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Chemical reactions in plasma state
I am wondering about chemical reactions when the reactants are in a plasma state. Consider hydrogen and oxygen. Heated, these would react to form water.
If hydrogen and oxygen plasma were combined, would the separation of electrons from nuclei prevent normal chemical reactions? Or would the reaction take place in the plasma state?

Of course it prevents normal reactions. In temperatures like 10^5 K there are no molecules.

Most plasmas are not fully ionized. In fact, many have very low levels of ionization (maybe 1% or so). However, ionizing atoms or molecules, and splitting molecules, certainly makes it easier to achieve a number of chemical reactions (and break up things too!). What, really, is your question?

The broader question is about chemical reactions with reactants in plasma state.  Combining hot oxygen and hot hydrogen would produce water.  Would combining these two in plasma form also produce water, or would the plasmas mix and not react.  Assume 100% ionization of these plasmas.  That is really my question.

Does this answer your question? Is it possible to combust a plasma?

Yes, chemical reactions do take place in high-temperature plasmas. In fact, these reactions present some problems in the analytical technique known as Inductively Coupled Plasma - Mass Spectroscopy (ICP-MS), which typically uses an argon plasma.
As noted in the comments, the conditions within a plasma are very different from those in a laboratory at STP, for example. But that just means that the chemical reactions that are observed are different, not that they don't exist. In ICP-MS for example, one is trying to detect the masses of very low concentrations of various elements in the presence of relatively huge quantities of argon, oxygen and hydrogen (the latter assuming that an aqueous sample is being analyzed). Species that are formed within the plasma include $\ce{Ar2+}$, $\ce{ArO+}$ and more. When one of these species has the same mass as the trace isotope of interest, it only takes a very small percentage of these products to interfere with the analysis.
This example of argon plasma can largely be extended to your specific question "If hydrogen and oxygen plasma were combined, would the separation of electrons from nuclei prevent normal chemical reactions?". Depending on the conditions of the plasma (temperature, concentrations, degree of ionization, etc.) the answer is that the reaction would only form some small degree of transient water molecules, as compared to say molecular hydrogen and oxygen gas at high temperatures or in the presence of a catalyst, which would overwhelmingly form water. In other words, the former (plasma state oxygen and hydrogen) would form trace amounts of water, and probably hydrogen peroxide and other oxygen-hydrogen species, where under "normal" conditions the formation of water would be strongly favored (thermodynamically anyway). It's just the nature and definition of a plasma that whatever species you have in your plasma will want to be broken into it's elemental constituents and the formation of molecules will be strongly unfavored.
Summary:
The conditions within a plasma are very different from conditions typically encountered in a non-plasma environment, and thus the chemical reactions are different. The formation of otherwise exotic species like $\ce{Ar2+}$ are relatively commonplace within an argon plasma for example ("common" being a relative term of course, as concentrations of $\ce{Ar2+}$ may still be relatively low, just much higher than would be seen outside of the conditions present in the plasma). Plasma conditions themselves can also vary widely, from conditions where only a small portion of the gas molecules are charged to conditions where the majority are charged. Regardless, the high energy state of a plasma tends to result in the destruction of molecules to their atomic precursors, but occasionally there are reactions to form products that would not be seen outside of a plasma.

That's nice answer, not what OP was expecting though.

Yea, I left out a specif part of the OP question. I made an edit to address this, though it looks kind of awkward, but I think it more directly addresses the OP anyway. As usual, thx for the input ;)

An excellent answer, with the general and the specific.  Thank you!

Glad it was helpful :)

How do you form species like Ar2 and ArO if you're in an plasma? Doesn't being in a plasma imply that everything is ionized--in which case how would you form the covalent or ionic bonds necessary for molecules? Or is your example with the ICP-MS just a case where you have a distribution of atoms, some of which happen to be below the ionization point which allows these species to form in small amounts?


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How is benzene prepared from acetylene?
The mechanism of this polymerization reaction which finds its utility in the preparation of benzene in a red hot iron tube piques my curiosity.

I've read this Wikipedia article on cyclotrimerization, but it's too high a level for a high-schooler to comprehend. Can you please explain this to me in simpler way?

It isn't prepared this way. Its derivatives are though.

@Mithoron I'm not sure, but my best guess is cyclotrimerization of propyne yielding mesitylene is what you're talking about.

Rarer derivatives than that. And I don't see an explanation without talking about details of organometallic compounds' reactions.

Walter Reppe developed a process for the "safe" trimerization of acetylene to benzene and cyclooctatetraene. Although completely superseded by more modern alternatives, this process was used before/during WW2 because starting from only coal, calcium carbonate and electricity one could synthetise calcium acetylide and therefore acetylene, having no need for oil derivatives, scarce during war.
The mechanism (in simplified terms - hold onto the stones!) is that on the surface of the catalyst (iron, copper or nickel depending on the variation of the process) there would be formation of transient "clusters" of three or four molecules of Fe/Cu/Ni-acetilyde. That close arrangement facilitates one of the electrons of the triple bond of the acetylide "attacking" one of the carbon atoms of one of the nearest acetylide neighbours (again, simplifications). That molecule with an excess of one electron would in turn atack its other in vicinity and finally this one would attack the first one (that became deficient in the first step).
Since the aromatic structure of benzene is rather stable compared to ethyne, and the product is 1/3 (or 1/4) the number of moles of reagents, the formation of product is favored under pressure.
Some sources:
Walter Reppe and the chemistry he developed: https://en.wikipedia.org/wiki/Walter_Reppe
Trimerization: https://en.wikipedia.org/wiki/Trimer_(chemistry)

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When water undergoes electrolysis where in the beaker does the water molecule break apart?
When I was thinking about it I remembered reading two things, one was that certain gases "form" on certain electrodes. The other was something like, such and such ion "travels" towards such and such electrode.
Does the molecule break up on the electrodes?
Or perhaps it breaks up at a random spot between?

The water molecule gets split up into a cation and an anion in the solution, the ions then rush to their respective electrodes and finally gain or lose electrons to form the gases you're talking about.

I think you do not know much about electrolysis.

The water molecules form long networks of hydrogens bonds; let's imagine one such H-bonded chain that stretches from the anode to the cathode. Reduction will occur at the cathode, so the nearest water molecule (on the end of our chain) will be deprotonated and that proton will be reduced to $H\cdot$ radical, which can combine with a nearby $H\cdot$ to form the reduction product, molecular hydrogen.
Now, the water molecule that was deprotonated can re-protonate itself with the hydrogen of its neighbor (the second $H_2O$ in our hypothetical chain), likely the very hydrogen that they were sharing as an H-bond. This leaves the second water molecule deprotonated, so it fully claims the proton the third molecule was offering in their H-bond.
The process continues, with the negative charge effectively traveling the length of the chain toward the anode as each water molecule in turn briefly becomes hydroxide. This "proton hopping" behavior is known as the Grotthuss Mechanism. This animation can give you a sense of how this phenomenon happens, but note that the animation shows an acidic mechanism (extra proton, positive charge traveling) rather than the basic mechanism I described (a missing proton, with the negative charge traveling). Both are likely at play during the electrolysis of water. (Actually, the two mechanisms probably cannot be distinguished due to the simultaneous nature of the redox process, which I'm about to describe).
At the anode, hydroxide will be oxidized to $O_2$ through a few steps that aren't necessarily relevant to your question. The big idea is that the hydroxide will be created from a water molecule at the anode end of the chain of H-bonded molecules, and it will probably happen nearly simultaneously as a water molecule at the cathode end is deprotonated to form the proton reactant. Most probably both ends of our H-bonded chain of water molecules will react at more or less the same time.
If you like metaphors, think of it as a line of people each holding a basketball. Everybody passes their basketball to the right, and received a basketball from the person on their left... Except the people standing on the two ends. One will throw their basketball into the hoop, and the other will be torn to shreds and evolve molecular oxygen.

That makes so much sense! Now I can see the circuit is a full loop.

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Sunlight degradation of butyric and propionic acid
As I understand from Wikipedia article sweat is metabolized by skin flora primarily to butyric and propionic acid. What will sunlight do to these acids? Will my cloth smell less if dried in sunlight?
My concerns is that if drying in sun will remove odour fomr clothes as sunlight will cause degradation of odor causing chemicals(what ever they may be) and bacteria.
Sunlight should break bind in organic chemicals and make them odourless

Hello and welcome to Chemistry.SE. You might want to do a couple things to help clarify your question. First, you could [edit] in a link to the Wikipedia article. Also, it might be helpful if you expanded in a couple sentences why you think your clothes would smell less if "dried in tropical dry heat", and also just what that phrase means (i.e. dry heat with sunlight or without sunlight). Just some suggestions. Best of luck!

FYI: There are actually quite a few rather complex molecules responsible for body odor. According to Wikipedia: "The main components of human axillary odor are unsaturated or hydroxylated branched fatty acids with E-3M2H (E-3-methyl-2-hexenoic acid) and HMHA (3-hydroxy-3-methyl-hexanoic acid), sulfanylalkanols and particularly 3M3SH (3-methyl-3-sulfanylhexan-1-ol), and the odoriferous steroids androstenone (5α-androst-16-en-3-one) and androstenol (5α-androst-16-en-3α-ol)." And as you said, butyric and propionic acids also definitely play a role in body odor.

@airhuff edited question


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Could stabilized aqueous ozone be bottled to prevent half life/sanitization window?
Could stabilized aqueous ozone be bottled to prevent half life/sanitization window?
Typically upon creating stabilized aqueous ozone, it has a shelf life and a sanitization window of 24 hours from the time of production. Is this timeline due to oxidation, or would proper bottling prevent the degradation from occurring as fast? If not, is the process of it converting back to oxygen and water something that is inevitable due to time?


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Li2O + 2KCl =2 LiCl +K2O: Is it a feasible Reaction?
$$\ce{Li2O + 2KCl <=> 2LiCl + K2O}$$
I got this reaction on a test question paper. With three other options, it was asked whether this reaction is feasible or not?
I googled it and found the reverse reaction of at every webpage I looked. So, it seems to me it isn't feasible.
I think that the reverse reaction is feasible because $\ce{LiCl}$ is quite covalent in nature(Fajan's rule) So, for becoming more ionic in nature $\ce{Li+}$ makes bond with oxygen ion. (I now think it's a bogus reasoning)
What could be the reason behind it? Is it feasible?

Although every reaction is reversible in nature, let's try to understand where the equilibrium might be biased, i.e. whether the reactant side favoured or the product side.
A rule of thumb while predicting feasibility is to remember, a reaction where you can get a weak acid from a strong acid is feasible.
$$\ce{Strong Acid -> Weak acid}$$
In your case we're dealing with $\ce{KCl}$ and $\ce{LiCl}$. The acidic or basic nature or salts is determined by the extent of hydrolysis of its ions. Ions from a weaker species undergo hydrolysis while ions from a stronger species don't. Another way to look at this is to compare the acidic and basic strengths of the acids and bases which formed them in the first place.
$$\ce{LiOH + HCl -> LiCl + H2O}$$
$$\ce{KOH + HCl -> KCl + H2O}$$
Notice that $\ce{HCl}$ is common to both of these species, so we have to just compare the basic strengths of $\ce{LiOH}$ and $\ce{KOH}$. Down the group (in alkali metal hydroxides) basic strength increases. So $\ce{KOH}$ is a better base. And thus $\ce{KCl}$ is a weaker acid. Since we're making a stronger acid from a weaker acid, the reaction can be termed as unfeasible.


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Reactivity of phenols with carbonates
Why do phenols not react with carbonates? I read that stability and position of equilibrium have a part to play, but I can't seem to elucidate a rigorous explanation.

Related: https://chemistry.stackexchange.com/questions/67382/why-is-phenol-soluble-in-sodium-carbonate-but-not-in-sodium-bicarbonate

Carbonic Acid ($\ce{H2CO3}$) and its conjugate $\ce{HCO3-}$ have a pKa of 6–7. So bicarbonate can deprotonate functional groups with a pKa < 8, but not those with a pKa > 8. The $\ce{HCO3-}$ ion is not strong enough to pull the hydrogen out of Phenol.
Phenols have a range of pKa which lies between 8 to 14, on average 10. So Phenols don't react with $\ce{NaHCO3}$ despite being an acid.

$\ce{HCO3-}$ has a pKa of 10.3, according to Wikipedia.

Organic carbonates such as dimethyl carbonate, or inorganic such as sodium carbonate? Heat phenol with the latter and pKa will not matter so much as the equilibrium will be driven to the right through CO2 release


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Why left and right circularly polarized light have different index of refraction in chiral media?
This is a followup question of (Dependence of the angle of rotation on the wavelength of plane polarized light).
Ron's answer tells about the difference between left and right circularly polarized light. But the explanation isn't in the answer.
So, what's the reason behind different index of refraction?

The answer to this is really a combination of the two answers given in the question you linked.
Namely, chiral molecules are unique in that they make a distinction between left and right circularly polarized light.
As porphyrin pointed out, the index of refraction is dependent on the ability of a dipole to be induced in a molecule by electromagnetic radiation. Linearly polarized light is itself achiral, thus any given wavelength of linearly polarized light has only one magnitude of dipole it will induce in a molecule, all else being the same. Circularly polarized light, on the other hand, is itself chiral. Thus, when we bring two chiral objects together, the molecule and the circularly polarized light, there is no reason at all to suspect the dipole induced in the molecule should be the same in both cases. The symmetry necessary to make such a claim simply is not present.
Thus, if you imagined tabulating the index of refraction for some molecule as a function of wavelength using circularly polarized light, you would need to have two columns. One would be for right-polarization and the other for left-polarization.

Also, because it is relevant to the answer, I will reiterate what Ron was saying in the comments. The photon is an inherently quantum object. It is a boson and thus has two possible values of spin angular momentum, $\pm\hbar$. These two values correspond to two allowable states the photon can be in, one which is polarized to the right and the other which is polarized to the left. That is, the electric field vector spins in a circle one way or the other. There also allowable solutions to the state of a photon which are some linear combination of these two polarization states. A 50/50 linear combination gives linearly polarized light. Thus, it seemed like you might have been picturing two photons whose polarizations were cancelling each other out. This is incorrect. Rather, you have to think of a single photon in a superposition of two polarization states.

"Rather, you have to think of a single photon in a
superposition of two polarization states." I have thought in this way. Not as photon, but the plane polarized light to be superposition of left and right circularly polarized light.


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How do the silicon-hydrogen bonds in polydimethylsiloxane react with a H2PtCl6 catalyst
How does $\ce{H2PtCl6}$ react with $\ce{Si-H}$ to give $\ce{[PtCl4]2-}$?
I know $\ce{H2PtCl6 -> 2H+ + [PtCl6]2-}$ but I'm not sure what is next.
This question comes from this laboratory assignment (PDF link), in which I am asked to write the redox reaction between the silicon-hydrogen bonds in polydimethylsiloxane and the metal compound $\ce{H2PtCl6}$.
I found a source which confirms that $\ce{H2PtCl6}$ reacts with $\ce{Si-H}$ to give $\ce{[PtCl4]2-}$, but I don't understand how the $\ce{Si-H}$ interacts with the $\ce{H2PtCl6}$.

$\rm H_2PtCl_{what?}$

$\ce{H2PtCl6}$ sry typo

OK, then it simply reduces. What else do you want to know? The mechanism may be pretty complicated; it may even be unknown.

hmm i see, because I need to write the redox reaction between the silicon-hydrogen bonds in polydimethylsiloxane and the metal compound
This is the link to the information i have: http://imgur.com/cfsTdh1
and im having trouble doing it

Are there any Si-H bonds in PDMS? I thought there weren't; and the structure at Wikipedia concurs. (checks OP's imgur link) Ah, you meant polymethylsiloxane, that makes more sense. (looks again) Or, well, no -- that figure doesn't make a whole lot of sense to me.

Sry, this is my first time using this site and I don want to seem like im just looking for answers to my question. This is the source http://drgregsmaterialsweb.com/Nanocomposites%20Lab.pdf of my question and all the info i have is in there and im having trouble with the redox reaction question :(

@ycY This doesn't seem like it's entirely your fault -- the lab instructions / assignment seem to have been confusingly written. Please don't take offense at the community activity around your question -- we're trying to help you edit it into a condition that'll make it a good fit for the site. Thanks for working with us!


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Can isotopes of a given element be represented by different symbols?
Can isotopes of any given element be represented using a completely different chemical symbol? What's the IUPAC's take on this?
Sure, ordinarily you would add the isotope's mass as a superscript to the element's symbol to differentiate it from other isotopes: For example, carbon-12 ($\ce{^{12}C}$) and carbon-14 ($\ce{^{14}C}$); however the base-symbol $\ce{C}$, for carbon, doesn't change.
But the isotopes of hydrogen don't seem to follow this strictly. Often, I see deuterium ($\ce{^{2}H}$) and tritium ($\ce{^{3}H}$) represented by $\ce{D}$ (I see this one in organic chem textbooks a lot) and $\ce{T}$ respectively. Does this "convention" fit in with IUPAC norms? If so, can isotopes of other elements be represented differently as well?

IR-3.3.1 Isotopes of an element
The isotopes of an element all bear the same name (but see Section IR-3.3.2) and are
designated by mass numbers (see Section IR-3.2). For example, the atom of atomic number
8 and mass number 18 is named oxygen-18 and has the symbol $\ce{^{18}_{}O}$.
IR-3.3.2 Isotopes of hydrogen
Hydrogen is an exception to the rule in Section IR-3.3.1 in that the three isotopes $\ce{^{1}_{}H}$, $\ce{^{2}_{}H}$ and $\ce{^{3}_{}H}$ can have the alternative names protium, deuterium and tritium, respectively. The symbols D and T may be used for deuterium and tritium but $\ce{^{2}_{}H}$ and $\ce{^{3}_{}H}$ are preferred because
D and T can disturb the alphabetical ordering in formulae (see Section IR-4.5). The
combination of a muon and an electron behaves like a light isotope of hydrogen and is
named muonium, symbol $\ce{Mu}$.⁵
These names give rise to the names proton, deuteron, triton and muon for the cations $\ce{^{1}_{}H+}$, $\ce{^{2}_{}H+}$, $\ce{^{3}_{}H+}$ and $\ce{Mu+}$, respectively. Because the name proton is often used in contradictory senses, i.e. for isotopically pure $\ce{^{1}_{}H+}$ ions on the one hand, and for the naturally occurring undifferentiated isotope mixture on the other, it is recommended that the undifferentiated mixture be designated generally by the name hydron, derived from hydrogen.

Source:
N.G. Connelly, T. Damhus, R.M. Hartshorn, A.T. Hutton (eds) (2005). Nomenclature of Inorganic Chemistry (PDF). RSC–IUPAC. ISBN 0-85404-438-8.

Addendum:
The small subscript ⁵ present in the source is a reference to Names for Muonium and Hydrogen Atoms and Their Ions, W.H. Koppenol, Pure Appl. Chem., 73, 377–379 (2001) which can be viewed over at:
https://www.iupac.org/publications/pac/pdf/2001/pdf/7302x0377.pdf

$\cdots$ A particle consisting of a positive muon and an
electron ($\pu{\mu^+ e^–}$) is named “muonium” and has the symbol $\ce{Mu}$. Examples: “muonium
chloride,” $\ce{MuCl}$, is the equivalent of deuterium chloride $\cdots$

Additional historic information is contained in answers to https://chemistry.stackexchange.com/questions/35462/why-are-isotopes-of-hydrogen-the-only-ones-with-special-names/35509#35509

Muons are negatively charged. Presumably you mean an antimuon?

It's supposed to be an antimuon, https://en.m.wikipedia.org/wiki/Muonium.  @berry please check the original source.

@OscarLanzi Checked, the source reads muonium, also I've added the reference it was linked to.

@E.P. ^please see this addendum

So... you meant to say 'positive muon' instead of 'muon'?

While it is read like "carbon 12", and "carbon 14", etc. the atomic notation of the nuclides is the other way around, i.e. the mass number precedes the element symbol.  Hence it is not $\ce{C^{12}}$, but $\ce{^{12}C}$.
The special names and symbols seen for $\ce{^1H}, \ce{^2H}, \ce{^3H}$ (but not the other isotopes of neither hydrogen, nor the ones of other elements) follow a recommendation by IUPAC, published in Pure and Applied Chemistry in 1988 (doi 10.1351/pac198860071115, open access).  It is still in power, as the corresponding recommendation by 2001 in the same journal shows (doi 10.1351/pac200173020377).

Preceding the symbol with the mass is a relatively modern invention. When I was involved in chemistry (be it 40 years ago) the historic positioning after the symbol was the rule and I see no need to change - it's change for change's sake. If it's good enough for Linus Pauling (College Chemistry, 1964 edition) then it's good enough for anyone. A similar situation occurred in electronics where educational institutions flip-flopped from conventional current flow to electron current & back again, causing confusion that still reigns. Planets asteroids & stars are named systematically - except a few

@Magoo This is indeed a noteworthy hint.  Hence I looked up page 63/700 in the archive (https://archive.org/stream/CollegeChemistry/Pauling-CollegeChemistry#page/n61/mode/2up) in chapter "The structure of the Atomic Nuclei" and found it.  Interesting reading, with figures clearer than in many contemporary (colourful) chemistry books.

@Magoo - with many problems relating to ions and charge states being caused under that convention. I think various sub-fields in physics and chemistry arrived at slightly different conventions in the last century, and we might still be working out the bugs...


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Which is more acidic: Me3CCH2COOH or Me3SiCH2COOH?
Acidic strength of organic acid depends on two factors – inductive and resonance effect.
Now, electron donating groups have +I effect. As a result of which they increase electron density around carbon and adjacent atoms. It results in increased electron density in $\ce{C-O-H}$ bond of acid, making the release of proton difficult. So effect of electron releasing group is that it decreases the acidity of acid.
The trimethyl silicon ($\ce{Me3Si}$) is also electron releasing group but it has lesser donating tendency than trimethyl carbon, $\ce{Me3C}$. Therefore, $\ce{Me3Si}$ inductive effect is lesser than $\ce{Me3C}$ group. Consequently, $\ce{Me3SiCH2COOH}$ is more acidic than $\ce{Me3CCH2COOH}$.
Is this right? Actually, my book writes that $\ce{Me3SiCH2COOH}$ is lesser acidic than $\ce{Me3CCH2COOH}$.

I suppose the electronegavity of C is more than Si and hence C resists a bit in inductive effect than Si


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What is the exact meaning of the term 'isostructural' molecules?
Wikipedia:
Isostructural

Isostructural chemical compounds have similar chemical structures.

Chemical Structure

A chemical structure determination includes a chemist's specifying the molecular geometry and, when feasible and necessary, the electronic structure of the target molecule or other solid. Molecular geometry refers to the spatial arrangement of atoms in a molecule and the chemical bonds that hold the atoms together, and can be represented using structural formulae and by molecular models;[citation needed] complete electronic structure descriptions include specifying the occupation of a molecule's molecular orbitals.

My Doubt
Given that, while practicing, I encountered a question asking whether $\ce{NO3-}$ and $\ce{SO3}$ are isostructural or not. If we go as per the geometry, they are isostructural ($\ce{sp^2}$ hybridization). But if we consider the electronic structure, the former has bond order 4/3, the latter has 2 (assuming $\ce{S=O}$), and so may be, they are not isostructural. (In addition, they also have different total number of electrons.) This (that they are not isostructural) is the answer.
So, this question. Is the term defined or is it used without any strict definition?

Bonding is identical between those two.

I did read somewhere that in $\ce{SO_3}$, the bond order is 4/3, but am unsure of it, as well as unable to find the source. (Does formal charge +2 on S indicates this? Moreover, the diagrams I find over the internet containg three S=O bonds.) Therefore, I did mention "assuming S=O". The answer is given as 'not isostructural'. (The question intends to ask: Is the answer wrong? Can the definitions be worded better? Is there anything more to it?)

Ad question you mention you have encountered.. this type of questions not focused on the problem matter, but rather on vague terminology, and are surely made by evil professors. In the answer I would describe all my train of thought.

Two molecules are isostructural if they have a similar number of atoms arranged in a similar structure (spatial coordinates). In your example, both $\ce{NO3^-}$ and $\ce{SO3}$ have same number of atoms, and they have similar spatial arrangement as well. So they are isostructural. Sometimes using hybridization of the central atom works in finding out, but not always. For example $\ce{SCl3}$ and $\ce{BF3}$ are isostructural, even though the former is $\text{sp}^3\text{d}$, while the latter is $\text{sp}^2$.

Unable to 'simply' google the structure for SCl3; do the lone pairs occupy the axial orbitals? If yes, then is there any explanation why this is against the [atleast-as-I-know-it] usual configuration of lone pairs occupying equatorial position?

Lone pairs occupy axial positions because they're they face least repulsions.

My coursebook gives the same reason for the structure of ClF3, just that the lone pairs are equatorial (because they face less repulsion)! How do these two structures differ then?

$\ce{ClF3}$ isnt isostructural with $\ce{SCl3}$. Central atoms have different number of electrons.

As I understand, there are 2 orbitals containing lone pairs / lone electron.  (Tell me is if I am wrong.) Then why is it that, in one, the lone pairs are equatoria,l and axial in the other. (Or is the generalisation that equatorial lone pairs cause less repulsion wrong?)

In axial position, lone pairs are separated by 180° while in equatorial positions, they are separated by 60°.

Okay I am asking this question. Future readers may refer there.

This is incorrect. (1) $\ce{SCl3}$ does not exist as far as I can tell. (2) In a structure with 3 bp and 2 lp, lone pairs are best placed equatorial, because the total energy is dictated not just by lp-lp repulsions but also lp-bp repulsions. In the T-shaped structure (i.e. both lp equatorial) the lp-lp repulsion is slightly larger, but lp-bp repulsions are minimised.


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Project Ideas Mixing Chemistry with CS
I'm a high-school senior who has just taken the AP Chemistry exam and my teacher has just assigned me an end-of-year project exploring whatever aspect of Chemistry I want. I really enjoy computer programming as well (took the AP last year) and do it a significant amount in my free time, so I was considering finding a Chemistry project that involves some coding or computer modeling (or anything having to do with computers). Could anybody recommend some topics to check out? (I looked at molecular dynamics simulations, but a lot of that stuff contains multivariable calc./quantum mechanics, which might be a bit much).

Computational chemistry is pretty complicated. Even the classical stuff like MD or Force Fields might be too big a bite for your project. (I'll think about it a bit and maybe write an answer later)

As far as programming goes, Monte Carlo algorithms use fairly simple codes. Perhaps someone could design simple studies of finding equilibrium bond lengths in different diatomic molecules using a very simple force field. In such molecules, the only geometrical parameter to vary is the internuclear separation, so a Lennard-Jones potential could be used to find the most stable internuclear distance. This should be easily done in matlab, but I suppose you could use a range of different languages.

This seems like a "shopping question", which aren't a good fit for this site. If you have specific questions on computational chemistry, we can possibly answer those, but solicitations of a list of suggestions here is a little too broad.

If you want to learn something about quantum mechanics, simple model systems one could play around with are the Particle-in-a-Box or Hückel MO Theory. Nice applications are always calculating HOMO-LUMO gaps for dyes, although they do not perform very good.

R.M. is right, since there is no correct answer to this. But I think it's great that you want to learn more about computational chemistry! I'd be happy to be of whatever help I may be in brainstorming via e-mail or something.

Regarding MD, you can go pretty far with just a simple Verlet algorithm.

It's probably best in your case to use an already existing software and show a variety of results from the program with some description of how it works and what the results mean. As a shameless plug I'll share with you the code I've been working on for ~4 years in grad school, MCMD. It works on Mac and Linux; not tested on Windows. If you have questions about how to use and interpreting results feel free to get in touch. https://github.com/khavernathy/mcmd

Thanks for the suggestions — would it be possible to write a really simple modeler where you start with two atoms/ions (can customize each to any of a number of elements) and start a timer which shows how they interact and, if they attract, the lowest potential energy state/distance they come to?

These exercises (written in Latex)  take you through writing a simple MD program.  There's some python intro at the beginning of each exercise that you can skip https://github.com/jensengroup/molstat/tree/master/handout


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Which nuclei are NMR active?
I thought that only way a nucleus can be NMR active is when the atom has an odd mass, which means that there is an odd number of protons or neutrons and an even number of the other particle. I thought that for an atom with even mass, the number of neutrons and protons are equal, and therefore the magnetic moments would cancel.
Now I reading that if both the protons and neutrons are odd numbered (giving even total mass) the nucleus is still NMR active. What is the reason for this?

NMR usefulness is also determined by the resulting effective magnetic moment and isotope abundance.

The spin of atomic nuclei can be predicted based on the knowing the number of protons and neutrons in the nucleus:

Even/Even. Nuclei that contain an even numbers of protons and an even number of neutrons have $I = 0$ and are NMR silent. Examples include $\ce{^{12}C}$, $\ce{^{16}O}$ and $\ce{^{32}S}$.

All other nuclei are NMR active:

Odd/Odd. Nuclei that contain an odd number of protons and an odd number of neutrons have I that are positive integers. Examples include $\ce{^2H}$ $(I=1)$, $\ce{^{14}N}$ $(I=1)$ and $\ce{^{10}B}$ $(I=3)$.
Odd/Even & Even/Odd. All other nuclei (odd/even and even/odd) have spins that are half integral. Examples include $\ce{^1H}$ $(I=1/2)$, $\ce{^{11}B}$ $(I=3/2)$, $\ce{^{13}C}$ $(I=1/2)$, $\ce{^{17}O}$ $(I=5/2)$, $\ce{^{19}F}$ $(I=1/2)$ and $\ce{^{31}P}$ $(I=1/2)$.

A precise formulation is far more complex, and arises from the fact that the key components of the nucleus (proton and neutron) are made up of quarks. The neutron, although neutral in charge, is actually made of 3 quarks (udd), which themselves are charged, and not uniformly distributed about the neutron 'sphere'. Protons also consist of 3 quarks (duu). The nuclear spin comes from the total angular momentum of all these quark components. To delve deeper is probably outside the realm of chemistry, and is probably better suited to a physics forum (at least it is outside my ability to explain).

Okay. So it not as simple as 1 neutron canceling magnetic moment of 1 proton, so that odd/odd or even/even is nmr inactive right? I used to think this is the case because neutron and prton have same magnetic moment. So just like plus and minus charge they cancel each other out.


text:
How do thiol groups act as reducing agents?
In my biochemistry practicals, we used reducing agents such as beta mercaptoethanol:

and dithiothreitol (DTT)

Both of these have S-H groups, and I am sure that these are involved in the reduction process. However I am not sure as to how. I know that when you have two S-H groups close to one another in cysteine residues in amino acids that form a disulfide bridge, which must be oxidation because I know they are reduced in the reducing environment of the cytosol (hence are not found in cytosolic proteins, in general):

The only way I can think of that S-H groups act as reducing agents is when they form disulfide bridges between them, and an electron (and H+ ion) is released. Is this what happens? If so, it would result in two molecules of the beta mercaptoethanol becoming bonded together, and with the DTT it could either form a long chain with S-H groups between molecules reacting, or it could form rings of any size in which the end S-H groups form a disulfide bridge. I think this should result in the solution becoming viscous and 'gooey' due to the long chains and larger molecules.
Is this how S-H groups can act as reducing agents?

I guess you're more or less right.


text:
Drawing the condensed structure of ethylcyclobutane
The IUPAC name for this is ethylcyclobutane. I know that the formula is $\ce{C6H12}$.
I have no problems drawing the skeletal structure, but I have a little difficulty drawing the condensed structure. I know this is basic, but I just started learning how to name them.
I tried it below. For $\ce{CHCH2CH3}$, am I right to write it in a straight chain?

You don't  draw condensed structures for rings. Draw the ring like a square and condensed the side chain.

Related: What is the meaning of parentheses in a condensed molecular formula?

$\ce{CH3CH2CH(CH2)3}$ is the best you can do.
If you want to draw the cycle, then you should make it pendant, drwing a bond from the ring to the ethyl group.

There is no exact definition, what condensed formula means. One might be able to condense the formula up to the following

UPDATE

Note that I use nonrecommended parentheses type for repeating unit (see e.g. here),
and see Andselisk's answer for the graphical bond indicator recommendation (in a different branch of chemistry, though).

Is it possible to condense it like $\ce{(C4H7)C2H5}$

@PrittBalagopal, (C₄H₇)C₂H₅ formula is not bad. It is not totally unambiguous, but other compounds it might mean, like "ethylbutene" would have different more appropriate condensed formula.

TL; DR: In order to unambiguously denote both branching and cyclicity in a condensed structural formula, a linear formula should include a notation for connectivity, like such:

There is a special notation (external linker, like in mykhal's answer) for the bridging atoms with connectivities of three and higher in IUPAC's Nomenclature of Inorganic Chemistry [1, pp. 26–27]:

IR-2.3.4 Special bond indicators for line formulae
The structural symbols
  $\require{HTML} \style{display: inline-block; transform: rotate(90deg); font-size: 3em}{\Large ~~~[~~}$
  and
  $\require{HTML} \style{display: inline-block; transform: rotate(-90deg); font-size: 3em}{\Large ~~[~~~~}$
  may be used in line formulae to indicate bonds between non-adjacent atom symbols.
Examples:

With this in mind, I'd like to propose the following condensed formula for ethylcyclobutane:

I used a dirty hack by utilizing a redox macros from the $\mathrm{\LaTeX}$ package chemformula without arrows which are supposed to denote electron flow:
\documentclass{article}
\usepackage{chemmacros}
    \chemsetup{
        modules = {all},
    }

\begin{document}

\ch{\OX{a,C}H(CH2CH2\OX{b,C}\redox(a,b)H2)CH2CH3}

\end{document}

References

IUPAC “Red Book” Nomenclature of Inorganic Chemistry, 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.

Nice catch. Unfortunately no such bond indicators are defined in IUPAC recommendations for organic chemistry. I guess if they were, maybe the slightly confusing parentheses would be optional for such simple cases.

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


text:
Why basicity of oxides increases down a group?
Take group $13$ oxides as an example. The elements in this group are $\ce{B,Al,Ga,In,Tl}$.
Boron oxide $\ce{B2O3}$ is acidic and $\ce{Al2O3}$ and $\ce{GaO3}$ are amphoteric while $\ce{In2O3}$ and $\ce{Tl2O3}$ are basic in nature.
What is the reason for this?

Metallic character increases down the group.

But are they really basic? I mean the last two. Aren't they amphoteric? I think they are less acidic because the electronegativity gets lower as you get lower down the group.

As suggested in comments, major factors play a part

Atomic size

As the atomic size of the element increases, its electron affinity decreases consequently. It turns out that when moving vertically in the periodic table, the size of the atom outdoes its electronegativity with regard to basicity. The atomic radius of gallium, for example is approximately twice that of boron, so in a gallium ion, the charge is spread out over a significantly larger volume.
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.

Electron affinity

The ability of an atom “to accept an electron” decreases down the group therefore increases the tendency of the elements to be good Lewis bases (recall a Lewis base is an electron-pair donor). Thus the decrease in electron affinity down the group means the elements readily donate an electron rather than accept them.

Electronegativity

The more electronegative an atom, the better it is able to bear a negative charge. Down the group, the electronegativity of the elements decrease down the group.  B is very electronegative and doesn’t readily release electrons but instead holds on to it, additionally is able to stabilise a negative charge, and consequently has good acidic properties unlike proceeding elements down the group.

Can you tell how polarizability and basicity are related?


text:
How do I name this alkane with a branch that has bromine?

I identified the orange portion as the longest carbon chain which is the first step.
Next, I can't call it 3-ethylhexane as the bromine is a branch from a methyl branch.
This is the problem I am facing. Since the methyl branch is on the 3rd carbon, and bromine branch is in the "branched" 2nd.
So, is it called as 3-(2-bromomethylhexane)? Or have I got it wrong?

3-(bromomethyl)hexane
You can't have a 2-position on a methyl group, there is only a 1-position and we omit redundant 1-enumerations of position.

This type of formulae can be confusing. Might be helpful to redraw it as $$\ce{CH3(CH2)2CH(CH2CH3)CH2Br}$$ or $$\ce{CH3CH2CH2CH(CH2Br)CH2CH3}$$ There is only one carbon atom in methyl, so 2-bromomethyl is nonsense. The locant means on which carbon atom is the functional group attached. There exists a type of nomenclature, where the locants designate the position, where the carbon atom was replaced, it would be theoretically 2-bromaethyl, but it is never used for halogen compounds; sometimes is for compounds containing oxygen, nitrogen, ... (oxa-, aza-, ...).
When there is only one possibility where to place the the substituent, it is possible to exclude the locant, so 1-bromomethyl becomes just bromomethyl.
The composed substituent name must be in parentheses so the final name is 3-(bromomethyl)hexane, as Lighthart has answered. Without them, there might be confusion whether it might mean (though not completely defined) 3-bromo(methylhexane).
UPDATE: Why the correct name is NOT 1-bromo-2-ethylpentane? There is no principal functional group with suffix, the name is composed using substitutive nomenclature, which is the preferred naming method for halogenated hydrogarbons. The principal chain is just the longest one. Bromo group then has no priority over alkyl substituent (besides alphabetic order).

The IUPAC name for this compound would be 3[ 1' - bromomethyl] - hexane
//Edit:
Where 1' refers to the carbon in the methyl group attached to the 3rd carbon in the primary chain.
Also if you had an ethyl group attached and if the Bromine was attached to the 1st or 2nd carbon in that chain then it would be 1' -bromoethyl.... And 2' -bromoethyl.... . Try drawing this structure of this compound.
3-[1'-bromo - 2'- chloro- 3' - flouro- 4' - [1"- methyl ethyl] hexyl] hex- 3 - ene.
Apply the same concept, i.e., the carbons in the longest chain of the substituent group will be numbered 1', 2', 3' and so on and the carbons in substituent group of the substituent group would be 1", 2", 3" and so on.

This info is unfortunately wrong. The apostrophe(like) symbol is sometimes used, e.g. in biphenyl, symmetric systems (oxydi...), but never in such a case.

Mykhal you are wrong. I don't know where you are getting your information but this is the correct way of naming such compounds. But I was also wrong  when I answered the question.

now I am loosing you. I am getting the info from official IUPAC books, and naming software. Please edit your answer and cite a rule that would prove you right.

Please refrain from vandalisation of your own posts, and from language that does not comply with our be nice promises. Consider this a warning.


text:
When is a ligand strong field or weak field?
I have seen many examples where a strong field ligand is written for a particular ion .
For example, we have $\ce{Co(H2O)6^3+}$. What is the hybridization of $\ce{Co^{3+}}$? Now, we know $\ce{H2O}$ comes later in the spectrochemical series, thus it should be considered a weak field ligand. But for $\ce{Co^3+}$, $\ce{H2O}$ acts as a strong field ligand and pairs up the electron causing the hybridization to be d2sp3. This doesn't make sense. If we can see that a ligand is weak with respect to the spectrochemical series and it turns out that its not just for a particular ion, why is it so?
Also, another example is where $\ce{NH3}$ is supposed to be a strong field ligand but comes out to be a weak field ligand just for a particular ion.
Why does this happen? How do I know when to consider a ligand strong or weak?

https://chemistry.stackexchange.com/a/73027/43942

Instead of asking when is the ligand strong- or weak-field you should be asking whether the complex is high- or low-spin, as there are obviously more factors than the ligand. Ligand X can produce a high-spin complex with metal M whereas it might produce a low-spin complex with N. Do you then call X a strong- or a low-field ligand? It's ambiguous - as for the reasons, a typical undergrad inorganic chemistry textbook will describe them. Also you can't get a good answer by using hybridisation or CFT - a proper rationalisation will need MOT

Nope its d2sp3.


text:
Adding ethanol to louche
I'm watching MIT chemistry by Donald Sadoway. In one of his lectures devoted to solutions and phase separation, he performs experiments with absinthe.
First he mixes absinthe with 5 $\times$ water which turns into milky louche.
Then he adds some cognac and the mixture becomes transparent again. I don't grasp his explanation of this.
Here is a timestamped link to the video.
I'll also briefly summarize his argument below:

... and why did he add cognac? ... it's cause if you have a fat phase here and you've got an aqueous phase here and if you add alcohol, you got
$(fat)\:\textrm{CH}_3 \textrm{CH}_2 \textrm{COOH} \: \dot{}\dot{}\:\textrm{OH}_2$
... this can bond to the water by hydrogen bond and this aliphatic tail can stab the fat and bring them into solution. This is why you have these recipes...

What does "stabbing the fat" mean?
Does it mean that the ethanol somehow makes the surface between fat and water disintegrate thus turning the mixture into a single phase solution?
Or maybe ethanol + water + fat somehow  combine to form a single molecule so that, again, there is a single phase solution now?

Absinthe is a strongly alcoholic beverage containing anethole, which is insoluble in water and very soluble in ethyl alcohol.  When water is added to the absinthe, the alcohol becomes too dilute to keep the anethole dissolved, and it appears milky. When more ethyl alcohol is added in the form of cognac, the solution again becomes strong enough in alcohol to dissolve the anethole and the solution becomes clear again.
The phrase "stab the fat" is a bit odd, but I'll do my best to explain. Anethole is a fat-like or fat-loving oil, meaning it dissolves well in other fatty sorts of compounds, but not in water (which is lipohpobic, or fat-avoiding/fat-fearing). Ethyl alcohol has two parts to it. One end of it (labeled (fat) in the chemical formula in the question) is lipophilic (attracted to fat) and the other end is hydrophilic (attracted to water). If there is enough alcohol in the mixture, the lipophilic end of the alcohol molecules can surround the anethole ("stabbing the fat") which leaves the other end of the alcohol free to be dissolved in the water (and also any excess hydrophilic portions of the alcohol).
I hope the "stab the fat" explanation was clear. Don't hesitate to ask for specific clarifications in the comments below.

Thanks for the answer! I'd like to clarify a bit about dissolving. As I understand, the milky colour of the original mixture is caused by the light scattered from the fat-phase/aqueous-phase boundary. So after more cognac is introduced into the solution does this boundary disappear? and we get a single phase without separation?

To make a kind of illustration to my question: in the original mixture we have [... w w w |f f f| w w w w w |f f f f| ...] After the cognac is added to the solution we get [... wef wef wef wef ...] here w stands for water molecule, f stands for fats (whichever are there), e for ethanol and | represents a phase boundary. Is this correct?

First off, of course you are welcome ;) You're correct about the cause of the milky color, but we need to clarify what the milk is. In this case, it is a suspension of macrosopic droplets of anethole. As anethole happens to have nearly the exact density of water, it's tendency to settle out is very small. As the solution as a whole (water + alcohol) has a density that is less than anethole, I suspect that eventually there would be a layer of anethole on the bottom, with the alcohol/water layer on top. My point is, the anethole is not dissolved at all when you see the milky stuff (to be cont.).

So, when there is not enough alcohol to surround the anethole with it's lipophilic tails (stabbing the fat) there is truly a phase boudary. Once there is enough alcohol to surround each anethole molecule, it becomes dissolved, there is just a single phase, and it appears clear. I think the point of confusion might be one of scale. When the anethole is dissolved, we're talking about each individual molecule being surrounded by a layer of the alcohol. When it's milky (undissolved) the droplets are trillions of trillions of molecules in size, just holding onto each other to escape the solvent.


text:
Why does 2-chloropropane's mass spectrum have a peak at 27?
I have noticed that all halogenoalkanes of the form 2-Xpropane seem to have a mass spec peak at 27, implying that a $\ce{C2H3+}$ ion is being formed.
Forming a $\ce{C2H3+}$ ion, however, would mean breaking three bonds and, given its abundance, that wouldn't make sense - especially in the case of 2-chloropropane where it has an abundance of 25.36 according to this
 website.
Why exactly does the $\ce{C2H3+}$ ion form?
EDIT: You can also see this peak in the mass spectrum of 2-bromopropane, 2-iodopropane and 2-fluoropropane, but it is less pronounced.
EDIT 2: as pointed out by @PLD, it appears that other molecules of this form have peaks at 27 for a different reason to that of 2-chloropropane, so my previous edit is irrelevant.

Maybe alpha cleavage with H2 elimination or whatever; can't see unlikeliness in here.

Nice question. I am not sure that 2-bromopropane and 2-iodopropane m/z 27 share the same originating process though. Referring to Rosenstock et al. (http://pubs.acs.org/doi/pdf/10.1021/ja00373a001) and Ikuta et al. ( http://www.journal.csj.jp/doi/pdf/10.1246/bcsj.49.66) the m/z 27 ion is described as a fragment of either m/z 41 (allyl cation) or m/z 29 (ethyl cation), but neither of these ions do appear in the fragmentation spectrum of 2-Chloropropane, nor do these two articles reference 2-chloropropane. I did not find any studies specifically on this issue.

The m/z 42, 41, 39, 27 pattern looks like that of the EI spectrum of propene / cyclopropane. This suggests loss of HCl as first step, followed by fragmentation of m/z 42. Why so abundant?  According to Maccoll & Mature (http://onlinelibrary.wiley.com.proxy.scd.u-psud.fr/doi/10.1002/oms.1210150912/pdf) : the appearance energy for m/z 42 (ionized cyclopropane or propene) from 1-chloropropane is  lower (about 2.7 eV) than that for 2-chloropropane. Thus the m/z 42 ion would be formed with much higher energy, leading to more fragmentation in the latter case. I am not completely convinced though.

Here is a little mechanism I made up about it. Not quite sure if its correct, seems to fit the mass spectrum though.

Definitely have a look at McLafferty classical textbook. Maybe you will find the answer there.

I guess that the ethene cation is higher in energy than the allyl cation. But maybe, it is just about in range to be formed in equilibrium, and as the neutral methylene is removed from the equilibrium, the reaction is effectively drawn towards the side of the ethene cation. Maybe two methylene units even recombine to ethene, though I am not convinced that this could be possible in the HV of the mass spectrometer.


text:
Why is nitrous acid prepared in situ?

Why is nitrous acid prepared in situ?

I am not sure about the above question. First of all, I don't exactly understand what is meant by "in situ". Secondly, how would preparing nitrous acid in situ, in the diazotisation of amines with nitrous acid, make it less toxic or unstable?

When you're completely lost, Wikipedia can give you a good start on an amazing array of topics. In situ, with respect to chemistry, nitrous acid, preparation. These don't answer your questions (especially the HNO2 one), but they're a good place to start.

Well, if you don't make it in situ, you would have to add the nitrous acid you have already kept in your laboratory shelf, and surprise! It's not there! Because, it has already decomposed.
Nitrous acid ($\ce{HNO2}$) has an intermediate oxidation state of +3, and hence, it tends to disproportionate into more stable oxidation states:
$$\ce{2HNO2 -> NO2 + NO + H2O}$$
$\ce{NO2}$ itself dissolves in water to give $\ce{HNO3}$ and $\ce{HNO2}$. This gives an overall reaction in aqueous solution as:
$$\ce{3HNO2 -> HNO3 + 2NO + H2O}$$
This should give you a good idea on why we make nitrous acid within the reaction mixture. You can't store it beforehand. It must be made when it's time to use it.

Yes, HNO2 is unstable, however, per Wikipedia the gaseous anhydride N2O3 is described as a:

deep blue solid[1] is one of the simple nitrogen oxides. It forms upon mixing equal parts of nitric oxide and nitrogen dioxide and cooling the mixture below −21 °C (−6 °F):[2]
$\ce{NO + NO2 ⇌ N2O3}$
Dinitrogen trioxide is only isolable at low temperatures, i.e. in the liquid and solid phases. At higher temperatures the equilibrium favors the constituent gases...

And further:

Nitrite salts are sometimes produced by adding N2O3 to solutions of bases:
$\ce{N2O3 + 2 NaOH → 2 NaNO2 + H2O}$

So, equivalently, one may be able to employ stable stored cold N2O3 and not work with in situ formed nitrous acid.
So, on the question, "Why is nitrous acid prepared in situ?", one possible answer is that it is actually an elective (not to employ N2O3), and not always a requirement.

You would have to chill the N2O3 as you say, including purchasing it and having it transported chilled.  It is much easier logistically and economically to use the in-situ route.

In the language of chemistry,in situ means The reaction mixture.Here,in case of HNO2,it is too unstable.Here,if we see the reaction,
HNO2→HNO3+NO+H2O
The oxidation no of central atom N:
in HNO2-+3
in HNO3-+5 (oxidation,as oxidation no increased)
in NO-+2(reduction,as oxidation no decreased)
If we observe,we can see that,here at the same time, the oxidation and reduction of N occurs.When in any compound,the oxidation and reduction of any element occurs at the same time,we call it Disproportionate reaction instead of oxidation-reduction reaction.
HNO2 has an intermediate oxidation state of +3.Thats why,it tends to be disproportionate into more stable oxidation state.
Instead of this,to prepare  Nitrous acid,we take the mixture of NaNO2 and HCl.
NaNO2+HCl→NaCl+HNO2
Through this process,HNO2 is prepared in the labratory in need

No, this is not correct. Firstly, disproportionation reactions are still redox reactions, much like how a Toyota is still a car. You don't stop calling it a car because it is a Toyota. Secondly, "intermediate oxidation states" are not always unstable. If that was the case, then $\ce{N2}$ itself should not be stable, as it should then (given appropriate amounts of water) disproportionate into $\ce{NH3}$ (OS -3) and $\ce{HNO3}$ (OS +5).


text:
Application of the Woodward-Hoffman rules to a [14+2] cycloaddition
I am just doing some pericyclic practice problems and I am just trying to understand the Woodward Hoffman rules.
The attached picture is a problem I attempted that was checked against the answers but I am confused how the symmetry is allowed.
When I add the (4n + 2) and 4n together I get 4. Doesn't n have to be an integer and the sum an odd number for the symmetry to be allowed?
EDIT: I copied the answer below with incorrect stereochemistry. Please refer to the second image for the correct stereochemistry

EDIT: Correct stereochemistry below

General rules for cycloadditions
Before getting into the actual Woodward Hoffman 'analysis', one can quickly check any proposed cycloaddition reaction against the following table of general results.
The table is essentially an overview of the Woodward Hoffman rules that doesn't require any consideration of orbitals/orbital symmetry. It can often be useful to check against a table like this to ensure you've done the real Woodward Hoffman analysis correctly.
\begin{array}{ccc} \hline
{\text{Number of }\\\text{Electron Pairs}} &
  \text{Photochemical} &
  \text{Thermal}\\[2ex]\hline
\text{even}~(4n)    &
  \text{suprafacial-suprafacial}  &
  \text{suprafacial-antarafacial} \\[1ex]
\text{odd}~(4n + 2) &
  \text{suprafacial-antarafacial} &
  \text{suprafacial-suprafacial}  \\ \hline
\end{array}
From your example, we can count $14 + 2$ electrons in the π system, giving a total of $16$ which is a $4n$ integer (where $n = 3$). We also know from the reaction scheme given that the cycloaddition is taking place under photochemical conditions. We can, therefore, conclude that the reaction is allowed photochemically if both components react suprafacially.
The actual Woodward Hoffman analysis
Note: I'm doing this for the thermal $[14 + 2]$, firstly because its a little more interesting in terms of orbital overlap, but secondly to give you the opportunity to do the analysis properly for your problem. The approach taken here is the one taken by Ian Fleming in his Molecular Orbitals and Organic Chemical Reactions book.
First: Draw a diagram of the reaction showing the orbitals involved. In this case we are interested in the π system of the heptafulvarene and the π system of the tetracyanoethene.

When drawing a diagram, no orbital phases are needed- this is already taken into account by the generalised Woodward Hoffman rule. (Also note that its not important how you define the components, the 14π system could be considered as a series of seven 2π systems).
Second: Consider how each component is reacting. In a cycloaddition, each component may either react suprafacially (the reaction takes place on the same side of the π system) or antarafacially (the reaction takes place from opposite sides of the π system) (also note that although this is theoretically possible, it is geometrically impossible in some cases due to the inability to get into the correct orientation for orbital overlap).
Third: Apply the Woodward Hoffman rule, as given below.

A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is odd.

In the example above, the 14π component is taken to be reacting antarafacially (denoted π14a), and the 2π component suprafacially (denoted π2s). When the Woodward-Hoffman rule is applied, this gives an odd number of components, making the reaction thermally allowed.
We could, however, also have drawn the reaction as having the 14π component reacting suprafacially (denoted π14s), with side on overlap of the 2π component reacting antarafacially (denoted π2a). Both are equally valid.
From that,you should now be able to consider your photochemical reaction.  One slight difference exists in the formulation of the Woodward-Hoffman rules for photochemical reactions, taking account of the excited state that a component exists in.  Namely, the total is now even, not odd, for an allowed reaction.

A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is even.

Further reading
All of the diagrams were taken from Fleming's Molecular Orbitals and Organic Chemical Reactions, student edition, 2009. The book contains many more examples and a slightly more rigorous explanation of the Woodward Hoffman rules.


text:
What is the mathematical basis behind the Jahn-Teller effect?
Both first-order and second-order Jahn-Teller distortions play a very important role in chemistry.
It is often said that the Jahn-Teller effect is based on symmetry arguments, and hence nothing can be said about the extent or the exact form of the distortion.
What are these "symmetry arguments", and how can they be derived?

1. Perturbations
As already mentioned, the Jahn–Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent of distortion". There is one condition on the vibrational mode: it cannot transform as the totally symmetric irreducible representation of the molecular point group, as such a vibrational mode cannot bring about any distortion in the molecular geometry (it may lead to a change in equilibrium bond length, but not in the shape of the molecule).
$\require{begingroup} \begingroup
\newcommand{\En}[1]{E_n^{(#1)}}
\newcommand{\ket}[1]{| #1 \rangle}
\newcommand{\n}[1]{n^{(#1)}}
\newcommand{\md}[0]{\mathrm{d}}
\newcommand{\odiff}[2]{\frac{\md #1}{\md #2}}$
In the undistorted geometry (i.e. $q = 0$), the electronic Hamiltonian is denoted $H_0$. The corresponding unperturbed electronic wavefunction is $\ket{\n{0}}$, and the electronic energy is $\En{0}$. We therefore have
$$H_0 \ket{\n{0}} = \En{0}\ket{\n{0}} \tag{1}$$
Here, the Hamiltonian, wavefunction, and energy are all functions of $q$. We can expand them as Taylor series about $q = 0$:
$$\begin{align}
H &= H_0 + q \left(\odiff{H}{q}\right) + \frac{q^2}{2}\left(\frac{\md^2 H}{\md q^2}\right) + \cdots \tag{2} \\
\ket{n} &= \ket{\n{0}} + q\ket{\n{1}} + \frac{q^2}{2}\ket{\n{2}} + \cdots \tag{3} \\
E_n &= \En{0} + q\En{1} + \frac{q^2}{2}\En{2} + \cdots \tag{4}
\end{align}$$
In the new geometry (i.e. $q \neq 0$), the Schrodinger equation must still be obeyed and therefore
$$H\ket{n} = E_n \ket{n} \tag{5}$$
By substituting in equations $(2)$ through $(4)$ into equation $(5)$, one can compare coefficients of $q$ to reach the results:
$$\begin{align}
\En{1} &= \left< \n{0} \middle| \odiff{H}{q} \middle| \n{0} \right> \tag{6} \\
\En{2} &= \left< \n{0} \middle| \frac{\md^2 H}{\md q^2} \middle| \n{0} \right> + 2\sum_{m \neq n}\frac{\left|\left<m^{(0)} \middle|(\md H/\md q)\middle|\n{0} \right>\right|^2}{\En{0} - E_m^{(0)}} \tag{7}
\end{align}$$
The derivation of equations $(6)$ and $(7)$ will not be discussed further here.1
Distortions that arise due to the $\En{1}$ term are called first-order Jahn–Teller distortions, and distortions that arise from the $\En{2}$ term are called second-order Jahn–Teller distortions.

2. The first-order Jahn–Teller effect
Recall that
$$E_n = \En{0} + \color{red}{q\En{1}} + \cdots \tag{8}$$
Therefore, if $\En{1} > 0$, then stabilisation may be attained with a negative value of $q$; if $\En{1} < 0$, then stabilisation may be attained with a positive value of $q$. These simply represent distortions in opposite directions along a vibrational coordinate. A well-known example is the distortion of octahedral $\ce{Cu^2+}$: there are two possible choices, one involving axial compression, and one involving axial elongation. These two distortions arise from movement along the same vibrational coordinate, except that one has $q > 0$ and the other has $q < 0$.
In order for there to be a first-order Jahn–Teller distortion, we therefore require that
$$\En{1} = \left<\n{0}|(\md H/\md q)| \n{0}\right> \neq 0 \tag{9}$$
Within group theory, the condition for the integral to be nonzero is that the integrand must contain a component that transforms as the totally symmetric irreducible representation (TSIR). Mathematically,
$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_{(\md H/\md q)} \otimes \Gamma_n \tag{10}$$
We can simplify this slightly by noting that the Hamiltonian, $H$, itself transforms as the TSIR. Therefore, $\md H/\md q$ transforms as $\Gamma_q$, and the requirement is that
$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_q \otimes \Gamma_n \tag{11}$$
In all point groups, for any non-degenerate irrep $\Gamma_n$, we find that $\Gamma_n \otimes \Gamma_n = \Gamma_{\text{TSIR}}$. Therefore, if $\Gamma_n$ is non-degenerate, then
$$\Gamma_n \otimes \Gamma_q \otimes \Gamma_n = \Gamma_q \neq \Gamma_{\text{TSIR}} \tag{12}$$
and the molecule is stable against a first-order Jahn–Teller distortion. Therefore, all closed-shell molecules ($\Gamma_n = \Gamma_{\text{TSIR}}$) do not undergo first-order Jahn–Teller distortions.
However, what will happen if $\Gamma_n$ is degenerate? Now, the product $\Gamma_n \otimes \Gamma_n$ contains other irreps apart from the TSIR.2 If the molecule possesses a vibrational mode that transforms as one of these irreps, then the direct product $\Gamma_n \otimes \Gamma_q \otimes \Gamma_n$ will contain the TSIR.
In a rather inelegant article,3 Hermann Jahn and Edward Teller worked out the direct products for every important point group and found that:

stability and degeneracy are not possible simultaneously unless the molecule is a linear one...

In other words, if a non-linear molecule has a degenerate ground state, then it is susceptible towards a (first-order) Jahn–Teller distortion.
Take, for example, octahedral $\ce{Cu^2+}$. This has a $\mathrm{^2E_g}$ term symbol (see this question) - which is doubly degenerate. The symmetric direct product $\mathrm{E_g \otimes E_g = A_{1g} \oplus E_g}$. Therefore, if we have a vibrational mode of symmetry $\mathrm{E_g}$, then distortion along this vibrational coordinate will occur to give a more stable compound. Recall that the vibrational mode cannot transform as the TSIR, so we can neglect the $\mathrm{A_{1g}}$ term.
What does an $\mathrm{e_g}$ vibrational mode look like? Here is a diagram:4

It's an axial elongation, which happens to match what we know of Cu(II). However, there is a catch. The vibrational mode is doubly degenerate (the other $\mathrm{e_g}$ mode is not shown), and any linear combination of these two degenerate vibrational modes also transforms as $\mathrm{e_g}$. Therefore, the exact form of the distortion can be any linear combination of these two degenerate modes. It can also involve negative coefficients, i.e. it might feature axial compression instead of elongation; there is no way to find that out using arguments purely based on symmetry.
Therefore, in this case, it is simply a coincidence that the $\mathrm{e_g}$ mode displayed is exactly the same as the form of the distortion seen in Cu(II). Nevertheless, it is encouraging to see that the axial elongation indeed transforms as $\mathrm{e_g}$ - which lends credence to the group theoretical analysis above.
On top of that, there's also no indication of how much distortion there is. That depends on (amongst other things) the value of $\En{1}$, and all we have said is that it is nonzero - we have not said how large it is.
This is what is meant by "impossible to predict the extent or the exact form of the distortion".

3. The second-order Jahn–Teller effect
Pearson has written an article on second-order Jahn–Teller effects.5
For the second-order term, the energy correction is of the form
$$E_n = \En{0} + q\En{1} + \color{red}{\frac{q^2}{2}\En{2}} \cdots \tag{13}$$
Here, the $q^2$ term means that $\En{2}$ has to be negative if we want to see a second-order Jahn–Teller distortion. Unlike the first-order case, if $\En{2} > 0$, there will not be any distortion.
The second-order correction to the energy comprises two terms. The first term, $\left<\n{0}|(\md^2 H/\md q^2)|\n{0}\right>$, is always positive. (The proof of this is left to the reader. Hint: use the fact that $\md H/\md q$ is hermitian.) It may be interpreted as a restoring force that tries to bring the nuclei back to their original positions, and it is related to the fact that if the electronic state remains unperturbed (i.e. $\ket{n} = \ket{\n{0}}$), the unperturbed nuclear positions represent the most stable nuclear configuration.
The second term has the form
$$\sum_{m \neq n}\frac{\left|\left<m^{(0)} \middle|(\md H/\md q)\middle|\n{0} \right>\right|^2}{\En{0} - E_m^{(0)}} \tag{14}$$
which may seem like a slight monstrosity, but it is actually much easier to analyse than it looks. The summation over $m$ indicates that we are going to count every single electronic state $\ket{m}$ that is not the ground state $\ket{n}$. Since the numerator is a square modulus, it is either zero or positive, and since $\En{0} < E_m^{(0)}$, the denominator is always negative; therefore, this term is necessarily either zero or negative.
If $\En{2}$ is to be negative, then we need the second term to dominate the first. For this to occur, there are two prerequisites:

The denominator must be small, i.e. $\ket{m}$ must be a low-lying excited state such that the energy gap $\Delta E = \En{0} - E_m^{(0)}$ is small;
The numerator must not be zero, i.e. there must be a low-lying excited state of the appropriate symmetry $\ket{m}$ such that $\Gamma_m \otimes \Gamma_q \otimes \Gamma_n$ contains $\Gamma_{\text{TSIR}}$.

The first condition usually means that it will suffice to consider the first few excited states. In many of the examples I know of, the excited state that mixes with the ground state is the first excited state. In such cases one can even simply get rid of the sum and set $\ket{m}$ to be the first excited state.
These symmetry requirements are much less restrictive than previously, and second-order Jahn–Teller distortions tend to be much more widely seen than first-order distortions. A small selection of compounds in which second-order Jahn–Teller distortions are important are: p-block hydrides, $\ce{PbO}$, $\ce{Hg^2+}$, $\ce{WMe6}$, $\ce{R2Sn=SnR2}$, and anti-aromatic compounds such as cyclobutadiene. Let us use octahedral $\ce{Hg^2+}$ as an illustration.
In undistorted $O_\mathrm{h}$ symmetry, $\ce{Hg^2+}$ has a closed-shell $\mathrm{d^{10}}$ configuration and therefore its electronic ground state is $\Gamma_n = \mathrm{A_{1g}}$. However, upon excitation of one electron from the 5d orbitals (specifically the $\mathrm{e_g}$ set) to the 6s orbital (which transforms as $\mathrm{a_{1g}}$), the term symbol changes to
$$\Gamma_m = \mathrm{E_g \otimes A_{1g} = E_g} \tag{15}$$
Therefore, a vibrational mode transforming as $\mathrm{E_g}$ will facilitate a second-order distortion, since
$$\Gamma_m \otimes \Gamma_q \otimes \Gamma_n = \mathrm{E_g \otimes E_g \otimes A_{1g}} \tag{16}$$
contains the TSIR. Again, there is no way of knowing the exact form or the extent of the distortion; we only know that it transforms as $\mathrm{E_g}$. In the case of Hg(II), the distortion is manifested as an axial compression to give a "2 short, 4 long" coordination geometry, which is often described as "linear".
A second factor that favours the distortion is the extremely small 5d–6s gap in mercury, due to relativistic 5d destabilisation and 6s stabilisation. To see the importance of the small $\Delta E$, consider $\ce{Zn^2+}$, which has a larger 3d–4s gap; linear Zn(II) compounds are rare, while linear Hg(II) compounds are the norm.
Most of the time, the relevant excited state $\ket{m}$ arises from promotion of an electron from the HOMO to the LUMO. It is easy to show that if this is the case,
$$\Gamma_m \otimes \Gamma_n = \Gamma_{\text{HOMO}} \otimes \Gamma_{\text{LUMO}} \tag{17}$$
The second-order Jahn–Teller distortion can then be viewed as a reduction in symmetry, such that the HOMO and the LUMO, which transformed as different irreps in the undistorted geometry, now transform as the same irrep and therefore mix with each other. In the case of $\ce{Hg^2+}$:

This interpretation using the symmetry of individual orbitals, however, only works when the relevant excited state $\ket{m}$ is derived by excitation of an electron! In some (admittedly very rare) cases, it is possible that both $\ket{n}$ and $\ket{m}$ are derived from the same electronic configuration. This is the case for cyclobutadiene, and the Jahn–Teller effect in cyclobutadiene cannot be rationalised using orbital mixing. $\endgroup$

Notes and references
(1) For more details, look up perturbation theory in your quantum mechanics book of choice. In such treatments, the perturbation is usually formulated slightly differently: e.g. $H$ is taken as $H_0 + \lambda V$, and the eigenstates and eigenvalues are expanded as a power series instead of a Taylor series. Notwithstanding that, the principles remain the same.
(2) There is a subtlety in that the symmetric direct product must be taken. For example, in the $D_\mathrm{\infty h}$ point group, we have $\Pi \otimes \Pi = \Sigma^+ + [\Sigma^-] + \Delta$. The antisymmetric direct product $\Sigma^-$ has to be discarded.
(3) Jahn, H. A.; Teller, E. Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy. Proc. R. Soc. A 1937, 161 (905), 220–235. DOI: 10.1098/rspa.1937.0142. n.b. Considering that I don't have a more elegant proof for it, I don't have much of a right to call it inelegant.
(4) Albright, T. A.; Burdett, J. K.; Whangbo, M.-H. Orbital Interactions in Chemistry, 2nd ed.; Wiley: Hoboken, NJ, 2013.
(5) Pearson, R. G. The second-order Jahn–Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.

Excellent answer! Little addition, the famous "unless the molecule is a linear one" only applies for the bending distortion. $D_{\infty \mathrm h}$ to $C_{\infty \mathrm v}$ would be a perfectly possible JT-distortion.

Since the question was about the mathematical basis, the actual mathematical contents of the JT theorem is a group theoretical statement. It says,there is in general no symmetry reason why a degenerate molecular electronic state necessarily should split under some distortion of the molecule in exclusively tangentially touching states.This and the presumably the "simplest" (don't know about elegance) proof is here: V.I.Pupyshev, Int. J. Quant. Chem., 107(2007)1446–1453. In fact the terms can be still all zero for various reasons even special symmetry reasons.


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Colour in vanadyl sulfate
What is the reason that vanadyl sulfate shows blue colour, even in its anhydrous form? If it's d-d transition, can you explain how it works in this compound?

Vanadyl sulfate or $\ce{VOSO4}$ has $\ce{VO^2+}$ ion. It exists as a blue hydrated crystalline salt, known as acid vanadyl sulfate, $\ce{VOSO4 \cdot H2SO4 \cdot xH2O}$. If you're to heat it at $\pu{533 K}$ with concentrated sulfuric acid, you obtain anhydrous vanadyl sulfate, which is rather a greyish-green crystalline powder.¹
The ion $\ce{VO^2+}$ is more accurately written as a complex $\ce{[VO2(H2O)4]+}$.² Inductive effects of ligands and different symmetries of orbitals causes the d-orbitals to split up and thus they become non-degenerate.
The electrons possibly absorb and emit electromagnetic radiation in the visible spectrum. A rough estimate leads me to conclude that this complex would've been absorbing radiations towards the higher-wavelength (possibly red) region of the visible-spectrum thus emitting radiations corresponding to a blueish tone of light.
The colour is possibly due to d-d transition as the central metal ion ($\ce{V+4}$) has one electron in its d-orbital.

References:

B.Sc. Chemistry–II (UGC), R.L. Madan
Chemguide – Vanadium


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LeChatelier's Principle - Phosphofructokinase in Glycolysis
In glycolysis, PFK is allosterically regulated by ATP, F6P, etc.
When the concentration of F6P increases, the concentration of ATP increases too. I understand that high levels of ATP shift the eq towards the T-state, and this decreases the affinity of PFK for F6P.  But is there a principle in Le Chatelier's which explains how the increase in conc of one reactant affects the conc of another reactant?
The reaction regulated by PFK is: $F6P$ + $ATP$ <-> $F1,6BP$ + $ADP$
Thanks!

PFK plays a central role in the control of glycolysis because it
  catalyses one of the pathway’s rate-determining reactions. The control
  of Phosphofructokinase (PFK) is exquisitely complex.

To analyse this complicated enzymatic reaction pathway, we have to revisit important aspects of metabolic control mechanism:

Thus the flux (rate of flow) of intermediates through a metabolic pathway is constant; that is, the rates of synthesis and breakdown of each pathway intermediate maintain it at a constant concentration.
Regulation of the steady state (homeostasis) must be maintained in the face of changes in flux through the pathway in response to changes in demand. In an open system, such as metabolism, the steady state is the state of maximum thermodynamic efficiency.

This explanation provides a somewhat agreement of Le Chatlier’s principle. (i.e. where a dynamic equilibrium is affected by changing conditions, the position of equilibrium moves to counteract the change)
Consider this simple hypothesis:

J = vf – vr
Since a metabolic pathway is a series of enzyme-catalysed reactions,
  it is easiest to describe the flux of metabolites through the pathway
  by considering its reaction steps individually. The flux of
  metabolites, J, through each reaction step is the rate of the forward
  reaction, vf, less that of the reverse reaction, vr:
At equilibrium, by definition, there is no flux (J=0), although vf and
  vr may be quite large. At the other extreme, in reactions that are far
  from equilibrium, vf>> vr, so that the flux is essentially equal to
  the rate of the forward reaction, J~vf.

The flux throughout a steady-state pathway is constant and is set (generated) by the pathway’s rate determining step (or steps).
Consequently, control of flux through a metabolic pathway requires:

that the flux through this flux-generating step vary in response to the organism’s metabolic requirements and
that this change in flux be communicated throughout the pathway to maintain a steady state

Phosphofructokinase is a non-equilibrium enzyme of glycolysis
One important thing to take note of is that the enzyme phosphofructokinase, a highly controlled enzyme functioning far from equilibrium (but ironically itself operates via an equilibrium of two conformational states), evidently is the major target for regulating glycolysis.
PFK is a tetrameric enzyme with two conformational states, R and T, that are in equilibrium. The substrate site binds ATP equally well in either conformation, but the inhibitor site binds ATP almost exclusively in the T state. The other substrate of PFK, F6P, preferentially binds to the R state. Consequently, at high concentrations, ATP acts as a heterotrophic allosteric inhibitor of PFK by binding to the T state, thereby shifting the T <=> R equilibrium in favour of the T state and thus decreasing PFK’s affinity for F6P.
Increase in concentration of PFK does not increase the rate of glycolysis.
In this scenario I believe the Le Chatlier’s principle applies little here. The reason is because the control is very complicated Although PFK has a major role in regulating the flux through glycolysis, it is controlled, in vivo, by factors outside the pathway.
An increase in the in vivo concentration of PFK will therefore not increase the flux through the pathway because these controlling factors adjust the catalytic activity of PFK only to meet the needs of the cell.
Let’s consider another illustration showing enzyme activity versus substrate concentration:

The inhibition of PFK by ATP is relieved by AMP. This results from AMP’s preferential binding to the R state of PFK. If a PFK solution containing 1 mMATP and 0.5 mM F6P is brought to 0.1 mM in AMP, the activity of PFK rises from 10 to 50% of its maximal activity, a 5-fold increase. when [F6P] =0.5 mM (the dashed line), the enzyme is nearly maximally active, but in the presence of 1 mM ATP, the activity drops to 15% of its original level (a nearly 7-fold decrease).
However, measurements of [ATP] in vivo at various levels of metabolic activity indicate that [ATP] varies <10% between rest and vigorous exertion. Yet there is no known allosteric mechanism that can account for a 100-fold change in flux of a nonequilibrium reaction with only 10% change in effector concentration.
Thus, some other mechanism, or mechanisms, must be responsible for controlling glycolytic flux. Let’s discuss the action of adenylate cyclase; adenylate cyclase catalyses the following reaction:
2ADP <=> ATP + AMP (K=0.44)
as a result of the adenylate kinase reaction, a 10% decrease in [ATP] will cause over a 4-fold increase in [AMP]. Consequently, a metabolic signal consisting of a decrease in [ATP] too small to relieve PFK inhibition is amplified significantly by the adenylate kinase reaction, which increases [AMP] by an amount sufficient to produce a much larger increase in PFK activity.
In summary the control of the PFK (non-equilibrium enzyme) has limited Le Chatlier’s principle application as other crucial regulation mechanism exist through the reaction pathway. Even if you were to consider temperature changes, these control enzymes operate far from equilibrium (these have very negative delta G°) and the actual free energy changes in vivo differ from experimental ones making it difficult to simulate actual effects in equilibrium states. It is more likely based on supply-demand basis. One evidence of this is seen through the variation of metabolic flux  throughout glycolysis (may vary by 100-fold or more), depending on the metabolic demand for ATP.
However, for pathways which operate close to equilibrium it may be applied, for example, the delta G° for aldolase is endergonic, whereas under physiological conditions in heart muscle it is close to zero, indicating that the in vivo activity of aldolase is sufficient to equilibrate its substrates and products. Their forward and reverse rates are much faster than the actual flux through the pathway (although their forward rates must be at least slightly greater than their reverse rates). Consequently, these near-equilibrium reactions are very sensitive to changes in the concentration of pathway intermediates.
References
Biochemistry Voet and Voet


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Reduction of oxides by carbon
When reducing an oxide with carbon what would be the product? I think it would produce $\ce{CO2}$ and then the $\ce{CO2}$ would react with $\ce{C}$ to give some amount of $\ce{CO}$.  For example:
$$\ce{2CuO + C -> 2Cu + CO2}$$

You can see ellingham diagram which precisely tells what would be formed under what conditions.

If the conditions of the reaction (depending on temperature, pressure and any other gases present) are such that you are producing $\ce{CO2}$, then $\ce{CuO}$ should readily oxidize any $\ce{CO}$ that gets produced all the way to $\ce{CO2}$.
The main reaction should be as you said:
$$\ce{2CuO + C -> 2Cu + CO2}$$
and if you have any formation of $\ce{CO}$, it will be oxidized as:
$$\ce{CuO + CO -> Cu + CO2}$$

It depends a lot on the reaction conditions. You might have heard of extraction of iron from ore, which involves using carbon to reduce it. In that process $\ce{CO}$ is formed, while in your example of copper, $\ce{CO2}$ is formed. It depends on the extent to which it gets oxidized.


text:
Grignard vs organolithium reagent opening cyclohexene oxide

Could someone please explain the reason for the two different products formed? One with RLi and the other with RMgBr?

$\ce{RLi}$ is much more reactive than $\ce{RMgBr}$ and attacks through the SN2 process, thus inversion of configuration occurs. But in case of $\ce{RMgBr}$, $\ce{MgBr-}$ coordinates with epoxide O atom and forms carbocation on the adjacent carbon. Then a $\ce{C-C}$ bond migrates and the above product is formed.

Do you mind posting a curly-mechanism for this? Ty.


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Difference of reaction between NBS/CCl4 and NBS/AcOH
With NBS/$\ce{CCl4}$ allylic bromination occurs. But with NBS/$\ce{AcOH}$ bromination occurs on the alkene double bond. Why does this difference occur and how? I.e what will be the mechanisms of both?

I do not have the full the mechanisms, but with the $\ce{CCl4}$ there will be a radical mechanism, while in AcOH it involves ionic intermediates.

Why does AcOH support ionic intermediates?

Might be because AcOH is a polar protic solvent thus will solvate the ionic intermediates favorably.


text:
Why does increase in pressure favor formation of water in water-ice equilibrium (at 273K)?
The question:
Ice and water are placed in a closed container at a pressure of 1 atm and a temperature of 273 K. If the pressure of the system is increased to 2 atm while keeping temperature constant, why does the system's volume decrease?

First have a close look at the phase diagram.

More ice will be formed when pressure increases

@ Archis Welankar, are you sure? this is water/ice at constant temperature initially very close to the triple point.

To answer this question, it is best to look at a phase diagram of $\ce{H2O}$.
Note the axes are kilopascal and celsius, but we know that $0^{\circ}\text{C}=273\text{K}$ and $101\text{kPa}=1\text{atm}$. We can see that if the pressure goes up at a constant temperature, the system will move from the phase boundary into the liquid phase.
The generally cited reason for why this occurs is that hydrogen bond network of ice makes it less dense than liquid water, and since higher pressure will favor the more dense state, the ice will melt to become liquid. This leads to a lower overall system volume.


text:
metallic bonding in transition metal sulfides
I searched on the internet but failed to find a clear solution for this question:
Regarding transition metal sulfides like NiS or FeS, is there metallic bonding within their crystal structure?
If yes, how does it influence the macroscopic properties of the crystal and the possible reactions of the ions? Does the S atom affect the construction of the energy band and how do we calculate the certain effect?
If no, then how do we explain the low reflect rate, metallic luster and (semi-)conductivity?
My thoughts: there is metallic bonding in the crystals.

The polarization effect is most apparent in compounds of transitive metal which represents the covalent bonding in this type of compound. In other words, it signifies the extent to which the electrons are localized between the S atom and the metal atom. However, the formation of metallic bond needs the elections to be localized between metallic atoms. The paradox means that either the metallic bond can stabilize the S atoms, or the S atoms are involved in the formation of energy band, or the S atoms can form a special structure that stabilizes themselves.
Due to the structure of NiS crystal, this compound may have a layered structure. Hence, its fragility may be explained by a layered metallic bonding structure in the crystal.

modern chemistry does not consider metallic bonding a thing, it is a variation of electron-defiicient covalent bonding. Metallic behaviour is exactly this: a macroscopic behaviour, manifesting in a relatively high electrical conductivity. LIsting of some data on elecrical conductivitly of sulfides can be found here, together with bits of relevant theory http://www.ela-iet.com/IronGlen/PearceMagSulfides127.pdf

Metal-like delocalized bonding exists in scandium monosulfide, $\ce{ScS}$. Despite the formula, which comes with a halite-type structure (cf. magnesium sulfide, with a similarly sized cation), the scandium exists as $+3$ ions so that one of the compensating negative charges per $\ce{ScS}$ unit goes to a delocalized electron in the conduction band (in effect like an alkali metal).


text:
Why is the standard enthalpy of formation of black phosphorus not zero?
Black phosphorus is the most stable allotrope of phosphorus. Then why is its standard enthalpy of formation not taken as zero?

Reproducibility is more important than stability. To obtain black phosphorus, you have to heat your sample under high pressure for quite a while, and even then it may still contain a significant number of random crystal defects, so its properties are not quite the same each time you make it. White phosphorus, on the other hand, is prepared by sublimation and hence considerably pure.
That's why white phosphorus, rather than black, is assumed standard and has zero enthalpy of formation.


text:
How does Le Chatelier's principle apply to a fuel tank full of hydrogen?
A nano-tube packed fuel tank is full of Hydrogen, and there is a steady pressure of Hydrogen in the tank. While hydrogen gas is being removed from the fuel tank to power the car, the pressure in the fuel tank drops very little for some time. Explain this observation using Le Chatelier's principle.
My attempt at a solution: We look at the equilibrium $\ce{H2 (\text{absorbed by car}) <=> H2 (\text{in tank})}$. Initially, this equilibrium lies to the right, so it moves to the left. But this just means that the hydrogen in the fuel tank is being used up, and so the pressure should drop.

I'm not sure but perhaps the equilibrium of liquid hydrogen in the tank and hydrogen vapours in the tank should be considered.

The official answer considers the equilibrium H2 ( absorbed ) <---> H2 ( gaseous ) . One way I can think of to make this work is that H2 (absorbed ) is actually the hydrogen absorbed by the nano-tube. Initially, the two are in equilibrium, but as H2 (gaseous ) is used up, the equilibrium shifts to the right, and thus the pressure changes only a little.
Two questions come to mind :
What exactly is meant by the 'hydrogen absorbed' by the nano-tube?
Why does the pressure change at all? Shouldn't there be absolutely no change?

I think that your explanation just above is essentially correct. The adsorbed H2 can be considered as equivalent to  liquid H2 (should such a thing exist at normal temperatures). Then the H2 gas pressure is equivalent to the vapour pressure and is constant as long as some 'liquid' still remains in the tank. The equilibrium is between adsorbed (liquid) H2 and H2 gas.

Why would some hydrogen be adsorbed though? And shouldn't there be no change at all in the pressure?

For practical reasons for hydrogen to be stored in large quantities it is best to do so adsorbed onto a substrate rather than have to cool it and to use enormous high pressures.


text:
Why should the salt being used in salting out process be neutral?
When doing salting out it's been said that the salt being used should be neutral, meaning both acid and base which make the salt should be categorized as strong acids or bases. Why is that?
For example, $\ce{NaCl}$ could be used as a salting out salt in principle, because $\ce{NaOH}$ is a strong base and $\ce{HCl}$ is a strong acid but using $\ce{NaCH3COO}$ is not right because $\ce{CH3COOH}$ is not a strong acid.

If you add the salt of a weak acid or weak base, what new equilibrium will result? How could this new equilibrium affect the proteins present in the solution?

Please don't use irrelevant tags.

"Its [sic] been said" is inadequate contextualization. Many organic solutes are potentially capable of being ionized, especially in aqueous solution. Maintaining aqueous media at a neutral pH will for most solutes minimize this (undesired) effect. For instance, in the following reference (note date and distinguish between what is observational (and thus probably "true") and theoretical (based on what is likely to be incorrect chemistry)), read the last sentence on page 531 over to 532. http://www.jbc.org/content/33/3/531.full.pdf


text:
Why are glues/adhesives made up of polymers?
Why are glues/adhesives made up of polymers?
Can monomers be used to make an adhesive.
What is is about the fact that the molecules are stuck together in a chain that make them better/more useful as an adhesive?
Is it like its harder to push people over that are linking arms with other people?

As a side note, various monomers are known to promote adhesion. These monomers contain functional groups such as phosphate/carboxylic acids, silyl ethers, etc.


text:
Why does the standard enthalpy of formation diverge so far from the standard Gibbs free energy of formation for some substances?
If you look at a Table of Thermodynamic Values for chemical substances, most substances have very close values for their standard enthalpy of formation ($\Delta H_{\mathrm{f}}^\circ$) and their standard Gibbs free energy of formation ($\Delta G_{\mathrm{f}}^\circ$). This makes intuitive sense because differences in enthalpy, which in chemical reactions is mostly just a signifier of internal energy because so little work is done by reactions, under normal circumstances correlates with the change in free energy of that system. When heat is lost through a reaction, that system has both less internal energy but also less free energy because the production of heat and then exporting of heat to the surroundings requires the 'use' of free energy.
My question: why do some substances have widely diverging values for $\Delta H_{\mathrm{f}}^\circ$ and ($\Delta G_{\mathrm{f}}^\circ$)? It appears that some substances containing nitrogen have the widest divergence for standard conditions of 25 °C and 1 bar, whereas most substances have a $\Delta H_{\mathrm{f}}^\circ$ and ($\Delta G_{\mathrm{f}}^\circ$) differing by no more than 30%. For example, $H_\mathrm{f}(\ce{AgNO3}) = −124.39$ kJ/mol and $G_\mathrm{f}(\ce{AgNO3}) = -33.41$ kJ/mol, and $H_\mathrm{f}(\ce{N2O4}) = 9.16$ kJ/mol and $G_\mathrm{f}(\ce{N2O4}) = 97.89$ kJ/mol.
A mathematical explanation would be helpful, but I'm most interested understanding this conceptually and why these nitrogen-containing compounds exhibit these properties. Are these properties more common among other substances at extremely high or low temperatures and pressures?

Welcome to Chemistry Stack Exchange. The derived quantities H and G measure different things; have a look at their definitions. The difference is the energy $-T\Delta S$ , when this is large so is the difference between enthalpy and Gibbs energy. Nothing particular to nitrogen.

Thanks. However, saying that H and G measure different quantities doesn't explain why they are so extremely similar for most substances on the standard data tables.
Thinking about it more myself, this is what I have: given that the system in question is at constant temperature, where does the difference in $\Delta S$ come from?
Mostly, the work done by a chemical reaction is minimal, but some nitrates are used in explosives, which do work by expanding rapidly. This is what separates some nitrogen compounds from other substances and leads to differences in H and G on thermodynamic tables.

I think I didn't say that right, but is the overall idea correct?

It is not due to any reactivity a molecule may have but just the entropy. The absolute entropy $S^{\mathrm O}$ is normally obtained from calorimetry  as the sum of the entropy contributions from $0$ K to $298.16$ K and this means integrating as $\int C_P/T$  between phases plus $\Delta H/T$ for any solid phase transitions as well as melting and vaporisation transitions as necessary.

I realize I'm being rather stubborn, but I'd like a conceptual explanation for why some substances have widely varying degrees of ΔH∘f and ΔG∘f. Since the difference reduces to −TΔS, what is it about some substances that makes the −TΔS special as compared to 95+% of other substances?
Merely citing equations is one thing, but bereft of any conceptual explanation to illuminate these anomalies and give flesh to the mathematical skeleton, I still feel the need to understand why. If there truly is no conceptual explanation that exists, I guess it's my loss. Thanks for your patience.

Continuing from the comments above.
The difference is due to the entropy appearing in $\Delta G$ as the energy $-T\Delta S$. Classical thermodynamics cannot tell us what the entropy is in molecular terms. This is discovered using statistical mechanics via Boltzmann's entropy equation $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements of particles into their energy levels, sometimes called 'complexions'.
Different types of molecule have different natures, because they contain different number of atoms, which means that there are different numbers of molecular vibrations, or they contain atoms of different types, which means that vibrations are of different frequencies to one another. Hence molecules of different types have a unique set of energy levels (ignoring for simplicity degenerate vibrations).
As energy is added, the number of ways of filling up the energy levels is clearly different between different types of molecules because the stack of energy levels differs one to another. Single bonds between heavy atoms, for example, have lower frequencies, than those between lighter atoms, and so more levels are populated at the same temperature than in the heavier atom molecule. In this case the entropy is greater than that for a molecule with widely spaced levels.
The same type of argument can be made for rotational and translational energy levels, and can also be made for phase changes, such as solid-solid, or for melting or vaporisation. Which of these are to be counted depends upon whether the substance is gas, liquid, or solid at the required temperature.
Notes:
Technically the entropy is calculated as
$$S=R\ln(Z)+RT\left ( \frac{\partial \ln(Z)}{\partial T}  \right)_V$$
where Z is the partition function and $Z=\Sigma _i g_i \exp(-\epsilon _i/(kT))$ where $\epsilon_i$ is the energy of level i with degeneracy $g_i$ and k is the Boltzmann constant.