text:
Why do Magnesium and Lithium form *covalent* organometallic compounds?
Lithium and magnesium are Group 1 and Group 2 elements respectively. Elements of these groups are highly ionic, and I've never heard of them forming significantly covalent inorganic compounds.
Yet these elements form a variety of organometallic compounds ($\ce{PhLi}$, the whole family of Grignard reagents, etc). Organometallic compounds have significant covalent character (i.e., the bond can be called covalent)  in the carbon–metal bond.
What's so special about carbon that makes these elements form covalent bonds?

Do they really? I saw indications that Li-C has more ionic component than covalent. It's not like it can be easily told, or matter all that much. "Ionic" doesn't preclude being molecular.

The character of the bond is determined by the electronegativity of the atoms.
Speaking of bonds as purely ionic or covalent is not always correct - usually it is more correct to say that a bond has ionic or covalent characteristics.
So comparing the difference in electronegativities gives us the following:
$$\begin{array}{cc}\hline
\text{Difference in electronegativity} & \text{Type of bond} \\ \hline
< 0.4 & \text{Non-polar covalent} \\
0.4 \mathrm{-} 1.7 & \text{Polar covalent} \\
>1.7 & \text{Ionic} \\ \hline
\end{array}$$
At the upper end of the polar covalent spectrum, the bonds frequently have both covalent and ionic characteristics.
For organometallic compounds, the difference in electronegativity between Li and C is $1.57$.  So while this is still in the polar covalent range, it is also close to ionic.  Similarly, the difference between Mg and C is $1.24$ - again, a very polar covalent bond.
Compare this to the difference between H and C ($0.35$) - a non-polar covalent bond.
So to answer your question, the thing that is "special" about carbon is that it has a fairly low electronegativity compared to the chalcogens and halogens.  Granted, bonds with carbon are also going to be weaker than in say LiCl, but that's what makes organometallic compounds actually work to form carbon-carbon bonds.
(The electronegativity values came from wikipedia's great chart)

"usually it is more correct to say that a bond has ionic or covalent characteristics."-->yeah, that's why I was careful to talk in terms of "covalent character", though I missed a spot. Great answer, now I feel stupid. I'll wait before accepting this, though--let others have a chance.

The character of the bond is not determined by the electronegativity of the atoms. There's only a loose correlation.

Why do magnesium and lithium form covalent organometallic compounds?

I think this question touches a more general question in chemistry: Why do some metallic elements form 'covalent' bonds at all? And the other version of this question would be why sodium or potassium does not usually form covalent bonds. The answer for these questions would become long and winding very easily since it is going to touch on how electrons move and shift in forming a chemical bond, covalent or otherwise.
So I think a similar question that can have an easier answer would be: When does an element form 'covalent' bond with another element? Answer: If the electronegativities of the two elements differ by less than 1.9, the formed bond would be covalent. But then again, the answer is not absolute and serves only as a guideline. The case in point is $\ce{Na-C}$ bond, which has a $\delta(E.N)=1.6$ $(E.N. = \text{electronegativity})$ but is considered an ionic bond.
Back to the original question. Why do magnesium and lithium form covalent organometallic compounds? A hand-waving answer would be that in forming a $\ce{Li-C}$ or $\ce{Mg-C}$ bond, the electrons are not moved all the way from metal to carbon, but only partially to somewhere between the metal and carbon.
I hope these had touched the topic a little bit.

The atomic and ionic radii of Li are very small. Therefore, the nuclear attraction is strong compared to the rest of the alkali metals, which leads to a greater polarising power in Li ion, and the formation of covalent bonds.


text:
Edges of graphite
I've been asked this question a few times, and while I think I know the answer, I'd like  to know more.
Graphite, as we know, is a sheet polymer. Since polymers are bound to be finite by physical considerations, graphite must have "edges". My question is, what happens at these edges?
I can cook up these possibilities:

A graphene sheet "folds back" on itself, thus forming a tube and taking care of most of the edge carbons. But this would make graphite less slippery, so I doubt this is the case.
Random elements/compounds from the environment at time of synthesis latch on to the edges, taking care of the valency of Carbon
The edge carbons form double/triple bonds amongst themselves

I'd like to know more about this. I'm also curious what happens to the resonance of graphene at the edges.

Are you sure "folding back" would make it significantly less slippery? After all, those bonds would only be present along lines of length of order $\mathcal{O}(\sqrt{n})$, while the forces between sheets act on a whole area: $\mathcal{O}(n)$, where $n$ is the number of atoms per sheet. Of course these forces are much weaker than valence bonds, but only by some 2-4 orders of magnitude. What is the size of graphene sheets in graphite? I would have thought it's more than $10^8$ atoms, so $\sqrt{n} > \eta$ and the forces between sheets would still dominate.

@leftaroundabout: No, I'm not sure ;-) Yeah, that sort of makes sense. The slipperyness would decrease if the tubes had small diameter but large length.

It does appear that the graphene sheets fold back on themselves. This paper reports that the graphene sheets that make up both natural and synthetic graphite, double back over each other with "nano-arches".
This is an  from the paper.
There are also SEM and TEM micrographs of a graphite crystal that are pretty interesting.

Damn, my hunch was proven wrong :p. Pretty  interesting structures, I'll read the whole paper sometime soon.

@Manishearth - A quick read over the paper indicates that whilst this is a major phenomenon in graphite edges, it's probably not the whole story. Sheets also need to be terminated in the orthogonal direction to the arches which probably means that the 'dangling bonds' are passivated in other ways. Also, note figure 3d, which depicts a stepped, non-folded structure on account of the instability of the arches at high temperature. I suspect the most inclusive (but least helpful) conclusion is 'it's complicated'.

@Richard Terrett-Right, I should have been more specific in my answer. I did not mean to imply that this was the whole story, just that there might be fewer edges than he had originally envisioned.

@JaniceDelMar - No problem :) I imagine the folded tubes are possibly much longer than they are wide.

Defects in graphite result from oxidation. They are $\ce{ OH, COOH}$ and epoxy groups, as in  this image:

From http://cnx.org/content/m29187/latest/
I think the edges of graphite have the same chemical nature, and thus have these defects.

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text:
Why does sulfur, but not oxygen, catenate?
Oxygen is a rather boring element. It has only two allotropes, dioxygen and ozone. Dioxygen has a double bond, and ozone has a delocalised cloud, giving rise to two "1.5 bonds".
On the other hand, sulfur has many stable allotropes, and a bunch of unstable ones as well. The variety of allotropes, is mainly due to the ability of sulfur to catenate.
But, sulfur does not have a stable diatomic allotrope at room temperature. I, personally would expect disulfur to be more stable than dioxygen, due to the possibility of $\mathrm{p}\pi\text{-}\mathrm{d}\pi$ back-bonding.
So, why do sulfur and oxygen have such opposite properties with respect to their ability to catenate?

First, a note: while oxygen has fewer allotropes than sulfur, it sure has more than two! These include $\ce{O}$, $\ce{O_2}$, $\ce{O_3}$, $\ce{O_4}$, $\ce{O_8}$, metallic $\ce{O}$ and four other solid phases. Many of these actually have a corresponding sulfur variant. However, you are right in a sense that sulfur has more tendency to catenate… let's try to see why!
Here are the values of the single and double bond enthalpies:
$$\begin{array}{ccc} \hline
\text{Bond} & \text{Dissociation energy / }\mathrm{kJ~mol^{-1}} \\ \hline
\ce {O-O} & 142 \\
\ce {S–S} & 268 \\
\ce {O=O} & 499 \\
\ce {S=S} & 352 \\ \hline
\end{array}$$
This means that $\ce{O=O}$ is stronger than $\ce{S=S}$, while $\ce{O–O}$ is weaker than $\ce{S–S}$. So, in sulfur, single bonds are favoured and catenation is easier than in oxygen compounds.
It seems that the reason for the weaker $\ce{S=S}$ double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the $\mathrm{3p}$ orbitals overlap is small and the $\pi$ bond is weak. This is attested by looking down the periodic table: $\ce{Se=Se}$ has an even weaker bond enthalpy of $\ce{272 kJ/mol}$. There is more in-depth discussion of the relative bond strengths in this question.
While not particularly stable, it's actually also possible for oxygen to form discrete molecules with the general formula $\ce{H-O_n-H}$; water and hydrogen peroxide are the first two members of this class, but $n$ goes up to at least $5$. These "hydrogen polyoxides" are described further in this question.

Do the rest of the elements follow the expected order? I mean the order is: S-S > O-O > Se-Se > Te-Te ?

Expanding upon this answer, we may need to reconsider what we consider as "catenating". Does it mean it's easy to form bonds with itself to form polymeric substances? If so, easy at what temperature? I believe oxygen catenates just as easily as sulfur albeit at a lower temperature range.

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text:
Are there any examples of nuclear spin isomers having consequences for chemical reactivity?
Ortho- and parahydrogen are two forms of the $\ce{H2}$ molecule that are distinguished by their pairing or antipairing of nuclear spins, giving rise to metastable singlet (ortho-) and triplet (para-) states. This constitutes an example of nuclear spin isomerism.
Whilst the thermal properties of these two species are different, my understanding is that their chemical properties are essentially identical. Are there any examples of this kind of isomerism having nontrivial implications for chemical reactivity?

This grant proposal gives some information about chemistry of nuclear spin isomers… from which it seems the answer could be yes (molecules being in different rovibrational states allowing or forbidding certain reactions depending on the isomer)

Interesting question... I was always fascinated by the fact that there are more than one H2s. Regarding the chemical nature, I don't think so--I've never seen magnetic properties affecting chemical ones. But I've not seen much.

@Manishearth - magnetic properties (of electron configurations) affect chemical properties all the time. However nuclear magnetic properties are a whole other topic...

@rich any examples? (I probably already know examples of this, just can't remember any. Stupid brain.)

Yes!
Papers in the field of preparation or selection of nuclear spin isomers have introductions that justify the work by potential applications and some of them include chemical reactions, either under that name or under the more descriptive terminology of “reactive collisions”. While no specific cases are cited as examples, the potential exist but only in limited situations such as bimolecular reaction in molecular beams.
From what I have seen, the canonical reference is “Detailed symmetry selection rules for reactive collisions”, Mol. Phys. 1977, 34, 477–504. This is the theoretical paper laying out the foundations of the effect of nuclear spin isomerism on reactions. I don't have access to the full text, but here is the abstract:

It confirms what I suggested in my earlier comment: the different nuclear spin isomers have different rovibrational states, and accounting for effect changes the selection rules indicating which collisions are reactive or not.
One specific example which I found is “Observation of Ortho-Para H3+ Selection Rules in Plasma Chemistry”, Phys. Rev. Lett. 1997, 78, 3844–3847:

Large dependences of the intensity ratios on the parent gases have been observed, indicating the spin memory of protons even after chemical reactions. The results clearly demonstrate the existence of selection rules on nuclear spin modifications in chemical reactions theoretically anticipated by M. Quack.

Their results are obtained on the formation of H3+ ions from a H2 plasma.


text:
What makes a radical 'free'?
The qualifier 'free' seems to be ubiquitously attached to discussion of radicals as highly reactive species with unpaired spins. What, precisely (or imprecisely, as the case may be) does 'free' really mean?
Free as in free speech? Free beer? Freedom of movement? Kinetic accessibility?

I think that it just means that the radical is  unattached--not part of a bond.. A bit of redundancy, in a way. By this definition, there's no such thing as an "unfree radical".

Free to react with nearly anything?

I liked this question because I had never thought much about it. However, it's not such a mystery because the answer is on Wikipedia's “radical” page:

Historically, the term radical was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, methyl alcohol was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons. However, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons.

So, the free in free radical was originally used to distinguish it from a radical bound to a carbon skeleton.

Digging into historical chemistry texts reveals that “free radical” becomes really widely used around 1930:

However, some quotes for earlier works use it, though it does not appear so frequently that it may be considered a common expression. In other words, at that time simply the juxtaposition of the adjective free with radical, became synonymous with group (or today's noun substituent). For example:

“This hydrocarbon was formerly supposed to be the free radical of the methyl group, but no methyl compounds have in any way been derived from it, whilst by the action, of chlorine upon it, ethyl chloride is obtained as the product” (1870, Lessons in elementary chemistry)

or

“Although, however, the free radical methylene is unknown, many of its compounds have been prepared, of which some, such as the haloid ethers, &c.” (1884, A Treatise on Chemistry, Henry Enfield Roscoe and Carl Schorlemmer)

Though more quotes could be dug and provided, the ones I've read support this idea that free radical initially meant “the chemical group as a molecule by itself”.

I always thought that the radical was free-er if it was more stable. So dioxygen is a free radical (strictly a diradical) as it has two unpaired electrons.

Re-reading this I think the answer fails to give a clear explanation of current usage and confuses the issue by bringing in the evolution of the term. The old terminology is little help in understanding current usage.

TLDR; It depends who you are talking to (organic chemist, ESR spectroscopist, or physical chemist/chemical physicist), and when.
The way I read the quote below the fold, there have been three major senses of the definition, at least to the best of the knowledge of a giant in the field like Herzberg:

The sense as described by Terry Bollinger, where an entity that in modern terms is described as a functional group (HT: F'x) is found to be stable as an isolated entity.
Free in this case refers to the chemical stability of the species, while radical is used in the older sense of the term indicating a 'fundamental chemical unit'.

The sense as might be used by an ESR spectroscopist, encompassing any systems bearing nonzero electronic spin.
Free in this case carries the sense of 'available for measurement by spin resonance techniques,' and radical is used in the newer sense, indicating the presence of nonzero electronic spin.

The sense defined by the 'chemical transiency' of species that are chemically short-lived but physically stable (possess a nonzero dissociation energy), regardless of their electronic spin.
Free here refers to 'detached transiently from a more stable system,' whereas radical hearkens back to the older sense of the term, meaning a 'fundamental chemical unit.'

I'm pretty sure Gerhard Herzberg is a reliable source. I will quote (extensively!) from the introduction to his 1971 Spectra and Structures of Simple Free Radicals, since he says it far better than I could (all emphasis is in the original; bracketed ellipses are my elisions, un-bracketed ellipses are original):

The concept of a radical in chemistry is a very old one; it goes back to Liebig. Quoting an old [1884] text book of organic chemistry, "Radicals are groups of atoms that play the part of elements, may combine with these and with one another and may be transferred by exchange from one compound into another." Free radicals first came to be considered after Gomberg at the turn of the [20$^\mathrm{th}$] century observed triphenylmethyl to be a chemically stable system. However, simpler radicals like $\ce{CH3}$, $\ce{CH2}$, $\ce{CH}$ are extremely short-lived species, difficult to produce and study in the free state. They are chemically unstable even though in general they are physically stable; that is, if undisturbed by collisions they do not spontaneously decompose: they have a nonzero dissociation energy.
According to the quantum theory of valence a group of atoms (a radical) when split off a parent molecule often has one or more unpaired electrons—that is, has nonzero spin $\left(S\right)$. This circumstance has led many authors, particularly organic chemists, to define a free radical as a system with nonzero spin. Such a definition is particularly convenient for those working in the field of electron-spin resonance, since it implies that all systems and only systems that can be investigated by electron-spin resonance are free radicals. While such a definition is extremely simple and straightforward, it does have two drawbacks: according to it certain chemically stable molecules such as $\ce{O2}$, $\ce{NO}$, $\ce{NO2}$, $\ce{ClO2}$ must be considered as free radicals, while on the other hand quite a number of systems that are highly reactive and short-lived, such as $\ce{C2}$, $\ce{C3}$, $\ce{CH2}$, $\ce{CHF}$, $\ce{CF2}$, $\ce{HNO}$, $\ldots\,$, in their singlet states $\left(S=0\right)$ are not considered to be free radicals. Indeed, one and the same system, such as $\ce{CH2}$, would or would not be a free radical depending on the electronic state in which it happened to be. [...].
Therefore many physical chemists and chemical physicists use a somewhat looser definition of free radicals: they consider any transient species (atom, molecule, or ion) a free radical—that is, any species that has a short lifetime in the gaseous phase under ordinary laboratory conditions. This definition excludes $\ce{O2}$, $\ce{NO}$, $\ldots$ but includes $\ce{C2}$, $\ce{CH2}$, $\ce{CHF}$, $\ldots$ even in singlet states. It also includes atomic and molecular ions. [...] While most of the free radicals that we shall be discussing have lifetimes of less than a millisecond, we must realized that there is no sharp boundary; indeed, some of the radicals we are including have lifetimes of about $0.1$ sec.

A radical is a group of atoms that tend to stay together and act as a persistent chemical unit in reactions. The methyl group $CH_3$ is a common example. My own favorite radical is ammonium $NH_4$, which behaves strikingly similarly to an alkali metal such as lithium or sodium, so much so that you can even create amalgams of ammonium with mercury.
A free radical is what you get if the radical group is capable of existing on its own. The ammonium radical is an example, and one of the more stable ones. Methyl can be free also, but tends to be far more reactive and so shorter lived.
Free radicals always have "dangling bonds," since they are by definition parts of larger molecules that have been broken off of those molecules. However, that doesn't mean they are any more reactive than are free elemental atoms, ammonium again being a good example.

This is an obsolete usage of the word. Moreover, the idea that "radicals" have dangling bonds is not correct on your definition (ammonium doesn't, for example).

Heh! My theory is that the original question was so old that "radical" was still being used that way... Seriously, no disagreement, it's just that the question itself was about the phrase "free radical," so it seemed appropriate to answer in the same way.


text:
How can I perform structure alignment on an inorganic species?
I'm attempting to perform atomistic optimal structure alignment on an inorganic species. Can anyone point me in the right direction to software or a technique for performing this?
There exists a wide range of structural alignment software, but this really appears to be geared towards aligning protein backbones, whereas I'm interested in aligning unlabelled Cartesian coordinates of small molecules.

Inorganic structures – that could be proper solid or molecular metalloorgnic. For the later one you can use xp (program written G.M.Shedrick - comes with Bruker diffractometers). If you have running license of Cambridge Crystal Data Base there is program called IsoStar, and I think that it should do the job. I haven't used it my self but apparently Olex2 can make aligning of fragments (this program is free so that's the place were you can start). I'm not quite sure but I think that old GROMOS code could also make a fit of orthogonal coordinates.

I'm sure VMD (although also geared towards protein alignment) is flexible enough to align any two sets of atoms. You will almost definitely have to learn some TCL scripting, but getting past the (perhaps steep) learning curve is well worth the effort. Besides VMD has excellent documentation.
 If you post a more specific problem I'll be happy to post a more specific answer:)

OpenBabel using SMARTS. pymol also does it.


text:
How are non-carbon stereogenic centers named (S/R)?
The R/S naming system for stereocenters relies on the Cahn–Ingold–Prelog priority rules to rank the four substituents of a stereocenter; the R or S name is then attributed depending on the spatial orientation of the ranked substituents. While this is mostly used for asymmetric carbon atoms, it holds true of other stereocenters with four substituents.
However, nitrogen and phosphorus atoms can also act as stereocenters if the inversion of their configuration is blocked (e.g. because they're in a cyclic molecule) or slowed enough (e.g. because of bulky substituents). Apparently, these stereocenters are also labelled in R/S fashion. Examples I found include Tröger's base:

(above is the (5S,11S)-enantiomer, below is the (5R,11R)-enantiomer),
and DIPAMP:

Having said that, I can't find any reference as to how the R/S naming scheme is applied to these tetrahedral stereocenters which don't have four substituents, but a lone electron pair and three substituents instead. I assume that the same scheme can be followed if one considers the lone pair as the lowest-ranked substituent, but I can't find any canonical reference as to that.
So, how are such stereocenters named? What authoritative reference can I find on this topic?

Note that it need not be "non-carbon".. Carbanions are chiral as well, but I don't think they're ever needed to be named

Now I wonder what the rules are for an ionic organic compound (Eg a carbanion with an ionic bond with $\ce{K+}$ )

but I can't find any canonical reference as to that.

Look harder ;-)
It seems to be that the lone pair has lowest priority.
OK, I can't find any authoritative reference to that (I don't have access to the original paper), but I've got plenty of others:
Alright, IUPAC says:

Except for hydrogen, ligancy, if not already four, is made up to
  four by adding "phantom atoms" which have atomic number zero
  (0) and are thus always last in order of preference.
This has
  various uses but perhaps the most interesting is where nitrogen
  occurs in a rigid skeleton, as for example in a-isosparteine (14);
  here the phantom atom can be placed where the nitrogen lone pair
  of electrons is; then N-i appears as shown alongside the formula,
  and the chiraiity (R) is the consequence; the same applies to N-i6.
  Phantom atoms are similarly used when assigning chirality symbols to chiral sulfoxides (see example to Rule E—4.9).

When ligancy is less than four, we obviously have some lone pairs hanging around (except in Hydrogen--which has been excluded, and some other cases like carbocations--which are planar anyway)
So the lone pair becomes a "phantom atom" and has 0 priority.
Now I wonder what happens to these rules if you have two lone pairs, one of them muonic ;-)
Check the first revision of this post for my original, non-authoritative references

"Phantom atoms are similarly used when assigning chirality symbols to chiral sulfoxides" - one particularly famous example is the proton pump inhibitor (S)-omeprazole (esomeprazole, Nexium), whose only chiral center is the sulfoxide sulfur. So, for chiral centers that are pyramidal, turn the lone pair away from you and check for the clockwise/anticlockwise order from the other three pendants.


text:
Can an atom have more than 8 valence electrons? If not, why is 8 the limit?
According to some chemistry textbooks, the maximum number of valence electrons for an atom is 8, but the reason for this is not explained.
So, can an atom have more than 8 valence electrons?
If this is not possible, why can't an atom have more than 8 valence electrons?

2017-10-27 Update
[NOTE: My earlier notation-focused answer, unchanged, is below this update.]
Yes. While having an octet of valence electrons creates an exceptionally deep energy minimum for most atoms, it is only a minimum, not a fundamental requirement. If there are sufficiently strong compensating energy factors, even atoms that strongly prefer octets can form stable compounds with more (or less) than the 8 valence shell electrons.
However, the same bonding mechanisms that enable the formation of greater-than-8 valence shells also enable alternative structural interpretations of such shells, depending mostly on whether such bonds are interpreted as ionic or covalent. Manishearth's excellent answer explores this issue in much greater detail than I do here.
Sulfur hexafluoride, $\ce{SF6}$, provides a delightful example of this ambiguity. As I described diagrammatically in my original answer, the central sulfur atom in $\ce{SF6}$ can be interpreted as either:
(a) A sulfur atom in which all 6 of its valence electrons have been fully ionized away by six fluorine atoms, or
(b) A sulfur atom with a stable, highly symmetric 12-electron valence shell that is both created and stabilized by six octahedrally located fluorine atoms, each of which covalently shares an electron pair with the central sulfur atom.
While both of these interpretations are plausible from a purely structural perspective, the ionization interpretation has serious problems.
The first and greatest problem is that fully ionizing all 6 of sulfur's valence electrons would require energy levels that are unrealistic ("astronomical” might be a more apt word).
A second issue is that the stability and clean octahedral symmetry of $\ce{SF6}$ strongly suggest that the 12 electrons around the sulfur atom have reached a stable, well-defined energy minimum that is different from its usual octet structure.
Both points imply that the simpler and more energetically accurate interpretation of the sulfur valence shell in $\ce{SF6}$ is that it has 12 electrons in a stable, non-octet configuration.
Notice also that for sulfur this 12- electron stable energy minimum is unrelated to the larger numbers of valence-related electrons seen in transition element shells, since sulfur simply does not have enough electrons to access those more complex orbitals. The 12 electron valence shell of $\ce{SF6}$ is instead a true bending of the rules for an atom that in nearly all other circumstances prefers to have an octet of valence electrons.
That is why my overall answer to this question is simply "yes".
Question: Why are octets special?
The flip side of whether stable non-octet valence shells exist is this: Why do octet shells provide an energy minimum that is so deep and universal that the entire periodic table is structured into rows that end (except for helium) with noble gases with octet valence shells?
In a nutshell, the reason is that for any energy level above the special case of the $n=1$ shell (helium), the "closed shell" orbital set $\{s, p_x, p_y, p_z\}$ is the only combination of orbitals whose angular momenta are (a) all mutually orthogonal, and (b) cover all such orthogonal possibilities for three-dimensional space.
It is this unique orthogonal partitioning of angular momentum options in 3D space that makes the $\{s, p_x, p_y, p_z\}$ orbital octet both especially deep and relevant even in the highest energy shells. We see the physical evidence of this in the striking stability of the noble gases.
The reason orthogonality of angular momentum states is so important at atomic scales is the Pauli exclusion principle, which requires that every electron have its own unique state. Having orthogonal angular momentum states provides a particularly clean and easy way to provide strong state separation between electron orbitals, and thus avoid the larger energy penalties imposed by Pauli exclusion.
Pauli exclusion conversely makes incompletely orthogonal sets of orbitals substantially less attractive energetically. Because they force more orbitals to share the same spherical space as the fully orthogonal $p_x$, $p_y$, and $p_d$ orbitals of the octet, the $d$, $f$, and higher orbitals are increasingly less orthogonal, and thus subject to increasing Pauli exclusion energy penalties.
A final note
I may later add another addendum to explain angular momentum orthogonality in terms of classical, satellite-type circular orbits. If I do, I'll also add a bit of explanation as to why the $p$ orbitals have such bizarrely different dumbell shapes.
(A hint: If you have ever watched people create two loops in a single skip rope, the equations behind such double loops have unexpected similarities to the equations behind $p$ orbitals.)
Original 2014-ish Answer (Unchanged)
This answer is intended to supplement Manishearth's earlier answer, rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell.
I call it donation notation, and it dates back to my high school days when none of the chemistry texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate.
The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are "borrowing" each electron. A dot with an arrow means that the electron "belongs" mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case, the electron is no longer attached to the arrow at all but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow.
Here are examples using table salt (ionic) and oxygen (covalent):

Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indicating that it has "donated" its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.)
Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms.
Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative):

In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and "borrow" entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulfates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, some of the pairs in the central octet can remain unattached. This results in less oxidized anions such as sulfites and nitrites.
Finally, a more ambiguous example is sulfur hexafluoride:

The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is some controversy even today about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases.
2014-04-04 Update
Finally, if you are tired of dots and arrows and yearn for something closer to standard valence bond notation, these two equivalences come in handy:

The upper straight-line equivalence is trivial since the resulting line is identical in appearance and meaning to the standard covalent bond of organic chemistry.
The second u-bond notation is the novel one. I invented it out of frustration in high school back in the 1970s (yes I'm that old), but never did anything with it at the time.
The main advantage of u-bond notation is that it lets you prototype and assess non-standard bonding relationships while using only standard atomic valences. As with the straight-line covalent bond, the line that forms the u-bond represents a single pair of electrons. However, in a u-bond, it is the atom at the bottom of the U that donates both electrons in the pair. That atom gets nothing out of the deal, so none of its bonding needs are changed or satisfied. This lack of bond completion is represented by the absence of any line ends on that side of the u-bond.
The beggar atom at the top of the U gets to use both of the electrons for free, which in turn means that two of its valence-bond needs are met. Notationally, this is reflected by the fact that both of the line ends of the U are next to that atom.
Taken as a whole, the atom at the bottom of a u-bond is saying "I don't like it, but if you are that desperate for a pair of electrons, and if you promise to stay very close by, I'll let you latch onto a pair of electrons from my already-completed octet."
Carbon monoxide with its baffling "why does carbon suddenly have a valence of two?" structure nicely demonstrates how u-bonds interpret such compounds in terms of more traditional bonding numbers:

Notice that two of carbon's four bonds are resolved by standard covalent bonds with oxygen, while the remaining two carbon bonds are resolved by the formation of a u-bond that lets the beggar carbon "share" one of the electron pairs from oxygen's already-full octet. Carbon ends up with four-line ends, representing its four bonds, and oxygen ends up with two. Both atoms thus have their standard bonding numbers satisfied.
Another more subtle insight from this figure is that since a u-bond represents a single pair of electrons, the combination of one u-bond and two traditional covalent bonds between the carbon and oxygen atoms involves a total of six electrons, and so should have similarities to the six-electron triple bond between two nitrogen atoms. This small prediction turns out to be correct: nitrogen and carbon monoxide molecules are in fact electron configuration homologs, one of the consequences of which is that they have nearly identical physical chemistry properties.
Below are a few more examples of how u-bond notation can make anions, noble gas compounds, and odd organic compounds seem a bit less mysterious:

I regret to have to make a critical comment about such a highly rated answer, but this is not an answer to the question, but rather a missive on an alternative graphical representation of resonance structures.

I have to second @Eric's comment. It is unfortunate, that this is such a highly voted answer as promotes a concept, which is far too simple. Especially after the update, the "u" notation for carbon dioxide makes no sense whatsoever. This is a highly complicated molecule and the so called "u bond" is indistinguishable from the traditional bond.

There is, of course, a name for the "u bond" which is well-known and widely taught in chemistry: it is a dative bond. This answer, however, merely skirts the issue: dative bonds are indistinguishable from ordinary covalent bonds and as such, the identification of a bond as being "dative" cannot be an explanation for the phenomenon of hypervalency in the last few compounds mentioned. In other words: drawing a problem in a different way doesn't make it go away.

Thank you orthocresol, that is a much more specific term than simply saying "coordinate bond"! My u-bond graphical notation is of course nothing more than a simple graphical accounting technique for keeping track of which electrons come from where in a dative bond, one that simultaneously emphasizes by using only one bond-pair line that it is indistinguishable from any other single covalent bond. As with any line notation, it of course does not address the delightfully deep physics issue of how SU(2} half-spin (which should not exist in 3D space!) plus Pauli exclusion creates bonding.

Alchemists, and @ericbrown from years ago: Yes, an atom can have both more and less than 8 electrons in its valence shell, and still form a compound with low enough total energy to be stable. One with fewer than 8 is called a free radical, but I am not aware of any universal term for ones with more than 8. Eight is a universal low due to our living in 3D space with electron spin pairing, which give strong orthogonal separation of the energies of 3 pairs, plus the special spherical-symmetry case of the same pair with no angular momentum. If we were 4D, the whole-shell number would be 10!

@TerryBollinger An example of an atom that has more than 8 valence electrons is a transition metal. Others include the actinides and lanthanides.  We really don't need a 4D universe to have atoms with more than 8 valence electrons.

Eric, I agree completely! That was an overly quick answer on my part, and I am currently writing two more carefully done updates to add to my original answer. As you aptly noted, and as Manishearth addressed in detail, the real issue is overall energy minimization of orbitals, not magic numbers. Eight is just a particularly pronounced member of a series of valence sets that starts with 2 (hydrogen and helium), goes to 8 next, and then on to numbers like 18. While the energy minimums do get a bit "squishier" (technical term) higher up, they are still very much part of an overall sequence.

The edit doesn’t really help this answer; it still considers an outdated concept (the so-called twelve electrons shell around sulphur in $\ce{SF6}$ which is actually a combination of 2e2c and 4e3c bonds indistinguishable by mesomery perfectly in line with the octet formalism) and builds itself around that. Yes, the octet is, formally speaking, a minimum but the octet plus one case is at such an astronomically high level it won’t be attained. Also, not all electron-deficient compounds are radicals; see singlet carbenes, boron compounds etc.

I actually cannot believe that this answer got any worse that it originally was. With your 12 electron valence case you must also include the prerequisite for that: having electrons in d-orbitals of sulfur; which has been disproved quite a few times. (Expanded octett, hypervalency to name the trigger words.) Plus it is absolutely unnecessary to describe bonding in that way, as is resonance, and the combination of 3c2e and 3c4e bonds is also an unnecessary crutch, but at least that's not completely wrong. Bonds can have covalent and ionic contributions.

Thanks for the feedback, but seriously, if you disagree so strongly with me (and Jan, and no doubt others) and have not already done so, stop adding comments that few will read  and instead add or update your own clearly stated answer describing what you consider to be the correct solution. That way everyone can understand precisely why the amswers from me and others are wrong. The whole intent of Stack Exchange is, after all, to provide top-notch answers that blow away the other ones. If you can do that, please do so! That is a good thing, not a bad thing.

There are currently at least three answers that already do that. Unfortunately they are not written as deceivingly simple as yours, because it simply is not as simple as you present. It is answers precisely like this, that keep debunked scientific myths alive. The only way to combat that is to tell you: You are wrong.

The answer by Gavin Kramar can be considered the answer here, according to me.

Yes, it can. We have molecules which contain "superoctet atoms". Examples:
$\ce{PBr5, XeF6, SF6, HClO4, Cl2O7, I3- ,  K4[Fe(CN)6], O=PPh3 }$
Almost all coordination compounds have a superoctet central atom.
Non-metals from Period 3 onwards are prone to this as well. The halogens, sulfur, and phosphorus are repeat offenders, while all noble gas compounds are superoctet. Thus sulfur can have a valency of +6, phosphorus +5, and the halogens +1, +3, +5, and +7. Note that these are still covalent compounds -valency applies to covalent bonds as well.
The reason why this isn't usually seen is as follows. We basically deduce it from the properties of atomic orbitals.
By the aufbau principle, electrons fill up in these orbitals for period $n$:
$n\mathrm{s}, (n-2)\mathrm{f},(n-1)\mathrm{d},n\mathrm{p}$
(theoretically, you'd have $(n-3)\mathrm{g}$ before the $\mathrm{f}$, and so on. But we don't have atoms with those orbitals, yet)
Now, the outermost shell is $n$. In each period, there are only eight slots to fill in this shell by the Aufbau principle - 2 in $n\mathrm{s}$, and 6 in $n\mathrm{p}$. Since our periodic table pretty much follows this principle, we don't see any superoctet atoms usually.
But, the $\mathrm{d,f}$ orbitals for that shell still exist (as empty orbitals) and can be filled if the need arises. By "exist", I mean that they are low enough in energy to be easily filled. The examples above consist of a central atom, that has taken these empty orbitals into its hybridisation, giving rise to a superoctet species(since the covalent bonds add an electron each)
I cooked up a periodic table with the shells marked. I've used the shell letters instead of numbers to avoid confusion. $K,L,M,N$ refer to shell 1,2,3,4 etc. When a slice of the table is marked "M9-M18", this means that the first element of that block "fills" the ninth electron in the M(third) shell, and the last element fills the eighteenth.
Click to enlarge:

(Derivative of this image)
Note that there are a few irregularities, with $\ce{Cu}$, $\ce{Cr}$, $\ce{Ag}$, and a whole bunch of others which I've not specially marked in the table.

I feel obliged to add a disclaimer to such a highly upvoted answer. While commonly taught as such in introductory chemistry, the involvement of d-orbitals in hypervalency is not true, as they are, in fact, not low enough in energy to be filled. Gavin Kramar's answer to this question describes hypervalency in a more accurate fashion.

Something worth adding to this discussion that I'm surprised hasn't been mentioned about such "hypervalent" molecules like $\ce{SF6}$.
One of my professors at university informed me that the common explanation (that the empty d-orbitals are empty and are thus accessible) is actually most likely incorrect. This is an old-model explanation that is out-of-date, but is for some reason continuously taught in schools. A quote from the Wikipedia article on orbital hybridisation:

In 1990, Magnusson published a seminal work definitively excluding the role of d-orbital hybridization in bonding in hypervalent compounds of second-row elements.
  (J. Am. Chem. Soc. 1990, 112 (22), 7940–7951. DOI: 10.1021/ja00178a014.)

When you actually look at the numbers, the energy associated with those orbitals is significantly higher than the bonding energy found experimentally within molecules like $\ce{SF6}$, meaning that it is highly unlikely that the d-orbitals are involved at all in this type of molecular structure.
This leaves us stuck, in fact, with the octet rule. Since $\ce{S}$ cannot reach into its d-orbitals, it cannot have more than 8 electrons in its valence (see other discussions on this page for definitions of valence etc, but by the most basic definition, yes, only 8). The common explanation is the idea of a 3-centered 4-electron bond, which is essentially the idea that sulfur and two fluorines 180 degrees apart share only 4 electrons between their molecular orbitals.
One way of comprehending this is to consider a pair of resonance structures where sulfur is bonded covalently to one $\ce{F}$ and ionically to the other:
$$\ce{F^{-}\bond{...}^{+}S-F  <-> F-S+\bond{...}F-}$$
When you average these two structures out, you will notice that sulfur maintains a positive charge and that each fluoride has a sort of "half" charge. Also, note that sulfur only has two electrons associated with it in both structures, meaning that it has successfully bonded to two fluorines while only accumulating two electrons. The reason they have to be 180 degrees apart is due to the geometry of the molecular orbitals, which is beyond the scope of this answer.
So, just to review, we've bonded to two fluorines to the sulfur accumulating two electrons and 1 positive charge on sulfur. If we bonded the remaining four fluorides from $\ce{SF6}$ in the normal covalent way, we'd still end up with 10 electrons around sulfur. So, by utilizing another 3-center-4 electron bond pair, we achieve 8 electrons (filling both the s and p valence orbitals) as well as a $+2$ charge on the sulfur and a $-2$ charge distributed around the four fluorines involved in the 3c4e bonding. (Of course, all of the fluorines have to be equivalent, so that charge will actually be distributed around all of the fluorines if you consider all of the resonance structures).
There actually is a lot of evidence to support this style of bonding, the simplest of which is observed by looking at bond lengths in molecules such as $\ce{ClF3}$ (T-shapes geometry), where the two fluorines 180 degrees apart from each other have a slightly longer bond length to chlorine than the other fluorines do, indicating a weakened amount of covalency in those two $\ce{Cl-F}$ bonds (a result of averaging out a covalent and ionic bond).
If you are interested in the details of the molecular orbitals involved, you may wish to read this answer.
TL;DR Hypervalency doesn't really exist, and having more than $\ce{8 e-}$ in non-transition metals is much harder than you would think.

As I mentioned earlier this is the answer here.

In chemistry, and in science in general, there are many ways of explaining the same empirical rule. Here, I am giving an overview that is very light on quantum chemistry: it should be fairly readable at a novice level, but will not explain in its deepest way the reasons for the existence of electronic shells.

The “rule” you are citing is known as the octet rule, and one of its formulations is that

atoms of low (Z < 20) atomic number tend to combine in such a way that they each have eight electrons in their valence shells

You'll notice that it's not specifically about a maximal valence (i.e. the number of electrons in the valence shell), but a preferred valence in molecules. It is commonly used to determine the Lewis structure of molecules.
However, the octet rule is not the end of the story. If you look at hydrogen (H) and helium (He), you will see that do not prefer an eight-electron valence, but a two-electron valence: H forms e.g. H2, HF, H2O, He (which already has two electrons and doesn't form molecules). This is called the duet rule. Moreover, heavier elements including all transition metals follow the aptly-named 18-electron rule when they form metal complexes. This is because of the quantum nature of the atoms, where electrons are organized in shells: the first (named the K shell) has 2 electrons, the second (L-shell) has 8, the third (M shell) has 18. Atoms combine into molecules by trying in most cases to have valence electrons entirely filling a shell.
Finally, there are elements which, in some chemical compounds, break the duet/octet/18-electron rules. The main exception is the family of hypervalent molecules, in which a main group element nominally has more than 8 electrons in its valence shell. Phosphorus and sulfur are most commonly prone to form hypervalent molecules, including $\ce{PCl5}$, $\ce{SF6}$, $\ce{PO4^3-}$, $\ce{SO4^2-}$, and so on. Some other elements that can also behave in this way include iodine (e.g. in $\ce{IF7}$), xenon (in $\ce{XeF4}$), and chlorine (in $\ce{ClF5}$). (This list isn't exhaustive.)
Gavin Kramar's answer explains how such hypervalent molecules can come about despite apparently breaking the octet rule.

This may be a definition issue if the asker is in high school or a recent graduate. The first three current editions of high school text books I pulled from the shelf (AP and beginning chemistry) use the definition for valence electrons as "electrons in the highest occupied principal energy level".

Note that the 18electron/EAN rule isn't always followed.. Paramagnetic, octahedral complexes never follow it. They can't. Neither can tetrahedral/square planar complexes. These are usually  still superoctet, though.

@ManishEarth I'm very worried about some of the answers given on SE that cover electronic structure concepts. I am wondering if it would be useful to start a meta discussion on how to answer "why" questions from 1900's chemical bonding theory -- should the answer be in terms of the old chemical rules or in terms of quantum mechanics?

This question may be difficult to answer because there are a couple of definitions of valence electrons. Some books and dictionaries define valence electrons as "outer shell electrons that participate in chemical bonding" and by this definition, elements can have more than 8 valence electrons as explained by F'x.
Some books and dictionaries define valence electrons as "electrons in the highest principal energy level". By this definition an element would have only 8 valence electrons because the $n-1$ $d$ orbitals fill after the $n$ $s$ orbitals, and then the $n$ $p$ orbitals fill. So, the highest principal energy level, $n$, contains the valence electrons. By this definition, the transition metals all have either 1 or 2 valence electrons (depending on how many electrons are in the $s$ vs. $d$ orbitals).
Examples:

Ca with two $4s$ electrons would have two valence electrons (electrons in the 4th principal energy level).
Sc with two $4s$ electrons and one $3d$ electron will have two valence electrons.
Cr with one $4s$ electron and five $3d$ electrons will have one valence electron.
Ga with two $4s$ electrons, ten $3d$ electrons, and one $4p$ electron would have three valence electrons.

By the other definition, they could have more since they have more "outer shell" electrons (until the $d$ shell is filled).
Using the "highest principal energy level" definition for valence electrons allows you to correctly predict the paramagnetic behavior of transition metals ions because valence electrons (the $d$ electrons) are lost first when a transition metal forms an ion.

Can we consider (the d electrons) in transition element  as a valence electrons according to "highest principal energy level" definition for valence electrons ?

There is a big difference between a "rule" and a law of nature. The "octet rule" is a turn-of-the-last-century concept that somehow managed to get into introductory chemistry books and never got kicked out with the advent of modern quantum mechanics. (Circumstantial proof: it is impossible to identify individual electrons to label them "valence" or "not valence".)
Therefore, you won't find any answer based on physical evidence as to why/why not a rule that is not based on physical evidence will hold.
Atoms take their spatial configuration because it happens to be an electrostatically-favorable circumstance, not because electrons avail themselves like "slots".

It probably got in because it was easy to explain a lot with the concept, and it doesn't get kicked out because it can still explain a lot in a very simple fashion, being close enough to the truth while doing so. Also, while it might not be possible to identify electrons, it is possible to calculate orbitals, i.e. electron pairs and by juggling with hybridisation and mixing, assigning them to either be core orbitals or valence orbitals, the latter usually centrable on a bond/atom, and giving a four-per-atom type of solution.

Why 8? has not really been addressed by the above answers, and while tangential to the question, it is somewhat important to be considered. In general, but not always, atoms react to form complete quantum 'shells', with electrons interacting with all their orbitals.
The principal quantum number ($n$) determines the maximum azimuthal quantum number ($l$), in the sense that $l$ can only take values between $0$ and $n-1$.  Thus for the first row, $n = 1$ and $l = 0$.  For the second row, $n = 2$ so $l = 0,1$.  For the third row, $n = 3$, so $l = 0, 1, 2$.
The azimuthal quantum number $l$ determines the range of possible magnetic quantum numbers ($m_l$), which lies in the range $-l \leq m_l \leq +l$.  So for the first row, $m_l = 0$. For the second row, when $n = 2$ and $l = 1$, then $m_l = -1,  0, 1$. For the third row, $n = 3$, $l = 0, 1, 2$, $m_l = -2, -1, 0, 1, 2$.
Finally, the spin quantum number $m_s$ can be either $+1/2$ or $-1/2$.
The number of electrons that can fill each shell is equal to the number of combinations of quantum numbers.  For $n=2$, this is
$$\begin{array}{cccc}
n & l & m_l & m_s  \\ \hline
2 & 0 & 0   & +1/2 \\
2 & 0 & 0   & -1/2 \\
2 & 1 & +1  & +1/2 \\
2 & 1 & +1  & -1/2 \\
2 & 1 & 0   & +1/2 \\
2 & 1 & 0   & -1/2 \\
2 & 1 & -1  & +1/2 \\
2 & 1 & -1  & -1/2 \\
\end{array}$$
for a total of 8 electrons.
The second row contains 'organic compounds', of which millions are known, so there is frequently a bias in teaching chemistry to focus on the "octet rule". In fact, there is a duet rule to be considered for hydrogen, helium (and lithium which dimerizes in the gas phase), and the 'rule of 18' for transition metals.  Where things get 'wonky' are silicon through chlorine. These atoms can form a complete quantum shell via the octet rule, or 'expand' their octets and be governed by the rule of 18. Or situations in-between, such as sulfur hexafluoride.
Bear in mind, this is a gross simplification, because these atomic orbitals mix to form molecular orbitals, but the counts of the atomic orbitals influence and directly correlate with the counts of the resulting molecular orbitals, so the combination of atomic quantum numbers still provides some interesting information.

Fun fact: Lewis never coined the term "Octet rule". He called it simply the rule of two and stated, that for many elements four electron pairs are used for bonding.

Let's take a look at the periodic table: in the first row, there's only two elements: Hydrogen and Helium. They do not follow an octet rule. Hydrogen can only have a maximum of two electrons on the valence orbital. It turns out that the octet rule is not exclusive, meaning it is not the only rule that helps understand Lewis structure and electron configuration. Why do we use the octet rule, then?
Every period in the periodic table represents an energy shell of an atom. The first period represents the shell K, the first energy level, which has only the s orbital. Every orbital can only be filled with 2 electrons, both with a quantum spin towards opposite directions. Thus the maximum number of electrons possible for the first energy level shell, K, is 2. This is reflected in the fact that Helium is a noble gas, yet only contains 2. The second energy level shell, L, has the s orbital and the extra 3 p orbitals. Those add up to four orbitals or 8 electrons. Because the elements most commonly used are in the second and third period, the octet rule is in frequent use.
Elements of the third energy level are very similar. They still follow the octet rule, because even though now the have 5 d orbitals, no orbital needs to be filled. The electronic configuration shows that 4s is filled before 3d, so they don't need to fill the d orbital, thus they usually also obey the octet rule. However, third energy level shell elements, unlike second-row elements, (see Gavin's comment fir reference) are not limited to the octet rule. They can form hypervalent molecules in certain cases where the use that d orbital and fills — this is not the case with all apparent hypervalent molecules, SF6 is not hypervalent, it uses weak ionic bonds and polarity, but there still are hypervalent molecules out there. It will always depend on which state is more convenient in terms of electrostatics.
At the fourth energy level shell, there are f orbitals introduced, but we are not even close to filling them at that point because we first need to fill the d orbitals. The 5 d orbitals signify 10 electrons, plus the previous eight from the octet rule, sum up to 18. This is the reason why there are 18 columns in the periodic table. Now, a new rule superposes, and this is the well known 18-electron rule, which was mentioned above. Transition metals obey this rule with more frequency than not, though there are occasions in which they still obey the octet rule. At this point, with so many orbitals to fill, and with electrostatics playing a role in electronic configuration, we can obtain different cations from the same element with certain metals. That is also why they don't discuss oxidation state numbers with transition metals like they do with the first three rows of the table.

Welcome to Chemistry SE, your answer unfortunately doesn't add much to earlier stuff. And your explanation of SF6 is wrong imo.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is circularly polarized light ever used to isolate particular chiralities?
Chiral molecules tend to absorb one circular polarization of light more than the other. Is this ever used to isolate a particular enantiomer?

Yes. For some chiral compounds, each enantiomer in a racemic mixture absorbs certain frequencies of light differently depending on the circular polarization type of the light.
That means that if you can pump in sufficiently intense and energetic (e.g. ultraviolet) circular polarized light pulses into the racemic mix of such compounds, you may in some cases increase the rate of breakdown of the enantiomer that absorbs that particular circular polarization (left or right), resulting in an increase in the relative concentration of the other enantiomer.
You can view the first page of a paper on asymmetric photolysis using circularly polarized ultraviolet light here.

I stand corrected...interesting paper.

@JaniceDelMar, not corrected, please! I liked your answer, but in digging around I found to my own surprise that there's been some interesting work on this topic. The connection of some of this research to people pondering natural amino acid chirality is particularly fascinating, but to me doesn't answer much. After all, intensely chiral sources of circular light are to the best of my knowledge extremely rare in nature, so demolishing amino acid enantiomers via such light only defers the mystery a bit.

As I said, interesting paper...but I must admit, having worked in industry, I was thinking more along practical lines and this doesn't seem a particularly practical separation technique. I wonder what scale they've done this on? (Which I could probably find out with further reading.) This amino acid chirality connection seems a "reach" to me.

I think your comments nailed it: The paper I quoted is the most successful example I could find of people trying to develop some sort of general model for using chiral light (cheap!) to isolate chiral enantiomers (costly!). I don't think anyone has succeeded in finding a method that provides that level of generality, given that this "best I could find" example depends on too many happy coincidences and so does not generalize well.

The paper seemed to indicate that part of the difficulty in performing this experiment was getting a good source of circularly polarized ultraviolet light of sufficient intensity. Laser physics has come a long way since 1977; this might be easier now than it was.

@Dan, that's an intriguing point. Hmm. Maybe someone should explore that one? More generic methods for chiral enantiomer isolation would likely be of real pharmaceutical interest.

The optical rotation is certainly used to follow the separation of optical isomers (enantiomers if they are mirror images, or diastereomers if you have more than one stero-center and the two isomers are not mirror images). You can see how the optical rotation changes as you use the normal separation techniques (chromatography, distillation, crystallization, etc.). If you know the true value for the substance, you can determine the purity. But I don't know how polarized light would be used to do the actual separation.


text:
What does "reaction coordinate" mean? What does it mean when a reaction has TWO reaction coordinates?
I'm familiar with the concept of a reaction coordinate from high school chemistry as some generic conformational parameter that all of the intermediate states of a chemical reaction lie upon. Recently, though, I've seen some presentations where chemists presented plots of the free energy with respect to two reaction coordinates, and my understanding is that the plot was of a measurement (I could be wrong, though).
What exactly does the reaction coordinate mean? What does it mean to have more than one reaction coordinate for the same reaction? Is the free energy as a function of the reaction coordinate a measurable quantity?

I suggest to you the book "Computational Chemistry", especially the second chapter, in which you will find an interesting discussion on the potential energy surface, which is the concept underlying plots such as the ones you're referring to.

@CHM: I'm familiar with the potential energy surface, but I was curious as to how one could generate a potential energy surface with respect to two reaction coordinates when the precise definition of those coordinates is unknown.

Well, you don't seem to be too familiar with the concept of PES. What you're seeing in Ashu's graph is a slice of a potential energy surface, so that only two dimensions are present. The "reaction coordinates" really are the geometry of atoms in the system, when they follow this particular path. Also, that book will probably be in your university's library, you really should read it.

@CHM: I have read it. I understand that a "normal" reaction coordinate plot is a slice of the potential energy surface. What I don't understand is how one could measure or calculate the PES as a function of two reaction coordinates without knowing what either reaction coordinate represents in terms of the spatial configuration of the molecules. I was under the impression that a given reaction could only have one reaction coordinate.

Hi Dan, I wonder if you might help us launch Materials Modeling Stack Exchange, it would be great to have you as one of the users in the private beta!

The reaction coordinate is the progress of a reaction from reactants to products with various intermediates and transition states in between. It is an abstraction. It has no relation to time. Rather it is the progress of bond-forming and bond-breaking reaction steps. The free energy change of partially formed and partially broken bonds cannot be measured. The free energies of transition states (local maxima) cannot be directly measured. These states are transient and cannot be directly observed. For a simple reaction with few mechanistic steps, the activation energy (transition state energy) can be calculated from kinetics measurements. The free energy of intermediates can often be determined directly or indirectly if they are long-lived. Most of these diagrams are assembled from a few measured data points and either extrapolation or theoretical computation.
If two reaction coordinate axes are presented, this often means that there are two separate pathways the reaction can take, and each pathway is presented on one of the two reaction coordinate axes. Perhaps the reaction involves two different steps, but those steps need not happen in the same order every time. As a generic example, consider nucleophilic substitution reactions in organic chemistry: A generic nucleophile ($\ce{Nu-}$) reacts with an organic molecule R-X, where X is a leaving group, and R is the rest of the molecule (which is unchanged or nearly unchanged). The overall reaction would be
$$
\ce{Nu- + R-X -> Nu-R + X-}
$$
This process involves two steps: forming the bond between Nu and R, and breaking the bond between R and X. These two steps can happen in three different orders: 1) the R-X bond breaks first and the R-Nu bond forms second (this is the dissociative or $\mathrm{S_{N}1}$ mechanism); 2) The R-Nu bond forms first and the R-X bond breaks second (this is the associative or $\mathrm{S_{N}Ac}$ mechanism); and 3) The R-X bond breaks and the R-Nu bond forms at the same time (this is the intermediate or $\mathrm{S_{N}2}$ mechanism). All three of these cases occur in reactions that are covered in any undergraduate organic chemistry text. These three processes would look like the following. The items in parentheses are transition states with partially formed/broken bonds ($\ce{\bond{....}}$).
Dissociative:

Step 1: $\ce{Nu- + R-X -> (Nu- + R\bond{....}X) -> Nu- + R+ + X-}$
Step 2: $\ce{Nu- + R+ + X- -> (Nu\bond{....}R + X-) -> Nu-R + X-}$

Associative:

Step 1: $\ce{Nu- + R-X -> (Nu\bond{....}R-X) -> [Nu-R-X]-}$
Step 2: $\ce{[Nu-R-X]- -> (Nu-R\bond{....}X) -> Nu-R + X-}$

Intermediate:

Only step: $\ce{Nu- + R-X -> (Nu\bond{...}R\bond{....}X) -> Nu-R + X-}$

A reaction coordinate energy diagram for a substitution process could then be graphed over two different reaction coordinates: breaking the R-X bond and forming the R-Nu bond.  For ease of two dimensional representation, these are often presented as contour diagrams. Various trajectories could be taken through this space based on what real species R, Nu, and X represent. For example, if $\ce{R^+}$ is a stabilized cation, then the dissociative pathway becomes more likely. These plots are called More O'Ferrall-Jencks plots. This link will take you to the Wikipedia article, which is not a very good article. The references in that article to the papers by R. A. More O'Ferrall and W. P. Jencks are probably better places to go to find out more. Most advanced undergraduate or graduate level textbooks of physical organic chemistry, particularly Advanced Organic Chemistry, Part A: Structure and Mechanisms by Carey and Sundberg, will cover this topic, at least as it relates to organic chemistry.


text:
Why is fabrication of ordered graphite substrates difficult?
While reading this paper (suggested by @Janice here), I came across the following line:

Surface reconstruction is known to be extremely important for Si, the material of  20th century electronics. However, surface reconstruction has been much less studied for graphite. One of the reasons might be graphite's limited use in electronics because fabrication of ordered graphite substrates is difficult.

(emphasis mine)
I've always believed that graphite isn't much used in semiconductor tech since it isn't a metalloid and conducts in an entirely different manner, probably making doping not work. This reason is new to me, but quite interesting.
I can sort of guess why "fabrication of ordered graphite substrates is difficult" — I'm not too sure though.
At any rate, I'd like to know more about it. Exactly why is such fabrication difficult?

Also, a good demo for the semiconductor aspect of it is to draw out a fairly thick (5cm x 1cm) pencil line on a piece of paper.  If you have a voltmeter handy, check the resistance across your "block" lengthwise.  I would guess it would be on the order of a few hundred ohms, but don't quote me on that -- you can probably calculate it from the resistivity of graphite and the length.

What do You mean with "metalloid"?

@georg metalloids conduct via a conduction band--same as metals. They can be doped easily. Graphite works by the delocalization voodoo, so it's completely different. Doping may only work if the dopant can participate in the delocalization. (What I think, anyway)

@chris oh, that's my mistake--the pdf is locked to prevent copy-pasting, so I manually copied it. Yeah, I know that graphite isn't a metalloid.

I don't really agree with their reason, but it might have to do with reaching a high enough purity... perhaps a higher mp of carbon over silicon is a basis for this??

@chris hmm, makes sense--I'm not fluent in the semiconductor-fabrication processes. If you can expand that into an answer, I'll accept it :)

@Manishearth This site  offers quite a lot of information about the steps used for fabrication of graphite (presented in slide form). It's long enough that I don't really want to summarize it, but even paging through it will give you some idea of the complexity of the steps. A promising avenue still under development (from my memory of various article) is vapor phase deposition to put the carbon where you want it in the shape you want. They cover this a bit at the end of these slides.

Graphite is a semimetal, and no more mambojumbo than with metals, but it doesn't do anything with the problem.
Silicon/silica can be melted, purified, doped, can single crystals grown, can oxidize the surface, making the silicon semiconductor an silica insulator. By "can" I mean that engineeers know well tested methods, optimized for half a century.
Graphite cannot be reasonably melted, cannot grown to single crystals, cannot doped with similar techniques, highly anisotropic, so cutting maybe an issue, and when you make thin layers, it is hard to oxidize in a controlled way to make an insulator out of it. The resulted oxides are semiconductors, and there is a large variety of local structures that can appear with different electronic properties. By cannot I mean that no one really though about these problems until the last decade, so there might be equivalent solutions with e.g graphene, nanotube or other nanocarbon structures.


text:
Retention of surface texture after combustion
Just some while ago I was burning a box of cardboard in my backyard and I saw a very curious phenomenon.
I saw that as the cardboard box started burning it started to curl.  Up after a while as I came back to check if the fire was out I saw that though the cardboard box wasn't a box any longer but individual sides/faces of the box had retained their ridges and texture though they had tuned greyish white and crumbly.
My questions:

Why did the cardboard start to curl up as it began to burn?
Why did it retain its texture if not shape and size?

The curling up has a simple reason: the burning was not uniform. Paper contains water (really!), and heating it makes it lose the water.
Now, paper is a bad conductor of heat. When you burn it, inevitably the flame will be mainly on one side(the side that's pointing up). The flame will transfer heat to nearby regions of paper, and evaporate some (or possibly all) of the water. The loss of water leads to a contraction of the cellulose fibers in the paper (I'm not too sure of this, comments appreciated). But, this only happens on one side of the paper. Even though paper is razor-thin, the fact that it still has two "faces" is important here. Now, due to the contraction, the paper curls in that direction (similar to how a bimetallic strip works). That's it.
The reason for retaining texture, is something I'm not too sure about. The burning reaction throws up a lot of soot, but a lot of the products stay right where they are. This results in the charred/burnt paper occupying the previous position of the cellulose. While charred paper isn't a polymer and thus isn't too strong, it can "mimic" the initial structure just by occupying the position of the cellulose. But, unlike cellulose, if we try bending it, it will crumble since there are no long polymer "chains"/intermolecular forces holding it together. It's like the difference between a house of cards and a similar "house", made of a single piece of rubber. The structure is the same, but the former case is only staying together because it had the "luck" to exist in a stable configuration. Disturbing it will destroy it.


text:
Why don't trigonal S and P compounds undergo inversion at room temperature?
Most molecules containing nitrogen atoms in trigonal pyramid configuration undergo a relatively fast process of inversion at room temperature. On the other hand, the free energy barrier for phosphines, sulfoniums and sulfoxides are high enough that they are optically stable: the rate of racemization is slow at room temperature.
I wonder what effect is responsible for this difference in behaviour (the larger energy barrier). I expect that it has to do with the size of the central atom (N being one row higher than P and S), but is that all there is to it? And how does size impact the free energy barrier for the inversion: energetically or entropically (or both, of course)?

IIRC, quantum tunnelling is significant in trigonal nitrogen inversion.

related http://chemistry.stackexchange.com/questions/50106/ http://chemistry.stackexchange.com/questions/38599/ http://chemistry.stackexchange.com/questions/16190/

Azane all by itself has an inversion barrier of about $6\ \mathrm{kcal/mol}$, which is  low compared to phosphine's approximately $30\ \mathrm{kcal/mol}$ barrier. Pyramidal nitrogen inversion requires a planar transition state. One can use a classical theory to model the thermal rate and we could write $$k \propto \mathrm e^{-E_\mathrm a/(RT)} $$ and using models like these we would predict that ammonia at $300\ \mathrm K$ would have a rate on the order of $10^8\ \mathrm{s^{-1}}$, instead what is observed is something on the order of $10^{10}\ \mathrm{s^{-1}}$
Assuming the potential energy surface has two minima equal in energy and a higher energy TS, two routes are possible: a thermal process going over the maxima or quantum-mechanical tunneling. Determining a tunneling frequency requires writing down vibrational wave functions, finding their linear combination and overlap. If we were to do all this for a particular vibration, we'd see that the tunneling frequency decreases exponentially with a dependence on increasing $\mu$, thickness of the barrier/shape, and the barrier height. As an example, $\ce{ND3}$ has approximately an order of magnitude slower rate of inversion than $\ce{NH3}$ at a sufficiently low temperature – this has been attributed to a decrease in its tunneling frequency.
But why the difference in the energy barriers and corresponding rates depending on the atomic position in the periodic table? Perhaps the easiest/quickest explanation rests on considering what the TS requires for interconversion. Ideally the classical conversion requires passing through a trigonal planar structure with bond angles on the order of 120 degrees. If you consider the series of azane, phosphane, arsane then you'll see bond angles of 108, 94, 92 degrees respectively. In the absence of other effects, the barrier goes from smaller to larger. Indeed, asane-like molecules were the first of this series resolved, followed by phosphane-like. To the best of my knowledge the fastest have yet to be resolved, because rates on the order of $10^{-5}\ \mathrm{s^{-1}}$ are required at RT for a decent shot at isolating them. Naturally, this isn't the whole story, as I've previously alluded to above. Let me know if you would like to know more.

Neither of the two enantiomers is actually an energy eigenstate of the molecule. If the right-handed and left-handed enantiomers are states $|R\rangle$ and $|L\rangle$ respectively, then the ground and first-excited energy eigenstates are going to be something like $|g\rangle=\left( |R\rangle +|L\rangle \right)/\sqrt{2}$ and $|e\rangle=\left( |R\rangle -|L\rangle \right)/\sqrt{2}$.
Because each enantiomer is in a superposition of energy eigenstates, a molecule initially in one chiral state will oscillate back and forth from one to the other, with a frequency determined by the difference in energy between $|g\rangle$ and $|e\rangle$.
In general, the difference in energy between  $|g\rangle$ and $|e\rangle$ is determined by the barrier between the $|R\rangle $ and $|L\rangle$ states: the higher and wider the energy barrier, the slower the oscillation. Since P and S are much larger than N, the energy barrier between the $|R\rangle $ and $|L\rangle$ states is larger, and the frequency of oscillation between $|R\rangle $ and $|L\rangle$ is much lower.

Sorry, my question is: why is the barrier higher for larger atoms?

I don't know why (or even if) larger atoms would have a higher barrier, but they would have a wider barrier by virtue of just being larger. Part of the point was that a wider barrier is enough to slow down the oscillation.

Indeed, that's a good point (not the answer to my question, but something I hadn't considered). As you said, the wider the barrier, the slower the oscillation… but between N and P, the energy barrier actually is much higher for phosphine (132 kJ/mol) than ammonia (24 kJ/mol). I would expect that contribution to be more important than the barrier widening.


text:
Are these two space groups in a group–subgroup relationship?
Given two crystal space groups, how can one determine if they are in a group-subgroup relationship? The specific case at hand is P 21/m 2/m 2/a (aka P m m a, #51) and P m m 2 (#25), but knowing how to solve this question in a general manner will sure prove helpful in the future as well…

I personally don't know too much about crystals or group theory, but it occurs to me that this question is already clear on chemistry and needs help in the math. Migrate to math.SE? (of course, the groups may be only used by chemists, in which case it may be a better fit here)

@Manishearth I considered it, but I think it has better chances of being answered here than on math.SE… If I don't get an answer here, I’ll follow your suggestion.

I'm not a master of group theory or crystals in general either, and I had a hard time attempting to work it out from the linked descriptions, but handily, there's a site which will tell you.

The Pmma is a minimal non-isomorphic supergroup of Pmm2; or you can say that Pmm2 is one of maximal non-isomorphic sub-groups of Pmma.
How to solve the question:
(1) fast and dirty: take International Tables for Crystallography vol. A go to your group (in mine edition Pmma starts on p.274). On second page describing space group you can find Maximal non-isomorphic subgroups. The third listed is [2] Pmm2 (25) 1; 2; 7; 8.
You can see that the operation in  #50 and #25 have a lot of similarities. No.1 you can ignore as it's [x,y,z] but 2, 7 and 8 for #50 are [-x+.5,-y,z], [x,-y,z] and [-x+.5,y,z]. As one can see the #25 has following general operators: (2) [-x,-y,z] (3) [x,-y,z] and (4) [-x,y,z].
(2) The hard way: Mathematics is explain in the same volume in chapter 13.1. Isomorphic subgroups. Of course the knowledge of group theory and matrix calculus is to your advantage ;-)


text:
What material are heat bags made of?
I just bought a heat bag from store and it has special property.
It is made up of transparent plastic bag and transparent unknown liquid inside when sold. There is a tiny circular metal piece in there. If you bend the metal piece, crystalization will occur and release quite a lot of heat. The crystalization will spread out and change all the liquid into crystals.
When the hardened plastic bag cool down, you need to put it into boiling water and the crystals will dissolve and become liquid again.
What is more interesting, I tried to heat it up but not letting all the crystal dissolve, take it out of the boiling water and assume it is ready for the next use. However, the half-dissolved liquid/crystal turns into all crystals again (without me bending the metal piece). I am guessing the transformation was not complete and the process was reverted.
Did the Japanese invent this material?

The bag contains a supersaturated solution of some salt with a highly negative enthalpy of crystallisation, which could be something like sodium acetate ($\ce{Na^{+}CH3CO2^{-}}$). The dissolved salt would really like to come out of solution but lacks a nucleation site. Bending the metal disk produces a transient site where crystallisation can begin, and from there are chain reaction results, causing all of the solute to crystallise and release the heat associated with this crystallisation.
Your guess regarding the half-dissolved crystals is correct - the crystals that have not dissolved act as seed crystals for further crystal growth. You need to dissolve all of the crystals to prevent the process from spontaneously reversing.
Despite the popularity of this kind of reaction for modern self-heating food/drink packaging in Japan, I suspect the discovery of this kind of reaction is lost to antiquity, as for instance sodium acetate can be made by simply mixing salt and vinegar. I recall reading that the highly exothermic reaction of calcium oxide and water was used in self-heating cans of food back in World War I.

The phenomenon of sodium acetate trihydrate (not sodium acetate!) to "melt in its own crystal water" is known since long. Who had the idea to market those bags, I don't know. BTW this is not an invention, one cannot get a patent on such a bag.

I refer you to US patents 4077390, 4587950, 5645749, etc. ad nauseum.

Incidentally, if you search for "hot ice" on YouTube you can find some rather neat demos of this outside the heating pads. (I particularly like the instant stalagmite formation ones.)

My guess is that producing Na-Acetate is not the difficult part in producing the heat bag but rather to obtain solution (or salt) and container that are clean enough so that the crystallization does not start spontaneously. Note also that the patents deal not with the material but with suitable construction of containers with a trigger for the crystalliztion.


text:
Is Blue Silica Gel more harmful than the Orange ones?
I am currently searching for a portable dehumidifier but notice that most of them come with a indicator that actually using the Blue Silica Gel, which I believe that it may contain Cobalt Chloride.
So, I would like to ask,
(1) if that is the case, would the portable dehumidifier become a harmful product?
(2) would an Orange Silica Gel contain other harmful chemical that is currently not found?

I had a poke around in an effort to determine the dye used in orange silica gel and was able to find two material safety datasheets (MSDSes) for orange silica gels which both gave different empirical formulae for the dye ($\ce{C25H30ClN3}$ and $\ce{C20H14O4}$), which suggests that there are multiple formulations of orange/yellow silica gel. Both of these dyes are organic molecules (as opposed to transition metal salts), and now their possible identities are given at the bottom of this response. Suffice to say that orange silica gels were designed specifically to avoid the use of cobalt chloride, which is fairly toxic.
As a document1 from the US National Parks Service indicates, the chief danger of cobalt chloride is its environmental toxicity if it enters soil or water through improper disposal. Cobalt salts are often rather evil, and this is actually reflected in the name, which is ostensibly an alteration of 'kobold'. However as long as the beads are kept in a suitable enclosure such as a filtered dessicator plug, they should not present a hazard. Inhalation of dust from the silica gel beads, direct handling of beads or cobalt chloride-impregnated materials or complete inundation of the dessicant (which could dissolve some of the cobalt chloride) are three possible ways the cobalt chloride could present a direct hazard, but the disposal of blue silica gel-containing products in landfill and subsequent leaching of cobalt chloride is probably the chief concern.
As far as the orange beads go, as I was not able to determine the specific indicator molecule(s) I can't ascertain their toxicity, suffice to say that they are by design less harmful than cobalt chloride and appear to be generally considered safe under normal use.
P.S. (finally, after chrome crashing multiple times)
F'x points out that $\ce{[C25H30N3]^{+}Cl^{-}}$ is crystal violet, which is not super toxic but for which there is evidence of carcinogenicity2 and which is considered an environmental pollutant.
$\ce{C20H14O4}$ may very well be phenolphthalein, which is considered to be a carcinogen by some authorities, most notably appearing on the California Prop. 65 listing3.
I guess the idea is that cobalt chloride is less desirable than either of these dyes. Don't eat the beads, and wash your hands!

(1) U.S. National Park Service; Cobalt Indicating Silica Gel Health and Safety Update; 2005, pp. 1-3
(2) Littlefield, N.A.; Blackwell, B.-N.; Hewitt, C.C.; Gaylor, D.w.; Chronic toxicity and carcinogenicity studies of gentian violet in mice; Fundamental and Applied Toxicology, 2004, pp. 902-912
(3) Chemicals Known to the state to cause cancer or reproductive toxicity; State of California EPA; March 16 2012, p. 17

Wikipedia says “An alternative indicator is methyl violet which is orange when dry and green when hydrated”. Isn't that simply it?

@F'x - Yeah I noticed that C25H30ClN3 is the empirical formula of crystal violet but didn't mention it because, well, it's typically violet. Additionally, C20H14O4 could be phenolphthalein, but that's violet as well! I'll add that.

Wrt. toxicity: 1. Wikipedia says: "antibacterial, antifungal, and anthelmintic properties" (and has a link to a proper paper). This implies that it is toxic for quite a number of organisms. 2. It is used as (non-toxic) DNA dye. I.e. it does stain DNA, but cells do not (immediately) die from that. While this does not imply carcinogeneity, it shouldn't be too surprising, either.

@cbeleites - Hi, I can't access the referred paper and there's no abstract available. Ref. 2 in my answer points to a paper that indicates that crystal violet is carcinogenic in mice. As always, the medicinal use of substances such as crystal violet and phenolphthalein go to show that there are many degrees of carcinogenicity and that substances can have markedly different acute and chronic toxicities.

@RichardTerrett, Hi Richard, It good to hear that "as long as the beads are kept in a suitable enclosure such as a filtered dessicator plug, they should not present a hazard." from a expert like you. Now, I can feel safe when buying a portable dehumidifier.

@RichardTerrett: Abstract is at pubmed, if you let me know an email address I could get you the paper. But already the abstract states that crystal violet has been used for such medical purposes for a long time.

@RichardTerrett: My point is that "common (bio)chemical sense" says: anything that is useful for killing bacteria, fungi and worms has to be toxic for them... And anything that penetrates cell wall and attaches to DNA is at least suspicious in terms of carcinogenic potential. (Which is not what the abstract I glanced at suggests: your paper points to de-methylation in the liver after eating the substance; and the berkely data base doesn't list other papers; skin application may be quite different). So maybe the surprise is the rather low carcinogeneity considering that it does stain DNA...

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What are the mechanisms for capture of As(III) and As(V) by magnetite from water?
Magnetite, $\ce{Fe3O4}$ is able to capture $\ce{As(III)}$ and $\ce{As(V)}$ from drinking water through the following mechanisms.

Bidentate-binuclear complex

Monodentate-binuclear complex (both from here)

I'm assuming these are standard reactions in inorganic chemistry, but are there known mechanisms (I don't have a text available)? The first looks like an attack on the $\ce{As}$ by the $\ce{O}$, but the second seems to take place in two steps.

Regarding inner sphere complexes, I wonder: have you read the Wikipedia page and if so, what is your specific question? I like your question overall, but this part looks like a general reference…

in short, there is no sphere… in coordination chemistry, the term coordination sphere is used to denote the atoms and ligands directly bonded to the central atom, regardless of the actual shape of the complex (which is never a sphere… most often a tetrahedral or octahedral coordination pattern).

I don't have the time to review some of the literature I found, but you may try to find the information you seek here.

Unfortunately there are no (or only very hard to find) studies to the mechanism of the actual sorption of arsenic (acids) to magnetite. Most literature does describe it as a simple Ligand exchange, although I find that hard to believe. From what I have read[1-4] I will try to compose some hints on what may happen.
First of all lets have a look at arsenic acid. It is a medium strong acid and a quite well oxidising agent.
\begin{align}\ce{
H3AsO4 &\equiv AsO(OH)3\\
AsO(OH)3 + H2O &<=> H3+O + AsO(O^- )(OH)2\\
AsO(OH)3 + 2H3+O + 2e- &<=> As(OH)3 + 3H2O
}\end{align}
In presence of any electron rich material and acidic environment it will therefore not be unusual that arsenous acid will be formed.[5] We may therefore also focus on this, amphoteric molecule:
\begin{align}\ce{
H3AsO3 &\equiv As(OH)3\\
As(OH)3 + H3+O &<=> As+(OH)2 + 2H2O\\
As(OH)3 + {}^{-}OH &<=> AsO^{-}(OH)2 + H2O
}\end{align}
Depending on the pH of the solution one would expect to find many of different of these  species. Arsenic is also somewhat oxophil, meaning that the $\ce{As-O}$-bond is quite strong.
The magnetite in aqueous solution reacts basic, also forming complex like structures or hydroxides, for example this fully hydrated cluster (ref. [2], stoichiometry omitted):
\begin{aligned}\ce{
 Fe3O4 + $x$\, H2O &<=>  Fe2(OH)2(H2O)8^4+
}\end{aligned}
Of course one has to pay attention, that the used ferric nano particles for sorpting arsen(o)ic acid(s) are far larger than the used model complex. However, we can see, that there are functional groups like $\ce{-OH}$ or $\ce{-OH2+}$ (or maybe even $\ce{-O-}$) covering the surface.
Using this as prerequisites we can think about what might happen next. I generally would suggest to start with physisorption, i.e. the cluster and the acid molecules come in close range. Since we are dealing with contaminated water it is maybe safe to assume, that we can deal with soluble salts of arsenoic acid, hence focusing on $\ce{AsO3^{3-}}$.
When the complexes come into close range hydrogen bonds will form, with
$
\newcommand{\nanoFe}{\bbox[lightgrey,3pt,border:1px solid black]{\ce{Fe}}}
\nanoFe$ being the nano particle:
$$\ce{\nanoFe-OH\cdots{}O-AsO2^{3-}}$$
Since the whole surface is covered with hydroxide groups, several of these hydrogen bonds may form. As we can see, with this we may now enter the realm of chemisorption. Once these complexes have formed, it is from an entropic point of view favoured to release smaller molecules, water in this case. Also the oxygen atoms bound to arsenic are now in close range to iron atoms. Since we are dealing with nano particle, there will be non-coordinated irons in proximity. This allows for a more permanent complet to form:
$$\ce{\nanoFe-O-AsO^3- -O\cdots{}HO-\nanoFe}$$
With this monodentate complex we have an already stable complex of a bound arsenic to an iron. It is furthermore possible, and probably favoured, to form a bidentate complex with a somewhat similar mechanism.
Let us also consider a more basic scenario. In this case the surface of the magnetite will not be entirely covered with hydroxide atoms, but more freely accessible oxygen atoms will be present. Direct bond formation between $\ce{\nanoFe-O\cdots{}AsO3^{3-}}$ may occur.

After writing all this I also found a credible source that somehow proves my theory. In “Understanding Arsenate Reaction Kinetics with Ferric Hydroxides”[6] they used a model system of $$\ce{HAsO4^2- + Fe2O10H15+ -> Fe2O10H15-HAsO4-}$$ to explain the mechanism.
In this they describe that physisorption is barrierless, leading fast to the adduct that is $\Delta G^\circ_\mathrm{r}=-58~\mathrm{kJ/mol}$ more stable in energy. However, chemisorption and developing the equilibrium is a more lengthy process.
When the formed complexes become neutral, the activation barrier to release arsenate is very high, compared to adsorption. I will stop right here, please read the publication for further details.

Notes, references and further reading:

Dimitri A. Sverjensky, Keisuke Fukushi, Geochimica et Cosmochimica Acta 2006, 70 (15), 3778-3802.
Nianliu Zhang , Paul Blowers , and James Farrell, Environ. Sci. Technol. 2005, 39 (13), 4816–4822.
Arnold F Holleman; Egon Wiberg; Nils Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. Walter de Gruyter & Co., Berlin, New York, 2007. (german)
Egon Wiberg, Nils Wiberg: Inorganic Chemistry. Academic Press, 2001. (english)
Thompson, A. & Goyne, K. W. (2012) Introduction to the Sorption of Chemical Constituents in Soils. Nature Education Knowledge 3(6):15
Heavier halogens ($\ce{X=I, Br, Cl}$) should do the job and they will most certainly be present in usual ground water (more or less acidic solution) :
$$\ce{AsO(OH)3 + 2H3+O + 2KX <=> As(OH)3 + 3H2O + X2 ^ + 2K+}$$
James Farrell and Binod K. Chaudhary, Environ. Sci. Technol. 2013, 47 (15), 8342–8347.
Also interesting: (a) Suvasis Dixit and Janet G. Hering, Environ. Sci. Technol. 2003, 37 (18), 4182–4189 (b) Dimitri A. Sverjensky and Keisuke Fukushi, Environ. Sci. Technol. 2006, 40 (1), 263–271.

You are talking about standard acid base reactions and formation of a salt (more covalent than ionic), so you can have an educated guess about the mechanism just thinking through what happens here:
As(III) and As(V) are simple inorganic (weak) acids, where a proton jumps up and down from the O. In literature, you may find estimations about the frequency about this jump. If not I would just use pKs and similar data of e.g. phosphoric acid to have an idea about the magnitudes.
The magnetite surface is a (weak) base, so the OH can accept a proton somewhere. The formed water stays there as a water ligand on Fe, or something kicks it out and dissociate.
Since both are weak, my guess would be that neither the unprotonated As acid nor the protonated magnetite surface is a dominant species. More probably the undissociated proton of the As-acids would attach to the OH of magnetite, and help to kick out the water while forming a salt. This salt is unlike NaCl or similar, and the Fe-O-As has a strong covalent character.


text:
What are the border definitions in the Ramachandran plot?
I've heard some people mention that this or that program isn't "up to date" with respect to the borders used in the Ramachandran plots to classify φ/ψ as being in the most favorable/acceptable/disallowed regions.
How are the accepted regions defined?  What is the current accepted paper that covers this?

I wish I could answer your question outright, but this is outside my area of expertise. The best I can give you is a place to start. I expect any recent book on proteomics would cover this topic. For example Introduction to Proteomics: Principles and Applications (Methods of Biochemical Analysis) by Mishra and Blobel, Wiley 2010.
I found some promising papers with a little searching with SciFinder. Neither I nor my institution subscribe to Protein Science, but the abstract looks promising. I have skimmed the article in PNAS. Hopefully one of these articles has your answer.
Zhou, et al, "Revisiting the Ramachandran plot from a new angle", Protein Science 2011 1166-1171.
Porter and Rose, "Redrawing the Ramachandran plot after inclusion of hydrogen-bonding constraints", PNAS, *2011, 108, 1, 109-112