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How do you separate enantiomers?
There are some stereochemical reactions that result in the presence of enantiomers. When moving forward with a practical organic synthesis, how does one usually separate them in order to continue with one of the enantiomers?

If you're a bad-ass like Louis Pasteur, you do it with a microscope and a pair of tweezers.

There are several ways that enantiomers can be separated, but none of them are particularly simple.
The first way to separate them is chiral chromatography.  In chiral chromatography, silica gel is bonded to chiral molecules to form what is called a chiral stationary phase.  The enantiomers will then separate as they run down the column because one of the enantiomers will interact more strongly with the column and "stick" in place.  Chiral sugars (ex. cellulose) are frequently used in chiral chromatography.
The second common method is to react the enantiomers with another chemical to form diastereomers.  While enantiomers are identical in terms of chemical properties, diastereomers are not.  Diastereomers can be created by reacting a mixture of both the enantiomers with another chiral molecule, such as s-brucine, which is commonly used because it is cheap.  Diastereomers have different chemical properties (for example melting points), so it is much easier to separate them.  Then, after separation, the enantiomers can be recovered from the single diastereomer.

Brucine seems like a scary chemical to work with in large quantities. Obviously not used for the same substrates, but tartaric acid sure seems less scary.

The physicochemical properties of enantiomers being the same (except for optical activity), one needs to make use of the chirality of the molecules to separate them. This is done by asymmetrizing the environment of the molecules. I'll explain.
Lets say you have a racemic mixture of a chiral carboxylic acid. To effectively differentiate both enantiomers, you can use an enantiopure amine to form a salt with your carboxylic acid. You've then created two diastereomers from the combination of chiral molecules. Since the physical properties of diastereomers are different, you can then separate them.
One way to do this would be to crystallize your salts, extract/wash them, then pass them through a flash chromatography column, and then get rid of the amine.
Another option would be to use a chiral chromatography column, which would effectively have the same effect.
There are also analytical techniques that can differentiate between enantiomers, for example MEKC used with chiral cyclodextrins.

I think you've gotten the point that, to separate enantiomers, you need to create diastereomers.

This isn't best solution to this problem, as its usage would be quite limited, but nonetheless it is still a solution: using enzymes specific to the enantiomer you wish to remove. As an example, using enzymes specific for D-amino-acids would leave you with its L-counterpart untouched. They can then be separated as the "product" form of the D-amino-acid is likely to have changed its physical properties.


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Why is it that the bigger the side-chain is, its hydrogen atoms tend to be more equatorial?
The most stable cyclohexane form is the chair conformation but on the other hand, the bigger the side-chain of the cyclohexane is, its hydrogen atoms become more equatorial rather than axial, which brings them closer to the cyclohexane's hydrogens.
Shouldn't they arrange in a less energetic, but a more stable conformation?

The general rule is that larger substituents prefer the equatorial positions. I'm not sure I understand what you're asking here, which hydrogens do you mean?

""Shouldn't they arrange in a less energetic, but a more stable conformation?"" The "but" is nonsense.

Consider rewriting your question and choosing a more appropriate title, as both are a bit confusing. Also, @MadScientist, he seems to be talking about the side-chain's hydrogen atoms.

@CHM: but then "equatorial" is not exactly an appropriate adjective for those hydrogens, seeing that they're not attached to cyclohexane... yes?

It's not exactly clear from your question what distances you consider, but I have taken snapshots of a methylcyclohexane with the methyl group in the equatorial and axial positions:
      (equatorial)
    (axial)
The distance of the methyl $\ce{H}$ atoms to the nearest cyclohexane $\ce{H}$ atoms are displayed, and you'll see that indeed the axial position has a shorted $\ce{H–H}$ distance of ~ 2 Å, while equatorial position has longer $\ce{H–H}$ distances.
Also, as others have remarked in the comments: only the groups (and $\ce{H}$ atoms) directly on the cyclohexane ring are said to be in axial or equatorial position. Atoms of side chains do not follow this terminology, in particular for side chains that can rotate freely as it means nothing there.

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Availabilty of magnesium ions in phosphate buffer
Phosphate buffer is a common buffer in biological applications, it's especially popular for NMR. Magnesium ions are necessary for many biological systems to work, though the amount of magnesium ions you can add in a phosphate buffer is limited by the relatively low solubility of magnesium phosphate.
What I'm curious about is whether the phosphate buffer can affect the availability of magnesium ions even in concentrations where no visible precipitation occurs? Common phosphate buffer concentrations are around 10-100 mM, and magnesium ions are usually 5 mM or below. At those concentrations I see no visible precipitation, but is the magnesium actually fully available or is the phosphate sequestering it and thereby lowering the effective magnesium ion concentration in the solution?
Even just calculating the solubility of magnesium salts in phosphate buffer is not straightforward, as there are multiple species of magnesium phosphate and possibly even other salts like $\ce{MgKPO4}$.
Does phosphate buffer affect the availability of magnesium ions in solution compared to a non-interacting buffer? If that is so, is there a way to calculate the size of that effect?

Could you provide a link to what phosphate buffer is and what it contains?

@F'x Just potassium phosphate, not one of the fancier biological buffers.

Technically, phosphate can and probably do form associates with magnesium ions. To find how strong, you should search databooks for relevant equilibrium constants, it is not such common info to keep it in mind if not needed.

I am using a buffer which has approximately 50 mM Phosphate and 7 mM Mg, There is no visible precipitate. But my experiments are Mg sensitive and at least 5 mM Mg is needed for proper response. My preliminary experiments are indicating that in the presence of phosphate free Mg is limited.

If you're still interested in this, fluorescence may be a good method to test the phosphate dependence on magnesium activity. First record a fluorescence spectrum of a solution of magnesium and fluorescent dye, then add small volumes of concentrated phosphate and note any changes in the fluorescence. Of course, you would have to do a control experiment to see if the phosphate is directly perturbing the fluorescence. This write-up describes some magnesium-chelating dyes that you could use, though if you don't want to spend a lot of money on a dye, you may already have some chemicals in your lab that will work. Phenol looks like promising alternative as it fluoresces, is decently water soluble (83 g/L), and forms a complex with magnesium which breaks planarity, which would certainly change the fluorescence spectrum.
It would only take a day to get accurate quantitative results if you have access to a fluorometer.

I don't have a source (at the moment) but it is accepted that high phosphate buffers do chelate/sequester Mg2+ ions. Pyrophosphate is particularly infamous for this. During RNA transcription experiments, the solution will get cloudy which is due exactly to this phenomena. It is also seen in cell-free protein synthesis which requires a steady concentration of Mg2+, that the increased concentration of the phosphate buffer requires an increased amount of Mg2+ to maintain the same effective free Mg2+ concentration [1].

Do you have a source for plain phosphate? I have no pyrophosphate in the buffer.

If you deal with cell culture its irrelevant; because the magnesium channels located at the surface of the cell will be forcing magnesium ions to flow into the cell, and if the balance between free and chelated magnesium ions is disturbed in favor of chelate – it will produce free magnesium ions due to the chelate constant.

I'm not dealing with any cells here, just simple, purified macromolecules

So you should be more specific asking question – what do you mean by "availability of magnesium ions"? availability to what? if simple solution is involved you can calculate and compare the solubility ratio of your possible reaction

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Are there any major exceptions when comparing electron affinity?
I was tasked with figuring out whether carbon or nitrogen has a more negative electron affinity value. I initially picked nitrogen, just because nitrogen has a higher $Z_\mathrm{eff}$, creating a larger attraction between electrons and protons, decreasing the radius, causing a higher ionization energy, and therefore decreasing the electron affinity value, but I was actually wrong, and the solutions manual explains it as this:

"As you go from C to N across the Periodic Table you would normally expect N to have the more negative electron affinity. However, N has a half-filled p subshell, which lends it extra stability; therefore, it is harder to add an electron."

Are there any major exceptions to the rules when comparing electron affinity? I'm hesitant to use nitrogen as an exception, because I don't know how far it extends. If nitrogen has a more positive EA than carbon, does that also extend to boron, aluminium, or phosphorus?
I later found that this also applies when comparing silicon and phosphorus. The explanation given was the same.
What exceptions should be noted when comparing electron affinities? Are there any at all? And how far does the exception with atoms with half-filled p subshells extend?

This exception rule is actually orbital filling rule. For two electrons to be in same orbital they need to have different spins (Pauli exclusion principal). This electron pairing requires additional energy and thus it is easier to add electrons if there are free orbitals. When element has a half-filled p sublevel all 3 orbitals have one electron and pairing takes place (difference between energy levels of 2p and 3s is greater than electron pairing energy).
Electron pairing effects have significant impact to physical properties of coordination complexes (like color and magnetic properties).

I say the reason why the electron affinity of fluorine is not as negative as chlorine and that of O is not as negative as S is because of the electron repulsions in the small compact atoms keep the added electrons from being tightly bound as we might expect. . .

Exceptions abound in electron affinity. Another case is in that of $\ce{F}$ versus that of $\ce{Cl}$. You would think that $\ce{F}$ being far more electronegative, would have the more negative electron affinity, but actually, that is not the case. The small size of $\ce{F}$ makes another electron energetically unfavorable due to electron-electron repulsion. $E\ce{(F)} = -328 kJ/mol$, while $E\ce{(Cl)} = -349 kJ/mol$
In general, exceptions arise when new subshells are being filled/half-filled, or in cases where the atom is too small. In the first case, $\ce{Be}$ and $\ce{Mg}$ are interesting examples: they have a positive electron affinity (just like $\ce{N}$, in fact) because of the energy difference between the s and p subshells. This is no longer the case in $\ce{Ca}$, which as a low-lying 3d orbital; $E\ce{(Ca)}$ is $-2 kJ/mol$.


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Why are there two Hydrogen atoms on some periodic tables?
Most periodic tables only feature one Hydrogen atom, on the top of the first group. But some, like the one I was given, also show Hydrogen in the 7th group, to left of Helium.
Why are there two Hydrogen atoms? What's the difference between the two, and why do we work with the left one more often than the right?

The two hydrogens are the same, but some periodic tables show hydrogen in both places to emphasize that hydrogen isn't really a member of the first group or the seventh group.
Hydrogen is a diatomic gas in it's elemental state, which is different from the other group one metals (and similar to the group seven elements).  At the same time, hydrogen usually loses its electron, similar to sodium and the other group one metals.

I agree with this statement. I just want to add that hydrogen can (occasionally) form the hydride ion H-, much similar to the group seven elements.

Also hydrogen has metallic properties if you put it under enough pressure.

""Also hydrogen has metallic properties if you put it under enough pressure."" This is not a chemical reason. One should avoid the terms metals or nonmetals when discussing chemical properties. Metal or not is a property of the elementary state, after some reaction it is no langer inmportant.

@Cardtrick Alkali metals form singly charged anions, too: alkalides

I have seen a few tables where the hydrogen is slightly lifted off the row of alkali metals.


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What makes an epoxide stable?
Overall ring strain seems to be a big issue when it comes to organic chemistry. That is why cyclopentane may be in an "envelope form" or why cyclobutane may be in a kinked, kite form. Both of these example molecules are not planar. So, why is epoxide (oxacyclopropane) a stable molecule, if at all?

Chemists talk about conformation, not "form". What do you mean by "stable"? Epoxides seem pretty reactive to me :)

""a stable molecule, if at all?"" Do You doubt the existence of epoxides? Some molecule being stable without giving conditions means "You can fill that in a bottle and it will suvive for some days at least" in lab slang.

You could say it is kinetically stable, but thermodynamically unstable.

Well, you have to ask: how stable? and compared to what?
Epoxides, including ethylene oxide, are generally considered relatively unstable molecules, with a high chemical activity and involvement in numerous reactions, including polymerization and thermal decomposition. However, ethylene oxide does exist as a molecule, as does cyclopropane, which also contains a three-membered ring.
So, let's compare for example the ring strain in ethylene oxide and cyclopropane. I found some computation data on their relative stability from these lecture notes, which indicate that “ethylene oxide has less ring strain than cyclopropane”. I would attribute this to the fact that the C–O–C angle has less strain because its “relaxed” value would be 104.5°, compared to the “relaxed” C–C–C angle of 109.4°.

Epoxides are stable because, first and foremost, they are ethers. Ethers are an exceptionally unreactive functional group. Why? Because alkoxides are poor leaving groups that do not participate in many common nucleophile-electrophile reactions (e.g., SN2). Of course, there are ways to force ethers to react, such as through an SN1 pathway utilizing strong acid catalysis.
Among the ethers, epoxides are (among) the most reactive due to the ring strain, which enables chemists to perform reactions with epoxides that do not work with other ethers. For example, the reaction of an epoxide with a Grignard reagent is one of the first carbon-carbon bond forming reactions introduced in introductory organic chemistry.
I believe it's for this reason, their early importance in organic chemistry education, that epoxides have a reputation as exceptionally reactive. However, most epoxides are very readily prepared, isolated through aqueous workup, and purified on silica gel with no deleterious effects. In fact, epoxides can be survive being subjected to strong nucleophiles if a more reactive group is present in the molecule. For example, an ester can be reduced with DIBAL in the presence of an epoxide, and recently I was able to carry out a synthesis that used butyllithium in the presence of an unhindered epoxide with no trouble.

You can't quite compare rings consisting only of carbon atoms with rings with oxygens in them. In the latter, the $\ce{C-O}$ bonds are polar, the electrons are closer to the oxygen atom which means the acuteness at the $\ce{C}$ atoms is not as much of a problem as it would be in a hydrocarbon ring.


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Are buckminsterfullerene-corannulene complexes actually stabilised by π stacking?
According to wikipedia and the references given therein, $\pi\cdots\pi$ stacking interactions are the result of interaction between the quadrupole moments of two aromatic rings, rationalising the stabilisation of perpendicular and offset-parallel association modes of the benzene dimer.
Conventional wisdom is that the supramolecular complexes of buckminsterfullerene and functionalised corannulenes (the so-called buckycatchers) are stabilised by $\pi\cdots\pi$ stacking. How can the quadrupole interaction explanation be reconciled with the pseudo-spherical symmetry of buckminsterfullerene? Surely the interaction must be entirely Van der Waals in nature?

In short, I think the answer is that although the overall symmetry of the buckminsterfullerene is almost spherical, the quadrupole–quadrupole interaction with other molecules with aromatic cycles are local (in addition to the ever-present van der Waals interactions).
I tried to think of a way to exemplify that, and drew the 2D picture below:

In this 2D model, the “double bonds” of the circular structure represent the individual aromatic rings of your buckminsterfullerene (which isn't superaromatic, i.e. electrons are not delocalized over its entire structure; think may be a key point), bearing each a local quadrupole moment (symbolized in the red/blue thingy). The other particle with a quadrupole is close to one of the ball's quadrupoles, and this local interaction dominates. In fact, other interactions maybe be stabilizing or not, depending on the orientation of the individual quadrupoles, but they are weaker that the local quadrupole–quadrupole interaction.

This sounds reasonable, and I didn't appreciate the lack of superaromaticity in $\mathrm{C}{60}$. Your suggestion is similar (but with crucial differences) to that of a researcher (focused on graphite intercalates) I got the chance to speak with, who speculated that the dihedral symmetry elements of the $I{h}$ symmetry may permit a molecular quadrupole. I won't accept the answer just yet to give others a chance to chime in.


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Why is distillation not a viable way to separate ammonia from water?
Since the steam pressure of ammonia is higher than that of water, I would expect distillation to be a reasonable way of separating a mixture of both.
However, in industrial applications known to me there's always a step of stripping the ammonia chemically from the vapor. Why is this necessary?

Whilst I can't answer the question, I can confirm that the water-ammonia system is not azeotropic. Do you have any idea of how much ammonia ends up being scrubbed?

I'm thinking about 5-10kg$NH_4$/t in the input, "pretty much all" of it will be scrubbed (no exact figures). The curve you show for ammonia looks qualitatively like the curves in my textbook examples for distillation.

What industrial process is this? Distillation is expensive.

More often than not, in industry, something is done with economics in mind. In this case it appears that for the concentrations and volumes, a chemical strip is the more economically viable option. An exercise for you might be to check that.

Ammonia stripping from digestate of a Biogas plant, the stripped digestate is to be used to dilute the feedstock for the plant, ammonia has to be stripped because of the inhibition of the biological process.

I don't see where a distillation more or more complicated components than the stripping system - but I may be wrong.

It has little to do with how complicated the system is (initial cost), but instead has to do with the amount of power consumed during the process (operating cost): distillation requires bringing the entire solution to boil and further providing enough heat to boil away your entire mixture; stripping, on the other hand, only requires elevated temperatures, a little base and a supply of hot air.

Additionnally, depending on the state in which one wants to obtain ammonia, the process might prove tedious: fractionnal distillation will require either cooling the condenser to -33°C or increasing the pressure of the setup to 10 atm. The former requires refrigerating systems, the latter will require more sample heating and handling of a pressurized system.

Having embarrassed myself with my first answer, let me try this again. Ammonia and water hydrogen bond exceptionally well. In fact, the $\ce{O-H...N}$ value is 29 kJ/mol while the $\ce{O-H...O}$ value is 21 kJ/mol. (Ref) So, if there is any water around, ammonia would be happy to become solvated as they recondense.
Also, ammonia is a weak base and reacts with water (which is not usually the case when you're doing a distillation). $$\ce{NH3 + H2O <=> NH4+ + OH-}$$ So, if there's any water around as the ammonia is distilling and they recombine, they would also react which would make it difficult to remove the last of the water. (Some reaction would occur because the lowest energy point of the equilibrium is not at reactants or products.)
So, as noted in the comments, distillations to prepare dry ammonia use a scrubber to remove the water.

(pure) dissolution is always some kind of physical reaction - a change of physical state, combined with a change in energy. In this sense the reaction (while very different to ethanol for example) is integral in any distillation. If it were not for ammonia-water interaction the mixture would be inseparable by distillation! (it would follow raoult's law).

I keep hoping someone will jump in and answer this question to save me from thinking about it. Odds and ends from researching this question that might help someone else answer it.

Clusters of water and ammomia, such as $\ce{NH3(H2O)20}$ as well as various other ratios, are known and have been investigated. I didn't find any references that reported them in vapor phase in distillation columns.
But, there is a reference that reports that water clusters of varying sizes are found in steam during distillation, so it seems possible that water-ammonia complexes could exist during distillation since hydrogen bonding is strong between water and ammonia in solution. (I did see reports suggesting their existence as atmospheric pollutants.)
Clusters involving both positive and negative ion complexes of ammonia and water have been reported. (Such as $\ce{(H2O)20NH4+}$ (Ref) or $\ce{(H2O)p(NH3)q-}$ (Ref)
It's difficult for me to tell from the abstracts (or the papers) whether these species are important in regular laboratory conditions, since the researchers are normally looking at them by MS.
A single $\ce{H2O}$ bonded with $\ce{NH4+}$ forms a very strong hydrogen bond 92.5 kJ/m.
Why doesn't ammonia form an azeotrope with water? Of the common solvents that can form effective hydrogen bonds with water, only methanol and ammonia don't form azeotropes. (ref)

The comments for the question mention stripping ammonia from a water in a digestate of a Biogas plant. Ammonia can be removed from waste streams by steam or air stripping after adding NaOH to suppress formation of ammonium ions. Reports suggest that this is fairly efficient (>99% removal). Perhaps the remaining ammonia in the effluent could come from the water-ammonia clusters that are not separated in the distillation. There also seem to be other options for treating the waste stream discussed on this forum.
That's it, done thinking about this one (and not looking for the bounty, but it would be nice to have an answer).

Its not clear if you are trying to make pure ammonia.  If your just trying to remove it or concentrate it then the process is rather simple. You raise the pH over 10 so that all the ammonia in solution is NH3 rather than a mix of NH3 and NH4.  Heat it up to boiling temperature ( which is lower that 212 F for the ammonia/water mixture) and exhaust the vapors into a cooled condenser.  Subsequent distillations will increase the concentration of ammonia.  I have found that each pass increases concentration by about a magnitude of 3 to 5 times. The concentration of the distillate will drop as the concentration of the liquid being distilled drops.

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Why do different substances have different boiling points?
For example, why does for example oxygen turn into gas at a much lower temperature than water?
Does it have anything to do with the molecular structure? A water molecule does have a more complex structure than oxygen, though the R-410A (a mixture of two gases commonly used in heating pumps) is much more complex than water, and it boils at -48.5 degrees Celsius.

The boiling point of a liquid depends on the intermolecular forces present between the atoms or molecules in the liquid since you must disrupt those forces to change from a liquid to a gas. The stronger the intermolecular forces, the higher the boiling point.
Two oxygen molecules are attracted to each other through London dispersion forces (induced temporary dipoles between the molecules) while water molecules are attracted to each other by hydrogen bonding (attraction of the + dipole on H in one molecule to the – dipole on an oxygen in an adjacent molecule) that is relatively strong. (Hydrogen bonding is an important intermolecular force for molecules where H is directly covalently bonded to F, O or N, which are quite electronegative and thus form bond with H with a relatively strong dipole.) London dispersion forces become more important for atoms and molecules with more electrons. Dipole–dipole attractions are also important in some molecules.

This is a good, to-the-point answer, so this comment is just supplemental for anyone wanting to explore the question further. This link provides a "dancing molecules" analogy that explains relationship between intermolecular charge bonding, gases, surfaces, and evaporation. Further down, it also addresses why crystals form at lower temperatures. It's heavy on analogy and requires no specific science background, but keeps the analogy scientifically accurate.

I would like to add that molecular mass may have something to do with boiling point. It generally takes more energy to move more massive objects, and so a higher mass may lead to a higher boiling point.  (It seems difficult/impossible to separate out the effects intermolecular interaction from mass.)

@Eric - Boiling point does increase with molecular mass - but not because of the increase in mass. It's a coincidental increase. The reason is that temperature is already a measure of kinetic energy - two systems at the same temperature have the same average molecular kinetic energy. The reason that boiling poin increases has to do with the increase in induced polarizability of larger atoms, and the increase in the number of contact points for larger molecules.

Because boiling point of different materials depend on the intermolecular forces present between the atoms. The stronger the intermolecular forces the higher the boiling point and the weaker the intermolecular forces the lower the boiling point.

Welcome to the site! Um, this better be as a comment since it doesn't add anything to the accepted answer and isn't explained enough about a question that has lots of room for the answer to be explained.


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Is aqueous medium necessary for complex formation? if so, why?
If dry ammonia gas is passed through anhydrous copper sulfate, will it turn blue (due to the formation of tetraamminecopper(II) complex)? Or will silver chloride form diamminesilver(I) complex in liquid ammonia?
If the above complexes do not get formed in non-aqueous mediums, why is it so?

It should be noted that ammonia is a mildly polar solvent and that (for instance) you can prepare $\ce{[Cu(NH3)6]^2+}$ by reaction with liquid ammonia1. The description of the geometry (two pages earlier) of $\ce{CuSO4.5H2O}$ (namely containing a square planar $\ce{[Cu(H2O)4]^2+}$ unit with two additional $\ce{[SO4]^2-}$ above and below the plane) suggests that ammine ligands could fulfill a similar role to water.
Bear in mind that passing ammonia gas over a solid will probably mean that if the reaction happens it will only happen near grain surfaces.
To answer your second suggestion, Housecroft and Sharpe1 identifies a reaction producing $\ce{[Ag(NH3)4]+}$ from $\ce{Ag2O}$ in liquid $\ce{NH3}$.
Hope this is informative.
[1] Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 2nd Edition; Pearson Prentice Hall, 2005; pp. 635–637, 693.

I cannot answer concerning the $\ce{[Ag(NH3)2]+}$ complex, because I don’t know enough about it.
I do know, however, that the tetraamminecopper(II) complex is better considered a tetraamminediaquacopper(II) complex $\ce{[Cu(H2O)2(NH3)4]^2+}$. Structurally, the four ammine ligands form an almost square planar geometry around the copper cation and have rather short bond lengths. The two aqua ligands are on the axial positions (trans if you wish) and have a significantly larger bond length due to Jahn-Teller distortion. The complex is in-between octahedral and square planar geometry. Professor Klüfers’ web scriptum to his general and inorganic chemistry course gives $\ce{Cu-N}$ bond lengths of $203~\mathrm{pm}$ and $\ce{Cu-O}$ bond lengths of $251~\mathrm{pm}$. This is consistant with $\ce{[Cu(H2O)6](ClO4)2}$ where the equatorial oxygens are $195~\mathrm{pm}$ from the central copper ion and the axial ones are $238~\mathrm{pm}$.
In liquid ammonia, there are no water molecules present that could be used to generate this complex. Instead, as Richard correctly stated, a hexaamminecopper(II) complex is formed. You could also argue that the concentration of ammonia is high enough in liquid ammonia (compare to ammonia concentrations in aquaeous solutions) that all six coordination sites can be filled.
Generally, there is no restriction for coordination compounds to form only in aquaeous solutions. Well-known examples would include every single palladium-catalysed organic reaction in organic sovlents, the Sharpless epoxidation which uses a $\ce{[Ti(tartrate)(OR)2]2}$ catalyst and many organometallic compounds that exist in coordination clusters in solution. In one lab course, I was assigned to generate $\ce{[Ni(DMSO)6][NiCl4]}$ (termed $\ce{NiCl2 . 3 DMSO}$) from anhydrous $\ce{NiCl2}$ in DMSO by dissolving and filtration.


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Pros and cons of Cartesian vs. Z-matrix representations of molecules?
Over the course of my studies, I have switched largely from using Z-matrix representations of molecular geometries in calculations to Cartesian representations.
The software that I use now makes it easy to add the sorts of constraints/restraints/transits that I would have previously used Z-matrices for, and I know that Z-matrix geometries can be problematic in large molecules* where minute changes in a bond angle or dihedral (due, for instance, to rounding errors/low-quality gradients) can result in large movements in peripheral atoms.
What pros or cons exist for either geometry definition that I don't know about? What circumstances recommend one representation over another?
*Or small molecules with silly Z-matrices.

Cartesian Space
In Cartesian space, three variables (XYZ) are used to describe the position of a point in space, typically an atomic nucleus or a basis function.  To describe the locations of two atomic nuclei, a total of 6 variables must be written down and kept track of.  The general ruling is that for Cartesian space, 3N variables must be accounted for (where N is the number of points in space you wish to index).
Internal Coordinates
Z-matrices use a different approach.  When dealing with Z-matrices, we keep track of the relative positions of points in space.  Cartesian space is 'absolute' so to speak.  A point located at (0,0,1) is an absolute location for a coordinate space that extends to infinity.  However, consider a two atom system.  The translation of the molecule through space (assuming a vacuum) will have no affect on the properties of the molecule.  An H2 molecule centered around the origin (0,0,0) is no different from the same H2 molecule being centered around (1,1,1).  However, say we increase the distance between the hydrogen atoms.  We now have altered the molecule in such a way that the properties of that molecule has changed.  What did we change?  We simply changed the bond length, one variable.  We increased the distance between the two atoms by some length R.  With Z-matrices, we keep tabs on internal coordinates: bond length (R), bond angle (A), and torsional/dihedral angle (T/D).  Using internal coordinates reduces our 3N requirement set by the Cartesian space down to a 3N-6 requirement (for non-linear molecules).  For linear molecules we keep tabs on 3N-5 coordinates.  When performing complex computations, the less you have to keep track of, the less expensive the computation.
Symmetry
Consider the following molecule, H2O.  We know from experience that this molecule has C2V symmetry. The OH bond lengths should be equivalent.  When using some sort of optimizing routine, you may want to specify symmetry in your system.  With a Z-matrix, the process is very straightforward.  You would construct your Z-matrix to define the OH(1) bond as being equivalent to the OH(2) bond.  Whatever program you use should automatically recognize the constraint and will optimize your molecule accordingly giving you an answer based off a structure that is constrained to C2v symmetry.  With Cartesian space this is not guaranteed.  Rounding errors can cause your program to break symmetry, or your program may not be very good at guessing the point group of your molecule based on the Cartesian coordinates alone.
Picking the Right One
As a preface, programs like Gaussian convert your Cartesian coordinate space (or your pre-defined Z-matrix) into redundant internal coordinates before proceeding with an optimization routine unless you specify it to stick with Cartesians or your Z-matrix.  I warn you that specifying your program to optimize using Cartesian coordinates makes your calculation much more expensive.  I find that I will explicitly specify 'Z-matrix' when I know I'm dealing with high symmetry and when I know my Z-matrix is perfect.
You will want to use Z-matrices on systems that are rather small.  If dealing with systems with high symmetry, Z-matrices are almost essential.  They can be rather tricky to implement and you will likely spend some time figuring out the proper form of your Z-matrix through trial-and-error.  If you wish to scan a particular coordinate, Z-matrices are also very helpful as you can tell a program to scan across a bond length, angle or torsion with ease (as long as you've properly defined that coordinate in your Z-matrix).
I use Cartesian coordinates for large systems, systems with very little or no symmetry, or when I'm in a hurry.

This seems to be a pretty comprehensive answer! Regarding your comment about the reduction of degrees of freedom in the Z-matrix specification w/r/t Cartesians, I would have thought the smaller number of variables would result in virtually meaningless performance improvements for nontrivial molecules.

Richard, the problem is that there can be very many specific scenarios where Cartesian space can actually be more efficient than using internals.  My post generalizes some rules-of-thumb so to speak.  Efficiency within a certain application is not as straightforward as you may think (see http://jcp.aip.org/resource/1/jcpsa6/v127/i23/p234105_s1 for an example).  I just thought I should clarify this point.

Ring systems (like benzene) are the canonical example of when Z-matrices go awry. A Z-matrix cannot contain all the bond coordinates of the ring. One either has to suffer an intrinsically asymmetric description of a highly symmetric system, which is both intellectually unsatisfying and can lead to practical numerical convergence issues arising from broken symmetries, or otherwise define one or more dummy atoms in the Z-matrix, which is then no longer a minimally redundant description of the system.
The choice of coordinate system really depends on the intended calculation. There are more than two choices of coordinate system that can be used, by the way. While internal coordinates are often erroneously considered synonymous with Z-matrices, there are in fact many other internal coordinate systems that are not Z-matrices, such as pairwise distance coordinates or the various redundant internal coordinate systems.
Some specific examples:

Redundant internal coordinates are the most efficient known coordinate systems for carrying out geometry optimizations. Roughly speaking, the redundancy is useful for avoiding singularities in nonredundant systems like Z-matrices, and minimizing the correlations (lack of independence) between coordinates that occur in coordinate systems like Cartesian coordinates that result in large off-diagonal cross terms in the Hessian matrix. You can find more details in the original literature which is cited in any quantum chemistry package's user manual.
If you are coding up analytic gradients, these tend to be simplest in Cartesian coordinates because you don't have to worry about curvilinear effects in the Hessian matrix. Non-Cartesian coordinates have extra terms in gradient expressions arising from Jacobians; these can be fairly expensive to calculate.
Z-matrices themselves are often useful when creating interpolations along a specific internal coordinate like a specific torsional mode, because they are an internal coordinate system that is not redundant and hence allow various internal coordinates to be varied independently.

Very helpful answer! I'm especially interested in your mention of pairwise distance coordinates. Is this just a distance matrix? Wouldn't it be very inefficient given its maximum redundancy?

Yes. I don't think I said anything about whether it was actually found to be useful for any particular application...

Distance matrices are actually quite handy in certain situations. It is very simple to convert from Cartesian coordinates to a distance matrix, and relatively easy to transform back into Cartesians. The distance matrix also has the very useful property of being translationally and rotationally invariant.


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How is the electron "trapped in the crystal" in irradiated NaCl?
Pennsylvania State University provides irradiated $\ce{NaCl}$ (180,000 rads of gamma radiation) to teachers for a very entertaining demonstration. The salt is orange after irradiation. When it is placed on a hot plate, you see a flash of visible light and the salt returns to it normal color. The explanation provided with the salt is that "The heat allows the electrons that were trapped in the crystal structure (when the energy was deposited) to return to their original position." This language evokes a picture of a particle electron in a crystal jail for me. The explanation goes on to say that the energy is released as a photon of visible light. (I understand this part of the explanation.) So, my question is, how are electrons trapped in a crystal structure? What keeps them in an excited state?

What you're describing is thermoluminescence, which is used as a paleochronology method (aka Trapped Electron (TE) dating, afaik). The wikipedia article has a good synopsis. Simply, electrons get bound to potential wells generated in the radiation-defected structure. When you heat the crystal up you allow it to anneal and return to a more regular structure. The defected crystal is stuck in an energetic local minimum.
You can also do this trick with X-rays. Calcite (or was it aragonite?) exposed to X-rays will turn light blue for a while.

@JaniceDelMar - also, 180,000 rads is an astonishing amount of irradiation!

I think it is amethysts that turn from a (very cheap) brown to a (very expensive) bright blue under neutron radiation.

Could you explain how the crystal is defected? Or provide some reference? Are these the color center defects?


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Living-cationic vs cationic polymerization
To my understanding, cationic polymerization is a type of living polymerization but my teacher's notes and Wikipedia seem to suggest they're different. I can't see the difference.
The same goes for anionic polymerization. From what I've studied so far, the difference must be subtle, if there is one.
Is there such a difference?

This difference between living and non-living polymerizations is small in words but large in effects. In principle cationic and anionic polymerizations can be living. In practice, it is not as easy as it looks on paper. See below.
A living polymerization is any chain (or addition) polymerization that is prevented from terminating, i.e. the ends of the chains are still reactive.  In a living polymerization, chain initiation has one initiation and propagation continues as long as there is monomer. Termination (almost) never happens.  If your polymerization stops because your monomer runs out, adding more monomer restarts the process and the existing chains get longer. The number of chains formed matches very closely with the number of equivalents of initiator, and the length of the chains depends on the ratio of monomer to initiator.
Most chain polymerizations (radical, anionic, cationic, transition-metal insertion, and ring-opening) do not behave this way. There is a mechanism for termination that becomes increasingly kinetically favored as the concentration of free monomer decreases. Sadly, the wikipedia article for chain termination is pretty sad. Most chain polymerizations propagate and terminate quickly, with termination usually initiating the next chain. Thus, in a non-living polymerization, new chains are constantly being initiated and old chains are constantly terminating until their is no more monomer.  In these cases, adding additional monomer later would not accomplish anything because the chain ends are "dead". Experimental evidence for this phenomenon includes the formations of more chains than there were equivalents of initiator. The length of the chain is more dependent on the relative rates of propagation and termination than on the initial ratio of monomer to initiator.
For radical polymerization, termination mechanisms include fragmentation by β-hydrogen abstraction or by recombination. Both of these mechanisms are inherent in the radical nature of the polymerization, and so are not preventable by exclusion of other species. Radical polymerizations are made living by the addition of species that "protect" the radical at the end of the chain to prevent it from terminating. Examples include ATRP and RAFT. The wikipedia articles for both are pretty good.
For cationic polymerization, termination mechanisms include β-hydrogen abstraction, chain transfer, and attack by nucleophiles like water. In principle, a cationic polymerization could be a living polymerization if great pains were taking to exclude all species which might induce termination. In practice this is doable, but difficult, as termination by chain transfer is always possible. See the wikipedia article on living cationic polymerization. The article is not great, but suggests that cationic polymerizations done in nonpolar solutions (not in bulk) with nonnucleophilic Brønsted or Lewis acids at low temperatures can become living polymerization.
For anionic polymerization, termination mechanisms include protonation by protic molecules, chain transfer, and reaction with electrophiles, which can include O2 in some reactions. Like cationic polymerizations, anionic polymerizations can be made living by rigorous exclusion of terminating species. Again, this is difficult, due to the possibility of chain transfer termination. However, with the correct choice of conditions, it is doable, as suggested by the wikipedia page for living anionic polymerizations. Another way to accomplish some anionic polymerizations in a living fashion is group transfer polymerization, even though it does not  follow a strictly anionic mechanism.
I know that Contemporary Polymer Chemistry by Allcock, Lampe, and Mark, Prentice Hall 2003 covers this topic well, but the book is a little intense to read. In particular, the book derives the kinetic expressions for various polymerization types in the living and non-living cases. I expect other textbooks focusing of the chemistry of polymers might also cover this topic in one fashion or another.


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Why is benzyne an intermediate?
I've always seen benzyne (benzene with a triple bond) classified as an "intermediate". I really don't see why it needs to be an intermediate, though. The possible reasons I can come up with are flimsy:

It has an unstable sp2–sp2 $\pi$ bent bond: we have bent bonds with more angle strain in cyclopropane, which is not an intermediate. Though those bonds are $\sigma$ bonds.
It immediately reacts with itself to form a dimer, biphenylene, in the absence of other reagents: there are many other molecules which spontaneously dimerize (eg $\ce{NO2}$ at certain temperatures), yet they are not classified as "intermediates".

So, what makes benzyne an "intermediate"?
More generally (but possibly too broad), what makes a molecule an "intermediate"?

As Juha says, o-benzyne (to distinguish from e.g. p-benzyne which is generated by the Bergman reaction) is an intermediate, in the sense of o-benzyne itself not being the product of interest; rather it is subsequently reacted with a trapping agent (e.g. a diene for Diels-Alder reactions) within the reaction mixture, and it is the adduct that is isolated.
However, there have been isolation experiments conducted, where o-benzyne is trapped in such a way that it is barred from reacting; for instance, Chapman and coworkers isolated benzyne in an argon gas matrix at 8 K for subsequent spectroscopic study. More recently, Warmuth studied benzyne by trapping it within a so-called hemicarcerand; the benzyne complex, kept at 173 K, is stable enough that NMR experiments can be performed on it.
See this survey article by Wentrup for more references.

Intermediate is a middle step in the chemical process: reactants -> intermediate(s) -> products. Benzyne is formed from benzene and is highly reactive. Thus, it is never the first or last step of the process. To my knowledge benzyne is rarely observed as a product, because its reacting fast with other molecules because its triple bond.
What makes a molecule an intermediate? It is a highly reactive "middle product" of a reaction.
See here and there.


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What benefits do higher-rated NMR tubes provide?
You can buy NMR tubes in a huge variety of qualities, with an equally huge difference in price between the cheapest and the most expensive NMR tubes. They are usually rated for a specific spectrometer frequency, e.g. 300 MHz+ or 600 MHz+.
What is the difference between these different grades of NMR tubes? And how large is the effect on the quality of the spectra if a lower-grade tube is used?

Bad NMR tubes can also have curves in them, which can bring them into contact with the probe, damaging them. This is usually a result of heating the tube, rather then it being purchased through; I can give details on how/why this happens and how to avoid it if people want.

A general rule is: crap goes in, crap comes out. A large-sample low-field 1D NMR at room temperature is usually only minimally affected by using a cheap NMR tube. There are important differences though and I’ll highlight a few.
The first distinction between prices is what the tube is constructed from: quartz obviously costs more than borosilicate. Why would a chemist ever use the more expensive quartz? You can heat/cool quartz faster (nice for thermal studies), the UV cutoff is lower (think 190 nm opposed to 320 nm) which is important for photolysis, you can work with quartz at higher temperatures (around 1300 °C instead of 250 °C), and the purity of quartz is better controlled than your typical Pyrex. There are different grades of quartz, fused and synthetic, and there are different grades of borosilicate, such as the high-quality Pyrex or the lower-quality Class B, each comes with its own limitations as far as purities are concerned and so forth.
Three more important parameters have to do with the manufacturing of your tube are: concentricity, camber and wall thickness. Lower quality tubes will tend to have less precision and accuracy over each of these parameters and as a result your sample may wobble while spinning (introducing problems such as modulation sidebands). A particularly bad tube can hit your RF coils and cause damage to your probe over time slowly or quickly if it is ignoring any reasonable standard – even more apparent for a tube at this level of “quality” is that it may be easier to break while acquiring your sample and we all should be aware of how much fun that is for everybody involved.
Shimming can deal with impurities present in the glass (such as ferric oxide) and increased impurities in the glass/inhomogeneities will result in taking longer to get a good shim. Time is money.
A lot of these things have lower tolerances in more complex experiments and at higher fields. It really does depend on your particular experiment and what you’re hoping to get out of it.

Really nice answer. I was wondering if you could comment on whether or not it is possible for adsorbates to quasi-permanently contaminate tubes and whether tubes exist with surface coatings for reactive species? (just casual interest)

It's quite possible and is something to seriously consider when doing trace analysis, for example iron, silver, indium, uranium, cobalt, rubidium will all adsorb into pyrex in varying quantities. These are usually ppm changes though and I'd be surprised if NMR were sensitive enough. For cases where glass is incompatible, go to fluoropolymers.

I do not agree with your statement that shimming can deal with paramagnetic impurities in the glass. These impurities would create strong local gradients, which will be impossible to shim out. Furthermore using a modern gradient shimming system the length of the shimming routine is not as severely affected.

the overall shimming quality is certainly reduced with an increasing impurity of paramagnetic species, but I expect that the impurities of interest are not aggregated and largely homogeneously distributed throughout the NMR tube. I'm not an expert in NMR though. I've used these newer techniques for acquiring 1D 1H NMR spectra before, but the synthetic chemists I know (many are inorganic) stay away from these auto-shimming techniques.

There are additional aspects concerning NMR tubes. Besides their precision of wall thickness and concentricity the material itself can be of great importance.
1H and 13C-NMR
Borsilicate (pyrex) glass will be fine for standard applications. For high throughput measurements these NMR-Tubes rated for 400 MHz can be disposed after usage avoiding any cross contamination. These tubes have de facto no background signal for these nuclei. Good quality tubes rated at 400 MHz can be used at higher field machines, too, however one has to be aware that some degradation in signal quality can occur. This has to be cross checked with higher rated tubes.
29Si-NMR
In this case both borsilicate and quartz tubes will produce a glass hump, a broad signal resonating at the position where typically Q-groups are found. A possible solution is the usage of PTFE (teflon) liners, which are tubes made of plastic. However one has to be aware that the probe head itself contains quartz tubes which are built in to support the RF coils. Only changing the tube material could still lead to a significant background signal.
In addition special care has to be taken concerning gradient-equipped probes as the gradient assembly can contain silicon as well.
Changing the NMR-tube can help, however it is worthwhile to measure just the empty probe to get an impression how big the background of the tube is.
There are spectroscopic means to at least partially suppress these background signals like relaxation filters or using a DEPTH sequence.

And for some reason if you need to do boron NMR, then using the quartz tubes is a good idea... http://chem.ch.huji.ac.il/nmr/techniques/1d/row2/b.html

The coil on the probe is usually fixed on a tube of borosilicate glass (duran/pyrex), which will show up in your spectrum as well. Bruker will happily sell you a boron-free probe, however. ;)

According to Bruker the coil support is made of Quartz. At least at the recent probes.

High-priced tubes are guaranteed to be cylindrically symmetric. This is very important for the shimming, because otherwise the susceptibility differences between air, glass and solvent will make local gradients* that are hard to shim out (i.e. you need a thorough re-shim for every sample).
This used to be partially counteracted by spinning, but modern probes don't spin the sample any more, plus that gave you spinning sidebands.
The high-priced "high frequency" tubes are also supposed to be made of higher rated glass, which is more homogeneous and has low and constant susceptibility. Not sure if that's true or if it really makes a difference.
Don't dry your expensive tubes in the oven standing tilted in a beaker at 200°C! The thin glass bends very quickly.
(*If your sample tube is not overfilled and located at the correct height in the probe, you will get very bad suceptibility effects from the upper and lower end of the sample. See e.g. Olivier Buu, J Mag Res 257, 2015)


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What is the difference between cross-linked bond and secondary bond in polymers?
In solid materials, there are two kinds of bonds between polymers: primary and secondary. Primary bonds can be also called cross-linked bonds.

What is the difference between the two (strength, type, response to stress/strain, etc)?
How do they compare to bonds inside a polymer?

I've never heard of primary or secondary bonding. What I assume you're talking about is chemical vs physical reticulation?

Maybe this can help you http://books.google.co.in/books?id=_ad2mQ-b5cUC&pg=PA31&lpg=PA31&dq=What+is+the+difference+between+cross-linked+bond+and+secondary+bond+in+polymers&source=bl&ots=RBYxwPcEqt&sig=tS8NdOwk1Uj6npljAjdljYsdFBs&hl=en&sa=X&ei=9oCiT6fJKsbKrAf5nJW8Bw&ved=0CB8Q6AEwAA#v=onepage&q=What%20is%20the%20difference%20between%20cross-linked%20bond%20and%20secondary%20bond%20in%20polymers&f=false

@CHM Yes, I think I mean that. You may want to check out: http://chemistry.stackexchange.com/questions/230/what-is-the-difference-between-physical-and-chemical-bonds

This reference indicates that primary bonds in a polymer are the intramolecular bonds (the covalent bonds) that hold the polymer together. Cross-linking different chains of polymer together does occur through a covalent bond, such as a disulfide bond.
The secondary bonds that help to give the polymer its physical properties are intermolecular forces, such as hydrogen bonding for polymers that contain hydroxyl or primary or secondary amine groups. These secondary bonds can be disrupted by changing solvent conditions or heating, for example, which wouldn't disrupt the primary bonds.
So, I suppose that you could say, loosely, that there are "two kinds of bonds between polymers in solid material" if you mean between the polymer chains before they have an opportunity to cross-link (by a change in pH, for example). After they cross-link, the physical properties of the bulk sample of the polymer will still be affected by secondary bonds, the intermolecular interactions, between the cross-linked chains.
The primary bonds will be stronger and can only be affected by chemical changes, the secondary bonds (intermolecular forces) can be affected by changing the physical conditions.


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General rules for deciding volatility
Given an organic compound, is there any way to decide if it is volatile (or compare volatility--everything is volatile in the end)?
Volatility is due to the tendency to evaporate. The conclusion I can draw from this is that a more volatile compound will be lighter and/or will have less intermolecular forces. But this doesn't seem to work--I recall that methanol is less volatile than ethanol ([citation needed]).
So are there any general guidelines for this or is it another experimental property?

This answer to this question applies to your query. Many physical properties depend on the intermolecular forces experienced in the liquid or solid. These intermolecular forces are hydrogen bonding (strongest), dipole-dipole attractions (in polar molecules), and London dispersion forces (weakest of the forces when comparing molecules of the same size, but dependent on the number of electrons in the substance). Stronger intermolecular forces would make the substance less volatile.
As noted in the first response, methanol is more volatile than ethanol. Methanol and ethanol would both have both hydrogen bonding (a relatively strong type of dipole-dipole attraction) and London dispersion forces. Because ethanol has more electrons (because it is a bigger molecule, but not necessarily because it is a heavier molecule) it would have more London dispersion forces in comparison to methanol, so, with stronger intermolecular forces, it would be less volatile. So, this example follows the general guidelines--figure out the intermolecular forces present in the molecules you want to compare and the molecule with the weakest cumulative intermolecular forces will be the most volatile.
Intramolecular bonding (covalent bonds, when you're talking about molecules) does not affect physical properties.

In chemistry and physics, volatility is the tendency of a substance to vaporize. Volatility is directly related to a substance's vapor pressure. At a given temperature, a substance with higher vapor pressure vaporizes more readily than a substance with a lower vapor pressure.(Taken from Wikipedia)
But this doesn't seem to work--I recall that methanol is less volatile than ethanol
I think you have got this wrong. According to the wikipedia page of Methanol and Ethanol the vapour pressure of Methanol > Ethanol thus Methanol is more volatile than Ethanol
Volatility can also be predicted on the type of intermolecular and intra molecular bonding and of course on the BP of the compound.

Volatility depends on four factors for organic compounds:
1) Branched chained hydrocarbons are more volatile than straight chained hydrocarbons
2) The more branched an hydrocarbon is the more volatile it is
3) Depending on the bonds i.e hydrogen bond, dipole-dipole intermolecular bond and van der Waals' forces. Volatility increases from hydrogen bond to van der Waals' forces. Therefore compounds like ethanol would be less volatile in comparison to ethane.
4) Volatility decreases with increase in relative molecular mass. That is, it decreases down an homologous series. E.g methane would be more volatile than ethane.

Volatilty is the tendency of a substance to vapourize.
It is directly related to vapoure pressure of a substance(pressure exerted by the vapourse in equilibrium with the liquid at a given temperature).
more voletile substance have low BP because of weake intermolecular forces.


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What is the meaning of the '2' on some Compressed Gas Hazard signs in the lab?
The main chemistry prep room at my institution (and also at the ambulance station) is labelled with the standard green compressed gas sign.  However, there is a small '2' included within the diamond.  None of the chemistry technicians, tutors or ambulance staff seem to know what it is referring to and I can't find any reference of it on the internet.  Would anyone here happen to know?

The related products of this online provider show that other hazards have different numbers but some also have two (e.g. flammable).  Is there a standard that is being referred to here?

The 2 refers to DOT (Department of Transportation) hazard class 2. This reference describes the different DOT hazard classes.
As summarized in the comments:

Class 2 (compressed gas) is divided into three subclasses (2.1 - flammable, 2.2 - non-flammable, non-toxic and 2.3 - toxic). I imagine the diamonds for Class 2 don't contain the subclass number because there is a different diamond colour/symbology for each subclass (as opposed to for instance 5.1 (oxidising agent) and 5.2 (organic peroxide), which have the same symbol and colours).

If you look at the wiki article on dangerous goods, there are good explanations of many of the various hazmat signs we see daily and take for granted. If my edit is accepted, you'll notice a slight change in the wording, but the icon, sign color, shape, and safety number are all the same. This is because these have been standardized internationally, including the fact that your example must have a green background. Crazy, huh?

Another great resource for searching and browsing data about various hazardous materials (more geared towards professionals) is ADR Dangerous Goods.


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What makes banana bonds possible in diborane?
Diborane has the interesting property of having two 3-centered bonds that are each held together by only 2 electrons (see the diagram below, from Wikipedia).  These are known as "banana bonds."
I'm assuming there is some sort of bond hybridization transpiring, but the geometry doesn't seem like it is similar to anything I'm familiar with Carbon doing.  What sort of hybridization is it, and why don't we see many (any?) other molecules with this bond structure?

Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.

As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since hybridization is just addition of wavefunctions, we can always change the coefficients to give proper geometry). I'm not too sure of this though.
$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing  $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.
It seems that there are a lot more compounds with 3c2e geometry. I'd completely forgotten that there were entire homologous series' under 'boranes' which all have 3c2e bonds (though not the same structure)
And there are Indium and Gallium compounds as well. Still group IIIA, though these are metals. I guess they, like $\ce{Al}$, still form covalent bonds.
So the basic reason for this happening is due to an incomplete octet wanting to fill itself.
Note that "banana" is not necessarily only for 3c2e bonds. Any bent bond is called a "banana" bond.
Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar.
Sneaks off and checks Wikipedia. Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.
I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.
*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.

I see how it could have a tetrahedral shape, but it seems like it wouldn't due to the strain.

@jonsca: Distorted tetrahedral. Yes, one can say that the hybridization isn't exactly $sp^3$

For that matter: a good number of the compounds formed by the boron group elements display 3c2e bonds... Apart from the multitudes of boron (the boranes already show a rich diversity!) and aluminum compounds, there are gallium and indium compounds that display 3c2e bonds; e.g. in here, here, here, here and here. I'm sure there are more...

One should be careful talking about hybridisation. It may be used as a concept explaining a certain bonding situation that results from a certain geometrical composition of a molecule. Also most covalent bound molecules have more-centre bonds.

AFAIK, Banana bonds happens in hydrides of metal carbonyls and in metal carbonyls themselves. They are accompanied by direct sigma-bond, though.

Something I found really interesting about B2H6 is that the B-H bond polarity is inversed. Resonance makes the 3c2e bridging hydrogen partially positively charged even though its electronegativity is greater than that of boron.
https://upload.wikimedia.org/wikipedia/commons/8/80/Diborane_resonance.svg
This came up in a recent paper where they demonstrated a new type of H-bond. In the introduction, they showed that B-H-pi interactions could occur with B2H6 and benzene due to this inversed polarity (however the complex could only exist at very low temperature)

One thing I wish to ask: is the banana bond to be considered under classification of sigma or pi? If so which type is it and why? Or is it unreasonable because those classifications only exist for 2c-2e bonds?

Just as a note, it seems that AlCl3 (well, it's dimer, which Im presuming what you're referring to) isn't completely accurately described with dative bonds. It would seem that it is best described with just an odd bond with a mix of ionic and covelant character (as all bonds are). Dative bonds in this case would only be a useful interpretation tool https://chemistry.stackexchange.com/a/64600/74326 ^

Here is a plot of the Quantum Theory of Atoms in Molecules answer to you question. I have shown the bond paths of $\ce{B2H6}$.  Indeed, they are "banana-like" but interestingly they are curved inward, unlike the case of cyclopropane which are curved outward.
(Hybridization does not exist. Also, I'm not sure if there is a point of ascribing "number of electrons"--like they are aliquots--to any bonding interaction.)
Also, notice that I have drawn the bond paths between the B's and the four similar hydrogens as solid (covalent), and the set of bond paths along the "bridge" as dotted (not covalent). This is because the sign of the Laplacians of the electron density at their respective bonc critical points (yellow spheres) are opposite.

By bond paths, I assume you mean a curve of maximum electron density between atoms?

Technically, the steepest ascent path through the electron density connecting the two atoms.

Could you add level of theory, please. I am not sure what other kind of bond there could be between boron and hydrogen, certainly not ionic.

@Martin I can't recall what the level of theory is, probably B3LYP/6-31G*

Hydribization 'does not exist' may be true, but it also does not not exist.  The concept is useful for explanations, so this answer could be greatly improved by addressing why the view of hybridization leads to an answer disjoint from the physical chemistry of the situation.

Also, where are there 16 atoms to describe a 10-atom system here?

There are 8 purple-pink spheres indicating the atomic nuclei. There are 8 yellow spheres indicating the "bond critical points" - essentially a mid-point between the atoms. These are useful to demonstrate there is no shared B-B electron density.

Multicentre bonds like these (in this case, a 3-centre-2-electron or '3c2e' bond) are pretty fascinating and there is a lot of theoretical work that goes on to understand these. I am a chemist as well, and had to go through learning Wade's rules and all of that. There will be a time for an answer such as that.
But for now I want to post a very, very, simple answer that aims to be more intuitive towards the amateur chemist; because I don't see anything that fills this role, and I don't think it is that hard to explain.

What makes banana bonds possible in diborane?

A "normal" covalent bond is two atoms sharing two electrons. Think about $\ce{F2}$. Isn't it a bit weird how those electrons can belong to both atoms "at the same time", to fulfil both their octets? But you can believe that, because you were taught it at an early stage in chemistry. That's what we could call a 2-centre-2-electron, or '2c2e' bond: there are two electrons being held, or shared, between two centres (two nuclei).

Now, fluorine has a lot of valence electrons to form covalent bonds with. Consider what happens if there aren't enough electrons to go around: you need to start sharing the electrons with more atoms at once. So, the 3c2e bond in diborane is just the same, except that it's three atoms sharing two electrons, and somehow those two electrons belong to all the atoms at the same time.

Just like that, everybody can fulfil their complete octet (or duet):

Each boron has 2 "normal" covalent bonds to the outer hydrogens (4 electrons), plus 2 orange electrons, plus 2 green electrons, which makes 8.
The hydrogens at the top and the bottom each have one pair of electrons.

What's the catch? Why can't we share the electrons between even more atoms? Well, the more you share the electrons, the weaker the bond gets. This is already evident in the case of diborane: the inner ('bridging') B–H bonds are longer, and weaker, than the outer ('terminal') B–H bonds.
It's like only having two slices of cake to feed three people. Because you don't want anybody to throw a tantrum, you cut the slices up and give everybody 2/3 of a slice, and everybody still gets to have cake, but it's just not as much as they would have liked to have.
But to return to the original question of what makes banana bonds possible: it's not as magic as it seems to be at first glance. The idea behind its formation is exactly the same as a normal covalent bond: all the atoms would prefer to have a complete octet. The only difference is that we've got three bonding partners participating in the bond, instead of just two.

What sort of hybridization is it?

You could figure this out from the bond angles and the geometry. However, IMO, it's not worth finding this out: even if you calculated it and found that it was $\mathrm{sp}^{2.7182818}$ in one bond and $\mathrm{sp}^{1.5}$ in the other, you aren't going to do anything with that information. Note that these are B–H bonds, not C–H bonds, and anyway the two types of bonds (2c2e / 'terminal' and 3c2e / 'bridging') are different in terms of their strengths, so you can't use the hybridisation info to compare it against something like methane.

why don't we see many (any?) other molecules with this bond structure?

Most elements have enough valence electrons to not have to resort to these weaker bonds. For example, carbon already has four valence electrons and is happy to just form four normal 2c2e covalent bonds; but boron only has three valence electrons. Even if it uses these to form three 2c2e bonds, that's only six electrons: not a complete octet.
Sure, it could hope to get a dative bond from somewhere else, but that's contingent on it bumping into a molecule that could give it a dative bond, like $\ce{NH3}$ (see https://en.wikipedia.org/wiki/Ammonia_borane). But if it's trapped a container without any ammonia in sight, then it does the next best thing, which is to try to stretch out the electrons between more bonding partners.
In fact, boron is pretty notorious in this regard: it doesn't have enough electrons to form bonds, but still wants to fill up some kind of 'octet'. The problem of not having enough electrons leads to very interesting bonding patterns in boranes; diborane is merely the tip of the iceberg. For more info, see https://en.wikipedia.org/wiki/Boranes.

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What is the difference between physical and chemical bonds?
If you characterize the chemical bonds to two categories physical and chemical bonds, how do you do it? Aren't all bonds chemical and physical?
From the freedictionary.com, chemical bond:

Any of several forces, especially the ionic bond, covalent bond, and metallic bond, by which atoms or ions are bound in a molecule or crystal.

What is the physical bond then? How should the atom be bonded to molecule if these bonds are not the ones. What are these "several forces" excactly?
In the answer I would like to see the lists of chemical and physical bonds.

Makes you think... the ionic bond in salts and the bond you get when charged paper bits stick to a comb are both essentially electrical attraction, yes? But one is a physical bond, and one is a chemical bond...

In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is the sharing of electrons between two or more nuclei. In contrast, other interactions are often said to be intermolecular (which is somewhat more specific than the term “physical”.

In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).

Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In The Nature of the Chemical Bond (1960), he gives the following definition:

The Chemical Bond Defined.—We shall say that there is a chemical bond between two atoms or groups of atoms in case that the forces acting between them are such as to lead to the formation of an aggregate with sufficient stability to make it convenient for the chemist to consider it as an independent molecular species.

A bond is what links atoms into molecules, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…
The complete opposite: consider all interactions as chemical bonds, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like this one:

A chemical bond is the physical phenomenon of chemical species being held together by attraction of atoms to each other through sharing, as well as exchanging, of electrons and is a phenomenon that is fully described by the laws of quantum electrodynamics. In general, strong chemical bonds are found in molecules, crystals or in solid metal and they organize the atoms in ordered structures. Weak chemical bonds are classically explained to be effects of polarity between molecules which contain strong polar bonds. Some very weak bond-like interactions also result from induced polarity London forces between the electron clouds of atoms, or molecules. Such forces allow the liquification and solidification of inert gases. At the very lowest strengths of such interactions, there is no good operational definition of what constitutes a proper definitional "bond".

This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).
You can classify interactions by energy: decide that chemical bonds are those interactions that have an energy higher than a certain threshold, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.
Finally, what I believe is the most common description is to look at the nature of the interaction, and classify it following a certain convention. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.
I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the New Oxford American Dictionary as well, which I quite like:

chemical bond a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.

Short and powerful. What I like in that is that it gives a general prescription, allowing one to argue individual cases and not based too much on convention. What are the features of a chemical bond? Well, it has to be an attractive force between atoms, sure… but I think the most relevant criterion of all is sharing (or transfer) of electrons. That is, after all, what chemistry is about: description of electronic clouds around two or more atoms. And I think when this criterion is applied to the list of interaction types commonly classified, it works quite well (whithout being dogmatic).
Also, what I like in it is that a given interaction type can be considered one way or another depending on its strength. The best example for that may be the hydrogen bond, which is the archetype of the almost-chemical-bond…

Now, 'What explicitly defines a Hydrogen bond?'  3.. 2... 1... GO! :D

@LordStryker easy! ;-)   IUPAC Recommendations published in Pure Appl. Chem., 2011, Vol. 83, No.8, pp 1637-1641, doi:10.1351/PAC-REC-10-01-02

Ah yes.  I knew that the H-bond definition was going to be updated but I had not actually seen the paper.  Very good!  I am still somewhat dissatisfied though.  One example I can give is the criteria regarding H-bond angle.  The closer to 180 it gets, the stronger the bond.  So controversy then lies in, how acute can the angle be before it is no longer considered an H-bond?  I've seen so many people (professors and students included) argue about what is and isn't an H-bond due to the angle being 'too far' from 180.  Sure one can say 'H-bond like' but that starts to sound like hedging to me.

+1 for such an extensive discussion. Number 5 is indeed a nice compact definition, though to be pedantic one would still need to put in a quantitative threshold for what's meant by "strong" and crystal: otherwise a group of many oppositely-charged ballons would classify as a chemically bound crystal.

What is the justification for drawing the "dividing line" at e.g. 50 kJ/mol?

@F'x PS not an attack against you--you did a great job cataloging the different types of bonding descriptions commonly used in chemistry.  I meant to discredit that manner of classification.

The answer is that there is one (or perhaps two) types of bonds. A bond occurs when two atoms are attracted in a net-electrostatically favorable way.  (Of course, the electrons and protons are subject to their quantum nature)
Why two? In the Quantum Theory of Atoms in Molecules, the "procedure" is:

Ascertain whether a bond path exists between two atoms (atomic basins).  This is a yes/no answer -- the is a bond or no bond -- because it is electrostatically favorable or unfavorable.
If a bond path does exist, then the sign of the Laplacian of the electron density at the bond critical point tells you whether the interaction is "covalent" or "not covalent" though these terms are not strictly applicable. +/- yeilds only TWO possible types of bonds.

BTW: The Laplacian of any field is a measure of whether the electron density at a particular point is a "sink" or a "source." The reality is: there is an entire, continuous spectrum of values of the Laplacian. None of them present themselves as being called "dipole" or "van der Waals"
In summary, there is one language of bonding, and that is couched in Quantum Physics.  You will never hear of van der Waals forces in a physics [quantum electrodynamics] class, because there is no such thing.  There is no van der Waals term in the Schroedinger equation.
The difference between chemical and physical bonds as taught in class is simple:  chemistry as a whole has not shaken the pre-quantum revolution description of a bond.  It is disastrously complicated for students and professionals in the field, as there is a never ending squabble as to whether something is a "dipole-dipole" or "dipole-induced dipole" or "three center,  two electron"  But these are all arguments built on shaky scaffolding.
The fact is that the rules of physical bonding are almost "boring."  The second fact is that most bond descriptions in chemistry have only certain elements of reality. To the credit of chemists, they need to quickly and efficiently describe certain oft-appearing motifs in bonding--without resorting to a quantum chemistry calculation--and they have done a pretty good job.  It's just that these descriptions are somewhat ambigious and always open to interpretation, hence not purely quantum mechanical.

All bonds are physical, of course. But not all bonds are chemical – as a somewhat stupid example consider a magnet attached to your fridge, which is in a way a physical bond but doesn't have much to do with chemistry.
The definition you quoted is fine, and leaves quite clearly those kinds of bonds that are not included as "the physical bonds". For instance, Van der Waals forces that do not bind molecules into a rigid crystal structure, particularly London dispersion forces, are usually not considered chemical bonds, neither are the forces that hold together atomic nuclei or neutron stars.
Of course, none of these bind molecules together; if you restrict it to that you're right that every bond is physical and chemical at the same time.

Bond means force of attraction. Matter is made up of molecules and and atoms. The physical force of attraction which holds atoms and molecules in a matter is called physical bond. Vander waal's forces, coulombic forces are physical forces. The bond which is made due to the behaviour of atom to become stable is called chemical bond. The physical bond is due to property of particles called mass,electric charge(positive or negative).


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What is the geometric configuration of the four fluorine atoms during the synthesis of xenon tetrafluoride?
Neil Bartlett (1932–2008) first synthesized $\ce{XeF4}$ (and $\ce{XeF6}$) in 1962.  In the synthesis, a nickel chamber is used, and heated to 400°C, causing the formation of $\ce{NiF4}$, which Wikipedia claims is not part of the reaction.
Due to the nature of the noble gas, xenon, one might imagine that the fluorines would need to  could be "forced" onto the $\ce{Xe}$ atom at the same time, otherwise with each addition, the molecule would be even less energetically likely.  I was trying to ascertain in what geometric form the four fluorine atoms would configure themselves to make this addition possible.
I had first assumed that the nickel was being used as a catalyst, and the fluorines could be plucked from its surface accordingly, but this does not jive with the description of the synthesis above.  According to the response below, one synthesis of the molecule does rely on forming fluoride radicals on a catalyst, but it's mysterious that the synthesis above does not mention this.
Assuming there is no catalyst, how are the fluorines at 400°C oriented to allow the addition of 2 molecules at presumably the same time?
F - F
.   .
.   .
F - F

(and is that even possible?)

The answer appears to be right there in the linked synthesis details - the nickel is non-catalytic and is only there to generate a passivated surface that won't degrade in the high-temperature fluorine atmosphere.
[Edit]
The plot thickens! this paper suggests that xenon may in fact abstract fluorine off the passivated nickel fluoride surface due to observed zeroth order kinetics, which lends some serious credence to jonsca's original inference of Ni being catalytic. This will take some more investigation...

You are correct about that, I have edited the question.  Of what form does the $F{_2}$ take, then?

$\ce{F_{2}}$ is a diatomic gas. Bear in mind that the $\ce{Xe}$ is in the +4 or +6 oxidation state in the corresponding fluorides. Housecroft and Sharpe pp. 496 describe a synthesis of xenon fluorides involving UV irradiation or electric discharge. This would suggest to me homolytic cleavage of $\ce{F_{2}}$ followed by $\ce{F^{\cdot}}$ radical attack on $\ce{Xe}$.

Ah, okay, I was assuming that the fluorines all came as a package.  I know nil about inorganic, so thanks for pointing that out.  If you edit your answer to reflect that, I'll accept it.

@jonsca - I'd have to confirm that that's how the reaction actually goes but it makes sense to me.

@jonsca - I did a bit of poking around about the mechanism and found an interesting paper (I can only access the abstract from here). I've added an addendum to my answer. Note that nickel is not required to make the reaction go.

The F-F bond is 159 J/mole, much weaker than Cl-Cl. So You do not need a catalyst at all at 400 °C. There will be a lot of F atoms (as "fluoride radicals" are named corrcetly) And last not least, reaction if more than two molecules in gas phase is not possible.


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Non-integer hybridization
I've known that hybridization in distorted geometries is not exactly $sp^3$ or $sp^2$ or whatever. For example, $\ce{PH3}$ has nearly pure $p$ orbitals in the $\ce{P-H}$ bond, and the lone pair is in a nearly pure $s$ orbital.
Basically, since hybridisation is an addition of wavefunctions, instead of a perfect symmetrical addition of the kets, we get something else.
While answering this question, I realised that it's not that easy predicting the numbers.
Take the same image of $\ce{B2H6}$:

At first, seeing the 97°, I thought "well, the inner $\ce{B-H}$ bonds will be almost pure $p$". But, with that, I couldn't figure out where the 120° came from, because that is for perfect $sp^2$.
I then realised that I was being stupid and it wasn't that simple--just because you have a 120° doesn't mean pure $sp^2$. But, I was at a loss trying to find out approximate hybridizations for $\ce{B2H6}$.
How does one generally go about predicting such "nonuniform hybridizations" if one knows the bond angles?
Approximations are OK--I believe the exact mixture ratios will require some knowledge of the exact wavefunctions.

I never encountered non-integer hybridization, and I don't think hybridization is so much useful as a concept…

@Manishearth With the bond angle of 120, I think sp2 and then proper MO theory on the P orbitals and bridging hydrogens makes sense.

@Nick: Yeah, maybe this system is outside of VSEPR's capabilities due to the banana bond (even I think so now). But that was just an example.

Classical hybridization theory does not allow for noninteger hybridizations. However, ab initio calculations can be interpreted using a bond order analysis method such as NBO, where the MO coefficients are used to provide the closest analogue possible to a classical hybridization picture.
For example, one of the pure $sp$ orbitals in Pauling's valence bond theory (where hybridization was first introduced) has wavefunction
$$ \phi_{sp} = \frac{1}{\sqrt 2} \phi_{s} + \frac{1}{\sqrt{2}} \phi_{p_x} $$
Taking the square of the coefficients, this orbital is $\frac 1 2$ s character and $\frac 1 2$ p character, i.e. it is an $sp$ orbital.
The basic idea of a bond order analysis method is to reexpress a molecular orbital into a form similar to
$$ \phi = c_1 \phi_{As} + c_2 \phi_{Ap_x} + c_3 \phi_{Ap_y} + c_4 \phi_{Ap_z} + \dots$$
If the other coefficients are very small, the ratio
$$ n = \frac {c_2^2 + c_3^2 + c_4^2}{c_1^2} $$
would yield a number that could be used interpret $\phi$ as a $sp^n$ orbital on atom $A$.
Edit: As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR. The former is an electronic property whereas the spatial arrangement of atomic nuclei are not, and the relationship between the two becomes much more complicated.

"As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR." I would have hoped you could explain that further. For example Bent's rule gives an explanation for correlation between geometry and hybridisation, and I find that by far more sophisticated than VSEPR.

Wave functions can be used (as described in the answer by AcidFlask) but for something that has just s and p hybridization (tetrahedral, trig planar, equatorial positions of TBP), I find it easier to use the formula $\cos x = \frac{S}{S-1} = \frac{P-1}{P}$, where $x$ is the angle between the central atom and two identical bonding atoms (in degrees), $S \times 100~\%$ is the percentage of s-character in each of the bonds, and $P \times 100~\%$ is the percentage of p-character in the bonds.
It works really well to explain things like why $\ce{PH3}$ and $\ce{NH3}$ act differently as bases, as it can be used to show that the bonds are mainly p-character (~$95~\%$), so the lone pair on $\ce{PH3}$ is mainly in an s-orbital, so is a weaker base.  It could also be applied to your molecule, like the bonds in the 93 degree angle have $13~\%$ s-character, so $86~\%$ p.
For angles less than 90 degrees, the formula does not work (negative answers), but angles less than 90 have to be all p because they represent bent (or banana) bonds when they appear in molecules (like yours) with no d orbitals open.

The formula only works for very symmetric molecules and it would assign the same hybridisation to the bond you are comparing the angle to. It completely ignores the other bonds in the molecule, see Bent's rule for a phenomenological explanation.

I have stumbled on this years later, and since this page comes up high on a google search for non-integer hybridization, I thought I should correct some things.
Non-integer hybridization has been around from its beginning, as its math falls out trivially from the coefficients of the orbitals used to generate the hybrid orbitals. It predates NBO by many decades.
Non-integer hybridization is a useful model for bonding because it provides simple explanations for many things. Examples include the IR stretch for cyclopropanes and cyclic ketones, the acidity of bicyclobutanes, the basicity of hindered amines, HCH bond angles in small rings.  It provides a simple orbital basis for ideas from VSEPR such as the HXH angle in H3CCl vs ammonia.  As with any model, it will fail outside of its purview, and your mileage may vary on whether you find it the easiest way to understand things.  One does not see the math of it often in recent textbooks but its ideas are implicit in the way that many things are described.
There is a very simple mathematical relationship between the angle between two orbitals and the hybridization. If you describe the hybridization of an orbital as sp^(lambda^2), then lambda^2 = -1 / cos (angle).  For example, the HCH angle in cyclopropane is 116° and you can find from this that the hybridization for the CH bonds is sp^2.28.
So for the original question, the 120° angles mean that the orbitals used for those BH bonds are essentially pure sp2.  The hybridization for the orbitals for the internal BH bonds is slightly more complicated.  One cannot use the angles in a ring to determine hybridization because one often has bent bonds when the atoms are constrained, but the total s character for each boron has to be 1.  Since the external BH bonds use up 2/3 of the s, the remaining 1/3 is split between the two orbitals and their hybridization is about sp5 (1/6 s).
There is a tremendous amount of old literature on all of this.