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How does the Freezing Point fall in a solution?
It is known that addition of a non-volatile solute to a volatile solvent(liquid) to give a solution reduces the Vapour Pressure of the solution (well solvent actually as only solvent is volatile).
This leads to the Elevation of Boiling Point and "Deppresion" of Freezing Point.
I am clear with elevation of BP. But it is said that a liquid freezes when the vapour pressure of the liquid phase attains the vapour pressure of its solid state.
So, how possibly does the freezing point falls? For instance let the Vapour Pressure of the solid state be at an arbitrary point "x"(at a temperature A Kelvin) and that of the liquid phase be at "y".
So when we add a solute to the liquid:
Vapour pressure of liquid(solution) = y - $\alpha $ (where $\alpha $ is the depression in the vapour pressure)
So now the vapour pressure of the liquid(solution) reaches "x" rapidly that is at a higher temperature or a temperature greater than A which is an elevation in the freezing point.
I am always confused with this part.

Focus more on free energies rather than on vapor pressures (which derive, ultimately, from free energies after all).  For a mixture of B (solute) in A (solvent), the entropy of mixing is $RT(x_A ln(x_A) + x_B ln(x_B))$, and the enthalpy of mixing will go approximately as $x_A x_B \Omega$, with $\Omega$ as a measure of the interaction of A and B. The entropy term will always result in a reduction of free energy at small $x_B$ regardless of the sign of $\Omega$, but in the case of salt in water $\Omega$ is negative, driving further solubility.
Now, about those temperatures of phase transitions (which as you should recall occur when the free energies of the phases are equal, more fundamental than vapor pressures).  On the boiling end, water with salt in it has a lower free energy than water without salt - the boiling point of the salt water has to be higher than for pure water.  The presence of the solute makes it happy to stay liquid.
So, why isn't the freezing point raised in an analogous way? For pure A, of course, the freezing point remains that of pure water.  The entropy of mixing is equivalent for solid and liquid in this case, and it is not clear how different the excess enthalpy is going to be - so why does liquid continue to be stable?  Well, it only does while it has a higher concentration of salt in it than is in the solid.  It forms a classic eutectic point on the binary phase diagram (see, for example http://antoine.frostburg.edu/chem/senese/101/solutions/images/saltwater-phase-diagram.gif).  The only requirement of melting point depression is that water at a high concentration of salt has a lower free energy than the solid at a lower concentration of salt.  Given the outline of free energy above, this is likely to hold for some range of temperatures.
Again, remember that what is being lowered is the freezing point of water with salt in it - what starts freezing out is ice with a lower salt content in it.


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Can a mixture of water miscible solvent and water be separated by fractional freezing on a dry ice bath?
If a mixture of water with acetone, methanol, ethanol, or other water miscible solvent, is cooled below 0 degrees, would the water freeze and aggregate out as ice, separating out of the mixture. Or would the frozen 'water particles' be dissolved in the solvent, similar to how solid organics dissolve in solvents?
I suppose it really depends on the concept of solubility vs miscibility. I've been researching this for a while now and haven't been able to find a clear definite answer.

The short answer is

Yes

This is exactly how acetic acid is purified: Acetic acid freezing distilation
In this case, acetic acid (freezing point $16\ ^\circ \text{C}$) freezes first, leaving a water-enriched fluid behind. Conceivably you could remove water from methanol or acetone in the same way (except that water would freeze first). In practice, fractional distillation can be done more economically on a large scale. Fractional distillation requires 1 step (with lots of theoretical plates), while you need to repeat fractional freezing multiple times to get to high purity. Acetic acid is purified by fractional freezing perhaps because it may azeotrope with water.

Can you explain why you would need to repeat the freezing multiple times to get a high purity? Assume we want to purify the solvent and not the water. Wouldn't the freezing only need to be repeated to purify the frozen (water), not the solvent? Some solvent would be trapped in the ice, sure, which would be scooped out, but pretty much all of the water would freeze into ice, and float to the top (if solvent is dense), allowing it to be removed. If all ice is removed this way, how does water remain in the solvent? If the liquid isn't dense, the mass could be frozen and solvent decanted off.

It isn't as simple as the water simply freezing out. A mixture of miscible liquids typically doesn't behave as the separate components. At the point at which freezing begins (which may be higher or lower than that of either component), both the frozen material and the liquid still contain both components. The frozen is enriched in the higher melting point component and the liquid is depleted. Barring a eutectic point, one can remove the solid and freeze the liquid again to enrich it further, repeating until the desired purity is attained.

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Can electrons switch orbitals within a shell?
I know that electrons can move from say 2s orbital to an unoccupied 2p orbital, as in Carbon atom which can form 4 bonds this way. But I want to know is it possible for an electron say in orbital 2p to switch its place with an electron in 2s? I know only 2 electrons are allowed in an orbital, but is it possible for one electron to oust another and take its place so the ousted electron would have to take the other one's place? Is this theoretically possible?

Your question digs very deep into the quantum mechanics, namely exchange interaction ( http://en.wikipedia.org/wiki/Exchange_interaction) of fermions, which are indistinguishable particles ( http://en.wikipedia.org/wiki/Identical_particles )

The short answer to this is: according to our current understanding of quantum mechanics, we can't tell.
And now for the long answer:
First off, if you haven't already seen Richard Feynman's 60-second primer to the scientific method, you should definitely watch it now. It's short. I promise I won't go anywhere while you're watching it.
...
Alright. So the way science works is to test ideas by experiment. This means that in order to distinguish between which of two ideas is right, we'll need to come up with some difference in their behavior. This can sometimes be problematic in science because the difference in behavior between two models can be incredibly small. To take just one example, it wasn't until the 20th century that an apparent error in the orbit of Mercury was confirmed, which was eventually explained by relativity. Before that, if you had given someone the theory of relativity and the Newtonian theory of gravity and asked them to tell you which one was right, they would have shrugged their shoulders. There simply wasn't enough high-quality evidence to decide between one or the other.
On the other hand, what if there's a nightmare situation in which we simply can't tell the difference no matter how good our instruments are? Imagine that I give you a small cube which lights up when you shake it, but I've used magic so that no matter what experiments you try to use to figure out what's inside of the cube, you can't get any information. You can try to use x-rays to image the inside, or weigh it, or spin it around to see if the insides are unevenly distributed, or look at it with an IR camera...but every single time, the magic renders your results useless.
Now suppose I asked you how the box worked. Without being able to get any info, you're stuck! No matter how clever you are in making your guess or how many guesses you come up with, you can't really say anything about any of them, because you're being magically prevented from getting information about what's in the box.
This is a very important idea, so I want to make sure you really understand it. Without some sort of actual behavioral difference that can be looked at, science has no way of distinguishing between alternate ideas.1
Now, quantum mechanics is sometimes a lot like our magic box (to pretty much anyone who tries to study it, it is a magic box). Our current understanding of QM says that we have this wavefunction that tells us what the system is doing. We can't actually look at the wavefunction. The best we can do is to use some operators on it, which will give us the values of certain properties (don't worry if this all sounds very mysterious, it takes years of study to understand some of this stuff).

So how can we see if the wavefunction matches what we think it is? Well, we can look at what we should get from the operators, and what we actually do get from measuring it. For instance, by applying something called the Hamiltonian operator to the wavefunction, we get the total energy of the system. We can then measure the total energy of the system and see if theory and experiment agree. We can do that a whole bunch of times, measuring the momentum, and kinetic energy, and the average position of the system. If we do it enough times, we can say "eh, it's probably correct" and then move on to the next problem.
Now, to get back to your original question: can electrons ever switch places? Well, our current understanding of quantum mechanics is that, when electrons switch places, the wavefunction changes sign.2 "Aha!" I hear you say. "So we can watch the wavefunction and see if it changes sign, and if we ever see it flip, we can say that the electrons switched places!"
Well...not really. Because we can't look at the wavefunction directly, we can only measure things through the operators. And here's where the nasty trick comes into play: for all operators that we can actually observe, there won't be any change if the wavefunction flips its sign.
We're stuck! Just like our magic box, we can't get any meaningful information about what's going on inside, and so we can't really tell if the electrons switched places. Even if they did, it would look exactly the same to us (no matter how good our instruments are, we cannot see a difference, because there literally isn't one). It's an undecidable problem!
Now, if you're anything like me, this is a really crummy and disappointing answer. It took me a long time to accept the fact that there are some things in science that we just can't decide. So let's get a little creative: what would have to change in science for us to be able to tell?
First, and most directly, we would need an alternate theory to QM that allows us to distinguish between the two cases. Seeing how incredibly complex current theories are, I doubt that will happen anytime soon (though I could be pleasantly surprised). The second is that we would need some way to measure the wavefunction directly, instead of looking at it through operators. I've heard whispers of groups making progress on that front, though I honestly don't know if there's any truth to that or not. One way or the other, I don't expect we'll be able to answer this question tomorrow, or anytime relatively soon.

Some time back, I heard about a very clever paper that had to deal with this problem. There was an effect that could either depend on the temperature or density of water. Of course, you can't control the temperature and density of water independently (the stuff expands when you heat it up), and so everyone had just assumed that temperature was responsible. This group realized that, if you supercool water, it takes on densities similar to those found above 0C. By carefully supercooling the water and measuring this effect, they found that at similar densities of water, the effect was the exact same, even when the temperature differed by over 50C. It was density that was responsible all along!
This is one of the most mysterious things about all of physics. Even after two years of studying quantum mechanics, I have yet to find anyone who can offer a satisfying explanation for why fermions (half-integer spin particles) should flip their wavefunction upon exchange of (ostensibly identical) particles. If you know of a good read on the subject, please put it in the comments and I'll add it here!

The short answer to this is: the question is meaningless. :D And you actually do not need the wave function formalism to explain why.

I believe your answer was the one I was originally going for. This is why I shouldn't be allowed to write responses past midnight.

Firstly, you seem to have a misconception about atomic orbitals and molecular orbitals. When carbon takes part in bonding within a molecule the electrons don't move from its 2s orbital into an unoccupied 2p orbital. I think what you are referring to is hybridization: You take the 2s and 2p orbitals and build linear combinations from them to get a set of e.g. four energetically equivalent $\ce{sp^3}$ orbitals and then you use these orbitals to form bonds between carbon and, say, the 1s orbitals of four hydrogen atoms thus getting methane $\ce{CH4}$. But the problem is that the carbon $\ce{sp^3}$ or hydrogen 1s orbitals are not actually present in a physical sense in the $\ce{CH4}$ molecule. Those orbitals are merely a mathematical tool, a basis, used to describe the actual molecular orbitals of the methane molecule. A more in-depth explanation of that can be found here.
Ok, so there is no moving electrons from one orbital to another when forming bonds. But what about your question concerning the switching of electrons? Well, electrons are indistinguishable particles. That means if you have two electrons, one on the left and one on the right, you could exchange them without ever noting a difference in the whole system: it will still have the same energy, same momentum, same wavefunction, etc. Let's take the example of lithium: lithium has two 1s electrons and one 2s electron, so three electrons in total. If you'd label those electrons with the names Peter, Paul, and Mary you couldn't say whether Peter and Paul are hanging around in the 1s orbital while Mary has a good time in the 2s orbital or whether Peter and Mary have fun in the 1s orbital while Paul sits lonely in the 2s orbital because electrons are indistinguishable. Both situations would look absolutely the same to you. So, even if two electrons would switch their orbital "positions", you really couldn't tell.
A point that might raise confusion is that there is still the Pauli exclusion principle, saying that two electrons having the same spin can't occupy the same orbital. But that is a completely different matter that has nothing to do with Peter, Paul and Mary - to stay in the picture. You can't say Peter and Paul are both spin-up and Mary is spin-down, so Peter and Paul can never be together in the 1s orbital and one of them is always paired up with Mary. The only thing you can say is that among Peter, Paul, and Mary you must have two with spin-up and one with spin-down (two spin-down and one spin-up would be equivalent). But they can't all be spin-up (at least not in the ground state of $\ce{Li}$) because then you would have two electrons with the same spin in the 1s orbital which is forbidden by the Pauli exclusion principle.

Is this theoretically possible?

Theoretically your question from the title is meaningless so science would not even answer it and that is why.
Electrons are indistinguishable, or identical, particles in a sense that they all have exactly the same physical properties so they cannot be distinguished from one another, even in principle. Thus, there is no observable difference between the two situations you are trying to distinguish:

Electron-one is at $\ce{2s}$ orbital while electron-two is at $\ce{2p}$ orbital.
Electron-two is at $\ce{2s}$ orbital while electron-one is at $\ce{2p}$ orbital.

You can not tell these two different configuration apart. Imagine that we play the following simple game: I show to you (how do I do so is irrelevant) an atom with one electron at $\ce{2s}$ orbital and another one at $\ce{2p}$ orbital. I then hide it from you and either do nothing, or exchange the electrons. Finally, I show you the atom once again and ask you, did I exchange electrons or not? Would you be able to answer that question? Surely, no! And this is essentially what makes your question from the title meaningless.

I know that electrons can move from say $\ce{2s}$ orbital to an unoccupied $\ce{2p}$ orbital, as in carbon atom which can form 4 bonds this way.

Of course, here you can clearly tell the difference between the following two situations

Two electrons at $\ce{2s}$ orbital and two electrons at $\ce{2p}$ orbital.
One electron at $\ce{2s}$ orbital while electron-one is at $\ce{2p}$ orbital.

by just counting electrons on diferent orbitals (again, how do you do so is irrelevant).

I see your point, but there's a way to detect that switch if it happened. I think electrons had to jump an energy barrier to switch orbitals and probably change their spin. So if we could detect that energy change, then it would mean that it has happened. Like if two identical twins switch their places in a room, one may not be able to tell, but actually they have burned some energy while changing their places. Even if we assume they have exactly the same physical properties. We are not talking about the "states" here. The initial and final states may not be different at all.

So when I say is it theoretically possible, I mean is the barrier passable by electrons? Is there a state in an orbital no matter how much unstable and transient that 3 electrons exist in one orbital?

@Wise, of course, we are talking abous states here. And the thing you do not understand is that there is no such state as "e-1 on 2s orbital and e-2 on 2p orbital", as well there is no state "e-1 on 2p orbital and e-2 on 2s orbital". Electrons are indistinguishable, and consequently, there is only state "e-1 or e-2 on 2s orbital and e-2 or e-1 on 2p orbital". And this state is obviously the same as the state "e-2 or e-1 on 2s orbital and e-1 or e-2 on 2p orbital".

@Wildcat You do not understand the crux of the question. Wise (and now, myself as well) wishes to know whether such interactions can happen. We do not wish to know whether we can distinguish the final state of the system and the initial state of the system; that is beside the point.

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Phenyl vs phenol functional group?
Phenyl is a functional group with an aromatic ring bonded to another group. And, phenol is a molecule that is just a phenyl bonded to a hydroxyl group.
However, some sources consider phenol itself a functional group. Are they both functional groups, or is phenol just a molecule?

Phenol is both a molecule and a functional group (or at least a class of molecules). IUPAC has the definitive answer on the matter:
Phenols:

Compounds having one or more hydroxy groups attached to a benzene or other arene ring, e.g. 2-naphthol

So in addition to a molecule, are phenols only a class of molecules? Or are they also functional groups?

@epicfaace You can have a phenol moiety, which is a phenyl moiety with at least one hydroxyl group attached. (But I would still consider this kind of nomenclature a little odd.) This way it would/ could be considered a functional group. Going with compound class is more concise however.

The two links provided never mentioned anything about functional groups. Are classes = functional groups?

@Cheng Here is the IUPAC Gold book definition: Organic compounds are thought of as consisting of a relatively unreactive backbone, for example a chain of sp3 hybridized carbon atoms, and one or several functional groups. The functional group is an atom, or a group of atoms that has similar chemical properties whenever it occurs in different compounds. It defines the characteristic physical and chemical properties of families of organic compounds.


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HI/P reduction mechanism
I recently answered a question about Breaking Bad's initial methamphetamine production method (i.e. the reduction of (pseudo)ephedrine). The reaction is as follows:

(Source)
It is infamously known as the $\ce{HI/P}$ reduction and was (and still is somewhat) a real plague in the United States until cough medicine containing ephedrine came under tightened control. That for an introduction, now the question: what is actually the reaction mechanism of the $\ce{HI/P}$ reduction?
The first step seems like a simple $\ce{S_{N}2}$ substitution of $\ce{I^{-}}$, catalyzed by the protonation of the alcohol group by hydroiodic acid (making it a far better leaving group). But what is the mechanism after that? How is phosphorus involved? Does the $\ce{HI/P}$ reduction have any legal use?

I guess finding an actual reaction mechanism for this is on the edge of impossible. The first part is probably protonation/ activation and something between $\ce{S_{N}1/S_{N}2}$. The second step is most likely carried out with red phosphorous as an electron source and removal of $\ce{PI3}$. But as for all redox reactions, there are too many possible pathways to determine the mechanism. As for the legal use, there are possibly more effective and selective (and less dangerous) methods to reduce organic compounds.

Theoretical examinations of these reactions are fine, I think.

You may find this article illuminating. The authors propose a stepwise SET mechanism with $\ce{HI}$ being catalytic, recycled by $\ce{P_{red}}$.

This process is known as the Nagai reduction method. You can read more here: http://www.sciencemadness.org/smwiki/index.php/HI/P_reduction

According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $\ce{I-}$ to $\ce{I2}$ is proposed. Red phosphorus is required for the regeneration of $\ce{HI}$ from $\ce{I2}$. $\ce{P_{red}}$ reacts with $\ce{I2}$ to the phosphorus iodides $\ce{PI3}$ and $\ce{PI5}$, which are subsequently hydrolyzed to $\ce{H3PO3}$/$\ce{H3PO4}$ and $\ce{HI}$, which is reused in further reduction steps. Catalytic amounts of $\ce{HI}$ are therefore sufficient to run the reaction in the presence of phosphorus. However, a too low concentration of $\ce{I-}$ will lead to elimination instead of substitution in the first reaction step.

(Image source)
The SET mechanism of the reduction is not described in detail in the article, so the following is a bit speculative. As a resonance-stabilized radical is involved, homolytic cleavage of the $\ce{C-I}$ bond is one of the first steps, which gives the benzylic radical $\ce{R.}$ and an iodine radical $\ce{I.}$. A single electron is transferred from $\ce{I-}$ to $\ce{R.}$, forming the carbanion $\ce{R-}$ and $\ce{I.}$, which recombines with another iodine radical to $\ce{I2}$. Protonation of $\ce{R-}$ yields the final product.

I really doubt PI5 exists, PI3 is much more reasonable.

As for the second part of the question, this reaction is actually used commercially in the fine chemicals and specialty chemicals industry. An example that comes to mind is the reduction of 4-hydroxymandelic acid to 4-hydroxyphenylacetic acid. This intermediate is used in one route to the popular beta blocker "atenolol". Of course, as mentioned in the comments, other reactions are often favored when possible. In addition to the issues mentioned, phosphorus and iodine are both relatively expensive.

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

@paracresol Why not? I did ask this. Even though it was not the major part of the question. In addition Jannis Andreska did not answer this, so it is a useful contribution.


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Relativistic effect: d-electrons in metallorganic complexes
With higher period the d-electrons of the metal are less strong bonded and therefore oxidative addition is easier for $\ce{Ir(I)}$ than for $\ce{Rh(I)}$ and much easier than for $\ce{Co(I)}$.
For metal carbonyl hydrides, electrons with higher period are more stronger bound to the nucleus, so $\mathrm{p}K_\mathrm{a}$ increases in this order: $\ce{HCrCp(CO)3 < HMoCp(CO)3 < HWCp(CO)3}$
This explanation sounds logical!
I also wrote that for $\ce{CO}$-complexes because of the higher effective atomic number for higher period atoms (higher ionization energy) the $\pi$-backbonding is weak and therefore less stable as e.g. for $\ce{Cr(3d)}$.
This explanation sounds logical as well!
But is the written tendency of the oxidative addition not a contradiction to the other tendencies?

I don't think that d-shell electrons will have any relativistic effect. "Relativistic effect" doesn't mean relative to other elements in the same column of the periodic table, but to Einstein's theory of relativity. In simple terms 1S electrons of heavy elements are bound so tightly (energetically) that it is if the electron increases in mass. So in ab initio calculations of the binding energies of 1S electrons versus atomic number you could throw in a correction for  the effect.

I have some decent guesses.
(1) The square planar complexes with Ir(I), Rh(I), and Co(I) decrease in reactivity to oxidative addition. Suppose we add H2; here, this may be due to how iridium has the largest d orbitals and thus the least compact electron density. Thus, it is easiest to polarize its d electron density towards the added hydrogens and Ir(I), for instance, can become Ir(III) more easily than Co(I) can.
(2) For your m-C hydrides, since the hydrogen only has a bonding AO, maybe the larger radius of the metal increases the bond length and thus makes the W-C hydride most acidic.

For the CO complexes of transition metals, suppose we compared Cr, Mo, and W again. Cr has the smallest d orbitals, so it is worst at overlapping its d orbitals with CO's carbon pi*, making the Cr-C bond least stabilized and thus the weakest.

What do you think?

@MaxW: Relativistics affects chemistry. The main 'mechanism' is that due to the scalar (non-four component) effects in deed the innermost electrons orbitals get contracted ($e^-$ gets fast => heavy => radius smaller) thus screen much more effectively the nuclear charge, thus the outer shells expand/ get less strongly bound.

I'm pretty sure the answer to this question lies not in the general bonding trends (and EN trends) caused by relativistic effects but in the specific electronic configuration effects. At least this is the reason reductive elimination is easier for Pd (d10) than Pt (s1d9); however, the Co, Rh, Ir series has different electronic configuration than the  Ni, Pd, Pt series. I'll read some papers and try to expand on this...

I don't think your trends are correct. iridium is a stronger π-donor than rhodium, for instance, despite it being in a "higher" period. Is this what you mean by higher?

Echoing the previous commenters, I also don't see why relativity should play any role in this. I'm not sure how accurate trend #3 is, either. I was at least taught that backbonding increases from 3d < 4d < 5d because of more diffuse d-orbitals (hence stronger overlap with CO $\pi^*$, as @timaeus222 already noted quite a while ago. The C-O stretching frequency is not very diagnostic, however: Elschenbroich 3rd ed. gives them as 2000, 2004, and 1998 wavenumber units for Cr(CO)6, Mo, and W respectively.

I was wrong about the relativistic effect. For example see paper Relativistic Effects and the Chemistry of Gold on high Z atoms. I was thing only of direct effect. Rather the effect is indirect. Because of the tighter binding of the inner shells the outer shells are more shielded and hence less bound.

First off, be aware that oxidative addition occurs by several mechanisms which are governed by different factors.[1]
As you move down a period, the outer electrons become more easily accessed, as the outer electrons are increasingly screened from the nuclear charge by the inner electrons.[2] This is why both ionization energy and electron affinity decreases down a period.
Classical organometallic chemistry uses metal carbonyl stretching frequencies as a direct measure of π back-bonding and electron density.[3] For a recent discussion, see: Chem. Sci. 2016, 7, 1174. Compare the stretching frequency of the group VI metal hexacarbonyls [4]: $\ce{Cr(CO)6}$: $\pu{2000 cm^-1}$ and $\ce{W(CO)6}$: $\pu{1987 cm^-1}$. Tungsten donates more electron density back into the carbonyl than does $\ce{Cr}$.
The importance of relativistic effects in structure calculations is frequently the subject of reviews.[5]

References:

Crabtree, R. H. The Organometallic Chemistry of the Transition Metals, Ch. 6
Waldron, K. A.; Fehringer, E. M.; Streeb, A. E.; Trosky, J. E.; Pearson, J. J. Screening Percentages Based on Slater Effective Nuclear Charge as a Versatile Tool for Teaching Periodic Trends. J. Chem. Educ. 2001, 78 (5), 635. DOI: 10.1021/ed078p635.
Tolman, C. A. Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis. Chem. Rev. 1977, 77 (3), 313–348. DOI: 10.1021/cr60307a002.
Clark, R.; Crociani, B. Solvent effects on the infrared spectra of chromium, Molybdenum and tungsten hexacarbonyls. Inorganica Chimica Acta 1967, *1undefined 12–16. DOI: 10.1016/S0020-1693(00)93131-1.
Pyykko, P. Relativistic effects in structural chemistry. Chem. Rev. 1988, 88 (3), 563–594. DOI: 10.1021/cr00085a006.

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What determines the strength of reducing and oxidizing agents?
The best reducing agents are located at the bottom left of the periodic table (low electronegativity) and the best oxidizing agents are located at the top right of the periodic table (high electronegativity), excluding noble gases.
Is electronegativity the factor that determines the strength of reducing agents and oxidizing agents?

The strength of reducers and oxidizers depends upon the thermodynamic favorability of their reactions.  The strongest elemental reducing agent is lithium, which is not the least electronegative element.  When Li acts as a reducing agent, metallic bonds are broken and one electron is removed from each Li atom.  These processes are endothermic.  The Li is solvated (exothermic).
For an oxidizer like chlorine, the Cl-Cl bond is broken (endothermic) then the Cl gains an electron (exothermic).  The chloride is solvated (exothermic).  All this adds up to how favorable the reduction is.

There are ionic oxidizing agents that are not at the top right of the periodic table that are good oxidizing agents.  For example, silver and gold ions.

It's all about electron energy levels. Populated orbitals with high energy mean the substance tends to be a reducing agent, while unpopulated orbitals with low energy mean the substance tends to be oxidising.

this might help

Because compounds can be oxidizing agents like potassium permangenate (KMnO4) and reducing agents LiH4, what makes a compound an oxidizing or reducing agent is oxidation & reduction tables. As oxidation is the giving off of electrons and reduction is the acceptance of electrons, like plus vs minus relationship, if you have an oxidation table you can turn it into a reduction table by putting the flipping the table, changing the signs and reversing the equations. Anyway, reduction tables are more standard, where the strongest oxidizing agents have the most positive/largest standard reduction potentials and the strongest reducing agents have the most negative/smallest standard reduction potentials.
The standard reduction potentials are determined with a volt-meter connecting two cells together as electrons pass through the salt-bridge.
This site explains it  rather nicely
But there are some patterns such that
$\ce{2M(s) + 2H2O(l) -> 2M+(aq) + OH^{-}(aq)}$ $\ce{ M = Li, Na, K, Rb, Cs}$
Cesium reacts more violently (in an explosive reaction which occurs as hydrogen gas is ignited from the heat of the strongly exothermic reaction) than all the metals above it because it is a stronger reducing agent reducing agent being oxidized itself more than the metals above it because it has the the lowest ionization energy due to shielding as the many shells of electrons around the nucleus of Cesium reduce the pull of the positive charged nucleus on the electrons due to the electrons in the electron shells repelling the valence electrons further from the nucleus by like charge repulsion. Ionization energy is simply the measurement of the heat energy required to cause an atom to lose an electron in the gas phase.
This makes sense, but there are anomalies to consider not going by the electrochemical series from experiments and just trying to rationalize off of stuff like ionization energies, electronegativities, etc that I started off my answer with. The reason that you might find anomalies is because the reduction potentials in voltages of elements are calculated in aqueous solution, while ionization energies are calculated in the gas phase, though many oxidation-reduction reactions occur in liquid solutions, one potential difference being the failure to consider enthalpy of solvation when considering ionization energies. Furthermore, Pauling's electronegativites can be calculated from running some equations through elements considering the physics of ionization energies, so ionization energies and electronegativites give you a similar perspective.

You say: "The standard reduction potentials are determined with a volt-meter connecting two cells together as electrons pass through the salt-bridge."
Note: Electrons pass through the external metal wires, and NOT through the salt-bridge. Only ions pass through the salt-bridge.

Oh, thank you for the correction. Sometimes I get excited when I'm typing and I make errors. My apologies.


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How much does water expand when heated, and does salt affect that?
For this question, assume all things not mentioned are perfectly controlled, all at 14.7lbs of atmospheric pressure, and water is 100% pure of anything not mentioned, with no dissolved air.
Given a quantity of water that is 1 liter at 20°, what would the volume be when heated to 30°?
Secondly, the same question, except the water being heated is fully saturated with salt.

It would be useful if scientists would translate this data into something more meaningful to lay people.
For example, for 1 litre of sea water at the average salinity, the average global temperature and average air-pressure in millibars, what would be the volume be at increments of 0.5°. e.g. at what temperature would it be 1.1 litres?

The volumetric expansion of a liquid is given by the following equation
$$\Delta V=V_0\beta\Delta T$$
where $\beta$ is the coefficient of thermal expansion and $\Delta T$ is the change in temperature.
Here's a link to a nice table containing coefficients of expansion for water, both pure and salinated.  Wikipedia shows the maximum solubility of salt in water to be around 36% (w/w).
From the first link we find that
for pure water at $25\ \mathrm{^\circ C}$: $\beta = 257\times10^{-6}/\mathrm K$
for water with 35% $\ce{NaCl}$ dissolved at $25\ \mathrm{^\circ C}$: $\beta= 297\times10^{-6}/\mathrm K$
Now you should be able to calculate the values you are interested in.

Thanks so much. This is very interesting. For whatever reason, this morning I came up with a theory that perhaps more of the see rise we are experiencing is due to expansion than is due to new water in the oceans. Immediately after asking this question, I Googled and found that this theory is nothing new, and that expansion might actually account for 70% of sea rise.

For a starting point I would suggest http://web.mit.edu/seawater although this will not help much on the 'saturated' end of things since the salinity values in the tables do not reach the 20C saturation values. I leave finding the saturation value to the requester, since google brings it up quite easily.


text:
Exceptions to motif: nucleophile attacks electrophile
I hear this motif - nucleophile attacks electrophile - all the time with regard to organic chemistry. Are there any exceptions?
I think this statement is always true in general because it's a restatement of Coulomb's law. But then are there any organic reactions that cannot be characterized as nucleophile attacking electrophile?
What I learned: Radical termination does not involve nucleophiles attacking electrophiles.

there is nice describtion about this here

That gives me no exceptions @Freddy

What about rearrangements and fragmentations? In these reactions you have only one reactant and thus no nucleophile attacking an electrophile.

@Philipp With respect to rearrangements, I would probably want to consider them on a case by case basis. For fragmentation reactions, we might consider the reverse reaction where the nucleophile and electrophile could be more easily identified. I definitely would caution against looking only for reactions with one reactant. Nearly every "polar" reaction has been carried out intramolecularly where one part of the reactant acts as the nucleophile and another acts as an electrophile.

I believe that you are asking if there are any reactions that are essentially "non-polar". There are two classes that immediately come to mind: (1) radical reactions and (2) pericyclic reactions.
A common radical reaction studied in introductory organic chemistry is the radical halogenation of alkanes. Instead of cationic and anionic reactants/intermediates/products, there are radical intermediates that do not fall into the nucleophile/electrophile classification.

Pericyclic reactions involve the flow of pi-electrons and do not require any charged reactants/intermediates/products. Reaction 1 below is a Diels-Alder reaction and reaction 2 is an electrocyclization. Depending on your definition of nucleophile and electrophile, a Diels-Alder reaction might fit the motif of "nucleophile attacks electrophile." That is, the HOMO of the diene (1,3-butadiene in the example) interacts with the LUMO of the "dienophile" (ethylene). If you define a nucleophile as the reaction component providing the HOMO, then a diene is essentially a nucleophile in the Diels-Alder reaction (and vice-versa for the "dienophile"). However, since these reactions don't use the same types of nucleophiles as say SN2 reaction, they are often considered to fall outside of the usual "nucleophile attacks electrophile" motif. That's the way they were first introduced to me, at least.

I should add that many organometallic mediated reactions fall outside of the "nucleophile attacks electrophile" motif as well. My answer was confined to organic reactions.

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


text:
Can you make napalm out of gasoline and orange juice concentrate?
According to the novel Fight Club by Chuck Palahniuk:

The three ways to make napalm: One, you can mix equal parts of
  gasoline and frozen orange juice concentrate. Two, you can mix equal
  parts of gasoline and diet cola. Three, you can dissolve crumbled cat
  litter in gasoline until the mixture is thick.

The napalm is originally called napalm because is mixture of naphthenic acid  and palmitic acid. This makes me very skeptic about it being created by gasoline and juice. Also I see no reason for it to be frozen.
The only thing that sounds realistic to achieve the sticky effect is the use of cat litter - but you could as well just use dry dirt, couldn't you? And dirt doesn't really burn so you're weakening the burning effect.

Modern napalm (Napalm B) is polystyrene-based, it has a completely different chemical composition, but it is still called napalm (as described in the Wikipedia article you linked). The typical recipe for homemade napalm is styrofoam and gasoline, I would suspect that the recipies from the movie are all fake.

The nitroglycerine recipe in the book is exact to the letter - and the movie gives you some ideas as well, though they don't explain how to nitrate the glycerine made from fat.

Orange juice or coke doesn't even mix well with gasoline. Considering that is coming from a  fiction book, it may have been chosen only for the shock value. Average reader do not understand what palmitic acid is.

I modified the order of the text a bit so that someone reading the questions page does not think the recipe is something of your creation.  I have removed the home-experiment tag because this isn't really something that the average person should be making at home.  Please take precautions.

And you added safety tag, like I was asking something about safety. Also, the fact that you're not supposed to make an experiment at home doesn't mean you will not. In the Fight Club, they also used gloves and glasses when working with NaOH.

This reminds me of Payday2's Meth recipe that just makes salty water.

Here's an excerpt from an interview with the author of Fight Club, Chuck Palhniuk.

How to make Napalm with Frozen Orange Juice and Gasoline?
Well, Ed Norton changed one ingredient in every one to make them useless. So, that really pissed me off because I really research those really well. Actually its styrofoam and gasoline - it make the most incredible explosive.

http://www.dvdtalk.com/interviews/chuck_palahniuk.html
So no, Fight Club does not accurately convey any of the recipes. All were clearly modified. And most likely for safety reasons - i.e. you wouldn't want someone to walk out the theatre with a "bright" idea. On the other hand I would still expect a mixture of gasoline and orange juice concentrate to be rather flammable ...

and personal research into styrofoam and gasoline has indicated that it is not explosive, but it is very flammable.

gasoline+styrofoam: flamabe AND STICKY! The whole point of napalm is to be sticky, so under normal fight conditions it makes the most possible damage both on buildings and on personal.

Did the author publish the book with the correct recipes somehow?

I don't understand this interview, why would Palhniuk claim that Ed Norton altered the recipes when he had nothing to do with writing the book?  See this Google book search (https://books.google.com/books?id=-qgAgg91C9wC&q=gasoline#v=snippet&q=gasoline&f=false) for reference, as this edition has the altered recipes... echoing @TomášZato, is there a version of the book with unaltered recipes?

@silverjam - he meant Ed Norton modified the recipe in the movie.

Chuck Palahniuk did actually add fake things into the book because the FBI I think it was contacted him and told him that it was too dangerous to publish them in his novel, he also left out certain parts about making nitroglycerine for the same reason.


text:
Differences between chemical physics and physical chemistry?
What makes these two seemingly identical topics separate? What does each field more focus on, like do chemical physics researchers study more the atomic/molecular interactions while the physical chememists study more macroscopic properties? I assume they have quite a bit of overlap.

see this as well http://pubs.acs.org/doi/10.1021/acs.jpca.7b09596

You are of course right, there is a lot of overlap. There are also the scientific publication series ChemPhysChem and Physical Chemistry Chemical Physics amongst others, that deal mainly with subjects from the both areas.
In principle they are most likely dealing with almost the same subjects. And you often find physicists working in chemistry and vice versa. However, there is a little difference in the general approach to problems.
Physical chemistry usually approaches the problems at hand from a chemist's point of view, focussing (initially) mainly on macroscopic properties of an ensemble and use physical laws to determine them. It is often considered the parent field for many other subjects, like physical organic chemistry, chemical kinetics, and spectroscopy, among others. Most common publication titles are Zeitschrift für physikalische Chemie and The Journal of Physical Chemistry A, B, C, and letters. And many others...
Chemical physics on the other hand studies chemical ensembles and reactions from a physicist's point of view. They focus more or less, like you assumed, on ions, atoms, clusters, surfaces, free radicals, and molecules interacting with each other. This does include, but is not limited to, the study of solvation effects and single entities like quantum dots. Here are fewer specialist titles, like The Journal of Chemical Physics and Chemical Physics Letters. Of course publications are often widely accepted also in other journals of neighbouring areas. (Sometimes it is also treated as a subfield of physical chemistry.)
In modern times the various fields are not separated by a sharp line, they most likely blur into each other. You could probably ask the same question for Biochemistry and Chemical Biology, or Computational Chemistry and Quantum Chemistry, or Polymer Chemistry and Material Sciences. And there are more. Interdisciplinary cooperation is encouraged by many researchers and institutions, it is important not to care too much about the labels anymore.


text:
What will happen when a metal reacts with an acid which is neither dilute nor concentrated?
We have always studied about metals reacting with dilute and concentrated acids, but I do not understand what will happen if the same metal reacts with an acid that is neither concentrated nor dilute?

I'm voting to close this question as off-topic because it's vague.

This question is a bit vague, but briefly, the reason some acids behave differently at high and low concentration is because changing the activities of components in the solution changes the redox potential of the various possible reactions.
Let's take the reaction of iron with sulfuric acid as an example. This is not necessarily the only factor and I'll make a few assumptions that are likely quite inaccurate to simplify the problem, but it will illustrate the situation:
Let's assume that the half-reaction with iron is always the same—iron is being oxidized:
$\ce{Fe_{(s)}<=> Fe^2+_{(aq)} +2e-}\ E^\circ=+0.44\ \mathrm{V}$
Now, if you look at a table of standard reduction potentials, you'll see that there are a couple of potential half-reactions that could be reduced to drive the iron oxidation:
$\ce{2H+_{(aq)} +2e- <=> H2_{(g)}}\ E^\circ=0\ \mathrm{V}$
$\ce{SO4^{2-}_{(aq)} +4H+_{(aq)} + 2e- <=> SO2_{(g)} +2H2O_{(l)}}\ E^\circ=+0.17\ \mathrm{V}$
If all these reagents were at unity activity, then the second reaction would be more favoured and we would see mostly $\ce{SO2}$, but the activities of the substances in solution are also important. By the Nernst equation, $E=E^\circ -\frac{RT}{zF}\ln Q$, we see that the potential of a complete cell or a half-cell reaction is dependant on the reaction quotient $Q$. Because we're assuming that only the one reaction happens to the iron, we can just look at the reduction half-cell. If we assume that the sulfuric acid dissociates completely to $\ce{SO4^2-}$, then $[\ce{H+}]\approx 2\cdot[\ce{SO4^2-}]$. If we assume the gaseous products of the reaction don't affect the equilibrium, for our two reactions, $Q$ is $\frac{1}{[\ce{H+}]^2}=\frac{1}{(2[\ce{H2SO4}])^2}$ and $\frac{1}{[\ce{H+}]^4[\ce{SO4^{2-}}]}=\frac{1}{(2\cdot[\ce{H2SO4}])^4[\ce{H2SO4}]}=\frac{1}{16\cdot[\ce{H2SO4}]^5}$, respectively.
Now, if we substitute these into the Nernst equation for different concentrations of acid, we can see which reaction is more favourable: In dilute acid (1 mM), we get −160 mV for hydrogen formation and −238 mV for sulfur dioxide. The picture changes as the acid gets more concentrated. In 1 M acid, the hydrogen formation half-cell has a potential of 18 mV, but sulfur dioxide is now up to 206 mV.
So you can see how the concentration of acid changes which reaction is more favourable. To answer your question about what happens in the middle, one need only do the calculation for a moderate concentration and find that at some point, the two half-reactions become close in potential and we may see a significant mixture of both reactions occurring.
This is a very simplified picture and the Nernst equation is only technically valid under conditions of no current flow, meaning that other factors may influence what is observed experimentally, but I hope it's apparent that there's not really a solid line between dilute and concentrated. Many different reactions can occur simultaneously, even if a single reaction is dominant under certain conditions.

Some metals react explosively with water (Alkali metals, Group 1, sodium, potassium), and water is neutral. Some metals do not react even with very strong and consentraded acids (platinum).
Some acids are strong, some are weak. Even vitamine C is an acid, but so also are sulferic acid and nitric acid. Consentration also plays a role in this picture.
Magnesium reacts slowly with water, and is oxidized. Gold does not react. Iron reacts. and so on. The nobler the metal, the harder it is for an acid to oxidize it, and the stronger and / or more consentrated the acid will have to be to break down the metal. Alkali metals are 'weak' enough to be oxidized by clean air.
So this question is far from simple to give a short answer to.


text:
What is the cause for thermodynamic reversible and irreversible process?
My book defines

Thermodynamic irreversible process is defined as the process in which the changes take place rapidly, instead of occurring in successive stages with infinitesimal changes in quantities so that the system does not get a chance to attain equilibrium and there is an appreciable difference between the driving and the opposing force.

I have some problems in understanding it.

Which equilibrium are they speaking of?
Why does rapidity of the change cannot help it to attain equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?
What does they want to mean by driving and opposing force?

Your question is a good one. But I think the definition of irreversibility in your book is very poor. It is better to define an irreversible process as a process where the overall entropy (of system + environment) increases. An irreversible process cannot be brought back to its initial state, because that would
require an entropy reduction which would go against the second law of thermodynamics. Nevertheless the entropy of a subsystem can decrease at the expense of
an entropy increase of the environment, which at least offsets the entropy reduction of the subsystem.

But your second question (What does the rapidity of a process have to do with reversibility?) is much more difficult to answer in a satisfactory manner that doesn't simply lead to another question.

Let me start by reiterating what @Phillip said in his comment: this is not a good definition for an irreversible processes - a better definition is:

An irreversible process is one in which the net entropy generation term is greater than zero.

All real processes are irreversible. The "process speed" comes into play when we want to make approximations for idealized systems. It turns out that one way to convert a process into a theoretically reversible one is to set the condition that it moves very slowly between thermodynamic states, so that at any given step, it is in equilibrium. The reason that this isn't a good definition for a reversible process is that there are cases where a rapid change is reversible, and cases where a very slow change is irreversible. It's just that in general use, the "very slow process" approximation is commonly used to obtain a reversible process.

Which equilibrium are they speaking of?

Typically they mean equilibrium between the system and surroundings, but it can also mean within the system itself. Equilibrium is another way of saying "maximum entropy" - the matter and energy are as "spread out" as they are going to get. If the entropy is maximized, no more can be generated. In "real life" this means the system is well-mixed and that matter and thermal energy has dissipated throughout the system - there are no temperature or concentration/density gradients. This is easier to visualize using particles instead of energy - imagine an ideal gas in one side of a partitioned chamber, where the partition has a hole in it to allow the molecules to cross into the other section:

This is an example of an internal equilibrium - the particles cannot cross the system boundary, but they can reach an equilibrium within the system. An example of a system in equilibrium with its surroundings would be a container filled with water at room temperature - there are no thermal gradients across the container boundary.

Why does rapidity of the change prevent it from attaining equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?

Whenever a system is perturbed, meaning moved away from an equilibrium state, it takes some amount of time to reach a new equilibrium. This amount of time is called the relaxation time. Different systems have different relaxation times, and if you make changes quickly enough, the system won't have time to relax, and therefore won't be in equilibrium. Using a piston in a cylinder as an example, if you compress the piston by a very small amount, then the pressure inside will be in equilibrium with the pressure outside almost immediately.

What do they mean by driving and opposing force?

Using the piston as an example again, in this case the driving force would be the external pressure on the piston. The opposing force would be the pressure of the gas inside the cylinder. For a very slow process, at any given time the outside pressure and inside pressure would be the same - the volume would change to make sure this is true. For a very rapid change, the gas inside the cylinder would take longer to reach equilibrium, and so the pressure outside would not equal the pressure inside - the driving and opposing forces would not be equal.
It is important to remember that these are approximations that are intended to allow us to convert really difficult (usually unsolvable) differential equations that describe non-equilibrium or dynamic systems into something that is solvable. The trade-off is that we go from a realistic, fully dynamic system to one with constraints - we treat it like a series of tiny jumps between systems that are in equilibrium. In many cases, these approximations don't affect the accuracy of the solution. However, whether or not it will is determined by the time scale of the change relative to the relaxation time of the system.

I'll try and somehow answer this backwards.
3) Driving and opposing forces: Each process is driven by some kind of 'force' (please do not understand 'force' in the traditional physical sense; it's more of 'reasoning, why something would want to do something'). For example, when having a saturated sodium chloride solution, the sodium and chloride ions that are still bound in their ion structure in the precipitate would like to dissolve, the driving force being the increase of global entropy by a greater disorder of the entire system.
On the other hand, every process also has an opposing force, that attempts to force the process back into its original state. In the case of the saturated sodium chloride solution, the opposing force would probably be something along the lines of releasing water molecules out of a solvation shell, and thereby increasing local entropy.
Finally, for an (essentially) irreversible process, like the reaction of hydrogen and chlorine to form hydrogen chloride gas: This is a radical chain reaction. For each reaction step (formation of chlorine radicals, homolytic cleavage of hydrogen by a chlorine radical to form hydrogen chloride and a hydrogen radical, homolytic cleavage of chlorine by a hydrogen radical to form hydrogen chloride and a chlorine radical, and finally radical recombination in whichever combination) there are of course driving and opposing forces. However, the driving force is usually much stronger than the opposing force. For example, while it is energetically favourable to cleave a chlorine molecule with a hydrogen radical to form hydrogen chloride and a chlorine radical, a chlorine radical would never attack hydrogen chloride to liberate a hydrogen radical as HCl is far more stable than either hydrogen or chlorine.
2) The smaller and less rapid a change happens—in our case, dissolving one pair of sodium and chloride is a very small change—the easier it is to just go the other way. If a change is 'rapid' (I don't really like the word here, but I'm not good enough to find a better one), then obviously the driving force is stronger than the opposing force, creating a bias towards the product side.
But I would agree that this part of the definition is especially weak and hard to understand …
1) Any process can be understood as a dynamic one, with the change happening in both forward and backward direction. So even if you were to burn methane—clearly an irreversible process—you can still consider very minute quantities of carbon dioxide and water recombining to form a methane molecule or two (and oxygen). You won't, however, observe a proper equilibrium between methane/oxygen and carbon dioxide/water.
So while reversible processes are always at equilibrium between one side and the other side, irreversible processes are shifted so far to one side, that it doesn't really make sense to define an equilibrium for them, i.e. they 'do not get a chance to attain equilibrium'.
I hope I could help in some way, and I think you should get a better book ;)


text:
What are the standard math prerequisites for physical chemistry?
I'm currently a chemistry major and I know that physical chemistry requires lots of calculus.  Which mathematics courses should I take so that I am well prepared for my physical chemistry courses? Specifically, I need to take courses in thermodynamics, quantum chemistry and statistical thermodynamics and beyond, as I want to specialize in physical chemistry.
Is it true that the more mathematics courses that one takes, the easier physical chemistry becomes? I'm currently taking a multivariate calculus course now, and I wonder which is the next level of math courses I should take.
I plan to take as many mathematics courses as possible. I'm thinking of taking ordinary differential equations after i finish multivariate calculus. What additional courses beyond ordinary differential equations should I take? My school offers courses in complex analysis, real analysis, and partial differential equations. Will they be useful?

consider: for quantum mechanics: linear algebra, function approximation, Fourier series and transforms, numerical methods (eigenvalue problem, integration, error estimation) for statistical thermodynamics: series summation.

Well my school offer ODE, real analysis, Complex analysis and PDE. Will they be useful?

Hm. Take all of them, maybe except complex analysis (it is not absolutely necessary, but you need concept of a complex number for QC. you don't, however, need to concern yourself with deep analysis of functions of complex argument, but it sometimes suddenly useful in integration of ordinary functions). Search for linear algebra and numerical methods books. The topics are quite simple in nature and may be self-taught, since you don't have to work within - only to apply them.

I got through statistical thermodynamics with calc 3 and linear algebra alone. But it was tough. Diff. Eq. should be helpful for that class, but in my experience if you have a decent math background ( calc 1, 2, 3 ) you can usually get enough guidance to survive the physical chemistry courses even if you don't have the exact math training. If you plan on doing physical chemistry professionally, learn all the math you can.

In my experience, the most helpful advanced math courses you can take for Physical Chemistry are Differential Equations and Linear Algebra. Multivariate Calculus is helpful, but not as helpful as Diff. Eq. would be, especially in the beginning. Linear Algebra is very helpful for quantum chemistry, Diff. Eq. is also helpful for quantum, but more so for statistical mechanics. In the comments you mentioned several different Diff. Eq. options - as @permeakra said, taking all of them is a good idea, but ODEs and PDEs will probably be the most useful.

How about Complex and real analysis? In my school though, the prerequisite for PDE is real analysis. That means i have to take real analysis before i can take PDE.

Well, then I guess you need to take it ;). I can tell you that you will definitely be solving PDEs in chemical thermodynamics, quantum mechanics, and statistical mechanics. I'm actually a little surprised that these aren't prerequisites for Physical Chemistry courses in your program, in schools I am familiar with (I'm in the U.S.) "Differential Equations" gives an overview of all of this - maybe the courses at your school go more in-depth?

My undergrad only required calc II for p-chem, and I was taking calc III the same semester as p-chem, it was much harder to understand some of the math. But I think the math department was tired of having classes mostly full of chemistry majors.


text:
What is Bent's rule?
I'm all bent out of shape trying to figure out what Bent's rule means. I have several formulations of it, and the most common formulation is also the hardest to understand.

Atomic s character concentrates in orbitals directed toward electropositive substituents

Why would this be true? Consider $\ce{H3CF}$.
Both the carbon and the fluorine are roughly $\ce{sp^3}$ hybridized. Given that carbon is more electropositive than fluorine, am I supposed to make the conclusion that because carbon is more electropositive than fluorine, there is a great deal of s-character in the $\ce{C-F}$ bond and most of this s-character is around the carbon?
Or is this a misunderstanding of "orbitals directed toward electropositive substituents"? The fluorine is $\ce{sp^3}$ hybridized and these orbitals are "directed" toward the carbon in that the big lobe of the hybrid orbital is pointing toward the carbon. So does electron density concentrate near the fluorine? Because that would make more sense.
And this s-character concentrated toward the fluorine has the effect of what on the bond angle? I understand that the more s-character a bond has, the bigger the bond angle - consider $\ce{sp}$ vs $\ce{sp^2}$. But since the $\ce{C-F}$ bond now has less s-character around the carbon, the $\ce{H-C-F}$ bond angle can shrink, correct?

Fluorine is definitely not sp3 hybridised. Usually none of the terminal atoms uses this hybridisation, they are usually always sp hybridised.

To elaborate on Martin's point:  Geometry drives hybridization.  Hybridization does not drive geometry.

That's a good, concise statement of Bent's rule.  Of course we could have just as correctly said that p character tends to concentrate in orbitals directed at electronegative elements. We'll use this latter phrasing when we examine methyl fluoride below. But first, let's expand on the definition a bit so that it is clear to all.
Bent's rule speaks to the hybridization of the central atom ($\ce{A}$) in the molecule $\ce{X-A-Y}$.
$\ce{A}$ provides hybridized atomic orbitals that form $\ce{A}$'s part of its bond to $\ce{X}$ and to $\ce{Y}$.  Bent's rule says that as we change the electronegativity of $\ce{X}$ and \ or $\ce{Y}$, $\ce{A}$ will tend to rehybridize its orbitals such that more s character will placed in those orbitals directed towards the more electropositive substituent.
Let's examine how Bent's rule might be applied to your example of methyl fluoride.  In the $\ce{C-F}$ bond, the carbon hybrid orbital is directed towards the electronegative fluorine.  Bent's rule suggests that this carbon hybrid orbital will be richer in p character than we might otherwise have suspected.  Instead of the carbon hybrid orbital used in this bond being $\ce{sp^3}$ hybridized it will tend to have more p character and therefore move towards $\ce{sp^4}$ hybridization.
Why is this?  s orbitals are lower in energy than p orbitals.  Therefore electrons are more stable (lower energy) when they are in orbitals with more s character.  The two electrons in the $\ce{C-F}$ bond will spend more time around the electronegative fluorine and less time around carbon.  If that's the case (and it is), why "waste" precious, low-energy, s orbital character in a carbon hybrid orbital that doesn't have much electron density to stabilize.  Instead, save that s character for use in carbon hybrid orbitals that do have more electron density around carbon (like the $\ce{C-H}$ bonds).  So as a consequence of Bent's rule, we would expect more p character in the carbon hybrid orbital used to form the $\ce{C-F}$ bond, and more s-character in the carbon hybrid orbitals used to form the $\ce{C-H}$ bonds.
The physically observable result of all this is that we would expect an $\ce{H-C-H}$ angle larger than the tetrahedral angle of 109.5° (reflective of more s character) and an $\ce{H-C-F}$ angle slightly smaller than 109.5°  (reflective of more p character).  In terms of bond lengths, we would expect a shortening of the $\ce{C-H}$ bond (more s character) and a lengthening of the $\ce{C-F}$ bond (more p character).

S-character must be conserved?

Bear in mind that Bent's rule speaks to hybrid orbitals - a part of a bond, not the entire bond.  I would say, 1) the atom will re-hybridize its hybrid orbitals so as to partition s character wisely (e.g. to follow Bent's rule, use more s character (stabilize orbitals) in orbitals with more electron density as opposed to orbitals with less electron density) and 2) the s character content in all of the hybrid orbitals for a given atom must sum to 1.

(1) There is no such thing as rehybridisation, as it is a mathematical concept that is a result of a certain bonding situation. (2) Bent's rule does not specifically apply to hybrid orbitals, but to the linear combination of atomic orbitals. (It is a tiny but all defining difference.) (3) It is not a rule a priori, much more a generalised observation.

@Martin 1) I "understand", but it is a handy word. When a water molecule vibrates and the HOH angle changes, there is a change in the electron distribution around oxygen.  I would say that the oxygen rehybridizes during the vibration. For me, it's a shorthand way to denote a change in electron density with a consequent change in physical properties. 2) I thought that a carbon sp3 hybrid orbital was a linear combination of the carbon 2s and 2p AOs. If I'm wrong could you elaborate a bit? 3) I agree, I meant to say that it was a tendency in my original post, but forgot.  Edited and added it.

A hybrid orbital that is denoted as $\ce{sp^3}$ is shorthand for $\ce{s^{$1/4$}p^{$3/4$}}$ and is a fixed linear combination of the respective mother orbitals. Bent's rule specifically refers to losing this fixation, therefore accounting for any linear combination of atomic orbitals. It is also important to state, that Bent's rule specifically refers only to a framework in bonds. So you are of course not wrong.

About the last paragraph: Does this mean that the $C-F$ bond length of ${CH_3}F$ will be more than that of $CF_4$ ... and similar argument for $C-H$ bond length in methane and ${CH_3}F$?

@AnuragBaundwal Yes, the $\ce{C-F}$ bond length in $\ce{CH3F}$ is longer (139 pm) than that in $\ce{CF4}$ (132 pm). The $\ce{C-H}$ bond length in $\ce{CH3F}$ (108.7 pm) is slightly shorter than that in methane (109.1 pm).

Why can't it be sp5......how did you decide it to be sp4?

@HydrousCaperilla  It could be $\ce{sp^5}$.  Like I said, I would expect it to "move towards $\ce{sp^4}$ hybridization", it might move even further in that direction as you suggest.

why is smaller angle reflective of p character?

@Buraian as you move towards orbitals with pure p-character, the angle between them decreases towards 90 degrees.

What would the angle approach as the s character increased?

@Akash With 2 substituents (otherwise there is no angle) it will approach 180° (sp hybridized).  Take a look at some of my posts on Coulson's theorem.

Have you read the Wikipedia article to Bent's rule (especially the Justification paragraph). I think it explains the things rather well. In the example of $\ce{H3CF}$ the $\ce{H}$ is more electropositive than $\ce{C}$ and the $\ce{F}$ is more electronegative than $\ce{C}$. So, using the assumption that like in $\ce{CH4}$ the $\ce{C}$ atom is $\mathrm{sp}^3$ hybridized as a starting point, Bent's rule tells us that the $\ce{C}$-orbitals that are used to form bonds between $\ce{C}$ and $\ce{H}$ will not be "pure" $\mathrm{sp}^3$ orbitals but will contain a higher $\mathrm{s}$ character whereas the $\ce{C}$-orbital that is used to form the bond between $\ce{C}$ and $\ce{F}$ will contain a higher $\mathrm{p}$ character than a "pure" $\mathrm{sp}^3$ orbital. As for bond angles: the consequences of Bent's rule for the bond angles are also explained rather well in the Wikipedia article.

Higher s character for the C-H bonds because the H's cannot stabilize electron density and higher p character for the C-F bond because the fluorine can better stabilize electrons?

@Dissenter Essentially yes. Instead of directing equivalent sp³ orbitals towards all four substituents, shifting s character towards the C-H bonds will stabilize those bonds greatly because of the increased electron density near the (more electronegative) carbon, while shifting s character away from the C-F bond will increase its energy by a lesser amount because that bond's electron density is more localized on the more electronegative F anyway and thus further from the carbon.

How do we decide the orientation of the orbitals - i.e. where they are directed?

@Dissenter It is more about the bonding partners. When you have different bonding partners, say electropositve A and electronegative B, for a central atom C, then the bonds/bond orbitals from C to A will have more s character than expected and the bonds from C to B will have less s character than expected. So, the s character concentrates in bonds/orbitals directed toward electropositive substituents. That is the sense in which the word "directed" is used in the definition.

So the main idea is to put s-character around the more electronegative atom?

@Dissenter Yes, because s orbitals "penetrate" closer to the nucleus such that the higher electronegativity affects them more strongly than p orbitals.

@Dissenter "So the main idea is to put s-character around the more electronegative atom?"  If we are applying Bent's rule, then we are focused on (discussing) the hybridization of the central atom in X-A-Y


text:
How can I work out what reaction will happen?
I have recently started learning chemistry, and I've been reading a GCSE (UK high school level) textbook on the stuff :D
Much of it has been fairly straightforward, though there has been a reoccurring theme that has been bothering me.
I've encountered many instances where the author simply presents a reaction with little explanation, as if to suggest that the reason as to why it would occur is obvious.
Let me give you an example I have come across just today. Ethene in bromine water:

I think to myself, 'why would this happen?'. There are no 'free electrons' that I can see in ethene. OK, bromine is highly reactive, but what will it do? Will it react with H, or C, and how? What about the bromine bond? It's all very uncertain in my mind.
I always try to reduce it to an energy problem. We just want the most stable state, right? But, there are many possible outcomes. Am I supposed to apply some logic, follow some flowchart, or work out all possible outcomes and calculate which has the lowest energy? Do chemists just remember all simple reactions, and reduce problems to those? Or, am I just overthinking it?
A side problem to this is why, if we place an alkane in bromine water, we don't get the same thing happening with ethane.

There are no 'free electrons' that I can see in ethene

That's correct, but...  In ethene there are two bonds between the two carbons that hold the two carbons together.  A chemist calls this a "double" bond.  The important thing about a double bond (that's not obvious from the drawing) is that the two bonds are not the same.  One is a strong sigma bond, just like in ethane; the other is a weaker pi bond.  While the electrons in the pi bond are not free, they are not held as tightly

as the electrons in the sigma bond and are, therefore, more available for reaction.  This is why ethene is so much more reactive than ethane.

Am I supposed to apply some logic... Or, am I just overthinking it?

Yes, there is logic and that body of logic is what organic chemistry is all about.  In the case of bromine reacting with ethene, the reaction belongs to a class of reactions known as electrophilic additions.  Most reactions in organic chemistry involve a molecule that has electrons available to share (a nucleophile) and a molecule that would like more electrons (an electrophile).  These two molecules react (form a chemical bond) by the one donating\sharing its electrons and the other molecule accepting\sharing those electrons.  Here, the pi bond in ethene has electrons that it is willing to share (they're not free, but they're not tightly held either), while the bromine molecule is highly polarizable, much like $\ce{Br^{\delta +}-Br^{\delta -}}$, with the positively polarized bromine atom wanting electrons.  The reaction proceeds as pictured below, to yield 1,2-dibromoethane.

A side problem to this is why, if we place an alkene in bromine water,
  we don't get the same thing happening with ethane.

I bet you can answer that one now.  Ethene has a pi bond with electrons loosely held and available for reaction.  Ethane, on the other hand, only has a sigma bond where the electrons are more tightly held and less available for reaction.

You must learn some basic things in organic chemistry to understand mechanisms for such simple reactions and these are: functional groups, nucleophiles, electrophiles, leaving groups, curly arrows. You should easily find explanations of these in any organic chemistry textbook. Then you could use this link http://www.chemguide.co.uk/mechanisms/eladd/symbr2.html#top to understand what is happening in this reaction.
Here are the other basic types of reaction in organic chemistry:
http://www.chemguide.co.uk/mechmenu.html#top


text:
The walls of a plastic vial with Mercury inside are darkening. What is causing this?
I have a sample of Mercury in a plastic vial,
The vial is a BD Vacutainer (genuine), made of polyethylene terephthalate. It is used for blood samples, and the one I used was new.
When the vial was new, it was almost as transparent as glass, as seen in the image:
vacutainer http://mm.claflinequip.com/catalog/product/cache/2/image/750x750/9df78eab33525d08d6e5fb8d27136e95/i/2/i20885.jpg
However, after about a month, here is what the vial looks like:

As you can see, the vial has darkened.
The vial was sealed with a small initial layer of hot glue about 1 centimeter from the top, then the last centimeter was filled to the brim with epoxy, to ensure it won't leak.
I have been handling the vial quite a lot, but it has been stored in the padded box as seen in the image.
What is causing the plastic to darken? Once again the vial is made out of polyethylene terephthalate if that could be useful.
Thanks!

My guess (only a guess, not an answer!) is that it is mercuric oxide forming on the walls.  Hot glue is not a hermetic seal, so oxygen will leak in, and react with mercury vapor. Mercuric oxide is a solid at normal room conditions.

I've only ever seen mercury in glass containers, but sometimes the glass got dark too.

Mercury doesn't react with PET under normal conditions, but mercury vapor will diffuse into plastics.  The diffusion will change the optical density of the plastic in the area where diffusion occurred.
Here's a link to a nice study where the mercury was doped with a radioactive isotope of mercury.  This doped mercury was placed in a glass chamber which also contained plastic discs held at a known distance above the mercury.  The plastic discs were removed after various times and then microtomed.  The various layers of the microtomed plastic disc were measured for radioactivity.  Analysis of the resulting data allowed the investigators to measure the rate constant for diffusion of mercury into the plastic.
One final point, plastics are sometimes contaminated with small amounts of various impurities.  Sometimes these impurities were intentionally introduced, such as radical chain terminating materials (sulfur is an example) in order to produce polymers with a certain molecular weight distribution.  So while the plastic itself might not react with mercury, an impurity might.  The above link also provides analysis of plastic contaminated with sulfur and finds a much higher mercury diffusion constant in such materials.


text:
Reaction Pathways - Substitution and elimination

This one was a bit involved. The alkyl halide turning into an alcohol with retention of stereochemistry suggested that an SN1 mechanism was at work. The addition of NaOH and the inversion of stereochem followed by "only" suggested again an SN2 mechanism. No E2 products; hydroxide isn't basic enough to deprotonate an alkane to a significant extent. The transformation of the alcohol into a nitrile group with inversion of stereochemistry is SN2; attacked with a non-protic solvent for good measure. The transformation into a nitrile group with retention of stereochem is SN1; treatment with tosyl chloride turns the -OH into a good leaving group.

Sorry, but I don't see what your question is? Are you asking whether your proposed reaction pathways are correct?

@Dissenter Reactions involving an SN1 mechanism don't proceed with retention of configuration, rather both enantiomers are usually formed.

Correct, I meant they proceed with both a retention and inversion. @Philipp - that is my question.

@Dissenter So in the TsCl cyanation shouldn't you also show the other enantiomer?

Oops. Does the resulting -OTs group leave on its own accord, creating a planar carbocation - hence the racemization?

Yes, if it's an SN1 reaction. I thought you were saying it's an SN1.

Can you please rework your post to actually include a question and while you are doing that, give it a more descriptive title.

Since Greg E. focused on the first parts of the roadmap, I'll focus on the cyanation reactions.
What you have shown for the sequence from (R)-2-butanol to (R)-2-cyanobutane (retention of configuration), actually gives inversion of configuration. It is probably more clear if we look at this as two discrete steps. Step one is reaction with TsCl (and pyridine typically) to give the tosylate with retention. Step 2 is reaction with NaCN to give the cyano compound (S)-2-cyanobutane with inversion. Perhaps that can all be done in one step, but I've never seen it or tried it.

Going from (R)-2-butanol to (S)-2-cyanobutane (inversion of configuration) by direct reaction with NaCN will not give the target compound. Hydroxide is not a suitable leaving group for SN2 reaction. The hydroxyl must be converted into a better leaving group first, such as the tosylate above.
That leaves the problem of how to synthesize (R)-2-cyanobutane (retention of configuration). Since the SN2 reaction with NaCN is going to invert configuration, you need a method of converting -OH into a leaving group with inversion. Two successive inversions give overall retention of configuration. The two most common ways taught for this transformation are SOCl2 (-OH to -Cl) and PBr3 (-OH to -Br). Both of these reagents install a halide with inversion of configuration, allowing SN2 reaction with NaCN to invert back to the original stereochemistry.

+1, very good answer, IMO. I do want to point out that if $\ce{SOCl2}$ is used, it's important to add pyridine, not just to neutralize evolving $\ce{HCl}$ but also because reaction of the pyridine with the $\ce{ROSOCl}$ intermediate prevents the reaction from taking place by $\mathrm{SN_i}$ instead.

@GregE. Thanks. I've never used SOCl2 for the reaction from an alcohol (only an acid), so I wasn't aware of that detail.

Yes, it's rarely mentioned in most textbooks I've seen and even many professional organic chemists I've encountered aren't aware of it. It's an uncommon mechanism; I'm aware of it only because I happened upon it mostly by accident while thumbing through March's Advanced Organic Chemistry some time ago.

In theory, one can substitute a leaving group with retention of stereochemistry by performing two successive $\mathrm{S_N2}$ reactions. The first one inverts stereochemistry, and the second one inverts it again, thus restoring the original configuration. The Finkelstein reaction might be applicable: reaction of your 2-bromobutane with sodium iodide in acetone should yield 2-iodobutane via $\mathrm{S_N2}$, which can then be reacted with sodium hydroxide. Reaction of a secondary alkyl halide with $\ce{NaOH}$ will definitely yield competing elimination products, however. Note that, with secondary substrates, some competition will exist between $\mathrm{S_N1}$ and $\mathrm{S_N2}$, but the latter can be favored by use of an aprotic solvent and an excess of nucleophile.
Reaction with $\ce{H2O}$ is going to be slow and would yield a racemic mixture of $\mathrm{S_N1}$ products, as well as the competing $\mathrm{E1}$ product (with possibly some minor side-products from bimolecular pathways as well, depending on conditions). If you're satisfied with a racemic mixture, I would think elimination followed by acid-catalyzed hydration or oxymercuration-demercuration would be faster and more reliable in practice.
Cyanation with inversion can be done by direct reaction of your 2-bromobutane with, e.g., sodium cyanide. Cyanide is less basic basic than hydroxide, so competing eliminations are much less of a problem. If you want retention of configuration, replace the bromine first (as described above), then perform the substitution with $\ce{NaCN}$. The same ideas apply if starting from a tosylate instead.

Oops, so the only part I erred on was thinking that hydroxide couldn't deprotonate the alkyl halide and give me some elimination products?

@Dissenter, yes, in general you would see a large proportion of elimination products.


text:
What is the bond angle of water?
I have been trying to find out the bond angle of $\ce{H2O}$, but every site I visit has a different answer.
So far, I have found the following angles listed:

Site 1: 104.4º
Site 2: 107.5º OR 104.5º, depending on where you are in the article.
Site 3: 104.5º

Your second link does not say that water has a 107.5 degree bond angle. Rather, 104.5.

It actually says both... I suspect that 107.5 was a typo.

First, there is no precise value due to Heisenberg uncertainty principle - there is only an average value.
Second, under what conditions?  The average bond angle will be different in liquid water at room temperature than in a vacuum at OK.

Hmm... I guess I was looking for the average angle at standard temperature and pressure in liquid water.

In your comment, you say you are interested in "liquid water".
The values in your self-answer are for gas phase water, and the second source is citing the first source, so it is really just one source.  The value is determined from gas phase rotation spectroscopy.
For liquid water, the value is not as precisely known:
105.5° (calculated) and 106° (experimental) as reported in Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572.

While researching the above, I found some reliable sources that suggest the bond angle is 104.48º.

Source 1
Source 2

EDIT:
As DavePhD says, these values are for water in the gas phase. For liquid water, the bond angle is around 105.5-106º.

Huh, I"ve never seen it written out to two decimal places. I guess this means that one of David Brin's books (which have a plot point involving 104 degrees) are technically correct and it would round down.

Source 1 actually lists a theoretical value to three decimal places, but it lists the two decimal place version in the abstract.

authors: القرشي ابوستيت ابوستيت


text:
How can I clean solid sulfur and its oxides from test tube?
I was experimenting with sulfur few years ago. The test tube I used for that purpose has still sulfur stuck onto it. I was searching for a way to dissolve it.
I found an idea to try warm acetone. Sounded like cheap solution, but doesn't work - or it will take ages to dissolve.
I'm not a chemist, so I don't have any strange chemicals at home.

You don’t have sulfur oxide, because it’s a gas.  2. If your sulfur is in the form of sulfate salt, use water. Many sulfates are soluble in water. 3. Just dump the test tube. Easier, cheaper, and healthier.

There are a couple of ways to do it depending on what chemicals are available to you. Since sulfur melts around $115~\mathrm{^\circ C}$, you might first melt it and let as much liquid drip out of the test tube as possible in order to make the following step easier.

Perhaps the most straightforward method involves toluene or xylene
which you can probably purchase at a hardware store.  Put some
toluene or xylene in the test tube and heat it up (note: be careful,
toluene and xylene are flammable), the sulfur should slowly dissolve.
Pour the waste liquid out while hot.  Repeat the process if
necessary.  If you can find both toluene and xylene, pick the xylene
because it has a higher boiling point - it can get hotter and
dissolve the sulfur better.
Chromic acid (sulfuric acid and potassium dichromate) will remove sulfur.  Pour the chromic acid into the test tube and let it sit for a couple of days, then pour it out, and with the help of a spatula, you should be able to remove the sulfur as a solid chunk.
Carbon disulfide dissolves sulfur, but $\ce{CS_2}$ is toxic and extremely flammable

The chrome VI species in cromic acid are highly toxic/cancerogenic. Don't.

It would no doubt be cheaper to buy a new test tube than the chemicals needed to dissolve the sulfur. SAFETY I wouldn't heat toluene, xylene or carbon disulfide over a open flame. // Chromic acid is a common cleaning solution in a chem lab, but it is really nasty since it oxidizes "most" things. Not really something for a typical home lab in my mind.

Since you stated:

I'm not a chemist, so I don't have any strange chemicals at home.

you will neither have facilities for their proper storage and  waste management.
Consequently, the only reasonable solution is: Get some new test tubes (about 0.20 Euro for a 160 x 16 mm tube), but don't risk your own health or pollute the environment by messing around with carcenogenic ($\ce{CrO3}$ in $\ce{H2SO4}$) or neurotoxic ($\ce{CS2}$) agents.

If you were a chemist, and had the facilities and chemicals: Still dump that test tube.

I once had this problem in the lab. I put some NaOH (I guess it can be found in some bleaches or cleaning supplies for declogging pipes) and water in it. It warms in the process, but it's not a problem. After an hour or so, and after bit of scratching (watch out, the brush might decompose in such basic solution), the sulfur was gone.

About the sodium hydroxide (NaOH) option: I've had sulfur residue attached to my condenser. Letting the glassware 'bathe' in high concentration NaOH, then washing it of with 10% hydrochloric acid (HCl) only partly removed the sulfur. It was unfortunately still detectable, by smell, when doing a simple water distillation.

And I am not entirely sure the glass is immune from bathing in such concentrated sodium hydroxide, either.

Hot sodium percarbonate solution can mildly oxidise some sulphur to SO2. It's safe and non toxic. Strong hot hydrogen peroxide will aslo oxidise some to H2SO4. And if its a test tube burning it off with a bunsen burner or buying a new test tube is probably the most cost effective choice. I have cleaned a sulphur dirty sinter using percarbonate then nitric acid.


text:
How to produce hot water-soluble soap balls? What is their chemical background? What about medical capsules?
There are water-soluble balls for soap baths. Then there are medical capsules that disappear in mouth. Is it possible do this kind of capsules and water-soluble containers at home? If not, what do I need to do them? What is the chemistry behind them?

Empty gelatin capsules are readily available at drug stores (or used to be when my kids were smaller).  They dissolve pretty quickly.


text:
The difference between heat of reaction at constant pressure and that at constant volume

The $K_\mathrm{p}$ value of a reaction is $\pu{10 atm}$ at $\pu{300K}$ and $\pu{4atm}$ at $\pu{400K}$. The incorrect statement about the reaction is-
a. The reaction is exothermic
b. The rate of forward direction is more than that of backward
      reaction.
c. The rate of backward reaction increases more than that of forward
      reaction with increase of temperature
d. The difference between heat of reaction at constant pressure and that at constant volume is $RT$.

The correct answer given is option b.
But then how can we say that the difference between the heat of reaction at constant pressure and that at constant volume is $RT$ for the above reaction?
How do we go about approaching this problem? Do we need to use the van't Hoff  equation? I tried using the ideal gas equation also but didn't get this $RT$. Also I think since change in enthalpy is negative so the reaction is exothermic.

What do you mean by "ko value" and "change in enthalpy is -ve"? Maybe this would be clearer if you used the math formatting this page provides. What are the reasons for you to "feel that this reaction is endothermic" if you haven't specified what reaction you are talking about?

Sorry, but your question is very hard to understand. You clarified the "ko value" issue but the rest of my first comment remains. What reaction are you talking about and where does your claim that "the difference between the heat of reaction at constant pressure and that at constant volume is RT" come from?

@Philipp please read the question again. This time I have give the actual question also. The correct answer to the actual question is the b part(which means that it is an incorrect statement)

Kp decrease with temperture so change in enthalpy is -ve so reaction is exothermic.

Again sorry, but your question still is very imprecise. I'm sure that it is not true for every possible reaction that the difference between heat of reaction at constant pressure and that at constant volume is RT. It might be true for some reaction involving ideal gases but the way the question is asked it is difficult for me to answer it.

In retrospect, the body of your question contains a lot of unnecessary information. Would you consider changing it to something a little bit more concise?

So the question is: Why is the difference between the heat of reaction at constant pressure and that at constant volume equal to $RT$?
First Part: heat of reaction
From the first law of thermodynamics we know
\begin{equation}
  \mathrm{d}U = \delta Q + \delta W = \delta Q - p \mathrm{d} V
\end{equation}
where $U$ is the inner energy, $Q$ is the heat, $W$ is the work, $p$ is the pressure and $V$ is the volume.
So, for a reaction at constant volume you get
\begin{equation}
  \mathrm{d}U = \mathrm{d} Q \ .
\end{equation}
Now, for a reaction at constant pressure you can invoke the relationship
\begin{align}
  \mathrm{d}H &= \mathrm{d}U + p \mathrm{d}V + V \mathrm{d}p \\
              &= \delta Q - p \mathrm{d} V + p \mathrm{d}V + V \mathrm{d}p \\
              &= \delta Q + V \mathrm{d}p \\
              &\overset{\mathrm{d}p = 0}{=} \mathrm{d} Q
\end{align}
where $H$  is the enthalpy.
This gives you the relationships that describe the change in heats released during reactions at constant volume and constant pressure.
The so called heat of reaction is defined as
\begin{equation}
  \Delta Q = \biggl( \frac{\partial Q}{\partial \xi} \biggr)
\end{equation}
where $\xi$ is the extent of reaction.
Second Part: temperature dependence
Now using the equality of mixed partials and the definitions of the heat capacities at constant pressure $C_{p} = \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p}$ and constant volume $C_{V} = \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V}$ you get for constant volume ($\mathrm{d}U = \mathrm{d} Q$):
\begin{align}
  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta U}{\partial T} \Biggr)_{V} &= \left( \frac{ \Bigl( \frac{\partial U}{\partial \xi} \Bigr)_{T, V}}{\partial T} \right)_{V} = \Biggl( \frac{\partial^{2} U}{\partial \xi \partial T} \Biggr)_{V} \\
  &= \Biggl( \frac{\partial^{2} U}{\partial T \partial \xi} \Biggr)_{V} = \left( \frac{ \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{V}}{\partial \xi} \Biggr)_{T, V} = \Delta C_{V}
\end{align}
and for constant pressure ($\mathrm{d}H = \mathrm{d} Q$):
\begin{align}
  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta H}{\partial T} \Biggr)_{p} &= \left( \frac{ \Bigl( \frac{\partial H}{\partial \xi} \Bigr)_{T, p}}{\partial T} \right)_{p} = \Biggl( \frac{\partial^{2} H}{\partial \xi \partial T} \Biggr)_{p} \\
  &= \Biggl( \frac{\partial^{2} H}{\partial T \partial \xi} \Biggr)_{p} = \left( \frac{ \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{p}}{\partial \xi} \Biggr)_{T, p} = \Delta C_{p}
\end{align}
where $T$ is the temperature and $\Delta C_{i}$ is change in the molar heat capacities of the reactants and products during the reaction at constant $i$ (where $i$ is either $p$ or $V$):
\begin{align}
  \Delta C_{i} = \sum_{k} \nu_{k} c_{i_{k}}
\end{align}
where  $\nu_{k}$ and $c_{i_{k}}$ are the stochiometric coefficient and the molar heat capacity at constant $i$ of the $k^{\text{th}}$ component in the reaction, respectively.
When you integrate the equations for the temperature dependences of the reaction heats you get:
\begin{align}
  \text{constant volume:  } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} \, \mathrm{d} T = \left( \Delta Q \right)_{V} = \int \Delta C_{V} \, \mathrm{d} T \\
  \text{constant pressure:  } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{p} \, \mathrm{d} T = \left( \Delta Q \right)_{p} = \int \Delta C_{p} \, \mathrm{d} T \ .
\end{align}
So, the difference of the reaction heats at constant volume and constant pressure would be:
\begin{align}
  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T \ .
\end{align}
Third Part: ideal gases
Now, why would this boil down to $RT$? This works when you consider reactions that involve only ideal gases. For ideal gases you have (see here):
\begin{align}
  c_{p} - c_{V} = R \ .
\end{align}
So, the integral simplifies to
\begin{align}
  \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T &= \int \sum_{j} \nu_{j} \bigl( \underbrace{ c_{p_{j}} - c_{V_{j}} }_{= \, R} \bigr) \, \mathrm{d} T \\
  &= R \int \sum_{j} \nu_{j} \, \mathrm{d} T \\
  &= R \sum_{j} \nu_{j} \int \mathrm{d} T \\
  &= R T \sum_{j} \nu_{j}
\end{align}
where $\sum_{j} \nu_{j}$ will be some rational number but not necessarily equal to 1 - it depends on the reaction.
This is basically the result you were looking for although the reaction heat difference will only be proportional to $RT$ with the proportionality constant being $\sum_{j} \nu_{j}$:
\begin{align}
  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = RT \sum_{j} \nu_{j} \ .
\end{align}


text:
Using Hydrates as a skin moisturizer
I was just in my AP Chemistry class earlier today learning about hydrates and their properties. It occurred to me that a compound with so much water could aid in moisturizing the skin. Is this a valid thought? And if so, could hydrates be irritating to the skin?
Through research, I personally thought Natron may be an option in a solution added to a cream or lotion.

Hydrates have a high affinity for their water, and probably will not give it up for your skin. Some may even draw water out of the skin.

@user137 I think that is the correct answer, if you'd care to expand on it a bit.

I do not think that hydrates will serve as good moisturizing agents, because they usually do not loose their crystal water unless when being heated. Magnesium sulfate heptahydrate, for example, starts to loose water at temperatures above 50 °C, and complete dehydration of the salt is only achieved at temperatures above 310 °C. Some hydrates can even have a dehydrating effect when they react with moisture to hydrates with higher $\ce{H2O}$ content. Calcium sulfate hemihydrate, for example, reacts with water to form the dihydrate in an exothermic reaction. The stability of hydrates is the reason why many anhydrous salts which are easily hydrated are used as drying agents, for example, sodium sulfate and magnesium sulfate.

This isn't necessarily true (that hydrates will hydrate your skin). For example, $\ce{MgSO4*7H2O}$ is Epsom salt. Do salts sound particularly hydrating? No.
Furthermore, dehydrated hydrates have a high attraction for water, as seen in $\ce{CaSO4}$. When hydrated, it forms plaster of Paris, which can be dried to make drywall.