the addition and substraction of Ψs. In a normal wave it would be the position of particles in a point in time affected by one or two waves, and in Ψ?

@user3116958, $\Psi$ is a mathematical object used to describe the state of a system. As we usually say, it does not have any physical meaning. The meaning of this is that it does not correspond to any physical quantity that can be measured even in principle.

@user3116958, the thing (which I feel from the conversation you do not quite understand yet) is that quantum waves are not classical waves, or, in other words, they are not physical waves like sound or electromagnetic waves from the classical theory. To realise that just try to answer the question, what is waving in the case of quantum waves? In classical theory wave are periodic oscillations of some physical quantity, but what oscillates for a quantum wave?

@Wildcat, why did chemists then decide to add (or substract) the two?

@user3116958, that is the essence of the LCAO-MO model: molecular orbitals are expressed as linear combination of atomic orbitals. Again, at the fundamental level that is purely mathematical trick (expansion of a function over a complete set of functions), so any naive way of looking at it as some kind of a physical process is doomed to failure.


text:
Which elements tend to form the most phases?
Some combinations of two elements have very complicated binary phase diagrams across the weight% horizontal axis. Others are rather simple. Is this a function of only one or both of the elements involved? I would think both. However, surely there are some elements that tend to form more phases than others. Has this tendency to form more phases (or less) been quantified for the various elements? If no, is it quantifiable?
Obviously elements form only a given number of known allotropes. However, it is not known if the number of known allotropes is exhaustive. Therefore the number of pseudomorphs in a binary phase diagram seems like a more thorough method of investigating an element's "binding flexibility."

Much empirical data is needed to either produce or validate phase diagrams.  Any phase diagram would likely need to be of practical consequence to justify spending the energy and resources to produce it.
An example of a binary mixture with practical consequence would be a metal oxide formed during corrosion. The many forms of iron and oxide are well studied since iron is commonly used to construct tools.
In 2015 a group claimed to use the computer code CALYPSO (Crystal structure AnaLYsis by Particle Swarm Optimization) to discover a new phase of beryllium.  Beryllium is used in aerospace and the construction of nuclear reactors, so the expenditure of effort was justifiable.
So to answer the  sub-questions:
Is this a function of only one or both of the elements involved?
Practically .. No. In the case of iron oxide in an aqueous solution, multiple iron oxide phases can form at the same temperature and pressure if the pH of the solution or the redox potential value of the solution were changed.
Has this tendency to form more phases (or less) been quantified for the various elements?  - Yes, phases of practical consequence have been quantified at great effort; however, discoveries of new phases are likely to continue.
It is possible to estimate the existence of new phases using computer algorithms; however, these phases must be experimentally validated.  A computer algorithm could be constructed around pseudomorphs; however, if a new phase were discovered it would need practical validation by experimentation.
"Which elements form the most phases?" is an impossible question to give and absolute answer as we are limited to certain pressures and temperatures with which to experiment. The way experimentation is done is likely to bias any verifiable answer toward more practical materials.


text:
Can atomic number change the type of electromagnetic radiation being emitted when a electron drop from a energy level to another?
Without getting into too much/any math, I just wanna talk about intuitions if possible.
If atomic increase, the attraction increase, and consequently the potential energy increase. More energy leads to higher frequency, different radiations.

I guess the reason if this diagram is universal is that the gap between 1 and 2 is too large too that the difference in energy cannot simply be compensated by adding more protons.

Sorry, but it is very hard to understand what you would like to say. Can you be a little be more detailed about what is your question, what atoms you are talking about, how does protons come into the picture, and in general, more specific? Maybe giving an example? atomic what? Potential energy of what? Frequency of what? In what state? Why do you think that diagram is universal? Those energies are only for hydrogen...

Absolutely, the differences in potential energies will be different for every element. This is because the potential energies will be different for every element. Notice that the potential energy for each energy level can be determined from the ground state. Let $E_g$ be the potential energy of the ground state (with the sign). Then for energy level $n$ the potential energy is $\frac{E_g}{n^2}$. Thus Differences between energy levels $n$ and $m$ can be written as $E_g\left|\frac{1}{n^2}-\frac{1}{m^2}\right|$.
Clearly it will be different for a different ground state energy which is precisely what the number of protons affects.
Note that the only important part of the mathematics is that all differences rely on the ground state potentials.


text:
Strength and magnetic charge of KNO3
A question asks for physical properties of potassium nitrate, and specifically asks for "magnetic charge" and "strength" (that is how it is phrased). I have properties such as white in color, vitreous, odorless, and melting point of $\pu{334^\circ C}$. It is relatively stable, and a strong oxidizer, so it acts as a good fuel source. However, is strength asking about bond strength? And I'm not sure how to find the magnetic charge since none of my research has found anything.

Surely there is some mistake, as nothing in the Universe that we know of has been measured to possess a magnetic charge.

That's why I specifically stated that they phrased it that way. If anyone has an idea about what they could be referring to I would appreciate any help. Or help with the strength portion? It's ionic, so it's strength is greater than covalent, but is it really that simple?


text:
Calculating how many grams will I get from a reaction
Question:

$$\ce{2NH3 + CO2 -> (NH2)2CO + H2O}$$
If we have $637.2\ \mathrm{g}$ of $\ce{NH3}$ and $1142.0\ \mathrm{g}$ of $\ce{CO2}$ how many grams of $\ce{(NH2)2CO}$ will we get?

This is the explanation I was given:

First, convert reactives to moles:

$637.2\ \mathrm{g}$ of $\ce{NH3}$ are $37.5$ moles.
$1142.0\ \mathrm{g}$ of $\ce{CO2}$ are $26$ moles.

Second, define the stoichiometric proportion between reactives and
  products...

From $2$ moles of $\ce{NH3}$ we get $1$ mol of $\ce{(NH2)2CO}$
From $1$ mol of $\ce{CO2}$ we get $1$ mol of $\ce{(NH2)2CO}$

Third, calculate the number of product moles we would get if each
  reactive was completely consumed:

Consuming all the $\ce{NH3}$ we would get $18.75$ moles of $\ce{(NH2)2CO}$
Consuming all the $\ce{CO2}$ we would get $26$ moles of $\ce{(NH2)2CO}$

The limiting reagent is $\ce{NH3}$, so the most we will get of
  $\ce{(NH2)2CO}$ is $18.75$ moles. Now convert that to grams, we get
  $1125.0\ \mathrm{g}$.

Why are we ignoring the existence of $\ce{H2O}$? Does it not matter?
IF the formula is unbalanced, do I have to balance it first before making this calculation? What if I can't balance it?

Here, $\ce{H2O}$ is a product and you are assuming that the reaction goes to completion, so it does not affect the yield of the reaction in any way. Maybe a suitable analogy is to think of the reaction system as a car - you give the car a certain amount of petrol (say 2 gallons) and it converts it to a certain number of miles (say 70 miles), as well as a lot of exhaust gases. Now the amount of exhaust gases that your car is putting out doesn't affect the fact that if you put in 2 gallons of petrol, you get 70 miles out of it.
If you were to treat the reaction as a reversible one, where you have an equilibrium between reactants and products, then in general, the amount of products present would also matter. But don't worry about this for now, I'm sure you will come across this in your study of chemistry at some point in time.
Good question - if the reaction is not balanced, then yes, you do have to balance it by finding the appropriate stoichiometric coefficients. If you don't know exactly how many miles per gallon (35 in this case) your car consumes, then you won't be able to work out how many miles you can drive if you feed your car 2 gallons of petrol. Not all reactions can be balanced, but if a reaction can't be balanced in one way or another, it can't happen in real life. For example, you can't balance the reaction $\ce{Pb -> Au}$, but as we all know, alchemy was (mostly) a load of rubbish. For this reason, any reaction that you come across in your study of chemistry, as well as in real life, can be balanced.

The product $\ce{H_2O}$ does not matter to the question.  You have some reactants and the reaction proceeds until one is used up.  As you are asked about the mass of one product, you don't care about how much of the other product there is.
Yes, you need to balance the reaction.  You need to figure out which reactant is the limiting one.  Your reaction is balanced, so you use two molecules of ammonia for each of carbon dioxide.  In the answer, under Third, they assume first that $\ce{NH_3}$ is the limiting factor and that there is plenty of $\ce{CO_2}$, finding you can produce $18,75$ moles.  Then they assume there is plenty of $\ce{NH_3}$ and you are limited by $\ce{CO_2}$, finding you can produce $26$ moles.  The first assumption is correct, so that is how much you can make.

Did anyone explain why you convert to moles? It is because moles is a certain quantity of molecules of and to balance the equation you need to work in molecules or molecule multiples.
So if it goes to completion you will use up all of the $\ce{NH3}$ as you say, but be left with 7.25 moles of $\ce{CO2}$, 18.75 moles of $\ce{(NH2)2CO2}$ and 18.75 moles of water. By the way, the original equation is already stoichiometricly balanced
6 $\ce{H}$ on both sides, 2 $\ce{O}$, and 2 $\ce{N}$ - It's just like algebra.
But if it wasn't you would have to balance it before calculating the rest and of course we are simplifying by assuming the reaction goes to completion.


text:
How can I change the bond type (single, double) of an existing bond in Avogadro?
I have a molecule extracted from the PDB. When I load it into Avogadro, it has some misplaced bond types. I want to correct this, by setting the bond type (single, double) for these bonds. How can I do this?

Have a look at this sample PDB file with annotations here. It tells you how you can set single and double/triple/etc.  bonds in the file.

I'd suggest e-mailing the Avogadro mailing lists with such questions. That's the best forum for Avogadro help. :-)

As discussed in the Avogadro website about the draw tool you can change the bond order:

By clicking on the bond - the bond order will go from single, to double, to triple, back to single.
By setting the bond order and "drawing over" the bond. The new setting (e.g., double) will replace the existing bond.

That link is now dead, do you have an up-to-date link to it? Clicking on existing bond does nothing for me.


text:
Can I disable "Display visual cues" by default in Avogadro?
These huge arrows are very annoying to me. I disable them every time I open Avogadro. Is there a way to disable them permanently?

Since you seem to have many questions about Avogadro, is there a reason you aren't using the Avogadro mailing list?

Sure. There's a big checkbox in the Navigate Tool settings:

When you quit, the settings should be remembered for next launch.
If not, please e-mail the Avogadro mailing list and we can figure out what's going on.


text:
Super-cooled liquids
I understand that it is possible to cool a liquid past its freezing point, and still maintain its liquid state because crystallisation would not happen in the absence of a seed crystal.
My question, was marked as a duplicate and I was linked to to Can all substances be supercooled?
One of the answers over there does address my question but only in part.
Here's what it says:

"However, when there is only liquid present, you will first have to form a small crystal nucleus which is not as stable as the bulk solid i.e. has a higher free energy. This is basically a kinetic barrier which prevents forming the solid.
  The formation energy of such a nucleus will depend on the interfacial energy of the crystal-liquid system. In a substance where this interfacial free energy is very little there won't be too much supercooling."

My question still remains. Why is the lack of a nucleus a kinetic barrier to the formation of a bulk solid? If you continue past the glass transition temperature, we form an amorphous solid anyway. How do the mechanics of vitrification and crystallisation differ in this regard?

possible duplicate of Can all substances be supercooled?

Duplicate only in the sense that it asks a question that happens already to have been answered by the answer of your linked question. Bigger problem is that it has multiple sub-questions. @getafix, each of these three questions, if not a duplicate of another already out there, is worthy of its own separate posted question.

@Brian and santiago I read through that thread and it doesn't sufficiently answer my question. I did, however, edit my question to make it a bit more clear.

Whatever weight my opinion carries, that was a great edit.  Given your use of 'glass transition temperature', are you specifically referring to polymer melts? (I may  be betraying some ignorance here, but... do solids composed of small molecules undergo a glass transition in the same fashion?)

When a liquid is supercooled its structure doesn't change significantly, only the particles have less kinetic energy available. As you cool even further, it could occur that particles get caught in local minima and eventually you'll end up with a glass, which can be thought of as a frozen liquid structure. As particles are trapped it will resist shear in an elastic fashion so you have a solid. In this regard it is a barrierless transition. There is no interface whatsoever.
Crystal formation is fundamentally different as the local structure changes first to form a nucleus. This has both energetic and entropic consequences and the aforementioned kinetic barrier is observed. I would also like to add that crystal nuclei are formed all the time but they don't reach the critical size if the temperature is too high.

Supercooling (and superheating) occur mainly due to the difference between surface energy and volume energy. An intersection of two phases/compounds has higher energy than a bulk material.
For example, the interface of ice and water has higher energy because frozen liquid molecules are favored by entropy to mix with the liquid. The only factor keeping the phase boundary in existence is the lack of (or excess of) energy to break the equilibrium between the phases. For any molecule to enter the liquid phase, one must enter the solid phase.
Boiling and freezing come into this when you consider that the temperatures tabulated for boiling and freezing actually measure these phenomena for when a phase boundary already exists. That phase boundary lowers the energy required for phase change, thus lowering the boiling point (or raising the freezing point).
Adding/removing energy when a phase boundary already exists simply shifts the phase equilibrium I already mentioned to the appropriate side. If no boundary exists, however, it becomes more difficult for molecules to escape. When a vessel is heated from the bottom, the hottest portion of the liquid is trapped by other liquid. Any single molecule, even if it attains the required energy, is quickly collided with other molecules and pushed back to the average.
Substances will spontaneously change phase if enough energy exists to favor that change, even without a phase boundary. Going even further, enough energy in the right form can cause a breakdown in the phase boundary.
I do not feel I can adequately answer the later part of your question, as I do not have as much background with solid chemistry. However, I will note that solids typically can exist in different configurations of the molecules. A common example is the difference between steel that has been quickly cooled and steel that has been slow-cooled: cooling more quickly, in this case, results in a harder, more brittle material.


text:
How is it possible for a substance to have a high heat of vaporization but a low boiling point?
The final paragraph of Dissenter's question here is worthy of standing alone:

[H]ow does one square a high heat of vaporization with a low boiling point? If it takes a lot of energy to vaporize something, then how can that something have a low boiling point?

(It's similar to this old question by Friend of Kim, but different in that it emphasizes an apparent contradiction.)

This seeming contradiction can be reconciled by examining the thermodynamic quantities involved.  First, per Wikipedia, the enthalpy of vaporization is "the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure."  Written symbolically:
$$
\Delta H_{\mathrm{vap}} = H_{\mathrm{gas}} - H_{\mathrm{liq}}
$$
For this situation, it is most helpful to consider $H_i$ as the enthalpy per mole ("molar enthalpy") of the species of interest in phase $i$.
Formally, again per Wikipedia, the boiling point for a substance is "the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into a vapor."  This is not so helpful for answering this question, however.
Another way of defining the boiling point is the temperature at which the liquid and the vapor of the substance are in equilibrium, for a closed system (no mass can get in or out) containing only the pure substance of interest at a given pressure$^*$.  The equilibrium condition here can be expressed in terms of molar Gibbs free energies (free energy per mole$^{**}$) as:
$$
G_{\mathrm{liq}} = G_{\mathrm{gas}}
$$
By definition, the molar free energy of a phase is equal to the molar enthalpy minus the temperature times the molar entropy:
$$
G = H - TS
$$
So, the boiling point equilibrium condition can be rewritten as ($T_b$ is the boiling point):
$$
H_{\mathrm{liq}} - T_bS_{\mathrm{liq}} = H_{\mathrm{gas}} - T_bS_{\mathrm{gas}}
$$
Subtracting $H_{\mathrm{liq}}$ and $-T_bS_{\mathrm{gas}}$ from both sides gives:
$$
T_b\Delta S_{\mathrm{vap}} = T_b\left(S_{\mathrm{gas}} - S_{\mathrm{liq}}\right) = H_{\mathrm{gas}} - H_{\mathrm{liq}} = \Delta H_{\mathrm{vap}}
$$
Here, $\Delta S_{\mathrm{vap}}$ is the molar entropy of vaporization, which is analogous to $\Delta H_{\mathrm{vap}}$.  The boiling point is thus:
$$
T_b = \frac{\Delta H_{\mathrm{vap}}}{\Delta S_{\mathrm{vap}}}
$$
So, what does this result mean?
Qualitatively, as noted in answers to the linked questions, $\Delta H_{\mathrm{vap}}$ is a measure of how much energy must be added to one mole of molecules in the liquid to free them from their associations with the other liquid molecules.  The physical meaning of $\Delta S_{\mathrm{vap}}$ is somewhat harder to describe, but corresponds roughly to the increase in the 'freedom to wiggle' afforded by the transition to the gas phase from the liquid.
So, in order to have a large $\Delta H_{\mathrm{vap}}$ (lots of energy required to drive molecules into the gas phase) but also a low $T_b$ (low temperature at which molecules "like" to transition into the gas phase) a chemical must have a large $\Delta S_{\mathrm{vap}}$, which means that each molecule gains a significant amount of 'freedom of movement' from entering the gas phase.  This makes sense in the case of ethanol versus water in Dissenter's question: ethanol is bulkier than water, so even though they both have similar values of $\Delta H_{\mathrm{vap}}$, ethanol gains more 'freedom to wiggle' in the gas phase and thus has a lower $T_b$.
This also helps to provide a qualitative explanation for why $T_b$ is proportional to pressure: as the pressure of the gas phase increases, the 'freedom to wiggle' gained by entry to the gas phase decreases, because higher $P$ means more collisions with gas-phase molecules.  Thus, one can expect $\Delta S_{\mathrm{vap}}$ to decrease with increasing pressure ($S_{\mathrm{liq}}$ isn't affected nearly as much by pressure changes as is $S_{\mathrm{gas}}$), which translates directly to the experimentally observed increase in $T_b$.
More details on the relationship between boiling point and pressure can be found in the Wikipedia articles on the saturation temperature and the Clausius-Clapeyron relation.  More information on equilibrium and Gibbs free energy can be found on www.chem1.com.

$^*$ In this model system, the "knob" one has available (see Gibbs' phase rule) to change the equilibrium pressure and temperature is to add to or remove from the system some number of moles of the species of interest.
$^{**}$ As orthocresol rightly notes in a comment, the equilibrium must be expressed in terms of the chemical potentials of each species present. However, in this case there is only one species present and thus examining the total molar Gibbs free energy is sufficient.

Brian, I disagree with your definition of boiling point as "the temperature at which the liquid and the vapor of the substance are in equilibrium". In fact, in those conditions the liquid and vapor of a substance will always reach an equilibrium at any temperature: this is governed by the vapor pressure of that substance and has nothing to do with being at the boiling point. Fortunately this does not affect the rest of the answer, i.e. your "boiling point equilibrium condition" happens to be correct for any temperature, as long as both phases are present, including at the boiling point.

To be pedantic, the condition for equilibrium is that the chemical potentials (which is, in this case, equal to the molar Gibbs free energies) of both phases are equal. Most of the equations should be written in terms of molar quantities, actually.

@orthocresol Pedantry appreciated; edit made. While $\Delta H_{\mathrm{vap}}$ was described obliquely as being intensive, it was definitely a point worth making more explicit throughout the explanation.

@MarcoB This was something that bugged me when initially writing up the answer, actually.  I have revised the answer somewhat to indicate the system considered in the analysis to be closed and composed solely of the chemical species of interest. I believe this makes the argument accurate, though I welcome critique as errors may well remain -- thermo is not my strongest subject.

Am accepting as answer since dissent tapered off.  Will gladly revisit if critique resumes in the future.


text:
Pourbaix diagram – writing reactions
How would you go about writing the reaction between water and chromium metal under standard acid and base conditions?
I know you would have to look at the vertical lines at pH 0 and 14 but I don’t know how you would know which species to look at and whether or not you write the reduction or oxidation of water.
Also, at pH 14 how would you calculate the ${\Delta G}$ value for the reduction between water and chromium metal?

related: https://chemistry.stackexchange.com/q/32509/7951 and https://chemistry.stackexchange.com/q/124897/7951

The red line in your diagram corresponds to the redox potential of hydrogen, which depends on $\mathrm{pH}$:
$$\begin{alignat}{3}
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\
\ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14)
\end{alignat}$$
The blue line in your diagram corresponds to the redox potential of oxygen, which depends on $\mathrm{pH}$, too:
$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\
\ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad\quad &&E^\circ = +0.401\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14)
\end{alignat}$$
Therefore, only areas between these lines are stable in aqueous solutions.
Your initial species is elemental chromium ($\ce{Cr}$). It is shown below the red line; therefore, it is unstable with respect to oxidation to $\ce{Cr^3+}$ at $\mathrm{pH}=0$
$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\\
\ce{Cr^3+ + e- \;&<=> Cr^2+}\quad\quad &&E^\circ = -0.408\ \mathrm{V}\\
\ce{Cr^2+ + 2e- \;&<=> Cr}\quad\quad &&E^\circ = -0.913\ \mathrm{V}\\\\
\ce{2Cr + 6H+ &-> 2Cr^3+ + 3H2}
\end{alignat}$$
and to $\ce{Cr(OH)3}$ at $\mathrm{pH}=14$
$$\begin{alignat}{2}
\ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\\
\ce{Cr(OH)3 + 3e- \;&<=> Cr + 3OH-}\quad\quad &&E^\circ = -1.33\ \mathrm{V}\\\\
\ce{2Cr + 6H2O &-> 2Cr(OH)3 + 3H2}
\end{alignat}$$


text:
Methylation of perchlorate
I am exploring (non-experimentally!) the methylation of a perchlorate precursor, either the acid or pre-existing salts, to produce the alkali methyl perchlorate. The methylating agent would be methane itself, and I would like to understand whether:

The product is possible from these (or more) reactants,
Whether the methane, as an organic combustible, would not simply deflagrate with perchlorate, leading to no other products (beyond fire or explosion),
If methyl perchlorate is deliquescent (as I believe that perchlorate esters hydrolyze rapidly with water), and
If low temperature conditions (< 5 °C) might permit '1' and avoid '2'.

Reference to any answers would be much appreciated, my interest being in recent events in the Siberian tundra...

The stated danger (at: http://chemistry.stackexchange.com/questions/32336/perchlorate-salts-explosive-reactivity-with-methane) of "...subsequent thawing, is almost assuredly a guarantee of detonation" is particularly interesting this Siberian regard...

https://sites.google.com/site/energeticchemical/alkyl-perchlorate-esters

How would perchlorate form and accumulate in large enough quantities to cause the Siberian divots? Though some occurs naturally in Desierto de Atacama, through accumulation of guano in the driest region on Earth,  it does not seem you'd have enough forming below the permafrost.

perchlorates occur naturally in nodularised evaporites and as aeolian deposits in dry, tundra environs, but it is also possible that the Siberian perchlorate is allochthonous, these 'divots' occurring in a region of discontinuous permafrost with through-going circulation that is continuous with the contemporary groundwater table.


text:
What's the solid product of the reaction of liquid ammonia with excess sulfur?

What would happen when you mix liquid $\ce{NH3}$ and excess of sulphur?

The answer key states that a solid with $\ce{N}$ and $\ce{S}$, and a gas which gives a black precipitate with lead acetate is evolved.
Surely, $\ce{H2S}$ gas will evolve. What about the solid? Is it ammonium sulphide or ammonium sulphite?

Are you sure they meant excess sulfur and not sulfur dioxide? The reaction of sulfur dioxide with ammonia is a famous industrial reaction, so that might be the reaction in the test. Though I reckon the products would be the same.

Yes they meant excess sulphur.

If you’re sure that you only have sulphur and liquid (i.e. cold, pure, not dissolved in water) ammonia, then only one of your two suggestions is possible.

I don't think this question has a simple answer.
This paper's first page (all I can see without a subscription) indicates that sulfur can dissolve in liquid ammonia, producing hydrogen sulfide (as the question correctly states) and... complicated stuff: S7N-, which disproportionates into S4N- and "an" ammonium polysulfide (maybe (NH4)2S6, maybe another composition).
The paper's first page also refers to the system's light sensitivity, so if you don't keep it in the dark, you may get different products.

If you really mean sulfur plus ammonia, then the solid cannot contain oxygen which is not present in the system.
There could be several products, but we would expect some sort of sulfur-nitrogen compound such as $\ce{S4N4}$, a reaction referred to here.

We could write any reaction of N and S such that $\ce{x N + y S -> N_xS_y}$. I wish to ask how "we would expect" that only $\ce{S4N4}$ would be formed, and not any other compound (like $\ce{S5N5}$)?

There are multiple sulfur-nitrogen compounds, really.  But $\ce{S4N4}$ is a relatively common and well known one.

Okay, I was only worried, even though $\ce{S4N4}$ may be a common and well-known compound, if the reaction conditions allow for its formation or not?

Hard to say.  The problem stayement is not very specific.


text:
How to read TINKER's force field parameters
I'm going to parameterize AMBER force field for the simulation of some small molecules. I have found the parameter sets provided by TINKER at here.
I'm stuck here because I couldn't find any explanations about the parameter values. They should be the parameters that appear in the formula of AMBER force field. However, I have no idea which numbers mean $k_b$, $l_0$, $k_a$, $θ_0$, or the rest parameters.

Have you looked at the Generalized Amber Force Field (GAFF) which is already parameterized for many small molecules?

Thank you for your comment. I found the GAFF parameter set, but still don't know which column stands for which parameters. Also there's no units given.

Oh I think this link (ambermd.org/formats.html) is exactly what I was looking for.

Okay I will take a stab at this.
Consider the following bond stretching parameters in the TINKER FF file:
      ##################################
      ##                              ##
      ##  Bond Stretching Parameters  ##
      ##                              ##
      ##################################

bond          1    1          310.00     1.5260

This tells you that this is a bond parameter between atom 1 and atom 1.  The index of atoms is given in the FF file as
   #####################################################
   ##                                                 ##
   ##  TINKER Atom Class Numbers to Amber Atom Types  ##
   ##                                                 ##
   ##    1  CT      11  CN      21  OW      31  HO    ##
   ##    2  C       12  CK      22  OH      32  HS    ##
   ##    3  CA      13  CQ      23  OS      33  HA    ##
   ##    4  CM      14  N       24  O       34  HC    ##
   ##    5  CC      15  NA      25  O2      35  H1    ##
   ##    6  CV      16  NB      26  S       36  H2    ##
   ##    7  CW      17  NC      27  SH      37  H3    ##
   ##    8  CR      18  N*      28  P       38  HP    ##
   ##    9  CB      19  N2      29  H       39  H4    ##
   ##   10  C*      20  N3      30  HW      40  H5    ##
   ##                                                 ##
   #####################################################

So clearly this is a bond parameter between CT and CT.  What do these atom notations stand for?  They are outlined in Table 1 of this paper.  The value 310.00 is $K_r$ (in kcal mol$^{-1}$)  and 1.5260 is $r_{\mathrm{eq}}$ (in Ang).

For angle bends:
      ################################
      ##                            ##
      ##  Angle Bending Parameters  ##
      ##                            ##
      ################################

angle         1    1    1      40.00     109.50

angle denotes what you'd expect.  The last two parameters, 40.00 ($K_{\theta}$ in kcal (mol radian$^2$)$^{-1}$) and 109.50 ($\theta_{\mathrm{eq}}$ in deg.) are for CT-CT-CT (atom1-atom1-atom1).

For torsional angles:
      ############################
      ##                        ##
      ##  Torsional Parameters  ##
      ##                        ##
      ############################

torsion       1    1    1    1            0.156    0.0  3

This is clearly a torsion between CT-CT-CT-CT.  Here is where it gets a little tricky.  If anyone finds this to be erroneous, please let me know and I will correct it.  The parameter, 0.156 is simply determined by the following formula:
$$\frac{\frac{V_n}{2}}{\mathrm{no.~of~paths}}$$
where $V_n/2$ is the magnitude of torsion in kcal mol$^{-1}$.  In Table 14 of the previously cited paper, for Torsional Parameters you find for X-CT-CT-X:

no. of paths = 9
$V_n/2$ = 1.40

Therefore, $1.40/9$ $\approx$ 0.156 which is exactly what is given in the TINKER FF file.  I've done this with a few other torsional parameters and it seems to add up. I believe footnote $f$ in Table 14 explains this.
The 0.0 is simply $\gamma$ (phase offset in deg.) and 3 is $n$ (the periodicity of the torsion).
Functional Form of Amber94 FF
It may be useful (and convenient) to spell out the functional form as presented in the cited paper here.
$$E_{\mathrm{tot}} = \sum_{\mathrm{bonds}} K_r(r-r_{\mathrm{eq}})^2 +$$
$$~~~~~~~~~~~~\sum_{\mathrm{angles}} K_{\theta}(\theta - \theta_{\mathrm{eq}} ) ^2 + $$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{\mathrm{dihedrals}} \frac{V_n}{2}[1+\cos(n\phi - \gamma)] + $$
$$~~~~~~~~~~~~~~~~~~~~~~~\sum_{i<j} \left[ \frac{A_{ij}}{R_{ij}^2} - \frac{B_{ij}}{R_{ij}^6} +
\frac{q_i q_j}{\epsilon R_{ij}} \right]$$

Reference: Cornell and co-workers. J. Am. Chem. Soc. 1995, 117, 5179.


text:
how to destroy excess iodomethane in reaction mixture?
I have used iodomethane to make a phenolic $\ce{OH}$ group to $\ce{OMe}$ group in Aq+Methanol medium. I have added water to precipitate the product of mine and will evaporate the solvents but I cant use rotavap until I am sure that any residual iodomethane is not contaminating the condenser of rotavap. How shall I destroy excess iodomethane? will aq. solution of $\ce{KOH}$ work?

Yes, you can destroy small amounts of methyl iodide by hydrolysis with an aqueous solution of sodium hydroxide. You can just add some sodium hydroxide solution to your reaction mixture, if your molecule is stable to basic conditions.
The reaction is: $\ce{NaOH + MeI -> MeOH + NaI}$
From: "A Comprehensive Guide to the Hazardous Properties of Chemical Substances" by Pradyot Patnaik, 2007


text:
Equilibrium constant in terms of partial pressures
$K_p$ is defined as $\Pi_j(\frac{P_j}{P^{\Theta}})^{\nu j}$ where $P_j$ is the partial pressure of component j and $\nu$ is the stoichiometric coefficient. I have a couple of questions already. Firstly, $K_p$ is meant to be dimensionless so therefore, $\frac{P_j}{P^{\Theta}}$ must be dimensionless. However, surely $P_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?
Given that, suppose I want to write the expression for $K_p$ in terms of partial pressures for the following reaction:
$$\ce{\frac12 O2_{(g)} + C_{(s)} <=>CO_{(g)}}$$
Do I assume that $\frac{P_{c}}{P^\Theta}=1$ leaving me with:
$$K_p=\frac{P_{\ce{CO}}}{(P_{\ce{O_2}})^{\frac12}}\frac{1}{P^\theta}$$

Actually, the definition of partial pressure is the pressure of the gas that it would have if it was the only gas in the container. So the partial pressure ($P_j$) does have units.

However, surely $P_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?

You are wrong to assume that partial pressure has no units.  Partial pressure has units of pressure, such as mmHg, atm, bar, psi, etc.
For ideal gas mixtures, $y_j=\frac{P_j}{P_{tot}}$ or equivalently $y_j P_{tot}=P_j$, i.e. the partial pressure $P_j$ of a gas is equal to its mole fraction $y_j$ times the total pressure $P_{tot}$.
The expression you gave for $K_p$ is dimensionless because $P^\Theta$ has the same dimension as $P_j$.

Do I assume that $\frac{P_C}{P^Θ}=1$....

That part sounds good.

leaving me with $K_p=\frac{P_{CO}}{(P_{O_2})^{1/2}}\frac{1}{P^\Theta}$

Not sure about that one.  How is $K_p$ dimensionless in what you wrote?

The standard change in free energy, which is used to determine the equilibrium constant Kp is based on the pure reactants at 1 atm going to pure products at 1 atm.  Therefore, the partial pressure must be expressed in atm in order to be consistent with Kp.  This essentially normalizes the partial pressure by 1 atm.


text:
Does pressure affect non-gaseous reactions?
Pressure holds a great significance when it comes to reactions involving gaseous species. Changing pressure can affect the kinetics and even mechanisms of such reactions.
However, does pressure hold any significance in reactions involving solid and liquid reactants only?

Examples, if any, would be appreciated.

Yes. Pressure, as a concept, is simply the force exerted by random molecular motion per unit area, and since all matter is composed of molecules, and all matter above absolute zero has some random motion, everything has molecules exerting some force over some area. (This pressure is the vapor pressure of the solid, which is typically extremely low. If the external pressure on the solid is higher than the vapor pressure of the solid, the solid molecules will not fly off and escape. Things with high vapor pressure exert a high "pressure" on their surroundings and thus escape easily - jet fuel or nail polish remover have very high vapor pressures.)
If you have a liquid reaction between molecules A and B, bringing the liquid to a higher pressure will increase the frequency of collisions between molecules of A and B (by the definition of pressure). Since there is a finite probability of the reaction proceeding for any given collision, increasing the collisions increases the rates.
Practically, for you to see that kind of effect in a laboratory would require a very pressure-sensitive reaction, large molecules whose shape changes with pressure, or a reaction with a large change in the number of moles. Biological chemistry is rich with examples pressure-sensitive non-gaseous reactions, mainly due to pressure changing the shape and behavior of large complex molecules.
There's also the effect of osmotic pressure, which is a pressure induced not by random molecular motion but by the tendency of the substance to seek a lower free energy state. Osmotic pressure will affect concentrations (the system will seek to reduce the concentrations of ions), which will affect reaction rates.
Increasing the pressure of solids and liquids can also change their physical properties (heat capacity and thermal conductivity are two excellent examples). These physical properties are innately tied to the reaction mechanism. Since most reactions occur in sequence, if the physical properties are different (and if they change over the course of the reaction), pressure will exert an effect there.
Finally, there is an effect technically excluded by your question, but important nonetheless, and that is the effect of pressure on solid-gas reactions (sublimation or deposition). The manufacture of integrated circuits involves depositing materials onto chips in layers, and pressure is an important way to control that rate.
In general, the more nonlinear the reaction mechanism, the more likely it is that the small effects that pressure creates will cascade into larger changes that are what you would call "significant."

Very simply,  if you exert pressure on a liquid reaction, you can use temperatures in excess of the RTP/STP boiling points. So, your reaction is now no longer bounded by the particular boiling point of a solvent and new regions can be investigated.  e.g.  over 100 deg C is easily achievable for THF or DCM at 20 bar.
In addition, you can restrict off-gassing during a reaction by applying pressure.  This may or may not be desirable.


text:
hydrogen environments of benzophenone
The hydrogen in $\ce{C4}$, $\ce{C6}$, $\ce{C10}$ and $\ce{C8}$ belong to same hydrogen environment.
The hydrogen in $\ce{C3}$, $\ce{C1}$, $\ce{C11}$ and $\ce{C13}$ belong to the another hydrogen environment.
Hydrogen in $\ce{C12}$ and $\ce{C2}$ belong to the once again, another hydrogen environment.
$\ce{C3}$, $\ce{C1}$ , $\ce{C11}$  $\ce{C13}$, $\ce{C12}$ and $\ce{C2}$ all have two hydrogen on the neighboring carbon atoms.
So my question is why aren't they classified as one hydrogen environment? Instead, why are the categorized under two different environments?

Whether environments are the same or not is based on symmetry considerations and not the number of hydrogens attached to the adjacent carbon atom.
Molecules and atoms within a molecule can be of three types:

identical - they can be interconverted by simple rotations or translations.
enantiotopic - they can be interconverted by reflection in a mirror or about a center of symmetry
diastereotopic - they cannot be interconverted.

Molecules or atoms within a molecule that are found to be identical are said to be in the same environment.  Molecules or atoms that are identical, or in the same environment, must be identical with respect to all chemical and physical properties.  These properties would include chemical reactivity, spectroscopic environment, etc.
In benzophenone, I can rotate the benzene ring about the $\ce{C5-C7}$ bond. This rotation interconverts $\ce{C4}$ and $\ce{C6}$.  If I told you to look away and I performed this rotation (and we removed the numeric subscripts), there is no way that you could tell if the original carbon or the other carbon was now in the $\ce{C4}$ location.  These two carbons (and their attached hydrogens) are therefore said to be equivalent by the above definition.  The same applies to $\ce{C1}$ and $\ce{C3}$, they are also equivalent to each other, or in the same environment.
I can also rotate the molecule around the $\ce{C7=O}$ carbonyl axis.  If I do this I can interconvert $\ce{C4, C6, C8, C_{10}}$.  This same rotation also (separately) interconverts $\ce{C1, C3, C_{11}, C_{13}}$, and also (separately) $\ce{C2}$ and $\ce{C_{12}}$.
Therefore, these 3 sets of carbons

$\ce{C4, C6, C8, C_{10}}$
$\ce{C1, C3, C_{11}, C_{13}}$
$\ce{C2, C_{12}}$

are different (diastereotopic) from one another (I can't interconvert $\ce{C4}$ and $\ce{C_{12}}$, for example).  So we have 3 different chemical environments, but within each environment there are several identical atoms. The same methodology and reasoning applies to the hydrogens attached to these carbons.  They also exist in 3 different environments, and within each of these 3 environments there are several equivalent hydrogens.


text:
Reaction-enthalpy for synthesis of 40 g hydrazine

Calculate the reaction-enthalpy for the synthesis of 40 g hydrazine (rocketfuel): $$\ce{4NH3(g) + Cl2 (g) -> N2H4(l) + 2 NH4Cl(s)}$$

Attempt at solution:
First we have to account for the decomposition of four moles of ammonia into its elements.
Since $\Delta H_\mathrm f^o(\ce{NH3})$ = $-46.1 \mathrm{~kJ~mol^{-1}}$, we do:
$$\Delta H_1 = 4(+46.1) = 184.4 \mathrm{~kJ}$$
$\ce{Cl2}$ is in its standard state so we can ignore that.
We know that $\Delta H_f^o(\ce{N2H4})$ =  $50.6\mathrm{~kJ~mol^{-1}}$. Therefore, $\Delta H_2 = \pu{50.6 kJ}$.
Then, $\Delta H_\mathrm f^o(\ce{NH4Cl})$ =  $-314.2\mathrm{~kJ~mol^{-1}}$ to give
$$\Delta H_3 = 2(-314.4) = -628.8~\mathrm{kJ}$$
Finally,
\begin{align}
\Delta H_{\mathrm{rxn}} &= \Delta H_1 + \Delta H_2 + \Delta H_3\\
&= 184.4 + 50.6 + (-628.8) = -393.8~\mathrm{kJ}
\end{align}
However, my textbook says the correct answer is $-492.3~\mathrm{kJ}$.
Can someone tell me what I did wrong, please? Also, I didn't account for the $40\mathrm{~g}$. How does that affect the problem?

For the 40g issue, consider that any energy values you calculate are "per mole", i.e. they refer to one mole of product. You have to take into account how many moles of hydrazine correspond to 40g of hydrazine.

You are on the right track. Basically, you are just missing the last step. Your answer for the enthalpy of reaction should be $ΔH_{rxn}=−393.8\frac{\mathrm{kJ}}{\mathrm{mol}}$. So all you have to do is find the moles of hydrazine
$n=\frac{40}{32.0452}=1.248\space\mathrm{moles}$ and multiply it by your $ΔH_{rxn}$.


text:
Electron affinities of the chalcogens and halogens
Here are the electron affinities of the 16th and 17th groups.

The general trend for electron affinity down the group is that it decreases because of the increase in atomic radius.The exception of $\ce{Cl}>\ce{F}$, I can understand is because fluorine has a high electron density and it is unfavourable to add more electrons as it would only increase the electron -electron repulsion.
The same reasoning can be applied to $\ce{S}>\ce{O}$. But oxygen's electron affinity is lesser than all the elements down group 16. Why is this so? And how is it so different from group 17 or why doesn't group 17 have $\ce{F}$ being less than the rest?.
I tried to apply the reasoning that the electron-electron repulsion in oxygen is so much that it is less favourable for $\ce{O}$ than any other element in group 16 to gain an electron. But in that case $\ce{F}$ is an even smaller atom hence with even more repulsion. So why are these two groups so different?

The amount of screening is the same in the two groups but the effective positive charge that the incoming electron feels is stronger in Group 7.
For instance, oxygen has the electronic structure of
$$\ce{1s^{2} 2s^{2} 2p_{x}^{2} 2p_{y}^{1} 2p_{z}^{1}}$$ and 8 protons. The effective positive charge that pulls the incoming electron (which approaches the second level) is $8 - 2 = 6$.
In fluorine we have,
$$\ce{1s^{2} 2s^{2} 2p_{x}^{2} 2p_{y}^{2} 2p_{z}^{1}}$$ and 9 protons. The effective positive charge is $9 - 2 = 7$.
Therefore, the net pull from Group 7 is stronger than Group 6.


text:
What is the correct Lewis structure of diazomethane?
I tried to google and watch YouTube videos on how to write Lewis structures properly, but I can't seem to understand how to do this correctly.
My teacher, when class was in session, gave me a set of instructions to complete Lewis structures:

Sum Valence electrons
Add electrons to outer atoms, while obeying octet or duet rule.
Then add electrons to center atom, obey octet rule, unless they're third row element or below.
Do formal charges for each atom valance charge - (unbonded electrons + 1/2(bounded electrons))
Once formal charges have been determined, minimize formal charges by using unbound electrons to make double or triple bonds
Redo formal charges to check if formal charges are as low as possible, for central atom.

So I attempted to do this for the diazomethane, but my answer (circled in green) does not match the book's answer (in red).

What did I do wrong?
After further considerations, I realized that my structures didn't obey the octet rule. So then would this revised structure be also correct?

I think both images in the red box are showing the "formal charge" on their respective atoms. Formal charge is similar to oxidation state used in inorganic chemistry as a bookkeeping scheme. The charge in either system is not to be taken literally, but merely gives us a handle on the electrons to track their movement.

The image above depicts three mesomeric structures of diazomethane ($\ce{CH2N2}$). When you count the valence electrons for every atom, you will find that only the left and middle structure satisfy the octet / duet (for hydrogen) rule, i.e. that every $\ce{C}$ and $\ce{N}$ atom has 8 valence electrons, and every $\ce{H}$ has two valence electrons. The structure on the right side does not obey the octet / duet rule because the positively-charged nitrogen only has an electron sextet. The same is true for the two structures you have drawn with an $\ce{N-N}$ single bond. In both cases, at least one nitrogen has only 6 valence electrons.


text:
Why does milk overflow when boiled?
I was wondering that like whenever I boil milk I just have to keep standing in front of it to make sure that I don't waste any milk but then I was wondering like Why does it even overflow ? In case of water it doesn't happen. What makes it to behave like this ?

Milk contains surfactants (primarily lipids), which allow stable bubbles to form. In intense boiling, these bubbles can become mixed with liquid, supporting the liquid above them and forming what I'll call for want of a better term a "wet froth." Imagine a bunch of sponges made out of liquid. They take up more space than they would if you compressed them down and forced out the air. The inclusion of the water vapor in the bubbles lowers the density of the mixture, causing it to expand and overflow the pot.
The state is only stable because the added energy keeps creating new bubbles to replace any that have popped and keeps forcing liquid upward, while it would normally fall.
Water does not experience this mostly because its bubbles are far less stable. In fact, pure water does not form bubbles at all. The bubbles formed in water is typically due to impurities.  If you were to add soap to a boiling pot of water, you would see the exact same effect, if not worse.

This is not a third of the problem.

Milk proteins denaturate in the heat and form a stable film on the surface of water bubbles. Bubbles on the surface burst in the beginning, and so the film becomes more and more stable there, stablilising the "foam" below.
The correct analogy is a pot of boiling water with pasta: As long as you leave the lid on, it will keep boiling over.

The reason is because milk is an emulsion of fat and water. When you boil milk, the fat separates to the top and forms a layer on top. The water below boils and vaporizes to form steam, but is trapped by the layer of fat above. It pushes the layer up, causing milk to rise.


text:
What are the conditions for liquid metal to be corrosive to another, different solid metal?
I've had already been aware of the fact that metallic aluminum can be degraded by coming in contact with mercury, but also came across another question more recently here explains a similar reaction with liquid gallium.
So my question is more general regarding corrosive action of any liquid metal on other metals.
What are the more general conditions for a liquid metal to be corrosive to another, different solid metal?
Perhaps the word corrosive is not the most suitable choice. Corrosion refers to a chemical reaction. If I understand it right what's happening, at least in the case of $\ce{Ga}$ and $\ce{Hg}$ on $\ce{Al}$, is a weakening of the grain boundaries by displacement and amalgamation.
Another interesting thought occurred to me. Liquid sodium is often used as a coolant in nuclear power plants. It's pumped through metal pipe (material unknown). So here is a good example of a combination that doesn't lead to corrosion. Are there examples other than I've mentioned that lead to corrosion? So what makes the difference?

I think you're looking for solubility. Although interstitial inclusions might play a role in some contexts, like iron exposed to high carbon environments.

In addition to solubility, which is important, the gallium corrodes aluminum by speeding oxidation: it removes the protective surface film passivating Al, allowing air or water to attack the metl.

I have inserted the comment into the question body. If you want to add more information, request or clarify anything of your question, you can [edit] it any time. This way it won't be overlooked.

There are no general conditions- indeed there no general conditions for the any of the forms of corrosion. Solubility is a factor and many other factors but it just comes down the actual interactions between a given liquid metal and another metal and the service conditions such as temperature, velocity, impurities.
Liquid-metal attack has the following categories:

simple solution,
alloying between liquid metal and solid metal,
intergranular penetration,
impurity reactions,
temperature-gradient mass transfer,
concentration-gradient mass transfer, or dissimilar-metal
mass transfer.

With that many means- determining what liquid metal damages what solid metal  is very difficult.
http://web.ornl.gov/info/reports/1956/3445603500959.pdf