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Is there a general consensus on the causes of the alpha-effect?
There have been various explanations posited for the α-effect. The α-effect refers to a phenomenon wherein nucleophiles with lone pairs on atoms adjacent (i.e., in the α- position) to the atom bearing the reacting lone pair sometimes exhibit dramatically higher reactivity than similar nucleophiles without α-electrons. This effect is especially adduced when no associated increase in Brønsted basicity occurs. For example, hydroperoxide ($\ce{HOO-}$) experimental reaction rate constants are orders of magnitude greater[1] those of hydroxide ($\ce{HO-}$) with various electrophilic substrates, despite the former exhibiting lower Brønsted basicity. There is also a thermodynamic α-effect, in which equilibrium constants are enhanced[2]. It is currently on the list of unsolved problems in chemistry on Wikipedia, but, due to a lack of references to that effect, I'm not entirely convinced it really should be listed there. Here's the summary of my research on the topic thus far:

I read Ren, Y. & Yamataka, H.[3], "The alpha-effect in gas-phase SN2 reactions revisited". In it, they claim that explanations based on ground-state destabilization (presumably due to repulsion between the electrons of the nucleophilic atom and the α-electrons) are not correct. Their reasoning is that this would result in a difference in the $\Delta G$ between reactants and products, leading to thermodynamic equilibrium effects. They argue that a correct explanation should be one exclusively involving stabilization of the transition state (i.e., minimization of $\Delta G^{\ddagger}$), and go on to offer some explanation for how this may occur (along with experimental data). Intuitively, their conclusion seems reasonable to me, and it also (at least to my naive comprehension) seems eminently testable. I don't know whether equilibrium effects consistent with ground-state destabilization have actually been observed or not; however, if they haven't, shouldn't that put the nail in the coffin of that theory? Or is it simply that the authors are searching for a purely kinetic α-effect, so that a distinction between a thermodynamic one needs to be made?
Fleming devotes a section to the effect in his book, Molecular Orbitals and Organic Chemical Reactions. He notes that the presence of the α-lone pair should raise the energy of the HOMO of the nucleophile, but also points that experimental results don't correlate sufficiently well with the HOMO energies of various α-nucleophiles. In particular, certain soft electrophiles (per HSAB theory), such as alkyl halides, apparently show an anomalous low preference for α-nucleophiles. In the context of SET mechanisms, Fleming says that the higher energy of the HOMO and the availability of α-electrons (which can stabilize a radical intermediate) ought to have a highly favorable effect on the rate of reaction, and notes that experimental results have borne this out. My interpretation of this is that, while the picture is perhaps murky for anionic mechanisms, transition-state stabilization clearly seems to be operative in SET mechanisms.

I've also read the original 1962 paper by Pearson and Edwards[4], which also largely argued for transition-state stabilization as the primary explanatory mechanism.
Overall, from my reading thus far, it seems that transition-state stabilization has been most consistently invoked and has the largest wealth of evidence and the most plausible arguments supporting it. What I'd like to ask is, (a) are there flaws in my reasoning or understanding of the material, and (b) is this truly a fundamentally unsolved problem, or is there actually some emerging consensus among experts?

Notes and References

Fleming provides a small table with relative rates ($k_\mathrm{rel} = k_{\ce{HOO-}}/k_{\ce{HO-}}$) in his book. For example, he gives $k_\mathrm{rel} \approx 10^5$ for reaction with $\ce{PhCN}$ and $k_\mathrm{rel} \approx 50$ for $\ce{PhCH2Br}$, while $k_\mathrm{rel} \approx 10^{-4}$ for reaction with $\ce{H3O+}$. The rate of reaction correlates in the expected way with Brønsted basicity only in the case of proton transfer.
Again, citing Fleming, he gives the example of the reaction of N-acetylimidazole with hydroxylamines, in which both rate and equilibrium constants are positively affected. Qualitatively, he explains this by noting that the α-electrons raise the energy of the lone pair conjugated to the π-system, making overlap of said lone pair with the π* LUMO more effective. Additionally, he claims both ground-state stabilization and transition-state destabilization as being factors in the reduced electrophilicity of oximes and hydrazones relative to (most) other standard imines.
Ren, Y.; Yamataka, H. The α-Effect in Gas-Phase SN2 Reactions Revisited. Org. Lett. 2006, 8 (1), 119–121. DOI: 10.1021/ol0526930.
Edwards, J. O.; Pearson, R. G. The Factors Determining Nucleophilic Reactivities. J. Am. Chem. Soc. 1962, 84 (1), 16–24. DOI: 10.1021/ja00860a005.

I really like your question and the research you have done. I also think your reasoning is quite solid. However, as the definition is quite vague and attempt on finding a 'true'explanation can also only be vague. I agree with you that transition state stabilisation might be the strongest clue so far, but I am also afraid, that it remains an rather unsolved question so far. (Anyway, I'd be delighted if you could post doi or isbn into your question.)

Transition state stabilization is mentioned in couple of new papers published about alpha-effect: 1 "solvent interactions for HOO– are quite different from those with the normal nucleophiles at the transition state, indicating that differential solvation may well contribute to the α-effect." And 2 " Differential solvation effects have been suggested to be responsible for the α-effect in this study,"

This is the n-th time I've read this question, first time commenting.  What does the literature say (if anyting) about quantum chemistry simulations of the alpha effect?  If first-principles quantum calculations don't capture the alpha effect, then I would think it is truly an unsolved problem in chemistry.    If they do however, then it's more a question of which hand-wavey heuristic theory/approximation to QM you want to invoke to "explain" it.  If no one has tried modern compuchem methods on the problem, I guess its still unsolved, but seems like a good problem to work on.

I am not a kineticist, and my quantum chemistry is long, long out of date, but what I was about to say was that I'd guess the reason the "effect" is "unsolved" is that it's not real.
That is, it is not a property of a single reactant while disregarding its environment (gas phase, solvent interactions). Then I saw that the two recent articles both were about solvation, so my comment is redundant (and certainly only a partially/inadequately educated guess). I'd also comment that comparing $\ce{HO-}$ with $\ce{HOO-}$ is apples and oranges. You should compare it with a species with an alpha atom which is electronegative but doesn't have a lone pair.
If it doesn't really have a published DFT model, then it might be good for an MS student to work on. I suspect answering it is like "curing cancer", it doesn't have just one  'reason', rather the cures depend on the exact nature of the reaction (including solvation).


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What is the most rare element in periodic table
What is the most rare element in periodic table? I have a lot of confusion about this.

I presume you mean "rare"? You need to be more specific, i.e., do you mean rare in terms of relative abundance, economic value, or some unusual physical or chemical properties? Are you including only naturally-occurring elements? Whatever your specific criteria, it's all easily answerable by a Google search.

An overview of Big Bang and stellar nucleosynthesis would be very fitting for this question.

Are you excluding elements that only have shortlived radionuclides?

Well, the first question to answer would be: where?  The rarest element in the universe? the sun? in meteorites? on earth? in the ocean? in humans?  We can tap in to the curated datasets provided by Mathematica to get the answer:

Elements up to atomic number 98 have been found in trace quantities; however, the dataset does not contain any information about the trans uranium elements which were once thought to be synthetic.
Here's a plot of abundance as a function of atomic number:

We can see that the elements beyond about Z = 82 (lead) have a marked decrease in abundance, and this is consistent with all of the elements with atomic number above 82 being radioactive.
Where do we come from?
We can compare the elemental abundance of these individual "spaces" to speculate on where humans come from.  Below are plots showing the correlation of the elemental abundances.  The plot labels give the best fit equation:

The best slope is observed when comparing our abundance to that of the Earth's crust; however it does appear as if the scatter is smallest for the Us vs. the Ocean plot.

source: Stack Exchange

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Will a Galileo's Thermometer containing liquid paraffin explode in my car if the temperature is above 35 degrees Celsius?
I'm driving from Sydney to Melbourne (approximately 12 hours trip) and I'm wondering if it's safe to have this Galileo's Thermometer in my car during the trip if it's going to be a scorcher of a day. It could easily reach 35 degrees Celsius or over during the trip. I'm a little worried that it might explode. Any help much appreciated!

Uhh... it shouldn't explode? Are there any precursors to your concern?

Hi Jeremy, no, no precursors, just mildly paranoid and ignorant. I saw on wikipedia that the flash point of kerosene (paraffin) was 38–72°C. However, I suspect that "paraffin" may be one of those names that refers to various substances of which kerosene is only one. I do not know enough chemistry to be sure, so I thought I'd ask the experts!

Well I doubt anything in the thermometer will explode, but you'd have to know exactly what's inside the thermometer to make sure.

Ours exploded in the recent heat wave in Melbourne (jan '14). It was inside the house which may have reached somewhere in the 30s but we were away so I can't be exact with the temperature.

Do not confuse the flash point with the autoignition temperature.
I don't know what atmosphere you have inside the thermometer, but either way I think it's hardly the case that you reach the autoignition temperature of paraffin (gasoline undergoes autoignition at 246-280 °C).

Thanks for your response, tschoppi. According to your first link, the flash point for Kerosene (paraffin) is >38–72 °C (100–162 °F), while the autoignition temperature is 220 °C (428 °F). I agree it would be highly unlikely that the paraffin in the thermometer would vaporise to form an ignitable mixture in air and then come into contact with an ignition source in my car! However, it can get awfully hot on the road here in Oz and I worried that a spark could be all it takes. Nonetheless, it became a moot point as we decided to keep the thermometer for ourselves :-)

Don't forget that the spark has to be in the thermometer, which makes this scenario all the more unlikely ;-)

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Which d-orbitals split according to CFT?
According to CFT, the degenerate d-orbitals split into $eg$ and $t_2 g$ groups. Are these the outermost nd orbitals which are vacant(4-d in case of 4th period) or the penultimate filled (n-1)d orbitals?
If they are the (n-1)d orbitals, how can they be degenerate if they are not completely filled? If they are (vacant) nd orbitals, how do they split into $eg$ and $t_2 g$ groups if they have no electrons to be repelled by the incoming ligands (CFT is based purely on electrostatic forces)?

All d orbitals. E.g. Ti$^{3+}$(aq) has only one occupied d orbital, but still splits. CFT is an approximation of Hamiltonian. It is similar with the degeneracy of hydrogen atom.  http://zh.scribd.com/doc/188210540/Crystal-Field-Theory-Tight-Binding-Method-And-Jahn-Teller-Effect Eq. (18) around.

The $d$ orbitals that split are the $(n-1)$ $d$-orbitals because they are the relevant ones that do chemistry (because they have the electrons).
If the orbitals are degenerate, it doesn't matter whether they're filled or not, the energy level is the same for all the degenerate states. So it also doesn't matter if they're filled or not, the energy states will split up according to CFT.
The compound only "feels" this raising or lowering of energy depending on how many electrons are filled in into the different orbitals after splitting.
To maybe clarify some confusion: The existence of an orbital doesn't mean that the electron has to be in there, it is only a mathematical construct.

If (n-1)d orbitals have electrons, how can the ligands fill it up with its own electrons? According to VBT, vacant d-orbitals(hybrid?) are needed. Does CFT use hybrid orbitals like VBT?

Using MO-LCAO you see that you mix the ligand group orbitals (LGO's) with the $d$ orbitals with the same symmetry. This gives you twice as many orbitals, where the ligand electrons go into the strongly stabilized ones and the metal electrons go into the destabilized ones (which in turn is of the form that we know from CFT).

In CFT, the ligands are approximated as external fields, without no electron in its description.


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What is going on in this reaction?
I was trying to remove the electroplating from a aluminum rod using inverse electrolysis and everything went wrong. I'm definitely not a chemistry expert, so I'd like to know what happened here.
I used hydrochloric acid (30%) as the electrolyte, the aluminum rod as the cathode and a common copper wire as the anode. I ran it through a 12V 1A power supply, in a small glass jar.
When I turned the power on a lot of bubbles started to appear at the surface of the rod, as expected. After aproximately 30 seconds the bubbling was too strong, so I decided to turn off the power supply, however, the bubbling didn't stop. As I know that electrolysis of hydrochloric acid releases chlorine, I decided to put the jar to outside, with both electrodes still inside it. After aproximately five minutes, to my surprise, the jar was bubbling extremely strong, spilling over and the electrolyte had turned black, and quite frankly it was kind of scary.
I removed the electrodes and the reaction ceased, but I still have no clue of what happened.

You used 30% hydrochloric acid as the entirety of the electrolytic bath, or did you just add a few drops of the concentrated acid to an amount of water? If it's the former, then it's pretty clear what happened.

As I recall, $\ce{Cu}$ is inert to $\ce{HCl}$, however $\ce{Al}$ will react with it to produce $\ce{H2}$ and $\ce{AlCl3}$, resulting in a turbid, dark gray solution that bubbles vigorously as the $\ce{H2}$ gas is evolved. The reaction may not have been instantaneous, especially if the $\ce{Al}$ had a passivation layer of oxides on it.
Once initiated, that reaction would've continued even after removal of the external current, and is probably the source of the black precipitate you describe.

Aluminium has absolutely no protection against hydrochloric acid. Its only limited by hydrogen formation. Its slower at lower temperature- the electrolysis heated the water and increased the potential to speed the reaction.


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Is pure water very corrosive?
On Outdoors StackExchange on a post about the safety of distilled water, someone claimed that pure water is very corrosive.

Water distilled for some laboratory uses (like the Z-pinch reactor) is distilled to 0.00000009% purity, and is caustic enough to burn through most things.
Water, purified enough, cannot be distributed through stainless steel pipes. Instead, they use glass pipes to transport it within the plants where it's made.

For sure, "caustic" is not the correct word, since water is neutral.
But, is it that corrosive? Or is it even less than water with salt?

That is entirely false. Water is certainly not caustic; that term is mostly reserved for strongly basic substances that will readily attack organic matter (e.g., human flesh). I've been drinking water (including, gasp, distilled water) for my entire life without my internal organs dissolving. As for its corrosiveness, while water will certainly react with various poor/common metals, it will not react easily with stainless steel, which has a passivation layer of chromium oxide that makes it highly resistant to corrosion. Furthermore, water of higher salinity is more corrosive, not less.

It's been claimed (in a comment on the linked site) that the ultrapure water is too corrosive for stainless steel pipes and has to be transported in glass tubing instead. It seems to me that the more obvious explanation is that at ultrapure levels, they don't want the water picking up contaminants from metal piping.

Here is another variant of the "highly corrosive" rumour: http://www.symmetrymagazine.org/article/july-2015/underground-plans :  In this article ultrapure water dissolves a hammer.

Water is extremely corrosive to potassium...

Once ultra-pure water starts reacting to yield a solute, it is not ultra-pure anymore, so it can't be that bad. If you try to dissolve a highly insoluble chloride, and your water contains traces of chloride, the reaction will not go forward. However, if you have a continuous flow of ultra-pure water (or just water not containing any chloride), traces of the insoluble chloride will dissolve. Not that that would be dangerous.

That quote had false statements all over it.
Type I ultra pure water (Milli-Q water, or others) are fairly common among all labs.  They are not corrosive to stainless steel.
As @chipbuster said, purified water is never distributed through stainless steel partly because concern of contamination.  Metal reacts not only with oxygen, but also many pollutants in the air.  Overtime on metal surface will build up a layer that has all kind of compounds in them.  When water flows through, ions get picked up.
However, the most major reason would be that glass is better in all respect.  The reason glasswares are commonly used in the lab have little to do with pure water.   Glasswares are relatively stable across wide range of pH and very hard to be oxidized or reduced.   Plus, glasswares are cheap, extremely easy to manufacture, and transparent, so you can see through.  You will really only need metal if extreme pressure or temperature is required.
As a matter of fact, it is also unwise to use glassware to distribute ultrapure water--you will get ion contaminations from glass as well.  Purified water are actually almost always stored in plastic containers and distributed through plastic tubes.

Water is corrosive at very high temperatures and pressures (in the sense that it will react with a lot of common materials), but generally speaking, pure water is less corrosive than water with impurities. Some power plants use purified water in their cooling systems for this reason.

If we define corrosivity as impurification of the water, ultrapure water eats nearly everything, including Pyrex. Plumbing is usually polyvinylidene fluoride (PVDF, Kynar), polyperfluoroalkoxy (PFA), and ethylene chlorotrifluoroethylene (ECTFE, Halar).
Stuff still grows in ultrapure water, slowly. Chip fabs' water systems must be hard-UV irradiated, ozonized, or otherwise disinfected plus ultrafiltered against bio-particulate generation.
The product tastes awful - bitter and flat.

Looking at corrosion from the point of view of the solvent is misleading here. The amounts of dissolved substances are very very small and insignificant with respect to the piping or the thing being corroded. So your answer is just wrong.

The water is ruined - and the devices it washes - because "very small" is NOT "insignificant."  The SR-71's original titanium alloy was stress corroded by chloride.  Read Ben Rich's book "Skunk Works" on what a marking pen did to parts, and summertime chlorination of tap water.

The quote in the original post has nothing to do with reality. Ultrapure water will not dissolve metals like acids. Ultrapure water can be identified by its resistance of 18.2 Megaohms and it is free from organics impurities.
Now the punch line is that ultrapure water is "contaminated" very quickly. This is all at what we analytical chemists call "ultratrace level" contaminants. Merck did a study (not published, I saw that in a conference) that it just takes 30 minutes in a lab for the resistance to decrease 10-12 Megaohms. The most common contaminant is surprisingly NH4(+) ion in a typical laboratory atmosphere along with chloride from HCl traces. CO2 lowers its pH to around ~ 6. Most likely UPW will have a very unpleasant or bland taste.
Yes supercritical water is extremely corrosive. It dissolves silica, attacks most metals, and so on.

Remember your equation for free energy
There is a germ of truth to this claim.  The thermodynamics should be familiar, and is already covered in another Chemistry.SE post, "Reaction quotient and Gibbs free energy at the start of a reaction."  The gist is that dissolving something in theoretically (hypothetically) pure water has a fabulous entropy advantage.  If it's in a container with a mole of surface ions exposed, you have Avogadro's number of ions trying to get into it from the walls of its container, and then that one ion has to go back for the reverse reaction to occur.  Equilibrium will favor the dissolution of substances even when the standard free energy is highly positive (endergonic).
The catch, of course, is that infinitely pure water doesn't stay perfectly pure for long!  The glass pipes are to prevent the pipes from ruining the water, not the water ruining the pipes because it only takes a minuscule amount of solute to reduce the free energy difference substantially.  Additionally, the reaction has to occur at a measurable rate by kinetics, so any compound not normally at risk of dissolving in water at all (such as well crosslinked polymers) won't dissolve no matter what the delta G is.
In the biological context this started with, we can be sure that the water is absorbing some substances from the body.  For example, a glucose transporter uses free energy from ATP hydrolysis to establish a very strong concentration gradient across the cell membrane - nonetheless, some net molecules of glucose are sure to move out of the cell when absolutely none are available to enter.  But this is true whenever you drink water completely lacking in glucose, regardless of whether it contains electrolytes or not.

This "question" was just trashed out several days ago, I am flabbergasted that it even got one upvote let alone 19.
Pure Water has no dissolved ions except equal amounts of H3O+ and OH-from self ionization, no dissolved gases, no organics, no VOCs, and no bacteria or viruses. Can this be attained? No! Can it be approached? Yes! Talk to Millipore, Barnstead, Culligan, and other companies in the super-pure water business for details and specs.
What can [pure]Water do: It can dissolve anything that is soluble in water, Ionic materials, gases, polar organics, and even nonpolar organics, to their limits of solubility. How is this controlled? by limiting contact with anything and constant recycling thru the purification system. Water will dissolve ions from any metal, ceramic, borosilicate glass [I know I worked for Bausch and Lomb Soflens; isotonic saline in borosilicate vials ruined lenses by leaching ions from vials], organics from any plastic, gases from any atmosphere, even ions from teeth. The only substance that I think might resist Water is fused silica. This is what water does, the universal solvent. To prevent this the solution is to limit contact or the common ion effect [the latter means the water is no longer pure].
Corrosiveness of water depends on an anode and anode reaction, a cathode and cathode reaction, an electrical connection between the anode and cathode, and ion transport between the anode and cathode. Remove any one of these and corrosion stops whether it is galvanic or electrolytic. Removing ions from water reduces the ion transport to the limit of the 18.2 Mohms so Water will never be completely noncorrosive or nonconductive, but the purer it is the less corrosive. Also, water reduces the possible anode and cathode reactions.
The idea that Water causes greater single electrode potentials seems specious. The charge separation on the metal will be inhibited by just that and to say that it is lowered by ions in solution is true but circular to the discussion because if they come from the metal the concentrations will be lower. Charge separation in a conductor is unlikely to be positive in entropy change.
Water is not biologically dangerous unless it contains contaminants, poison, etc. Drinking too much water will cause problems and relying on drinking to supply necessary minerals is doubtful. Reverse osmosis water and distilled water, to my taste at least, have a crisp, clean taste. In general, the cleaner water is the better to drink. The only caveat here in RO or DI water is that fluoride ion is removed or at least lowered and that is usually added and not natural. Fluoride in water has made a huge reduction in dental caries. [Again personal experience I [BF] have cavities in almost all my teeth, My children [AF]have only 2 or 3 between them.
Bacteria and virus contamination is nonexistent in good reverse osmosis water and is removed by ultrafiltration and UV sterilization in ultrapure Water systems. Pure Water will not support bacteria unless enough is added to feed those not killed and nutrients are added.
Water will leach monomers, oligomers, plasticizers, and initiators from plastics. Since it is a small molecule Water will diffuse thru some plastics. Pure water is probably better at these feats than less pure water. It is better at osmosis.

As stated by others, the original post if full of falsehoods.  There are also falsehoods in the replies.  I have a lot of experience with ultrapure water, my company makes very large ultrapure and bench top labwater systems.
1) UPW (ultrapure water) is not caustic.  Conversely, in it's purest form it because acidic when exposed to atmosphere because it readily absorbs CO2, which converts to carbonic acid.  However, by default, UPW can only have a pH of 7 (again when not exposed to air).  So proper pH measurement can only be done inline or with proper buffers. We actually wrote a paper on this topic because researchers were always complaining that their water was acidic.
2)  UPW will corrode iron, brass, poor quality steel.  However, it will not corrode stainless steel as long as the welds are proper, even then hard to make SS conducive to corrosion by UPW.  Example, resistivity meters that we use are made from SS (need to be to conduct electricity across 2 metal probes to get a reading).
3) UPW will not corrode Pyrex as stated above.  After all, HPLC and LCMS grade water (not going to define so look it up) is placed in glass containers. Distilled water is even made in glass stills for laboratory use.  Having said this, it will leach some silica and boron, so they actually should use special glass.  But it won't corrode it.  I had a bottle of LCMS grade water sitting in the lab for years.  Also, exotic plastics don't need to be used, such as PFA, PVDF, Halar, etc. The possible exception is in a fab of a semiconductor plant.  We use polypropylene, PEX (cross-linked polyethylene), and even PVC (but not ideal because in some cases the glue can leach TOC's - total organic/oxidizeable carbon that are undesireable, for example in a waste water treatment plant.).
4) the comment that "pure water is less corrosive than water with impurities", is not completely true.  Put a brass or iron bolt in distilled water and see what happens compared to tap water.  Also, power plants don't use UPW because tap water is more corrosive.  All steam powered plants (coal, gas, nuclear, etc.) use UPW, because residuals will form on the turbine blades and throw them off balance and destroy them.  Also, impurities from the water will leave residue on the evaporators in the cooling towers and render them useless.  One can't compare the corrosiveness of tap water and UPW.  I've seen very corrosive tap water (classic example is the city water issue in Flint, MI.  The water is corroding the pipes and leach lead into the water because it contains chlorates).  UPW can be corrosive to other metals because it forms carbonic acid when exposed to air. etc......
By definition, ultrapure water is defined as ~17.0-18Meg-ohm or type I water.  Meaning does not contain dissolved ions to conduct electricity.  In theory, if you throw your TV into a tub full of UPW it won't short out.  If swimming in a lake of UPW and lightening hits the water you won't be electrocuted, because it contains no ions to conduct electricity.  The theoretical limit for UPW is ~18.26 Meg-ohm.
DO NOT DRINK UPW, it will corrode metal fillings, leach calcium from your bones, eat the enamel in your teeth, even can cause your cells to explode (due to osmosis).  Fortunately, before you can do to much damage you likely will feel sick to the stomach.  The irony is that the purer the water is the worst it is for you to drink.  Though, you can denature UPW by mixing it with Tang, powdered drink mix, tea, coffee, etc.  It will make very strong tea and coffee.  Even though distilled water is not considered UPW (by definition, it is a low grade Type II), it isn't good for you to drink because it has no minerals.  Also tastes insipid.   While on the subject, home RO or 'purified water' you buy in the store, is considered as ASTM Type IV, but in our industry we call it Type III.
Yes, UPW is one of the strongest solvents known to mankind.  But that doesn't mean it will eat through everything.  It just means it's always trying to reach equilibrium by leaching out ions, VOC's and CO2.

Through what mechanism could water that you drink "leach calcium from your bones"?  Further, do you have sources for the claims that it will dissolve fillings, eat the enamel of teeth, or cause cell lysis in a human body?
People drink RO and various qualities of distilled water all the time with no ill effects - only the slight loss of mineral content that tap/well water contains.  A mouthful of 100% pure water at 0ppm TDS only has to mix with about one milligram of saliva in order to become 1 ppm level (RO - comparable) water that huge numbers of people have in their kitchens.

This was a great answer until the relatively dubious claims that highly pure levels of water are somehow dangerous to the human body. I'm shocked that somebody seemingly well versed in the subject of water purity would consider that some easily consumable amount of water would have any significant affect on an organism containing perhaps 100 times that amount of water. Also, suggesting severe damage to an organism by fact that the water is missing trace amounts of minerals requires proof that the organism is solely dependent on water as their source to begin with. These things you cite are in

@JimmyWilliam This was really a comment, not an answer. Please use the answer field only for actual answers. With a bit more rep, you will be able to post comments. For the moment I've converted your post to a comment.

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Electrochemical cell-SRP value
On what basis can the SRP value of different elements in the periodic table be explained?
For example why does Cu have a greater tendency to accept an electron and Zn have a tendency to lose electrons in a Daniel cell?

There are many factors and the ideal way to understand this would be with the quantum mechanical description of atoms, which gets too complicated for answering such questions. But in general, following are few simple-to-understand face that helps in understanding the SRP values-

Electronegativity and Electron gain enthalpy: Higher the electronegativity, greater is the tendency to accept electrons, be reduced and hence greater (more positive) is the Reduction potential. For example, reduction potential of fluorine is greater than chlorine and that is greater than bromine. As we move down the halogen group, electronegativity decreases.
Ionization potential: This is a consideration for metals, which usually loose electrons and hence electronegativity is not the right quantity to consider. The lesser the IP, easier is the oxidation, and more negative (lower) is the SRP. In gas phase, the SRP of cesium is the most negative (in the group) and it becomes less negative as we move upwards.
Hydration: This is an important fact to consider since most electrochemical reactions occur in aqueous phase and hence the commonly cited SRP values corresponds to the aqueous phase. Greater the hydration of the ion formed after oxidation/reduction, greater is the tendency to lose/gain electrons and hence lower/higher is the SRP values. Lithium ion has very high hydration enthalpy and hence lithium has a strong tendency to lose an electron to form an ion which can be excellently solvated. This lends it one of the most negative SRP in the periodic table. The same holds true for fluorine, which is excellently solvated in the ionic form by water and hence has a high SRP.
Size, EC etc: Several factors like ionic size, it's electronic configuration, the inner orbital structure and the nuclear force of attraction determines how the incoming electron will be received and consequently the SRP values. Higher atomic size, higher inner electron repulsion, almost-stable configuration(just 1 electron more) and lower nuclear force, all favour lower SRP values.

All of these consideration, in varying proportions can be applied to the transition elements, but they act differently and to different extent in different compounds. Hence, predicting the SRP order without experimental data is very difficult and not trustworthy.


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What is chemical free energy?
What is chemical free energy?  How can we practically explain its existence?  How can it be used to do some useful work like in an electrochemical cell?
I couldn't understand clearly about it when I referred a few books

What is your understanding of it thus far?  Putting what you've learned into the question will help someone answering to fill in the gaps.

Up to my knowledge, chemical free energy can be referred to Gibbs free energy. It can be termed as the energy available to do useful work. The existence of Gibbs free energy can be known by knowing whether the process is spontaneous or not. If the process is spontaneous, we can say that, there is chemical free energy, if process is non spontaneous, we can say that, there is no free energy. In electrochemical cell, it can be used to do useful work by converting it into electrical energy.

Chemical Free Energy or Gibbs Free Energy is best understood as potential energy for reactions.
If I have a lot of Gibbs Free Energy, then I have a lot of potential energy.
Kind of like if I were on top of a hill, it's very easy for me to roll down that hill.
Often what's talked about in books is dG or the change in Gibbs Free Energy. If we think of it is simply being on top of a hill (High Gibbs Free Energy/Potential Energy) G(initial)
 going to the bottom of the hill (Low Gibbs Free Energy/Potential Energy) G(final)
Then the change in Gibbs Free Energy will be negative for something spontaneous (like the rolling down a hill example).
Change is always calculated as: change = final - initial
Gibbs Free Energy is used in the same was as potential energy. But instead of translating to kinetic energy, it translates to Enthalpy (heat) and/or Electrochemical Energy (being able to move electrons). Both Enthalpy and Electrochemical Energy can then be converted to mechanical work.
The way to think about it is not so much that you can use Gibbs Free Energy, but rather it's just the innate property of whatever situation you're working with. Molecules have a tendency to be reduced or oxidized under different conditions, the Gibbs Free Energy just tells you which way is spontaneous.
dG = dH - TdS
This gets kind of complicated, since dG is affected by many factors, but it gives you an idea, that the Change in Gibbs Free Energy can become positive or negative under different conditions.
Keep in mind that Gibbs Free Energy isn't a real thing, but rather a concept that helps explain why things happen.
 Why do gases expand?
 I don't know, but the universe has a tendency to do this. So we say that it is spontaneous, just like how we would say things roll down a hill because of gravity.
 What's gravity?
 It's our way of explaining why things roll down hills. :p
I hope that helps a little bit.


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At what amount is digesting soap unhealthy?
Several of my friends, when they do the dishes manually, do not rinse the dishes after cleaning but merely dip them in the dishwater and then dry them (using a towel or a dish rack). I myself on the other hand always rinse the dishes with fresh (hot) water from the tap before drying them, mostly because the idea of having a dose of soap with my next meal freaks me out.
But is there some guideline as to how much of the soap you can ingest until something goes wrong, be it the taste of the next meal or some poisonous reaction? The soap used for doing the dishes manually doesn't have any warning labels on it... I've found this describing poisonous effects of automatic dishwater soap, and this pertaining to a dish soap that is used for manual dishwashing.

I think you're missing a link in the deductive chain: you've gone from rinsing and drying versus drying, to amount of chemicals ingested; you've just assumed that drying without rinsing has as a consequence the ingestion of more detergent. (and there's a third and fourth way: rising, then air-drying; and just air-drying)

While true, I think that the basic assumption "less rinsing: more soap" still holds. Even if you don't rinse and towel-dry, the towel will accumulate soap that may be deposited on the next item.

@tschoppi I tried that in the shower this morning.  My towel needs washed and my skin wasn't smooth.

I'm less familiar with safety regulations elsewhere in the world, but in the US, "chemical manufacturers" (which includes soap, sugar, salt, and just about everything else you can imagine) have to produce Materials Safety Data Sheets (MSDS) and make them publicly available.
Out of curiosity, I did some digging for soaps. Proctor and Gamble is a huge consumer producer, including dish detergent. So there are MSDS sheets for soap.
The MSDS gives oral LD50 values. That is, the lethal dose required to kill half the population.
The LD50 for soaps seem to vary by animal, but judging from a few MSDS I pulled up, they're ~2 g/kg or so. (That value was for rats.)
So if I had to guess, given, say a ~70 kg adult, we'd be talking about 140 g of soap, minimum to kill a person.
Now.. that's acute doses. I have no ideal if anyone has established long-term toxicity of soaps, and I can't find anything obvious online.

Incidentally, I'm with you. I rinse with clean water before drying -- if nothing else, it might affect taste.

To add to this, Primo Levi (The Periodic Table) relates that while he was a prisoner in Auschwitz he prepared fatty acids from paraffin (he was a slave labourer at the Bunawerke) and ate the product.  He claims these extra calories saved his life.

@AbelFriedman Yes, I'd think the long-term toxicity effects are pretty unlikely since "normal" soaps are based off fatty acids and are thus edible. Other ingredients might be less clear (e.g., artificial colors).

Did you find anything pertaining to taste? I.e. is it really necessary to rinse or are the amounts likely to be left behind negligible to taste? I see some potential for saving water here...

@tschoppi Actually, the biggest way to save water is to use an efficient dishwashing machine. I think the amounts of soap are likely to be negligible - consider it's a dilute wash solution and only a film will be left on the plate (if at all). I'd guess in the milligrams or micrograms of soap per plate. My preference for rinsing though is to make sure it goes through very hot water.

But I've also switched to a small countertop dishwasher, which saves water and time. One load uses only ~10L of water, including rinsing.

Well, I have to say that I don't have the funds necessary to finance a dishwashing machine. I am however aware of the high water efficiency that these machines exhibit nowadays.

One more thing to consider in addition to this fine answer is that the MSDS LD50 is for the pure active ingredient. Most products intended for contact with human skin are somewhere south of 20% SLS/SLES, so multiply the LD50 by about 5 to determine the raw volume of product required to be acutely dangerous.

As far as long-term toxicity, the components of SLS/SLES are animal fats and various relatively common ions that your body deals with every day. The chronic long-term effects would lie largely in the denatured alcohol (translation; added methanol) common in most liquid detergent products, which over time will damage the nerves.

@KeithS Agreed. Except in this case by the OP, the methanol is not likely to remain on the plate. I believe the poster wanted to know about soap residue remaining. After drying, the amount of methanol should be negligible.

I am new here, however I have an Opinion.
Soap from the dish water would not be the only concern for me. Food particles, oils,bacteria, etc., would be what would come to my mind.
So, "for me" to leave soap ( or anything else on the dishes) negates the act of washing them in the first place.
Just my opinion but I believe a thorough analysis of the rinsed and air dried dish as opposed to the un-rinsed and Toweled dish would show a significant difference in how "Clean" they were.
In terms of Ingesting soap ( dish soap in this instance )being a health hazard I personally would consider ingesting ANY soap to be a health risk.
I would think ( without being a Doctor) that such ingestion could Quite easily lead to Diarrhea, Irritable Bowels, the killing of essential enzymes and probiotics in the digestive tract just to mention a few things off the top of my head.
There are probably many others but this Is what I thought of first.
http://www.netwellness.org/question.cfm/19193.htm
http://www.merckmanuals.com/pethealth/special_subjects/poisoning/household_hazards.html
Detergents, Soaps, and Shampoos
Exposures to products containing anionic and nonionic detergents generally cause mild gastrointestinal irritation that responds well to symptomatic care. These products include human and pet shampoos, liquid hand dishwashing soaps, bar bath soaps (except homemade soaps, which may contain lye), many laundry detergents, and many household all-purpose cleaners. Some of these, such as electric dishwasher detergents, are also alkaline corrosives (see Poisoning: Corrosives). All animals are susceptible.
These agents are not well absorbed by the body, and toxicity is limited to irritation of the eye, mouth, or gastrointestinal tract, which is usually mild and resolves on its own. Vomiting and diarrhea are the most common signs. Dehydration and metabolic abnormalities can develop in rare cases after prolonged vomiting or diarrhea.
Dilution with milk or water may reduce the risk of spontaneous vomiting. Vomiting usually resolves on its own after short periods of food and water restriction. In severe cases or in animals with sensitive stomachs, medications that prevent vomiting may be required. Rarely, fluid treatment is needed. Eyes should be flushed with water or saline for 5 minutes.
www-ehow-com/facts_5802351_harmful-chemicals-dish-soap-html

The devil's in the dose. Those unpleasant symptoms seem to be referring to actually drinking the stuff. If this was the result every time someone left a few milligrammes of detergents on dishes, the OP's friends probably would have stopped doing it long ago.


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Determination of products in reactions involving carbocation rearrangement?
How do you determine the "migratory aptitude" of various groups during carbocation rearrangements? Is there a experimentally determined order?
For example, what will be the product in case of the following pinacol-pinacolone type rearrangement:

Will the methyl group be transferred or the cyclopentyl moiety? The same question arises if the cyclopentyl is replaced with phenyl group? In general, which shifts, methyl, hydrogen, phenyl or any other group, will take place? Or equivalently, what determines which group has a greater tendency to undergo shifts during carbocation rearrangements?

All else equal, the smaller group migrates preferientially. The energy required to more the group is related to the mass of the group:
$$W=\int_0^\vec{x}\vec{F}\cdot d\vec{x}$$
$$\frac{d\vec{x}}{dt}=\vec{v}   \ \ \ \ \ \therefore \ \ \ \ \ d\vec{x}=\vec{v}dt$$
$$W=\int_0^t\left(\vec{F}\cdot \vec{v}\right)dt
$$$$\vec{F}=m\vec{a}$$
$$W=m\int_o^t \left(\vec{a}\cdot\vec{v}\right)dt=m\int\left(\vec{v}\cdot\vec{a}dt\right)$$
$$\vec{a}=\frac{d\vec{v}}{dt}$$
$$\vec{a}dt=d\vec{v}$$
$$W=m\int_0^\vec{v}\vec{v}\cdot d\vec{v}=\frac{1}{2}m\vec{v}^2\vert_0^\vec{v}=\frac{1}{2}m\vec{v}^2$$

The group best able to sustain a positive charge migrates according to the following order: $$\ce{H > Me3C > Me2HC > Et > Me}$$

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


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How to liquefy or extract water from superabsorbent polymer?
I am looking for a good way to liquefy or extract water from superabsorbent polymer. From what I remember it was salt that can liquefy it. Since salt when dissolved will have sodium cation and therefore naturalize the polymer chain making it contract. Is that true?

Yes.Add some salt on the super absorbent polymer gel and salt can break super absorbent polymer's absorption ability.
For more details, google "soco polymer" or visit socochem.com

This may be true, but I suggest another type of explanation. If a super absorbent polymer absorbs its maximum in distilled water it is called a fully swollen hydrogel. There are two ways of decreasing the amount of water stored without using heat.
The first one is to increase the salt load of additional water and place the hydrogel within this salted water. The salt in the water will decrease the total water absorbed by the hydrogel and the hydrogel will shrink, giving up some of its water.
A more complete and full way to make the hydrogel to give up its water is the addition of an acid.  Lowering the pH of the hydrogel to below 3.5 will convert all the sodium carboxylates to the free acid form which does not absorb water.  These hydrogels are pH-sensitive, and if you plot pH vs absorbed fluid, you would find a bell curve that starts around 4.5 or 5.0, tops out around 7.5 and decreases as you increase the pH until around 10.5 to 11.0.

Now I understand, so adding acid is the best way. Where do you get the information from, I want to read more about hydro gel but I can't find much information about it. Thank you.


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What is the priority for numbering alkyl side chains?

I'm wondering why the above image is named 5-ethyl-3,3-dimethylheptane rather than 3-ethyl-5,5-dimethylheptane. I thought an ethyl substituent has a higher priority (I know this is true when arranging the final name, but what about numbering the parent chain)?

The process for naming this molecule correctly follows the steps listed here:

Identify the parent hydrocarbon, which will be the one with the longest chain. It is correctly identified as the 7-carbon chain, and as such is a derivative of heptane.
Identify the side-chain functional groups. We have 2 methyl groups and an ethyl group.  When listing them, they will be written alphabetically.  So, we have:
X-ethyl-Y,Y-dimethylheptane and not Y,Y-dimethyl-X-ethylheptane
Numbering (which is what you are asking about) is determined by numbering the carbon atoms consecutively from the end of the chain nearer the functional group of the highest priority. In terms of molecule numbering, alkane groups have the same functional group priority. We would only need to consider relative priorities of the ethyl and methyl groups if there was only one methyl substituent, in which case you would have 3-ethyl-5-methylheptane.
In our case, the process is to count in from the ends of the primary chain. Here, the first substituents occur at an equal distance from each end of the chain, both the ethyl and the methyl occur at position 3. We now need to invoke one of the IUPAC sub-rules for nomenclature. If the first substituents occur at an equal distance from each end of the chain, we need to number from the end nearer a second substituent. The second substituent is the extra methyl group, and so this is the end that gains numbering priority.

There are many nice examples of nomenclature on the internet. One very good example, which highlights your question is this page. Have a look at Rule 2 and sub-rule 2

"In terms of molecule numbering, alkane groups have the same functional group priority." sounds like it could be misleading since there is actually no priority order based on the kind of functional group for the numbering of locants for substituents that are expressed as prefixes at all.

That molecule would be 5-Ethyl-3,3-dimethylheptane based on the first point of difference rule. Starting from the left side of the parent chain, you will have a substituent at the 3,3,5 positions while on the right, you'll have your substituent on the 3,5,5 positions. 3,3,5 is lower than 3,5,5 so it should be the priority in naming the compound.
cite: http://www.chemrtp.com/chemical-info.ce?ID=FLLHBJKGZIOYAA-UHFFFAOYSA-N
It is also important to take note that the two substituent (ethyl and methyl) have the same priority when considering functional groups. When alkanes are in equivalent positions, the first point rule overtakes. However, if a double bond is present in the molecule, the double bond gets the priority. This will hold unless the double bond is present in the parent chain and not as a substituent.


text:
Does aldol condensation give different products in different conditions?
I know the mechanism of aldol condensation and general method to determine products in crossed aldol reactions, or reactions between dissimilar molecules. But the following statement in my class notes baffles me:

In presence of a base (alkaline medium) the $\alpha$ hydrogen of the lower aldehyde is more acidic and hence is removed in the first step, while in presence of an acid, the $\alpha$ hydrogen of the higher aldehyde is more acidic and hence migrates.

According to this, on reaction of propanal with ethanal, basic medium yields 3-hydroxy pentanal while acidic medium yields 3-hydroxy 2-methyl butanal.
Why is this statement true (if it is)? And can aldol condensation proceed in an acidic medium, since the presence of the base is necessary for the removal of the slightly acidic $\alpha$-hydrogen to start the reaction?

I think the issue is much more complicated. For one thing, the particular reaction conditions are relevant (i.e., thermodynamic vs. kinetic control). I'm dubious as to whether the premise that, under basic conditions, the lower aldehyde will form the enolate is accurate, unless conditions are selected for kinetic control (i.e., use of low temperatures and strong, sterically hindered bases). The enolate with the more substituted double-bond is more thermodynamically stable due to hyperconjugation. On the other hand, the smaller enolate may be better stabilized by solvation.

I believe your conception of the need for deprotonation under acidic conditions is misconceived.  Whenever you are under acidic conditions, your mechanisms must necessarily start with protonation.
The key concept here is the enol tautomerization.  The enol is more favored than for propanal as compared to ethanal because it results in more substituted alkene.  Thus there is likely a higher concentration of this nucleophile in the reaction mixture.  The mechanism of tautomerization is difficult to elucidate, but presumably starts with protonation of the carbonyl in acidic conditions.  Then, likely another carbonyl deprotonates the the recently protonated carbonyl at its alpha position, forming the enol.  This transformation has produced a nucleophile (enol -- 1-propene-1-ol ) and also a strong electrophile (a protonated carbonyl -- protonated ethanal).  The reaction occurs, leading to a protonated aldol product (proton on the carbonyl) which then is deprotonated by the next carbonyl and the process continues.
As far as the base catalyzed case, ethanal is more acidic because it does not have the extra methyl group donating electron density disfavoring formation of a negative charge.  Hence ethanal becomes the nucleophile and propanal becomes the electrophile.
In each case, these are only major products.  I don't expect the energy differences are large enough to give great selectivity.


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Hybridisation of Mn in potassium permanganate
I'm clear with the concepts of crystal field theory. But I can't figure out the exact reason why the hybridisation of manganese in potassium permanganate ($\ce{KMnO4}$) is $\mathrm{d^3s}$. Can anyone explain please?

This is formally a manganese(VII) compound and hence there are no 3d electrons.
The four $\ce{O^2-}$ ions are considered to be donating two electrons each to the atomic orbitals. Tetrahedral "hybridisation" can be achieved by using the $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$, and $\mathrm{3d}_{xz}$ AOs. Since in Mn(VII) the $\mathrm{3d}$ AOs are lower in energy than the $\mathrm{4s}$ and $\mathrm{4p}$ AOs, the "hybridisation" is best considered as $\mathrm{d^3s}$.
I have put "hybridisation" in quotes because it is not used in advanced chemistry (it is an artifact!) but it is correct in saying that the bonding MOs contain mostly $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$ character but they'll also probably mix some $\mathrm{4p}$ character in there for good measure!

If anyone, like me, had trouble with the "3d AOs are lower in energy" part,  here are some places to read up on it. The second link's from the page linked to by the first, but reading it should be enough, so I added it separately.

I feel a bit like a broken record: Do not use hybridisation to describe coordination compounds; it is not helpful.
However, the case of permanganate stands out as something where hybridisation is even more unhelpful than in the traditional ‘do not use it’ cases. Refer to the qualitative molecular orbital scheme of permanganate in figure 1.

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.
This complex ion cannot be understood well without π contribution of the four oxido ligands. The p-type orbitals of the oxygens transform, as the image shows, as $\mathrm{a_1 + e + t_1 + 2 t_2}$ which means that all of the orbitals of manganese that are not core orbitals ($\mathrm{3d, 4s, 4p}$) will participate in bonding to some extent. Especially the $\mathrm{t_2}$ orbitals are frustrating since oxygen has two energetically degenerate sets of them while manganaese offers another two sets (however non-degenerate; consisting of $\mathrm{3d}$ and $\mathrm{4p}$ orbitals) of them.
On the other hand, however, the analysis of populated orbitals will tell us that all occupied orbitals are ligand-centred orbitals. This again makes it a lot weirder to attempt to define any specific hybridisation for manganese: after all, it is in its $\mathrm{+VII}$, valence-electron less oxidation state.
If you really had to determine a hybridisation, anything short of $\mathrm{d^5sp^3}$ is not in line with the actual picture. That in turn cannot explain the tetrahedral geometry. You lost, go back.
Tl;dr: do not attempt to explain coordination complexes with hybridisation. Ever!

Everything here is true, but I disagree somewhat with "it is not helpful." For students who have not yet learned about group theory and delocalized MOs, introducing $d^3s$ hybridization is a way to teach that 1) $d_{xy},d_{yz}$ and $d_{xz}$ can combine with $s$ in a tetrahedral $\sigma$ bond network just as well as $p_x,p_y,p_z$ can, and 2) empty $3d$ orbitals contribute more to $\sigma$ bonding with ligands than $4p$ in tetrahedral metal complexes. The fact that the filled "bonding" orbitals are largely ligand in character and that the $p$ orbitals can mix in is a second-level consideration.

@Andrew I want to argue quite the opposite: if the students haven’t even learnt about MO theory, what the heck are they attempting to learn about the bonding situation in these complexes. Don’t teach half-facts which are wrong or lead to false conclusions more often than not. Group theory, on the other hand, isn’t really needed and can be added later as a shortcut technology.

-It's typical to teach valence bond theory (ie bonds formed by electrons in localized hybrids of AO's) long before any discussion of delocalized MOs and actual symmetry-allowed orbital mixing. Hybridization is deemed useful primarily to explain how structurally equivalent bonds are electronically equivalent and to connect AO shapes to molecular geometries (eg linear = sp, Td = sp3 or sd3, Oh = sp3d2 or sd5, etc). Where you draw the line between pedagogically useful oversimplification and misleadingly false is subjective, but it might be a good discussion for this forum.

yes because there is d3s hybridisation in this complex
and in d3s hybridisation only dxy, dyz and dzx atomic orbitals can only participates in this hybridisation so only three AO's out of five are utilised in this
and according to CFT energy difference between t2g and next s orbital is less so hybridisation can be easily occurs

Please see the other answers as to why hybridisation should not be used with d block complexes -and arguably not at all


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How does Walter White make pure crystal meth using a non-stereospecific reaction?
In the highly-rated TV series, Breaking Bad, Walter White, a high school chemistry teacher recently diagnosed with cancer, takes to making the illicit drug, crystal meth (methamphetamine), by two main routes.
First, along with his RV-driving accomplice, Jessie Pinkman, Mr. White uses the common small-scale route starting with (1S,2S)-pseudoephedrine (the active ingredient in Sudafed®).  This method features the use of an optically active starting material to make an optically active end product, (S)-methamphetamine.  However, making (S)-methamphetamine on a large scale is limited because it is hard to get sufficient quantities of (1S,2S)-pseudoephedrine.
In the second route, Mr. White uses his knowledge of chemistry to move to an alternative synthesis starting with phenylacetone (also know as P2P or phenyl-2-propanone):

Racemic methamphetamine was obtained by the Winnebago-based chemists by reductive amination of P2P using methylamine and hydrogen over activated aluminum.
While his blue-colored product is considered by his customers to be exceptionally pure, Mr. White clearly knows about the issue of producing the correct enantiomer.  In fact, he raises this topic more than once in the series.
Since the show might not want to tell us the answer, I am wondering what what other possible methods Mr. White could have used to obtain an enantiomerically pure product?

The simpler answer is that TV/movie/book depictions of illegal actions are often simplified/sabotaged in an effort to discourage people from trying to take these as instructions. Anyone whose expertise overlaps any such area -- in my case, computers and locksmithing -- gets used to seeing deliberately wrong depictions and simply saying "that's OK, it's only a plot device anyway". In other words, if you're asking this question you're asking the wrong question. It's entertainment. Let it go.

While you are right, I think of this as an excuse to discuss some really interesting chemistry.

http://en.wikipedia.org/wiki/Chiral_resolution

WW hijacked the methylamine from the train replacing it with water to compensate for the density difference. How did they store methylamine (bp 21 degrees F) in Albuquerque no less?

@keshlam, One has to especially laugh (or cry) when they correctly show the use of the two tools (torque and pick) but do it on an antique warded door lock.

@user55119 That was an aqueous solution. I think that practically only this recurring "got so great purity" is weird in the show, when it's hardly difficult to get and not really important in illegal distribution.

Somewhat related

Intriguing question.
First, the best yield would be achieved by selectively producing one enantiomer instead of the other. In this case, White wants D-methamphetamine (powerful psychoactive drug), not L-methamphetamine (Vicks Vapor Inhaler). Reaction processes designed to do this are known as "asymmetric synthesis" reactions, because they favor production of one enantiomer over the other.
The pseudoephedrine method for methamphetamine employs one of the more common methods of asymmetric synthesis, called "chiral pool resolution". As you state, starting with an enantiomerically-pure sample of a chiral reagent (pseudoephedrine) as the starting point allows you to preserve the chirality of the finished product, provided the chiral point is not part of any "leaving group" during the reaction. However, again as you show, phenylacetone is achiral, and so the P2P process cannot take advantage of this method.
There are other methods of asymmetric synthesis, however none of them seem applicable to the chemistry shown or described on TV either; none of the reagents or catalysts mentioned would work as chiral catalysts, nor are they bio- or organocatalysts. Metal complexes with chiral ligands can be used to selectively catalyze production of one enantiomer, however the aluminum-mercury amalgam is again achiral. I don't remember any mention of using organocatalysis or biocatalysis, but these are possible.
The remaining route, then, is chiral resolution; let the reaction produce the 50-50 split, then separate the two enantiomers by some means of reactionary and/or physical chemistry. This seems to be the way it works in the real world. The advantage is that most of the methods are pretty cheap and easy; the disadvantage is that your maximum possible yield is 50% (unless you can then run a racemization reaction on the undesireable half to "reshuffle" the chirality of that half; then your yield increases by 50% of the last increase each time you run this step on the undesirable product).
In the case of methamphetamine, this resolution is among the easiest, because methamphetamine forms a "racemic conglomerate" when crystallized. This means, for the non-chemists, that each enantiomer molecule prefers to crystallize with others of the same chiral species, so as the solution cools and the solvent is evaporated off, the D-methamphetamine will form one set of homogeneous crystals and the L-methamphetamine will form another set. This means that all White has to do is slow the evaporation of solvent and subsequent cooling of the pan, letting the largest possible crystals form. Then, the only remaining trick is identifying which crystals have which enantiomer (and as these crystals are translucent and "optically active", observing the polarization pattern of light shone through the crystals will identify which are which).

I always assumed WW would use a chiral resolution column for the separation of enantiomers. That's why I always thought his product dripped slowly out of those extrusion like tubes in sense that he would be applying pressure like a column chromatography

Meth doesn't have to be optically pure to be "pure".  A mixture of d,l-methamphetamine is still pure, but I get where you're going with this.
He has a few options:

Chiral resolution - he could make the racemic meth and they resolve it by selectively crystallizing out the desired enantiomer.  Chiral acids like tartaric acid can be used to do this.
He could perform a stereoselective reduction of the intermediate imine.  You couldn't do this with Al/Hg, but there are other methods to do it (stereoselective hydrogenation using a chiral catalyst).

Also, the thorium oxide furnace was not used to reduce the imine intermediate, it was used to make the P2P.  Phenyl acetic acid + acetic anhydride over thorium oxide will give you P2P.

Very good point as to the possibility of asymmetric imine reduction. If I remember my reading correctly, some of the best catalysts are relatively pricey and exotic Rh, Ru, and Ti-metallocene complexes. I think there are various cheaper and more mundane chiral auxiliaries that can effect similar asymmetric reduction, though probably with poorer yield and %ee.

Expensive yes, but possibly still practical, as the catalyst isn't consumed and so theoretically can be reclaimed and reused; also, most chiral complex catalysts are only needed in relatively small doses.

@KeithS, yes, that's true. I suppose the fact that catalysts of this type (e.g., the Noyori and Knowles Ru chiral complexes) are routinely used in industry is evidence enough of their cost-efficiency.

In a tube furnace phenylacetic acid and acetic acid are heated run through a catalytic bed (a number of different catalytic substances can be used). Reduced using aluminum. The product cleaned up then using d- tartaric acid crystallizing out the desired enantiomer.

That's very incomplete. You might want to include more details about what happens (equations/structures etc.) and some idea of the remaining steps.

Regarding user55119's comment about methylamine: about the train scene, they didn't steal methylamine per se from that train, they stole aqueous methylamine, i.e. a 40% methylamine solution in water.
This solution has a different density than water, thus explaining the need to compensate with a slightly different volume of water, and explaining why WW talks about "China sending a marginally weaker batch".
However, for meth synthesis most underground chemists prefer using the hydrochloride salt of methylamine rather than an aqueous solution, as it is easily prepared in small batches with precursors that are way easier to obtain than methylamine in any form -which is a surefire way of attracting you close attention from the authorities. Just like P2P, phenylacetic acid and acetic anhydride (which is needed for total conversion of the morphine contained in opium to heroin), methylamine isn't a restricted compound per se, but it is a watched chemical...
Which brings me to another gaping plot hole in the show: many of the items WW & Jesse have to obtain for their production (e.g. 10 L round-bottomed glassware and the right-sized mantle-heater, catalytic bed for thorium oxyde, chemicals etc) are actually very conspicous items that are usually only found in specific places: student or semi-industrial -testing the scaleability of a reaction before scaling it up to industrial production for example- chemistry labs or clandestine drug labs. Even a chemistry teacher would raise suspicion by buying these items for himself rather than on behalf of a school (and with proper paperwork and clearance from the DEA), so a tattoed street punk like Jesse showing up at a chemical or labware retailer and asking to buy them (in cash, no less!) would be sure to raise a lot of red flags at all levels, and would probably be woken up at 6 AM by a bunch of assault troops one of the following mornings. It just doesn't add up in the real world. :)


text:
How to compute the electron configuration of an atom?
How to compute the electron configuration of an (electrically neutral) atom ?
So when given the atom number how does one compute the electron configuration ?
I know the sum of electrons must equal the atomic number.
I also know the first shells fill up as $2n^2$.
I do not know how to compute quantum numbers when given the atom number.

run Hartree-Fock

similar http://chemistry.stackexchange.com/questions/901/what-are-alternates-to-the-periodic-table-of-the-elements

To calculate electronic configurarion you need to know several thing first.
Electron Shell

In chemistry and atomic physics, an electron shell, also called a principal energy level may be thought of as an orbit followed by electrons around an atom's nucleus. The closest shell to the nucleus is called the "1 shell" (also called "K shell"), followed by the "2 shell" (or "L shell"), then the "3 shell" (or "M shell"), and so on farther and farther from the nucleus. The shells correspond with the principal quantum numbers (1, 2, 3, 4 ...) or are labeled alphabetically with letters used in the X-ray notation (K, L, M, …).
  Each shell can contain only a fixed number of electrons: The 1st shell can hold up to two electrons, the 2nd shell can hold up to eight (2 + 6) electrons, the 3rd shell can hold up to 18 (2 + 6 + 10), and the 4th shell can hold up to 32 (2 + 6 + 10 + 14) and so on.

Subshells

Each shell is composed of one or more subshells, which are themselves composed of atomic orbitals. For example, the first (K) shell has one subshell, called "1s"; the second (L) shell has two subshells, called "2s" and "2p"; the third shell has "3s", "3p", and "3d"; the fourth shell has "4s", "4p", "4d" and "4f"; the fifth shell has "5s", "5p", "5d", and "5f" and can theoretically hold more but the "5f" subshell, although occupied in actinides, is not filled in any element occurring naturally.

See 'Number of electrons in each shell' section under this article.
Atomic Orbitals

An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. Each orbital in an atom is characterized by a unique set of values of the three quantum numbers n, ℓ, and m, which correspond to the electron's energy, angular momentum, and an angular momentum vector component, respectively. Any orbital can be occupied by a maximum of two electrons, each with its own spin quantum number. The simple names s orbital, p orbital, d orbital and f orbital refer to orbitals with angular momentum quantum number ℓ = 0, 1, 2 and 3 respectively. These names, together with the value of n, are used to describe the electron configurations.

Understand electron configuration notation.
Electron configurations are written so as to clearly display the number of electrons in the atom as well as the number of electrons in each orbital. Each orbital is written in sequence, with the number of atoms in each orbital written in superscript to the right of the orbital name. The final electron configuration is a single string of orbital names and superscripts.
For example, here is a simple electron configuration: $1s^2 2s^2 2p^6$. This configuration shows that there are two electrons in the 1s orbital set, two electrons in the 2s orbital set, and six electrons in the 2p orbital set. 2 + 2 + 6 = 10 electrons total. This electron configuration is for an uncharged neon atom (neon's atomic number is 10.)
Memorize the order of the orbitals.
Note that orbital sets are numbered by electron shell, but ordered in terms of energy. For instance, a filled $4s^2$ is lower energy (or less potentially volatile) than a partially-filled or filled $3d^{10}$, so the 4s shell is listed first. Once you know the order of orbitals, you can simply fill them according to the number of electrons in the atom. *The order for filling orbitals is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s. *
An electron configuration for an atom with every orbital completely filled would be written: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6 8s^2$.
Note that the above list, if all the shells were filled, would be the electron configuration for Uuo (ununoctium), 118, the highest-numbered atom on the periodic table - so this electron configuration contains every currently known electron shell for a neutrally charged atom.
Now you can fill in the orbitals according to the number of electrons in your atom. For instance, if we want to write an electron configuration for an uncharged calcium atom, we'll begin by finding its atomic number on the periodic table. Its atomic number is 20, so we'll write a configuration for an atom with 20 electrons according to the order above.
Fill up orbitals according to the order above until you reach twenty total electrons. The 1s orbital gets two electrons, the 2s gets two, the 2p gets six, the 3s gets two, the 3p gets 6, and the 4s gets 2 (2 + 2 + 6 +2 +6 + 2 = 20.) Thus, the electron configuration for calcium is: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$.
Note: Energy level changes as you go up. For example, when you are about to go up to the 4th energy level, it becomes 4s first, then 3d. After the fourth energy level, you'll move onto the 5th where it follows the order once again. This only happens after the 3rd energy level.
For 'How to find quantumn numbers?', see this.


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Equilibrium: Equations with complex roots
I'm studying civil engineer in my first year. We have a course called "General Chemistry". One of the chapters was about chemical balance of a reaction (called equilibrium in English, I think). We got exercises on this were we had to calculate the amount of chemicals got converted to others, using acidity constants Ka.
For example, the equations we had to solve were easy and looked like this:
$$K_a = \frac{(x)^2}{(0.1 - x)(0.2 - x)}$$
These exercises were easy. But then came the point were we combined two reactions at the same time and had to calculate the balance for both at the same time. The equations we get were:

The tutor used to make simplifications by getting rid of the "$-x$" and "$-y$" in the denominator, assuming x and y would be so small it would almost make no difference. I don't like making such simplifications: I want the correct answer. So, what I did was putting the equations in WolframAlpha and let it compute it, since I don't know how to solve such a system of equations. The result shocked me: the solutions for x and y were complex (so, with a real and imaginary part).
WolframAlpha says this:

Whereas the tutor found (due to simplification):

The tutor's solution is in the complex plane not really far away from the correct solution to the equation (WolframAlpha's solution #3).
So, my question is: How do you explain these roots are in fact complex? I don't think a complex amount of chemicals can react right?

I generally like to use $a + b \approx a$ if $\frac{a}{b} > 10^{2}$. Note that this is not the case for the values of $x$ and $y$ obtained, so depending on other relationships, the tutor's answer may be up to ~5% off. In that case, substituting with the new value of $x$ and $y$ in the denominator may be called for. Note that with a tool like WolframAlpha, simplifications are largely unnecessary. But if you're trying to compute this without the aid of such a total, or you're looking for a quick approximation which is frequently good enough, that is the way to go.

Complex roots to mass/concentration balance equations should be discarded as they have no physical interpretation. If there are only complex roots, then odds are there's some error in the manipulation of variables that lead to the equation.
Your case seems different though. Notice the very small magnitude of the imaginary part of the complex roots; the number is "almost real". This happens at times simply because of Mathematica's numerical root search algorithms, which aren't always guaranteed to reach the exact value. I'm actually a bit surprised that it got caught up in such a simple calculation.
As proof that the imaginary part of the root is nothing but an issue with the calculation, I simplified the equations until I found a better representation for Wolfram Alpha to parse, hoping the complex root would disappear.
First, let $A=1.77\times 10^{-4}$ and $B=3.3\times 10^{-4}$ for concision. Isolating $x$ in the second equation gives:
$$x=\frac{B(0.1-y)-y^2}{y}$$
Replacing $x$ for the above expression in the first equation then yields, after some tinkering:
$$(A+B)y^3+(B^2-0.1A-0.1B-AB)y^2+(0.1AB-0.2B^2)y+0.01B^2=0$$
Aha, a simple polynomial. Alpha should be able to deal with that just fine. Replacing the values of $A$ and $B$ produces the following equation, which Alpha solves with no imaginary part. The physically meaningful solution out of the three roots is $y\simeq 0.00458505$. Replacing that value in the equation for $x$ completes the problem, with $x\simeq 0.00228225$, very close to the approximate solutions.

Oh... my... god... I can't even rely on WolframAlpha anymore... :/ Thanks man. I'm recalculating in meanwhile.

I want to add this just as a supplement to Nicolau's answer. As he mentioned, the imaginary numbers arise due to imprecision in a computational method. However, even without resorting to computation, there are ways to see that the result is incorrect from a purely mathematical perspective.
The Fundamental Theorem of Algebra tells us that a degree $n$ polynomial will always have $n$ roots if you take into account multiplicity. From this, one can also prove that an odd degree polynomial (with real coefficients) will have at least one real root. As Nicolau showed, your equation can be written as a third degree polynomial, meaning that at least one of the roots had to be real. In fact, in your case all the roots have to be real because, with real coefficients, complex roots come in pairs $a\pm bi$ and you can see that none of your roots have the same magnitude for the imaginary part.


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"Number of moles of electrons involved in the reaction" for hydrogen cell
I understand we can calculate the maximum work done by the electromotive force (driving electrons through a cell) by multiplying the potential of the cell with the charge of one mole of electrons (Faraday's constant) and with "the number of moles of electrons involved in the reaction". I thought that in the case of a hydrogen cell: the reaction $\ce{H2 - 2e- -> 2H+}$ involves two electrons but I can see on wikipedia that in the Nernst equation for the standard hydrogen electrode $z$ (the number of moles involved) is considered to be 1.
So what does this $z$, or expression "electrons involved" actually mean?

Nernst equation is used for finding the cell potential when the various reactants, i.e. anodic and cathodic electrolytes are not in their standard states ($298\text{K}, 1\text{mol l}^{-1}$). Thus, in such cases, you first write down any balanced form of equation, which for our hydrogen electrode, I choose to write down $\ce{3H2 -> 6H+ +6e-}$ and these electrons are used in a standard $\ce{H2}$ cell on the other electrode. Hence the Nernst equation would give the cell potential for this concentration cell as ($E^\circ=0$ since the the standard potentials of both sides is the same since the materials are same, and the non-standard hydrogen cell is the anode)
$$E_{cell}=-\frac{RT}{6F}\ln\frac{[H^+]^6}{1}$$
By using the properties of logarithm, you can show this is equivalent to $-\frac{RT}{1F}\ln[H^+]^1$, and this is the equation if you took only 1 electron transfer in the reaction.
It does not matter what equation you write, since in the final analysis, the number of electrons taken are compensated by the power of the concentration term in the logarithm.


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Resonance stability: ester vs. thioester

My problem set solution cites:

"Although sulfur is less electronegative than oxygen, overlap is a
  more important consideration.  Sulfur is a member of the third period.
  As a consequence, a sulfur atom is larger than an oxygen  atom and the
  C–S bond is longer than the C–O bond, contributing to poorer overlap."

However, I don't quite understand why a longer bond length would contribute to poor orbital overlap, and how greater overlap contributes to more stability.

possible duplicate of Advantages of thioesters over esters in fatty acid catabolism

@CurtF. I don't think that's a duplicate. This question is about orbital overlap and its relation to bond length. The linked question is about the reactivity of thioesters vs. esters which is a different question.

Try to understand it in a layman's way, greater the extent of overlap between the atoms, the closer their nucleus will be. The stability of the molecule will thus increase as stability of a molecule is inversely proportional to the bond length of the atoms in the molecule.
However this is applicable only upto some extent like your situation but if the nucleus of two atoms get too close a stronger force which is nuclear force starts acting on them which repels them, it is a very short length force.

There are three factors in determining whether a molecular orbital will form: symmetry, energy, and distance. The orbitals need to be oriented in space in a way that would make them overlap, they need to be close in energy (2pz of atom a and 2pz of atoms b as opposed to 2s of a and 2p of b) and the distance between the nuclei has to be short enough, but not too short to cause repulsion between the positive charges in the respective nuclei. S-C double bond definitely exist however its contribution is small due to the poorer orbital overlap caused by Sulfur's big size that both increased the energy difference and distance between the individual atomic orbitals.


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Rate of electrophilic bromination of toluene and ethylbenzene
My book says that electrophilic bromination of toluene (methylbenzene) is faster than that of ethylbenzene. How is that possible?
I am doubting it because ethyl is a better $+\ce{I}$ group than methyl, so ethyl would increase the electron density in the benzene ring more than methyl. As a result, bromine, which is an electrophile, should react with ethylbenzene faster than methylbenzene.
So, why is the bromination of methylbenzene faster than ethylbenzene and please give the reaction mechanism, if possible?

First, I think it's probably important to note that the difference in $\text{+I}$ effect of a methyl vs. ethyl is very small. In principle, the ethyl is more electron-rich, however the strength of the inductive effect decays very rapidly with increasing distance. Ultimately, I suspect variance in steric factors, solvation, and electron-donating ability to the $\pi$-system of the ring by hyperconjugative overlap are the decisive factors, not the difference in induction.
If we consider sterics, the ethyl is obviously the bulkier group, and that may partially explain the difference in reactivity. The ethyl, moreso than the methyl, will hinder the approach of the electrophile, and may potentially raise the energy of the transition state and $\sigma$-complex intermediate slightly by:
(a) electrostatic repulsion toward the electrophile, particularly in the case of ortho- attack, and
(b) inhibit potential stabilization of the $\sigma$-complex intermediate by solvent molecules.
From the standpoint of electronic effects, I suspect the differences are marginal. To the extent of my understanding, the energies of $\ce{C-H}$ and $\ce{C-C}$ bonding $\sigma$ orbitals are very similar, and the efficiency of hyperconjugative overlap should consequently be similar as well. I've occasionally encountered claims that $\ce{C-H}$ bonds are slightly better $\pi$-donors, but I'm not aware of any conclusive evidence to that effect. If that is true, however, it might contribute to the observed disparities in reactivity.
For a quantitative approach, you may want to take a look at the substituent constants derived for various alkyl groups from the Hammett equation. The equation was originally developed with the aim of analyzing substituent effects on the thermodynamic equilibria of benzoic acid derivatives. However, since the reactivity of substituted benzoic acid is also affected by the mesomeric and inductive properties of ring substituents (albeit in the opposite sense from a benzene ring in an EAS reaction), I think the Hammett equation furnishes a reasonable proxy measure of at least the relative strengths of substituents as either donors or acceptors of electron density. I found one table of said substituent constants. (Keep in mind that those table values are for ionization of benzoic acid, so a negative value is actually indicative of an electron-donating effect, which is favorable for an EAS reaction but destabilizing to an acid.)
Finally, as to the reaction mechanism, I'm not going to reproduce it here. Every EAS reaction follows a very similar, schematic mechanism. For the mechanisms, take a look, for example, at the Wikipedia articles on electrophilic halogenation and electrophilic aromatic substitution more generally.

Reactivity of alkylbenzenes towards electrophillic substitution reactions is decided only on the basis of steric hinderence that alkyl groups provides to aromatic ring.if steric hinderence increases reactivity of aromatic ring decreases .steric hinderence by alkyl groups increases in the order -CH3 < 1¤ < 2¤ < 3¤ . Therefore reactivity decreases in the order CH3 > 1¤ > 2¤ > 3¤ .
E.g. toluene > ethylbenzene

This doesn't really add much that hasn't already been said in the previous answer.


text:
Notation of beta decay
Is it an ion or a neutral atom that is created through beta decay?
For example: $_6^{14}\mathrm C \to ~ _7^{14}\mathrm N + \mathrm e^- + \mathrm{v{_e}}^{-}$
Isn't $_7^{14}N$ an ion since neutron gives us only a proton and an electron, not a proton and 2 electrons? If so, why isn't it stated in the example above (it's from Wikipedia)?

@MelanieShebel there's one more "Beta" to decapitalize in the text, and the minus sign on the neutrino needs to be removed. Also thanks for fixing the spelling error in my question earlier!

An ion is created. Usually this is not noted down in the nuclear reaction because while talking of nuclear reactions we only concern ourselves about nuclei, not the entire atom. The reaction is to be read as "A 14-$\ce{C}$ nucleus decays into a 14-$\ce{N}$, an electron, and an antineutrino."
Also, note that the neutrino has no charge.

Further to @ManishEarth's answer, note that the distinction is important in doing mass-energy calculations.  In the above example, you can find the energy released by subtracting the tabulated mass of a $N^{14}$ atom from the tabulated mass of the $C^{14}$ atom, and ignore the electron mass.  Why?  Because the actual Nitrogen atom produced only has 6 orbital electrons, while tabulated values include the mass of 7 orbital electrons.  So, in using the tabulated mass, you already include an extra electron...


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How to separate KCl from NaCl and MgCl₂?
Having mixed powder of $\ce{KCl}$, $\ce{NaCl}$, and $\ce{MgCl2}$ (and, perhaps, some crap), how can the components be separated?

Do you want to recover all the compounds as separate, pure chlorides, in a quantitative manner? That is probably impossible without specialized equipment/reagents. Why do you need to perform this separation?

Because I need potassium chloride and have the described mix (a dietary salt replacement) is the only available source of it to buy at my location, @NicolauSakerNeto.

If separating the three into separate pieces is too complex then I can consider it enough to get rid of the Sodium and leave Magnesium in place. Perhaps there is a way to gather Sodium ions from a water solution?

Purity requirements are very liberal, I don't need very pure substances, getting KCl concentration from 30% to, say, 80% or even 70% can be considered better than nothing.

By the way, if you're doing this to try and get very low sodium table salt, I should warn you that actual potassium chloride isn't very pleasant. Also, this separation is really difficult because sodium and potassium are chemically extremely similar, so it usually relies on steric effects AFAIK. If you have problems getting hold of pure potassium chloride, you'll probably have problems getting hold of what you need to separate sodium ions and potassium ions.

No, @Aesin, the taste does not matter, but thank you for your commentary.

As previously said, this will be very hard and costly. Under what circumstances do you need the potassium chloride? Maybe context will help. Chances are you'll end up with less magnesium chloride, but equal amounts of potassium chloride and sodium chloride.

As I mentioned in the comments, a quantitative separation of the three compounds will probably not be possible outside of a proper laboratory, however a coarse separation is possible. A quick search on the internet yields this link for the separation of $\ce{NaCl}$ and $\ce{KCl}$. You didn't give us the exact composition of the table salt substitute, so I assume $\ce{MgCl2}$ is present in much smaller amounts than the first two compounds.
The simplest strategy is to use the difference between the solubilities of the salts in water or ethanol. Using water is much more convenient, so we'll stick to that. I'll outline the method to obtain the highest purity $\ce{KCl}$. First, add a very large amount of table salt substitute to a container with boiling water (you'll want to add some 200-250 g per 100 g of water) and let the mixture dissolve as much as possible (though it probably won't even come close to dissolving completely). What you have then is a boiling water solution saturated with both $\ce{NaCl}$ and $\ce{KCl}$. Now, while the water is still very hot, quickly filter off the excess salt (might be doable with a fine sieve, or a funnel with some cotton would work better) and reserve the liquid in another container. Let the liquid cool down (using an ice bath is optional but increases yield and purity), and it will start to precipitate the salts. Now you perform another filtration, but this time you keep the solid retained on the filter, not the liquid. This yields purified $\ce{KCl}$.
This strategy works because it turns out that $\ce{NaCl}$ and $\ce{KCl}$ behave rather differently in how their solubility in water is affected by temperature.
$\ce{NaCl}$ soubility in water:

35.6 g/100 g water (0°C)
35.9 g/100 g water (20°C)
39.0 g/100 g water (100°C)

$\ce{KCl}$ soubility in water:

28.0 g/100 g water (0°C)
34.2 g/100 g water (20°C)
56.3 g/100 g water (100°C)

Evidently, as the filtered liquid cools down, $\ce{NaCl}$ remains almost as soluble in cooler water as in hot water, while $\ce{KCl}$ is only about half as soluble. This means that a large amount of $\ce{KCl}$ will precipitate as the filtered liquid cools, while only a bit of $\ce{NaCl}$ will follow. Hence, ideally, from 100 g of water saturated with both salts at 100°C (and containing nothing else), you will be able to extract 25.0 g of a mixture of salts containing 87.6% $\ce{KCl}$ if you perform the second filtration at 20°C. If you filter at 0°C, then the yield increases to 31.7 g of a mixture containing 89.3% $\ce{KCl}$. If for some reason you don't want to do the second filtration, you can just let the water evaporate completely after the first filtration, producing 95.3 g of a mixture containing 59.1% $\ce{KCl}$ from 100 g of saturated boiling water.
In practice your yield and the purity of the $\ce{KCl}$ will likely be somewhat lower due to losses in manipulation and the presence of other impurities. The procedure may be complicated by the very large initial amount of table salt mixture that you have to dissolve in water. You can reduce the amount of the mixture, but at some point the purification yield will start to go down.

Thank you very much, @nicolau-saker-neto. For those seeking to use the method described in the answer I must note that although I have approved it (as it looks reasonable) I have not tested it myself yet.

@Ivan How would you even test it without a lab?

Because the procedure is basically just to boil water...

Add $\ce{MgSO4}$, until a saturated solution of magnesium salts has been obtained. Upon evaporating down, it will be found that all the potassium salts have been precipitated more or less completely, and with kainite $\ce{KCl. MgSO4.3H2O}$ being the essential potassium-containing salt.
Add the kainite to a boiling solution of $\ce{NaHCO3}$ with moles equal to those of $\ce{Mg^{2+}}$. Then, freeze the mix until $\ce{Na2SO4.10H2O}$ is visible along with the visible $\ce{MgCO3}$ suspension.


text:
Can bromide ion reduce ferric ion?
Can bromide ion reduce ferric ion ?
Just a query.
In my opinion it can since bromide ion can undergo oxidation.
Is my answer correct?

You can predict this with the help of electrochemical series.

On going up the series reducing power increases, for you case it can be seen that Fe is a better reducing agent than bromide ion as Fe is above Br2 so bromide can't reduce ferric ion.

According to you the reaction will be:
2Br- + 2Fe(III)  ---> Br2 + 2Fe(II) by balancing.
Spontaneity can be checked out by using standard EMF of the cell formed
Hence standard emf E(cell)=E(Reduction) for {Fe(III)/Fe(II)}  +  E(oxidation) for {Br-/Br2}
Now using standard values we get  E(Reduction) for {Fe(III)/Fe(II)}=+0.77 and  E(oxidation) for {Br-/Br2}=-1.07
hence the E(cell) we get is =.77-1.07=-.3(negative)
We have (Delta)G=-n*F*E(cell) where (Delta)G is change in Gibb's Free Energy change
n is no. of electrons(here n=2),F is Faraday's constant and E is Emf of cell(Standard)
If (Delta)G is negative then reaction is spontaneous and if (Delta)G is positive then non spontaneous.
Clearly here E(cell) is negative hence (Delta)G is positive hence reaction is non-spontaneous and will not occur[NOTE THAT HERE CONCENTRATIONS OF IONS AND PRESSURE OF GAS IS TAKEN UNITY IN ITS SI UNITS.WE CAN MAKE IT SPONTANEOUS BY VARYING THE CONCENTRATION OF ION OR PARTIAL PRESSURE OF GAS]

Can't there be a more obvious way to tell this ?

I guess no.U can't tell directly if the reaction is spontaneous or not.U have to take catch of thermodynamics.


text:
Non polar molecular and attractive forces
How come when a non polar molecule increases in size, the attractive forces between the molecules of a substance also increase?
I thought that when a molecule increases in size, the atomic radius increases, which creates a bigger space between particles; thus, the force of attraction would be weaker.
But, why does the force of attraction increase when a non polar molecule increases in size?

Your argument would be right if the force of attraction would have been due to the nuclear electrostatic attraction on surrounding atoms. But in non-polar molecules, the dispersive (London) forces are the major forces, which are not due to the nuclear attraction, but due to the attraction arising from random fluctuations of electron density around the atoms, and this fluctuations increases as the size of the electron cloud increases.

There is more than one factor that matters in the forces between molecules. The tradeoff between these different factors is complex and can't be reduced to a single simple equation or factor. So some of the factors become more important because of molecular or atomic size counterbalancing the factor of distance between the entities.
The collective force that cause molecules and atoms to attract are often referred to as van der Waal's force (though some sources use a more narrow definition that doesn't include all of the components below). The key non-bonding forces involved are:

forces between permanent dipoles
forces between permanent dipoles and induced dipoles
forces between instantaneously induced dipoles caused by quantum fluctuations

In non-polar molecules it is the third of these that dominates. Crudely stated the force arises because of uneven distribution of the electron cloud in a molecule or atom that happens as a result of quantum fluctuations in the cloud. An uneven distribution of electron density is the equivalent of a dipole moment. These dipole moments create induced dipoles in neighbouring atoms and can interact with other quantum fluctuations in the neighbouring atoms. These create small electrostatic forces between the atoms or molecules.
The reason why larger atoms or molecules seem to have larger forces is two fold. Larger molecules, crudely, have larger electron clouds and more ways to interact (shape matters too and large flat molecules "stick" together more than irregular lumpy ones: compare the melting points of benzene and toluene).
In larger atoms the explanation is simpler. Larger electron clouds are less tightly held by there nuclei than those in smaller atoms. Crudely, larger, floppier clouds can have more scope for the quantum fluctuations than tighter-held smaller clouds (also they have a larger area and that maters to the total force). So the size effect increases the strength of the force rather than reducing it as might be expected from simpler arguments.

I'm aware that this is a slightly superficial explanation, but a better one would probably involve enough quantum mechanics to make the average reader's head explode. I think i've captured the key intuitions that matter.