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Why do atoms generally become smaller as one moves left to right across a period?
It seems to me that the addition of electrons and protons as you move across a period would cause an atom to become larger. However, I'm told it gets smaller. Why is this?

A nice periodic chart showing the effect you mention can be found here. It's worth noting that compared to the halogens, the noble gases are quite large. If you think in terms of satisfying shell requirements, that's actually a bit counter-intuitive. Also, while in chemistry one usually speaks in terms of filling shells, the deeper reason why electrons sometimes seem to repel each other more than they are attracted to protons is because of Pauli exclusion of identical-state fermions.

Where did You read about that fact, without finding the explanation for it?

Chemistry and physics have developed different ways of describing phenomena that are really the same thing. Pauli exclusion, played against charge attraction by a point-like positive nucleus and the low mass (high quantum location uncertainty) of electrons, is what produces the complex patterns and geometries that chemistry summarizes as "shells." Shells are very convenient, since they exhibit simple rules (e.g. 8 electrons "fill" a shell) that are much easier to deal with than the lower-level exclusion analysis. Nonetheless, for a question like this, you need to drop to that deeper level.

Also: I'm serious about the sudden jump from the very small size of a fluorine atom to the far larger size of neon -- sort of like a baseball expanding to beach ball size -- and yet becoming hugely more stable is quite surprising in may ways. If electrons are extremely happy to reach 8 (vs 7) in a group, shouldn't they show it by doing something like getting into a tight little sphere? Instead, just the opposite happens! I've never seen that explained. I may even ask that one over in Physics just to see if someone knows why "stability = bloat" and "incomplete = tiny".

""the sudden jump from the very small size of a fluorine atom to the far larger size of neon -"" I see a big problem in masurement and definition of "atomic" radius of a fluor atom.

Georg, you are correct. The table I used apparently was based on binding radii, which are bogus in this context. The discussion of this over in Physics S.E. can be found on this link. The less visual but correct neutral atomic radii table can be found here. Note that in this table, element 10 (neon) is correctly listed as smaller in radius than element 9 (fluorine).

As you move from left to right across a period, the number of protons in the nucleus increases.  The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons).
As you move down a column, there are more protons, but there are also more complete energy levels below the valence electrons.  These lower energy levels shield the valence electrons from the attractive effects of the atom's nucleus, so the atomic radius gets larger.

Can you explain why "this attraction is much stronger than the relatively weak repulsion between electrons"? Why are the repulsions weak?

Just as a side note, there can also be more complete energy levels as one moves left across a row (the $d$ and or $f$ shells can get filled) but these will NOT be the outer shell.

@Yann, I believe that it has to do with the basic principles of magnetism, where the distance between "individual" electrons in the cloud is large, and their individual charges are small, where the protons can "combine" their charges, allowing it to have far more strength over the distance.

The forces are the same. the fierce repulsion of electrons is why you cannot put your hand easily thru a door. The arrangement of electrons in an atom minimizes distance to the nucleus while maximizing distance between the electrons the two electrons in a 1s2 configuration, hydride ion, each think that they have the nucleus to themselves, but we know better. That is why hydride ion is larger than the H atom and a halogen ion is larger than a halogen atom

Remember, the 'size' of an atom has nothing to do with the size of the nucleus. It has to do with the size of the valence shell (which itself is not well-defined*).
So, if we neglect change in electrical attraction, the size should stay the same—a shell is a shell and it need not 'expand' to accomodate electrons.
Now, as we add more protons and electrons, the attraction between the nucleus and shell increases and the shell contracts. Thus the atom gets smaller.
*Shells reach 'til infinity, so it's better to define the size on the basis of 'the electrons of the outermost shell will be within this region x% of the time.' This only changes this answer a tiny bit. Now, we say 'the probability of finding the electrons closer to the center increases due to increased nuclear charge.'

Your ultimate and penultimate paragraphs are trying to emulate electrons one cannot enlarge and should stay the same the other is contracting. which is it? perhaps a bit of both. the additon of d and f electrons result in reasonably constant sizes so rather than constant E level sizes the two effects almost balance.

There is an important tool developed to answer this question, which is effective nuclear charge, which is the effective nuclear charge experienced by an electron, as an electron in an outer shell will always be shielded by inner electrons.
So, as we move from left to right the number of electrons in outer shells increases whereas the number of electrons in inner shells remains the same.
Hence, effective nuclear charge increases, forcing electrons to come closer.

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How does NaCl maintain its crystalline structure?
My understanding is that $\mathrm{NaCl}$ is an ionic compound, in which $\mathrm{Cl}$ becomes (effectively) $\mathrm{Cl^-}$ and $\mathrm{Na}$ becomes $\mathrm{Na^+}$. So I understand why I would get a "sea" of particles that would stick together.
But why does the above mean that it will have a face centered cubic structure with the ions held in place so rigidly?

If the question is generic, and is “why do ionic crystals crystallize?”, I would think it is too basic and too broad…

Crystals have inspired a great many chemists because they are fascinating for a good reason. Not only are they aesthetically pleasing, but they serve as an excellent subject to tour a variety of theoretical subjects important for understanding high-level chemistry.
Crystalline materials are made up of periodic structures. We’re only going to primarily focus on binary compounds where there is not a high degree of covalency. There are several ways to think about this problem, but let’s start with the melting of a crystal.
We say that at some definite temperature a highly ordered crystal will melt into a liquid. Those of us familiar with the language of equilibrium thermodynamics might recognize that the change in free energy for this phase change can be written, at constant temperature, as,
$$ G_\text{liquid} - G_\text{crystal} = H_\text{liquid} - H_\text{crystal} - T ( S_\text{liquid} - S_\text{crystal} ) $$
$$ \Delta G = \Delta H - T \Delta S $$
If we suppose that this process is spontaneous then we would say that the change in Gibbs’ free energy is negative, i.e. $\Delta G < 0$. This is true if and only if,
$$\Delta H < T \Delta S$$
Traditionally we interpret this as saying that there is a thermally-driven increase in entropy when we melt a highly ordered crystal into a liquid which more than offsets the energy cost associated with the enthalpies of the interactions holding that crystal together. A chemist tends to learn early on that the reverse is not necessarily true: at some definite temperature a perfect crystal rarely forms from the liquid. This inability to just heat up any substance and always produce a perfect crystal by cooling illustrates how crystal formation is a case of kinetic- rather than thermodynamic- control. So the process by which you form your crystal could possibly result in a different crystal structure. Sometimes crystal structures change just by altering the temperature of the chamber you’re measuring the crystal structure in! Now neither of these cases apply to sodium chloride to the best of my knowledge.
The formation of an ionic crystal such as sodium chloride is a delicate balance between electrostatic attraction and Pauli repulsion. Electrostatic attraction says that between two different charges, $q_+$ and $q_-$, there is a Coulomb force given by, $$F= \frac{k q_+ q_-}{r^2}$$
where $r$ is the distance between the two charges. If one plays with the numbers then it’s easy to see that at short distances the force is strongest, but there is a limit to how close they may come together. Eventually a repulsive force due to a quantum mechanical principle called the Pauli Exclusion Principle overpowers the attraction. An equilibrium results in which the atoms sit a certain distance from one another so that, if you will humor me, the “forces” between them balance out. This is why we traditionally represent crystal packing using marbles with a unique radii. The radii of the hard marble represents where the Pauli repulsion overpowers the attraction.
You might say, “Sure, we have these kinetic, electrostatic and quantum mechanical factors to consider, but how do these help with the final crystal structure?” Hold your horses, we’re getting there.
A famous mathematician and scientist thought about the most efficient ways to pack spheres of the same size together. By most efficient I mean this in terms of what FedEx considers efficient, fitting things together into the smallest possible volume. This is also what electrostatics want. Kepler suggested that the best way to pack spheres with this in mind can be maximized in two ways: fcc and hcp.
This conjecture, however, rests on all the spheres being uniform. We can’t assume this is the same for the atoms in table salt because the cation, sodium, is said to be smaller than the chloride anion. A tool that is useful as a guide for helping predict the structure is based off of the relative size of the cation, $r_\text C$, to the anion, $r_\text A$. This radius ratio, $r_\text C / r_\text A$, is mostly only useful for simple binary species. These $r_\text C$ and $r_\text A$ are tabulated as ionic radii in many chemistry books; they are contrived by various rules proposed by researchers, such as Linus Pauling, applied to the experimentally determined interionic distances. In this system we find it useful to discuss how many neighbors an anion or a cation has of the opposite charge.
The coordination number ($\mathrm{CN}$) tells us how many neighbors a type of atom has around it. A certain $r_\text C/r_\text A$ gives us a feel for the probable CN. From one source I was able to calculate the ratio for $\ce{NaCl}$ as $r_{\ce{Na+}}/r_{\ce{Cl-}} = 0.564$. Most textbooks will say that if your ratio is $0.414{–}0.732$, then you have $\mathrm{CN}=6$. I’ll show you how to calculate the minimum value for $\mathrm{CN}=6$. The easiest way to obtain the maximum is to obtain the minimum for $\mathrm{CN}=8$.
Briefly, if we put a small cation on a $XY$ plane and surround it in the manner shown below then we could also place two circles in the $Z$-axis above and below the central cation for $\mathrm{CN}=6$. (I won’t show those two in the $Z$-axis for clarity.) We would say that a given $\mathrm{CN}$ is stable only if the spheres are all touching each other. We would say that for the ratios drawn that a $\mathrm{CN}=8$ is not stable because our cation’s sphere would be too small and wouldn’t touch all eight of its neighbors. A higher density is always preferred, so $\mathrm{CN}=4$ is not preferred when $\mathrm{CN}=6$ is stable.
Clearly, $\angle \mathrm{DAC} = 45^{\circ}$. Moreover, we can say that $\overline{\mathrm{DC}} = r_\text A$ and $\overline{\mathrm{AC}} = r_\text A + r_\text C$. Trigonometry will tells us that $$\sin 45^{\circ} = \frac{\overline{\mathrm{DC}}}{\overline{\mathrm{AC}}}$$ Substituting in those values just determined,
$$\frac{\sqrt{2}}{2} = \frac{r_\text A}{\overline{r_\text A + r_\text C}}$$
Solving for $r_\text C/r_\text A$ gives,
$$r_\text C/r_\text A=\frac{2}{\sqrt{2}} -1 = 0.414$$
which is the lower bound desired.
So we can say that the $\mathrm{CN}=6$ for the cation. Similarly you can figure out that $\mathrm{CN}=6$ for the anion. This is why we say it has a 6:6 coordination pattern, which is indicative of a sodium chloride type structure. Cesium chloride, on the other hand, is 8:8 and indicative of its structure type. We tend to say that these patterns are consistent with a certain crystal structure because all the pieces fit together consistently. This geometric concept is based purely on sphere packing and is not without its limitations.
A more quantitative way to figure out which geometry is preferred is to calculate the electrostatic energies for the different geometries. A Russian by the name of Kapustinskii derived a formula to do this,
$$E_\text{lattice} (\mathrm{kJ/mol}) = \frac{\alpha}{r_\text C + r_\text A} $$
From here you need to put it in terms of the radius ratio with a constant $r_\text A$,
(I should note that this is evil algebra: $ \frac{a/b}{d/b + 1}= \frac{a}{b+d} $ where $a=\alpha$, $b=r_\text A$, $d=r_\text C$.)
$$E_\text{lattice} = \frac{\alpha/r_\text A}{\frac{r_\text C}{r_\text A} + 1} $$
Basically $\alpha$ is different for a change in the crystal structure type and from there you compare which has the preferred energy. You can reproduce the geometric trends using this more quantitative model. There are problems with this approach too, such as the neglected dipole and quadrupole and covalency. Yes, even in $\ce{NaCl}$ there is some covalent character. I think we say that $\ce{NaCl}$ $67\ \%$ ionic character. These become important when two different crystal structures are close in energy or when more exact calculations are required.

If I'm reading your question rightly, I think the answer to your question is "geometry."
That is, sodium and chlorine atoms form the particular structure of face-centered cubic because the relative size of the ions -- small for sodium large for chlorine -- makes that arrangement less stressful (energetically favorable). However, if you change the relative sizes of the ions, face-centered cubic is no longer guaranteed.
Cesium chloride is an example. It's not really body-centered since it has two different kinds of ions, but the larger size of cesium ions gives it the same form as body centered, only with cesium and chloride ions alternating.

A crystal forms to balance the forces of repulsion and attraction amongst the ions. Some ionic compounds form different structures depending on how the are solidified (whether they are created through rapid or slow cooling, for example). Silica (SiO2) is an example of an ionic compound found with many different structures. When there is enough energy available to the ions, say at higher temperatures, they can move from these ideal, symmetric and balanced geometries and temporarily maintain a less favorable geometry. An ionic liquid is constantly doing this.
A crystal's specific structure depends on the size of the ions, because their size affects how the ions pack together.

So why does it crystallize at all?

@soandos Due you mean why does it crystallize as opposed to an amorphous solid? It crystallizes initially due to the coulombic effects, as Yann has pointed out, and these form energy barriers following the Bohr model that direct exactly what distance each ion should be from one another. To maintain equidistant bonds in the solid, it will have to form the packing order that fulfills that distance requirement. This is where crystallographic structures, such as the primative cell, face-centered cubic, etc. come in.

How does $\ce{NaCl}$ maintain its crystalline structure? It is a good question.
Because in crystalline structure of $\ce{NaCl}$  6 electrons of $\ce{Na+}$ ion are surrounded by 6 electrons of $\ce{Cl-}$ ion and 6 electrons of $\ce{Cl-}$ ion are surrounded by 6 electrons of $\ce{Na+}$. That's how $\ce{NaCl}$ maintains its crystalline structure.

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Gibbs free energy-minimum or zero?
A reaction proceeds towards the direction of lesser Gibbs free energy (at constant $T$ (temperature) and $P$ (pressure)). So, we could say that Gibbs free energy at equilibrium is minimum.
On the other hand, we have $$\Delta G=\Delta G^\circ + RT\ln Q$$, where $Q$ is the reaction quotient. At equilibrium, $Q=K_\text{eq}$, and we already know that $\Delta G^\circ =-RT\ln K_\text{eq}$.
Substituting, we get $\Delta G=0$ at equilibrium. But, we know that $G$ minimized itself—thus there was a change in $G$ and $\Delta G < 0$.
What am I missing here?

I think your question really arises from some confusion about what $\Delta G$ represents. In general, $\Delta X$ for a thermodynamic quantity $X$ is the change of $X$ along some process. You could make it clear by actually writing $\Delta G(\text{A}\rightarrow\text{B})$ where A and B are before and after states. (We'll note that, in the general case, $\Delta X$ depends on the path take from A to B, making this notation improper. If $X$ is a function of state, though, you're good to go.)
However, in the equation you quote:
$$\Delta G = \Delta G^0 + RT \ln Q$$
the $\Delta G$ is a free energy of reaction and should thus be denoted $\Delta_\mathrm r G$, with the correct equation being:
$$\Delta_\mathrm r G = \Delta_\mathrm r G^0 + RT \ln Q$$
The free energy of reaction is defined as $\Delta_\mathrm r G = G_{\text{products}} - G_{\text{reactants}}$.

Thus, this $\Delta_\mathrm r G$ is not the variation of $G$ over the entire reaction, which would be the $\Delta G$ of the system between the start of the reaction and the equilibrium.

PS: I think this link is the online resource I found with the clearer use and explanation of notations. Notations are important in thermodynamics!

Hmm, but isn't G_prod-G_reactants negative as well?

@Manishearth At equilibrium, it'll be zero.

Wait.. But doesn't a system go towards the lesser G? If G is nonzero elsewhere, this means that  it had to increase at one point.

@Manishearth yes, the evolution of the system is such that it minimizes the free energy of the whole system. But what is your question? You have to be very clear about what system and what process/path you are talking about, otherwise it is confusing…

I'm talking about a simple A-->B reaction.whole system is what I'd missed (which is a plus not a minus) thanks :)

Gibbs free energy is a measure of how much "potential" a reaction has left to do a net "something." So if the free energy is zero, then the reaction is at equilibrium, an no more work can be done.
It may be easier to see this using an alternative form of the the Gibbs free energy, such as $\Delta G = -T\Delta S$.

But that's the point. $G$ is not zero by the formula, $\Delta G$ is. Or is my interpretation of the formula wrong?

Just because $G$ is at its minimum does not make it zero. (see here last part of introduction)

It might be helpful if we defined the changes in free energy in these equations in a more precise manner.  Let's just consider an ideal gas reaction.
$\Delta G^\circ$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:
State 1:  Pure reactants (in separate containers) in stochiometric proportions at 1 atm pressure and temperature T
State 2:  Pure products (in separate containers) in corresponding stochiometric proportions at 1 atm pressure and temperature T
To directly measure $\Delta G^\circ$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G^\circ$ for that path.  This path might involve the use of constant temperature reservoirs and semipermeable membranes.
$\Delta G$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:
State 1:  Pure reactants (in separate containers) in stochiometric proportions at specified pressures and temperature T
State 2:  Pure products (in separate containers) in corresponding stochiometric proportions at specified pressures and temperature T
If the specified pressures just happen to correspond to the partial pressures of the gases in a reaction mixture at equilibrium, then $\Delta G = 0$.  For small excursions of the partial pressures from the equilibrium values,  $\Delta G$ will increase as the squares of the partial pressure increments.  That is what we mean when we say that it is at a minimum at equilibrium.
To directly measure $\Delta G$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G$ for that path.  This path might involve the use of constant temperature reservoirs, small weights to be added or removed from a piston, and semipermeable membranes.

Ignore my previous comment, sorry. Can you explain why $\Delta G = 0$ when moving from state 1 to state 2 provided that the pressures correspond to the partial pressures of the gas in equilibrium? I had always (mistakenly) assumed that $\Delta G = 0$ at equilibrium meant that $dG = 0$, i.e the slope $dG/d\xi$ where $\xi$ is the extent of reaction is zero, which is clear because $G$ of the whole system is minimum at equilibrium. Your answer gives a much more clear definition of what we mean by $\Delta G = 0$, but I can't come up with any reasoning as to why State 1 and State 2 have the same $G$

@Ashish Ahuja Are you familiar with the van't Hoff equilibrium box chemical reactor?  Are you familiar with the open system (control volume) versions of the 1st and 2nd laws of thermodynamics?

I was not aware of the van't Hoff reactor but I had a look at your answer here. Because in the reactor everything moves reversibly i.e theoretically the reactor could operate at an infinitely slow speed, the reaction must not be spontaneous in either direction and hence $\Delta G=0$; is this logic correct? I am not familiar with the open system versions of the 1st and 2nd law. I looked at the relevant wikipedia sections but not knowing what I was looking for it was hard for me to understand everything due to my current level of knowledge.

@Axhish Ahuja The logic is correct.  But I wish you had experience with the open system versions of the 1st and 2nd laws so that you could see how it is verified mathematically.  For a better understanding of these, and how to apply them to the Van't Hoff equilibrium box, see Fundamentals of Engineering Thermodynamics by Moran et al.  I think this book is available online.

Could you let me know where the Van't Hoff equilibrium box is in the book, since I couldn't find it in the index at the back?

I don't think you can find it in this book.  Look in Introduction to Chemical Engineering Thermodynamics by Smith and van Ness.  I gave you Moran et al for good coverage of the open system versions of the 1st and 2nd laws.

Ok, I will have a look to see if I can understand it. Nevertheless, your answer has cleared up a long time doubt of mine and the reversibility logic while not rigorous, at least makes intuitive sense to me, so thanks.

I suppose you are confused between $∆G$ and $G$. After equilibrium is achieved, $∆G$ is indeed zero, as no more reduction of free energy is possible (keeping conditions intact). What this also means is that $G$ is minimum.
Throughout the reaction, $∆G$ will have been negative until equilibrium is attained. This means the free energy will keep on reducing until no more reduction is possible. This indicates that $G$ is minimum, since $∆G$ is zero, i.e. the value of $G$ at equilibirum cannot be further diminished.

What $ΔG$ means in equilibrium? Doesn't $Δ$ refer to change between two states?

@adosar look at the framing of the question. The answer lies within the words of the question. Consider time t = zero when equilibrium has been achieved. Throughout the undergoing of the reaction before equilibrium, i.e. for all $t<0$, $∆G$ was negative. At $t=0$, $∆G$ is zero. What this means is $G$ minimized itself for the entire time t was less than zero. At $t=0$, it finally attains its minimum value and is not subject to any further reduction. Tl;dr - $∆G$ WAS negative for $t<0$ , $∆G$ FROM NOW ON is zero. $G$ WAS reducing itself, $G$ NOW cannot reduce anymore.


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Why does nitration of N,N-dimethylaniline occur at the meta position?
The nitration of N,N-dimethylaniline with $\ce{H2SO4}$ and $\ce{HNO3}$ gives mainly the meta product, even though $\ce{-NMe2}$ is an ortho,para-directing group. Why is this so?

https://en.wikipedia.org/wiki/Tetryl - reaction can proceed in other way

The best way around this issue, from a reagent/yield point of view, is to use acetic anhydride to N-acetylate the primary amine. This gives acetanilide, the acetyl derivative of aniline.

Essentially a dupe: http://chemistry.stackexchange.com/questions/31644/how-is-anilinium-ion-meta-directing-for-electrophiles

In the presence of these strong acids the $\ce{-NMe2}$ group is protonated, and the protonated form is electron-withdrawing via the inductive effect. This discourages attack at the electron-poor ortho position.
Under the conditions I know for that experiment, you get a mixture of para- and meta-product, but no ortho-product due to steric hindrance.

In these acidic conditions, the lone pair on the nitrogen will abstract a proton of the acid in an acid-base fashion. This protonated amine will then be considered a meta-directing group. It is a deactivator that will draw electron density away from the ring (because nitrogen, a relatively electronegative atom, will have a positive charge).
The best way to get around this is to N-acylate. This can be done by first reacting the mono-substituted amine with acetic chloride ($\ce{CH3COCl}$) with pyridine to get an amide. That way, the ring is still activated and the lone pair on the amide nitrogen will not interfere with acids. It will direct ortho-para (para favored due to sterics).


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Are there any general rules for choosing solvents for recrystallization?
Re-crystallization is a nice way of purifying a product, but choosing a suitable solvent, if you can't rely on the literature, seems like a lot of trial-and-error.

Are there any general rules on which kind of solvents could be used for re-crystallization?
Which criteria should one use when trying to recrystallize a compound for which no literature on usable conditions are available?

I would argue that we should say it is more of an art than just brute force, but sometimes the distinction between the two is very arbitrary.

Generally speaking, the best solvent will be dependent on the impurity that you are trying to remove.
The solvent must dissolve both the desired compound and the impurity at a high temperature, but only the desired compound at lower temperatures.  The solubility product of the impurity, as well as the common ion effect should both be taken into consideration.

At the worst, one might have to resort to a mix of solvents; if you can find a pair of mutually miscible solvents where your impure solid is soluble in one and insoluble in the other, then this might be the way to go. Of course you have the risk of oiling out...


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What is the origin of the bactericidal properties of silver in water?
I often hear that water gets purified by being in a silver vessel, which sounds plausible because of bactericidal feature of silver. What doesn't sound plausible, though, is the way it's explained: that silver releases ions into the water. Since silver is a noble metal, why would any "reaction" at all occur with something as neutral as water?
Is the above explanation nonsense? Does the "disinfection" of water happen only on contact with the metal?

Silver forms many compounds with little effort (think silver nitrate, or even just tarnishing in air). Why do you assume it is not reactive?

I thought so because it is a noble metal

This is a typical example for beginners "black or white" thinking. Noble metals are more or less noble. Have a look in Nernst's equation.

Silver is not as inert as gold. Tarnish is the name we give to the phenomenon when silver metal is oxidized and becomes a salt. Surfaces made of silver tend to disinfect themselves pretty quickly. As for disinfecting water poured into a silver cup, I imagine that would take a little longer since you have to wait for silver to diffuse away from the surface and into the solution. But even very trace levels of silver can have strong antimicrobial effects.

This doesn't explain the bactericidal property of silver.

That is not the question that is being asked. The question seems to accept that property of silver as a given, but then asks how the silver does that if it is inert. This answer simply corrects the fallacy that the OP made with regard to the inertness of silver.

It should also be noted that at one point in time, there were quite a lot of silver-based antiseptics in use (e.g. Argyrol, a complex of colloidal silver in egg protein).

What does diffuse into water? Ions or "stable" atoms?

In an oxygenated environment, Ag+ ions are actually more stable than silver metal. The reason why silver can exist as metal at all is because the oxidation reaction is very slow for a solid piece of metal, even though tarnish is thermodynamically favorable. So Ag+ would do the diffusing, not Ag(0).

As far as I know, a chunk of solid silver will not spontaneously react with water.  But if you pass an electrical current through silver electrodes immersed in water,  the silver will be oxidized according to the following equation:
$$\ce{2H2O(l) + 2Ag(s) -> 2Ag+(aq) + H2(g) + OH- (aq)}\qquad E^\circ=-1.63\ \mathrm V$$
That will get you the ions you need.  Incidentally, Russia's MIR space station and the ISS use electrolytically dissolved silver ions to purify drinking water.

What do they use it for?

They use the elecrolytically dissolved silver solution for water purification (though I'm not sure if they use water or another oxidizing agent).

As to the above answers I also want to include the mechanism of action of silver as an antimicrobial agent. The exact mechanism of action of silver as an antimicrobial agent is not known and the current hypothesis is silver will converted to silver ions and this positively charged ions will attack the cell membrane, DNA or proteins which are negatively charged thus inhibiting the growth of microorganisms.
See also this link.


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Stability of organophosphorous vs organophosphates?
In environmental chemistry, most of the organophosphorous compounds we worry about are actually organophosphate compounds.

Is there something about the $\ce{C-P}$ bond that makes it less stable (and thus less persistent in the environment) than a $\ce{C-O-P}$ bond?

Is persistence the only criterion to "worry" about?

No, it's not. Other factors include toxicity and extent of release into the environment. But regardless, I am curious about the relative stability of these two kinds of compounds.

I guess the toxicity is the main problem.

Setting the scene for your question
In Environmental Chemistry there are several pathways a chemical might travel through including:

being subjected to the biochemistry of an animal
photodegradation
chemical degradation (perhaps it slowly reacts with water, etc.)
uptake as a metabolite in plants
degradation by soil microorganisms

and this is ignoring aquatic and air pathways.
Say, within a mouse or a fly, a possible pathway for degradation for some insecticide is cleavage of the P-O bond. How they have determined this in the past is noteworthy; a synthetic chemist would isotope label an atom in the molecule, say $^{33}$P-labeled or $^{14}$C-labeled, and he purified it before finally determining its specific activity. Then a biochemist would dose a mouse with a solution of the compound dissolved in an appropriate solvent, such as olive oil. Urine and feces were collected after administration. An aliquot of the collected urine, for example, would have its radioactivity measured using scintillation counting. The organo-soluble species would be extracted from the rest of the urine and the distribution of radioactivity could then be determined between the aqueous and organic layers. Then one could investigate what were present by TLC if you're old school -- they would actually run, via cochromatography, all the possible metabolite products they could independently think of and synthesize to identify a TLC spot! (That is a lot of work.) These days we can cheat with HPLC-MS. So, you might ask, "are results from this generalizable?" I have to tell you that they generalize in a very rough way. Enzymes are responsible for all the nifty biochemistry and they tend to be rather selective towards a given substrate and, in many cases, chirality even matters! What they are picky about is a tough problem that is tailored to each individual enzyme. One can make all kinds of generalizations about the chemical nature of the bonds and some possible predictions from it, but those won't necessarily hold up in an animal model with a different molecule.
So nitty-gritty mechanisms might not be what we want to look at. Instead we'll step back and take a broader view of this landscape. We need to look at bigger items in terms of physical properties, namely:
1) Persistence: what is its half-life in soil, etc. This is a matter of degradation, whereas the next is all about movement of an intact molecule. A simple chemical way to test this is how they resist hydrolysis.
2) Mobility: over time, does the compound tend to sit or spread itself around?
For example, factors that influence this are:

sorption: how much does our chemical bind to our soil?
water solubility: how easy it is to move in water for our chemical?
vaporizability: how likely is the material to evaporate away from its source?

A famous organophosphate is malathion. The partition coefficient, according to the wiki page, is 2.36. This implies its solubility is greater in organics than water, but some of it will most certainly dissolve in water. I think, based on its size, it is reasonable to assume it has a negligible vapor pressure too. These factors suggest it will have a low mobility: it will tend to stick to organics, won't appreciably evaporate away and will take quite a lot of water to wash it away. Contrast this with a famous organochlorine, DDT, which has an even higher partition coefficient (on the order of 10, so even more sticky and less soluble in water) and an even higher mass (so probably a lower vapor pressure) and thus an even lower mobility. Indeed, organophosphates largely replaced organochlorines as pesticides, in part, because they persist for shorter periods of times in the environment.
Many of these organophosphates were tested by us and their history is relevant. World Wars I and II ushered in a new era of chemical warfare agents and pesticides, notably of the classes of organochlorines, organophosphates and carbamates. Some organophosphates hardly persist at all, such as tetraethyl pyrophosphate, but they are pretty toxic to mammals. Many other organophosphates were screened for low toxicity towards mammals and increased persistence. We actually sought longer-lasting half-lives for these compounds! It turns out that we see alkyl phosphates with shorter lives than the longer-lived aryl species.
We tend to see certain trends within the different organophosphates, such as the water solubility being greater for the less complex and more polar compounds. An ethyl- is less soluble than a methyl- derivative and an aryl-substituted species get partition coefficients approaching those of the organochlorines.
"All this is fine and interesting, but it doesn't have much to do with my direct question," you might say. And most of the above just sets the scene for my answer to your question. Your specific query, I believe, requires us to go back to a nitty-gritty detail which is really only relevant in the case of your comparison between a phosphate (P-O-C) and an organophosphorus compound (P-C).
My answer to your question
In the biochemistry of most life, phosphates are pretty ubiquitous. As a result, many enzymes have evolved to manipulate them in many different forms; one class of these enzymes is called phosphatases and they assist dephosphorylation reactions. They certainly cannot manipulate just any ol' phosphate (hence why more complex phosphates have longer residues). We could say organophosphates are more "natural" than your typical organophosphorus compound. There is, therefore, a greater chance of having a tool -- an enzyme -- that can pick your more "natural" looking compound apart than your harder to find P-C bond.
As you've indicated, these phosphates are actually pretty stable. It's why DNA is built using them! But when you compare the bond lengths between P-C and P-O and O-C, there's a decent chance of finding that a P-C bond length is longer than a P-O or O-C. This leads me to believe that there are many cases in which P-O and O-C are actually stronger than P-C! The same could easily be said in the other way too! This, I think, is why we have to step-back and look at the bigger picture: the exact physical properties vary considerably from one compound to another and paint a landscape that says the issue is complicated. Perhaps, instead of it being a mere thermodynamic issue dealing with a comparison of the general bonding energies, it has more to do with the kinetic advantage of having enzymes equipped to slice the P-O-C rather than P-C. Of course, your mileage may vary depending on your specific examples.

You raise an excellent point about organisms being better equipped to metabolize phosphates. But the close similarity to endogenous substrates is a double-edged sword; it both raises the likelihood that organisms have the right enzymes to metabolize the xenobiotic, and raises the odds that the xenobiotic can disrupt biological systems by mimicking endogenous phosphates. In some circumstances this might lead to a model wherein at low concentrations the organophosphate is rapidly degraded, but at high concentrations it is more persistent until physically dispersed or abiotically degraded.

This is certainly true. Interestingly many of the commercially developed molecules were screened to minimize disruption to mammalian systems.


text:
Is activated carbon classified as organic or inorganic?
Organic compounds are typically defined as “molecules containing carbon”. Wikipedia states that there for some historical (read: non-logical) reasons,

a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon, and cyanides, as well as the allotropes of carbon such as diamond and graphite, are considered inorganic.

I thus wonder: is activated charcoal (also known as activated carbon) typically classified as an organic or inorganic material?
If I follow the list of exceptions given by Wikipedia, it should be organic (it's not an allotrope of carbon, in particular), but I get the feeling that most people in the field of porous materials would classify it as inorganic. So, I'm looking for an authoritative reference on this question.

Asking for charcoal might be interesting (in fact it is some highly disturbed graphite) but activated charcoal is a complicated industrial product, containing more than just carbon. And, as Chris says below, this kind of question is not a question among chemists.

It's an important question when these categories relate to compositions of products protected by patents. These terms are frequently used to cover a wide range of potential inclusions, for example.

While nomenclature is of particular interest to organic chemists to specify an exact compound, the classification of X into broad category Y or Z isn't a precise science, and not really of practical use.  The article cites a textbook by Seager to this effect, stating

The distinction between "organic" and "inorganic" carbon compounds, while "useful in organizing the vast subject of chemistry... is somewhat arbitrary"

Even if you find a source that says "charcoal is (in)organic", you may just as well find one stating the opposite.  Just like the coal from which it may have been produced, it was once biomass and decidedly organic, but so was graphite and diamond, or CO2 and CO32−. I think it's overly pedantic and unproductive to try to come up with definitive judgements for these decidedly edge case scenarios.
After all, it's just a chemical on the shelf, what one does with it is far more relevant.  I don't use it on a daily basis, but it seems more like a tool than a reagent.  The fact it contains carbon seems beside the point; it's value isn't in the chemical composition but rather its extraordinary adsorptive properties.

+1. Arbitrary distinctions really serve no purpose. Cutting things into different categories is really only useful for a bureaucracy. Then you end up with things like Health Insurance not being the same thing as Dental. (Meant to be a funny!)

Charcoal is the product of burning, dehydrating, and evaporating the volatile substances from formerly living matter.  Since activated carbon is made by processing charcoal, it would be considered an organic product.
There's a bit of a grey area here, admittedly.  Iron is typically considered inorganic, but it's absolutely critical to life.  In the context of studying hemoglobin, some experts speak about this iron as organic.
Also, the distinction between "organic" vs "inorganic" is less meaningful than it used to be.

I like the reasoning… do you have any reference to back that up?

Charcoal is carbon. As such it is normally classified as inorganic.
Consequently, you will find the chemical and physical properties of charcoal being discussed in textbooks about inorganic chemistry but rarely in textbooks about organic chemistry.


text:
What software can calculate aqueous solution equilibria?
What software is available out there to calculate the equilibrium in a set of reactions in aqueous solution? In particular, I'm interested in software general enough to simulate things like titration curves involving acido-basicity, complex formation, precipitation…
A web search turns up plenty of academic papers giving algorithms to find the equilibrium of such systems, but I'd like to find software that actually performs it. In particular, it would be really helpful if there existed such software that included a decent database of reaction constants for some well-known reactions (pKa for common acide-base couples, complexation and precipitation constants, etc.).

Edit: Apparently I was unclear: I'm not looking for molecular simulation codes, but for software that can calculate chemical equilibrium in solution from initial concentrations and a database of equilibrium constants for each possible reaction. The kind of thing teachers can use to predict titration curve and demonstrate the effect of various initial concentrations to students.

ChemEQL is pretty nice in some instances and free, but it looks like it has been mostly abandoned.
I'm aware of Titrator, but it only runs on Windows and I've never had  a chance to run it. It's also free.
I've heard good things about Visual MINTEQ, but it is Windows only and I've never had a chance to run it. Supposedly well maintained.
There is also PHREEQC, which sort of works on stuff other than Windows. I hear it is harder to use, but probably worth it.

I've not tried it out, but implementations of ReaxFF may be suitable for your purposes. It performs atomistic finite-temperature simulation of reaction mixtures using a forcefield and as such is more general than your request but it seems like it could do the job.
It's available in ADF as an addon (note the nice demonstration of the time evolution of a combustion process), in LAMMPS (again, a cool demonstration of the concentration of species changing over time as a shockwave propagates through a crystal) and probably a lot of other packages.

Sorry if I was unclear, I'm not looking for molecular simulation codes… See my edit to the question.

@F'x - Ah, so ReaxFF is probably not so great as a didactic tool. You might want to look at Yenka chemistry, which might fit the bill either now or in the future.

After some more search, I have found two pieces of software who do titration simulation, but are (as far as I can tell) only available in French:

Xem and (I can't believe what I'm about to do)… Google translation of the same page
Dozzzaqueux and translation

While it is primarily constructed to perform titration and equilibrium calculations on metal coordination complexes, once you get used to its syntax HySS can be effectively used to simulate acid/base equilibria and titrations.  One bonus of HySS is that it also treats solids (precipitates), which can be useful for, e.g., poorly soluble phenol derivatives.
Windows-only, unfortunately.


text:
Precedence of 1,2 carbocation rearrangement
How can one choose which group has more shifting tendency in 1,2 carbocation rearrangement? The obvious order is via the stability of the carbocation of the group. But, phenylic groups have high shifting tendency--and a phenylic carbocation is unstable.
Is there a way of predicting these? I was thinking that it will be related to the delocalization in the triangular intermediate formed.
Update: I'm talking about comparing "migratory aptitude" of the $R$ groups in a system similar to $>C(+)-C(R_1R_2R_3)$

@CHM: The first case.. Eg if we have $C(+)-C(R_1,R_2,R_3)$, which $R$ will shift?

"migratory aptitude" is the more established term, IIRC.

The obvious order is via the stability of the carbocation of the group.

I think you meant the migrating groups stability?
This is not what textbooks say.
Reactions are ruled by the $ΔG$ of the activated complex.
This may be close  to educts, or to products, or some halfway
state or a real intermediate minimum is on the reaction path.
So, in some cases the model of a separate carbenium ion
(in a ionic pair maybe) is useful, sometimes you need to
think of a three-center bond state.

So there is no general rule? I thought maybe there might be a way to predict effectiveness of delocalization of the carbocation...

Of course there is a rule, You mentioned it, but You took the stability of the hypotheticallly migrating group's carbcation as a criterion. It is the real cation that makes up the activated complex, that is all.

oh, I never took that--that was the first thing that popped into my mind, thats all. Many times stability of a system translates to stability of a single ion. In this case, its slightly more complex. Though I'm beginning to realise what it could be... Probably a mixture of +I and +R.. Not sure about hyperconjugation, does that come into play?


text:
What is the mechanism of APTES mono-layer formation on glass substrates?
The reagent APTES is a fairly common "ink" for microcontact printing, a technique that forms covalent bonds between the silanols found on the surface of the glass and the silane in the the APTES. It's also been demonstrated that the mono-layer then polymerizes somewhat, forming bonds between neighboring silanols.

What is the mechanism by which the printing and polymerization occur?
What is the reason for using APTES instead of a more reactive trichlorosilane?

Here's two links (1, 2) that show the simple scheme on forming monolayers. The essential figure is also shown here.

For the second part of your question... I can only point out this article that describes which properties change, if you make the molecules more bulky or add chloride, etc. I would guess that APTES has well known properties and they are "neutral" regarding common applications.


text:
Color of chromate and permanganate
I've heard quite a few times that the chromate and permanganate have a $d^3s$ configuration. Also, their colors arise due to a rapid switching of electrons between the oxygen and metal atoms.
I don't really understand the 'rapid switching' part--it's obvious why it can give color, but I fail to see why there is a need for such a switching-what's so special about $\mathrm{Cr}$ and $\mathrm{Mn}$? (I also do not know what the switching exactly is)
An explanation of $d^3s$ would be appreciated, though not necessary.

By "rapid switching" they technically mean Ligand-to-Metal Charge Transfer (LMCT). A more modern framework is Ligand Field Theory. I would have to teach a class to fully explain it in those terms, but I'll try to explain it in terms of this hybridization you brought up.
A chemical bond implies a higher probability of finding electrons between two bound nuclei. Atomic orbitals describe the electron density for individual nuclei. A bond between two nuclei necessitates some physical overlap of the relevant atomic orbitals.
It is in this way that we say that the linear combination of atomic orbitals, say two consisting of a $2s$ and $2p$ orbital, results in a molecular orbital that describes the molecule. We would write the wave function for the complex as:
$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p)$$
The amount of "mixing" is just a matter of adjusting the coefficients, $C_n$. And so we might say that the bond in this made-up molecule is $sp$ hybridized. $sp^3$ hybridized would mean the following kind of wave function:
$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p) \pm C_3 \psi(2p) \pm C_4 \psi(2p)$$
Similarly, $sd^3$ could mean a wave function of the following nature:
$$\Psi = C_1 \psi( (n+1) s) \pm C_2 \psi(n d) \pm C_3 \psi(nd) \pm C_4 \psi(nd) $$ where $n=3$ in the case of Mn or Cr.
If one performs the proper molecular orbital calculations on the valence $3d$, $4s$, and $4p$ for manganese or chromium and the $2s$ and $2p$ for oxygen in tetrahedral symmetry then you can draw the molecular orbital energy diagram for your complex once you match your eigenvalues to the correct trace of the appropriate transformation matrix... and you can hear a bunch more jargon on it, or I can explain the main point of our models: why we think MnO$_4$ $^-$ is purple.
The bonds in, specifically permanganate  (I have never actually done the calculations on dichromate, although they should be similar in principle) are between Mn$^{+7}$ and four O$^{-2}$ in the geometry shown here. The electronegativity of oxygen clearly dictates most of the electron density and, as a result, has them sitting on ligands (oxygen) instead of the metal. An electronically excited state can be reached with light absorption in the range of 500-600 nm light due to the relative weakness of the interactions between the ligand and metal. If they were stronger interactions then it would take more energy to promote electrons into a higher energy state and the color would be shifted into the UV. A colorwheel tells you that light absorption in the 500-600nm range should be approximately violet, which is what we see for the color of this particular complex. Anyway, this excited state means some electrons temporarily move from the ligand to the metal, resulting in a LMCT band.

Makes sense, thanks! I guess my main confusion was how "rapid switching" has a meaning if electrons are shared anyway. As a ligand, it makes sense :)

"Color" is what we see when electrons transition from a high-energy orbital to a less energetic one, if the energy associated with that transition happens to produce a photon within the visible spectrum. It might be helpful to stop thinking of electrons as being on either the metal or the oxygen, and to consider that in these cases electrons occupy molecular orbitals, which can encompass more than one atom. In this case, it seems likely that one of the orbitals in question might have higher electron density around the metal, while the other could have higher electron density around oxygen.
Why are Cr and Mn special? I would interpret it as coincidence that the oxides of these metals are prone to electronic transitions that generate photons which register as vibrant colors to our eyes.

You're confusing stimulated/spontaneous emission with absorption. The classic example is chlorophyll a: it doesn't absorb green photons, but red and blue ones. We perceive it as green because the light that reaches our eyes, the light that is not absorbed but reflected, is green (as well as other wavelengths that we can't perceive).

So the transitions are occurring between which energy states exactly? Between BMOs and ABMOs?


text:
Can recent developments improve the total synthesis of B-12?
The  total synthesis  of vitamin B-12 by Robert Burns Woodward and Albert Eschenmoser, is over 30 years old. At its time, it was considered a landmark in the field.

With current developments (e.g. olefin metathesis, palladium catalysis, etc.) is it possible to optimize this synthesis?

By "optimize" I mean reduce the number of steps in the synthesis.

Almost certainly, yes. Applications of modern organic synthesis methods have drastically shortened the synthetic routes for such previously "daunting" targets such as strychnine see Vanderwal, tetracyclines Myers  and pyrroloindole alkaloids Movassaghi, to name but a few - and recent efforts by Baran and White on C-H activation of alkanes have already yielded more efficient syntheses of many complex polycyclic terpenes. There is no reason to think that significant improvement could not occur for chlorophyll or B12. In fact, the absence of chlorophyll or Vitamin B12 from modern synthetic efforts represents a bit of a blind spot; it could well be that new chemistry could be discovered in the process. Although that 1,16 hydride shift by Eschenmoser will be hard to beat :-)

I would not call that a blind spot. The synthesis was done, looking for a better route would be done if there were some interest in synthesis industrially. Just being the second one for fun is not promising. Do You know the second man to climb Mt Everest?

Total synthesis works this way: be first or be best. Sometimes (often!), the first synthesis is not always the most efficient. Academic organic chemists use total synthesis as a "sharpening stone" for the development of new methods. Strychnine has been synthesized >15 different ways; morphine likewise. What I'm saying is that methods development & total synthesis research is not directly tied to industrial synthesis needs. Indirectly, however, it is basic research that furthers the field.

A bibliographic for total syntheses of Vitamin B12 or any other cobalamin turns out to be fruitless. In fact, all references turn out to be to the famous historical synthesis, but I could not find much work (any work, in fact) on total syntheses, or even proposals for new synthesis routes. Not much was achieved since then.
This is partly due to lack of practical need. There are many molecules out there that are challenging to synthesize, and Vitamin B12 is currently produced at the industrial level by bacterial fermentation. From Wikipedia:

The species Pseudomonas denitrificans and Propionibacterium shermanii are more commonly used today. These are frequently grown under special conditions to enhance yield, and at least one company […] used genetically engineered versions of one or both of these species. Since a number of species of Propionibacterium produce no exotoxins or endotoxins and are generally regarded as safe (have been granted GRAS status) by the Food and Drug Administration of the United States, they are presently the FDA-preferred bacterial fermentation organisms for vitamin B12 production.

A good recent reference on the subject of its synthesis would be: K. C. Nicolaou and E. J. Sorensen, “Vitamin B12” in Classics in Total Synthesis, VCH: New York, 2003.

This is good to know, but it does not answer the question directly. At the same time, I don't think there can be a definite answer to this question. You'd have to focus on a particular series of steps in the synthesis to find out if they can be done better/faster with current methods. To me, the question is not too localized but actually too broad.


text:
Why do impurities lower the melting point of an isolated substance?
It is known that impurities in a desired isolated product lower the melting point of the mixture, even if the impurities' melting point is much higher than the desired product. Why is that so?

I think your question is too vague. F'x gave you a great answer on the subject of metal mixtures, but you could get one equally good on colligative properties of solutions, or on thermodynamics. What particular case are you thinking of? A synthesis product not pure enough? A solution of glucose in water?

It's what a TA told me after my observation in the organic lab; indeed a synthesis product wasn't exactly pure.

Indeed this is a very common phenomenon, not limited to alloys. This is basis for use of mp as a kriterion for substance indentity and purity. There are exeptions (very few in organic chemisty) eg when the impurity and the substance make up a compound (salt). Nick T 's explanation by thermodynamics of mixed phases is the right one.

It's a very general statement, but it's not always true. I'll explain why it's often true, and give a counter-example at the end.

Your majority component B and the impurity (let's call it A) form a binary system. In most cases, such binary mixtures exhibit a solid–liquid phase diagram as follows:

(image taken from these lecture notes).
This binary phase diagram has pure A on the left, pure B on the right. A and B form, somewhere, a eutectic. It is the point here at concentration e and temperature y. Because the existence of a eutectic point is guaranteed for any A/B binary system, and because the eutectic corresponds to a lower temperature, your liquidus curve decreases with increasing impurity concentration, and the impurity thus lowers the melting point.
However, not all binary mixtures form a eutectic. In the words of Wikipedia:

Not all binary alloys have a eutectic point; for example, in the silver-gold system the melt temperature (liquidus) and freeze temperature (solidus) both increase monotonically as the mix changes from pure silver to pure gold.

The corresponding phase diagram is as follows:

Anyone have a copy of the lectures notes he links? It's now defunct.

Thermodynamically, you're considering the chemical potentials ($\mu$) of the liquid and solid(s), specifically the temperature where they're equal.  In a mixture, the potential is lower as the disorder (entropy) has increased, so all things equal, it will favor the liquid over a purer solid where there can be more disorder.  Chemical systems seek to lower their potential through spontaneous chemical changes (e.g. of phase), minimizing the free energy ($G$).  Equilibrium occurs where $dG = 0$.  In a constant temperature, constant pressure system,
$$dG =\sum_i\mu_idn_i$$
$\mu$ is the chemical potential of a given species (some compound in some phase) and $n$ is the amount of that compound.
Assuming our solid generated is pure (valid sometimes, not always), our reaction (freezing) is
 $$\require{mhchem} \ce{A_{(l)} -> A_{(s)}}$$
so the relationship between the two species is equal and opposite (generating one mole of solid consumes one mole of liquid).
$$-dn_{(l)} = dn_{(s)}$$
therefore,
$$\begin{align}
dG = 0 &= \mu{(l)}dn_{(l)} + \mu{(s)}dn_{(s)}\\
 &= \mu{(l)}dn_{(l)} - \mu{(s)}dn_{(l)}\\
0 &= \mu{(l)} - \mu{(s)}\\
\mu{(s)} &= \mu{(l)}\\
\end{align}$$
As the solid is pure (prior assumption, note: $\star$ denotes pure compound),
$$\mu{(s)} = \mu^\star{(s)}$$
In an ideal mixture (assumes interactions between all components are equal) with mole fraction $\chi_A$ and $T$ being the freezing point of the mixture,
$$\mu{(l)} = \mu^\star{(l)} + RT \ln \chi_A$$
(If non-ideal, a general $a$ term is used $\mu{(l)} = \mu^\star{(l)} + RT \ln a_A$).  All together:
$$\begin{align}
\mu^\star{(s)} &= \mu{(s)} = \mu{(l)} = \mu^\star{(l)} + RT \ln \chi_A\\
\mu^\star{(s)} - \mu^\star{(l)} &= RT \ln \chi_A\\
-\Delta G^\star_{m, fus} &= RT \ln \chi_A\\
-(\Delta H^\star_{m, fus} - T \Delta S^\star_{m, fus}) &= RT \ln \chi_A\\
\frac{\Delta S^\star_{m, fus}}{R} - \frac{\Delta H^\star_{m, fus}}{RT} &= \ln \chi_A\\
\end{align}$$
If the mixture is pure (freezing point at $T^\star$), $\chi_A = 1$, so $\ln \chi_A = 0$,
$$\begin{align}
-\frac{\Delta H^\star_{m, fus}}{RT^\star} + \frac{\Delta S^\star_{m, fus}}{R} &= 0\\
\frac{\Delta S^\star_{m, fus}}{R} &= \frac{\Delta H^\star_{m, fus}}{RT^\star} \\
\end{align}$$
Combined...
$$\begin{align}
\frac{\Delta H^\star_{m, fus}}{RT^\star} - \frac{\Delta H^\star_{m, fus}}{RT} &= \ln \chi_A\\
\frac{\Delta H^\star_{m, fus}}{R}\left(\frac{1}{T} - \frac{1}{T^\star}\right) &= \ln \chi_A\\
\frac{\Delta H^\star_{m, fus}}{R}\left(\frac{T^\star - T}{TT^\star}\right) &= \ln \chi_A\\
\end{align}$$
As $T \approx T^\star$ and if we define $\Delta T$ as the change in equilibrium temperature (melting point) from the pure substance,
$$\begin{align}
\frac{\Delta H^\star_{m, fus}}{R}\left(\frac{\Delta T}{T^{\star2}}\right) &= \ln \chi_A\\
\end{align}$$
This can be taken further to derive the coefficient for the freezing point depression, but from here we can see that because $\Delta H$, all $T$'s, and $R$ are positive and $\ln \chi_A$ is guaranteed to be negative ($\chi_A$ must be less than zero), $\Delta T$ must be negative.
To recap the assumptions, this is an ideal mixture and the solid formed is pure.  Addressing the first, if non-ideal, we can use the general $a_A$ instead of $\chi_A$.  The mole fraction can never exceed 1, but I'm not sure about $a$; if it does exceed 1, then there would be a freezing point elevation.
As for the second, if the solid formed is a mixture (e.g. a metal alloy), the that throws another wrench in the works that I don't definitively know how to address.  I believe it would then be based on the difference of interaction between the components in liquid phase versus solid phase as well as the relative concentration in each.

+1 very detailed answer. Read this answer and mine, you can extend your discussion of idealities into a discussion of why some binary mixtures don't form eutectics…

It really doesn't have anything to do with eutectics, although you are quite right that many systems don't have eutectics.  The derivation above is quite sufficient to show that the addition of any impurity in small quantities will lower the melting point.  Another reason is the entropy of mixing, which when one component amount is small is always positive.

The two above answers are academic/scholastic. I'll give an intuitive one.
When impurity is in a solid, it usually (not always, as said in Answer 1) weakens the connections/forces between molecules, and hence makes it more vulnerable to heat (read lower melting point). A solid is like an army in rank and file. When you put a civilian in the army, it doesn't matter whether that civilian is as strong as Arnold Schwarzenegger, the formation will be messed to some extent.
However, weaker intermolecular forces do not mean that solid will becomes softer. It may become harder and more brittle. An example is that copper is changed to a harder bronze if tin is added.


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How can I determine if there are π-π interactions between an amide and an aromatic ring in a protein?
In a crystal structure I've determined, a triazole ring on my ligand appears to be stacking with a tyrosine (top in picture):

However, there is also an amide, courtesy a glutamine, near it (bottom).  Is it likely this terminal amide is engaging in π-π interactions with the triazole or just weaker dispersion forces?  How would I quantify how significant the interaction is?

You could use the supermolecular approach.  Determine the energies for two isolated monomers [E(mon1)] and [E(mon2)] then the energy of the two interacting monomers [E(dimer)].  Then you just do [E(mon1)+E(mon2)]-E(dimer) = E(interaction).  Of  course you'd have to determine how you wanted to define your monomers.  For pi-pi stacking, you'll want to use something better than MP2 most likely as MP2 tends to overbind aromatic dimers by a significant amount.  I'd suggest CCSD(T).

@LordStryker DFT-D is going to be a much better option here...

It seems that NickT was looking for an experimental solution. My post deals, however, with a computational solution.

How would I quantify how significant the interaction is?

Determining the interaction energy between two defined monomers such as your aromatic triazole and amide is a rather straightforward process.  This process is referred to as the supramolecular approach.  I'll point you to a paper that analyzes the benzene dimer.  This method is strictly a computational one so a bit of knowledge in computational chemistry is necessary.
The Supramolecular Approach
The What
The supramolecular approach boils down to this.
$$E_{\mathrm{int}} =  E_{\mathrm{dimer}} - (E_{\mathrm{mon1}} + E_{\mathrm{mon2}})$$
Here we have some interaction energy ($E_{\mathrm{int}}$) determined from the difference of a dimer energy ($E_{\mathrm{dimer}}$) and the sum of the two monomers ($E_{\mathrm{monomer}}$).  If both monomers were equivalent (say, you were interested in the interaction energy of the benzene dimer where each monomer was a benzene ring), you could simplify the summation to two times the energy of one monomer ($2\times E_{\mathrm{mon}}$). In your particular case, you have two different monomers.
The whole idea is if I have two interacting molecules, I can determine the energy of each molecule individually (as if they were separated at infinite distance) as well as their complex.  So as you bring these molecules closer and closer together, the energy starts to go down (the interaction energy).
The How
NOTE: Your geometry is from a crystal structure.  Do NOT modify this geometry.  You will want to keep everything exactly as is.  This means you don't want to optimize your system.  You do not want to eyeball monomer placement.  Take everything from your known structure and be careful not to change it otherwise it can ruin this process.  We are modifying the structure by truncating and capping but intermolecular parameters must stay the same for whatever it is you are trying to model.  You will want to optimize your cap meaning, run an optimization on your capped-monomer but freeze everything but the thing you are using to cap.

Define your monomers.  You will need to determine what part of your 'dimer' system is important for describing this weak interaction.  I recommend keeping the aromatic ring and truncating the ring with something similar to what is being truncated.  You could cap your monomer with a hydrogen or a methyl group for example.  If the piece you've cut out is highly polarizable, cap with something with a similar property.  If your truncated piece is neutral in charge, cap with something that is neutral.  You get the idea.
Define your dimer.  Your dimer is simply a combination of your two defined monomers.
Determine the method you want to implement.  Wavefunction based or density functional theory (DFT) will work for this.  The former will be prohibitively expensive.
Define your basis set.  For aromatic systems your best bet would be to use Dunning-Hunzaga's correlation consistent family of basis sets.  I recommend using aug-cc-pVTZ for good results.  Whatever you decide, be sure your basis set includes polarization and diffuse functions.  The suggested basis set does this (augmented means diffuse on all atoms whereas pVTZ means 'polarized-valence triple zeta').
Determine the energies of your monomers and dimer.  You will want to run a single-point energy calculation on your monomers and your dimer. Optimize your monomers first but freeze all atoms except those you added to cap your monomer.
Determine the interaction energy.  Plug your energies into the equation given above and determine the interaction energy.  Convert to whatever units you wish to use (I prefer kJ/mol but most people use kcal/mol so you may want to use that).

I hope this helps.  Looks like an interesting project to say the least.
A quick note about the 'valence' basis set:
Typically core electrons do not participate in any sort of interesting chemistry (they are so much lower in energy than the valence electrons that they do not play a role in things like chemical bonding).  When you perform these energy computations, you will be invoking the frozen-core approximation (default on many quantum software packages but be sure to check the documentation first).  The pVTZ basis set expands the basis functions out to triple-zeta quality for valence electrons only.  The core electrons are given s-type functions and that is it.  If you ever find yourself in a situation where the core electrons are important to your system of interest, you would want to use the pCVXZ basis sets (where X = D,T,Q, etc.) which stands for polarized core-valence X zeta.  Here the core electrons are given the full set of X-zeta quality basis functions.  However this means you've just made your computation much more expensive.  The frozen-core approximation is typically used.

explanation is highly motivating. But I got some basic query. What if I use only the aromatic geometry here and truncate everything else, i mean why I need to cap ? Given that this is an amino acid chain, polarizibility   won't be so high in this case may be.

You cannot model an entire crystal structure, protein, etc. with a method that can reasonably quantify $\pi$-$\pi$ interactions with acceptable accuracy as this would be computationally intractable.  You would have to cap your system somewhere.

If I got a structure(chain) looks like, say, H3C-CH2-N-H...O-C-CH(CH3)-CH3. And if I want to get the energy between N-H...O-C. Then is it sensible to use only that concerned part(i.e. N-H...O-C) with the strict-geometry I have ?

Is your system an entire crystal structure or protein?  No its not though given the location of your comment it was easy for me to assume this.  Your system is small enough you would not have to cap anything and I would recommend that you characterize the full thing.

Actually I'm dealing with full amino acid chain from protein crystal structure. That's why I ask, if I truncate the whole thing except the particular section I'm intending to find i.e. N-H...O-C, then would it serve well ? (the example I put on there was just an analogy)

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


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Michael reaction or nucleophilic attack at the alpha carbon of an alpha-beta-unsaturated ketone?
What conditions promote a nucleophile to undergo the Michael reaction over the nucleophilic attack at the alpha carbon of the carbonyl group in an alpha-beta-unsaturated ketone? I'm looking for an answer that considers organic and inorganic nucleophiles (like organocuprates/organolithium compounds).

An $\alpha,\beta$-unsaturated ketone is electrodeficient at the $\beta$ position. This can be seen if you draw the resonance structures of such a molecule.
The $\beta$ carbon is thus a good site for nucleophilic attack. But, as you know, carbonyls are also prone to nucleophilic attack. To discriminate between the two, you need to look at how the reaction is controlled, either thermodynamically or kinetically.
In a kinetically controlled reaction, the product that is formed fastest predominates. In a thermodynamically controlled reaction, the predominant product is the energetically favored one.
A Michael addition is a 1-4 addition, where a nucleophile attacks the $\beta$ carbon, and produces the thermodynamically favored product. On the other hand, a 1-2 reaction (on the carbonyl) gives the kinetic product, and is obtained at low temperatures.
Why is the 1-4 product thermodynamically more stable? Because the resulting product benefits from keto-enol tautomerism, which results in lowering the energy of the system. Usually, the more resonance forms a compound has, the more its electrons are delocalized, the more stable it is.
Draw the resonance forms of the 1-4 and 1-2 products, and see.

You asked for specific affinities of different organometallics in 1-4/1-2 additions. My knowledge is that organocuprates ($\mathrm{R-CuLi}$) will perform Michael additions, and that organolithians seem to prefer 1-2 addition. Also, according to this source, Grignard reagents do not seem to have a preference.
My take on this is that the cuprate is less reactive, and therefore can form the thermodynamic product, whereas the lithium reagent is so destabilized that it reacts right away.

So given the right circumstances (e.g. low temperatures), one could make a Michael addition form almost exclusively an 1-2 product?

It must first be understood that Michael Addition is a special type of conjugate (1,4) addition where the nucleophilic species is an enolate. This will undergo 1,4 addition to the A-B unsaturated carbonyl.
Organocuprates add in a 1,4 fashion. The mechanism behind this is not fully understood, but it is known that radical intermediates are involved. Organolithium agents add in a 1,2 fashion. The reason for this is that the organolithium is very close to a carbanion and it will attack the most positive species in the molecule. In this case, the carbonyl carbon is the most electrophilic and so the organolithium will attack there. For reference, organomagnesium compounds are fickle and can attack either 1,2 or 1,4.
The alpha carbon is the carbon adjacent to the carbonyl carbon. The alpha carbon is actually quite nucleophilic (which is aldol condensation, alpha-alkylation, and alpha halogenation can occur). So, a nucleophile will attack either the partially positive berta carbon or the carbonyl carbon.

The nucleophile in a Michael addition can be an organometallic reagent. Also, I would think the $\beta$ carbon would be the most electrodeficient.

authors: Αντώνιος Κελεσίδης


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Acid catalysed synthesis of ethyl ethoxymethylenemalonate
One synthesis of quinolones begins with the formation of an ethyl ethoxymethylenemalonate, as seen in this Organic Syntheses paper.

I've been asked if the malonate derivative would be formed if methylmalonate was treated with trimethyl orthoformate with a catalytic amount of $\mathrm{H^+}$, without heating. The nature of the solvent isn't mentioned, but I guess the reaction is carried out in a non-aqueous medium with an ion-exchange resin, since I think trimethyl orthoformate acts as a water trap.
My hunch is that there shouldn't be a significant quantity of product formed. We're dealing with an activated methylene, why not simply use a base and then carry on with our addition? I can't imagine how, under normal circumstances, the elimination product ethyl ethoxymethylenemalonate will be formed.
How can this reaction happen in a mechanistically reasonable way?

You are using triethyl orthoformate, not the trimethyl orthoester. That's where the pendant group is coming from. If it had been just a water trapping agent, I presume the OrgSyn people would have had the sense to put the thing in an arrow instead on the left side of the reaction.

Ethyl, methyl, propyl, etc. doesn't matter here. I can devise a mechanism, but I'd need to provide heat for it to work, since it consists of an elimination. I'm just wondering if there's some unknown-to-me chemistry at work for the elimination to proceed without heating.

Ok, I will try my hand on the mechanism. In my opinion the reaction should work with a catalytic amount of $\ce{H+}$ since the key step for the reaction is the hydrolysis of the orthoester. This is an entropically very favourable reaction since from one reactant molecule you get two product molecules.

The resulting molecule is extremely reactive. So it can easily react with the enolized malonate. The enolization might be facilitated by the $\ce{ZnCl2}$ and/or the cat. $\ce{H+}$.
(I'm not entirely sure whether the $\ce{ZnCl2}$ really plays the role I've given it below in this reaction. So, if anyone knows good arguments against it, please let me know.)

The protonated product can then take part in an $\mathrm{E}1$ elimination reaction to yield ethyl ethoxymethylenemalonate. The carbocation which results when $\ce{EtOH}$ leaves the molecule is stabilized by the neighboring ethoxy group. The hydrogen on the central carbon atom is very acidic, since it is placed between three carboxylate groups, so it can even be removed by rather non-basic molecules, e.g. by an $\ce{EtOH}$, a malonate or an othoester molecule.

So much for the acid catalysed mechanism. Now for the question, why one doesn't simply use a base to facilitate the addition of the activated methylene compound. Well, this is due to the fact, that orthoesters are quite stable against basic hydrolysis. So, under basic conditions you don't get the carbonyl component of the addition reaction.

This looks plausible. I'm just surprised to see E1 occur without heat. Thank you sir.

@CHM: Ok, I will try to explain why I think it works. The rate determining step of the E1 mechanism is when the leaving group leaves the molecule thereby forming a carbocation. Since this step is very much facilitated (i.e. the activation barrier is significantly lowered) by the stabilization of the intermediate carbocation by the ethoxy group, less heat than normal will be needed to let the reaction happen at a reasonable rate. So room temperature might be enough.


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Are there any ionic amorphous solids?
This question on NaCl crystalization actually got me wondering: are there any ionic amorphous solids? Like ionic crystals are crystalline materials of electrostatically-attracted ions, can ions form an amorphous phase? I can see no reason why not, but I cannot think of any example either.

Do you mean are there any that will naturally form, or are there ways to take substances that would normally form crystalline solids and make them form amorphous solids instead (second is definitely possible)

@soandos examples of both would be nice… but I was thinking of the first. In fact, I think I do already have an example of the second, as you sure can freeze ionic liquids into a glassy amorphous state…

that is along the lines of what I was thinking. I would tend to doubt the existence of the first type.

I assume you meant amorphous solid phases, since you can always melt any ionic compound to form an essentially amorphous liquid phase.
Googling "amorphous salts" brings up a fair number of results in the primary literature, such as this and this and this. The defining characteristic of such amorphous solid phases is that they exhibit no measurable melting transition, i.e. no latent heat of fusion can be measured in a calorimetry experiment. There also examples of amorphous semiconductor oxides and amorphous metal oxides and amorphous metals - feel free to search for these and see for yourself. So yes, they do exist, and in fact are of research interest.

The link to springerlink.com is broken. I am not able to find any copy saved on the Wayback Machine, either.

The difficulty with creating an ionic bond glass is that creating true glasses depends on a combination of directional bonding between atoms (or molecules) and a rate of cooling that doesn't give those bonds time to move into optimal positions. The result is a bit like piling together sticky spheres that become locked before they can find an optimal stacking arrangement.
Pure ionic bonds in contrast are more like slippery spheres that are attracted towards each others' centers, but don't have any particular preferences about which parts of their surfaces make contact.
Given that, the most likely result by far from very rapid cooling of a molten ionic solid will be to produce micro or nanocrystalline ionic solids that may superficially resemble glasses, but which under X-ray or neutron diffraction crystallography would prove actually to have crystal structure at a very fine scale.
With that said, you nonetheless can come up with a recipe for trying to make a truly glassy ionic solid, especially since there's no such thing as a pure ionic bond. You would want to cool a very thin layer of molten salt at at an incredibly high rate, at least millions of degrees per second if I recall glassy metal cooling rates rightly. For ionic, you would also want the result to be pretty close to absolute zero, since whatever degree of "directional stickiness" you can get out of ionic solids is likely to be less even than metallic bonds.
Finally, if you really want to try to make such a thing as a serious project, you of course must check the literature in detail. My rule of thumb is that if you dig deep enough, there are almost no decently plausible ideas that you can think of in the physical sciences that someone has not at least written a paper on in some journal, and maybe even done some experiments.
Still, what you are asking is an intriguing idea for a real experiment, and it wasn't that many years ago that both metallic glasses and Penrose-tiled quasicrystals were new ideas.

On the contrary, there are numerous examples of amorphous salts, amorphous metal and amorphous metal oxides in the research literature that do not require cooling rates of "millions of degrees per second" to be formed.

Examples welcome! Organic salts will be much easier, but if you can find a reference to a simple elemental salt that shows no micro or nano crystalline structure under X-ray diffraction ("amorphous" can be a bit ambiguous, and often includes such not-real-glass states), that would be especially interesting.

Also, a "just in case" clarification: "millions of degrees per second" was just the unit used in some very early paper I read. The equivalent unit of "degrees per microsecond" captures the modest ranges a lot better. Such work used very thin films chilled very quickly indeed to keep the bonds cattywampus (technical term, sorry) down to the atomic level. Nanocrystals can be tough to prevent, so the earliest glassy metals were limited to very thin films. That got fixed as folks developed better alloys, ones that leave the atoms scratching their heads when deciding quickly who to bond to next.

Just add lots of different ionic units with complicated ratios to form an intricate salt lattice and the cooling rate can be reduced to 1000C/s form a glass. After all that's the basis of metallic glasses.

The following is not a proof, but an idea that would imply that it is not possible to have such a material.
The reasoning goes as follows:
In any ionic "sea" of atoms, there are two types of situations:

The atoms have enough energy to move freely, disregarding the attractive and repulsive forces of the other ions in the sea. This would cause a fluid, as the particles are not maintaining a position anywhere.
They do not have that much energy, and therefore they will be impacted by the attractive and repulsive charges in the "sea." It then follows that there is an optimal configuration that will stabilize all of these forces. While I suppose that it might be possible for the optimal configuration not to be periodic (and thus not crystalline) I would tend to believe that it is not possible (it would seem from tessellations that if one assumes some basic characteristics, they must be regular). In a similar vein, I believe that it is not possible to have degrees of freedom in such an optimized state (though there may of course be more than one locally optimal state possible) as each atom is at a local minimum of energy, and it lacks the ability to "move up" the gradient (if it had more energy, it would be a liquid).

I therefore believe that it is improbable that such things exist.

By your argument, how do you explain the existence of a glass?
As I'm sure you're aware (this is more for other people), the use of "proof" should never leave a chalkboard dedicated to mathematics. "Proof" does not belong in a physical science.

Glasses cool too quickly is my understanding. Yyou can create an amorphous ionic compound of any type by cooling it rapidly (see comment to question)

I would advise those who doubt in the existance of ionic solids to google ionic liquids which are widely commercialised.  They probably solidify at sub-zero temperatures.
http://pubs.acs.org/doi/abs/10.1021/ie101653n?journalCode=iecred
Many pharmaceutical salts can be obtained in the solid amorphous state by spray-drying, freeze-drying or milling.

The textbook example are crystals isotropized over geologic timescales through the effect of ionizing radiation.  Heat them up, to overcome the activation barrier, and they release the accumulated energy and turn crystalline again.
The relevant keyword is "Wigner effect".

One type of ionic compound that is likely to appear in amorphous form is silicates. Fu and Krawczynski 1 studied Raman spectra of natural and artificial silicate glasses. According to the authors:

Amorphous silicates are common in extraterrestrial chondrites of petrologic types 1 and 2. In addition, high percentage[s] of amorphous components and poorly crystalline phyllosilicates were found in mudstones at Gale Crater by the CheMin instruments on board of Mar[s] Curiosity rover...

Such compounds might be expected to be more common on bodies where less geological activity (which often involves high temperature processes) is available to promote crystallization.
Naturally occurring amorphous silicates tend to have polymerized or macromolecular anions, which hinder the crystallization process. Some naturally occurring examples given by Ref. [1], from the Earth and Moon, are below. Note that the anion has a low enough oxygen content to form polymeric/macromolecular structures in all cases. [Dear readers: I would like to make a table, but the syntax is not working for me.]
Obsidian
Principal oxides(*): $\ce{SiO2,Al2O3, Na2O,CaO}$
NBO/T(#): 0.01
Hawaii basalt glass
Principal oxides: $\ce{SiO2,Al2O3, CaO,MgO,FeO,TiO2,Na2O}$
NBO/T: 0.73
Apollo 15 brown glass
Principal oxides: $\ce{SiO2,MgO,FeO,Al2O3,CaO,Cr2O3}$
NBO/T: 1.66
Apollo 16 green glass
Principal oxides: $\ce{SiO2,Al2O3,CaO,MgO,FeO}$
NBO/T: 0.43
*Oxides listed account for at least 99% of the oxygen atoms in the material.
#Ratio of Non-Bridging Oxygen to Tetrahedrally coordinated atoms, the latter being $\ce{Si,Al,P}$; a value less than 2.0 allows macromolecular anions to form and lower values indicate more extensive polymerization.
Reference

Fu, X., A. Wang and M. J. Krawczynski (2017). "Characterizing amorphous silicates in extraterrestrial materials: Polymerization effects on Raman and mid-IR spectral features of alkali and alkaline earth silicate glasses". J. Geophys. Research Planets 122, 839-855, https://doi.org/10.1002/2016JE005241.


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How does chloroform degrade nitrile gloves?
Many of us have experienced the failure of nitrile gloves when exposed to chloroform. What's going on at a mechanistic level when this occurs?
I would guess that the chloroform dissolves some of the polymer into its constituent monomers, but I've never heard anything more definite than that. Is it a similar mechanism to what happens when chloroform is left too long in a plastic bottle?

+1 Interesting question. If the chloroform does dissolve the polymer, it most probably does not depolymerize it. I think dissolution is much more probable than depolymerization.

Its not a reaction. Its more like acetone and plastic. You get the plastic back when the acetone evaporates.

Nitrile gloves are made of nitrile rubber, or poly(butadiene/acrylonitrile). This polymer is highly soluble in chloroform, with some papers I found indicating that one can dissolve up to 18% in mass of nitrile butadiene rubber in chloroform. Moreover, it permeates easily through NBR, meaning we can expect the dissolution to be fast in addition to thermodynamically
favourable.
Finally, I would not expect the mechanism here to be any different from that of any polymer dissolution by a good solvent. The solvent will permeate through the polymer, intercalate between polymer chains, and solvate them (inducing swelling). Once solvated, the network of polymer chains looses its mechanical properties and they can fully separate. (I wish I could find a good existing illustration for that part, but I can't right now… If anyone can, feel free to edit!)

Close but no cigar. It does dissolve the polymer, but certainly not into its constituent monomers. That would be one heck of a chemical reaction.
Make sure you choose the proper gloves for whatever you are working with.


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How does the choice of metal (oxide) catalyst affect the range of unsaturated compounds that can be hydrogenated?
In the hydrogenation of unsaturated compounds with hydrogen gas and a catalyst, the choice of palladium on carbon is able to hydrogenate alkenes and alkynes, but is unable to hydrogenate aromatic compounds. Using rhodium or Raney nickel, however, allows one to hydrogenate aromatics. How does the choice of catalyst affect what kind of unsaturated compounds can be hydrogenated?

@Georg I think he wants a mechanistic explanation.

@Georg: Oh, I assure you I did. Unfortunately I seem to not be reading the right references, since a lot of them just mention "catalyst X works for hydrogenating substrate Y". Now, anything else you can offer, good sir?

The currently accepted hydrogenation mechanism for metal like palladium is the Horiuti-Polanyi mechanism. (image)  This mechanism works with good yield for alkenes and alkynes.  For an aromatic compound, the activation energy is much higher because the aromaticity of the target molecule is being broken by the first hydrogenation.  Indeed, 1,3-cyclohexadiene is very unstable and will revert to benzene and hydrogen instead of continuing to cyclohexane.
Addition of a Lewis Acid makes palladium an acceptable catalyst for the hydrogenation of benzene, however.  An intermediate compound with the benzene and the LA is first formed, and hydrogen is added to this intermediate.  This paper discusses the mechanism of hydrogenation in the presence of a Lewis Acid.
Rhodium catalysts work slightly differently.  They generally have ligands attached that help stabilize the intermediates.  You can see a simplified mechanism here and there are papers discussing specific catalysts here and here.  Basically, in the case of benzene, the relatively unstable 1,2-cyclohexadiene is held in place on the catalyst until it has been completely hydrogenated.

In that Horiuti- Link there is a big error. No D in the upper product, althouh the reaction is +D and -H ! In general, JM asked for heterogenous katalysts, and there is little to no knowlegde besides the phenomenology of results. The problem is the microscopic structure of the Pd or Pt or Ni preparations.


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If nothing sticks to Teflon, how does Teflon stick to frying pans?
The most notable characteristic of polytetrafluoroethylene (PTFE, DuPont's Teflon) is that nothing sticks to it. This complete inertness is attributed to the fluorine atoms completely shielding the carbon backbone of the polymer.
If nothing indeed sticks to Teflon, how might one coat an object (say, a frying pan) with PTFE?

Why do the fluorine atoms have to be on both sides?

""This complete inertness is attributed to the fluorine atoms completely shielding the carbon backbone of the polymer."" This is wrong, any atom will shield the carbons. Because the fluorine diameter is small, this shielding is even lower for PTFE than in other similar polymers. The true reason is the low polarsisability of the fluorine substituents. So, all other molecules try to "stay among themselves" resulting in minimum interface area.

@Soandos, care to elaborate?

It has to be so common a question that the answer is actually given in various places on Dupont's own website (Dupont are the makers of Teflon):

“If nothing sticks to Teflon®, then how does Teflon® stick to a pan?"
  Nonstick coatings are applied in layers, just like paint. The first layer is the primer—and it's the special chemistry in the primer that makes it adhere to the metal surface of a pan.

And from this other webpage of theirs:

The primer (or primers, if you include the “mid coat” in the picture above) adheres to the roughened surface, often obtained by sandblasting, very strongly: it's chemisorption, and the primer chemical nature is chosen as to obtain strong bonding to both the metal surface. Then, the PTFE chain extremities create bonds with the primer. And thus, it stays put.

While this is probably the best answer available publicly, it also doesn't seem very satisfying from a scientific viewpoint.

Before putting Teflon on a pan, the pan is scratched hardly leaving some tiny holes where the hole opening is smaller than the hole inner size (like a bottle). When Teflon is cast into that, it physically can't get out. That's also why when you scratch a Teflon pan, things start sticking to it.

they spray a primer first on the substrate to be coated (the frying pan) , then they spray the fluoropolymer powder on it. the primer is the tricky part. i think the idea is they make another copolymer of ptfe and another monomer that dont have all its C atoms saturated with F. this copolymer sticks to ptfe because it contains ptfe and at the same time can stick to other substances more readily vs pure ptfe polymers due to other c atoms not fully saturated with Florine.this copolymer is mixed with a polyamide or polyimide or other forms of plastic that would stick after melting to the frying pan. so to sum it up . the polyamide plastic sticks to the aluminum of the pan. then the copolymer of ptfe and perfluoro polymer sticks to the polyamide and finally the top coat of ptfe is applied over that copolymer and sticks to it . the idea here is this copolymer that is half ptfe and half perfloropolymer that sticks to ptfe above and other adhesives below. its a gradient from sticky substance up to a completely non stick pure ptfe. at least this is what i understood from reading some of the patent information about the primer used by chemours(dupont). for more detailed information you can refer to these patents.


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What can I do if I don't get a phase separation between my organic and water phases?
Solvent extraction in a separation funnel is a very common method in preparative organic chemistry. But sometimes you don't get a nice phase separation between the organic and the water phase.
What are the possible causes that can prevent the formation of two distinct phases, and how can you force a phase separation when you encounter such situations?

sometimes I try just waiting, acidifying the solution, heat, an ultrasonic bath, centrifuge, filtering through sodium sulfate... even gasp adding salt.

For "possible causes", I should think that having a surfactant present would up the chances of forming a stable emulsion, preventing separation. I don't imagine that anyone would be knowingly trying to purify a surfactant by solvent-solvent extraction, but it's something worth bearing in mind if some surface-active-agent were produced as a by-product / was an unreacted reagent / was present from an earlier reaction step and had not been fully removed.

One such cause, where separation does not take place is the nature of the organic solvent. Its better to check the miscibility of the solvent with water. There are several websites, where one can get this info. For example 1 and 2.
Even with hydrophobic solvent, if the problem persists, one can add NaCl and then check the separation.

Adding salt to the mixture augments the ionic strength of water, therefore increasing the "difference" between water and your other solvent. I've found that this usually works.

Yes, assuming your solvent isn't miscible in water, adding brine (concentrated solution of NaCl in DI water) usually breaks the emulsion.

Well, shaking a sep-funnel is a great way of creating a horrible intractable emulsion with no clear phase boundary, so don't do that (also, the stopper might pop off as soon as you set it down).
As for breaking an emulsion, if you have the time, just leave the sep-funnel somewhere and wait a few days. I seem to recall the first and last time I shook a sep-funnel the lab tech added some salt to the water/diethyl ether emulsion however I'm not clear on this.
Finally, consider sacrificing part of your product by drawing the funnel off to the very base of where the emulsion begins, draining the emulsion layer entirely and then taking the remainder. If you have blobs of one phase in the other, you may be able to remove them with a pipette.

I'm not sure why you made the comment about creating the emulsion/popping the stopper off by shaking the sep funnel. In order to get a good extraction, you need to shake the funnel. To prevent the stopper from popping off, you need to use proper technique and vent the funnel from the stopcock periodically as you shake it.

@scientifics - regarding venting the sep funnel, you are of course correct, and any incidents involving stoppers can be attributed to inadequate venting. However, as far as emulsifying phases are concerned, IMHO protracted swirling is preferable to shaking.

An additional option besides adding salt and having patience, is to place your funnel inside a heating oven at a temperature that is not harmful to your reaction. In my experience, it helped with the separation. To prevent buildup of pressure, remove the lid, and maybe place it upside down so it covers the funnel without really blocking vapor escape.
Also, if you plan on adding salt, it can help to empty the content in a beaker, add the salt there, and stirr with some heating for a while. Then put the content back in the funnel.