text:
How to derive the relation between gibbs energy and equilibrium constant?
I want to understand the derivation between gibbs energy and equillibrium constant $$\Delta G=\Delta G^o+RT\ln Q?$$
I have seen a similar post on CSE  Derivation of relationship between equilibrium constant and Gibbs free energy change   which seems to be incomplete and still confusing so I am again asking this question.
The derivation that was written in the post was as follows:

Using the fundamental equations for the state function (and its natural variables):_
  $$dG=-SdT+VdP$$
$$V=(\dfrac {\partial G}{\partial P})_T$$
$$\bar G(T,P_2)=\bar G(T,P_1)+\int_{P_1}^{P_2}\bar V dp$$
  Here $\bar x$ represents molar $x$, i.e. $x$ per mole
  $$\bar V=\frac {RT}{P}$$
$$\bar G(T,P_2)=\bar G(T,P_1)+RT \ln\frac {P_2}{P_1}$$
  Defining standard state as $P=1\text{bar}$ and $\bar G=\mu$
$$\mu(T,P)=\mu^o (T)+RT\ln \frac{P}{P_o}$$
  consider the general gaseous reaction $aA+bB\rightarrow cC+dD$
$$\Delta G=(c\mu_C+d\mu_D-a\mu_A-b\mu_B)$$
  for "unit progress" in reaction. Using $\mu_i=\mu^o_i+RT\ln \frac{P_i}{1bar}$
$$\Delta G=(c\mu^o_C+d\mu^o_D-a\mu^o_A-b\mu^o_B)+RT\ln \frac{P_C^cP_D^d}{P_A^aP_B^b}$$
$$\Delta G=\Delta G^o+RT\ln Q$$

I know the relation between Gibbs free energy,Enthalpy,Entropy and Temperature as $$G = H - TS$$
how is the relation $dG=-SdT+VdP$ derived from the above formula?
Why do we take a path where Entropy and volume are constant  in the equation $dG=-SdT+VdP$ as in a chemical reaction both volume and entropy can change?

When you are talking about the change in a state function like G, it is important to precisely define the initial and final thermodynamic equilibrium states of the system you are considering.  In your equation describing $\Delta G$, what is your understanding of the initial and final thermodynamic equilibrium states that are being considered?

@ChesterMiller It must involve some change in pressure and volume  under constant temperature.

Firstly, if you know the relation $G= H-TS$, you can reach at the differential form of it just by taking differential at both sides, i.e.$$dG= dH-SdT-TdS$$ Now, recall the definition of enthalpy as $H= U +PV$, So, we can write, $dH= dU + PdV + VdP $, and also recall the heat supplied to the system as $dQ= TdS $ and by the first law of Thermodynamics, $dQ= dU+ PdV$ .Thus combining we have $$dG= dU+PdV+VdP-SdT-dQ = VdP -SdT$$ Thus you derive the required relationship.

Now coming to your next question, it's not about taking path where Entropy and volume remains constant. A state in a thermodynamic system can be defined in terms of three parametrs which are interrelated i.e Pressure, Temperature and Volume. In fact, you can only independently vary any two out of these three parameters and third one will automatically get adjusted. Here the independent variables are Temperature and Pressure. By changing the Pressure and Temperature independently, the change in the Gibbs' Free Energy is calculated. The coefficients are volume and Entropy only but that doesn't mean they are kept constant. So, the interpretation of the equation $dG= VdP- SdT$ needs to be done correctly.

The initial state is one of pure reactants in separate containers and in stoichiometric proportions at temperature T and arbitrary pressures.  The final state is one of pure products in separate containers and in corresponding stoichiometric proportions at temperature T and arbitrary pressures.  That is what the derivation is analyzing.  The superscript "0" states in this equation are the same, except that here the pure species are at 1 bar.
For a pure species, the general equation for dG is $dG = -SdT+VdP$. This equation does not assume that S and V are constant. To get the free energy of each species at an arbitrary pressure and temperature T, we integrate VdP from one bar to the arbitrary pressure (using the ideal gas law); at a pressure of 1 bar and temperature T, the free energy of the pure species is equal to $\mu^0(T)$, the free energy of formation of the species (from its elements) at T and 1 bar.
That basically summarizes the initial and final states, and how $\Delta G$ is calculated for the change between these states.

The Boltzmann equation is actually derived from an entropy and energy argument. I think Banjaminson does a decent job, but I'll provide a brief overview as it's relevant.
Assume that we have some system with different energy levels, e.g. in the case of a gas particles may have different speeds (kinetic energy). In expectation, entropy always increases as a high-energy particle is much more likely to lose energy to a low-energy one than vice versa. Entropy is defined as
$$S = -\sum p_i\ln p_i,$$
summing over every energy level. Suppose the average energy is fixed to
$$G = \sum p_i G_i.$$
From Lagrange multipliers1, the entropy is maximized when
$$1+\ln p_i + \lambda G_i = 0,$$
which gives
$$p_i\sim e^{-\lambda G_i},$$
in particular
$$\sum p_i = 1\implies p_i = \frac{1}{\sum e^{-\lambda G_i}}e^{-\lambda G_i}.$$
Let $Z$ be that bottom summation, called the "partition function"---how many ways the particles can be assigned states given the proportions in each state. We can use the entropy to simplify it:
$$S = -\sum p_i\ln p_i = -\sum p_i(-\lambda G_i-\ln Z) = \ln Z + \lambda G.$$
Because people figured out temperature before the Boltzmann equation, we say that
$$\lambda = \frac{1}{N_a k_bT}$$
where $T$ is temperature and $k_b$ is the "Boltzmann constant", how to get from Kelvin to a physically meaningful value of temperature. Note that $R = N_a k_b$ is just the constant times one mole, because chemists multiply all their energies by Avagadro's number. Rearranging yields
$$G = RT(S - \ln Z).$$
Unfortunately this is incomplete, because I'm busy and unlikely to come back, but if someone with lots of karma wants to comment/edit this post and finish it, that'd be great.

Footnotes

The gradients $\nabla S$ and $\nabla G$ must be parallel, or equivalently, $\nabla S$ must be tangent to the surface of $G(p_1, p_2, \dots, p_n)$. Otherwise we can move along this surface, keeping the same energy but increasing entropy.


text:
Ozonolysis mechanism question

Is this mechanism will be similar to ozonolysis ?
My try :

My Question is will B be the answer ? And what does Me2S do ?

Definitely Yes.

Me2S reduces the ozonide and any remaining ozone. I think the reaction will take place on the more electron rich double bond. The product you have drawn is not one of the options offered among the answers.

@Waylander should the Answer be B ?

I highly doubt that the reagent in the first step is O2, and if that's the case, then it's oxidation reaction. But, I'm not sure!

@rv7 its O3 ,Ozonolysis will happen I think

B is the only product that has the terminal carbons at the correct oxidation state. If you want a diol work up with borohydride

As Waylander pointed out, you lost a carbon in your product.

The question could have been improved by modifying the reagents as follows: 1) 1 equiv. O3; 2) excess Me2S, -78 deg. C.

@Zhe, Ignoring the options, why didn't the ozonide form on the other alkene(the one with the methyl group?). Sorry for disturbing you, but I have encountered this question myself, and am confused regarding this.

@B.Anshuman I assume ozone, with 3 oxygens, is more reactive with electron-rich alkenes than the reverse.

The first part of your question has been answered in the comment, as for the part about what does $\ce{Me2S}$ do:
The mechanism used here is reductive ozonolysis and $\ce{Me2S}$ acts a reducing reducing agent itself getting oxidised to dimethysulphoxide $\ce{(Me)2SO}$ (DMSO).
As for the mechanism of reductive ozonolysis with $\ce{Me2S}$ here's a helpful link.

source: Stack Exchange

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

sort: votes


text:
Electrolysis of molten CaOH and CaCl
So recently I have been trying to get elemental Calcium but I only found a cheap solution in electrolysis of molten $\ce{CaCl2}$ (thats because calcium chloride is very cheap) and I am not sure If it would work Can you give me advise? I also wonder If it would work on $\ce{Ca(OH)2}$?
Would the reaction be like
$$\ce{Ca(OH)2 -> Ca + 2 OH}$$
and
$$\ce{CaCl2 -> Ca + Cl2}$$
or something else ( I am 13 so I dont know much about chemistry )

It's much more efficient to buy calcium than to obtain it yourself by electrolysis...

You'll get nowhere with calcium hydroxide. Try to melt it, and it gives off water, becoming calcium oxide. Try to melt that, and you'll melt your container first. And your electrodes.
Electrolysis of calcium chloride works commercially, but it's very difficult in a home lab. I've seen discussions from people who have tried, but I don't think any of them were successful.
Please don't try to do this if you don't have much experience with chemistry. Start with something like copper-plating; copper sulfate is easy to get, and you can do it with an aqueous solution at room temperature. Working with molten salts and alkali/alkaline-earth metals is a really bad idea if you're just starting to learn.

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


text:
Optical Isomerism in Coordination Compounds of type MA₂B₂(CC)
I was asked the following question:

How many optical isomers are there for $\ce{[Cr(NH3)2Cl2(en)]^+}$ ?

First of all I figured out the following four geometrical isomers of the above complex:

Out of which, (iii) and (iv) are optically inactive, due to a plane of symmetry (shown by faint line) passing through them as shown below:

For (iii), you can suppose: green as $\ce{en}$, orange as $\ce{NH3}$ and red as $\ce{Cl}$. Similarly, for (iv), you can suppose: green as $\ce{en}$, orange as $\ce{Cl}$ and red as $\ce{NH3}$.
Therefore only (i) and (ii) will show optical isomerism, which I've confirmed myself.

But, the following four isomers was provided as answer, in which I don't think that the second configuration of complex will show any optical activity due to plane of symmetry (as discussed above for the (iv) geometrical isomer).

What I am doing wrong?

I don't think it's you who's doing wrong.

Well, it's a valid mirror plane. It's just that it's the same compound after mirroring...

I came to the same conclusion as you, total 4 stereoisomers (2 of them create one enantiomeric pair).
In the 3rd party solution, they depict the same isomer at bottom, i.e. only 3. The missing one is that with two $\ce{NH3}$ ligands in the same plane with $en$, and $\ce{Cl-}$ ligands in straight line (trans), perpendicular to that plane (not that horizontal, which is depicted)

Here is my simplified schematic representation of the octahedral complex with ethylenediamine ligand bound by back-pointing bonds in horizontal plane, hidden; the rest is vertical axis and two front-pointing horizontal bonds (seesaw geometry, not to be confused with Fischer projection for tetrahedrals!) Dotted line is mirror plane, equal sign means identity (by 180° rotation around observer-to-model-line axis).

has B ligands in the (en)-defined plane, A ligands perpendicular (axis)
(it's their bottom left isomer, idenitical with bottom right, for A=$\ce{NH3}$, B=$\ce{Cl}$; note that they have en oriented vertically, everywhere, with one end in dotted horizontal plane, other one in upper vertical position, I have it somewhat simpler, rotated for en to be horizontal);
has the opposite, A's and B's swapped (and it's not mirror image to 1.)
(it's their missing isomer);

If A and B are in the plane then there are two possibilities how to relatively place A and B in the perpendicular axis (resulting in mirror images),

(it's their top right);
(it's their top left).

If the claimed 3rd party correct solution image you provided is complete - Yes, the solution is complete.

@rv7 OK, I did not want to visit unknown sites. Well, you have apparently posted someone's complete solution, that is unfortunately incomplete :)


text:
Can Hartree-Fock mix more orbitals than there are electrons?
I'm trying to understand Hartree-Fock well enough to write my own implementation. A point that I seem to be coming to that I'm not sure is correct: the method only produces one-electron orbitals that are linear combinations of the starting orbitals?
That is, if I have $n$ electrons, I start them off in some $n$ orbitals (say, atomic orbitals). Is it correct that the final result will only mix those $n$?
It seems very natural that I might want to include other orbital shapes -- maybe I want the first $m \gg n$ atomic orbitals in my molecule, and let all of those contribute to building lower-energy one-electron orbitals. As best s I understand it, standard HF does not allow for this, because at each step of diagonalizing the Fock matrix it's just using combinations of the existing orbitals -- with no room for the excited states.
Is my understanding correct? Are there other extensions of HF which allow broader mixings?
One possibility that I could imagine would be starting off with $m$ electrons, so that I have $m$ orbitals available. Then I run HF, and at the end I only use the $n$ lowest orbitals. This runs into the problem that I will still be computing Coloumb terms from the $(m-n)$ extra "fake" electrons, which will disturb the shape of my $n$ lowest orbitals. So perhaps the correct approach would be to use $m$ electrons mixing, but when I compute the Fock matrix $F$, I only include $J$ and $K$ terms from the $n$ lowest modes (but letting these affect the energies of all $n$ modes). But that seems messy, because now $F$ isn't symmetric... in the end this seems like a risky path to go down.
Help appreciated, thank you.

One note about wording: You used "excited states" in connection with "Fock matrix". However the Fock matrix is about single electron wave functions (=orbitals), while "excited states" in chemistry usually refers to electronic states which are many-electron wave functions.

First some general considerations
Using more orbitals than electrons is not just possible, but almost essential for Hartree-Fock. Although in practice one can do calculations in a so called minimal basis set, this approach is lacking a lot of accuracy. Usually one adds additional basis functions, which leads to basis sets classified as double zeta (triple, quadruple and so on are also possible).
For open shell systems even a minimal basis set includes more atomic orbitals than electrons are available. For example if you want to calculate a single carbon atom, you would have one basis functions for each of the $1s$, $2s$ and all three $2p$ orbitals. This gives a total of 5 atomic spatial orbitals (=10 spin orbitals), but only 6 electrons. (This is not really a good example, as occupying just one of the three $2p$ orbitals breaks the spherical symmetry of the atom, but I am leaving the scope of your question ...)
Virtual orbitals
Since all molecular orbitals (MOs) are expressed in this basis of atomic orbitals (AOs), it means we will get as many MOs as we put AOs in. If ordered by their energy, we first get a set of occupied MOs until all electrons are distributed. After that comes a set of virtual (or unoccupied) MOs. Those are still soluations of the Roothaan-Hall equations, i.e. they are eigenfunctions of the Fock-Operator, but they do not contribute Hartree-Fock energy.
However, virtual orbitals are important in correlation methods going beyond Hartree-Fock, e.g. Coupled Cluster or Pertubation Theory. Here the virtual orbitals become partially occupied and thus influence the ground state energy.
Equations
The ground state energy in case of closed-shell HF reads
\begin{equation}
  E_0 = 2\sum_a h_{aa} + \sum_a\sum_b 2J_{ab} - K_{ab}
\end{equation}
where the sum only goes over the occupied spatial orbitals, which equals half the number of electrons.
And for the Fock operator we have
\begin{equation}
  f(1) = h(1) + \sum_a 2J_a(1)-K_a(1)
\end{equation}
with the same restrictions to the summation. Since the Fock operator is applied to all MOs, this means the effect (coulomb repulsion) of the occupied orbitals is considered in each virtual orbital, but the virtual orbitals do not affect the occupied ones.
(Equations are taken from the book by Szabo and Ostlund, p. 134)

I see! Okay, this is what I was suspecting, but I felt very far from confident in it, and I hadn't seen any references actually clearly delineating how to treat occupied vs. non-occupied orbitals correctly. Thank you very much!


text:
Valence Electrons of Molecules
Why do we say that hydrogen has 1 valence electron? Hydrogen (H2) is a molecule so how does this make sense?

Don't misunderstand it. When you say Hydrogen has 1 valence electron, it means you are referring to Hydrogen atom (H) by default. Although, it's ambiguous saying Hydrogen has 1 valence electron. It should be said as Hydrogen atom has 1 valence electron.

Come to think of it, there is no ambiguity. Molecules don't have valence electrons; atoms do.


text:
Apple Cider vs. Cleaning Vinegar
My grandparents stumbled on some cheap  Heinz Apple Cider Vinegar while cleaning their house.

Ingredients: Made from the juice of apples diluted with water to a uniform pickling and table strength of 5% (50 grains) acidity.

Barring the origin and 1% difference in acidity, how does it differ from Heinz® All Natural Cleaning Vinegar? Is Apple Cider Vinegar equally as potent for cleaning clothes and the house?

Ingredients:
  Made from select sun-ripened grain, diluted with water to a special cleaning strength of 6% acidity.

Beverly Leestma purports that the brand beneath at Walmart didn't indicate "the acidity but did state it was “Not For Food Consumption." Some vinegars (like this Industrial Strength Cleaning Vinegar) are still 5% but aren’t refined or purified to be food quality."

Assuming that the industrial strength cleaning vinegar's acidity is also 5%, then why can't it be consumed?
Why must raw vinegar be refined or purified for human consumption?
Is Beverly Leestma correct in her wariness beneath?

Part of me thinks Heinz is just looking to get in on the popularity of homemakers who want natural cleaning.  I’m not convinced that the 1% of extra acidity warrants the higher price tag and that maybe they’re just trying to cash in on a trend at our expense.

Cider vinegar is made from apple juice so it has stuff from apples left in it. Thus it will not dry "clean". If you leave some out to evaporate until dry you'll get a gummy spot. //  "White" vinegar can be made from cider vinegar by distilling off  the acetic acid (vinegar) and leaving the apple gunk in the pot. So white vinegar is just acetic acid and water. It will dry "clean" without leaving a residue.

The acetic acid in "Cleaning vinegar" may come from an industrial process. One way to produce acetic acid industrially, is carbonylation of methanol. And methanol is toxic!
So unless the product has been analyzed for residues of methanol and approved for food consumption, it is probably unwise to use it for consumption.


text:
Why do Gas Laws only use Kelvin Temperature?
I didn't see this in any other question, although it seems to be a fundamental question of the gas laws. I understand that Charles' Law practically has to use Kelvin temperature, because you would end up with a negative number in some cases, but can someone explain clearly what Charles did with the three gases and why the volume can only be proportional to the Kelvin temp and not the Celsius temp? thanks.

Imagine the following scenario. You want to find a relationship between the number of people and the total number of hands. But to facilitate easier counting, your count of hands is the number of hands - 273. Are you going to find the simple relationship that you were supposed to find?

It is partially about definitions and partially an about making the formulae simpler
Consider how you work out the relationship between volume and temperature of a gas. You measure the volume of a fixed amount of gas at different temperatures and plot the results on a chart (volume on Y, temperature on X). The points show a linear relationship (for gases whose behaviour is close to ideal at the range of chosen temperatures).
Where does the line cross the Y-axis (temperature)? It isn't at 0 °C, is it? It is at about -273 °C. So the formula could look like this: V=x(T+273) with temperature in Celsius. Or, you could redefine the temperature scale and make it simpler: V=xT with T in Kelvin. This is far simpler in practice and keeps the maths neater.
But this is, essentially, how we know what absolute zero is. This is the very definition of the term (which works on other temperature scales as well but only Americans still care about Fahrenheit). We chose 0 as the original base for the Celsius scale because that is the freezing point of water not because of anything to do with gas laws. So that is an arbitrary point that happens to give a more complicated gas law. Hence the Kelvin scale where the zero relates to the gas law and makes it simpler.
So volume is proportional to temperature in Celsius only if you adjust the temperature because the scale has an arbitrary point chosen for zero. Kelvin keeps things simpler and neater.

First you must know how was Charles' discovered his law?
See on wikipedia Jacques Charles
Here is a quote from Wikipedia

Around 1787 Charles did an experiment where he filled 5 sites to the same volume with different adds. He then raised the temperature of the sites to 80 °C and noticed that they all increased in volume by the same amount. This experiment was referenced by Gay-Lussac in 1802 when he published a paper on the precise relationship between the volume and temperature of a gas. Charles' law states that under constant pressure, an ideal gas' volume is proportional to its absolute temperature. The volume of a gas at constant pressure increases linearly with the absolute temperature of the gas. The formula he created was $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

As all statements of experimental laws are completely based on the results of experiment performed and cannot be altered. Charles'law is also an experimental law, and it was discovered from the experiment performed by Jacques Charles and he found that the Volume of a ideal (close to Ideal) gas is directly proportional to absolute temperature.
You are saying that

Why volume can't be proportional to temperature in Celcius?

But how it can be if it is not the result of the experiment.
Suppose a cricketer just scored six and you are asking why can't we write his score as four instead of six. In the same way Charles discovered that Volume is directly proportional to absolute temperature and you are saying why it is not temperature in Celcius.
If you want to know why volume can only be proportional to absolute temperature - experiment is the proof.


text:
Why electrons are attracted by cathode in Voltaic/Galvanic cell

I studied that in the above setup electrons are attracted by Cu because of their reduction potential but they don't involve in reaction because of the d-bands of the electronic structure so Hydrogen accepts the electron and become as Hydrogen gas
My question is why don't the H+ ions absorb electrons near the Zinc strip itself, why the electrons are travelling through the external load to combine with H+ on the other electrode instead of combining to them at the Zinc strip itself
If the above scenario is impossible to happen How the H+ ions are combining with released electrons near the Zinc strip in the Activity of putting a Zinc strip inside a beaker with dilute sulphuric acid
I would also like to know why this is not as popular as Salt bridge experiment (Is this an inefficient way of producing electricity?) I was unable to find much resources on these kind of setups(Without salt bridge) on the internet.
Please correct me If I made any silly mistakes ;)
 There is a similar Question regarding the electrolysis process but I had some uncleared doubts

Please provide reason for downoting!

Changing a few words may make the situation clearer. There are three reactants: 1) Zn, 2)Cu and 3) H2SO4. Actually, the Cu does not change, so it is not really a reactant, but it is part of the overall reaction.
What happens when Zn is placed into acid: some Zn atoms dissolve to become Zn++ ions, and they leave their electrons behind, so the Zn electrode become negatively charged. H+ ions may be attracted to the electrode, but there are barriers to formation of H2...too complicated for a simple discussion. But if you lead the electrons to a different electrode (like Cu), H+ ions can come closer to the electrode, and more easily form H2 molecules which will rise to the surface of the solution. I would say not that electrons are attracted to the Cu, but rather that they are forced there by the negative voltage developing on the Zn.
Why don't the electrons combine with H+ near the zinc strip? Zinc (especially pure zinc) is said to have a high hydrogen overvoltage. This means essentially that H2 is not formed and evolved as easily as theory suggests; it could be that a monolayer of hydrogen is deposited, but is stuck to the Zn surface as atoms which do not combine to H2 easily - you have to force them by forcing on more H+ ions. And hydrogen evolution eventually  happens even with ordinary zinc because there are always tiny imperfections in the surface which can more easily permit 2 H atoms to combine to H2. In past days, a very thin surface coating of mercury was applied to zinc electrodes in order to coat these surface imperfections and reduce the ability of the zinc to evolve H2.
Perhaps the Salt Bridge experiment is a way to more clearly separate the two reactions of Zn dissolution and H2 evolution. In real batteries, salt bridges are not used. They would give extra resistance to current flow and reduce energy output.

The answer was really helpful but regarding the "but there are barriers to formation of H2...too complicated for a simple discussion.", Can you provide any external links(Wikipedia,etc.) for further understanding/reference, Thank You so much.

One (short) article is: https://en.wikipedia.org/wiki/Overpotential  ; a longer one is: https://en.wikipedia.org/wiki/Electrochemistry . There are lots of term in both articles that you can Google to get more information. In a way, it is simple, but it is complex because every element has its own peculiarities that need to be explained by different mechanisms. Do a little at a time, try to understand one system at a time; eventually you will understand it all. (Not that I do...) BTW, overvoltage and overpotential can be used interchangeably if you want to.


text:
Second order Jahn-Teller effect for d0 complexes
I was wondering why the second order Jahn-Teller effect is also said to happen for many d0-systems? It makes sense to me that due to relativistic effects and the stabilization of the s-Orbital and destabilization of the d-orbitals the energetic gap between the let's say eg in an octahedral complex and the above laying unoccupied a1g s-orbital becomes small enough that, for the right amount of electrons filled into the d-orbitals a mixing of both states is of some advantage leading to a second order Jahn-Teller effect like in Hg(II) compounds. But why does it also happen to d0-complexes?
My suggestion would be that we cannot discuss any field splitting in the first place. So first of all, how would any distortion here give us an energetic advantage and then also we would have to take the nephelauxetic effect of the host or ligand more into consideration if we do not consider a real crystal field splitting. I don't really know where a methyl ligand would be considered in the nephelauxetic series (since some heavy transition metal methyl complexes form trigonal prismatic complexes rather than octahedral ones, which is often explained with a second order Jahn-Teller effect) but if I consider some covalency in this metal-carbon bond then it may shift the d-orbitals down enough so they could perhaps mix with the lower occupied p-Orbitals. So that perhaps the mixing is not between occupied d- and unoccupied s-orbitals but between occupied p- and unoccupied d-Orbitals.
EDIT:
I was reading an article on the second order Jahn Teller effect yesterday and found the following line:

I am not sure if I got all those rules correctly but it seems like I have to consider the product of both states and the product has to be of the same symmetry. If I think about the next low lying occupied orbital in an octahedral environment it would be the filled t1g Orbital. So This would mean I have to consider a t1g and t2g mixing here. I have no clue about those irreducible representations and point groups but from the tables you can find on the net I found that a multiplication of T1 and T2 in Oh point-group would end up in an A2 + E + T1 + T2. As this contains both T1 and T2 I guess it will somehow obey those rules.
Perhaps someone with more experience in symmetry and character tables might have suggestion about this.

After quite some research I finally found the original work on the problem. This comes from:
S. K. Kang, H. Tang, T. A. Albright, J. Am. Chem. Soc. 1993 115 1971 - 1981.
I won't discuss the symmetry parts, simply because I have no clue about symmetry and point groups and transformations. I copied the original description for the necessary relationships for a second order JT-effect to happen in the question, so if you know more about this you can probably confirm that.
So to my understanding what is happening for d0 compleces is the following. You really get a mixing of the lower and filled $\mathrm{t_{1u}}$ Orbital with the higher unoccupied $\mathrm{t_{2g}}$ orbital. Both of them split in a set of one orbital that is unchanged in energy, $\mathrm{a_{2}''}$ (for the original $\mathrm{t_{1u}}$) and an $\mathrm{a_{1}'}$ (for the original $\mathrm{t_{2g}}$), and two degenerate orbitals with $\mathrm{e'}$ symmetry that is stabilized and therefore lowered in energy for the original $\mathrm{t_{1u}}$ and elevated for the original $\mathrm{t_{2g}}$. According to the paper this is done by a distortion of the octahedron by a normal mode of $\mathrm{t_{2u}}$ or $\mathrm{t_{1u}}$ symmetry. The first is said to equal a rotation along a threefold axis in the octahedron to get a trigonal prism and change the $\mathrm{O_{h}}$ symmetry to $\mathrm{D_{3h}}$.
You can also see this in the appendix of the book:
T. A. Albright, J. K. Burdett, M.-H. Whangbo, Orbital interactions in chemistry, second edition, Wiley, 2013.

The next question is, why does this stop at the trigonal prism and not continue in mixing the other states as well? And the answer is, for some hypothetical hydrides this seems to happen. While in a d2 complex the new $\mathrm{a_{1}'}$ is filled and therefore an elevation in energy quite unvavorable a d0 complex should further distort. Then $\mathrm{a_{1}'}$ and $\mathrm{a_{2}''}$ mix and distort along a normal mode of $\mathrm{a_{2}''}$ symmetry that changes the trigonal prism $\mathrm{D_{3h}}$ to something they call pyramidalization with $\mathrm{C_{3v}}$ symmetry. This step can also be achieved directly from the octahedron by mixing all orbitals at once and twisting a triangular face of the octahedron.

Now the final question was, why do we observe so many trigonal prisms then? And their final remark on this (unless I skipped something) that it's about the ligands. What ligands favor a second order Jahn-Teller effect for a d0 metal center? Basically only strong σ-donors. They mustn't have any π-donating effects. Since the HOMO LUMO gap has to be quite narrow the $\mathrm{t_{1u}}$ orbital has to be high. So no electronegative ligands. But also the $\mathrm{t_{2g}}$ orbitals have to be low, so a strong ligand splitting is favored but in a way that the $\mathrm{t_{2g}}$ is lowered. And most probably a strong nephelauxetic effect is advisable as well. This leads to alkyl and hydride as possible candidates. For the hydride many of the desired complexes are just unknown but the calculations show that is could distort to $\mathrm{C_{3v}}$. For the alkyl like many prismatic organometallic alkyl complexes we know the methyl seems to be to bulky to stabilize the pyramidal structure.
In the same sentence they also mention that for example alkoxides are not forming prisms. So this leaves me with the question how Mo-S or W-O structures in solid state still show the trigonal prism. But that has probably different reasons so I'll exclude them here.

What Mo-S or W-O systems are you thinking about? I'm nowhere near an expert on this, but MoS2 was mentioned in my undergrad lectures - it has edge-sharing trigonal prismatic [MoS6] units which maximises Mo-Mo bonding. But that is a d2 system, not a d0 system.

For Mo-S I was referring to the d2 system. There is a quite good summary on trigonal prisms in solid-state compounds and they also seem to favor a d2 system in general while another paper I read on MoS2 talked about band structures and related calculations to determine the origin of the trigonal prism.

About the W-O systems, I am not sure how much there is actually present / known in literature. Our group has some researchers that study these systems and trigonal prisms appear a lot but I have no idea how much I am actually allowed to tell here. It's just where I got the idea since they often mention second order Jahn Teller without knowing much about it.

To add those two papers, although I am not sure how to add a link in the comments:
DOI: 10.1002/chem.200903032 (for the prism in d2 systems)
DOI: 10.1021/ja00324a012 (for the MoS2 case)


text:
Relationship between the 2nd theorem of Hohenberg and Kohn and the Variational Principle
In Density Functional Theory, the 2nd theorem of Hohenberg and Kohn states:

The electron density of the system that minimises the ground-state energy is the exact electron density of the system.

Similarly, there is also the widely-known Variational Principle in quantum chemistry which states:

The wavefunction that minimises the ground-state energy of the system is the exact ground-state wavefunction.

They seem very similar to me and I would just like to ask if there are any similarities between the two, perhaps in terms of the proofs of these two fundamental theorems.

Meh, does it even take fluxional molecules into account?


text:
How long does a linear hydrocarbon chain have to be before we call it polyethylene?
Might be a really silly question here but here goes. Polyethylene is a polymer with a repeating unit of just 2 carbon atoms attacked to 2 hydrogens each which connect to other repeat units.
This structure is just an aliphatic hydrocarbon like butane or hexane or decane etc. My question is, at what carbon chain length does a hydrocarbon cease to be a normal alkane but instead becomes a polymer?

Actually, polyethylene is still an alkane as it fits IUPAC definition perfectly fine [1, p. 1313]:

alkanes:
  Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms.

Smaller alkane molecules you are referring to, are oligomers. IUPAC defines the prefix oligo- as follows [1, p. 1353]:

oligo:
  A prefix meaning 'a few', and used for compounds with a number of repeating units intermediate between those in monomers and those in high polymers. The limits are not precisely defined, and in practice vary with the type of structure being considered, but are generally from 3 to 10.

At the same time removing numerical restraints when referring to oligomer molecules [2, p. 2289]:

1.2 oligomer molecule
  A molecule of intermediate relative molecular mass, the structure of which essentially comprises a small plurality of units derived, actually or conceptually, from molecules of lower relative molecular mass.

Obviously, the boundary between oligomers and polymers is blurred. Usually for the emergence of "polymer" sufficient degree of polymerization $N$ starts with $10^2 - 10^4$ for synthetic polymers and about $100 - 200$ for globular proteins.
A significant feature of any polymer as compared to its molecular weight counterpart is that when one or several units are added or removed into the polymer molecule, the properties of the polymer material remain almost unchanged. Properties of oligomers substantially depend on molecular weight.
As a typical example, melting points of low molecular weight alkanes gradually increases with polymerization degree (Source):

reaching a plateau of approx. $\pu{400 K}$ for linear low density polyethylene.
References

Moss, G. P.; Smith, P. A. S.; Tavernier, D. Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995). Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375. https://doi.org/10.1351/pac199567081307.
Jenkins, A. D.; Kratochvíl, P.; Stepto, R. F. T.; Suter, U. W. Glossary of Basic Terms in Polymer Science (IUPAC Recommendations 1996). Pure and Applied Chemistry 1996, 68 (12), 2287–2311. https://doi.org/10.1351/pac199668122287.

It's rather molecular mass concept.
According to normatives, MM of polyethylene is from tens of thousands units to several million. So molecule must weight at least 10000 to be considered polyethylene (otherwise it can be named as oligomer of ethylene). But I should mention that there are no precise "threshold" like "10-100 monomers = oligomer, 100+ - polymer", though some textbooks give these (or other) numbers. Line between them is pretty vague to be explained with one simple number. Line is "drawn" by their properties.

The entanglement molecular weight of PE is in the range of a thousand g/mol. Sth with 4000 to 5000 is definitely a polymer already.


text:
Why is STP mol/liter so specific?
How do we know that one mole of an ideal gas occupies (exactly) 22.4 Liters? Why is it so precise a value? Many other constants/conversions in science have multiple decimal places, so why is this conversion terminated so soon?
Yes, I have googled it, and searched it up on StackExchange, and flipped though my chemistry textbook. I just want a nice clear explanation for the exact value of the the mol/liter conversion.
Finally, please don't use the ideal gas constant $\ce{R=8.314 J/K\cdot mol}$ to justify your answer, because we get $R$ by using this conversion fact.
Thank you for any information or insights!
EDIT: This has been suggested as a duplicate of What volume does one mole of an ideal gas occupy?, which does not answer my question because it derives the liter/mol conversion from the ideal gas constant.

How do we know that one mole of an ideal gas occupies (exactly) 22.4 litres?

It doesn't.
The current recommended value for the molar volume of an ideal gas at a temperature of $T=273.15\ \mathrm K$ and a pressure of $101.325\ \mathrm{kPa}$ is $V_\mathrm m=22.413962(13)\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$ (source). Note the given uncertainty; i.e. it is an experimental value  that is not exact.
Also note that these conditions do not correspond to STP. According to current IUPAC recommendations, STP corresponds to a temperature of $T=273.15\ \mathrm K$ and a pressure of $100\ \mathrm{kPa}=1\ \mathrm{bar}$. At this state, the molar volume of an ideal gas actually is $V_\mathrm m=22.710947(13)\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$ (source).


text:
What are radicals in chemistry?
I read somewhere that radicals are atoms, molecules or ions that have unpaired electrons, but how do ions have unpaired electrons when an ion has all its electrons paired either by loosing or gaining electrons to attain the nearest noble gas configuration?

Radicals arise from homolytical breaking of bond - thus, when one electron is given to each atom. Usually this depicted with a "dot".

There is a lot more to it, but... To put it simple, what if you simply won't let an ion loose or gain as many electrons as it needs to have the noble gas configuration?

@Ivan Neretin: so do you mean that ions, which do not attain the noble gas configuration, can be formed? If so,then that explains a lot thanks.

Of course they can. Then again, don't think of this as a universal explanation, because there are more.

but how do ions have unpaired electrons when an ion has all its electrons paired either by loosing or gaining electrons to attain the nearest noble gas configuration?

(Quote from your question, emphasis mine)
Your fallacy is the word I emphasised. Surely it is often taught like that—especially in introductory chemistry—that when ions are generated the noble gas configuration (vulgo: octet or dublet in the case of He configuration) is reached. That is not the whole truth, though.
Especially when you form a polycharged anion such as oxide $\ce{O^2-}$, you can (and should!) consider the electrons being added (or removed) stepwise. Thus, going from an oxygen atom you would initially generate an oxide radical anion $\ce{O^.-}$ before arriving at the classical oxide anion $\ce{O^2-}$. If you look at it energetically, the first electron addition is sometimes (but not always!) exothermic but the second one is always endothermic, so it makes energetic sense to separate like this.
These monoatomic radical ions are not commonly encountered as they typically have very short half-lifes. They often rapidly react to either give the full noble gas ion or the original, neutral atom. However, the vast majority of known ions are molecular or polyatomic ions such as phosphate. Within these, one very common species of radical anion forms rather stable compounds: the superoxide anion $\ce{O2^.-}$.
Structure-wise, it is similar to the peroxide anion $\ce{O2^2-}$ which can be written as $\ce{^-O-O-}$ except with one electron removed. However, it is usually generated  starting from dioxygen ($\ce{O2}$) with one electron added, which turns dioxygen’s double bond ($\ce{O=O}$ in Lewis formality) into a single bond. when the heavy alkaline metals (potassium, rubidium or caesium) are burnt in air, the corresponding superoxide is the product:
$$\ce{K + O2 -> KO2}\tag{1}$$
If lithium or sodium are reacted identically, the product is the oxide or the peroxide, respectivesly:
$$\begin{align}
\ce{4Li + O2 &-> 2Li2O}\tag{2}\\
\ce{2Na + O2 &-> Na2O2}\tag{3}\end{align}$$

From the Gold book http://goldbook.iupac.org/html/R/R05066.html
"A molecular entity such as .CH3 , .SnH3 , Cl. possessing an unpaired electron. (In these formulae the dot, symbolizing the unpaired electron, should be placed so as to indicate the atom of highest spin density, if this is possible.)"
Go to the Gold book for more details.


text:
Conditions for imine formation
Firstly, note that this is not a duplicate of Acid in Imine Formation. Clayden, Warren, & Greeves (2012) states explicitly on p. 231 that imine formation requires acid catalysis, and that the reaction takes place optimally under slightly acidic conditions, close to neutrality. This makes sense as we need to form the $\ce {-OH2^+}$ allowing water to leave. However, if we look at this list of literature, we would observe that most of the imine formation reactions do not necessarily require acidic conditions and some even take place under solely basic conditions. How do we explain this?
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

Most of the imine formation reactions do not start from an aldehyde and an amine. Furthermore, I believe that ‘if we look at this list of literature’ almost immediately constitutes a too broad question because each case is slightly different but I’m willing to leave it open.

@Jan The point I am trying to make from the reference to the list of literature is that acid catalysis is not a necessary condition for the reaction. I do not wish to discuss each particular case of the reaction mentioned in the list.

But in most of those cases involving carbonyls and amines, I can boil it down to ‘X supplies the acidic proton’ or ‘harsh conditions cause higher proton mobility’.

Formation of imines is usually a nucleophilic subsitution of oxigen with nitrogen. For this to happen, the oxigen usually must be bound beforehand as carbon-oxigen bond is quite strong. The most straightforward way is to use acidic catalysis.
Most reactions that you referenced use acidic catalysis, though sometimes it isn't immediately apparent because they use solid catalyst such as molecular sieves a.k.a. H-form zeolites. A few use Lewis acids, such as $\ce{Ti(OEt)4}$. They still count as acids.
The cases when no acid is used are not pure nucleophilic substitution. The reference you provided features various oxidative and other catalytic reactions, proceeding through different mechanism, that do not need acid, and a few reactions where reagent itself bears acidic function, like tosyl isocyanate, that is a strong electrophile and can bind with oxygen on its own.
TL;DR: the cases when acid apparently isn't used usually include it in non-obvious way. Other cases proceed through wild varying different mechanisms, usually involving oxidation.

Good observation regarding the use of solid catalysts. But clearly, there are some reactions that use only basic catalysis on the list. For example, the 1st (J.-M. Huang, J.-F. Zhang, Y. Dong, W. Gong, J. Org. Chem., 2011, 76, 3511-3514) and second (L. Paquin, J. Hamelin, F. Texier-Boullet, Synthesis, 2006, 1652-1656) ones. How do you explain that?

@TanYongBoon In this caset the reaction is performed in heterogeneous system (benzaldehyde has very low solutbility in water) and includes extraction with organic solvent. Unlike the authors, I would rationalize this as reaction proceeding in organic phase, where alkylammonium salt acts as a weak acid, while unsubstituted ammonia remains in water phase. This is in agreement with their finding that the yield of the reaction increases in the row Cl<Br<I; since solutbility of ammonium salts in organic solvents increases in the same row.


text:
What is density in density functional theory?
This may be dumb question. Please bear with me.
Density functional theory (DFT) is a successful theory for electronic structure calculations of materials. In DFT, electron density is the fundamental variable that is proven by the Hohenberg-Kohn theorems. However, I find it difficult to understand how density is used to calculate the energy of the system.
In general, number of electrons per volume will provide the electron density. Okay. Now, I consider Hydrogen atom (for simplicity). If I calculate the electron density, in fact, it depends on how much volume is chosen. If I chose a unit volume, its charge density is 1e. If I double the volume, the charge density halves ($\frac{1}{2}e$). What decides this volume? How to chose this volume?

You're thinking about it in terms of charge density = (finite charge)/(finite volume), and this does change depending on what volume you take (unless it is perfectly homogeneous), but usually what charge density means is (infinitesimal charge)/(infinitesimal volume).

There is no choosing. Density is defined at each point of 3D space, much like $\psi$ function.

The probability density is defined at each point, being the square modulus of the wavefunction. The probability of finding electrons(s) in a given volume is the integral of the probability density over that volume. I have heard this second quantity referred to as the electron density as it tells you how much electron-stuff is in a given volume, but unfortunately this ends up causing confusion due to two uses of the word density which a subtly different ...

"Probability is defined at each point": How these points are defined?

What is density in density functional theory?

The density in density functional theory (DFT) refers to electron density, which is the object of all DFT calculations. DFT differs from various other wavefunction-based methods which focuses on determining the ground-state $\ce {3N}$-dimensional wavefunction*, such as the Hartree-Fock (HF) method. Electron density, often denoted as $\rho$ or n(r), refers to the 3-dimensional function of 3 spatial coordinates describing the distribution of electrons in a given system. It is of interest in DFT as it allows us to determine all observable properties of the system without knowing the exact ground-state wavefunction, hence providing great computational ease. The following is the usual expression of electron density (for spin-restricted calculations, for a system of an even number of $\ce {N}$ electrons, which would have $\ce {N/2}$ different electronic energy levels since electrons in the same orbital are considered to have the energy, thus also meaning that we are summing only $\ce {N/2}$ of the different individual densities, and then multiplying by 2) that you would find in resources on DFT:

*$\ce{N}$ refers to the number of electrons in the given system. The wavefunction is a function of $\ce {3N}$ dimensions since each electron has 3 spatial coordinates. If spin is considered as a fourth coordinate, then the wavefunction becomes $\ce {4N}$-dimensional. Obviously, such a wavefunction would not be easy to solve for exactly.

However, I find it difficult to understand how density is used to calculate the energy of the system.

This is where the functional in the name DFT comes in. The 1st theorem of Hohenberg and Kohn tells us that:

Ground-state energy of the system is a unique functional of its electron density.

What exactly is a functional? A functional is a function of a function. A commonly used functional would be the definite integral. In Kohn-Sham DFT (which treats the system as comprising of non-interacting particles moving in an effective potential), notably the most widely-used version of DFT (another version being orbital-free DFT), the functional can be seen as comprising of three main contributions: 1) kinetic energy of electrons, 2) coulombic interactions between electrons and nuclei, and 3) coulombic interactions between pairs of electrons. However, there are some effects that we would need to take into account. For example, when we consider coulombic interactions of each electron with the electron density (in solving the Kohn-Sham equations), we also include a non-physical self-interaction (i.e. the interaction of the electron with itself). Particles in the system can also undergo exchange interactions and their motions are also correlated (i.e. they are not independent). Hence, another term called the XC (exchange-correlation) functional is also included into the overall density functional expression to account for all these additional effects that have not already been accounted for.
The following is an expression of the Kohn-Sham density functional taken from Wikipedia:

The first term is the electron kinetic energies, the second represents the electron-nuclei interaction energy, the third represents the electron-electron interaction energy, while the last term is the mysterious XC functional.
The site has more posts on the topic which you may find useful:
Are there any full worked examples of DFT calculations?
Density Functional Definition

In the definition of density $n(r)$, what is $\psi$? How do we know that for a given atom atom or molecule?

@phenomenon Psi refers to the one-electron wavefunctions of the molecule. They are obtained as solutions to the Kohn-Sham equations, which are a set Schrodinger-like equations each describing the individual one-electron wavefunctions. Please refer to DFT textbooks for more information.


text:
Effect of increasing product concentration on time of completion of reaction
This was the question asked in our chemistry exam:
In the experiment of determination of equilibrium constant of esterification of ethanol, if , instead of using
clean dry bottle, if a bottle already containing slight amounts of ethanoate and water is used to carry out the
reaction under identical (remaining) conditions, what would be the consequence?
According to our lecturer it's answer is "The reaction will take lesser time to reach equilibrium". But according to Le Chatelier's principle, increase in concentration of products shifts the reaction backwards. So it should take more time to reach equilibrium.
Please help!!

In a way, equilibrium is never reached at all.

@IvanNeretin Why?

https://chemistry.stackexchange.com/questions/57075/is-there-no-actual-equilibrium

In equilibrium reactions the equilibrium is approached exponentially in time. The lifetime of the exponential is the reciprocal of sum of the forwards and back rate constants multiplied by the equilibrium concentration of the species, thus for reaction $\ce{A + B \leftrightharpoons C + D }$ the lifetime of the reaction is $\tau = 1/[k_1 (A_e+B_e) + k_{-1} (C_e+D_e) ] $ where $k_1$ is the forward rate constant and $k_{-1}$ the backwards one, and subscript $e$ means equilibrium amount of a species. Thus if you add more material you change the equilibrium concentrations  and so the lifetime gets shorter as its a reciprocal in the concentrations
The lifetime is the time taken to change to $1/e$ of the initial amount, approx to 38%, whereas the half-life is the time taken to halve the initial amount. The lifetime is the reciprocal of the total rate constant for reaction which is why it is used instead of the half-life.
(The exception is the simple $\ce{A\leftrightharpoons B}$ reaction when the lifetime is independent of the equilibrium amounts ). In equilibrium reactions the lifetime is also called the relaxation time.


text:
Why does LDA solution gradually turn dark brown?
LDA (lithium diisopropylamide) solution in THF/hexane gradually turns red-brown liquid. Especially commercially available one is extremely dark red-brown. What makes it brown?
I've searched online journals, but I couldn't find possible theories.
When I synthesize LDA, at first LDA solution in THF/hexane is light yellow. Of course, I keep it under nitrogen/0-10 degrees celsius but it gradually turns brown.
Does LDA react THF or the THF stabilizer BHT? Do they make LDA solution brown? Please give me any plausible explanations.

BHT is a phenol. It would be deprotonated and unlikely to be in the solution. Butyllithium reacts with THF to form ethylene and the enolate of acetaldehyde. Perhaps LDA does this slowly. Traces of moisture could protonate some of the enolate leading to base catalyzed condensation products that are brown from oxidation.

It seems to be plausible. Thank you for your kindness!


text:
Decomposition vs Combustion reaction
I'm balancing chemical equations, and I came across the following:
$$\ce{O2 + C4H9NH2 ->  CO2 + H2O + N2}\tag{unbalanced}$$
My initial reaction was that it is a combustion reaction. According to my workbook, the definition is "an organic compound reacts with oxygen to form carbon dioxide and water." This equation seems to fit the bill. However, the solution states that in fact this is a decomposition reaction. In a YouTube video I saw, it said that decomposition reactions are a single reactant to multiple products. Where is the misunderstanding? Is it possible the workbook is wrong?

I would say your interpretation is more correct than the book's.

The workbook is wrong on some many levels. It is an oxidation, and you'd not get N2 but a nitrogen oxide of some sort depending on temperature and pressure.

@MaxW I think OP could benefit from you turning that into an actual answer ;)

You're right, I'd consider that a decomposition reaction is primarily a single chemical species breaking into other chemical species. Here is the IUPAC Gold Book definition for decomposition.
The IUPAC definition specifically mentions two phases, which implies that the decomposition is "irreversible." In other words, there isn't a significant backwards reaction.
The molecular formula $\ce{C4H9NH2}$ could be a number of different aliphatic primary amines, but all of them are stable under ordinary conditions. In other words, you'd have to "burn" them in a combustion reaction with oxygen to get $\ce{CO2}$ and $\ce{H2O}$. In such a reaction I would absolutely not expect to get $\ce{N2}$ as a product. Rather I'd expect some sort of nitrogen oxide $\ce{NO_x}$. Which one would depend on temperature, pressure, and the proportion of oxygen.
So the unbalanced reaction would better be written as:
$$\ce{O2 + C4H9NH2 ->[\Delta]  CO2 + H2O + NO}_x$$

I disagree. When you burn simple organics, you get most of your nitrogen as N2. Maybe a few percent go to NOx.

@IvanNeretin - For both n-butyl amine and for tert-butylamine PubChem says "Produces toxic oxides of nitrogen during combustion."

Well, the mentioned few percent are quite enough of a reason to bother about toxicity.


text:
Cs Point Group's "Horizontal" Reflection Plane
Why is it that $C_s$ character tables always list a $\sigma_h$ operation despite the definition of $C_s$ not necessitating the reflection plane be horizontal?
In fact, how can the reflection plane even be characterised without a principal rotation axis to base it on?

There is always a principal rotation axis, even if it's only a $C_1$ axis. As there are an infinite number of these one can always be found perpendicular to the reflection plane, and as horizontal planes take precedence over veritcal ones, the operation is listed as $\sigma_h$.


text:
EPDM tube resistant to ozone on the long run?
I am working with ozone produced with an ozone generator and pure oxygen.
The resulting O2 / O3 gas mixture is probably about 80-85% O2 and up to 15-20% O3.
The generator does produce about 500g of O3 per hour.
Currently I am using PTFE tubes to bring the O3 where it is needed.
However I am not happy with the mechanical properties of the PTFE tubes, as they are stiff and as a result they are uneasy to work with.
I have found now that EPDM tubes are way easier to work with, but I have doubts about their resistance to O3.
While many datasheets and Wikipedia do tell that EPDM is resistant to O3, I question if it is really resistant to extreme high concentrations of O3 on the long run.
Switching from PTFE to EPDM in the future would be a good thing as EPDM is also cheaper and easier to work with.
Does anyone here have experience with high concentrated O3 and EPDM?

The problem with EPDM is that it has tertiary carbon sites (and some double bonds for cross-linking). Forget the latter as let's assume the material is already cross-linking.
Normally carbon backbone polymers with substitution need anti-oxidants to prevent the C-H, where there is a side group, from oxidising. These delay the oxidation process and define the lifespan of the material. So after time, the anti-oxidants will be consumed and the polymer will be prone to oxidation, which leads to fragmentation.
Ozone is a great oxidant as you know. When we used high pressure ozone,  we always used fluorinated everything. By using fluorine saturated polymers, there's no sites for oxidation to occur.

The remaining double bonds from the (usually unconjugated) diene compound are not in the backbone, because otherwise the name would be EPDR, R for rubber (which is not a common material afaik).
That makes it resistant to oxonolysis, which cuts the backbone of butadiene or isoprene based rubbers, especially under mechanical load, because the latter exposes more and more new double bonds to the attack. This is what makes the well known cracks on the sides of older tyres.
Also EPDM products are always vulcanised, or otherwise crosslinked, so your tubing does not contain any significant amount of double bonds that could get ozonolysed.


text:
Endo/Exo for same length bridges
I'm looking at a cycloaddition that produces a bicyclic intermediate, there are two different ways of doing the addition which can be described as endo/exo. However, all three bridges are the same length. How do I assign priorities and endo/exo here?
Here's a simplified example of this reaction. In my opinion on the dienophile the $\ce{NH2}$ has the higher priority, on the pyridazine I would suspect the $\ce{N-N}$ bridge to be higher in priority which means the left product would be endo, the right one exo. But I cannot find information that confirms this.

To my knowledge the endo/exo rules have as yet not been incorporated into the CIP system. Your idea has merit though. [Although your DA reaction is offered as an example, it is unlikely that the enamine tautomer of the dienophile would prevail over the imine.]

@user55119 well that would explain why I cannot find any information on this. And yes I know about the hypothetical tautomer and the product will eliminate N2 right away, but since I'm doing calculations everything is possible. ;-) That was just a very simple system with all the key features I needed.

Endo/exo stereodescriptors cannot be used for the preferred IUPAC names anyway.

@mykhal I need it to describe the configuration of the transition states and I don't see an easier way to do this.

IUPAC not withstanding, a CIP-like nomenclature could be developed. Just as E/Z covers all double bonds and cis/trans does not, endo/exo could be replaced with a more inclusive CIP system. E.g., your NH2 group has precedent over C6H5 while N=N has priority over C=C. Assuming the NH2 is bold in the first structure and dotted in the second, so your first DA adduct would be Z-ish and the second structure E-ish.  ;)

@user55119 that "Z-ish" and "E-ish" is exactly endo/exo...


text:
Does concentration or pKa define acid strength?
For example, is 3M HCl a stronger acid than 1M HCl?
I would reason that the concentration of an acid/base does not influence its strength. Strength is determined by the pKa, and, as per Le Chatelier's Principle, the initial concentration does not influence the equilibrium constant.
It may shift the equilibrium concentrations (meaning that the pH is higher for the 3M HCl), but it will not change $K_a=\frac{[H_3O^+][Cl^-]}{[HCl]}$ at equilibrium.
Is this right?

Your reasoning is incorrect.

What's hard about this? 3 > 1 so there is more acid and hence it is stronger in that sense. Also HCl ionizes fairly completely so the pH is different too.

@MaxW, What then determines the strength of an acid? Is it the amount of H+ or H3O+ ions in the solution? How is that any different from the pH?

@MaxW, really what I am asking is: what does acid strength truly mean?

It means nothing, unless you define it to mean something. That's why, BTW, scientists tend to stick to the terms with precise meaning (concentration, pKa, etc.)

Whole solution is an acid - compound may act as acid or not. Calling HCl itself an acid is a sort of classification made with usual function.

The term ‘strong acid’ is sometimes used in a rather fuzzy way and you ran into problems doing so. I prefer to use the term ‘strong acid’ only with respect to an acid’s $\mathrm pK_\mathrm{a}$ value and disregard all other influences. This gives a clearly defined measure of acid strength and we can easily sort various acids by their strength into stronger or weaker acids.
However, this is looking at the acid as a molecule. In real-life applications you are typically more interested in the property of a solution. To give a real-world example, imagine a $\pu{1M}\ \ce{HBr}$ solution and a $\pu{12M}\ \ce{HCl}$. solution. Obviously, $\ce{HBr}$ is the stronger acid, but the concentration of $\ce{HCl}$—also a strong acid and thus fully deprotonated—is higher. Therefore, the $\ce{HCl}$ solution is more concentrated or, as some would say, more acidic. It can do greater harm and it is able to protonate more Brønsted base molecules than its $\ce{HBr}$ counterpart.
If instead of examining a $\pu{1M}$ and a $\pu{12M}$ solution I had examined a $\pu{e-3M}$ and a $\pu{1M}$ solution, we could even base the discussion around the solution’s resulting pH: the stronger acid is much more diluted and will result in a solution of pH 3 while the weaker acid results in a solution of pH 0.

Normally, when comparing Acid Strengths between two acids, the first criterion used is their respective value for Ka at the same temperature. Obviously the one with a higher Ka at the same temperature is "stronger".
If you have two different acids (let's say HF and HCl) with the same pH, their strength can be compared in terms of their conversion ratio (a ratio that represents acid dissociation in solution).
Conversion = ax/CAo
Where:
a = stoichiometry coefficient for the acid in its dissociation reaction;
x = change in acid concentration to reach equilibrium;
CAo = Initial concentration of acid
For both HCl and HF, a = 1 due to their corresponding dissociation reactions.
For HCl, it can be assumed that CAo = x, so its Conversion is 1 or 100%
For HF, CAo >> x, so its Conversion is << 1 or << 100%
Therefore, HCl is the stronger acid.
However, since you're comparing the same acid (HCl) at different concentrations (and same Temperature, I'm assuming), concentration should be used as a secondary criterion because both the Conversion and Ka remain constant.
Since 3M is higher than 1M, the former should be regarded as the stronger acid.


text:
What's the use of adding 73 mM H2O2 to a supertanant?

Enzyme assays
Determination of catalase specific activity
The CAT (EC 1.11.1.6) enzyme activity was measured according to Gallego et al.38 Plant samples were grinded in $\pu{0.1M}\ \ce{KH2PO4}$ buffer (pH $7.0$) and centrifuged at $\pu{4500 rpm}$ for $\pu{5 min}$ (Eppendorf AG bench centrifuge 5417 R, Hamburg, Germany). The supernatant was separated, added with  $\pu{73 mM}\ \ce{H2O2}$, and measured at $\pu{240nm}$ through UV/Vis (Perkin Elmer Lambda 14 Spectrometer).

The questions I want to ask are:

Why is $\ce{KH2PO4}$ used?
Why centrifuge it at 4500 rpm?
Why is $\ce{H2O2}$ added to the supernatant?

As an addition to iad22agp's absolutely correct answer: In catalase assays the decomposition of H2O2 is measured at 240 nm as this is a region where H2O2 absorbs while using rational concentrations. At the same time O2, H2O and CAT (in used concentration) do not absorb. (Primary literature see here)
One unit of catalase will decompose 1.0 µmole of H2O2 per minute at pH 7.0 at 25 °C. The rate of disappearance of H2O2 is followed by observing the rate of decrease in the absorbance at 240 nm. For further calculations see here.
You may also read the helpful literature for this assay which is actually given in your own text. See here: Gallego et al. 1996

It appears that grinding with phosphate buffer extracts the soluble enzyme catalase, centrifugation separates solid plant material to give a clear solution containing the enzyme to be assayed, and the hydrogen peroxide acts as the substrate for the enzyme. Catalase converts the hydrogen peroxide to oxygen and water. The spectrophotometer is then used to measure the rate of the reaction, and deduce the amount of enzyme present, but what exactly is being detected is not clear.


text:
confusion about exocyclic double bond

I believe that there are only two exocyclic bonds in this structure. But in some text its mentioned as three.
can someone clarify.

IMHO there are total zero exocyclic bonds.

I agree. Exocyclic means outside the ring structure; the double bonds are all within the ring structure

but, ring B has two exocyclic bonds

@RZZ …relative to that ring. yes, but they are still in another ring. It depends on the term exocyclic bond purpose and context.

ok. so for the purpose of calculating the Lambda max using woodward fieser rule. how many do you think are there?

@RZZ This is exactly a type of contextual info that should be included in the question text. There's an "edit" link below it (besides others).

By definition (in Woodward-Fieser rule), an exocyclic double bond is a double bond where one of the $sp^2$ carbon atoms of the bond is a part of a ring, while the other $sp^2$ carbon of the bond is not a part of same ring (it could be a part of another ring).
If both the $sp^2$ carbon atoms of the double bond is part of the same ring, then it is called endocyclic double bond. From these definitions, you can understand that exocyclic would stand for a double bond outside the ring and endocyclic would stand for a double bond within the ring. Following examples would illustrate the definition:

yup!! agreed. so now according to this definition how many exocyclic double bonds are there in the structure i posted?

You may count it by yourself. You'd learn by that.

Three is the correct answer. One double bond is exocyclic to both cyclopentyl and cyclohexyl rings.


text:
Using thermodynamics to predict the acid-base character of fluoride ion in water
I was working through the end-of-chapter exercises of the acid-base chapter in Shriver's Inorganic Chemistry when I came across the following problem:

4.7. The effective proton affinity $\ce{A^{'}_{p}}$ of $\ce{F-}$ in water is 1150 kJ $\ce{mol^-}$. Predict whether it will behave as an acid or base in water.

I was initially stumped by this problem since the $\ce{A^{'}_{p}}$ of $\ce{F-}$ is between the $\ce{A^{'}_{p}}$ values of water and hydroxide ion (1130 kJ mol- and 1188 kJ mol-), which tempted me to say $\ce{F-}$ is amphoteric in neutral water even though I know it's basic from general chemistry.
The answer that the book's solution manual provides is the following:

The proton affinities of H2O (l) and OH-(aq) are 1130 kJ mol- and 1188 kJ mol-, respectively. Consequently, the reaction below between $\ce{H3O+}$ and $\ce{F-}$ is exothermic by -20 kJ mol- (1130 kJ mol- minus 1150 kJ mol-):
$\ce{ H3O+ (aq) + F- (aq) -> H2O (l) + HF (aq)}$
The reaction below between $\ce{H2O}$ and $\ce{F-}$ is endothermic by 38 kJ mol- (1188 kJ mol- minus 1150 kJ mol-):
$\ce{ H2O (l) + F- (aq) -> OH- (aq) + HF (aq)}$
Therefore, in neutral water, $\ce{F-}$ will behave as an base.

I understand that the first reaction has an approximate enthalpy change of -20 kJ mol- and is favored because it is the reverse reaction of the following reaction between $\ce{HF}$ and $\ce{H2O}$ and reaction between acids and $\ce{H2O}$ are favorable when the $\ce{A^{'}_{p}}$ of $\ce{H2O}$ is greater than the $\ce{A^{'}_{p}}$ of the conjugate base:

$\ce{ H2O (l) + HF (aq) -> H3O+ (aq) + F- (aq)}$

However, I have a few points I don't understand:

Why does the text consider the second reaction with $\ce{H2O}$? Doesn't $\ce{F-}$ act as a base in both reactions?
If both reactions occur and $\ce{F-}$ acts as a base, then which one is dominant?
If the first reaction is dominant, then how does it proceed given the low concentrations ($\ce{1 x 10^{-7} M}$) of $\ce{H3O+}$ in neutral water?

The textbook walks us through the interpretation of these enthalpies (they also caution us not to use enthalpies as sole criterion for spontaneity of reactions):

So they are asking whether water reacts with the acid HA, and whether water reacts with the conjugate base $\ce{A-}$.
If we try that for the case at hand, we first consider whether HF reacts with water:
$\ce{H2O(l) + HF(aq)->[?] H3O+(aq) + F−(aq)}$
Here, water and fluoride are the bases "fighting for the proton" (reactants: fluoride has the proton; products: water has the proton). Fluoride wins (i.e. HF does not give away the proton) because it has the higher proton affinity, and the reaction does not proceed judged by enthalpy alone.
Second, we consider whether the fluoride ion reacts with water:
$\ce{H2O(l) + F-(aq) ->[?] OH-(aq) + HF(aq)}$
Here, fluoride and hydroxide are the bases "fighting for the proton". Hydroxide wins because it has the higher proton affinity, i.e. this reaction also does not proceed judged by enthalpy alone.
So HF is not a strong acid, and fluoride is not a strong base. Another way of saying this is that hydroxide is a stronger base than fluoride, and hydronium is a stronger acid than HF.
In the textbook walk-through, they conclude that HA is a strong acid and - for a different reaction - $\ce{A-}$ is a strong base. In our case, HF is a weak acid  and fluoride is its conjugate weak base. If you try the same question with HCl, you will see it is a strong acid and chloride is not a strong base. (I don't understand the conclusion of the solution manual - they might have gotten confused because they know fluoride is a weak base, but their calculation together with the textbook argument merely shows that it is not a strong base.)
Now to your questions:

Why does the text consider the second reaction with H2O? Doesn't F− act as a base in both reactions?

I think they ask "does HF react with water?", and "does fluoride react with water?". I don't know why the answer flips one of the reactions (maybe the solution manual wasn't written by the textbook author, and the original intent of the question got lost).

If both reactions occur and F− acts as a base, then which one is dominant?

There are three equilibrium reactions, the two shown, and auto-dissociation of water. I don't know the kinetics of the three, but they are all fast and the three equilibrium constants are linked because if you subtract the two reactions you show, you get the auto-dissociation of water. For the equilibrium state, it does not matter which set of two you consider.

If the first reaction is dominant, then how does it proceed given the low concentrations (1x 10−7 M) of H3O+ in neutral water?

Reaction rate constants vary by many orders of magnitude. The reaction of fluoride and hydronium is nice because the reactants are of opposite charge, so maybe it has a high rate constant. The reaction of fluoride and water is nice because water is the solvent, so it is present at high concentration and already at close proximity. I could not tell you which reaction is faster even if I know the concentrations of all the reactants.
Here is the table of proton affinities as a quick reference:


text:
Equivalent conductivity of barium chloride at infinite dilution

Find out the equivalent conductivity of barium chloride at infinite dilution given that the ionic conductivities of $\ce{Ba^{2+}, Cl-}$ are $\pu{127 ohm^-1 cm^2 equivalent^-1}, \pu{76 ohm^-1 cm^2 equivalent^{-1}}$ respectively.

Equivalent conductivity is defined as.

The equivalent conductance of an electrolyte is defined as the conductance of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.

One gram equivalent weight would be simply half of the entire compound so
$$\text{Equivalent Conductivity} = \dfrac{1}{2} \times 127 + \dfrac{1}{2} \times 2 \times 76$$
This comes out to be 139.5 however the answer is 203.

https://chemistry.stackexchange.com/questions/7833/what-is-an-equivalent https://chemistry.stackexchange.com/questions/26288/confused-on-normallity

One must be really careful with the units while dealing with conductivity, conductance problems.
If you read the question carefully, the equivalent conductivity of $\ce{Ba^{2+}}$ and $\ce{Cl^{-1}}$ are provided to you.
So the molar conductivity of $\ce{Ba^2+}$ is: $2 \times 127 ~\pu{ohm^-1 mole^-1} = 254 ~\pu{ohm^-1 mole^-1}$
and that of $\ce{Cl^{-1}} = 76~\pu{ohm^-1 cm^2 mole^-1} $
Now apply Kohlrausch's law of molar conductivity of solution at infinite dilution:
$\lambda^o_{\ce{BaCl2}}= \lambda^o _{\ce{Ba^{2+}}} + 2\lambda^o_{\ce{Cl^{-1}}} \\ \implies  \lambda^o_{\ce{BaCl2}} = 254 + 2\times 76 = 406 ~\pu{ohm^-1 cm^2 mole^-1}$
Now, as you say, equivalent conductivity is $\dfrac{1}{2}$ times the molar conductivity for $\ce{BaCl_2}$ so equivalent conductivity of $\ce{BaCl_2} = \dfrac{1}{2}\times 406 = 203 ~ \pu{ohm^-1 cm^2 equivalent^-1}$ which is the correct answer.
Alternatively, observe that $203 = 127 + 76$ so Kohlrausch's law for equivalent conductivity of strong electrolyte at infinite dilution may be stated as:

The equivalent conductivity of a strong electrolyte at infinite dilution is equal to the sum of the equivalent conductivities of the anions and cations.

So, ionic conductivity and equivalent conductivity are the same thing?


text:
Do microwaves heat polar molecules or molecules with polar bonds?
IR is only absorbed my molecules with polar bonds
regardless of the overall polarity of the molecule , what about Microwaves ?

Sure, water is one example.

It depends on the strength of the incident EMF. Pure $\ce{H2}$ can be ionized by microwave radiation in multi-photon collisions, as well as being heated. Even nonpolar molecules have quantum fluctuations in the electron cloud, so they interact to some extent with n electromagnetic field.

Microwave radiation can only excite rotational motion in molecules as this needs very little energy. IR excites vibrations which generally need more energy.  An oscillating dipole is needed for IR absorption thus HCl can absorb as can CH$_4$ as some of its vibrational motion can generate a transient dipole. Thus no permanent or oscillating dipole no absorption of the radiation, i.e. O$_2$ or N$_2$ molecules do not have IR or microwave absorption. (Raman scattering is used to measure their vibrational and rotational energy levels.)
In rotational motion a permanent dipole is required for photon absorption to occur. The rotating molecule is seen as an oscillating electromagnetic field by the radiation and if it is if the correct frequency energy is passed from the (microwave) radiation into the molecule, i.e. a photon is absorbed. The molecule now has one more quantum of rotational energy and the microwave photon no longer exists.


text:
Test for the dissolved iron
Ok there’s this question on my chemistry review that I can’t figure out. It states:
A sample of well water is known to contain a high concentration of iron. what solution could you use to test the water to get a positive precipitate test for the dissolved iron?
I don’t understand it because how can the iron dissolve in water?
I tried writing that NaOH would be able to create a precipate with iron because FeOH would be insoluble together but Na is more reactive than Fe so that wouldn’t displace.
Can someone explain the answer to me (by the way the teacher wrote that there are many solutions and NaOH is one of them)

Have a look here for a specific test for dissolved iron http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/qualitative-analysis#iron_ions

I'm voting to close this question as off-topic. You have multiple issues understanding what is being asked of you that would require a rather elaborate answer touching on too many topics. Furthermore, it is best if your teacher clears your misconceptions.

Iron dissolved in water is $\ce{Fe^{2+}}$. It can be present as $\ce{FeS,FeCO3,FeSO4.7H2O}$, but (for natural water sources) usually as $\ce{Fe(HCO3)2}$. In order to precipitate iron, you may use any method used to remove $\ce{Fe^{2+}}$ from water (so you should somehow obtain $\ce{Fe^{3+}}$, which will precipitate in $\ce{Fe(OH)3}$ form). Industrial water treatment usually offers two options:

Oxidation using $\ce{O2}$: $\ce{4Fe(HCO3)2 + O2 + 2H2O = 4Fe(OH)3 + 8CO2}$

Sometimes (especially when water is really "high" in iron), you can simply observe water in the bucket "rusting" with time ($\ce{Fe(OH)3}$ has brown-yellowish colour). Industrial methods use aeration columns - this is much faster process than "wait for water in the busket".

Oxidation using iron removers (they may contain $\ce{MnO2, KMnO4, H2O2, O3,}$ etc.)

One of examples is $\ce{4Fe(HCO3)2 + 3MnO2 + 2H2O = 4Fe(OH)3 + MnO + Mn2O3 + 8CO2}$.
So, basically, there are two options:

Open reservoir (or put $\ce{O2}$ there). Wait.
Add something.

There's elementary misconception in OP question - thinking iron was supposed to be metallic. Another thing is that counterion is pretty much irrelevant, you'd'n't make mistake with FeS without this discussion.


text:
What does 'a' stand for in 'a-C' according to research paper lingo?
I came across this paper but I cannot figure out what is a-C as it is not defined. The C here is obviously Carbon. Regarding 'a' prefix My best guess is that it stands for amorphous. Is this correct?

a for amorphous is totally correct. Also used in other publications like here.


text:
Does "higher symmetry" mean any symmetry group of higher order?
What is the precise definition of a higher symmetry in group theory? I see this term scattered around, such as in Cotton's Chemical Applications of Group Theory:

...In the cyclic groups each of the operations, Cn, Cn2, Cn3, ..., , Cnn-1, constitutes a class by itself and we continue to use this notations. However, in all other groups of higher symmetry, the number of classes spanned by these operations will be reduced in the following way...

Does higher symmetry strictly mean a supergroup? Or is it a quantitative statement about something like the number of symmetry elements or number of irreducible representations (which are not always ordered the same way, as in the case of C3v and S4).
I am failing to find intuition because, if it does mean strictly supergroups, we can't compare the "amount" of symmetry in C2 compared to C3 (or Td compared to Ci even!).

"Higher symmetry" per se does not mean anything, unless you define it to mean something.
Depending on the context, it usually means one of the following things:

Any symmetry that is a supergroup of the group in question. In this case $C_{4v}$ is higher than $C_4$, while $C_6$ is incomparable to both. Bear with it. Partially ordered sets are quite a thing in math.
(rarely used) Any symmetry group having more elements than the group in question, so that $C_{4v}$ is higher than $C_6$, which in turn is higher than $C_4$.
(apparently, the meaning used by Cotton) Any symmetry group which is not a cyclic rotation group. In this case $C_{4v}$ is higher, while $C_4$ and $C_6$ are lower.

See, there is an intuition, after all. Worse yet, there are quite a few different kinds of intuition.
So it goes.


text:
Is fluorine sp3 hybridized in BF3
Since BF3 has the shape of tetrahedral, and in the provided solution, Boron is sp2, why is Fluorine sp3 hybridized?

Consider the lone pairs and bonds of fluorine simulatensously for first ding the hybridization

Tetrahedral is one thing BF3 is not.

First things first: $\ce{BF3}$ is not tetrahedral. It cannot be, as a tetrahedral structure requires a total of five atoms (one in the centre and four in the corners).[1] Instead, it is trigonal planar. However, that is of minor importance.
Secondly, fluorine is most certainly not sp3 hybridised. In general, hybridisation is determined by geometry and not vice-versa, thus we only need to establish the trigonal planar geometry to arrive at boron’s sp2 hybridisation. However, this doesn’t help us for terminal atoms as there is no ‘geometry’—only a single binding partner. In some cases, constraints may exist due to multiple bonds which require π-type bonding orbitals. Again, this is not the case for single bonds which rely solely on their σ orbital.
Instead, the basic assumption should be to assume the hybridisation as low as possible. In that way, fluorine’s s type orbital can be as low as possible in energy—which means a higher energy gain. It is easy and possible to assume fluorine bonds solely with a correctly aligned p orbital and so it does.

Note:
[1]: This sentence depends on the system of classification you were taught. I was taught only to label those molecules as tetrahedral which have five atoms that correspond to the four corners plus the centre. If there are four atoms as in the case of ammonia ($\ce{NH3}$), it is possible for the overall structure to be a tetrahedron with one corner removed. We were taught not to call this tetrahedral but a trigonal pyramid. Your teacher’s mileage may vary.

I object. For example, NH3 has but four atoms, yet is tetrahedral all right.

@IvanNeretin I object! We used the term trigonal pyramidal for compounds where the electron lone pair is the ‘fifth atom’, since electrons are not visible in structures and thus the atom positions give a flatter trigonal pyramid, not a tetrahedron.

All right, I stand corrected.

This dispute about tetrahedral and trigonal pyramidal can be easily resolved. The former describes the molecular geometry (i.e. with respect to ligands around given central atom) while the latter describes the electronic geometry (i.e. with respect to regions of electron density around a given centre).

The answer is - they are not really sp3 hybridized. Usually (for low-level textbooks) reasoning goes like "since we have 1 bond and 3 lone pairs, fluorine has 4 electron-ish things => fluorine is sp3". But actually, the closer element is to 18 group (and F is in 17), the less preferable becomes hybridization of s and p orbitals. For fluorine, this is (almost completely) not preferable process, so you shouldn't consider it sp, sp2 or sp3 hybridized.

Why closer to 18 group tells that hybridization of s and p orbitals may not be preferable? @KellyShepphard


text:
Does electrolysis of aqueous tin(II) chloride produce Cl₂, SnCl₄ or both?
When applying a current to an aqueous solution of tin(II) chloride, tin crystals grow from the cathode. However, I'm not sure about the anode. It seems like there are two possible reactions:
$$\ce{2 Sn^2+(aq) + 4 Cl-(aq) -> Sn(s) + SnCl4(aq)}$$
and
$$\ce{Sn^2+ (aq) + 2 Cl- (aq) -> Sn(s) + Cl2}$$
Online sources that I've found — including an educational lab and YouTube videos of the experiment being performed — give inconsistent answers. I'm thinking that both products are produced; if that is the case, how can I predict exactly how much of each will be produced? Can other factors, e.g. current, electrode material, pH change the results?

Let's try to look at those electrons. C - cathode, A - anode.
$$
\begin{align}
&\mathrm{C(-):} &\ce{Sn^2+ + 2e- &→ Sn^0} \\
&\mathrm{A(+):} &\ce{2Cl^- &→ Cl2^0 + 2e-}
\end{align}
$$
So, your second reaction is right. But the first is wrong: $\ce{Sn}$ is reduced, but there are no oxidation. You should use
$$\ce{Sn + 2 Cl2 → SnCl4}$$
But I should mention that this reaction is for $\pu{115 °C}$, so don't worry, it won't go for your conditions.
Another way is
$$\ce{SnCl2 + Cl2 → SnCl4}$$
It requires a lot of work, bubbling $\ce{Cl2}$ through aqueous solution and distillation in the end, so you also shouldn't worry.
You can basically think that all you $\ce{Sn}$ that becomes a part of new structure goes into solid $\ce{Sn}$. It's acceptable since synthesis of $\ce{SnCl4}$ is not an easy thing.


text:
Can the oxidation of hydrogen peroxide be reversed?
Hydrogen peroxide is said to be unstable, for it undergoes auto-oxidation on standing/heating:
$$\ce{2H_2O_2 -> 2H_2O + O_2}$$
where $\Delta S=\pu{70.5 J {mol}^{-1}K^{-1}}$ and $\Delta H^{\Theta} = \pu{-98.2 kJ {mol}^{-1}}$.
I speculate if the decomposition can be reversed in some way under suitable conditions, particularly at infinite pressure and near absolute zero temperature. At once this seems to be a problem of entropy reversal. But I seek a more rigorous treatment of the problem with the help of quantum theory. Consider the molecular wavefunctions $\Psi_{1A}$ , $\Psi_{1B}$ of water (2 molecules) and
$\Psi_2$ of dioxygen respectively. By studying the interaction potentials of these three wavefunctions as a function of different thermodynamic coordinates, it can be possible to find the desired suitable conditions.
However, the problem begins here – this is not a case of just two molecules where we have a number of methods of analysing the system, I am talking about three molecules, so the calculation is not simple. Moreover, I doubt if I need to study the wavefunction of hydrogen peroxide too in some respect. Overall, I am lacking some theoretical and computational knowledge. Any help is greatly appreciated.

You could make a start by looking up  information about a related reaction such as OH+OH studied via what are called LEPS potentials.

@porphyrin, the LEPS potentials talk about spatial coordinates of the molecules, on the other hand I am interested in thermodynamic coordinates i.e. $P$ and $T$.

Yes, if you use some quantum chemistry tool like density functional theory to find the enthalpy of the compounds at many different conditions then you can determine where hydrogen peroxide is the ground state.
You would probably find this at low temperatures and high pressures since it is denser than the other two phases and oxygen is condensed as a liquid or solid so the entropic contribution is smaller.

I guess hydrogen peroxide is never the ground state. True, I didn't run the calculations at wide range of conditions, so it is just a guess. But then again, it is a somewhat educated guess.

@Ivan Neretin, in that case it must be possible to prove  that hydrogen peroxide is not the ground state.

Of course it must be possible, but it is a great deal harder than you seem to think, and even harder than that.

Density functional approximations are not nearly accurate enough to predict any of these values with any confidence. And then there is the problem of treating a condensed phase the same as a 'gas'... and other problems.

The scenario in question would probably occur when all three phases are at low temperature and high pressure, and all three crystallized. DFT would be fine in that case.


text:
A radioactive element that only emits beta rays
Are there any radioactive elements which only emit beta rays?
If not, are there any radioactive elements that only emit beta and gamma rays?

Why, many. $^{14}\rm C$ will do.

And how rapidly does it emit electrons? Also, it would be appreciated if you could give a comprehensive list of all such isotopes.

Have you tried searching for "list of pure beta emitters"?

@AbdulMoizQureshi "And how rapidly does it emit electrons?" – Do you know the concept of activity, decay constant, and half life?

Loong, no, only halflife. I also know that the more the halflife, the less tge radiation.

Ian Bush: Thanks! I tried a lot of other wordings that amounted to the same thing, but got no relevant results.

In the long list of nuclides that are considered for radiation protection (taken from ICRP, 2008. Nuclear Decay Data for Dosimetric Calculations. ICRP Publication 107. Ann. ICRP 38 (3)), we can find the following pure beta emitters and their half-lives:
H-3 (12.32 a)
Be-10 (1.51E+6 a)
C-14 (5.70E+3 a)
Si-32 (132 a)
P-32 (14.263 d)
P-33 (25.34 d)
S-35 (87.51 d)
Ar-39 (269 a)
Ar-42 (32.9 a)
Fe-60 (1.5E+6 a)
Ni-63 (100.1 a)
Ni-66 (54.6 h)
Se-79 (2.95E+5 a)
Rb-87 (4.923E+10 a)
Sr-90 (28.79 a)
Zr-93 (1.53E+6 a)
Ru-106 (373.59 d)
Pd-107 (6.5E+6 a)
Cd-113 (7.7E+15 a)
Cd-118 (50.3 min)
In-115 (4.41E+14 a)
Sn-121 (27.03 h)
Cs-135 (2.3E+6 a)
Re-187 (4.12E+10 a)
Pt-202 (44 h)
Pb-209 (3.253 h)
Bi-212n (7.0 min)
Cf-255 (85 min)

I'd upvote, but don't have enough reputation. Which one of those has a high rate of beta particle emission?

@AbdulMoizQureshi - you are given the link to the reference in the answer. Perhaps you should look through that to find the information you want.


text:
Does hydrazine crystallize when freezing?
When discussing problems of keeping hydrazine from freezing on spacecraft we stumbled upon a question: can pure hydrazine separate through fraction-freezing from the stabilizing substances like UDMH? That would largely depend on what form it takes. If it crystallizes, it's a risk - it's quite unstable when pure. If it turns amorphous, embedding the stabilizer within the structure, it's moot. Publicly available resources are quite skimpy on descriptions of freezing hydrazine (or I just suck at searching). So, -
Does hydrazine crystallize when freezing? Can it trigger fraction-freezing separation if it's the first freezing substance?

Pretty much everything crystallizes when freezing. The exceptions are relatively rare.

@IvanNeretin: non-polar substances tend to just grow more dense gradually turning into amorphous solids.

Even if this were mostly true (which it isn't), it doesn't apply to hydrazine, which is polar.

@SF. - There are a hundred-odd stable(ish) elements that all seem to have crystalline solid phases...

@JonCuster: And there are millions of complex hydrocarbon compounds with lengthy chains that don't....

@SF. - many very complex molecules (e.g. proteins) can be made into crystals - it just isn't easy (just ask any good x-ray crystallographer). The issue is, of course, giving the structure enough time at the perfect temperature to get the packing right.

@JonCuster: In this specific scenario we want to prevent crystallization by whatever means available.

I would take a different position on answering the question, and look at the water-hydrazine binary phase diagram. Looking at J. A. McMillan and S.C. Los, J. Chemical Physics 42(1) 160-161 (1965) one finds the experimentally determined phase diagram. With enough water in the mix, one proceeds smoothly from the pure hydrazine melting point down to a slight eutectic near the monohydride, but only 3K lower in temperature.
Further, the same authors have another paper, J. Chem Phys 42(3) 829-834 (1965)  that discusses the nonequilibrium phase transformations. The first sentence in the abstract says

Although neither hydrazine nor water supercool to a large extent, their mixtures do. Aqueous solutions of
  hydrazine may be strongly supercooled and stabilized as rigid glasses.

In the introductory paragraph, one also finds

Since the phase diagram of this system
  is known, it was tempting to look for a correlation
  between the irreversible processes of crystallization and
  the equilibrium phase transformations. The hydrazine-water
  system is particularly useful for this purpose,
  since crystallization can indeed be induced in the entire
  range of concentration. Samples of different hydrazine
  content were then supercooled until glass stabilization
  occurred and then warmed up at different rates. The
  processes of glass transformation and crystallization
  were studied, and values of the free-energy barriers
  could be obtained for the latter process.

The rest of this second paper is devoted to measuring and describing thermodynamically the crystallization curves at different water-hydrazine compostions.
So, yes, you can definitely get a glass, if cooled quickly enough, but it just might decide to phase separate on you if it warms up a bit.
For the case of (only) hydrazine-1,1-dimethylhydrazine, the phase diagram is covered in J. A. McMillan, Journal of Chemical and Engineering Data 12(1) 39-40 (1967). This shows a very slow decrease in the liquidus temperature as 1,1-dimethylhydrazine is added, with the eutectic occurring at 94% mole percent 1,1-dimethylhydrazine at 214K, only 61K below the pure hydrazine melting point. There are no compounds. Further, the article states

Unlike the systems of the components with
  water, the mixture hydrazine-1,1-dimethylhydrazine
  does not supercool

which indicates that phase separation and crystallization of the two components will readily occur. On the other hand, the samples were sealed in glass under nitrogen, and if kept in the dark were stable over a 6 month period.

Unfortunately hydrazine for use as rocket fuel is stabilized with UDMH, not water, which would be very bad for performance of the propellant.

@SF. - noted. additional information added specifically on hydrazine-1,1-dimethylhydrazine. It isn't a particularly good antifreeze...

It was good enough for the military :)  Huge thanks for the great answer!

I found an old research paper which states that hydrazine forms monoclinic crystalline structure in molecule pairs.
Given the ongoing discussions on whether a rocket-fuel mixture separates or not as it cools, I'm adding this paper which says hydrazine monohydrate forms trigonal crystals.   I admit to massive ignorance as to the exact compounds used in fuels.

The usual hydrazine fuel is stabilized with UDMH (Unsymmetric dimetylhydrazine, $NH_2 - N(CH_3)_2$; properties similar to hydrazine but much more stable,  −57 °C freezing point and acts both as stabilizer and antifreeze agent on plain hydrazine.)


text:
Reference conditions for balances on nonreactive processes
In one of the examples in my book (Elementary Priniciples of Chemical Processes) where I am asked to calculate the rate of heat input required to heat a stream of methane and air from 20°C to 300°C, two references conditions are chosen (see picture).
In the text below the box it says "The reference condition for methane was chosen so that $H_{in}$ could be set equal to zero, and that for air was chosen so that $H_{in}$ and $H_{out}$ could be determined from Table B.8."
Why isn't it that $H_{in}$ for air is set equal to zero since the specific enthaply for air at 25°C and 1 atm is 0? How did they get that $H_{in}$ for methane should be zero?

I'd start by looking at Table B8. What does it provide as a value for H of air?

I've added a picture of Table B8

It seems you are right. I share your confusion. Not sure why H is set to 0.

As long as there is no chemical reaction, you can use a different reference temperature for every species in the mixture.  Mathematically, it will still work out OK.