text:
Effect of increasing nitrogen concentration in the synthesis of ammonia
For the reversible reaction
$\ce{N2 + 3H2 <--> 2NH3}$
When $\ce{N2}$ is increased, the the forward reaction is favored, increasing the yield of $\ce{NH3}$ and decreasing the yield of $\ce{H2}$. I read from multiple textbooks that the $\ce{H2}$ decreases in concentration to a greater extent than $\ce{NH3}$ increases in concentration, such that the final concentration of $\ce{NH3}$ is greater than that of $\ce{H2}$. Why is this so?

The final concentration of H$_2$ is lower than the concentration of NH$_3$ because for each NH$_3$ you produce, you need 1.5 H$_2$.  So the rate of consumption of hydrogen is higher than the rate of production of ammonia.
Here you can read about the multiple steps in the synthesis of ammonia, also called the Haber-Bosch Process.  As you mention, by increasing the concentration of nitrogen, you can increase the rate.  The rate limiting step is separation of the nitrogen atoms in N$_2$

The first bit:

the $\ce{N2}$ decreases in concentration to a greater extent than NH3 increases in concentration

This is very easy to explain. If you look at the stoichiometrically balanced equation, you need 3 molecules of $\ce{H2}$  for every 2 molecules of $\ce{NH3}$. Consequently, when H2 concentration decreases, it will do so less than $\ce{NH3}$ increases (2 < 3).
To answer the second bit,

the final concentration of $\ce{NH3}$ is greater than that of $\ce{H2}$

To explain this part, we need to think about what it means to say "the forward reaction is favored." This is to say that energetically $\ce{2NH3}$  releases more energy into the system whereas $\ce{N2}$ and $\ce{3H2}$  store more energy in bonds.
To figure out the extent to which an equation favors one side or the other, we would need to look at the energy levels specifically. But for many gas equations, there's a nice way to guess -- which is related to gas laws. Looking at the ideal gas law (P = nRT/V), the amount of energy expressed in pressure is a function of "n" if Temperature and Volume are constant. Thus when n goes down, the amount of energy needed goes down. By reducing the number of molecules, we lower (2 is less than 4).

Thank you! Regarding your explanation to the second part, what did you mean by "energetically NH3 releases more energy into the system"? I'm confused as to how energetics might play a part in whether the reaction is favored or not. And if we were to "look at the energy levels specifically", which energy levels in particular would we have to consider?

bond energy of  N-N + N = N + N (triple bond) N  and 3x H - H vs. N-H bonds x 3 x 2 (see the chart here for general values https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf)


text:
How exactly does the oxidation of TEMPO by NaOCl/KBr work?
I would like to understand the exact process by which TEMPO is oxidized to the nitrosonium cation by NaOCl, as in the part of the mechanism shown in the diagram (1) surrounded by the red box:

The confusion for me is that TEMPO clearly needs to donate 1 electron to $\ce{OCl-}$ in order to form the neutral, nonradical nitrosonium cation, but we also get a $\ce{Cl-}$ anion, so there must also be a 2 electron process involving the $\ce{Br-}$ "co-catalyst" coming into play. How do I reconcile these 1 electron and 2 electron processes so as to draw a formally appropriate mechanism for this part of the reaction?
What exactly happens to the $\ce{OCl-}$ oxygen, and how does it happen?
(1) Org. Synth. 2005, 81, 195. DOI:10.15227/orgsyn.081.0195.

Are you referring to the total mechanism? One specific step within it? Not sure what you're asking here.

Just the oxidation of TEMPO to the nitrosonium cation.

Can you show me where you found this mechanism? This might be a case of a poorly drawn mechanism.

@MrFlux it doesn't matter. I just want to know the details of the oxidation of the TEMPO radical by NaOCl. Here is another picture: http://pubs.rsc.org/services/images/RSCpubs.ePlatform.Service.FreeContent.ImageService.svc/ImageService/Articleimage/2011/NR/c0nr00583e/c0nr00583e-f3.gif some of then say the stoichiometry is 1/2 NaOCl, but I don't really see how this works either.

A simple google search tells me you found this mechanism off Sigma-Aldrich page on TEMPO oxidation, not exactly a reliable source. Looks like poorly drawn mechanism.

Refer to this article "Detailed study about oxidizing alcohols to aldehydes by TEMPO free radical" for more information.

I would propose the following mechanism:

Note that the oxygen radical anion isn't that bad as a leaving group as the Cl-O bond is weak. It can participate in an electron transfer with chloride anion, or rather the bromide at the beginning of the reaction as it is more abundant (I have written the protonated form - hydroxyl radical as it is less likely that two anions will come close enough for electron transfer):

The bromine radical can in turn react with another TEMPO molecule. That is why two equivalents of TEMPO react with one equivalent of hypochlorite.
Keep in mind that there are countless possibilities and deviations from the mechanism I proposed.

This mechanism is good. In the first step, there is no change in enthalpy (if we assume the bonds between oxygen and chlorine are the same). In the second step, since TEMPO has very good electron-donating groups attached to the nitrogen it would have a greater tendency to donate electrons to the oxygen. Also the chloride ion is a good leaving group so it makes forming of the nitrosonium ion easier as well.

Thanks @Marko
Does it make sense yo use two electron arrows for that first step?

Sorry for that. It is a bimolecular homolytic displacement and fishhook arrows should be used. Fixed.

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text:
Work done under constant pressure

Work done under a constant pressure.

I know $$\Delta U = \Delta V \times \pi_T + C_V\Delta T $$
I thought I can give a try to find work done from above equation under constant pressure for an ideal gas.
$$q_p + w_p = C_V\Delta T$$
$$\implies C_p\Delta T + w_p = C_V \Delta T$$
$$\implies w_p = -\Delta T (C_p - C_V)$$
$$\implies w_p = - R\Delta T $$
Adiabatic work is $w_\mathrm{ad} = C_V \Delta T$
For a adiabatic work under constant pressure $w_\mathrm{ad} = w_{q} \implies C_V = -R \implies C_p = 2R$
Constant $C_p, C_V$ and negative $\gamma$ does not look correct to me.

Which of my steps are incorrect ?

You can't do adiabatic reversible work at constant pressure.  The pressure has to change.  If you want to do adiabatic irreversible work at constant pressure, say by suddenly increasing or decreasing the pressure load to begin with, and then holding it constant during the process, this is sometimes referred to as adiabatic work at constant pressure, but it really isn't.  The final pressure will not be equal to the initial pressure that was present before you changed the applied load.

So I should use Kirchoff's law to determine $q_p$ ?

I also did not understand why work can't be done on a adiabatic system under constant pressure.

In a closed system, the only way to compress or expand a gas such that the initial pressure is equal to the final pressure, and is constant along the entire path is to raise its temperature by adding heat.  But that's not adiabatic.  Try solving a reversible adiabatic path, and see how that plays out.

Ok I will try that out. Is my statement, $w_p = - R\Delta T$, correct for work done under constant pressure for an ideal gas ?

Yes, if that's the work done per mole by the surroundings on the gas.  Don't forget that, in an adiabatic reversible process, the pressure is changing, so you have to take the integral of PdV=RTdV/V to get the work.

Thank you, I will tell you what I got. I find $w_p = - R\Delta T$ fascinating , It is pretty amazing.

That's only if the pressure is truly constant throughout the process.

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text:
Why is the chemical formula for ethanoic acid CH3COOH and not C2H4O2?
Why does the hydrogen ion move to the front of the formula and not to the subscript of the H already in the formula?

The first one is preferred since its makes it easy to see that it has a carboxyl group and a methyl group, so its easier to draw the Lewis structure and figure out the molecule's properties. For example, from the first formula one can see that it is an acid. However, its more difficult to figure that out from the second one.

I would not call either formula preferred; see my answer to the duplicate question.


text:
Why does physisorption (physical adsorption) increase with ease of liquefaction?
This is what my textbook says about physisorption

The amount of gas adsorbed by a solid depends on the nature of gas. In general easily liquefiable gases (i.e. with higher critical temperature) are readily adsorbed. Thus $\pu{1g}$ of activated charcoal adsorbs more sulfur dioxide (critical temperature $\pu{630K}$) than methane (critical temperature $\pu{190K}$).

I couldn't figure out why this should be so. In fact, according to my reasoning it should be the opposite. A gas which is easily liquefiable has higher intermolecular forces (here van der Waals forces) which would bound the gas molecules together and reduce physisorption. Where is my reasoning wrong ?

It can be a probable answer but I am not sure about it.
You are correct to say that easily liquifiable gases have stronger intermolecular forces of attraction (van der Waals forces). As a result of which they are strongly bound to themselves. Consider physisorption, here, the forces between the adsorbate molecules and the adsorbent  are weak van der Waals forces. Once the surface area of the adsorbent is completely occupied, the next layer of adsorbate molecules will be bound weakly. So if the gas molecules have stronger van der Waals force of attraction, they would be adsorbed readily in preference to gas molecules having weaker van der Waals forces of attraction.

Your textbook (which one is it? by the way) is correct: gases with higher critical temperature are easier to liquefy. You can confirm this by comparing boiling and critical points of different gases.
For example:

for CO$_2$ (high critical temperature, $T_c=304$ K) triple point is 220 K.
for N$_2$ (low critical temperature, $T_c=126$ K) normal boiling point is 77 K.

This clearly means that CO$_2$ has stronger intermolecular forces than N$_2$.
This also means that the forces between CO$_2$ and another molecule are stronger that between N$_2$ and the same molecule.
A surface is a (almost infinite, in the nanoscale) set of atoms. Many orders of magnitude denser than the gas phase. If $CO_2 - CO_2$ interactions are stronger than $N_2 - N_2$ interactions, also $CO_2 - surface$ interactions are stronger than $N_2 - surface$, whereas the gas phase interactions become negliglible due to its much lower density.
This is why we observe stronger adsorption interactions for molecules with high critical temperatures. You can see this in the plot attached from this source.

The gas having lower critical temperature than room temperature will exist in gaseous form at room temperature. So to liquefy it we need to decrease the temperature.
While the gases with higher critical temperature will themselves be having strong van der Waal's attraction and thus they will get easily liquefied.
And as you said, easily liquefiable gases will get adsorbed more at the surface.

The thing is that we are talking about adsorption which is surface phenomena and we know gases with high critical temp are easily liquifiable which implies gas molecules can form strong van der waal forces but as forces on surface of gas are unbalanced then gas molecules tries to make van der waal forces with the surface of adsorbent [gas being adsorbate].And hence The amount of gas adsorbed by a solid depends on the nature of gas. In general easily liquefiable gases (i.e. with higher critical temperature) are readily adsorbed as the form large number of weak van der waal forces.
hope it helps!

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text:
How much NaOH required to change the pH of a weak acid solution by 2
If I have a $\pu{20mL}$ $\pu{0.1M}$ solution of sodium dihydrogenphosphate, how much $\pu{0.1M}$ $\ce{NaOH}$ do I need to add to raise the $pH$ by 2.
Here's what I've done so far. I'm doubt it's right and I get stuck with respect to the change in volume.
Edit: I realize that this wrong as sodium dihydrogenphosphate is polyprotic.
I found using the $K_a$ that a $\pu{0.1M}$ solution will lead to a $[\ce{H+}]= 3.47\times 10^{-5}$
Since $-log(3.47\times 10^{-5}) = pH = 4.46$, the pH of the final solution is $4.46 + 2 = 6.46$.
The change in $[\ce{H+}]$ is $3.47\times10^{-5} - 10^{-6.46} = 3.44\times10^{-6}$, so there must be a concentration of $3.44\times10^{-6}M$ $\ce{OH-}$ added to the solution.
Since my solution is $0.02L$, $0.02 \times 3.44\times10^{-6} = 7\times10^{-8}$ moles of $\ce{OH-}$.
The ratio of $\ce{OH-}$ to $\ce{NaOH}$ is $1:1$, so:
$\pu{0.1M}$ $\ce{NaOH} = 7\times\dfrac{10^-8}{x}$, $x = 7\times10^{-7} mL$ of $\pu{0.1M}$ $\ce{NaOH}$ added
I know this is a VERY small amount of $\ce{NaOH}$ but is there a way to account for the fact that this change in volume will affect the $pH$ of the solution?

Those calculations would be correct for a strong acid, not a weak one or a polyprotic one. // Quibble point -- You don't "add" a concentration of 3.44*10^-6M OH rather you change the concentration by that much.

The problem with your answer isn't really just that phosphoric acid is polyprotic. Your calculations would work fine for sodium hydrogen sulfate. The problem is about the pKa's of the polyprotic acid.

RANT - I hate problems that don't use significant figures properly. So is the answer supposed to be +/- 1 ml, +/- 0.1 ml, or +/- 0.01 ml?
First your overall approach is right. There are three parts to solving the problem:

(1) Calculate pHs
(2) Calculate how many moles NaOH
(3) Calculate how many ml of 0.1 molar solution.

So the problem is that there are four phosphate species, $\ce{H3PO4, H2PO4^-, HPO4^{2-},}$  and $\ce{PO4^{3-}}$. The precision needed for the answer is somewhat of a mystery. So what simplifying assumptions are valid?
I'm just going to guess that the volume of NaOH should be accurate to 0.01 ml.
So let's look at the problem another way. I start off with 20.00 ml of 0.1000 molar solution of phosphoric acid and titrate with 0.01 ml at a time with a solution of 0.1000 molar sodium hydroxide. As every aliquot of NaOH is added the pH is measured. So:

(1) when I have added exactly 10.00 ml I have effectively made sodium dihydrogenphosphate. I add 2 to this pH to get the target pH.
(2) I look to see volume of NaOH that gives the target pH and subtract 10.00 ml.

SIDEBAR

Just looking at the problem I'm guessing a few ml. Let's say 5.00 ml. Measuring the NaOH to 0.01 would then mean 1 part in 500 precision (2 parts per thousand). Let's say that the initial $[\ce{H+}]$ is $3.47\times 10^{-5}$. 1 part in 347 is about right. But a pH of 4.46 only has two significant figures, that is 1 part in 46 which is too little precision for my guess of about 2 parts per thousand being needed. So the pH should be noted as 4.460.

Looking at Wikipedia we can see that for phosphoric acid that pKa1 = 2.148,
pKa2 = 7.198, and pKa3 = 12.319.
(1) Given these pKa's we can't do better than 2 ppt.
(2) Only a small part of the $\ce{H2PO4^-}$ will ionize to $\ce{HPO4^{2-}}$ and $\ce{H^+}$. Thus the initial pH will be lower than 7.198 but quite above 2.148.
(3) Given (2) enough $\ce{H3PO4}$ might form to lower the pH when calculating to 2 ppt. This is right on the edge for 2 ppt precision...
(4) The species $\ce{PO4^{3-}}$ can safely be ignored in all the calculations. i.e.  $\ce{[PO4^{3-}] << 0.1000 molar}$
1.0 Quick and dirty method...
To avoid solving a 3rd or 4th order polynomial...
1.1 Calculate pHs
The reaction of interest is:
$\ce{H2PO4- <=> H+ + HPO4^{2-}}\quad\quad 6.339 \times 10^{-8}$
We'll let $\ce{[H+] = [HPO4^{2-}]}$ and assume that $\ce{[H2PO4-]} = 0.1000$ so
$ 6.339 \times 10^{-8} = \dfrac{\ce{[H+]^2}}{0.1000}$
$ \ce{[H+]} = 7.96 \times 10^{-5}$
pH = 4.099
Therefore end pH = 4.099 + 2.000 = 6.099

Checks
$\ce{[HPO4^{2−}]}$ of $7.96\times10^{−5}$ is just less than $1\times10^{-4}$ so the assumption that $\ce{[HPO4^{2−}]}$ is not appreciable to  $\ce{[H2PO4−]} \approx 0.1000$ is just ok.
We've also assumed that an insignificant amount of $\ce{H3PO4}$ will form which isn't true...  $\dfrac{\ce{[H3PO4]}}{\ce{[H2PO4^−]}} = \dfrac{\ce{[H^+]}}{7.11\times 10^{-3}} = 0.0112$

Da fix
We have 3 species of phosphate:
0.0008 $\times \ce{[H2PO4^{−}]}$ = $\ce{[HPO4^{2−}]}$
1.0000 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H2PO4^{−}]}$
0.0112 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H3PO4]}$
$\text{--------}$
1.0120 total $\ce{H2PO4^{−}}$
Normalizing $\ce{H2PO4^{−}}$ to 1.0000 total we have for the concentrations
0.0001 = $[\ce{HPO4^{2−}}]$
0.0988 = $[\ce{H2PO4^{−}}]$
0.001107 = $[\ce{H3PO4}]$
We rearrange the equilibrium for the reaction
$\ce{H3PO4 <=> H^+ + HPO4^{2−}}$
to solve for $\ce{[H^+]}$
$\ce{[H^+]} = \dfrac{\text{K}_{a1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = 7.97 \times 10^{-5}$
so everything now checks...
1.2 Calculate how many moles NaOH
Given:
(1) $\ce{[H^+]} = 7.96\times 10^{-7}$
(2) 0.1 molar = $\ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$
Assume:
(1) $\ce{[H3PO4]} \approx 0$
(2) $\ce{[PO4^{3-}]} \approx 0$
(3) moles change in $\ce{H^+}(\text{aq})$ is insignificant
$6.339\times10^{−8} = \dfrac{\ce{[H+][HPO4^{2-}]}}{\ce{[H2PO4^-]}}$
or
$\ce{[HPO4^{2-}] =} \dfrac{6.339\times10^{−8}\ce{[H2PO4^-]}}{\ce{[H^+]}} = 0.07962\ce{[H2PO4^-]}$
We also know that
$0.1000 = \ce{[H2PO4^-] + [HPO4^{2-}] = 1.07962[H2PO4^-]}$
$\ce{[H2PO4^-] =0.09263}$
$\ce{[HPO4^{2-}] =0.00737}$
$\ce{H2PO4^- + OH^- ->  HPO4^{2-}}$ so
Moles NaOH = $(0.020 \text{ liters})[0.00737] = 1.47\times10^{-4}$
1.3 Calculate how many ml of 0.1 molar solution.
liters 0.1 molar NaOH = $\dfrac{1.47\times10^{-4}}{0.1000} = 1.47\times10^{-3}$
ml 0.1 molar NaOH = 1.47

A different approach:
I am using the following pka values for the phosphate system: pka1 = 2.15, pka2 = 7.21 and pka3 = 12.36.
The pH of the original solution of 20,00 ml of 0,100 M sodium dihydrogenphosphate can be calculated to be 4.68. You want to add x ml of 0,100 M NaOH to increase the pH of the solution to 6.68.
The wanted pH = 6.68 is close to pka2 = 7.21. Only two protolytes of the phosphate system will be present at significant amounts at pH = 6.68 i.e. H2PO4(-) and HPO4(2-).
The charge balance can be written as:
[H3O(+)] + [Na(+)] = [OH(-)] + [H2PO4(-)] + 2[HPO4(2-)] + 3[PO4(3-)].
At pH = 6.68 the [H3O(+)] will be insignificant compared to the [Na(+)]. Similarily, [OH(-)] and [PO4(3-)] will be insignificant compared to [H2PO4(-)] and [HPO4(2-)]. These statements will be obvious when plotting a logarithmic diagram of the system.  Accordingly, the charge balance can be re-written as:  [Na(+)] = [H2PO4(-)] + 2[HPO4(2-)].
Please note, that the addition of NaOH will not change the concentration of the sodium ions, but will decrease the concentration of the phosphate protolytes.
=> [H2PO4(-)] + 2[HPO4(2-)] = 0,100 or, for our convenience Ca + 2Cb = 0,100.
The equilibrium equation gives: pH = pka2 + log Cb/Ca=> Ca = 3,3884* Cb.
=> 3,3884*Cb +2Cb = 0,100 => Cb = 0,018558 (M)
=> Ca = 0,100 - 2Cb = 0,100 - 2*0,018558 = 0,062883 (M)
=> Ctot = Ca + Cb = 0,081441 (M)
=> Ctot has decreased from 0,100 (M) to 0,081441 (M) by the addition of the NaOH.
Suppose we add x ml of 0,100 M NaOH in order to reach the target pH:
At pH = 6.68 we get:
Ctot = 20,00*0,100 /(20,00 + x) = 0,081441
=>
x = (2,00 - 20,00*0,081441)/0,081441= 4,557 ≈ 4,56 ml
Answer: You need to add 4,56 ml of 0,100 M NaOH.


text:
Why is E2 reactivity rate of 1-iodo-1-methylcyclohexane faster than (1R,2S)-1-iodo-2-methylcyclohexane even though the transition state is same?

Why is E2 reactivity rate of the second compound said to be faster than the first compound even though the transition state is same? My teacher said that in E2 more hindrance implies faster reaction. In the first compound H atom is more hindered due to methyl group. Even then why is the reactivity of the second compound in presence of a strong base, said to be faster?

See: E1 and E2 elimination reactions of isotopically labeled bromocyclohexanes

If the rates are different, the transition state must be different, no?

Why do you think the transition states are identical? The connectivity is different...


text:
Hydrolysis of salts containing amphiprotic anion
I am trying to understand the derivation of pH of salt containing amphiprotic anions, but I am facing some difficulty:
Ionisation:
$\ce{HCO3^- + H2O <=> CO3^2- + H3O+};\quad K_\mathrm{a_2} $
Hydrolysis:
$\displaystyle\ce{HCO3- + H2O <=> H2CO3 +OH-};\quad \frac{K_\mathrm{w}}{K_\mathrm{a_1}}$
Taking the assumption: $\text{Degree of ionisation} = \text{degree of hydrolysis}$, or, $[\ce{CO3^2-}]=[\ce{H2CO3}]$, an approximate calculation of pH can be done by the relation:
$$\mathrm{pH}(\ce{HCO3-})=\frac{\mathrm{p}K_\mathrm{a_1}+\mathrm{p}K_\mathrm{a_2}}{2}$$
I don't understand the assumption $[\ce{CO3^2-}]=[\ce{H2CO3}]$. We know that $[\ce{CO3^2-}]=[\ce{H3O+}]$ and $[\ce{OH-}]=[\ce{H2CO3}]$. But if $[\ce{CO3^2-}]=[\ce{H2CO3}]$ then $[\ce{H3O+}]=[\ce{OH-}]$. Hence pH should be 7. I don't know where my logic is going wrong and it would be great if someone could help me clarify the derivation.

If the degree of ionisation is equal to the degree of hydrolysis, then (assuming you take the same initial molar concentration of [HCO3-]):
Let's say the initial concentration was 'c' (with no [H2CO3] and no [CO32-]), the degree of dissociation to be α and the degree of hydrolysis to be η

At equilibrium:

[CO32-] = c (1 - α)

[H2CO3] = c (1 - η)

Since we know α = η, we can say that [CO32-] = [H2CO3]
Only because you made the assumption that degree of dissociation = degree of ionisation, do you get pH = 7.
But, since we assume water to always be in excess, even in a real-world situation, difference in pH (from 7) would not be significant.

Can you provide an answer using the Hydrolysis Equilibrium Constant ($K_h$)? The derivation given in my book uses that too, so if I could understand the use of Equilibrium constant in this equation, I'd be able to apply it better...

I ask you again, for an answer involving the Equilibrium Constant... I'd be really benefited by it

That's not too much of a problem.
K(h) = c eta^2 / 1 - eta  (Hydrolysis)
K(a) = c alpha^2 / 1 - alpha (Dissociation for the acid)
Since we know the relation b/w alpha and eta, it should not be too much of a problem to incorporate this into the answer.


text:
Why is polyacetylene an insulator?
Recently I have studied Huckel-theory and I appreciated how this approach is suitable to provide simple models to understand chemical behaviour.
However, it is highly approximate and qualitative at best.
According to Huckel's theory, for example, the homo-lumo gap decreases going on to add carbon atoms to the chain. And so, I could wrongly suppose that polyacetylene is a metal (the band gap shall be very small).
Why is not so? Which is the 'approximate' assumption in this theory that predicts an incorrect result?

Polyacetylene can be conducting (check here : https://www.nobelprize.org/nobel_prizes/chemistry/laureates/2000/macdiarmid-lecture.pdf). I think what you are trying to ask is why doesn't the bandgap become zero? Why does the bandgap settle into a constant value?

In the solid, while the polymer chain can be very long the conjugation is not due to twists & turns and interactions with other chains. Conjugation may only be a few $10$'s of carbons long . Furthermore odd number of C atoms leads to two possible conjugations with a non-bonding electron (also called 'topological kink' or soliton) between them. This can diffuse up and down the conjugation. When doped polyacetylene can conduct.

https://en.wikipedia.org/wiki/Peierls_transition I think that the part that Huckel theory misses, is the electron-electron repulsions. Localisation of electrons due to repulsion is what drives the Peierls distortion. (Disclaimer: I don't know much about this, though.)

The single e in each carbon p orbital to a chemist naturally forms alternating double and single bonds, physicists call this 'dimerisation' and 'symmetry breaking' and leads to Peierls distortion. i.e alternate double & single bonds. As above, odd numbers of  C atoms leads to a 'radical' or 'non-bonding' electron at the interface of single /double conjugation, a soliton.

Porphyrin's comments will explain most of what you need with this addendum: due to the Peierl 's distortion, the band gap is greater than zero due the the polymer chain not being a rigid, conjugated rod along the whole length (due to entropy, but I'm not a free energy expert). This means there is a band gap, leading to semi conductor properties. This is overcome by doping, to give rise to a polymeric conductor.

its due the peierls disortion. According to the huckels theory, the polyacetylene should be behaving like metal. but it is not.

When constructing the Hamiltonian for a Hückel system, we assume maximal orbital overlap for neighboring atoms, and zero overlap for all the others. This implies that the bonding at any bond and atom is perfectly symmetric, which would then lead to a conductor at infinite chain lengths.
In reality, this highly symmetric structure is obviously not the preferred one: the PAc forms isolated double and single bonds.
This formation of asymmetric sections on the molecule can be conceptually explained by Peierls distortion, as already mentioned by kiran, porphyrin and orthocresol.
So I guess the reason why Hückel theory fails at this molecule is its assumption of perfect symmetry. Simple Hückel thoery can never account for more complicated orbital overlap, bond length differences, twisted double bonds etc. etc.

I really think to answer this fully you have to at least mention Peierl's theorem and the associated distortion (see e.g. https://galileo.phys.virginia.edu/classes/752.mf1i.spring03/PeierlsTrans.htm ) and that for a half filled band, as in polyacetylene, the observed long-short ordering of the bonds is what is predicted.


text:
Finding formula of hydrated potassium chromium(III) sulfate through experimentation
I'm in Year 13 and this is a past paper question which I don't have a mark scheme for and I'm confused on the chemistry.

Hydrated potassium chromium(III) sulfate has the formula $\ce{KCr(SO4)2.xH2O}$
An experiment using heating can determine $x$.
Give a step by step description of how you would carry out the experiment
Include all measurements taken
Outline how you would use your results to find $x$.
You are provided with solid $\ce{KCr(SO4)2.xH2O}$ and apparatus usually found in a school laboratory

I think mostly I want to know how would begin with this type of question, like what part of my knowledge do you think this is testing me on because I don't know where to start.

I think the key lies in the word "heating" in line 2.  What happens to a hydrated compound as you heat it (assuming it doesn't decompose)?  It's simple if you knew there were one water of hydration, so imagine what would happen to the weight of the a mono-hydrated compound as you heated it?  Then if there were more waters of hydration, repeat?  That's a pretty big hint, bet let me know if I've just confused you.

All the clues you need to solve the problem have already been given. A key clue is the second line; the word heating. What do you know about heating and how it affects water content? Use this as a starting point.
Thenceforth, the question’s solution should suggest itself.


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Use of subscript cc for HCl and H2SO4
Does the subscript $\ce{X}_{cc}$ mean concentrated? I have never seen this notation. How often is it used and what are the general guidelines to denoting concentrated reagents?

It is unusual to me. Could you link a reference, or show a picture of it being used?

Is this link possibly relevant.  It's a long shot for sure, but this has me pretty stumped. In the link it basically uses Xcc where cc means "C Cluster".  I still have no idea what the first C means even in this context.

A professor used it in a lecture and I'm quite sure he meant concentrated by it. So, it is not really a part of common usage?

I found a very old book that used this notation to describe a plot where $X$ many cubic centimeters of $\ce{HCl}$ are added to a solution.

That would be my guess as well, but in general I think it would mean the concentration in mass per cc


text:
IChO problem involving the condensation point of NF3, NHF2, NH2F
This is a problem from the 2016 International Chemistry Olympiad (IChO):

Which of $\ce{NF3}$, $\ce{NHF2}$, and $\ce{NH2F}$ condenses at the lowest temperature?

The answer book says that it is $\ce{NF3}$. But I don't understand why. In $\ce{NF3}$, $\ce{N}$ should have the greatest positive charge of the three molecules; therefore, it should have the highest b.p., no?
Maybe the answer lies in somewhere in the geometry of ions. Can't figure out what. Any hint?

The interactions do not limit to only dipole-dipole, which means positive charge is not everything.
What's special about $\ce{NF3}$? It has no hydrogens. As a result, it cannot form hydrogen bond, therefore its boiling point (or condense point) is far lower than the others'.


text:
Effect of removing the product from equilibrium
For the reaction
$\ce{PCl3 + Cl2 <=> PCl5 }$
Upon removing some $\ce{PCl5}$ from the equilibrium, the equilibrium position will shift right and the forward reaction is favored, increasing the yield of $\ce{PCl5}$.
Is the information suggested by the attached graph true? (Do $\ce{PCl3}$ and $\ce{Cl2}$ decrease by the same extent?)
For an arbitrary reaction $\ce{2A + B -> 3C}$, how will the concentrations of each substance change accordingly when say, some $\ce{C}$ is removed from the system? Will $\ce{A}$ and $\ce{B}$ still decrease by the same amount, and why?

Concerning the title of the question. Removing $\ce{PCl5}$ would be removing a product, not a reactant.

Concentration is moles per volume and the volume remains constant so $\Delta x$ is effectively a value in moles. Obviously, moles behave like coefficients in the reaction equation.

First, let’s analyse the situation the graph presents. Obviously the reaction is (just restating for clarity):
$$\ce{PCl3 + Cl2 <=> PCl5}\tag{1}$$
We have already removed some $\ce{PCl5}$ from the mixture and are now looking at the self-reestablishing equilibrium. Obviously, for each molecule of $\ce{PCl5}$ we create we lose one molecule of $\ce{PCl3}$ and one of $\ce{Cl2}$ — this is the information we gain from the reaction coefficients (they are all $1$). This can then be expanded to the amount of substance reacting, namely $$\Delta n(\ce{PCl5}) = -\Delta n (\ce{PCl3}) = -\Delta n(\ce{Cl2})\tag{2}$$
Your graph uses concentration $c$ as $y$ axis. Concentration is defined as:
$$c = \frac nV\tag{3}$$
Since we can assume the volume $V$ to be constant, we see that concentration and amount are simply related by a constant factor; thus any differences in concentration will be analogous to differences in amount. Therefore, the labelled $\Delta x$ in the graph are indeed all identical as the represent the same concentration difference and thus the same difference in amount; and we have shown in equations $(1)$ and $(2)$ that the amounts are identical.

What about a case such as your hypothetical following case:
$$\ce{2A + B <=> 3 C}\tag{4}$$
Well, in general the same idea applies except that you must now take care of the different coefficients. Say we removed some $\ce{C}$ from the mixture. New $\ce{C}$ will be created and the creation of three molecules of $\ce{C}$ will use up two molecules of $\ce{A}$ and one of $\ce{B}$. Thus:

For each mole of $\ce{C}$ created we lose $\frac 13~\mathrm{mol}\ \ce{B}$
For each mole of $\ce{C}$ created we lose $\frac 23~\mathrm{mol}\ \ce{A}$

Therefore the relationship is:
$$\Delta n(\ce{C}) = -3\Delta n(\ce{B}) = -\frac 32\Delta n(\ce{A})\tag{5}$$

For your example reaction (2A + B -> 3C), the concentration of A and B will not change by the same amount. If you removed 3 mols of C, you would consume 2 mols of A and 1 mol of B to replace the mols of C removed. This can be generalized to say that you will always need to use twice as much of A as you do B to produce some C, therefore the concentration of A will decrease faster than B as you remove the product, C, and shift the equilibrium towards the product side.
The graph pertaining to the phosphorus trichloride and phosphorus pentachloride equilibrium shows both reactants reducing by the same amount because both reactants are consumed in a 1 to 1 ratio to produce the product. For that reaction, the graph properly represents the concentration profiles.


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What is the logic behind complex splitting?
Like, on a superficial level I know that due to different coupling constant value complex splitting follows a different rule than when splitting involves same hydrogen neighbors. But I don’t understand why and how complex splitting occurs at a intuitive level. Like in complex splitting, is the magnetic moment from the two different neighboring hydrogen adding or subtracting together to affect nmr peak of one of the peak? Is that what’s happening?

Spin coupled nuclei are affected by the spin states of their coupled partners. That is to say, they can detect whether their spin coupled partner is spin up or spin down (for spin 1/2 nuclei), and the energy level transitions are different, depending on whether that other spin is up or down. If a nucleus is coupled to more than one spin, it is affected by the spin state of each of those nuclei. 'Complex' coupling or splitting is a qualitative term used to describe first order couplings that involve more than two coupled nuclei. They can easily be described using standard splitting diagrams.
Consider the example below for the spin system AMX

In the absence of any spin coupling, spin A will appear as a singlet - it has only one energy level transtion available. If it is coupled to spin M, the energy levels of spin A will depend on whether spin M is up or down, and so there are two non-equivalent energy level transitons possible for spin, different in energy determined by the magnitude of the coupling constant. Now, if Spin A is coupled to M AND X, then each of the combinations of the AM transitions will also depend on the spin state of X, and so each of the AM transitions will also have two possible transtions - a total of 4 lines, or what we would call a doublet of doublets.
A useful reference for interpreting complicated, but first order, splitting patterns is by Hoye:
http://pubs.acs.org/doi/abs/10.1021/jo001139v


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Does chromium oxide form on Stainless-steel in 'deoxygenated' water?
Will a passive layer form on stainless-steel in pure (0.055 μS/cm) water with an oxygen concentration < 50 ppb. If so, why is this type of water generally called 'deoxygenated'? If a passive layer (chromium oxide) does not form with such low oxygen concentration why is this, since there IS still oxygen in the water.

There are too many unknowns for a definitive answer. For instance there are many kinds of stainless steel. In general water at room temperature wouldn't "attack" stainless steel. In steam plants however steam under high pressure is much more reactive. Having oxygen in the steam makes the oxygen/water combination even more reactive. Since there is oxygen in the air it is nearly impossible to get ALL the oxygen out of any water sample. Look at it this way. Air has about 78% nitrogen and 21% oxygen by volume. A mixture of 99% nitrogen and 1% oxygen would kill you.

From a practical standpoint, once the stainless steel item is made the passive layer immediately begins to form. Passivation with acids is a process used to accelerate this process and have parts ready to use in a shorter period of time. So if you have any high quality SS parts, they will come to you with the passive layer already developed.
Theoretically speaking, if a completely naked section of stainless was put into this "pure water", there would still be some oxide formation because the specification of <50 ppb still allows for some oxygen to be present. However, this should not have any practical implications. I suspect the name "deoxygenated water" is used to indicate there was a conscious process used to reach such a low amount of free oxygen in the water but not to claim o% free oxygen content.


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Why is the attached hydrogen of 1,3-diethyl-5-(1-phenoxyethyl)benzene homotopic?

The circled hydrogen is considered homotopic with reference to the hydrogen identified by the arrow. I predicted this as they are on identical substituents but am now confused because there orientations with respect to the molecule with the benzene are different. Am I misinterpreting something about the molecular structure?

I would start by replacing the hydrogens one at a time with a labelled hydrogen. Then, see if the two molecules generated are identical. If they are identical, then the two hydrogens are homotopic.

Which hydrogen is homotopic to which is rather hard to see in this compound without actually sitting down and drawing the relevant structures. Before I do this, I am going to get a few obvious points out of the way:

The compound is chiral (see the $\ce{PhO}$ group on the third side chain). This not only greatly reduces the number of enantiotopic hydrogens but also the symmetry of the molecule.
Further analysis shows that there is, in fact, no element of symmetry, i.e. the molecule’s point group is $C_1$. This means that we cannot resort to any ‘cheap hacks’ to determine homotopic hydrogens. We have to use the old-fashioned methods.
Homotopic hydrogens are those that can be transformed onto each other by rotations only (of the entire molecule or rotatable bonds) in such a way that the final depiction is identical to the initial depiction.

We have a few rotatable bonds at our disposal; I will be using the $\ce{Ph-CHXMe}$ bonds, where $\ce{X} = \ce{H, OPh}$ — i.e. I will be rotating around each carbon-carbon bond that connects a residue to the central aromatic ring. Following are the two rotation steps necessary to transform the arrow-hydrogen onto the circled hydrogen:

Scheme 1: The required rotations. The rotated bonds are shown by curly arrows; non-moving parts of the molecule are grey. The circled hydrogen (red) and the other hydrogen on the same carbon (blue) have been highlighted.
I chose to rotate both ethyl groups first and to rotate the entire benzene ring second. However, the order is not important. As we can see, the first rotation transforms the circled hydrogen behind the plane. (I need to rotate the other ethyl group, too, so that the final rotation will give me the initial product again.) The second rotation of the entire benzene ring with the two ethyl groups again puts the circled hydrogen in front of the plane.
Thus, your analysis was correct; the two hydrogens in front of the paper plane are homotopic as are the two behind it. The two hydrogens on a single carbon (red and blue in the picture) are diastereotopic as they cannot be transformed onto each other preserving the symmetry of the molecule.


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C14 Nuclear Diamond to AA Battery Apples-to-Apples Engery output
I was reading an article about nuclear diamonds and they made an apples-to-oranges comparison of a 1 gram C14 nuclear diamond to a single AA battery. Here is the part I'm referencing:

"An alkaline AA battery weighs about 20 grams, has an energy density
  storage rating of 700 Joules/gram, and [uses] up this energy if
  operated continuously for about 24 hours." Meanwhile, "A diamond
  beta-battery containing 1 gram of C14 will deliver 15 Joules per day,
  and will continue to produce this level of output for 5,730 years — so
  its total energy storage rating is 2.7 TeraJ."

I'm far from a chemist or any other applicable type of scientist otherwise I'd take a crack at the answer myself. The question simply is this: according to the data provided how many Joules per gram would a C14 nuclear diamond produce if it weighed 20 grams (to compare it to the quoted AA battery that weighs 20 grams)?
To visualize the desired apples-to-apples answer:
Conventional 20 gram battery: 700 Joules/gram, for 24 hours.
Nuclear ?? gram battery: ??? Joules/gram, for 5,730 years.

Note that the period of 5730 years corresponds to the half-life of C-14.

I'll point out that the battery output was normalized to 1 gram whereas the battery was noted as weighing 20 grams.

And of course the output will be decreasing over time (implicit in the half-life, but perhaps better made explicitly). And this ignores the problem of converting the beta in to useful electrical output. Much less finding a gram of C14 (or 20g for that matter!).

Let me address this part of your question first:

"[...] how many Joules per gram would a C14 nuclear diamond produce if it weighed 20 grams"

I think it's important to understand that the bolded/italicized phrase is redundant, because your are discussing (as did the article) the energy output per gram.  So the per-gram output of a 20 gram battery is the same as that of a 1 gram battery.  The 20 gram battery would have a total output of 14000 Joules in it's lifetime, where the one gram battery would have a total output of 700 Joules.  The fact that the article, and your question, normalize to a per-gram basis is how they've maintained an apples-to-apples comparison.
The semantics get tricky when they talk about the mass of the diamond battery, and by my thinking this is where they may fail to give an apples-to-apples comparison.  The diamond would not be pure $\ce{C^{14}}$, and at a minimum would contain a thin protective coating of non-radioactive $\ce{C^{12}}$ (protecting those on the outside from the radiation on the inside).  So when they say "A diamond beta-battery containing 1 gram of $\ce{C^{14}}$", it is not made clear what the total mass of diamond would be (e.g. we aren't given the ratio of $\ce{C^{14}}$ to the $\ce{C^{12}}$ that would also be present).
That said, I think that the apples-to apples answer to your question, based on the information given in the article, is as follows:
Conventional 20 gram battery: 14000 Joules/day, for 1 day.
Nuclear 20 gram $\ce{C^{14}}$ battery: 300 Joules/day, for 5,730 years.
or, equivalently:
Conventional 1 gram battery: 700 Joules/day, for 1 day.
Nuclear 1 gram $\ce{C^{14}}$ battery: 15 Joules/day, for 5,730 years.

I'm well aware of what half-life is as well as the fact that not all of the material is radioactive. It was more of the 15 versus 700 Joules of energy I was interested in. I suppose a secondary question would be, how large would those nuclear diamonds be in comparison? I imagine large enough to negate the size/energy efficiency of conventional batteries though the nuclear batteries certainly have a much longer shelf-life. Thank you.

I'm glad I was able to help.  Thanks for the very interesting question!


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London/dispersion forces
"London (dispersion) forces are responsible for the fact that non-polar substances can be condensed to form liquids and sometimes solids at low temperatures"._
I learned that what results in London fores between molecules is the "cloud of electrons" that  results in an instantaneous dipole and induced dipole, which eventually results into forming these weak London forces._
The aforementioned statement suggests that as we lower the temperature we form a solid or a liquid meaning that the inter-molecular forces (London forces in this case)became stronger._
My question is: if we lower the temperature we subsequently decrease the kinetic energy of the electrons,  i.e. we decrease the frequency/probability of these "electron clouds" moving to one side of the molecule and forming instantaneous and induced dipole, which means that London forces should be weaker and not stronger to form a liquid or a solid.
This is how I thought of it, can anyone suggest what's wrong with my way of thinking of it? Is it that what we relate to the strength of the London forces is how long the electrons cloud stay on one side rather than how often that happens?

Kinetic energy is about moving the molecule as a whole, and not the electron clouds inside it. The latter is almost unaffected by temperature.

why doesn't temperature affect the electrons movement/speed?  @IvanNeretin

@Bayan Consider the forces acting on a helium atom. There is gravity, which is really tiny (see chart on strength at https://en.wikipedia.org/wiki/Fundamental_interaction) and the mentioned dispersion forces, which are also rather small but more importantly, short-ranged ($r^{-6}$). These forces need to be counteracted by kinetic energy. Now consider that electrons feel the electromagnetic force, which is way stronger. So, when going to high temperatures, at some point, electron "movement" is affected by the temperature - eventually, you get a plasma. But that means really high temperature.

Related https://chemistry.stackexchange.com/questions/74780/why-do-alkenes-with-more-surface-contact-have-greater-london-forces/74792#74792

When we lower the temperature of a substance, we are decreasing the average molecular kinetic energy (they're slowing down). As mentioned in the comments on the question, this doesn't really affect the electron cloud.
So, the substance has been cooled down and molecules are moving more slowly. At this point, it's not that the dispersion forces because stronger, but rather that they become comparatively stronger. While in the gaseous phase, the molecules have velocities of several hundred meters per second. At those speeds, London forces don't really have much of an effect. But, as the molecules are cooled and slowed down (this seems to suggest this is around 10m/s), London forces can cause more interactions between molecules. If the substance is cooled far enough, those forces are able to overcome the kinetic energy of the molecules and they condense into a liquid. As the cooling continues, the molecules move around less and less until they take on a definite shape, at which point it is considered a solid.
The primary issue with your intuition is merely understanding what cooling the substance down really does.


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Magnetic Particles
My chemistry book mentioned magnetic particles today when introducing electromagnetic radiation. It stated that a magnetic field is a region of space where magnetic particles "experience a force." In the same way that an electric field influences the movement of electrically charged particles.
I've always had trouble understanding magnetic fields because of the ambiguity. Electrically charged particles have names (Protons and Neutrons among others) but what are the names for magnetic particles,  and what are they?
Are magnetic particles smaller than atoms?

Pretty much all particles are magnetic, including protons and neutrons. (The latter, BTW, are not electrically charged.)

I think that statement is misleading. Magnetic charge does not exist, so there is no equivalent to a magnetic electron or proton. Instead it acts on two major types of quantities:

Moving charges or current. Magnetic fields cause charges to rotate around the direction of the field.
Spin. Spin is a quantum phenomena where elementary particles like electrons and protons interact with magnetic fields as if they had tiny bar magnets inside of them. The spin aligns itself to try to be parallel to the magnetic field.

In the majority of cases, you can essentially think of the magnetic field as a force that acts on moving charges and permanent magnets (like spin). But there is no such thing as a fundamental magnetic particle with magnetic "charge".

Can you tell me why the spin of charged particles are influenced by magnetic fields? Is it the charicteristic charge itself that the field interracts with, or is there something else about the proton or electron that is inherent within it that interracts with the field. In other words,  what is the real science behind what you called the little bar magnets inside them?

Spin is an intrinsic property of elementary particles. Electrons, protons, and neutrons all have spin. There honestly isn't a very good reason for spin that isn't circular reasoning. But, it eventually comes down to being an essential ingredient in making Quantum mechanics compatible with special relativity.

I've never hear of "magnetic particles" in conjunction with basic science. The term is, however, used to describe very small pieces of iron or iron oxide used in industry for finding defects in metal parts .https://www.nde-ed.org/EducationResources/CommunityCollege/MagParticle/Equipment/Particles.htm
More importantly, from a basic science perspective, is that electrons, protons, ions and other charged objects will experience a magnetic force when moving through a magnetic field, exception being if they are moving directly along the field lines.  This is interesting because it shows a link between electricity, as in charge, and magnetism.
This is also important to a beginning Chemistry student because this relationship was instrumental in the discovery of the electron and developing early models of the atom.

Can you tell me why charged particles are influenced by magnetic fields? Is thethe charicteristic charge itself that the field interracts with,  or is there something else about the proton or electron that is inherent within it that interracts with the field.

@Ashton. Some folks can tell you more about this but I don't think anyone can say why at the most fundamental level. There are four force known in nature, a fifth if you include dark energy. Physisists have succeeded in showing that electricity and magnetism can be "unified" into one force, the electromagnetic force. Current moving in a wire creates a magnetic field. Magnetic fields exert forces on moving charges. Electricity and magnetism are "two sides of the sme coin". You might try your "why" question separately, see what you get.


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Can two electron occupy the same spatial spot in a statistical way?
My friend claimed that if we have a two-electron atom in the ground state, and somehow we get the "real" electronic wavefunction $\psi(\mathbf{x_1},\mathbf{x_2})$ of the system, the statistical probability of finding two electrons (spin-up and spin-down) at the same spot $\mathbf{x_0}$ is nonzero, and the repulsion potential between them looks like a cusp condition when an electron is near the nucleus.
I think if two electrons occupy the same place, then the repulsion must be infinitely large, which is not possible. So I believe the probability should vanish here. However, he argues that in the dynamic case, two electrons can never be in the same place, but statistically, the probability obtained by squaring the wavefunction is not zero.
I'm pretty confused now. Who is correct?
Update:
The link here seems to contradict the answers below. I do not quite understand the statement "while at large distances between the electrons the conditional probability is greater than half of the corresponding unconditional one". Can someone give a hint?
Update 2:
Some helpful references mentioned by DavePhD:
Fitzpatrick's Quantum Chemistry "On the other hand, in the spin-singlet state, there is an enhanced probability of finding the two electrons at the same point in space (because of the final term in the previous expression). In other words, the two electrons are attracted to one"
Advances in Quantum Chemistry volume 1, page 121: for the beryllium triplet P state two opposite spin electrons being at the same point is the most probable configuration
A statement in Atkins Physical Chemistry "The other combination [opposite spins] does not vanish when the two electrons are at the same point in space".

This paper refers to "the probability of finding
two electrons at the same point in space" https://scholarship.rice.edu/bitstream/handle/1911/71020/Capturing%20static%20and%20dynamic%20correlations.pdf?sequence=1

I'm fairly certain this paper (http://aip.scitation.org/doi/abs/10.1063/1.1727605) answers the question, but I don't really understand it fully... Basically, it seems like they're saying in the exact wavefunction, particles (electrons or nuclei) can be in the same location so long as they satisfy certain "cusp conditions" which they derive in the paper. I might be misunderstanding, but I think that's true...

@jheindel That paper cites to this paper http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160100201/abstract but I don't have the full text.

Electrons do not occupy places. Other than that, you are mostly right. Indeed, $\psi(r_1,r_2)$ must vanish at $r_1=r_2$ because of the spin-statistics theorem, if not for other reason. Upd. It was brought to my attention that this reasoning is not quite right; see the comment below. Still, I would expect at least a downward-pointing cusp, much like the upward-pointing cusp found at the nucleus, only the other way around.
This, however, is not easy to visualize. We're not used to imagining multidimensional surfaces. And if we try to focus on one electron at a time, then we're back with the one-electron approximation, and the accurate multiparticle $\psi$ is lost to us.
BTW, the argument of infinite repulsion does not work. After all, $\psi$ doesn't vanish near the nucleus; wouldn't that imply infinite attraction? No, it would not. You integrate over the entire space and come up with some figure for energy, which is surely finite.

I think the spin part of the wavefunction has already taken care of the antisymmetry of the total wavefunction. The spatial part should be symmetric.

I'm very curious about the first sentence of this. What might be a good place to get more information about that?

Any textbook on quantum chemistry would do.

It is necessary to consider the wavefunction for the two electrons as the product of a spatial and spin part, $\psi=\psi_{space}\sigma_{spin}$. As the electrons are spin half particles (fermions) the total wavefunction must be asymmetric to exchange of coordinates. The answer to your question then depends upon whether the electrons are in a singlet or triplet spin state.
In the singlet state ('spin-paired') the spatial part of the wavefunction is symmetric and spin asymmetric;
$$ \psi_S=\psi_{space}(sym)\sigma_{spin}(asym) $$
in the triplet state with 'parallel' spins it is the other way round.
The electrons can be combined into 4 possible spin states. Labelling the spins as $\alpha, \beta$ the combinations are $\alpha\alpha, \beta\beta, \alpha\beta, \beta\alpha$, of these the last two are neither symmetric nor asymmetric and so linear combinations are made to ensure proper symmetry. The four states are now $\alpha\alpha, \beta\beta, \frac{1}{\sqrt(2)}(\alpha\beta \pm \beta\alpha)$.
The arrangement is shown in more detail in the next figure, where a vector model of spin angular momentum (cones/arrows) is used to illustrate the difference between singlet and triplet spin states; ($1$) and ($2$) are used to label electrons.

As a consequence of the symmetry of the spin states the spatial states are determined to be symmetric for a singlet state and asymmetric for a triplet. This is shown in the next figure, which shows, just for illustration,  the electron density plotted for particles in a box. In the singlet the electrons crowd together but in the triplet they are 'forbidden' to occupy the same region as one another  by symmetry (as Pauli exclusion principle). In some texts the crowding together is called a 'fermi heap' and the lack of this a 'fermi hole' although its seems not to be common usage in chemistry.

I'm confused. This is just for one 2D particle?

Isn't it a 1D box, with the two coordinates coming from the 1D position of the two particles? And how, if at all, is electrostatic repulsion include? It seems like this is really two non-electrostatically-interacting particles in a 1D box.

The repulsion comes from the fact that electrons are Fermions, and so from the symmetry that this imposes.

Have corrected error in 2D/1D in text . Wavefunctions calculated for a particles in a box using ψ(x,n)=sin(nπx/L)ψ(x,n)=sin(nπx/L) and then plotting square of  Ψ=ψ(x1,1)ψ(x2,2)±ψ(x2,1)ψ(x1,2)

I still think you're leaving out the Coulombic electron-electron repulsion.  Compare to this paper, especially section IV Coulomb Integrals.  http://www-alavi.ch.cam.ac.uk/papers/twoelinbox.pdf

An interesting paper. I have only used symmetry and used simple wavefunctions to illustrate the effects. The energy calculation does indeed need evaluation of the coulomb and exchange integrals but the symmetry of wavefunctions is as I have written. So more accurate $\psi$ would lead to better figures but these will be essentially equivalent in shape to mine. See Atkins & Friedmann 'Molecular Quantum Mechanics' 3rd ed,  ch. 7.9 'Excited states of Helium'.

Given that the two electrons have antisymmetric spin, your friend is right.
The phenomenon is called Fermi heaps.

Great link! but I think the reference here is more about how spin affects the spatial position instead of the exact coalescence condition we are interested in here. Dave's link to the Rice paper is more relevant

The Wikipedia about Fermi heaps and holes article also links to some nice animations http://quantum.bu.edu/notes/GeneralChemistry/FermniHolesAndHeaps.html

Yes, the probability density is non-zero.
The book High-Energy Atomic Physics in the section "4.3 Kato Cusp Conditions" has a detailed explanation of why this is so for the singlet state.
("Kato" is Tosio Kato who in 1957 published a famous paper on this topic On the eigenfunctions of many-particle systems in quantum mechanics Communications in Pure and Applied Mathematics Vol. 10, pages 151-177.)
In section 4.3.2 an equation is derived from the Hamiltonian for a two electron atom.
Where $r_1$ and $r_2$ are the distances of the electrons from the nucleus and $r_{12}$ is the distance between electrons:
$\frac{\partial{\Psi}}{\partial{r_{12}}}$ (at $r_{12} = 0$) $= \frac{m\alpha}{2}\Psi $
Eventually in section "4.4.3 Approximate Wavefunctions Along Coalescence Lines" ("coalescence" meaning two or more particles being at the same point) an approximate wavefunction for two electron atoms when electrons are at the same point is derived:
$\Psi = Ne^{-2\eta R}$
where:
"R" is the distance of the two electrons from the nucleus
$\eta = mZ\alpha$
and for helium $N = 1.55(m\alpha)^3$

Can I add here the references mentioned along with your bounty, like Fitzpatrick's and Atkin's?

@James yes, that's fine

@James This is the third reference I mentioned in the bounty, page 121: https://books.google.com/books?id=jXcTObz5NW8C&pg=PA121&dq=two+electrons+%22same+point%22+helium+beryllium&hl=en&sa=X&ved=0ahUKEwia-P62janSAhWB5CYKHYFVDG8Q6AEIJjAC#v=onepage&q=two%20electrons%20%22same%20point%22%20helium%20beryllium&f=false

@James Fitzpatrick page 220: https://books.google.com/books?id=v947DQAAQBAJ&pg=PA220&dq=Fitzpatrick+two+electrons+%22same+point%22&hl=en&sa=X&ved=0ahUKEwjC1O-YjqnSAhXJeSYKHQiQA1MQ6AEIGjAA#v=onepage&q=Fitzpatrick%20two%20electrons%20%22same%20point%22&f=false

@James and various Atkins references such as page 119 here https://books.google.com/books?id=QbQJAgAAQBAJ&pg=PA119&dq=atkins+two+electrons+%22same+point%22&hl=en&sa=X&ved=0ahUKEwi60625jqnSAhWD5SYKHQnxBuYQ6AEILzAD#v=onepage&q=atkins%20two%20electrons%20%22same%20point%22&f=false and page 341 here https://books.google.com/books?id=lk2PzH9LmS8C&pg=PA341&dq=atkins+two+electrons+%22same+point%22&hl=en&sa=X&ved=0ahUKEwi60625jqnSAhWD5SYKHQnxBuYQ6AEIITAB#v=onepage&q=atkins%20two%20electrons%20%22same%20point%22&f=false

Thanks for the links! By the way, what is alpha here? It seems like a screen factor for the potential?

@James $\alpha = e^2/4\pi\epsilon_0$

Your are right.  If two electrons occupy the same place, the repulsion will be infinitely large in the real solution of Schrödinger equation.
But under One electron approximation framework (HF,DFT), probability of finding two electrons (spin-up and spin-down) at the same spot $x_0$ is nonzero.
As for the two electons with opposite spin , motion of electrons are uncorrelated. That means the probability of finding two electrons at $r_1$ and $r_2$ can be written as,
$$ P(r_1,r_2)= \mid \psi_1(r_1) \mid ^2 \times\mid  \psi_2(r_2) \mid ^2  $$
when $r_1 = r_2$, $P(r_1,r_2) \neq 0$

I understand in HF it is a different story. I am asking for the real case

In first paragraph I talked about the real case.  Maybe I should make an implementation later.

Please see Ivan's analogy to the near-nucleus case.


text:
How can I rapidly liberate CO2 from dolomitie and siderite in the presence of organic carbon without oxidizing organic carbon to CO2?
I have an analytical apparatus that measures the total carbonate in solid materials by acidification of the material under a stream of $\ce{CO2}$-free air, followed by quantitative coulometric titration of the liberated $\ce{CO2}$.  The coulometer has a detection limit of about a microgram of $\ce{CO2}$, and so long as the material can be finely ground and homogenized we will typically use a sample size of 2 mg.  The sample is acidified with 5 mL of 2 M $\ce{HClO4}$ and heated to $\mathrm{80^oC}$ while sparging with the $\ce{CO2}$-free air stream.
For materials like $\ce{Na2CO3}$ or $\ce{CaCO3}$, the carbonates are thoroughly liberated as $\ce{CO2}$ and analyzed with full recovery within 5 minutes.  For materials containing significant amounts of dolomitic limestone ($\ce{CaMg(CO3)2}$) or siderite ($\ce{Fe(CO3)3}$), the analysis times can approach one hour or more. This is problematic not just because of the reduced sample throughput, but also because the detection limit of the coulometer increases with the length of analysis due to the time-integrated noise at the detector.
Although the issue described above has been a problem with this analysis for decades, it had always been given low priority until now.  Over the years we have tried different non-oxidizing acids (nothing volatile like halo-acids, as they would be titrated as $\ce{CO2}$), propietary commercial catalysts, and organic wetting agents.  The configuration described above has given the best results (lowest time to complete recovery of carbonate), even though in worst-case scenarios (i.e. low total carbonate, predominantly present as dolomite or siderite) the results are essentially useless and all we can do is report high detection limits.
I suspect that we probably can’t come up with a better acid than perchloric: it’s a very strong acid, it is not oxidizing under the conditions of this analysis, it’s non-volatile and perchlorate is stable and it doesn’t tend to form precipitates that might occlude portions of the sample, etc.
My thoughts are that there should be some reagent that would displace then complex the $\ce{Fe}$ and $\ce{Mg}$ in siderite and dolomite, respectively. We do not want to oxidize any organic carbon that might be present, however (at least not all the way to $\ce{CO2}$).
So, how can I alter this analysis so that the carbonates of dolomite and siderite are more rapidly liberated as $\ce{CO2}$, without oxidizing any organic carbon that might be present, and without introducing acidic gases to the coulometric detector?


text:
What is the significance of increasing (and decreasing) the number of ionic releaxation steps in a plane wave DFT calcuation?
The number of ionic relaxation steps is one of the adjustable parameters while setting up a DFT calculation using a commercial code such as VASP. I was wondering what's the significance of this value?

Hello and welcome to Chemistry.SE! Regarding your question, it could be greatly improved by giving a more detailed description of your specific problem. Giving an example would go a long way toward this end.  It also reads like a homework-type question, and I suggest reading this discussion regarding homework-type questions.  Good luck!


text:
Substituting Ulexite
I'm working on ceramic glaze chemistry. One ingredient I don't have, but want to substitute is ulexite:
$$\ce{NaCaB5O6(OH)6.5H2O}$$
I  have borax, calcium borate, and calcium silicate:
$$\ce{NaB4O7 . 10H2O, Ca3(BO3)2, CaSiO3}$$
Since the hydroxides and waters will burn off at firing temperature, it is safe to ignore them. I have already worked out that I can use a mix of the three to supply the needed mount of calcium, boron, and sodium, but I am uncertain of the silicate ions I would be introducing to the mix. I don't know how silicate decomposes at 2200 degrees Fahrenheit. This is important because if it does decompose into silica, it will throw off the balance of silica in the glaze. At such temperatures, iron oxide decomposes to $\ce{FeO}$, so it could make sense that this is possible, but I have no evidence. There isn't much about the decomposition of silicate online, any help?


text:
Reagents and intermediates for synthetic route

What are reagents 1 and 2, and what compound is A?
I was trying to solve this problem, and some ideas popped up (reductive ozonolysis, then aldol condensation and ionic addition, maybe?) but I wanted to be certain.
Thanks a lot in advance! Sorry for possible bad english.

The direct product of the aldol addition is the beta-hydroxyketone shown there. As long as you don't eliminate H2O, then you'll be fine


text:
Notation of hydrogen
$$\ce{e- + H2O -> [H$\cdot$] + HO-}$$
I was a little confused on the notation of the hydrogen in brackets ([H]) in the product side of the reaction. Why is there a dot next to it and what does it mean? Is it saying Hydrogen gas?
Also, what will be the reduction potential of this reaction?

No, it means it's a really unstable free radical. https://en.m.wikipedia.org/wiki/Radical_(chemistry)

The notation $\ce{H^.}$ refers to the hydrogen radical (a hydrogen atom). The dot represents the unpaired valence electron.
An equivalent notation for the reaction would be
$$\ce{e- + H2O -> H + OH-}$$
However, adding the dot emphasizes the fate of the electron on the reactant side.

There is two atypical conventions in the equation. One of them, the dot, is the usual indication that a radical, with a single unpaired valence electron, exists. Elemental hydrogen could be written that way, H• although because it will rapidly react and is not stable near STP, other conventions indicating the atom's electronic state (energy level of the electron) are just as, if not more, common. The other unusual convention is the square brackets. Their meaning is context dependent. In fact, it's possible you, the OP, added them to give emphasis to the species in question, with no further meaning attached. OTOH, their use in a chemical equation might indicate that the enclosed species is nominal (for instance used for bookkeeping) or existing transiently or just a theoretical species which hasn't been confirmed. The equation HCl + H2O → H+ + Cl- + H2O could also be written as HCl + H2O → [H+] + Cl- + H2O because H+ doesn't exist in aqueous solution (rather it is shorthand for H+nOm where m = 1, 2, 3 predominantly and n = 2m+1. You'll note that the Cl- isn't in brackets, that is not to suggest it has no water molecules surrounding it, but rather that the distances Cl ←→ HOH are larger than covalent bond distances, and are considered to be simple solvation.


text:
Why and how are azeotropes formed?
Why and how are azeotropes formed?
I could not find a detailed explanation of how azeotropic mixtures are formed anywhere on the internet. Now, that I figured it out myself, I thought I'd share it.

If we prepare a liquid-liquid solution of equimolar concentration, we know that the more volatile liquid component will have a greater composition in the vapour phase.
Let us keep increasing the concentration of the less volatile component in the solution. Initially, even though the solution is richer in the less volatile component, the vapour phase will be richer in the more volatile component but there will come a time when the vapour phase will be richer in the less volatile component. It may so happen that the composition of the vapour phase becomes equal to the composition in the solution phase. At this point, azeotrope is formed.
Note that if azeotrope is formed at all, both the solution and the vapour phase will be richer in the less volatile component.
Take the example of ethanol-water solution.
Ethanol boils at 78.4 °C, water boils at 100 °C. Azeotropic mixture is obtained at about 95% percent of ethanol. Note that the less volatile component is present in greater amounts when the azeotropic mixture is formed.
I will keep adding more to it as it comes to me.

Nice answer. But why these solutions are used?

Let a mixture of alcohol & water be in equilibrium with their vapours.  x is the mole fraction of A in one phase and y that in the second. The azeotrope occurs when $x=y$, which is that there is a particular composition of one phase in equilibrium with the other phase which has an equal mole fraction at a particular value of x . Then $(\partial T/\partial x)_P=0, (\partial P/\partial x)_T=0$ .

But here water is less volatile. Shouldn't it be present in a greater amount?

In an Azeotrope or constant temperature boiling mixture, the vapour has the same composition as the liquid. This means that removing any distillate does not change the composition; the mixture thus acts as if it were a single phase.
To show this the Gibbs-Duhem (GD) equation is needed, which in its general form is
$$ SdT - VdP + n_a d{\mu_a}+n_b d\mu_b =0 $$
where S, P and V are the total entropy, pressure and volume, $n_a, n_b$ are the number of moles. At constant temperature $dT=0$ then
$$ n_ad\mu_a+n_bd\mu_b = VdP$$
It is easier to work with mole fractions $x_i$ and to express the chemical potential in terms of an ideal solution as partial pressures $p_i$ is
The definition of chemical potential allows us to replace $\mu$ with partial pressure $p_i$ :
$$\mu_i^{soln}= \mu_i^{vap} = \mu_i^0 + RT\ln(p_i)$$
(The dependence of chemical potential $\mu$ on composition is known only for  the special case of an ideal solution.) Differentiating gives
$$d\ln(\mu_i) = RTd\ln(p_i)$$
and substituting into the GD equation produces
$$RT \Sigma n_i d\ln(p_i)= VdP$$
Dividing both sides by $\Sigma n_i = N$ forms the mole fractions $x_i$ gives
$$ \Sigma x_i d\ln(p_i) =\frac{VdP}{NRT}$$
Letting $v_l=V/N$ be the molar volume of the liquid and using $Pv_g=RT$ where $v_g $ is the mean molar volume of the vapour then
$$ \Sigma x_i d\ln(p_i) =\frac{v_l}{v_g}d\ln(P)$$
(As $v_l << v_g$ the right hand side is very small and in some instances may be set to zero)
In the case of two components
$$ x_a d\ln(p_a) + x_b d\ln(p_b) =\frac{v_l}{v_g}d\ln(P)  \tag 1 $$
Returning now to the Azeotrope calculation, let x be the mole fraction of component A in the condensed phase, and y the mole fraction in the vapour. The partial pressure of components A and B are
$$\begin{array}{lcr}
p_a = yP \\
p_b = (1-y)P\\
\end{array} $$
which can be substituted into equation $(1) $ to give
$$ xd\ln(yP)+ (1-x)d\ln((1-y)P) = \frac{v_l}{v_g}d\ln(P)$$
which can be re-arranged using $d\ln(yP) = dP/(yP)$  and $d(yP)=ydP+Pdy$  to give
$$\left({ \partial\ln(P)\over\partial y }  \right)_T= {y-x\over y(1-y)(1-v_l/v_g)}$$
This equation describes how the total pressure varies with composition and as $v_l << v_g$ the right hand side is positive.
As the azeotrope occurs when the gradient of pressure vs composition is zero, the boiling point has a maximum or minimum value thus
$$\left({ \partial\ln(P)\over\partial y }  \right)_T= 0$$
and this can only be the case when $x=y$ which is when the composition of the condensed phase and vapour are equal to one another. In an alcohol/water mixture at $1 \pu{atm}$ the alcohol is present at a mole fraction of $\approx 0.46$.


text:
Why are the stoichiometric coefficients used as exponents in the chemical equilibrium constant?
From what I can ascertain, the formulation of the chemical equilibrium constant is somewhat arbitrary. I don't understand the motivation for putting the stoichiometric coefficients in the exponents. This seems like you are double counting. For instance, if A ->  4B, for every mole of A that decomposes to B produces 4 moles of B. So my reaction would already have a higher amount of B. Then I take those four moles and raise to the 4th power. So if I started with 2 moles of A and 1 decomposed, I would have one mole of A to 4 moles of B. Then I would have a equilibrium constant of 256. How is this superior to not having exponents and just having an equilibrium constant of 4? Or for that matter there are many arbitrary ways I can stick on or not the stoichiometric coefficients. Why not take the 4th root of B over the first root of A?
Edit: Many are saying that this is a duplicate of Law of Mass Action. However, in that question the stoichiometric coefficients just HAPPENED to be the exponents. And as pointed out in the answer the order in the rate law doesn't have to be the same as them. My question is dealing with the Eqilibrium Constant. The Equipibrium Constant is defined such that the stoichiometric coefficients are ALWAYS the exponents. Why?

Related: https://chemistry.stackexchange.com/questions/68195/law-of-mass-action-collision-theory

Sorry, but I don't understand how this could be considered a duplicate of that question, @KlausWarzecha.

It is definitely related @Mockingbird. Your question did not come up on my search though.

@klaus warzecha. That question has to do with rate laws, which do not use stoichiometric coefficients. The equilibrium constant does.

There are some good questions and answers here on how the form of the equilibrium constant is obtained. See: Is there a reason for the mathematical form of the equilibrium constant? Truthfully, you can define the equilibrium constant any way you wish to. However, if you wish to use the typical thermodynamics equations, you will have to define it in a specific way. If you define $K = [\ce{B}]/[\ce{A}]^2$ for a reaction $\ce{A -> 4B}$ then $\Delta G^\circ = -RT\ln K$ will no longer apply (...)

It is perhaps easier to think of it this way. Let's say you have a chemical reaction $a\ce{A} + b\ce{B} \to c\ce{C} + d\ce{D}$. When you run through all the thermodynamics proofs and whatnot, the result you get out of it is $$\Delta G^\circ = -RT \ln \left(\frac{[\ce{C}]^c_{\mathrm{eq}}[\ce{D}]^d_{\mathrm{eq}}}{[\ce{A}]^a_{\mathrm{eq}}[\ce{B}]^b_{\mathrm{eq}}}\right)$$ If, and only if, you define that entire thing as $K$ will you obtain the familiar relation.

Thank you @orthocresol. My general chemistry 2 text book did not allude to any of that. I guess things were dumbed down for us in the class.


text:
Potential wells and symmetry in absorption spectroscopy
In a course I am following, we are analyzing the potential wells in molecular absorption spectroscopy.
For molecules with more than one normal mode, we can study the potential wells involved in transitions for each mode and apply the Franck-Condon principle to determine the intensity of each possible electronic and vibrational transition. In these transitions, some of them will not change the geometry of the molecular equilibrium geometry, others will. This leads to think about how the transition change the point group symmetry of that molecule.
Here comes my question: my teacher said that

modes that are non totally symmetric wont maintain the symmetry of the molecule and therefore the potential wells associated with the electronic ground state and excited state will be one just above the other, with the same equilibrium position and just an increase in energy, while totally symmetric modes won't have this property.

I can't understand why this is true, if there are approximations underlying it and also I can't find any reference for it.
This has been used, later, to distinguish between totally symmetric and non totally symmetric modes in the expression of the transition dipole moment, because for wells not one above the other (totally symmetric) we need to calculate a superposition integral between the two vibrational states, while for totally symmetric modes we know it can be written as a Kronecker delta.

This is a bit tricky to explain, the way I imagine it is this: In a totally symmetric mode where, say, each bond extends in phase the potential minimum need not be the same in the excited state and in the ground state but it might be so 'accidentally'. All the forces are balanced. In a non-totally symmetric mode, one set of bonds stretches as the others compress so the potential has to be the same in both states otherwise there would seem to be an internal force being generated that would move the molecule, which is not possible.
edit:
Other points are
(a) The spectrum reflects the change in geometry between states and thus is that of totally symmetric vibrations only, but see (d) below.
(b) the excited and ground state have to have the same symmetry,
(c) the asymmetric stretch is a symmetric function of normal coordinates, which is what your lower picture shows.
(d) the non-totally symmetric excited state mode need not have the same frequency  as in the ground state, i.e. potential can be wider or narrower. In this case there are selection rules based on the wavefunction's symmetry; both must be the of the same symmetry in the FC factor. The symmetry of vibrational wavefunctions for non-totally sym. modes change with vibrational quantum numbers and it works out that these type of transitions can have $0 \pm2,\pm4..$ transitions only.

Thank you!!! I'm trying to get into this way of looking at it, but I find it a little difficult. Is there any rigorous demonstration of this? Do you have any reference for it, where I can find a more complete treatment of the subject in general and of this problem in details?

I don't have a reference at hand  unfortunately. Some extra points I now recall (a) the excited and ground state have to have the same symmetry and (b) the asymmetric stretch is symmetric function of normal coordinates , which is what your picture shows.

I've gone through your edit, thank you for your help. I'm continuously trying to understand it, but I can't figure out how can the transition from the electronic ground state to the electronic excited state change the equilibrium position of the nuclei according to the symmetry of the mode I am studying.

It has been hard, but thinking about your answer and applying it to an easy molecule like water (especially the requirement for the molecule to maintain the same point group symmetry) I've understand what my teacher mean. Thank you!

Clarifications

First, ungerade is german for odd and gerade means even.
Now, we should discuss energy scales :
An electronic transition requires much more energy than a vibrational transition (on the wikipedia figure, the electronic transition is the one that leads to the higher potential well, in which lie the vibrationnal levels)
In terms of electronic distance to the nucleus now, we can give some images :

From the electronic point of view : It is intuitive that an electron with a higher energy will be on a "higher orbit" (larger orbital) than one with less energy, and thus will be further away from the nucleus. Therefore the offset in the equilibrium position of the electron ($q_{01}$ on the wiki figure)
From the vibrationnal point of view : you can imagine that a molecule that vibrates more will have its electrons more prone to go far away from the nucleus.

Answer
These were only to be clear about what we are talking. Here comes the answer to you question:
We suppose in the following a transition from the ground state to whatever.
You have to look at the orange curves as the wavefunctions of the electrons. The transition probability from one level to another is proportional to the overlap of their respective wavefunctions : $\langle\Psi_{ground} | \Psi_{excited}\rangle$.
Odd (ungerade) wavefunctions tend to be more confined on the edges of the wells and even (gerade) ones more confined in the center of the well.
Because of the offset in the equilibrium position of the electron due to the electronic transition and because the electronic transition is "vertical" in this representation, then you obtain a much larger probability to end up in an odd excited state rather than in an even one. This is easy to visualise thanks to the beautiful wikipedia figure.

First of all, thank you for your answer! I'm sorry if my question wasn't very clear, I was referring to vibronic transitions, in which both electronic and vibrational states change. Referring to the image you linked, I know that the two big wells are two different electronic states, in which we have different levels corresponding to vibrational states. My issue is with the fact that, in my teacher explanation, the two wells should be one the vertical traslation of the other if they refer to a ungerade mode, while they are sort of "diagonal" (as your picture) for gerade ones. This is what I...

... actually can't understand!

@JackI mmmh, I see your problem and I do not understand that either, I'm no spectroscopy expert moreover. Did your teacher give an example of a simple molecule so that it's easier to think about ?

@JackI However, I must add that the Frank-Condon principle precisely states what I tried to explain. Consequently, if what you write your teacher said is indeed true, that has little to do with F-C principle I believe.

I know it is not strictly related to my question, but what I am asking is a way of simplifying the calculation of the Franck - Condon Factor using symmetry. Unfortunately, it has been told to us as really general propriety of every molecule, without a specific example.


text:
Major product of radical fluorination vs bromination

I understand what structure is formed by using $\ce{Br2}$ to brominate this alkane, and I know that $\ce{F2}$ will be less selective in its abstraction of hydrogen. However, how can I know where $\ce{F2}$ will primarily react with this alkane? I just know that $\ce{F2}$ will react more readily and be less selective.

"Less selective" is a bit of an understatement. F2 will react everywhere.

There is nearly no thermodynamic aspect to this reaction. The activation energy for any alkane hydrogen abstraction by a fluorine radical is essentially 0$^{[1]}$. This means that there is an equally likely chance of fluorinating at any given position. In this case, we need only know the proportions of homotopic hydrogen. Homotopic hydrogen are labelled with the same colors in the molecule:

The proportion in decreasing order is 9:6:1:1:1. That means that we should expect about the following yield of monofluorinated products:

As an aside, you would never realistically expect to obtain a monofluorinated product. The fluorination of alkanes is violent, extremely fast, and only gets faster upon fluorination.

$^{[1]}$ Chemistry StackExchange, Why are free radicals so reactive?


text:
What is the chemical reaction between caustic soda + salt + magnet wire?
Intro

Magnet wire(or enameled copper wire) is round insulated Copper wire.
The insulation is generally made of polymer film and this insulating films generally use polyvinyl formal(Formvar), polyurethane, polyamide...
Questions
I'm trying to find the easiest way to strip magnet wire (get the insulation off of magnet wire). I found this video but I want to know that:

What is the chemical equation of this reaction?

$\ce{NaOH + NaCl + Cu + Formvar -> ?}$
As you can see on the video, a gas is coming out.

What is the this gas? Is it dangerous?
Is copper affected by this reaction?


text:
Formal charge of oxygen in protonated alcohol
When ethanol is attacked by positive hydrogen ion, how can the oxygen atom acquire formal positive charge?
I tried calculation it as : $\frac{6-4}{2-1}$ which gives answer 3, but the answer should be 1. Can someone explain this?

I guess you need the answer using Lewis dot structure, so I used the same.


text:
Density of Interstitial compounds
I have read in my textbook (not very reliable) that density of interstitial compounds is lesser than parent compound. But how can this be true?
We add atoms to the lattice voids, so density should increase right?

Sometimes the atoms we add do not quite fit into the voids, so they push the atoms of the parent compound to make room for themselves. There is no universal answer; the resulting density may be greater, or it may be less.

I think you are right about your textbook. Mine clearly says that density is higher

Interstitial compounds are typically obtained when elements such as $\ce{H},$ $\ce{B},$ $\ce{C}$ and $\ce{N}$ are located within the interstitial sites of a metallic substructure. Nonetheless, the metallic substructure is not that of the pure metallic element in most cases.
For example let's consider $\ce{Nb}.$ We can dissolve some amounts of $\ce{N}$ inside the bcc structure of pure $\ce{Nb}$. On one hand, this dissolution increases the unit cell volume, thus reducing the density.
On the other hand, if the amount of dissolved $\ce{N}$ increases, a hcp structure is first formed $(\ce{Nb2N}).$ Further increase of $\ce{N}$ content leads to a fcc structure. In both hcp and fcc structure $\ce{N}$ occupy interstitial sites of the $\ce{Nb}$ substructure and in fact they are both interstitial compounds. But in this case the metallic substructure is not found in pure $\ce{Nb}.$
At the end you have the following densities for $\ce{Nb}$,
$\ce{Nb2N}$,
$\ce{NbN}$:
$$
\begin{array}{lc}
\hline
\text{Compound} & \rho/\pu{g cm^-3} \\
\hline
\ce{Nb} & 8.57 \\
\ce{Nb2N} & 8.25 \\
\ce{NbN} & 8.47 \\
\hline
\end{array}
$$
So, you can see that considering different interstitial compounds a general rule cannot be gained.

I took the liberty of correcting units for density (I'm not sure how to interpret original "Mg m-3"). I think you could improve your answer a bit by adding the source the densities have been taken from.

@andselisk The description of density based on crystallographic / diffraction data in $\pu{x Mg/m^{3}}$ with x a real/floating number is pretty typical e.g., for the supplementary information of IUCr's Acta Cryst E. It is somewhat odd, though kg (and not g) is the basis unit in the joint convention of IUPAC and IUPAP (IUPAC's Green Book).  For the number, it is the same as for $\pu{g/cm^{3}}$ - though one could say both are off by a factor of 1000 to $\pu{kg/m^3}$ which would use base units only.

It depends on what two components of the alloys.
As for your question on why adding atoms to the lattice void doesn't increase density, it depends on what the second component of the alloy is. If it is smaller, the decreased distance between electrons and the nucleus of the added component means stronger Coloumbic attractions of negative and positive. This increased attraction means that the atoms are actually closer together than before. On the other hand, if the atoms are bigger, the opposite occurs.


text:
Structure of Br3O8
What is the structure of $\ce{Br3O8}$? It has an odd number of electrons; does that make it a free radical?
The structure given in my book shows

Where did the 7th electron of the central atom go?

Picture from book (Pg 265, NCERT Chemistry Part II, class 11):

The electron didn't go anywhere. It's in an unhybridized p orbital on the central bromine, and yes, $\ce{Br3O8}$ is a free radical. That is why it decomposes above -80ºC.$^{[1]}$

$^{[1]}$ Cotton, F. A. Progress in Inorganic Chemistry - Volume 2; Interscience Publishers: New York, NY, 1960.

Thanks. That clarifies everything. Isn't it mandatory to mention a free radical to be so in the bond structure?

Yes, @ghosts. If the structure drawing in your book doesn't indicate in some way that there is a free radical, then it is very likely a misprint. Look closely and make sure there isn't a dot somewhere around the Br atom!

@CodyGray The asterisk has a different meaning indicated in the text. So no dot then :P

Well, now that you've shown us the picture, I can see that it does give the oxidation states for the bromines. So this is not a misprint, the data is there, you just have to understand it. If this is a textbook, though, I would say that it could be substantially improved by not expecting the student to understand how atomic oxidation states work and explicitly indicating that there is a free radical.

I see a number of basic copyediting errors in the text surrounding the structures - I think this is not a very good textbook.

@zwol There are many questions and answers on our site that use NCERT texts. I find them insufficient and outdated at best, sometimes they are even outrageously wrong; it often looks like they have been effortlessly copy & pasted together from already bad sources.


text:
Nomenclature of This State?
I've read this in a book

It says the state having three parallel spin is called triplet state .
But as far as I know it is determined by "2S+1", from this it comes out to be 4 . Then how it is "Triplet (3) " ?

You are correct in that if S is the total spin then $2S+1$ gives the spin multiplicity. As you note in the example $2(3/2)+1 = 4$, a quartet not triplet.


text:
Why do Cp2Mg and NaCp complexes turn pink when oxidized?
I am just wondering why the complexes sodium cyclopentadienide ($\ce{NaCp}$) and cyclopentadienyl magnesium ($\ce{Cp2Mg}$) turn pink when oxidized?
Where does the pink color come from?

With highly electroposive metals we may treat the cyclopentadienyl as being essentially an anion, $\ce{C5H5^-}$. When this is oxidized to form the radical, an extra electronic vacancy is opened up in one of the bonding orbitals, and another bonding orbital that is slightly lower in energy can transfer an electron into it with visible light absorption.
Let's play around with the Hückel model to get some estimates. For the anion, where all bonding orbitals are filled, the only possible transition in the ion is from a bonding orbital to an antibonding one, and the minimum transition energy is calculated as  $2.24|\beta|$. From the observed stabilization of benzene we may put in $|\beta|\approx 1.5$ eV, rendering the transition energy estimate above 3 eV and into the ultraviolet range. We are not likely to see color on this account.
Now let's consider the radical formed by taking an electron away, with a vacancy in the higher of the two bonding levels. An electron can transition into this position from the lower bonding orbital with an estimated energy of only $1.38|\beta|$, which is now between 2.1 and 2.2 eV. This is in the short wavelengths of visible light, so we could see a yellowish to reddish color.


text:
Discovery of optical activity of enantiomers - where did the idea of using polarized light came from?
According to Wikipedia, optical activity was first observed in 1811 in quartz. Louis Pasteur's contribution is also mentioned there and in other sources as a major advancement in establishing optical activity as a distinct characteristic of a compound, specifically as a differentiating factor between two enantiomers - a difference that became a standard descriptive characteristic.
My question is how did they even think of that? Why would Pasteur, Arago, Herschel, and other chemists would think of testing the effect of polarized light on chemical compounds? (vice versa actually, chemical compounds affect polarized light, but I assume they did not know that prior to that discovery...). Was it a standard lab procedure at the time?
Testing and determining physical characteristics (such as melting point, boiling point, etc.) seem reasonable and pretty straightforward, but I am wondering how did polarized light entered the chemistry lab as an identifying procedure.

Iceland spar was known for ages. Once you are used to the idea of polarized light, it is kind of natural to test it on other transparent minerals. Once you've discovered an interesting effect on some minerals, why not try it on something else?

I would absolutely believe it started with Iceland spar. Scientists of the time would have started with why does it do that?

Some minerals create polarised light so you don't need to have the idea first, just observe the properties of the minerals and try to figure out what is going on.

It wasn’t a standard lab procedure to test the effects of polarised light on a certain compound/solution/whatnot. However, back in the day it was standard lab procedure to re-run a certain experiment with practically everything your lab had: water, acids, solvents, table salt, cat hair, human hair, dust, fabric, … You would note if you observed an interesting effect and, where applicable, if that interesting effect would only occur in some cases.
So once you are aware that something like Iceland spar produced interesting phenomena (even though you might not be able to explain these phenomena just yet) you would start trying out all sorts of things to test what this interesting optical phenomenon is capable of doing. By chance, you would run across something that you would be able to observe as cancelling of light, even though all components of the system would let light through — and then you would attempt at understanding what you just observed.
They say science is merely grown men playing around for a reason.

"They say science is merely grown men playing around for a reason." Totally agree. +1 after 9 hours (used up all 40 votes)


text:
definition of oxidizing agent
For the reaction $\ce{SF4 + F2 -> SF6}$, since oxidation state of $\ce{S}$ increased from +4 in $\ce{SF4}$ to +6 in $\ce{SF6}$, can $\ce{F2}$ be referred to as the oxidizing agent?

Sure, it can and should.

As for definitions, a reducing agent is an electron donor, an oxidising agent an electron acceptor.

Oxidation means increase in Oxidation state or loss in electrons.
Example: $\ce{Fe^2+ -> Fe^3+ + e-}$
Oxidizing agent is a substance that oxidizes other substances and reduces itself.
In your example, $\ce{F2}$ will act as oxidizing agent as in the reactant side its oxidation state is $0$ but in the product side its oxidation state is $-1$, which means that its oxidation state in decreased, i.e. $\ce{F2}$ is reduced.

$$\ce{SF4 + F2 -> SF6}$$
Here, the oxidation state of $\ce{S}$ increases from +4 to +6, so $\ce{SF4}$ is the reducing agent and $\ce{F2}$ is the oxidizing agent.


text:
Calculating standard combustion enthalpy of dodecane from that of heptane & octane
These are the problem from UKChO-2015.

Given the standard enthalpy changes of combustion ($\Delta_\mathrm{c}H$) for heptane ($n = 7$) and octane ($n = 8$) are $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$ respectively, calculate $\Delta_\mathrm{c}H$ of dodecane ($n = 12$).

Any hint on regarding this problem? My efforts are not worth mentionable here

It seems it expects you to assume that the enthalpy is proportional to the number of CH2 units in the alkane.

It's just computed from changes in bonding. Write out the balanced equations. Figure out the number of bonds broken, and solve some algebra.

@ Zhe It's an Olympiad problem man. They don't expect you to do it so easily. The problem's needs to be done without bond enthalpy just using the information given in the question.

Heptane is $\ce{CH3-CH2(5)-CH3}$
Octane is $\ce{CH3-CH2(6)-CH3}$
Given combustion enthalpies for each of $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$  respectively, we now have a simple set of algeraic equations:
5A + 2B = $-4816~\mathrm{kJ~mol^{-1}}$
6A + 2B = $-5470~\mathrm{kJ~mol^{-1}}$
First we can solve for A as the combustion enthaply of a CH2 in a linear hydrocarbon and we get the value $-654~\mathrm{kJ~mol^{-1}})$.
We can now solve for B,
In the first equation we get 2B = ($-4816 - (654 * 5)~\mathrm{kJ~mol^{-1}}$, A = $-773~\mathrm{kJ~mol^{-1}}$
In the second equation we get 2B = ($-5470 - (654 * 6)~\mathrm{kJ~mol^{-1}}$ , A = $-773~\mathrm{kJ~mol^{-1}}$
dodecane  has CH3-CH2(10)-CH3 , for 10A + 2B.
10 * A  + 2 * B  = ($-6540$ + $-1546$)$~\mathrm{kJ~mol^{-1}}$ = $-8086~\mathrm{kJ~mol^{-1}}$

By way of comparison, the enthalpy of combustion that can be calculated from the enthalpy of formation given in the CRC Handbook is $-8087\ \mathrm{kJ\ mol^{-1}}$.

About $-8086~\mathrm{kJ~mol^{-1}}$.
Take the difference of the two givens and multiply by the difference of carbons between dodecane and octane.

So you mean the difference in enthalpy is proportional to difference of number of carbon atoms between two alkanes?

@Mockingbird I think this answer makes much more sense. Your answer neglects the fact that there are $\ce{CH3}$ groups at both ends which also get oxidised. To a first approximation those contribute a constant amount to $\Delta_\mathrm{c}H$. How much is this constant amount? You can work it out from the data given, but in fact you don't really need to know. All you need to know is each $\ce{CH2}$ group contributes 5470 - 4816 = 654 kJ/mol and hence dodecane is 5470 + 4(654) = 8086 kJ/mol. (With the negative signs, of course.)

@user41388 I'd recommend you edit your answer to include at least a minimal amount of working.

I'd put it this way. Take five octanes. Cut CH2 fragments from four of them and paste those into the fifth one. So, 5octane = 4heptane+dodecane. The same goes for enthalpies.

I wasn't first sure about this concept. But after seeing a gsurfer04's comment and working out a little bit, it became clear.
Alkanes combustion enthalpy is roughly proportional to the number of methylene(-CH2) groups it has. (You can simply work out the example given in the question)
So, number of methylene groups in dodecane 10
number of methylene groups in octane 6
So, dodecane's enthalpy= octane's enthalpy*(10/6)