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Am I using these spectra to identify the compound correctly?

What is the structure of the molecule whose data is shown below? Explain your answer [...] EI mass spectrum: The highest peak observed is $m/z$ 210 and its abundance is low. The only other significant peak is observed at 167 Th (base peak). [...]

Full question text: https://i.stack.imgur.com/MQoUD.jpg

Based on the IR value, I reasoned that it is a ketone. Nothing else in the IR spectrum really stood out to me, so I stuck to the idea that it must be a ketone. I assumed the mass of the compound is 210 g/mol so I subtracted the mass of oxygen and used the rule of 13:
Oxygen's mass = 16 g/mol
(210-16)/13 = 14 + 12/13
So the molecular formula is $\ce{C14H_{(14+12)}O = C14H26O}$
I noted that there were 7 signals recorded in the carbon NMR spectrum. I assumed that meant there were equivalent carbons in the molecule. The molecule I ended up drawing to match this is 7-tetradecanone.

Do you think this is the correct molecule?

Nice question, can't find a simple answer. I'd think, based on the NMR spectra, that the molecule has to have 2 equivalent phenyl groups, a -CH3 or something like that, and a single hydrogen bound to something interesting. I was thinking 2,2-diphenylpropanal, but the 1H-NMR shift does not quite fit an aldehyde. Also if you have phenyl groups attached to carbons, you are ought to have 91 m/z tropylium ions in the MS spectra. I am fairly certain that you have phenyl grops, however, the exercise states that there is nothing interesting around 91 m/z... hmm...

@KarstenTheis there is no CH in  2,2-diphenylpropanal

@Ezze The mass spectrum fits well to your suggestion of a diphenyl if it fragments as a unit.

@KarstenTheis right, and as I have said, the 1H-NMR shift simply does not fit an aldehyde. However, the hydrogen of a diphenyl is pretty well shielded... and a CH3-CO+ fragment has a nice mass of 43... Now time to let the asker to think a bit.

Retracted my close vote after the edit.

@Ezze 16 hours after posting, "time to let the asker [...] think a bit" about the question is up... It is still possible to think about the answers, though.

Structure elucidation from spectrocopic analysis needs some experience. I have feeling that you are novice to this field by your explanation of the structure. You cannot determine the molecular formula by the data given and, hence you cannot determine the degree of unsaturation of the molecule. Yet, based on $\ce{^1H}$-$\mathrm{NMR}$, you can propose there are at least two aromatic nuclei (probably mono-substituted) in the molecule, be case there is a 10-proton multiplet in the aromatic region ($\pu{7-8 ppm}$). This assignment is supported by the 4 resonances in the region of $\pu{110-160 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. That also suggest these two aromatic nuclei are identical. Two aromatic rings are accounted for 8 degree of unsaturation.
The given FT-IR spectrum suggest there is a carbonyl group in the molecule, which is confirmed by the resonance at about $\pu{205 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. The peak at about $\pu{205 ppm}$ would also suggest that the carbonyl is not an aldehyde, but a ketone by its chemical shift ($\gt \pu{200 ppm}$), which would be confirmed by the lack of aldehydic peak at about $\pu{10 ppm}$ in $\ce{^1H}$-$\mathrm{NMR}$. The carbonyl group  is also accounted for another 1 degree of unsaturation.
Now, let's analyze $\ce{^1H}$-$\mathrm{NMR}$ spectrum. There is a 3-proton singlet at around $\pu{2.2 ppm}$ for isolated methyl group (without a neighboring hydrogen), indicating it is a benzylic methyl or next to a carbonyl group. Also, there is another one-proton singlet at about $\pu{5.1 ppm}$ for a methine group (-$\ce{CH}$-), indicating its highly deshielding nature. That deshielding must come from two aromatic ring and carbonyl group together. Thus, based on these data, I can deduct the structure as 1,2-diphenylpropn-2-one as depicted in the diagram below (blue box):

The highest peak observed in mass spectrum of the compound at $\pu{210 m/z}$ with law abundance and the base peak ($100\%$ intensity) observed at $\pu{167 m/z}$ confirmed the structure. As depicted in the red box (vide supra), the fragmented cationic radical at $\pu{167 m/z}$ is very stable, which explained the absence of any other fragmentations (as stated by OP, it is the only other significant peak in MS).

My personal opinion is: screw IR, it does not give useful information – but NMR does.
The first thing to check in the proton NMR is the integrals of your signals (I used a ruler on my PC screen – which interestingly enough seems to be good enough for this case!). You have clearly aromatic protons, one singlet at around $\pu{5 ppm}$ and one larger one at around $\pu{2.2ppm}$. The signal at $\pu{2.2ppm}$ can safely be considerd a methyl group so I took its integral to be 3. That gives me an integral of 1 for the proton at around $\pu{5ppm}$ and 10 for the aromatic signals.
With that, I can assume that we have two phenyl groups, possibly chemically equivalent, one $\ce{CH}$ and one methyl group. Next, let’s take a look at the $\ce{^13C}$ spectrum. $\pu{206ppm}$ is clearly a ketone as we have no aldehyde proton. The next four carbon signals are probably the aromatic rings; since we only see one set of four but they seem rather large, we can conclude that the phenyl groups are indeed chemically equivalent. Finally, there are two more signals for the methyl and the $\ce{CH}$ group.
Thus, we have:

2 times $\ce{C6H5-{}}$
$\ce{-C(=O)-{}}$
$\ce{-CH3}$
$\ce{R3CH}$

We also know that the aliphatic protons are all isolated as they do not couple. Adding masses for the individual fragments gives us:
$$\underset{\text{phenyl}}{2\times 77} + \underset{\text{carbonyl}}{(12+16)} + \underset{\ce{CH}}{13} + \underset{\ce{CH3}}{15} = 154 + 28 + 28 = 210$$
Thus, we have all atoms accounted for and we just need to put them together.

 The solution is $\ce{Ph2-CH-CO-CH3}$ or sum formula $\ce{C15H14O}$. You were quite a bit off.

"screw IR, it does not give useful information": Given that the OP is obviously a beginner in interpreting spectral data, I would be careful with such statements. IR spectroscopy nowadays is often underestimated. It takes some practice, but a lot of information on specific (functional) groups can be obtained (also in this case), and proper structural clarification needs a combination of different methods, including IR.

@Alexander Your opinion may differ but I stand firmly by that first statement. I routinely do not measure IR because I consider it a waste of time. The last time I used IR data to determine samples was in the very first undergraduate organic chemistry lab where using IR was mandatory and we did not get NMR time.

From my side one concluding remark: What I wrote is not opinion-based. Often NMR spectroscopy can give you enough information to draw correct and well-founded conclusions, but often enough it does not (probably good for another question). I do not mean to critisise your methods and I assume you are successful without IR data. It is just about not biasing a beginner´s attitude towards the indeed very powerful IR spectroscopy too much.

Do you think this is the correct molecule?

No, for three main reasons:

As Ezze states in the comment, there are strong indications for phenyl groups in the NMR spectra. One phenyl group corresponds to 4 double bond equivalents, which is incompatible with your molecular formula (too many hydrogen atoms).
The H-NMR has too few multiplets for your proposed structure. You need an isolated methyl group (singlet) and a CH or OH group that is not coupled strongly to anything else.
Your suggested molecule lacks the symmetry to explain why there are only 7 signals in the C-NMR spectrum. Also, the alkyl chains in it don't show up in the H-NMR (would expect multiplets at 1-3 ppm).

Finally, the molecular formula for tetradecanone is $\ce{C14H28O}$, not $\ce{C14H26O}$, so the mass is off.
Ways to solve this

Fix the molecular formula. If there are less hydrogen atoms than in the proposed molecular formula, there must be more carbon or oxygen atoms. One simple change is to replace 12 hydrogen atoms by one additional carbon atom, giving the same mass ($\ce{C15H14O}$ rather than $\ce{C14H26O}$). This turns out to be correct, but at this stage it is tentative.
Consider the number of signals in the H-NMR in the aromatic region vs the XH and CH3 signal. See Jan's answer.
Find a structure where the CH3 signal can't couple with anything. Also, make sure you get seven signals in the C-NMR spectrum even though there are more than seven carbon atoms in the molecular structure, as you already realized. Having two phenyl rings helps to keep the number of signals low, and also fits with the large number of double bond equivalents deducted from the tentative molecular formula. See Mathew Mahindaratne's or Jan's answer for the actual structure.


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Empirical Evidence and Assignment of Orbital Energies (Electron Filling)
I see a lot of questions on electron configurations from students at various levels here. Students mechanically remember how to fill orbitals and sorry to say many teachers do the same. Not a single general chemistry or even undergraduate physical chemistry book discusses the "why" part of orbital filling and what is the experimental evidence behind the electron configurations? Yes, hand-waving discussion is always present. It all comes from spectroscopy and so on.
A long time ago I asked a question here, on the physics forum and ResearchGate as how one would come up with an electron configuration from the raw spectrum of a non-hydrogen element. Experimental Assignment of Electronic Transitions in Atoms (Grotrian Diagram)  However, even trained atomic spectroscopists did not have an answer. Is it an art which has been lost and all we have now are tables?
The question is that what is the point in arguing and teaching that electronic configuration of copper is [Ar] 3d10 4s1 or [Ar] 3d9 4s2? What difference does it make in the world when the teacher and the student have no clue of experimentally verifying it except both are memorizing the rules and exceptions like a parrot? Suppose a very good undergraduate asks this question that how do we experimentally know that configuration of copper is [Ar] 3d10 4s1 or [Ar] 3d9 4s2 vice versa? Most of us may not have an answer.
I would be very happy to know a resource that talks about the experimental evidence and determination of electron configurations of heavier metals. So far my efforts have not been very fruitful.
A similar question has been asked here Evidence of orbitals but it does not fully address my old query.

Your particular example of Cu (and many others) can be verified by a measurement of the Zeeman effect and the Landé $g$ factor. Since the ground state has a $^2S$ configuration and the excited state has a $^2D$ configuration the $g$ factors are rather different. Also by measuring fine- and hyperfine splittings one can tell a lot about the angular momenta and therefore the electronic configuration. In a similar way, spectroscopy of excited electronic states (e.g. via Rydberg series) also provides a wealth of information.

Is there any solved example for the determination of electron configuration using Zeeman effect confirming that copper is [Ar] 3d10 4s1 or [Ar] 3d9 4s2 vice versa? I am basically looking for an example (from historical point of view as well) that how electron configurations of heavier elements were determined. Thanks.

You can have a look at the paper by Sommer (Z. Phys. 39, 711, 1926), which is unfortunately in German. In his Table 1, he looks at the $g$ factor of Zeeman lines and compares that with predictions based on the $g$ factor for each level in the transition in order to determine the configuration of the term. He writes on page 729 (translated from German): "From the table of Zeeman effect measurements it is immediately apparent to what extent each term is supported by the observation of the Zeeman effect."

Thanks, I will try go thru it. DeepL translator is pretty good and I can find the relevant paragraphs.

I do believe I saw some explanation of this in, "Inorganic Chemistry: 5th Edition" by Gary L. Miessler. It's been quite a while since I took the course, but the content of that book seems fairly relevant to your question. I hope that helps.

Melanie, I had a quick look at the book. It is a nice book but it does not offer any experimental evidence of assigning spectral lines from an atomic spectrum.

As far as I know, it was a long and complex undertaking. You can't look at a single atomic species by itself. The understanding of level assignments to series slowly grew starting from the simplest cases.
Of course the first was hydrogen, where - at least initially - no series are discernible but one - there is almost perfect degeneration on l
. But that helped, as it allowed the first step, the one from wavelengths λ (or wavenumbers k=1/λ) to terms, later identified with energy levels. It was Ritz' combination principle:
k=Tm−Tn.
Already at this introductory level, a difficulty was to be overcome: the difference between emission and absorption spectra. The former is much richer. This was explained as a consequence of atoms in absorption initially being in the ground state alone, whereas in emission photons are emitted by excited atoms jumping to a lower state, not necessarily the ground one. For hydrogen this is especially notable, as there are no absorption lines in the visible region - the Balmer series has n=2
.
Then alkaline spectra were explored. In absorption spectra only lines Tn−T1
are observable, where T1 is what later would become the ground state energy (divided by hc). Note that here n=1, meaning the ground state, but this is not the principal quantum number of hydrogen-like classification, which is 2 for Li, 3 for Na, etc...
The main difference between hydrogen and alkaline spectra is that in the latter l degeneracy is broken. This required to separate terms in several series. Surely you know the origin of symbols S, P, D,... related to a different appearance of lines starting from different series towards the same final term. An instance is transitions S-P and D-P (in emission), In absorption only S-P transition is visible, giving rise to the principal series (therefrom P-terms).
But a second feature appears in alkaline spectra: the so-called "fine structure". It exists in hydrogen lines too, but is much less prominent, whereas the famous sodium doublet requires a modest resolving power to be seen. Its interpretation required the discovery of electron spin and of L-S coupling. The effect was a doubling of columns for all series, S excepted.
Well, this is a rough history, as I'm able to follow. Maybe a book could be of help: G. Herzberg, "Atomic Spectra and Atomic Structure". It's outdated as far as theory is concerned, but thanks to its publication date (first edition 1936) is nearer to the times when the studies were carried out.

Thanks for comments. Although Herzberg's Atomic Spectra is a lovely book, unfortunately he never discussed about spectral assignments of lines nor talked about electron configurations. I am waiting for other expert opinions as well.

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Non-monotonic rank of oxidizing agents
"Which oxidizing agent is more powerful" seems to depend on what is getting oxidized.
Consider for example oxygen vs chlorine. It is my understanding that chlorine is the more powerful agent for oxidizing sodium metal. The enthalpy of formation of sodium chloride is almost twice as negative as the oxide on a per-sodium atom basis (about -400 vs -200 kJ/mol Na).
However, if we are oxidizing hydrogen the picture reverses. Assuming everything stays in the gas phase and below about 600C at atmospheric pressure (see here) the equilibrium favors H2O rather than HCl.
It seems that when we are talking about dilute aqueous solutions we can define oxidant power based on standard electrode potentials, but when we generalize this concept (even if we standardize temperature and concentrations/partial pressures) there is no "right" answer unless we define a standard set of reducing agents to compare to. Is this correct?

In my opinion the most powerful oxidizing agents in the world are the cathode and the anode. They can reduce any metal ion to metal even under molten state and oxidize fluoride ion to fluorine. I don't think any chemical oxidation can achieve that. The problem with the set of reducing agents is that what would be the criterion for choosing them? We would be back to square one i.e., electrode potentials.

Yes, there is no definitive "right" order.

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Does Mie Scattering occur in Liquid or is Rayleigh scattering?
Rayleigh scattering occurs when the dimensions of the scatter is much smaller than the wavelength of the incident electromagnetic radiation.
Mie scattering occurs when the dimensions of the scattered is much larger than the wavelength of the incident electromagnetic radiation. An example is when light is scattered by small water droplets in clouds.
So does Rayleigh or Mie scattering occur in liquid or bulk water?
It seems the answer is Mie scattering. But in Raman spectrometry. The incident light say 532nm is rayleigh backscattered in liquid. This is why the equipment needs to have notch filter, etc. So why does Rayleigh scattering (?) occur in liquid/bulk water when the particles are much larger than the wavelength of light? Or is it Mie Scattering that the raman device is filtering in the incident light backscattered?

I am not sure that Mie is the right model in a continuous homogeneous medium. Raman works at molecular basis or so. Mie needs interfaces. For theoretical treatment you can ask on Physics as well. A solution or a pure liquid in different from a suspension of solid or liquid in a gas is what I wanted to tell.  You are looking for abrupt change from a phenomenon to another where things can go smoother or coexist. See my other comment.

So does Rayleigh or Mie scattering occur in liquid or bulk water? So why does Rayleigh scattering (?) occur in liquid/bulk water when the particles are much larger than the wavelength of light?

What is the approximate size of water molecule? It is on the order of 2.7 Angstroms or 0.27 nanometers. When we are talking about visible light, the wavelength is 500 nm/0.27 times larger than the scatterer! In case of pure water and with visible light, Rayleigh scattering is observed along with Raman effect. If you disperse larger "particles" in water such as milk, one sees Mie scattering.

How about water droplets in cloud? It is still water molecules on order of 0.27 nanometers.  So why is it Mie scattered instead of Rayleigh scattered?

Those are either ice particles or large (compared to wavelength) droplets suspended in air.

"However, the water vapor will not readily condense without help from other particles. The air making up our atmosphere is full of microscopic floating particles of dust, soil, smoke, sea salts, and other matter. These particles are called condensation nuclei  when they assist in cloud formation. Just as water particles condense on grass to form dew, the tiny airborne particles of water vapor condense into liquid or ice on the surfaces of dust particles in the air. As more water vapor condenses into water droplets, a visible cloud forms." https://ssec.si.edu/stemvisions-blog/what-are-clouds

But cant water scattering be caused by ES scattering. In the Einstein—Smoluchowski theory, scattering is considered to be caused by the random motion of molecules which in a sufficiently small volume causes fluctuations of density and, therefore, of the dielectric constant.

I don't remember the specific details but keep in mind those are all classical theories. At that time quantum mechanics was developing. The classical version of scattering is that the incoming electric field oscillates the charge in the molecule at the same frequency. An oscillating charge then emits light of the same wavelength.

@Jtl also the phenomena can be concomitant. Certainly the Raman spectrum of the water droplet is the same and involves Rayleigh sc. What gives the Mie sc. would be the droplet of water at its interface rather than water molecules.

Guys. So the following statement is wrong? I read it from someone in the net (what is he talking about here?): "When the molecules aggregate the situation changes: scattered waves coming from different parts of the aggregate interfere and partially extinguish each other. For aggregate sizes larger than a wavelength, Mie scattering predominates. And in case of a homogeneous, nonabsorbing bulk liquid you treat the interaction with light in terms of reflection and refraction."?

To elaborate my comment above, others say the interaction in liquid water will not be as simple as photon / molecule scattering because each water molecule is bound to its neighbours much more than when it's in the gaseous phase. So they don't believe rayleigh scattering is even possible in bulk water, what do you say?

@M. Farooq Could it be that raman spectrometer can induce rayleigh scattering in water because of the way the laser is focused? Meaning for ordinary light or sunlight, the molecules aggregate and the situation changes where scattered waves coming from different parts of the aggregate interfere and partially extinguish each other?

@Jtl, No there is nothing like that. If you hold a glass of water in sunlight, it will also do Rayleigh scattering. Read Wikipedia on Rayleigh scattering.

@Jtl, Would you mind sharing what do you do with Raman spectrometers?

I own a raman spectrometer just to study intermolecular forces, to be familiar with them. For those with personal raman spectrometer in their bedroom. What do they do with it? just curious.

Just curious, whats is the make and model. One has to be really lucky in order to $own$ a Raman spectrometer in their bedroom :-)

I wanted to see Raman effect visually but unfortunately the lasers required to demonstrate this are too dangerous.

This is the model R532 http://enspectr.com/ my practical use is really analyzing sugar in wine. Please go to this site https://www.physicsforums.com/threads/light-propagation-in-water-is-the-scattering-rayleigh-scattering.956775/ the arguments of Jestocost are all wrong and the OP is right? pls. let me know so I can let them delete it if it's wrong or answers so vague it can confuse people. Thanks

All the discussion there is very vague and perhaps a mix of truth and junk.

If you wish to understand Raman effect, learn it from the horse's mouth. Read Raman's discovery paper. All of them are freely available.


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Van der Waals forces
What exactly are Van der Waals forces? How do they arise and how can an instantaneous dipole cause another dipole? Shouldn't this make a chain reaction that causes all matter to eventually become polar?

Related: https://chemistry.stackexchange.com/questions/10210/whats-the-relationship-between-van-der-waals-forces-and-hydrophobic-interaction?rq=1

The post that was linked to you in comments contains  a nice, short description of the 'traditional' categories of van der Waals interactions. I suggest you take a look at that, it might just be enough to clear things up. The reason I am writing this post is that the categories introduced there are defined somewhat arbitrary, so I'll try to introduce some terminology that (as of today) is considered correct in the field.
So given a molecular system, it is quite easy to factorize the forces into intermolecular and intramolecular forces. Intramolecular forces are what hold the molecules together. These are (almost completely) covalent bonds, and these interactions would be same for one single molecule and the bulk phase of the matter. But we know, from the real life, that matter does not only exist as an ideal gas: we got liquids, solids, and even the gases are non-ideal. So evidently, there has to be forces between molecules. These forces are collectively called van der Waals forces for historical reasons.
Now there are three different types of interaction between molecules. There are electrostatic, induction and dispersion interactions.
Electrostatic effects are simple classical interaction between charged particles. Thesy simply come from the Coulomb law, can be either attractive or repulsive, and they are strictly additive (because Coulomb law is additive)
It is tricky to properly understand induction and dispersion. These two effect come from quantum mechanical perturbation theory. Without getting into details, perturbation theory means that the wavefunction of two molecules together is not simply the sum (or average) of the wavefunctions of the separate system, but there is an interaction between them, but this interaction can be expressed in terms of the interactionless wavefunction. The expression for the second-order energy correction is given by the Rayleigh-Schrodinger picture as follows (this is called the polarization approximation):
$$W^{(2)} = - \sum_{mn}' \frac{ <00|H'|mn><mn|H'00> }{W_{mn}-W_{00}}$$
In this expression, $|m>$ represents the mth excited state of the original wavefunction of molecule A and $|n>$ represents the nth excited state of the original wavefunction of molecule B. This expression could be read: the energy difference between the two separate systems and the interacting system depends on the excitations in the separate systems through an interaction operator $H'$.
The sum can be factorized into three components:
$$W^{(2)} = - \sum_{m\neq0}' \frac{ <00|H'|m0><m0|H'00> }{W_{m0}-W_{00}} - \sum_{m\neq0}' \frac{ <00|H'|0n><0n|H'00> }{W_{0n}-W_{00}} - \sum_{m\neq0,n\neq0}' \frac{ <00|H'|mn><mn|H'00> }{W_{mn}-W_{00}}$$
Here, the first term contains the energy contributions of the interactions of the ground state of A with the excited states of B. This is called the induction energy of B. The second term is the induction energy of $A$, which again, expresses the total response of the molecule A to the ground state of B. The third term is the so-called dispersion interaction, which is then a mixing of the excited states of both molecules.
So these expressions are the defining expressions for induction and dispersion. One should know that both of these interactions are always attractive , they are non-additive, and they are always present, between charged and neutral particles too. Also don't get confused that the energies are expressed in terms of excited states: there is no excitation going on in the system, it's just the mathematical tool that we used as definitions.
The total intermolecular force, or van der Waals force is then usually the sum of electrostatic, induction and dispersion terms. For the sake of completeness we might include resonant energy couplings and magnetostatic interactions too, but these are either rare or so small that we can neglect them.


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Why are two C-O single bonds more stable than a C=O double bond (hemiacetal)?
In my organic chemistry book it says that the formation of cyclic hemiacetals is energetically favorable in comparison to the open chain aldehyde as two C-O single bonds are more stable than a C=O double bond. Why is this the case?
To be more precise, I am specifically talking about monosaccharides.
I quote from "Organic Chemistry" by Clayden, Greeves, Warren (I am translating from German so it's not 100 % accurate):

[Translated from German] The change of enthalpy that goes along with the reaction results from the change of bonds: In this case a C=O double bond becomes two C-O single bonds, and those C-O single bonds are slightly more stable than the C=O double bond, resulting in  $\Delta$H being slightly positive.

Could you edit your answer to specify the textbook? This would give some more context to your question, and make an answer more meaningful.

a C=O double bond becomes two C-O single bonds, and those C-O single bonds are slightly more stable than the C=O double bond resulting in ΔH being slightly positive.

The tables I found (e.g. OpenStax Chemistry, Wikipedia) have the C=O bond with roughly twice the bond dissociation energy compared to the C-O bond, and in all cases the C=O bond is slightly more stable than the two single bonds. So the statement in the textbook is surprising. As these values are aggregates derived from large sets of molecules, their values will vary depending on which set of molecules they are based on.
If you go from less stable bonds (easier to break, smaller bond dissociation energy) to more stable bonds (harder to break, larger bond dissociation energy), the enthalpy of reaction should be negative (exothermic), not positive as claimed in the statement above.
For completeness, these are not the only changes. In addition, the O-H bond of the hydroxyl group is broken, and the O-H bond of the hemiacetal is formed. If you are estimating bond dissociation energies from a simple table, however, this will just cancel out because it is treated as the same kind of bond.
Estimating enthalpy of reaction with bond energies
Bond energies give you a rough estimate of reaction enthalpy, but will not pick up subtle differences between cases that are treated equally using this estimation. Here are two other examples where a C=O double bond is replaced by two C-O bonds, and the counts of all the other bonds are equal:

Hydration of a ketone, e.g. acetone and water reacting to form 2,2 propanediol. According to this web site, the reaction is slightly endothermic (about +6 kcal/mol). On the other hand, formaldehyde is mostly hydrated in water, so ketones and aldehydes behave slightly differently.
Butanone isomerizing to tetrahydrofurane. This is a hypothetical reaction, but NIST data suggests that it is also endothermic, again different than the estimate from bond dissociation energies.

So first, you have to look at the experimental evidence, and then you can try to rationalize it with a simple model, which sometimes will give the expected answer and sometimes will not.
"energetically favorable" vs major species
Just because a reaction is exothermic does not mean the product will be the major species. Not all monosaccharides form rings, and mixtures of ketones and alcohols typically do not form hemiacetals as major species. There are other energetic factors  (e.g. ring strain), and entropic factors (e.g. intramolecular vs intermolecular, conformational degrees of freedom) that also contribute to making one or the other species more prevalent.

Why are two C-O single bonds more stable than a C=O double bond (hemiacetal)?

This is what we observe. A simple model like adding up bond energies seems to support it, but maybe that is just luck. Computational methods are sufficiently sophisticated to predict that certain monosaccharides form cyclic hemiacetals. I'm not sure this answers the question of "why?", though.

More electronegative substituents turn the equilibrium unfavorable for aldehyde/ketone - all those hydroxy groups may be enough.

I presume they are talking about bond enthalpies? A C-O is worth about 84 kcal while a C=O is about 127 kcal, so 2 x C-O has a larger total bond enthalpy. In a monosaccharide there is also additional stabilization of the acetal form through the anomeric effect.
However I think "stable" is a misnomer here (or at least is too vague) - acetals are relatively unstable under aqueous acidic conditions, while aldehydes are relatively stable; conversely under basic conditions aldehydes will react to give aldol products (as well as Cannizzaro reactions and more) while acetals will be unreactive. "Stable" is largely a subjective term.

That's the problem, the bond enthalpy is the point where I lost it.
I found the following table and the values vary from the ones you have mentioned.
I'm relatively new to the concept of bond enthalpy so I'm sorry if I got that wrong.
http://cp.cm.kyushu-u.ac.jp/tools/IELSFOC/contents/virtual%20textbook/bond%20energy/bond%20energy.html

@jona173 127 kcal/mol for C=O given here is quite different from what is in the table you reference, and from what other tables have (~180 kcal/mol or ~740 kJ/mol).


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Resonance structures and mechanism
I wrote the following resonance structures and our teacher wants us to write the curved arrows showing the mechanism as well... I was wondering if it's okay to show in which direction the mechanism arrows are meant - I hope the question makes sense...
Thank you very much!

Sorry, but that doesn't make sense. Resonance has very little to do with a mechanism except that both might use the same arrows to indicate the electron flow. What you have drawn in both examples is one step of a mechanism to produce a resonance stabilized intermediate.

Related to what Zhe has said: What is resonance, and are resonance structures real? The depictions are fine, as long as you keep in mind that none of the structures within the brackets actually exist, therefore the arrows used are a mere derivation tool.

@Zhe - thank you both for your comments - it helped me and from now on I will distinguish between electron flow and mechanism :)

@Martin-マーチン thank you very much as well :)


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Kinetic Diameter: How to calculate it? Pressure influence?
Wikipedia states:

Kinetic diameter is a measure applied to atoms and molecules that expresses the likelihood that a molecule in a gas will collide with another molecule. It is an indication of the size of the molecule as a target.

And the following equation is given:

$$d^2 = \frac{1}{πln}$$
where,
$d$ is the kinetic diameter,
$r$ is the kinetic radius, $r = d/2$,
$l$ is the mean free path, and
$n$ is the number density of particles.

To me, this implies that the kinetic diameter can be altered by changing the number density of particles. Therefore, a change in pressure, volume, or concentration should would effect the kinetic diameter.
However, kinetic diameters are often discussed as if they were constants. Indeed, Wikipedia even gives a table of kinetic diameters for some common molecules.

Another definition of the kinetic diameter would be useful.

How is the kinetic diameter related to molecular structure? Can it be quantified as the distance between a pair of atoms?

If someone could also suggest a computational program to calculate the kinetic diameter of a molecule, that would also be of use.

Keep in mind that while altering the pressure, volume etc. changes the number density, it also changes the mean free path of the particles, and these effects are compensated. As for calculating the kinetic diameter by computational programs, well, it is a tricky business... Depending on your proficiency with theoretical chemistry, you might look up the pair potential expression for the second virial coefficient, and start from there...

Further to Ezze's comment, the reason that $d$ is expressed in that way is because it is in terms of measurable quantities, $l$ and $n$.  Most molecules are not spherical however, benzene, anthracene etc so $d$ would then be some odd sort of average. You can get good ideas of size from van der Waals radii, see the JSMOL image on the molecule's Wikipedia page.


text:
Deciding the order of heat of combustion of isomeric alkanes
I'm asked which one of the alkane has the lowest heat of combustion (note: the question says heat, not enthalpy) and according to the answers, the right one is d.

I'm trying to understand why d is the right answer but with no success.
Please help me understand the concept behind it for future problems.

In the bounty message above, I wrote that I have got an answer, but the reasoning involved in it was incorrect. So ignore the second part of the message :)

Branched-chain alkanes have lower values of ΔcH⊖ than straight-chain alkanes of the same number of carbon atoms, and so can be seen to be somewhat more stable. https://en.wikipedia.org/wiki/Alkane#Branched_alkanes
The heats of formation are given in https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation  and go from -40.0 kcal/mol for a to -41.8 for b to -42.5 for c to -44.5 for d. Compound d has the largest heat of formation, therefore will have the smallest heat of combustion.
The explanation is complicated. Sometimes you have to wonder why teachers spend so much time on effects that are so minor. The data are interesting, in a trivial sort of way. Show the data; it's interesting; move on.
Highly branched alkanes may stretch the bond angle. This can cause steric hindrance, and can substantially increase the reactivity. "Substantially". That's interesting. In addition, the reactivity of highly branched alkanes can be used to increase the octane rating of gasoline, because the stability of highly branched free radicals slows down the combustion and avoids explosions in the cylinder (pinging). Substantial is the key word, whether it it is reactivity or stability. The difference between 40 and 44.5 is hardly substantial. I would call that difference about half of substantial.
Wikipedia goes on: "However, in general and perhaps surprisingly, when branching is not extensive, branched alkanes are actually more thermodynamically stable than their linear (or less branched) isomers. For example, the highly branched 2,2,3,3-tetramethylbutane is about 1.9 kcal/mol more stable than its linear isomer, n-octane. Due to the subtlety of this effect, the exact reasons for this rule have been vigorously debated in the chemical literature and is yet unsettled. Several explanations, including stabilization of branched alkanes by electron correlation, destabilization of linear alkanes by steric repulsion, stabilization by neutral hyperconjugation, and/or electrostatic effects have been advanced as possibilities. The controversy is related to the question of whether the traditional explanation of hyperconjugation is the primary factor governing the stability of alkyl radicals." Surprisingly subtle and unsettled. Not really a great teaching topic.
The question regularly pops up in secondary school exams. Why? The reasoning seems unsettled. How to answer the question? The teacher must have given the answer if the teacher is at all competent and meant to provide the reasoning; so regurgitate the correct answer without reserving a special place in your memory for it. If the teacher did not provide the answer, then you must do your own research, in advance of the test - first, to discover what kind of questions your teacher is likely to ask (can you search out previous year's tests?) and second, to research the possible answers.
If you do well at this, you may be asked to teach the next class.

I have lost count of how many Chemistry concepts taught to me in school are actually incorrect, outdated models. Anyway, thanks for the answer :)

Wouldn't the increase in steric hindrance increase the amount of energy released when bonds are broken?

@Sid: Steric hindrance should make it more difficult to put the molecule together; that would mean a smaller heat of formation. The point is that the difference is small (from the data), therefore there must be other effects in process. Thank goodness teachers haven't found a new classification of energies that students have to remember! It's just shows how fuzzy data can be: we can't define everything to the first decimal, not even the first digit!

Although I am conversant with this aspect of thermochemistry, it is not my area of expertise. Nonetheless, your question as to why branching leads to a lower heat of combustion in a set of alkane constitutional isomers is one that has intrigued me for sometime. So much so that, after asking former peers steeped in physical organic chemistry, they were at a loss for a cogent, satisfying explanation. I have not researched this topic recently but in the years 2006-2009 a "dispute" between Gronert1,3 and Schleyer2 et al. was ongoing to account for the phenomenon.
The introduction to Gronert's first paper states the following:
"A new model based on 1,3 repulsive steric interactions (geminal repulsion) is proposed for explaining the variation in the C-H bond strengths of the alkanes. The model builds from the assumption that 1,3 repulsive interactions are the major factor in determining the stability of a C-C or C-H bond in an alkane. From this simple premise, the model successfully reproduces the effect of branching on the stability of alkanes, alkyl radicals, and alkenes. The results suggest that geminal repulsion can provide a simple, unified explanation for these fundamental stability trends..."
On the contrary, the Schleyer paper states:
"The most direct conclusion is that 1,3-alkyl–alkyl interactions stabilize hydrocarbons, not only the branched, but also the linear isomers." and "We define “protobranching” as the net stabilizing 1,3-alkyl-alkyl interactions existing in normal, branched, and most cycloalkanes, but not in methane and ethane."
Subsequently, Gronert's second paper counters with:
"Though expounding upon the potential ramifications of the protobranching
premise, the authors supply no clear explanation of the theoretical basis of the proposed interaction and provide virtually no data to support it despite the fact that the proposed interaction has been controversial for decades..."
If someone out there has a definitive answer, I'd be delighted to hear about it.

1) S. Gronert, An Alternative Interpretation of the C-H Bond Strengths of
Alkanes, J. Org. Chem., 2006, 71, 1209-1219.
2) M. D. Wodrich, C. S. Wannere, Y. Mo, P. D. Jarowski,
K. N. Houk, and P. von R. Schleyer, The Concept of Protobranching and Its Many Paradigm Shifting Implications for Energy Evaluations, Chem. Eur. J., 2007, 13, 7731 – 7744.
3) S. Gronert, The Folly of Protobranching: Turning Repulsive Interactions into Attractive Ones and Rewriting the Strain/Stabilization Energies of Organic Chemistry, Chem. Eur. J., 2009, 15, 5372 – 5382.

It is well known fact that branched-chain alkanes are more stable than the straight-chain alkanes with the same number of carbon atoms. That means branched-chain alkanes have higher values of enthalphy of formation ($\Delta H_f^\circ$) than that of straight-chain alkanes with the same number of carbon atoms.
During the complete combustion reaction with oxygen, all alkanes give off $\ce{CO2 (g)}$ and $\ce{H2O (g)}$ regardless of they are straight-chain or branched-chain:
$$\ce{C_nH_m + (n + m/2) O2 -> n CO2 + m H2O}$$
The heat of combustion, $\Delta H_c^\circ$, is equals to: $n\left(\Delta H_f^\circ \right)_\ce{CO2} + m\left(\Delta H_f^\circ \right)_\ce{H2O} - \left(\Delta H_f^\circ \right)_\ce{C_nH_m}$ (keep in mind that for this question, the difference between heat and entalphy really does not make a difference).
Thus, the heat of combustion is larger if the heat of formation of initial alkane is lower (less stable).
Estimation of heats of formation of organic compounds by additivity methods introduced by Benson and coworkers (Ref.1 & 2) is well suitable for OP's level of education to estimate heats of formation of given compounds. This method allows the rough calculation and prediction of heat of formation of organic compounds based on additivity. According to article in Wikipedia:

Starting with simple linear and branched alkanes and alkenes the method works by collecting a large number of experimental heat of formation data (see: Heat of Formation table) and then divide each molecule up into distinct groups each consisting of a central atom with multiple ligands. To each group is then assigned an empirical incremental value which is independent on its position inside the molecule and independent of the nature of its neighbors (in $\pu{kcal/mol}$ at $\pu{298 K}$):

P primary $\ce{C-(C)(H)3}: -10.00$
S secondary $\ce{C-(CH2)C}: -5.00$
T tertiary $\ce{C-(CH)C2}: -2.40$
Q quaternary $\ce{C-(C)4}: -0.10$
gauche correction $: +0.80$
1,5 pentane interference correction $: +1.60$

Let's apply these rules into following examples:
$$
\begin{array}{c|cc}
\text{Alkane} &\ce{C-CH3} &\ce{C-CH2-C} &\ce{C-CH-C2} &\ce{C-C4}  & gauche\text{ corr.} & \text{Tot.}& \text{expl.}\\\hline
n\text{-pentane} & -10 \times 2 & -5 \times 3  & \text{-} & \text{-} & \text{-} & -35 & -35.1\\
\text{2-methylbutane} & -10 \times 3 & -5 \times 1  & -2.4 \times 1  & -  & +0.8 \times 1  & -36.4 & -36.9\\
\text{2,2-dimethylpropane} & -10 \times 4 & \text{-} & \text{-} & -0.1 \times 1  & \text{-} & -40.1 & -40.1\\
n\text{-hexane} & -10 \times 2 & -5 \times 4  & \text{-} & \text{-} & \text{-} & -40.0 & -40.0\\
\text{2-methylpentane} & -10 \times 3 & -5 \times 2  & -2.4 \times 1  & \text{-} & +0.8 \times 1  & -41.6 & -41.8\\
\text{3-methylpentane} & -10 \times 3 & -5 \times 2  & -2.4 \times 1  & \text{-} & +0.8 \times 1  & -41.6 & -41.1\\
\text{2,3-dimethylbutane} & -10 \times 4 & \text{-} & -2.4 \times 2  & \text{-} & +0.8 \times 2  & -43.2 & -42.5\\
\text{2,2-dimethylbutane} & -10 \times 4 & -5 \times 1  & \text{-} & -0.1 \times 1  & +0.8 \times 2  & -43.5 & -44.5\\\hline
\end{array}
$$
By these calculations, it is evident that 2,2-dimethylbutane (d) has the largest heat of formation among hexane isomers, and hence least heat of combustion, which agrees with the experimental values (last column). Yet, keep in mind that this method is an empirical calculation, but suitable for OP's level of education.
References:

Sidney W. Benson, F. R. Cruickshank, D. M. Golden, Gilbert R. Haugen, H. E. O'Neal, A. S. Rodgers, Robert Shaw, R. Walsh, "Additivity rules for the estimation of thermochemical properties," Chem. Rev. 1969, 69(3), 279-374 (https://doi.org/10.1021/cr60259a002).
N. Cohen, S. W. Benson, "Estimation of heats of formation of organic compounds by additivity methods," Chem. Rev. 1993, 93(7), 2419-2438 (https://doi.org/10.1021/cr00023a005).

The reason I didn't mention GAV's is that they don't explain what the OP  is asking. The Benson values simply help predict the heat of formation. Good effort though.

This question has already had some pretty good answers by user55119, James Gaidis and so on. However, since the bounty setter was looking for a definitive answer in pertinence with some tests, and I am quite well-versed with the particular exam he is interested in(i.e. the JEE), I would like to provide my final insights along with a summary of the concerned parts so that this thread may be successfully closed.
As it has been pointed out above repeatedly, it's common knowledge that branched alkanes are somewhat more "stable" than their straight-chained counterparts and hence have lower heats of combustion(and it is the knowledge of this fact that this question is trying to judge). However, this variation is quite less, as James has nicely illustrated in his answer by quoting the relevant enthalpy values for an isomeric series.
The reason for this subtle variation in stability has been a matter of debate, and there is no unanimous answer for this. However, a few plausible theories have been discovered, and user55119 has done a remarkable job in providing excerpts and links to the research papers on this topic.
As far as I know, the people who design such exam questions usually do so by keeping any one of the possible explanations in mind. Most commonly, they will keep the idea of the greater steric crowding in case of branched alkenes and the somewhat greater bond energy of 1° C-H vs the 2° C-H.
Another theory which is slightly lesser common in such exam-setters opinion is a possibility of a hyperconjugative stabilisation in alkanes, the idea of "protobranching" like was stated above. But again, the lack of a collective agreement on one of these conjectures makes it tough to look for a logical rationale behind such questions.
Again, my personal take on this situation would be that, as far as these questions are considered, one can keep the fact I stated above as an absolute certainty for solving them. As for the underlying logic, one can marvel at the ingenuity behind these various explanations, because all of them present a good chemical argument, and present fascinating cases in themselves.
Hope this helps :)

Can you please elaborate more on the point regarding steric crowding? What does it mean? (I understand the second part of the sentence about number of 1°,2°,3° hydrogen atoms)

I was referring to the extra steric interactions that might occur in the branched alkanes that might impart stability to them(the 1,3 alkyl interactions as user55119 has put up in his answer)

For this question, you might ignore the difference between heat and entalphy, it really does not make a difference.
The question basically asks you to find the most stable alkane of the ones listed. The reason for this is that they all burn to produce the same $\ce{CO_2 + H_2O}$, so the heat of combustion is larger if the initial molecular energy is lower.
Now, for the answer: branching increases the stability of alkanes. The reason for this is not completely clear to me, I've heard several explanations over time. It might be helpful to think about the electron cloud being more 'packed', more 'spherical' in branched alkanes, and you may think that a 'spherical' cloud is not so favorable to break than an elongated one. If anyone got a more intuitive explanation, I'd be happy to hear it.
Regardless of the explanation, it is a known fact that the more branches one alkane has, the more stable it is. That is why the answer is d).

Thank you, but then comes the question--why is d more stable than c? Both have 2 branches, no?

The phenomenon is not clearly understood. A rule of thumb is to maximize methyl groups, i.e., maximize branching. See info on Group Aditivity Values of Benson: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/GAV.html

You wrote the heat of combustion is larger if the initial molecular energy is lower. Isn't it the opposite? That the more reactive (less stable) a molecule is, higher is the heat released in combustion. Please calrify.


text:
Why is dimethylglyoximate a bidentate ligand and not a tetradentate ligand?
Why can't the 2 oxygen atoms donate a lone pair and hence become binding sites? Is it because of the angle made by the oxygen and Nitrogen atoms?

You are right. I took this two pictures from Wikipedia to compare EDTA (a tetradentate) to dmg (a bidentate).
For EDTA you got this:

For dmg, on the other hand, you got this structure:

You can notice that for both of the cases, the metal complex is connected to the ligands through five-membered rings. This observation is general in chemistry: to have stable rings, you need five (sometimes six) members for those rings.
Now it is clear to see a general rule of thumb for multidentate ligands: the two donating atoms must be connected through 2 (sometimes 3) atoms. For dmg, you can see that the two nitrogens satisfy this rule, while the oxygen-nitrogen pairs do not.

For EDTA, you can see that going through the molecule, you can find an oxygen, a nitrogen, a second nitrogen and a second oxygen, all bridged by 2 atoms to each other (while the two oxygens of the carboxyl group are not connected by 2 bridge atoms, otherwise you might even have a hexadentate ligand).

Would it it be possible to scale the structures so that the font on the illustrations approximately matches the one used in main text? Two structures look too tiny, and two are tremendously large and take too much of screen real estate for no good reason.

I'm currently in the process of trying, but it looks like I am not skilled enough!

@andelisk would manually scaling the pictures using a 3rd party software and then uploading it help?

@TheJadeReaper Sure you can, but it will only give a satisfactory result for the larger ones. Upscaling a raster image will result in horrible quality, so someone has to either find appropriately scaled images, or just draw them from scratch.

Suggestion for scaling: if you replace the Markdown image syntax with <img src="<imgur link>" width="<number>" alt="alternate text">, you can play around with the width to scale. Note: the arguments have to be in exactly that order.


text:
Why CO is a weak pi-acid ligand?
This question was given by my chemistry professor, but I doubt that this question is not correct.
Following the lewis definition of the acid, any molecule or atom acts as a acid if it accepts an electron and according to Inorganic Chemistry by Weller CO is a good $\pi$ acceptor, then how it can be weak pi-acid ligand? In fact, CO is a poor $\sigma$ donor.
What am I missing?

Thing is there's no such thing as general scale of Lewis acid strength.

Considering $\ce{CO}$’s ability to stabilise low oxidation states e.g. in $\ce{[Fe(CO)4]^2-}$, I doubt the premise.


text:
I want to make NaOH from NaHCO3, what is the easiest way to do it without electrolysis?
I want to make NaOH from NaHCO3.
I can't use electrolysis as I lack the proper equipement.

You could heat it to turn bicarbonate into carbonate, and let it stand with Ca(OH2) and water mix. That's essentialy how potash lye was made for millenia

How hot though?

Look for decomposition temperature of nahco3 on wikipedia. You will see gas pops in the powder once you've reached the temperature, heat untill they stop appearing

The CO2 bubbles?

Wikipedia says 100°C


text:
Reaction mechanism of thiol to disulfide oxidation under basic conditions?
I have been looking everywhere for an explanation to a reaction mechanism concerning the oxidation of DTT (or just thiols to disulfides in general) using oxygen as the oxidant. I'm aware that oxidizing DTT with oxygen ($\ce{O2}$) will give me the oxidized form of DTT and $\ce{H2O2}$. However, I'm running my reaction in $\ce{MeOH}$ and $10\% \; \ce{KOH}\; (w/w)$ in $\ce{H2O}$ while bubbling it through with air. So, I get that the air is providing the oxygen source, but why run it in basic conditions? Why not just in water?
Below is how I think the oxidation might look in the presence of only oxygen as if DTT was left out in the open:

I have a hard time seeing why the basic conditions are needed besides it keeping the thiols de-protonated? Is the high $\mathrm{pH}$ to prevent the formation of $\ce{H2O2}$?
Does the mechanism below make sense?

I hope all this rambling made somewhat sense and I really REALLY hope that there is someone out there who can help me because I'm stuck.

This mech. shows what appears to be SET - not something I'd expect from thiol - it's quite possible that it rather abstracts hydrogen. pH is rater for tuning redox potential, so further oxidation won't happen.

Welcome to ChemSE! Given the low pKa of thiols, the bis-thiol certainly would be present as the dianion. Your second equivalents of water in place of 2 protons. Balance the right side with 2 hydroxides. While your diradical mechanism is viable, formally one would have thiolate add to oxygen. The second thiolate can displace on sulfur to eliminate peroxide.  These intermediates may be protonated at any stage of the process. Certainly, a thiolate is more nucleophilic than the neutral thiol in water.

Thank you so much for your answer! :) So I should replace the peroxide on the right side of the equation with 2 hydroxides? I'm not quite sure what you meant by 'your second equivalent of water in place of 2 protons'?

Did you use any catalyst in this case?

https://en.wikipedia.org/wiki/Dithiothreitol

No, this was all I used in the reaction.

Sorry for the confusion in the second sentence. Because you are working with aqueous alkali, you have no H+ but only water as a proton source. Replace H+ with H2O and place HO- on the right side to balance the equation. BTW, while I don't have to ping you, you as OP have to ping the person (@user55119) making the comment. I only came across your response by chance.


text:
Link between centrosymmetry, dipole moment and chirality
What is the relationship between the three concepts in the title?
I know all centrosymmetric molecules are achiral, but not all achiral molecules are centrosymmetric. Thus, there are no molecules that are centrosymmetric and chiral, but all other combinations exist.
What about having a dipole moment vs. centrosymmetry? And what about having a dipole moment vs. chirality? Is there a relationship between them, i.e. are some combinations possible and others not?
And if there is a link between these concepts, what is the proof? For example, for a centrosymmetric molecule, I could apply the inversion operation (which does not change the molecule as it has centrosymmetry) and then rotate it by 180$^\circ$, showing that there is a mirror symmetry. Thus, the molecule is achiral.

A dipole has a dipole moment, which is a vector. Arrows are not centrosymmetric. Thats a bit too obvious, so I have to ask: What is your actual question?

@Karl Looking at the answer by Tyberius, I think my actual question is molecules of what point group symmetry are chiral, and molecules of what point group symmetry have (or can have?) a non-zero dipole moment. That would allow statements like X is necessary but not sufficient for Y, where X and Y could be any of chirality, achirality, certain symmetry elements, non-zero dipole moment, a vanishing dipole moment.

So, in short, do all chiral molecules have a dipole moment? That`s an interesting question. ;)

A molecule can't be chiral if it has an $S_n$ improper rotation (note that $S_1$ is equivalent to a mirror plane and $S_2$ is equivalent to inversion). Chirality is only possible in the $C_n$ and $D_n$, as well the less common $T$, $O$, and $I$, point groups, which only have rotation axes as symmetry elements.
A dipole moment can only occur along the primary $C_n$ rotation axis of a molecule (if it has one) and if there are any perpendicular mirror planes or rotation axes, it will have to be zero by symmetry. A $S_n$ improper rotation with $n>1$ also precludes a dipole. This limits the point groups that can have dipoles to $C_n$, $C_{nv}$, and $C_s$.
Centrosymmetry is equivalent to having the $S_2$ symmetry operation. As mentioned above, this eliminates the possibility of chirality or a dipole. However, centrosymmetry is just a special case and  molecule won't be chiral and/or won't have a dipole under other conditions as well. That is to say, centrosymmetry is sufficient, but not necessary for eliminating chirality and/or a dipole.
To summarize:

Chirality
Polarity
Exact point group

Chiral
Nonpolar
$I,O,T,D_{n}$

Achiral
Polar
$C_{nv}, C_s$

Chiral
Polar
$C_{n}$

Where I'm taking "exact point group" to mean the highest symmetry group the molecule is contained in. This delineation has some minor exceptions. Technically, molecules of the appropriate symmetry still might be nonpolar (or at least too small to measure) by an accidental cancellation. For chirality, while all molecules of these symmetries are formally chiral, they can appear achiral in practice if the enantiomers can readily interconvert, as occurs with nitrogen inversion.

@BuckThorn I was basically trying to get across that centrosymmetry is sufficient for there to be no dipole or chirality, but not necessary.

Thanks for giving the entire picture! At http://symmetry.jacobs-university.de/, they list point groups O, T and I as chiral as well. To summarize your answer with that addition, is it correct to say: "Molecules of $C_n$ symmetry are chiral and polar, $C_{nv}$ and $C_s$ are achiral and polar, and $D_n$ as well as $T, O, I$ are chiral and not polar. All others are neither chiral nor polar."?

@KarstenTheis Careful. You'd need to specify that it is the maximal point group, since for example, $C_{n}\subset D_{n}$.

@KarstenTheis I had missed those cases, but they do fit the criteria of not having a $S_n$. I will add those in. Your summary looks correct, though there are minor exceptions. Technically, molecules of the appropriate symmetry still might be nonpolar (or at least too small to measure) by an accidental cancellation. For chirality, while all molecules of these symmetries are formally chiral, they can appear achiral in practice if the enantiomers can readily interconvert, as occurs with nitrogen inversion.

Concerning chirality, the best link is to remember that any molecule that has an improper rotation symmetry (an $S_n$ axis in the Schoenflies notation or $\overline n$ in Hermann-Mauguin notation) is achiral. $S_1$ or $\overline 2$ happens to be equivalent to a plane of symmetry while $S_2$ or $\overline 1$ happen to be identical to a centre of symmetry.
The reason behind this is that an $S_n$ axis is essentially rotation followed by mirroring; however, the order of operations is irrelevant (they are commutative) so you can also generate the mirror image first and then rotate the mirrored molecule to make it identical to the original one. The Hermann-Mauguin formalism is based on centrosymmetry rather than a plane of symmetry, which means that the angle of rotation is different but otherwise the two are equivalent. (Each Schoenflies improper rotation can be transformed into a Hermann-Mauguin one and vice-versa.)
As for dipole moment, this is essentially a vector. It doesn’t matter whether your vector points from the positive to the negative pole or vice-versa (physics uses one definition, chemistry the other); the important bit is that it has a value and a direction. As all symmetry operations a molecule has must transform all atoms onto an atom of the same kind, these symmetry operations must also transform the dipole moment and any other vector-type properties onto itself.
Thus, a dipole moment must always be fully contained within all planes of symmetry and if there is an axis of rotation the dipole moment must align with it. It follows that in all the following cases a molecule cannot have a dipole moment:

if it is centrosymmetric (the only vector transformed onto itself under inversion is the null vector)
if it has an axis of rotation that is not contained in a plane of symmetry (the axis and the plane only intersect in a single point which again confines the dipole moment to the null vector)
if it has more than two planes of symmetry that cannot be transformed into each other by consecutive rotations
if it has more than two axes of symmetry (which by definition can only and must intersect in a point)

These descriptions are mathematically very similar as they all boil down to a small set of common elements.


text:
Finding change in activation energy from change in rate constant in catalysis
I need to find by how much the activation energy decreases when the rate constant increases by a '$x$' factor when a catalyst is used in the reaction.
I'm just super confused about how to solve this. Using Arrhenius relation I have
$$\frac{k_2}{k_1} = \frac{A\mathrm e^{-E_\mathrm{a2}/(RT)}}{A\mathrm e^{-E_\mathrm{a1}/(RT)}}$$
$$x = \mathrm e^{(E_\mathrm{a1} - E_\mathrm{a2})/(RT)}$$
I took natural log on both sides to get rid of the exponent and then ended up finally with an equation I can't seem to get any relation from
$$RT\ln x = E_\mathrm{a1} - E_\mathrm{a2}$$
Now I need a ratio of the two activation energies, but I don't know how I get that from here as I have just a addition/subtraction relation between the two. Would be grateful if someone has a clue about how I could proceed.
The problem is I have come across a question in Atkins' Physical Chemistry [1, p. 841] and the answer is provided in percent, but I don't see how you could arrive at a solution like that with where I am:

Self-test 20D.4 Consider the decomposition of hydrogen peroxide, which can be catalysed in solution by iodide ion. By how much is the activation energy of the reaction reduced if the rate constant of reaction increases by a factor of 2000 at 298 K upon addition of the catalyst?
Answer: 25 per cent

References

Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.

Usually these questions ask you to find the difference between the two energies, not the ratio, because exactly as you said, you can only find the difference of those.

As Ezze has commented, you already answered your question (or rather, there is not much to add to your answer). You can express the required change in the activation energy as an absolute change Ea2−Ea1
, as a fraction change (Ea2−Ea1)/Ea1 or as percent change 100(Ea2−Ea1)/Ea1

It appears that you are correct in your thought process.
Determining by how much the activation energy of the reaction is reduced requires you to basically find
$$Δ(\%) = \frac{E_\mathrm{a1} - E_\mathrm{a2}}{E_\mathrm{a1}} × 100\% = \left(1 - \frac{E_\mathrm{a2}}{E_\mathrm{a1}}\right) × 100\%$$
There is a system of three independent equations
$$
\begin{cases}
k_1 = A\mathrm e^{E_\mathrm{a1}/(RT)}\\
k_2 = A\mathrm e^{E_\mathrm{a2}/(RT)}\\
\frac{k_2}{k_1} = x
\end{cases}
$$
with five unknowns $k_1$, $k_2$, $E_\mathrm{a1}$, $E_\mathrm{a2}$, $A$, which requires either one more equation, or to know one more parameter of the system in order to find $E_\mathrm{a2}/E_\mathrm{a1}.$
The textbook provides you with activation energy in the previous illustration right next to your problem [1, p. 841], so I assume you are supposed to use $E_\mathrm{a1} = \pu{76 kJ mol−1}:$

At this point the problem can indeed be solved to a certain numerical value:
$$E_\mathrm{a1} - E_\mathrm{a2} = RT\ln x = \pu{8.3145 J mol-1 K-1} × \pu{298 K} × \ln{2000} \approx \pu{18.8 kJ mol-1}$$
$$Δ(\%) = \frac{\pu{76 kJ mol-1} - \pu{18.8 kJ mol-1}}{\pu{76 kJ mol-1}} × 100\% \approx 25\%$$
The question whether it's a textbook flaw, or authors assumed the reader would read the material more carefully still remains open.
References

Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.


text:
What is the order of crystal field stabilisation energy for tris(ethylenediamine)chromium(III) chloride and hexaaquachromium(III) chloride?
In my book it is given that the crystal field stabilisation energy for $\ce{[Cr(en)3]Cl3}$ is more than that of the complex $\ce{[Cr(H2O)6]Cl3}.$ By what I have studied, the crystal field stabilisation energy depends on the number of electrons in $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ orbitals and the pairing energy.
The number of electrons in both the compounds is the same and both undergo octahedral splitting, so that term is same for both.
Now coming to pairing energy is the energy of $\ce{en}$ more than $\ce{H2O}$ since it is a chelating group? Or it depends upon the position of the ligands in the spectrochemical series?

it depends upon the position of the ligands in the spectrochemical series

Although the number of electrons are same in Eg and T2g orbitals though the ∆o is not same for both the ∆o for chelate ligands is > than that of normal ligands hence the CFSE OF FORMER IS MORE


text:
Stability order of chelate complexes of nickel and iron

Which stability order is correct?

$\ce{[Ni(en)2]^2+} > \ce{[Ni(dmg)2]}$
$\ce{[Fe(edta)]^2-} > \ce{[Fe(en)3]^3+}$

In the first case I think the stability depends upon the strength of the ligand so is $\ce{en}$ stronger than $\ce{dmg}$ to form a more stable compound? As the oxidation state of nickel is same for both the cases.
In the second case the oxidation state of iron is +2 and +3, respectively. But I am confused if the metal with higher oxidation state is more stable or the metal with lower oxidation state is more stable?
Does stability depends upon the number of chelating groups?

In general, metal with higher oxidation state forms more stable complex.

But that does not appear to be the case in the second example where +2 is more stable than +3 of fe


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Why does not aqua regia burn filter paper?
We were doing basic radical analysis in lab and we were asked to add aqua regia on the filtered PPT collected on the filter paper and collect the formed solution , my question is if aqua regia can dissolve even gold and platinum why didn't it burned the filter paper(or made holes in it )

Why would it? There is no "even". There is no universal rank of "strength" for all chemical reagents. HF eats glass, but is indifferent to gold. Aqua regia is the other way around. H2SO4 does neither of that, but would "burn" your paper all right. This is chemistry.

@IvanNeretin Well, there is always our trusty friend $\ce{ClF_3}$

For "burning" of paper by sulphuric acid, the better term would be "aggressive dehydration".
Note that aqua regia is rather a diluted acidic mixture without significant dehydrating or oxygen based oxidation behaviour, but manifests aggressive chloration behaviour, that dissolves precise metals.
Cellulose has quite good endurance against chlorine, which is used in various solutions for cellulose whitening.

Concentrated and strong acids like sulfuric acid are substances that extracts water anywhere and everywhere to react with and make H3O+ ions. They are even able to extract water from the cellulose of the paper, as cellulose is made of carbon plus water. So a drop of sulfuric acid on paper burns it, or rather transforms it into charcoal.  The word "burn" is not appropriated,
Aqua regia and concentrated hydrochloric acid will never do it, as these liquid substances are made of more than 50% water.


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Under what circumstances that Liquid-Liquid extraction would be good to use?
Under what circumstances that LLE would be useful ?
Are the 2 top reasons -
1. Difference between the boiling points of the components are very low or similar.
2. For Temperature sensitive compounds as there is no heating which may cause thermal decomposition.
are there any other reasons ?

You focus on why to use LLE instead of something else, while you should focus more for the primary reasons when LLE can be beneficial.

The above reasons to use LLE are focused on the compound thermal similarities and instabilities.
But it does not take into the account the amount of compounds. While distillation separation can be great for major components, it may fail for minor one, and definitely for trace ones.
The LLE is frequently used for separation and content enrichment of trace amounts, in wide range of grams to nanograms per litre.
Using the approach of chemistry of complexes, it may also lead to great separation of non-volatile compounds/ions/elements.
LLE is e.g used for mutual separation of  nuclear fission products and not spent fuel by the PUREX method.


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Feasibility of using ionic liquid fuels in an MHD type reactor
First off I would like to say that my knowledge on chemistry, even general chemistry, is basic if not poor.
That said, I have some intuituions on this matter. I haven´t found any paper regarding this kind of system, and that is pretty much a killer for me.
Even with that I´d still like to ask the question.
Ionic liquids, liquid salts, are regarded as having some conductive capacities, being low to moderate conductors due to their ionic nature.
I know that salts have been used in MHD (magnetohydrodynamic) systems due to their high melting points -the ionic bond being particularly strong- and their conductive potential.
I have an specific fuel reaction in mind, An hydrazinium based propellant + LOX (liquid oxygen) as oxidizer,
[HO-C2H4-N2H4]+[NO3]- + 2.25 O2 = 2 CO2 + 1.5 N2 + 4.5 H2O
for example.
I read that ionic liquid salts do not need to be dissolved for the anion and the cation to conduct electricity (https://www.researchgate.net/post/How_is_the_electrical_conductivity_of_ionic_liquids_in_comparison_with_metals)
Outside of that I´d like to know if hydrainium based reactions do have some electric potential, or if by any reason wich I did not manage to grasp, it is simply out of the question to use them in such machines.
I would be very pleased to know if something is just worng or has been misunderstood.
Source for the reaction: http://cdn.intechweb.org/pdfs/20222.pdf, page 9 (455 in the original text).
Overal description of how a MHD machine works: https://en.wikipedia.org/wiki/Magnetohydrodynamic_generator
*Edit:
The majority of the volume is occupied by the reaction products. Using just conventional fuels like liquid hydrogen is just more practical.
Still, we could use a thermal/catalytical decomposition reaction, like 6 NH3OHNO3(aq) = 2 N2(g) + 2 NH4NO3(s or l) + 6 H2O(g) + 4 HNO3(g), where both a salt and water are produced.

If fuel is oxidized at high temperature, the conductivity of the fuel itself is irrelevant -- only the conductivity of the combustion products, which are ionized because of the high temperature and are therefor conductive. Brittanica states that temperatures greater than ~2,500 K are needed for sufficient ionization. Alkali metals can be used to increase ionization, or higher temperatures have been used with inert gases.
Note that the temperature of ions and electrons in the gas are not necessarily the same, due to the difference in mass and time to reach equilibrium. This impacts on conductivity and the rate of erosion of the MHD components.


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Comparison between van der Waals radius and metallic radius

I was reading about different types of atomic radius. It was written that if any atom can possess all the three types of radius, then van der Waals radius will be greater than metallic radius. However from the above diagram, it appears that the van der Waals radius and metallic radius are equal.
Then how can the van der Waals radius be greater than metallic radius?

Metallic radius is half of the distance between the nuclei of two atoms when they are in crystalline structure and bond is formed between positively charged atoms, that means they are tightly packed in metallic lattice.
And van der Waals radius is half of the internuclear distance of two non-bonded atoms of the same element in the closest possible distance.
Don't compare both the situations using above diagrams.  The radii are calculated in two different situations.
As in case of metallic bond, the atoms are bonded tightly so their shells can overlap, but in case of van der wal  forces the bond is weak. Therefore, in case of van der waal force, the internuclear distance will be higher as compared to metallic bond.

Via:
https://clay6.com/qa/94625/covalent-radius-of-cl-is-more-than-its-van-der-waals-radius-true-or-false-
https://socratic.org/questions/why-is-van-der-waal-radius-greater-than-covalent-radius-for-covalent-compounds

If in metallic radius overlap takes place then how can you say that it will be greater than covalent radius


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Is it possible to use the laws of termodynamics for non-ideal systems?
For the reaction
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}\quad\Delta H = \pu{−560 kJ}$$
it is given that the reaction occurs at constant volume of $\pu{1L}$ and the initial and final pressures are $\pu{70 atm}$ and $\pu{40 atm},$ respectively, with an initial temperature of $\pu{500 K}.$ It is further given that all the gases show significant deviation from ideal behaviour.
I don't understand how we can apply the standard equation of
$$\Delta H = \Delta U + \Delta n_gRT \quad\text{or}\quad \Delta H = \Delta U +\Delta(PV)$$
As compared to
$$\Delta H = \Delta U + z\Delta n_gRT \quad\text{or}\quad \Delta H = \Delta U +\Delta(PV)/z$$
However, the only solutions I found to this problem use the standard equations.

Note that H = U + p.V by definition,no matter if the system is ideal or not. If we can neglect nonideality of a gas, we can use H = U + nRT

Yes, this can be done, at least up to moderate pressures like these if you know the 2nd virial coefficients for the pure gases.  You need to learn how to get the residual properties of gas mixtures at elevated pressures.  This is all explained in chapters 10-15 of Introduction to Chemical Engineering Thermodynamics by Smith and van Ness.  Do you at least know the mole fractions of the reactant and product mixtures?


text:
Use chlorophyll a Galvanic cell to create hydrogen and electricity
What if we want to take chlorophyll in a container and put it into light . Then chlorophyll will lose electrons and will become positive itself then we will have a solution of negatively flowing electrons and positively charged chlorophyll molecules.
But somehow we have to lose electrons and transfer them somewhere so what we can do is take a Galvanic cell and directly  place a wire between the solution and copper rod . Then maybe electrons can get transfer from chlorophyll solution to Galvanic cell and the Salt Bridge between Copper and zinc rods neutralize the solution .
Then once the chlorophyll becomes completely positively charged and electrons have transferred we will put water into it then the electrons from H2O will be gained from chlorophyll molecules and H2 will get separated into h+ two molecules and one molecule of o . Two molecules will form oxygen and get released into the air while h+ and will remain in the solution.
Then we will take another container with chlorophyll that has been put into the light and we will connect wire between these two chlorophyll solutions and electron from the the second container will flow to the first container and H + ions will get converted to H2 gas and we will capture this gas and will use all the electricity flowing between.
Will this method work?

Technical note: Sentences in scientific context are recommended to have 20-25 words in average, with recommendation not to cross the word limit 35. Additionally, elementary thoughts are recommended to put into separated paragraphs to make reading easier. Your post has 2 sentences, the first one quite giant, the 2nd one has 4 words. It is very difficult to follow for these formal reasons. It would be even worse in case of a speech.

Sorry,  I edited question to make it easier to read.  Please remove negative marking.

Try to create more than 1 paragraph. Divide the idea into several sub-ideas. You will see how easier it would be to read it.  It is very small effort and it helps readers a lot. For a test, grab any unknown multiparagraph text  and merge it into a single paragraph. Then read it. :-)

I have done it. Thanks for recommendation.

Good. But from chemistry point of view, it is much more complicated. There are no free electrons in the solution, but thay are bound to other organic molecules. But yes, there already is the shortcut to generate electricity, instead of glucose creation.

So there must also be a way to get h2 gas. And electricity.Please tell me the method.

The site policy supposes the user asking a question does his figural homework first, doing the basic research for the topic. Then coming only when he/she got stuck, providing info what he/she has already done. Perceived low effort at the user side may be considered as the reason for closing the question. The effort of the one who asks should not be highly disbalanced with the expected/requested effort of those who answers. That makes the SE sites unique with high quality answers AND questions.

Just beware that water splitting by the sun is ongoing research. Whatever you use as photosctivated species the concept is that. It might be TiO2, etc.I think you could search for something like "green or solar water splitting" etc...


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Comparison of C-Cl bond length in CH3Cl and CF3Cl
How can one compare the $\ce{C-Cl}$ bond lengths in chloromethane ($\ce{CH3Cl}$) and chlorotrifluoromethane ($\ce{CF3Cl}$)?

Bent's rule is also consistent with Gillespie's VSEPR model, and may provide alternative rationalisation for effect of electronegativity. So it is restated as: 'more electronegative atom not only prefers to stay in the orbital having more $\mathrm{p}$ character but also can increase the $\mathrm{p}$ character in its attached orbital from the central atom depending on the circumstance.'
With increase in $\mathrm{p}$ character in an orbital, bond length will increase while with increase in $\mathrm{s}$ character in an orbital, bond length will decrease. For example, $d_\ce{(C-Cl)}$ in $\ce{CH3Cl}$ ($\pu{1.78 \unicode{xC5}}$) $\gt d_\ce{(C-Cl)}$ in $\ce{CF3Cl}$ ($\pu{1.75 \unicode{xC5}}$).

Source: Concise Inorganic Chemistry for JEE (by J.D. Lee and Sudarshan Guha)
I understand this alternative statement to Bent's rule.
Can anybody explain how have the $\ce{C-Cl}$ bond lengths been compared here? I don't get this. I understand that fluorine is more electronegative than hydrogen and might induce small positive charge on the carbon atom. Also, the $\ce{C-F}$ bond will have more $\mathrm{p}$ character than the $\ce{C-H}$ bond. How do I proceed further to reach the conclusion?

The paragraph you cite does not define Bent's rule (see What is Bent's rule?, Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?).
If you compare $\ce{H3CCl}$ and $\ce{F3CCl}$, then the $\ce{C-H}$ bond has less carbon-p character than an idealised sp³ orbital, while the $\ce{C-F}$ bond will have more carbon-p character than the idealised sp³ orbital. This in turn means that in $\ce{H3CCl}$ the $\ce{C-Cl}$ bond will have more carbon-p character as an idealised sp³ orbital and is therefore longer, while in $\ce{F3CCl}$ the $\ce{C-Cl}$ bond will have less carbon-p character and will therefore be shorter.
Note of caution: I have read many questions about the concepts introduced in J. D. Lee's Concise Inorganic Chemistry; many of which revolve around outdated theories. I do not have a copy of that book, but from what I have experienced so far, I do not think it is a good book. Alas, in this instance it appears to be correct though.

Thanks for the explanation!
Regarding the note of caution: Even worse, mine is an adapted and highly modified version of the original book. I'm using it just because it is advised by my teachers for JEE (Joint Entrance Exam) preparation.

@SatvikPopli As a former JEE aspirant,I can tell you one thing,that many times the stuff we employ for JEE purposes  in chemistry is either not the complete truth,or is a mutiliated version of the original explanation so that you can solve a question. So best to accept that people on this platform know better :)


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Ratio of Water to Isopropyl Alcohol?
I am looking for a very simple formula to calculate the ratio of water to 99% isopropyl alcohol that I need to prepare a solution with a density of 0.925 g/cm3. I am assuming there is one.

You'll also need a specific temperature.

If performing an interpolation still passes as «simple» for you, you may find experimentally determined values about iPrOH in water tabulated for example in journals about (chemical) engineering.  The previews of this and this give you about 10 entries for free (which may be complemented by the entries in the CRC Handbook of Chemistry and Physics like here).  If more data are required, a more recent here suggests going to a library and having a look into Ullmann's Encyclopedia, too.
While keeping an eye on the temperature of the two, determine the volume of water and iPrOH to mix in separate containers in advance (1L of water and 1L of iPrOH don't add up to 2L of blend).

There are no formula for calculating the density of a mixture water + any alcohol as a function of the proportion of water.

No theoretical formula, but an empirical one fit to experimental data might exist.


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Why does adding salt to water help distillation?
I've been looking at instructions on how to get lanolin from fleece and a lot of the instructions mention that adding salt to the water before distillation will increase the lanolin yield. Why is this?
I would think that the distillation would separate the water but not the salt leaving the lanolin contaminated with salt.

My first thought is that vapor pressure has been lowered due to the colligative properties.

Good question. We frequently "salt out" semi-soluble organic compounds from water because the ionic nature of the salt is more compatible with the water and less compatible with the (less ionic) organic compound. This is used, for instance, to drive sodium stearate (soap) from water. Adding salt could help drive more lanolin out of the water, or what is probably the situation, making an emulsion of water and lanolin less stable so that you can get more lanolin and less "grunge". The lanolin is then purified by filtering, which would remove salt crystals that formed as the water evaporated. And filtering would remove other dirt.
One description of obtaining lanolin suggests driving off all the water, then purifying by adding olive oil and water to remove dirt. But then it raises the question: why not remove the lanolin before all the water is evaporated? The salt would remain in the water. https://permies.com/t/65837/fiber-arts/extract-lanolin-wool
It's also possible that the water pulled out by the wool extracts a great deal of the salt, leaving the lanolin with only a small portion of the original salt and water (depending on how much water you use in the first place).


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What is the mechanism for trans-stilbene to stilbene dibromide?
What is the mechanism for trans-stilbene to stilbene dibromide?

This is what I know:
The bond between the hydrogen and bromine is broken and the bromine is given a negative charge, while the $\ce{H}$ protonates the double bond between the two $\ce{Ph}$ groups. After that, I am unsure what the $\ce{H2O2}$ does

What is the role of the peroxide? What are your thoughts?  You need to make some effort to answer this or it will be closed.

@Waylander The bond between the hydrogen and bromine is broken and the bromine is given a negative charge while the h protonates the double bond between the two Ph groups. After that, I am unsure what the H2O2 does as well as the purpose of ethanol

Etahnol is just the solvent. What is hydrogen peroxide usually used for? If you look at the product stilbene dibromide you will see the trans configuration of the Br atoms, does this remind you of any other form of addition to a double bond?

@Waylander Doesn't hydrogen peroxide add an OH group to the molecule? Also, isn't addition of H2O2 a syn addition?

You know what syn addition is and don't know that H2O2 is oxidant?

The key feature in the product that shows what is going on is the trans configuration of the two bromines. This is characteristic of the addition of elemental bromine via a brominium ion mechanism here.
So where does the Br2 come from as we started with HBr? This is where the hydrogen peroxide comes in. It is well documented example here that hydrogen peroxide  oxidises hydrogen bromide to bromine. This transformation can take place in the presence of an activated alkene such as stilbene.


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How to do RPA (Random Phase Approximation) in VASP?
I'm new to VASP and as such don't know much of its intricacies yet. I want to find out the interlayer binding energy of a given molecule and to do that I already opted for the use of different functionals (DFT-D3, SCAN, etc.) to find the total energy of my system in case of the bulk material and in case of the isolated layer. From these two I calculated the interlayer binding energy but am still about an order of magnitude of.
Now, I thought of using RPA in VASP to maybe get closer results.
My question(s): How do I implement RPA into VASP? I've read the wiki entry on it but am still not sure how to modify the other INCAR tags. Do I add RPA while also utilizing some functional (i.e. one of those I already used) or is RPA sufficient on its own?
And are there other methods with which I could get better values?


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Why does Calcium have a higher ionization energy than Aluminium?
Given their places on the periodic table I'd assume Aluminium has a higher ionization energy, because it has fewer energy levels, and is on a "righter" row on the periodic table, but in reality it is the opposite. Does anyone here know why?

Note that instead of Al vs Ca comparison, it is better to compare Al vs Mg and Mg vs Ca.

Ionization energy is the minimum amount of energy needed to remove most loosely bound electron (valence electron).
The atomic number of Calcium  is 20. So it's orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$. This is the arrangement of electrons in the shells. All orbitals are completely filled. So, for removal of an electron from the outermost shell, higher amount of energy is required.
In case of Aluminium, the atomic number is 13. Its orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^1}$. One single electron is present in outermost p-orbital, so it can be easily removed as compared to Calcium.
So, ionization energy of Calcium is higher than the ionization energy of Aluminium.

The 3p electrons have higher energy than 3s electrons, mainly because of better nucleus charge shielding. See also Slater's rules for electron electrostatic shielding effect.
Therefore, 3p electrons have lower ionization energy.

The first ionization of calcium does not break the $3s$ subshell.

@OscarLanzi Sure, it does not, it is 4s.


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Interlayer binding energy: how to convert to meV/Å²?
I've noticed that in most papers the interlayer binding energy is given in units of meV/Å². But how do I get these units? Do I divide the interlayer binding energy I get by the area of my unit cell?

I'm not sure how area of unit cell is relevant here, normally it's just contact surface area. Also, it looks like the title asks to convert something to meV/Å², what is it, exactly? Or your question is solely the physical meaning of meV/Å²?

Typography tip: get to know compose key, which allows to almost instantly type a lot of Unicode symbols with a standard US-keyboard. Linux and macOS typically offer this out of the box, for Windows there is WinCompose: to insert an Ångström symbol, I type Meta + o + A (I assigned Meta to the right Alt key) and get Å. Similar for umlauts Meta + " + o → ö.

@andselisk Sorry, I meant the contact surface area I guess. So if I had $\vec{a}$ = (3, 0, 0), $\vec{b}$ = (1, 2, 0) and $\vec{c}$ = (0, 0, 3) for unit cell vectors, the contact surface area would be the area spanned by $\vec{a}$ and $\vec{b}$, right?

Yes, that would make sense to me.


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H-NMR with 10H integration?
I have the following two spectra (IR and 1H-NMR) see below.
The chemical formula is: C16H16O2
Thus DBE = 9
My approach:
From the IR, I can see that there's no broad peak around 2500-3300, thus I can rule out any carboxylic acids and alcohols.
Then I'll look at the sharp peak at 1734 which makes me suspect an aldehyde.
From the H-NMR I see the following:

This is where I run into some issues.
I see peak a and c as -CH2- groups next to another R-CH2-, but in two different chemical environments.
And I suspect that peak d might be caused by two benzene rings (2x5H=10H) in a symmetric molecule, like; Ar-R-Ar but I can't stitch these informations together to something useful.
I hope that someone can help me out.

There's no aldehyde, you do not have the NMR peak for CHO

Do you have an idea of what it could be then? Or at least what could cause the 10H peak?

Your signal at 4.4 could be ArOCH2R

Well shouldn't that R-group then be another CH2 for it to have a multiplicity of a triplet?
And do you have any idea of the peak at ~7.35?

A bis-substituted naphthalene could be consistent with the 10 aryl peaks.

Wouldn't that only give an integration of 6H?

Oops. Bad counting. Sorry... :(

Your 3.65 could be methylen between aryl and carbonyl

I'd suppose it's probably Ph-CH2-CH2-O-C(O)-CH2-Ph?

Exactly, I just went in to write that.

Looks a reasonably good fit with this predictor https://www.nmrdb.org/new_predictor/index.shtml?v=v2.103.0

This is not a hard question when compared to other questions of same kind. The first easy clue is the molecular formula, which is already given. That make your life much easier. The given molecular formula for the compound is $\ce{C16H16O2}$, hence you can determine the degree of unsaturation of the molecule ($\ce{C_nH_nO_x}$):
$$u= \frac{n \times 2 + 2 - m}{2}=\frac{16 \times 2 + 2 - 16}{2}=9$$
Based on $\ce{^1H}$-$\mathrm{NMR}$, you correctly proposed that there are two mono-substituted aromatic nuclei in the molecule, since there is a 10-proton multiplet ($2 \times 5\ce{H} = 10 \ce{H}$) in the aromatic region ($\pu{7-8 ppm}$). However, instead of your suggestion of two symmetric aromatic rings like $\ce{C6H5-R-C6H5}$, these two phenyl groups are not probably identical rings. That because, you can see a clear 2-proton doublet (not 4-proton) at around $\pu{7.2 ppm}$, separating from the rest of 8-proton multiplet. Regardless, these two aromatic nuclei would count for 8 degrees of 9 possible unsaturations and $\ce{C12H10}$ portion of the molecule leaving only $\ce{C4H6O2}$ to be accounted for.
The presence of strong peak at $\pu{1734 cm-1}$ in given FT-IR spectrum suggests that there is a carbonyl group in the molecule. As you suspected, it is neither a carboxylic acid group (absence of broad peak at $\pu{2500-3300 cm-1}$ in the given FT-IR), nor an aldehyde group (absence of 1-proton resonance at $\pu{9-10 ppm}$ in the given $\ce{^1H}$-$\mathrm{NMR}$). Nevertheless, it has counted for the last degree of unsaturation. Furthermore, that also counted for one oxygen atom and a carbon atom of the molecule as well. Now, we need count for only $\ce{C3H6O}$.
Now, let's analyze $\ce{^1H}$-$\mathrm{NMR}$ spectrum. There is a 2-proton singlet at around $\pu{3.6 ppm}$ for isolated methylene group (without a neighboring hydrogen), indicating it is most probably next to a oxygen atom in the molecule or in between one of phenyl group and the carbonyl group, based of its deshielding nature. Also, there are two 2-proton triplets present in the $\ce{^1H}$-$\mathrm{NMR}$: One is present at around $\pu{2.9 ppm}$ and the other at around $\pu{4.3 ppm}$, suggesting the presence of isolated (no other protons at vicinity) -$\ce{CH2CH2}$- group in the molecule. These 3 -$\ce{CH2}$- groups are accounted for the rest of $\ce{C3H6}$ portion of the molecule.
Furthermore, the highly deshielding nature of the resonance at around $\pu{4.3 ppm}$ strongly suggest that corresponding -$\ce{CH2}$- group is next to the final oxygen atom to be counted. Also, the other -$\ce{CH2}$- group corresponds to the resonance at around $\pu{2.9 ppm}$, which is only moderately deshielded, clearly suggests that it is neighboring the second phenyl group.
Based on these assignments, one can deduce the structure as 2-diphenylethyl 2-phenylacetate as depicted in the diagram below (inserted blue box) with suggested $\ce{^1H}$-$\mathrm{NMR}$ spectrum by ChemDraw, which gives a close resemblance to the actual one:

I will ignore the IR as I always do and work only with the NMR.

The ten aryl protons are probably two phenyl groups ($\ce{C6H5-{}}$). That leaves $\ce{C4H6O2}$ and takes care of eight of the nine double-bond equivalents.
The two triplets are very likely (examining the coupling constant would confirm that) part of a $\ce{-CH2-CH2-{}}$ fragment. On either side of these is either an oxygen or a quarternary carbon as there are no additional couplings.
The final singlet points to a $\ce{-CH2-{}}$ group with the same constraints as above.
We still have $\ce{CO2}$ and one double-bond equivalent to take care of. One functional group particularly suggests itself. Then, we have to play connect the fragments.

 $\ce{CO2}$ is likely an ester functionality since we do not have a carboxylic hydrogen to spare.

 One of the two triplet $\ce{CH2}$ is shifted downfield rather strongly. This suggests that it is directly bound to the ester group oxygen, giving $\ce{-CH2-CH2-OC(=O)-{}}$.

 Both of the remaining $\ce{CH2}$ groups will be benzylic but the singlet is more deshielded than the triplet. That suggests that the singlet $\ce{CH2}$ is not only benzylic but also α to the ester.

 Therefore, the final result is $\ce{C6H5-CH2-CH2-OC(=O)-CH2-C6H5}$.


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Energies of Virtual Molecular Orbitals (LUMO)
I just had a question about molecular orbitals. I was just wondering if the energies of the LUMO (virtual orbitals) derived from CCSD(T)/cc-pTVZ calculations could be trusted. I remember reading some time ago that these orbital energies could not be trusted since they contained no electrons. If they cannot be trusted, could you suggest any calculations I could perform to obtain the LUMO molecular orbital energies (perhaps TD-DFT or some other method)?

Related: https://chemistry.stackexchange.com/q/86469

I think a key question is what you mean by "trusted." Do you mean "does the LUMO eigenvalue correspond to the electron affinity" or "does the LUMO eigenvalue tell you about the first excited state" (among two somewhat similar properties). It might also help to know what kind of molecules you're considering (since some species have unbound LUMOs)

When thinking about orbital energy levels, it is not a good idea to think of them as fixed energy 'levels' that can be filled with electrons. This kind of picture is often found in introductory textbooks, but it is wrong. For example, it is known that even for the Hartree-Fock theory, the energy of a molecule is not a sum of the orbital energies. In this sense, you can never really consider orbital energies as the energies of the given electrons - which is really a meaningless quantity, if you think about it. Basically the only interesting quantity that you can get from an orbital energy is the first ionization energy based on Koopmans' theorem, which also is just approximately true.
So can you trust orbital energies? It really depends what you want to calculate. One thing that people often want is to calculate the UV-VIS spectrum of a molecule, which is, of course, the energy required to move an electron form an occupied orbital to a virtual orbital. But this energy should not be calculated as the difference of the energies of the two orbitals. The reason for this is not that virtual orbital energies can not be trusted, but the fact that when you move an electron, you now have a different electron configuration, that is, a different wavefunction. And with a different wavefunction, of course, you would get different energy levels. You should not use the energy levels of the original wavefunction to describe the energy levels of an excited state.
So in order to calculate UV-VIS spectra, you should rather use a multi-reference method with a large enough active space that includes both the ground and the excited states and also some 'buffering states' to expand the difference in the electron correlation. If you can afford a CCSD(T) calculation, then you might also be able to afford a CAS-SCF calculation with a formidable active space, maybe even add a bit of NEVPT2 to it. I also have some good experience with the SORCI method of Orca, if you want to check that out. And if you can not afford to do such large calculations, you might also fall back to TD-DFT - some good double hybrid methods have been made available this year.