text:
How closely should the energy components of ab initio SCF results at a stationary nuclear configuration preserve the virial theorem?
When running a SCF geometry optimization for water molecule, I found that the energy components of the optimized geometrical structure appear to deviate from Virial theorem sightly:
Kinetic energy = 75.424942151953331
Total energy = -75.585962823645531
How should this discrepancy be interpreted? Is it an indication of inaccuracy in the geometrical optimization algorithm of the software being used?

Please see my edits. I think the title is now a better reflection of the question, but please feel free to edit or roll back if you don't like it.

Thanks for the edits, the new title is better.
If I read correctly this book, Basic Principles and Techniques of Molecular Quantum Mechanics
By Ralph E. Christoffersen, pp 460, https://books.google.com/books?id=gO7iBwAAQBAJ&pg=PA456&lpg=PA456&dq=virial+theorem+trial+wavefunction&source=bl&ots=nwCbv51R1h&sig=HCmrsXEkCuJrUFDV-QChE2yBGNM&hl=en&sa=X&ved=0ahUKEwj82vyOp-7VAhXjz4MKHfi5DI44ChDoAQgpMAE#v=onepage&q=virial%20theorem%20trial%20wavefunction&f=false  virial theorem applies to non-exact wave function as long as the geometry is optimized, independent of the choice of basis set.

I hope I am not mistaken about this, but if I'm wrong then I'll learn so no big deal.
I believe this is a result of the fact that the electrons do not experience a true $r^{-1}$ potential due to electron-electron repulsion. Namely, we should expect that the electrons will repel each other which will cause the apparent attraction to the nuclei to be weaker than $r^{-1}$.
In other words, for a general potential,
$$
V(r)=r^{n}
$$
we have,
$$
\frac{dV(r)}{dr}=nr^{n-1}
$$
Because the Virial theorem can be stated as,
$$
\langle T\rangle=\frac12\langle\vec{r}\cdot\nabla V\rangle
$$
But for a potential which is less than coulombic, we have,
$$
\langle T\rangle=-\frac{n}{2}\langle V\rangle=-nE_{tot}
$$
That is to say, the electrons in water are feeling an effective potential of the form $V(r)=r^{-.99787}$ or so.
Or, in the language of the book you linked, the trial wavefunction of the form being tried for the hydrogen atom does not evaluate to a parameter $\zeta=1$.
Again, I am not completely confident that I nailed this answer, but for systems that have no obvious analytic form of the potential energy, there is no reason I see that we have should have $\langle T\rangle=-E_{tot}$ be exactly true. That is not to say the virial theorem is not satisfied, it certainly is, only that it is not as clean as we are used to.

I was a bit loose with some constants in my equations, please forgive me :)

Thanks for the insightful answer. Although SCF is single-body mean field theory, the wave function thus derived is an approximation to the exact wave function of the all-electron Hamiltonian, in which case n = 1 in your formula, even after electron-electron repulsion is taken into account. The above deviation is only significant when a small basis set, say 3-21G,  is used. For 6-31G deviation is rather negligible. You are probably right to the point that for smaller basis set,  it does not evaluate to a parameter ζ=1, but I do not fully understand how.

Yes I think you're correct. It makes sense. I'm completely guessing here but it could be that if smaller basis sets show a larger deviation, and they should follow the virial theorem exactly, that this is the result of some kind of intramolecular basis set superposition error... ?

@Wang-X-Y Ehhh... I ran some calculations on water with a number of different basis sets and there is very little change between them. In fact sometimes the smaller basis set had a ratio closer to one, but really the difference is quite small.
I think I'll post these sometime tomorrow because pretty much my answer above isn't right. I found it very compelling, but the fact HF is a mean-field theory invalidates my reasoning.

Thanks for looking into this, it's very nice to have someone to compare notes:-) Here is the input of the calculation, for the purpose of the project I worked on UHF rather than RHF chosen, not sure where it caused the differences.   molecule h2o {
O
H 1 0.96
H 1 0.96 2 104.5
}
set basis 3-21g
set scf_type pk
set g_convergence GAU_VERYTIGHT
set reference uhf
E, wfn = optimize('scf',return_wfn=True)

Originally, I was convinced that the molecular geometry has very limited influence on the virial ratio: only in so far as the effective basis set will change with the geometry as long as nuclei-centered Slater or Gauss (STO or GTO) basis functions are used.1
Hartree-Fock is accurate as full CI (within the given basis set and its postulates and earlier approximations) for one-electron systems. Let's look at $\ce{H}$:
$$
\begin{array}{lr}
\mathrm{Basis\ set} & |VR-2| \\
\hline
\mathrm{cc-pVDZ} & 0.00002250 \\
\mathrm{cc-pVTZ} & 0.00004588 \\
\mathrm{cc-pVQZ} & 0.00000153 \\
\hline
\mathrm{def2-SVP} & 0.00000000 \\
\mathrm{def2-TZVP} & 0.00000049 \\
\mathrm{def2-QZVP} & 0.00000404 \\
\hline
\end{array}
$$
$VR$ is the virial ratio, whose value I expect to be 2. At this point, we can conclude a) that the basis set has an influence and b) it's more complicated than simple basis set size, even in reasonably constructed basis sets/basis set families.
Let's look at $\ce{H2+}$ at different $\ce{H}-\ce{H}$ distances, indicated by the factor, i.e. as multiples of the equilibrium distance. The equilibrium geometry was found at the computational level: HF/aug-cc-pVQZ. The energy is given relative to the energy of the equilibrium structure.
$$
\begin{array}{lll}
\mathrm{factor} &\mathrm{rel.\ }E &|VR-2|\\
\hline
0.850 &0.120850254 &-0.11563802 \\
0.900 &0.048912238 &-0.07368802 \\
0.925 &0.026293496 &-0.05396290 \\
0.950 &0.011177260 &-0.03503389 \\
0.975 &0.002673918 &-0.01687637 \\
1.000 &0.000000000 &+0.00053214 \\
1.025 &0.002463880 &+0.01721247 \\
1.050 &0.009453714 &+0.03318366 \\
1.075 &0.020427030 &+0.04846357 \\
1.100 &0.034901950 &+0.06306886 \\
1.150 &0.072687778 &+0.09031731 \\
1.200 &0.119906546 &+0.11504445 \\
1.300 &0.233820556 &+0.15734817 \\
1.400 &0.363035528 &+0.19075111 \\
\hline
\end{array}
$$
We can conclude that at least for this system, the influence from the geometry is a lot larger than from the basis set.

In the comments, the suspicion that integral neglect thresholds and similar settings influence $VR$ was voiced. I decided to rerun the calculations from the first table using the tightest preset defined in ORCA, EXTREMESCF, which is described as close to machine accuracy. Given that some of the thresholds are then below $10^{-15}$ au, this seems credible. The result is:
$$
\begin{array}{lr}
\mathrm{Basis\ set} & |VR-2| \\
\hline
\mathrm{cc-pVDZ} & 0.00002250 \\
\mathrm{cc-pVTZ} & 0.00004539 \\
\mathrm{cc-pVQZ} & 0.00000156 \\
\hline
\mathrm{def2-SVP} & 0.00000000 \\
\mathrm{def2-TZVP} & 0.00000000 \\
\mathrm{def2-QZVP} & 0.00000000 \\
\hline
\end{array}
$$
The result surprised me, there actually is discernible influence on $VR$ from the thresholds. What's more, basis set families are different in this regard.
I also reran calculations from the second table when setting one of the atoms atomic charge to zero - i.e. with a ghost basis. The energy of the "dissociation" curve is not relevant for this crude investigation, and is thus omitted.
$$
\begin{array}{ll}
\mathrm{factor} &|VR-2|\\
\hline
0.850 &0.00027735 \\
0.900 &0.00027606 \\
0.925 &0.00027525 \\
0.950 &0.00027437 \\
0.975 &0.00027345 \\
1.000 &0.00027252 \\
1.025 &0.00027160 \\
1.050 &0.00027073 \\
1.075 &0.00026994 \\
1.100 &0.00026924 \\
1.150 &0.00026819 \\
1.200 &0.00026767 \\
1.300 &0.00026785 \\
1.400 &0.00026865 \\
\hline
\end{array}
$$
So the intramolecular BSSE can probably neglected for these investigations.
Further work would include extending post-HF methods to allow for calculation of the virial ratio. Application to methods that are regularly used for geometry optimization (e.g. DFT) is not trivial because exact calculation of the virial ratio is hard.2 A cursory evaluation of multi-electron systems on the HF level may also be useful.

1 I currently only have GTO programs at my disposal.
2 The kinetic part of the correlation energy is subsumed into the energy calculated by the XC functional [Jensen, Introduction to Computational Chemistry, 2nd Ed., Wiley 2007, p. 236]

The ratio of 2  is kept only by stationary geometrical state. For non-stationary there is another term: 2+ = R*dE/dR, in case of diatomic molecules.

Thanks for confirming the effect of basis set on the deviation, particularly for  the case of H atom. This is very convincing I think. As @jheindel mentioned in his answer, this 'complicated' issue might be related by BSSE. Or more likely this could be somehow  related to basis set incompleteness or BSIE. It would be interesting the deviation can be somehow formulated into a measure of BSIE.

@Wang-X-Y I was thinking the same thing. Truth be told I did those calculations because I thought it would be interesting if this could be formulated as a measure of intramolecular BSSE, but I think it has more to do with basis set incompleteness, and, honestly, how stringently you converge integrals. I would guess the likeliest thing is that this is from accumulation of error due to neglecting integrals. Although it seems weird this would manifest itself in the kinetic energy more than the potential energy.

@TAR86 you're first table is quite nice because it shows the trend does not necessarily depend on basis set size. The second table, however, is a bit artificial because the nuclear configuration needs to be at a minimum for VR-2 to theoretically equal zero. As Wang-X-Y says, another term must be included.

Also, I've become pretty convinced now that the only way we're gonna answer this question is if we get rid of all assumptions and do the calculation involving all integrals to be sure this isn't from screening out integrals.

It's very interesting that the deviation is zero for the def2 basis sets, could it be that for hydrogen atom these basis sets are actually "complete"? Just speculating as I am not family with these basis sets. Also I think the deviation might be much larger for  poly-atomic molecules at geometrical stationary state.

IIRC, the def2 basis sets, a.k.a. Karlsruhe or Ahlrichs sets, have non-polarization functions optimized for SCF/DFT-SCF (and take the polarization from Dunning's correlation-consistent sets, the cc sets).

I am posting a second answer rather than editing my first answer basically just because my first answer is wrong, and this answer is just going to present some data and then make the opposite claim of my first answer.
I ran some calculations using Molpro on $\ce{H2O}$ using a number of different basis sets. Molpro, conveniently, actually prints out the Virial ratio, and I believe it prints this as $\frac{\langle T\rangle}{\langle V\rangle}$ (as opposed to the other way). So, after optimizing the water using each basis set at the HF level, the ratio was found. Note that HF was used because finding this ratio for post-HF methods is much more complicated, and I believe that in the case of post-HF methods, my first answer actually would be correct. Namely, the ratio should no longer be -1.
\begin{array}{cccc}
\hline
\text{Basis}& \text{avdz} & \text{avtz} & \text{avqz} & \text{6-31g} & \text{6-31g**} & \text{6-31g++} & \text{STO-3g}  \\ \hline
\mathrm{Ratio} & -1.00080 & -1.00051 & -1.00008 & -0.99961 & -1.0008 & -0.99919 & -1.00600 \\
\mathrm{Deviation} & 0.00080 & 0.00051 & 0.00008 & -0.00039 & 0.0008 & -0.00081 & 0.00600 \\ \hline
\end{array}
Note that in the above avxz (x=d,t,q) is the aug-cc-pvxz basis set.
So, it does seem that with the correlation-consistent basis sets, the deviation from a ratio of one decreases as the basis set becomes more complete. The Pople basis sets don't follow any trend that I can see. STO-3g is the worst which makes sense cause it's the worst basis set.
I would suspect that these differences are not significant. I think this actually is a reasonable method of determining how converged a HF geometry is, but I am suspicious that something else is going on here. I don't know why there would be deviations. The only thing I can think of is intramolecular BSSE, which is rarely talked about because it usually doesn't matter, but maybe this measure is sensitive to it though. I kind of don't believe that though so I don't know what's going on. I guess one thing to try is to look at this ratio for a system which is known to suffer from intramolecular BSSE.
The only other thing I can imagine is the boring answer (but is probably right) that converging the total energy and the gradient, which is what are usually converged in geometry optimizations, does not converge the kinetic energy beyond the third-ish decimal place, so we see deviations.
As to your question of how much tolerance we should have for this, my guess is that the amount of tolerance we have now is sufficient because otherwise people would have figured out what was wrong and fixed it. That is, there is no obvious error in QM calculations that can be traced back to disobeying the Virial theorem. I would guess the reason for this is that usually post-HF methods are used anyways, so any effect of slightly disobeying the Virial theorem gets washed out because the geometry at higher levels is different from the HF geometry. Whether or not this same deviation is seen at higher levels of theory is something I don't know and would be trickier to answer.

The numbers of deviations for different basis sets in your calculation shows trend similar to what I saw but the slightly different numerical values.  These several explanation can all be valid, just difficult to evaluate to what extend they contribute, The question can be asked differently: seeing a deviation from Virial Theorem in one's optimized wave function, is it enough reason to infer that the basis set used is too small?


text:
How does the cyclopropyl group influence conjugation and aromaticity?
While answering the question Why isn't cyclobutane planar?, I recollected that cyclopropane and its derivatives behave like double-bond compounds, e.g. a cyclopropyl group can potentially be conjugated with neighboring $\ce{C=C}$ bonds and affect aromaticity, like in spiro[2.4]hepta‐4,6‐diene which has pronounced aromatic character (I remember this example from an early edition of March's Organic Chemistry):

Cyclopropyl group is generally known to be a poor $\pi$-electron acceptor, but a very good $\pi$-electron donor [1]. So, I'm wondering whether the cyclopropyl conjugation in general may result in altered aromatic properties, and to what extent; also, on how Hückel's $4n + 2$ rule is applicable here.
I've picked up existing cyclopropane derivatives which I think mimic electronic configuration of well-known aromatic and non-aromatic compounds, assuming each cyclopropyl group acts as a double bond:

[3]Rotane $\ce{C9H12}$ [2] (1) vs. benzene (1a);
Dispiro(2.2.2.2)deca-4,9-diene $\ce{C10H12}$ [3] (2) vs. cyclooctatetraene (2a);
9,10-bis(Spirocyclopropyl)-9,10-dihydroanthracene $\ce{C18H16}$ [4] (3) vs. dibenzo(a,e)cyclo-octatetraene (3a).

Crystal structure of (1), obviously planar triangle (1a is planar as well):

Crystal structure of (2), planar $\ce{C6}$ ring (2a is not planar):

Crystal structure of (3), all three $\ce{C6}$ rings are planar and are within a single plane (3a is not planar):

Provided with these three examples of planar systems for illustration (you can also pick up the systems of your interest), I would like to state a couple of questions:

Is there always a conjugation between a cyclopropyl group and an adjacent double bond? In other words, do cyclopropane's "bent" bonds always have extensive $\pi$-character and can therefore act as an alkene?
1a. Are electronic structures 1 and 1a; 2 and 2a; 3 and 3a similar, and to what extent?
Can we use Hückel's $4n + 2$ and $4n$ rules for cyclopropyl-containing monocyclic rings? If not, how to estimate aromaticity/anti-aromaticity in such systems?

Bibliography

de Meijere, A. Angew. Chem. Int. Ed. Engl. 1979, 18 (11), 809–826. DOI 10.1002/anie.197908093.
Boese, R.; Miebach, T.; De Meijere, A. J. Am. Chem. Soc. 1991, 113 (5), 1743–1748. DOI 10.1021/ja00005a043.
Haumann, T.; Boese, R.; Kozhushkov, S. I.; Rauch, K.; de Meijere, A. Liebigs Ann./Recl., 1997 (10), 2047–2053. DOI 10.1002/jlac.199719971007.
Rosenfeld, S.; White Tingle, C.; Jasinski, J. P.; Whittum, J. E.; Woudenberg, R. C.; Van Epp, J. J. Am. Chem. Soc. 1994, 116 (26), 12049–12050. DOI 10.1021/ja00105a054.

related https://chemistry.stackexchange.com/questions/33602/how-is-spiro4-4nonatetraene-antiaromatic

Cyclopropane often acts like a double bond in terms of stabilizing adjacent positive charge, when the 3-membered ring is properly aligned.  I've never see it discussed in the sense you suggest (removing one cyclopropane $\ce{CH2}$ group and then enlarging the other ring to include the remaining $\ce{CH2}$ group).  The appropriate analogy would be to replace the cyclopropane with an exo-methylene type of double bond (e.g. removing one of the cyclopropane $\ce{CH2}$ groups and creating a double bond between the remaining $\ce{CH2}$ group and the former spiro-carbon).

As to the "proper alignment", see here

Agree with ron. The analogy to compound 1 I expected was trimethylenecyclopropane.

If you go with Walsh cyclopropane, then there is a $p$-orbital right there. e.g., http://www.bluffton.edu/homepages/facstaff/bergerd/chem/walsh/derive.html

@M.A.R. I loved the edit summary ;)

About the molecule that you remember cited in your textbook, there's a study in ref.1. According to these authors:

The delocalization of the C1-C2 and Cl-C3 electrons of spiro[2.4]hepta-4,6-diene (1a) and partial rehybridization at C1, C2, and C3
  result in a contributing structure which is perhaps best represented as a $\pi$ complex (lb) of cyclopentadienylidene and ethylene.

From NMR studies they concluded that:

The comparison of the predicted downfield shift
  of approximately 6.6 ppm for the complex and the
  observed shift of approximately 1.0 ppm leads to the
  the conclusion that the $\pi$ complex may contribute as
  much as 15 % to the total electronic structure of 1.

The most recent reference I have found is from 1997(ref.2). They studied these structures:

In the series of the investigated compounds 1-5, spi- ro[2.4]hepta-4,6-diene (1) shows the strongest  influence of cyclopropyl  conjugation on molecular structure in the group of doubly  unsaturated   cyclopropyl  hydrocarbons

The conjugation properties of cyclopropane are subject of debate and other references don't support this. You can find a blog about this on the royal society of chemistry website(http://my.rsc.org/blogs/84/1009).
I have found another paper in ref.3 and according to this:

In one chemical attempt to test the transmission
  of conjugation by cyclopropane rings, Cannon, et al.were unable to alkylate or acylate the methyl group in diethyl 2-methylcyclopropane-l, 1-dicarboxylate and concluded that transmission of conjugative effects in the transition state for carbanion formation are small compared with analogous, unsaturated compounds.

Also:

Trachtenberg and Odian concluded from their review
  of the various conflicting reports that a cyclopropane
  ring is incapable of transmitting conjugation in the
  ground state but may be able to do so in some excited
  states. It is likely that, during chemical reactions,
  such transmission occurs only in a transition state,
  i.e., after bond breaking of the ring has commenced.

From what I have read, even assuming a conjugate effect, this would be much less evident than a C-C double bond.
About your questions:
Is there always a conjugation between a cyclopropyl group and an adjacent double bond? In other words, do cyclopropane's "bent" bonds always have extensive π-character and can, therefore, act as an alkene?
 Partially, the effect is not comparable to a true double bond. This hypothesized conjugation effect was found to be stronger in piro[2.4]hepta‐4,6‐diene(1) compared to the other molecules studied in ref.2(molecules 2 to 6 in the second picture).
1a. Are electronic structures 1 and 1a; 2 and 2a; 3 and 3a similar, and to what extent?
Considering that the supposed conjugation contribution of the cyclopropane ring is much less than a C-C double bond, I think they differ substantially. Unfortunately, I can't find any spectra or other additional data supporting this.
Can we use Hückel's 4n+2 and 4n rules for cyclopropyl-containing monocyclic rings? If not, how to estimate aromaticity/anti-aromaticity in such system?The Huckel rule is not valid in these complex cases. The Hückel's rule is based on calculations using the Hückel method and therefore has a limited validity. Here you need advanced MO calculations and as you have seen from all the references mentioned in such systems there's a lot of debate.
References:
1) Aromaticity via cyclopropyl conjugation. Electronic structure of spiro[2.4]hepta-4,6-diene, J. Am. Chem. Soc.197092154736-4738 https://pubs.acs.org/doi/abs/10.1021/ja00718a047
2) Structural Aspects of Cyclopropyl Conjugation: Experimental Studies and Ab Initio Calculations.  https://doi.org/10.1002/jlac.199719971007
3)Transmission of conjugation by the cyclopropane ring, Stewart JM, Pagenkopf GK, The Journal of Organic Chemistry, 1969. https://pubs.acs.org/doi/pdf/10.1021/jo00838a003?rand=vimssm6d

The paper from 1969 is an interesting find, however, it seems like authors give an overview of their observations and rather speculative results even for their own ether systems. I'm afraid neither question 1 nor question 2 is really answered here (not definitively, at least). I haven't really dive into the reference mentioned in the RSC Blog, but it also seems to be tangentially relevant here. Please correct me if I'm wrong, or feel free to expand your answer so that it's less broad and targets the systems mentioned in the question.

I have edited it. Unfortunately, I couldn't find more recent studies. your molecule 2 was mentioned in ref.2

Before going to your specific analogies, it's important to understand if the conjugation properties of the cyclopropane are a real thing, just the memory of a textbook that cite an example is not enough. About your analogies there isn't any study available. Finding an complete and exhaustive answer from the literature data available so far is impossible. I think it was a nice question that makes people reflecting about this topic, but the main question should be: Is really cyclopropane conjugated with neighboring C=C bonds?

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text:
Friedel–Crafts alkylation of benzene with tetrachloromethane
There was this question in an exam at our school which asked what the major product will be when excess of benzene is reacted with tetrachloromethane in the presence of aluminium chloride.
All of us got it wrong in the exam, but we realized that it's just Friedel–Crafts alkylation with extra steps. So later we came up with two competing answers: one group claimed it is tetraphenylmethane $(\ce{CPh4})$, while the other claimed it's triphenylchloromethane $(\ce{CClPh3})$ (aka trityl chloride).
Reason in favour of $\ce{CClPh3}$: After the formation of it, the negative inductive effect of the phenyl groups will supposedly be high enough not to allow the release of the last remaining Cl to form a carbocation. Also, there will be high steric hindrance in $\text{CPh}_4$.
Reason in favour of $\ce{CPh4}$: The phenyl groups are bonded to the central carbon through $\sigma$ bonds. So they can just rotate their planes to escape steric hindrance. Also, we do not know if the combined inductive effect of the phenyl groups is high enough not to allow the release of chloride. After all, the chlorine molecule is highly electronegative.
So, we're grappling with this problem and do not know the solution to it. Any help will be highly appreciated.

Fonken1 says that the reaction doesn't occur, but it is not due to steric hindrance or instability of tetraphenylmethane since the molecule is actually thermally stable. Instead, it argues that the formation of tetraphenylmethane due to alkylation of benzene using triphenylchloromethane is reversible in nature. The product readily dissociates into triphenylmethyl carbocation and benzene. Due to this unfavourable equilibrium, only insignificant amounts of tetraphenylmethane is formed.
Ben Norris also mentions that the reaction of benzene with tetrachloromethane is used to produce chlorotriphenylmethane, and provided an Organic Synthesis procedure2 that works on the kilogram scale.
References

Fonken, G. J. Alkylation of Benzene with Triphenylmethyl Chloride. J. Org. Chem. 1963, 28 (7), 1909–1909. DOI: 10.1021/jo01042a510.
Bachmann, W. E. Triphenylchloromethane. Org. Synth. 1943, 23, 100. DOI: 10.15227/orgsyn.023.0100.

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text:
The affect of effective nuclear charge on energy gap between subshells
A few days ago my teacher taught me about $\mathrm{d}$ orbital contraction. He said that in $\ce{SF6}$ the hybridization of sulphur is $\mathrm{sp^3d^2}$. He said that although the $\mathrm{d}$ orbital is higher in energy than $\mathrm{s}$ and $\mathrm{p}$ orbitals due to increase in positive charge on sulphur, due to high electronegativity of flourine the $\mathrm{d}$ orbitals contract more than $\mathrm{s}$ and $\mathrm{p}$ orbitals, hence the difference in energy decreases and $\mathrm{d}$ orbital can participate in hybridization.
There was no problem with the explanation as it seemed perfectly plausible. But today, while teaching MOT, he told that the gap between $\mathrm{2s}$ and $\mathrm{2p}$ orbitals increases as we move from lithium to flourine due to increase in effective nuclear charge and interelectronic repulsion, which is quite opposite to what he told some days ago.
What am I missing? Which of the following interpretation is correct? A qualitative explanation involving less of quantum mechanics is that I want and involving MOT and VBT only because I know only those theories. According to advanced theories d orbitals will not participate in bonding in SF6 but according to VBT they will. As I already mentioned that I know only VBT and MOT so I have asked the question in the context of these two theories.

d orbitals aren't involved in SF6 bonding

I have edited the question. I hope it's all right.

https://chemistry.stackexchange.com/questions/54737/how-are-the-hybrid-orbitals-of-sulfur-hexafluoride-shaped

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text:
Why is the standard reduction potential of an anode negative in a galvanic cell?
It makes not intuitive sense to me why something can have a negative voltage , I understand the concept of voltage drop , but voltage is not being dropped in the anode so why would it be negative ?

Sign of potential doesn't have any meaning, since zero is arbitrary anyway.

https://chemistry.stackexchange.com/questions/16785/positive-or-negative-anode-cathode-in-electrolytic-galvanic-cell

The anode is where the oxidation takes place, by definition, so the reduction potential must be more negative (or at least less positive) than the reduction potential at the cathode.
But, there is no guarantee that the reduction potential at the anode is negative.


text:
How to effectively induce the bending motion of carbon dioxide?
I would like to know how to effectively induce the bending motion of $\ce{CO_2}$ with femtosecond IR lasers. In other words, how to effectively populate bending vibrational states in $\ce{CO_2}$ neutral or $\ce{CO_2}$ cation?
The ground state of $\ce{CO_2}$ is linear, but the first electronic excited state is bent. Does it mean the bending motion can be more efficiently populated on the electronic excited state? What is the energy gap between the ground state and the first excited state of $\ce{CO_2}$? Is it possible to achieve efficient excitation with a strong IR laser pulse?

Electronic excitations are pretty much independent of the vibrational ones (though not vice versa). Also, they are way greater in energy than the latter, so you better make up your mind as to whether you want them or not. These would be two entirely different questions.

According to Herzberg (Electronic spectra of polyatomic molecules) $\ce{CO2}$ is transparent down  to at least $210$ nm and absorbs weekly thereafter. The excited state is bent at 122$^o$ so absorption could excite many vibrational quanta, but transitions from this state to the ground state will result in very highly excited ground state molecules. Your fs IR laser will not be of much use unless you  have enough energy and it is short enough (such as can easily produce a continuum) when absorption may be possible by multi-photon effects. However, you may more easily rip the molecule apart.


text:
Why do elements transition discretely between solid, liquid and gas forms, rather than existing as a spectrum?
I don't understand what makes a liquid entirely different from a gas, why wouldn't a liquid slowly become more gaseous, but instead transition completely from one form to the other?

Well, it would, if you'd walk around the critical point.

https://chemistry.stackexchange.com/questions/37088/why-must-both-the-critical-temperature-and-pressure-be-exceeded-to-achieve-the-s

The title is not very defining. Not all spectra are continuous; plenty are flat-line spectra.


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Nomenclature of Acetone
Why is the preferred IUPAC name of acetone "propan-2-one" and why not simply propanone? It doesn't make sense to specify the position of the carbonyl group as it can occupy only one position in acetone, i.e. on the 2nd carbon.

The IUPAC Blue Book defines rules where locants can be ommitted in IUPAC preferred names.
Unlike terminal locants in names for aldehydes that are omitted, there is no rule for homogeneous chains with three or more identical atoms that would allow for the ommission of the locant $2$ in propan-2-one.
This leads to the result that propanal is a IUPAC preferred name while propanone isn't.

You might also be concerned why ACETone sounds like it has two carbons and not three (2-PROPanone). It is because acetone was prepared early on from 1-1/2 acetic acids by dry distillation of the calcium salts via a Claisen condensation. The missing carbon is CO2.


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Rotten meat odor in plastic
I've been struggling for a few days with the cleaning of a fridge that was left off for a few weeks with raw chiken meat in the freezer compartment. When I found it, the meat had turned liquid with the smell you'd expect, spread in the entire fridge. I started cleaning thoroughly with bleach with little effect and then vinegar after stumbling on this piece of advice. But there is still a noticeable smell lingering.
My question is then: Is it possible that the plastic of the fridge has absorbed the cadaverine and/or putrecine that I assume are responsible for the smell? And if so will any amount of vinegar (or some other household acid) be able to remove the remaining smell?

Possible, but I'd guess that the fridge has some "internal" air channels that you can't clean. Also it would seem likely that liquid has seeped under the plastic liner.

My fridge is fairly small and basic: there's no active air circulation and the plastic is a single molded piece with only two holes for the cooling element at the very top that couldn't be reached by the liquid. I'll try to plug those holes to see if it makes any difference.

Yes, the odor does penetrate the plastic a bit. First,clean again with bleach (sodium hypochlorite, $\ce{NaClO}$ solution), disassembling some of the interior of the freezer, in particular. Use caution, because bleach can damage skin, clothing and flooring, as well as any aluminum parts left in contact. This will, hopefully, remove some of the really nasty microorganisms left by rotting poultry.
Then scrub multiple times with a thick slurry of baking soda ($\ce{NaHCO3}$) in water. The $\ce{NaHCO3}$ acts as a mild abrasive and it adsorbs or reacts with many organic odorants.
Next, leave the refrigerator off and open with a fan blowing through for as long as practicable... preferably outside.
When done, leave a thin layer of fresh $\ce{NaHCO3}$ on shelves and/or put a bit of it in a plastic bowl to help get rid of the odor as the surfaces outgas.
As the fridge chills down, the odor will decrease, but I'd store foods in closed containers to avoid absorbing any remaining odors.

I recommend you to put a large amount of activated charcoal into the fridge and left it several days. If you've still sensed the smell, replace the old activated charcoal with the new one. I hope it help you get rid of the smell.

That was the next thing I planned, I'll report as soon as I get my hand on this !


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Ordering compounds by boiling points
Here is a MC question from my chemistry professor:

What is the correct order of boiling points for molecules: $\ce{CO2}$, $\ce{LiOH}$, $\ce{CH3OH}$, $\ce{CH2O}$?

I am mainly confused about ordering compounds by boiling points by using dipole-dipole attraction/H-bonding (I did not consider using London dispersion forces because there were Os and Hs involved - do not know if I did the right thing).
How do I find out the dipole moment for polyatomic ions, especially complicated ones like the $\ce{CH3OH}$?

These are like simplest organic molecules...

This is more or less duplicate of https://chemistry.stackexchange.com/questions/15091/arrange-these-compounds-co2-ch3oh-rbf-ch3br-in-order-of-increasing-boiling

$\ce{CH3OH}$ is methanol. It is a molecular organic compound and not a polyatomic ion.

From the homework close reason: ‘should demonstrate understanding of the underlying concepts’. This is the case here as your question demonstrates non-understanding of a fundamental underlying concept.

Generally dipole moments of bonds and dipole moment of a whole molecule are different things. Dipole moment of a bond depends on the value of partial charge on both atoms forming the bond and since it has not only the numerical value but also direction, it can be represented as a vector.
The dipole moment of the whole molecule is basically the sum of all bond dipole moment vectors.
P. S. Methanol is not really an ion rather than a molecule.
P. P. S. The correct order for your professor's question is (from lowest to highest) CO2, CH2O, CH3OH, LiOH. Carbon dioxide is a covalent nonpolar molecule, so there are no dipole-dipole interactions (excluding induced dipoles), so it has the lowest BP. Methanal is polar, but has no hydrogen bonding. Methanol is known for H-bonds as most of the primitive alcohols. LiOH is an ionic compound with ionic lattice and has not dipole-dipole interactions but ion attraction.

It took me about 20 seconds to find the dipole moment of methanol on Wikipedia and about 3 minutes to find the BPs and DMs of all four compounds. It's generally a BAD IDEA to use a gross molecular property over such a variety of structures. (for instance LiOH exist in the gaseous state mostly as (LiOH)2 while it is ionic in the solid state (I'm not sure how it exists in its liquid state, I'd guess as a molten salt)). I'd have zero idea which had stronger H-bonding MeOH or LiOH but I know that ionic solids have high melting points, so I'd assume that LiOH had the highest BP as well. Since we also know that CO2 is a gas at STP, that leaves us with formaldehyde and methanol. MeOH is a classic example of h-bonding, so that leaves us with HC(=O)H and the question of whether C-H...O=C is going to be h-bonding or not. Well, H-C bond is pretty non-polar so it's not likely to be very good at h-bonding. Given that its also much lighter than MeOH, we've got a ranking.


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Best way to dissolve/emulsify beeswax and carnauba wax in water-based solution
What I can use to dissolve beeswax and carnauba wax and keep it in an aqueous solution? I do not want to use ethanol or acetone as they are too harsh.
I am adding about $\pu{5\%}$ wax to a $\pu{16oz}$ aqueous solution that also has some oil in it. I need to stop it from coagulating and be fine enough or in solution to pass through the nozzle of a spray bottle without clogging it ?

Well, you'd need proper emulsifier and right mixing procedure.

Thanks Mithoron, how do I identify the emulsifier and establish the mixing procedure? Do you think vegetable oil might work ?

No, and I have a feeling that you need to learn a lot do it. One google search and https://en.wikipedia.org/wiki/Emulsifying_wax pops up. You should have tried searching before asking.

I stand corrected. No excuse for lazyness! I'm not a chemist or student simply trying to find a solution (excuse the pun) and thought this might be a good forum for that

Well, we can help you to some extent, but if you'd knew more before asking question could be better. This link may be useful and maybe someone working with cosmetics winds-up.. who knows.. It's not like we can do entire part of project for you  though ;)

As a polysorbate maybe SCATTICS Castor oil might work, heating the wax and oil which can then be added to water, thoughts?

Well, I guess you need to wait for an expert in this stuff...

Try Oelic acid and TEA with Waxes.
1. Take 3% Oelic acid
2. Add 3% TEA in Oelic acid and stir it with continuous heating up to 50C.
3. Take 100ml Water in Above solution.
4. Take 20grm carnuaba wax and dissolve it in above solution with continuous mixing and heating up to 60C.
5. Again add 40 to 50ml water in above solution.

You need an emulsifier such as 1% Tween 80 and heat the wax with water to 65 °C with stirring, then let it cool down.
It will become a fat blur that dissolves in mixtures by agitating the mix.

Wondering what your use case was for. I'm doing a natural-ish waterless spray to clean your car with a quick spray solution. Beeswax and carnauba wax are both hydrophobic, meaning they are not soluble in water. To dissolve these waxes, you will need a solvent that is capable of dissolving them.
One option is to use a solvent like mineral spirits or turpentine. These solvents are commonly used to dissolve waxes and can be found at hardware stores or online.
Another option is to use a natural solvent such as citrus solvent or pine oil. These solvents are derived from natural sources and can be effective at dissolving waxes. However, they may have a strong smell and may be more expensive than synthetic solvents.
To make a waterless car wash spray with dissolved beeswax and carnauba wax, you will need to mix the dissolved waxes with other ingredients such as surfactants, lubricants, and water. It's important to note that the final product may not be completely aqueous due to the presence of the dissolved wax.
Carlos
caro1008@hotmail.com

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


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Silicone water repellent help
Is there a chemical or solution that can penetrate silicone water repellents in fabric? Need to break down or penetrate the silicone water repellent in selected areas.

Break down and penetrate are different - is either one OK? It might be helpful if you explain more what you want to happen. Are you trying to remove the silicone, or just get something to diffuse through it?

"Silicone water repellent" is barely helpful information. They are a category of materials which are used in an enormous variety of industrial, commercial and retail applications. Some products react and bind to the surface they are coating, others are simply dried onto the surface. Some cross-link, most don't. A typical consumer fabric silicone water repellent will be "penetrated" by solvents such as acetone, methy ethyl ketone (2-butanone) or toluene; mineral spirits might work. There's no way for us to determine whether these solvents can also damage the fabric or clothing item they are poured onto. Another possibility is to use a fairly harsh (alkaline) detergent, but again I can't say whether the fabric will be damaged. Many of the retail products gradually wear away, diminishing repellency slightly after each wash/dry cycle.


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Estimating standard Gibbs free energy of a molecule
In order to estimate the standard Gibbs free energy of a molecule at room temperature and normal pressure, we use Maxwell Boltzmann average over  rotational conformers.
Question
Should we consider cis/trans isomers besides rotational conformers? Is there any criterion for it?

Why, it's simple: those species which are there can contribute. Those which aren't, can't. Suppose you have a cis isomer; does it convert to trans at room temperature? No.

@IvanNeretin After a long time the two isomers trans/cis should equilibrate. So why we should no consider them?

@adosar Well, yeah, they will, but not before hell freezes over, so for all practical purposes we may safely assume they won't.

You will need to calculate the partition functions for each type of motion and then calculate the thermodynamic quantities from these.
As to internal rotation, if the energy for this is far greater than $k_BT$, such as with a C=C bond, then treat it as a torsion and use this frequency. If the barrier to rotation is small compared to $k_BT$ then treat it as a free rotor but reduce the partition function by n if there are n barriers. In the intermediate case you will have to find the barrier (say by spectroscopy) and fit the data to a model of the potential (usually of a cosine form)and then numerically solving the Schrodinger eqn. to find energy levels.
If your temperature is so high that cis-trans reaction occurs then you should probably have to treat this as a mixture.


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Graham's law of diffusion

The rates of diffusion of two gases $\ce{A}$ and $\ce{B}$ are in the ratio $1:4$. If the ratio of their masses present in the mixture is $2:3$, the ratio of their mole fraction is?

My tries:
$\frac{r_1}{r_2}=\frac{1}{4}=\sqrt{\frac{M_2}{M_1}}\to\frac{1}{16}=\frac{M_2}{M_1}$, $M_i$ is molar mass of $i$.
Also given $\frac{w_1}{w_2}=\frac{2}{3}$, multiplying this by above one gives $\frac{M_2\cdot w_1}{M_1\cdot w_2}=\frac{2}{3\cdot 16}\to\frac{n_1}{n_2}=\frac{1}{24}$, where $n_i$ represents number of moles of $i$.
But this is not a correct answer please help.

What is the correct answer ?

$$\frac{r_A}{r_B} =\frac{1}{4}\implies\sqrt{\frac{M_B}{M_A}}=\frac{1}{4}$$
$$\therefore \frac{M_B}{M_A} = \frac{1}{16}$$
On dividing the given mass by molar mass, we get the respective number of moles.
Let the number of moles be denoted by $n$
$$\therefore {n_A} = \frac{m_{A}}{M_{A}}$$
$$\therefore {n_B} = \frac{m_{B}}{M_{B}}$$
$$\therefore \frac{n_{A}}{n_B} = \frac{m_{A}\cdot M_{B}}{M_{A}\cdot m_{B}} = \frac{m_{A}}{m_{B}} \cdot \frac{M_{B}}{M_{A}} = \frac{1}{24}$$
I do not see what's wrong.
Do you want the mole fractions of A and B?


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Calculate activation energy of intermolecular displacement on a nano particle
I'm simulating a nano particle and wish to calculate the activation energy of the reaction where an atom is displaced from one point on the lattice to a neighboring void( might be at higher energy than initial).
Calculating this activation energy using energy difference will require me to calculate the complete trajectory to determine the point with highest energy, which is computationally very expensive.
Is there any other tested method to find this activation energy without having to perform so many computations each time.
Note: the new site is a neighbor of the old site and I expect( not assume) the atom to roll over two of its adjacent atoms to reach this new site.
Also all the atoms are identical.


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Maximum in Maxwell–Boltzmann distribution
I need to find the velocity at which the fraction is maximum in Maxwell–Boltzmann distribution
$$P(v) = \left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}4\pi v^2 \exp{\left(-\frac{mv^2}{2kT}\right)}$$
I know that occurs when $\mathrm dP(v)/\mathrm dv = 0$, but can anyone show the steps which lead to the answer?

Why, the steps are quite simple. (1) Take the derivative. (2) See where it turns to 0. Can you do that?

To make things easy for yourself, substitute $v^2=\frac{2kT}{m}x$ and then take the derivative of P(x) with respect to x.

In my opinion, this is fundamentally not chemistry, unless you plan to physically interpret the result. Are maths questions on topic, even if they're in a chemical context?

I don't think it's on topic, but it's no skin off my back to write a short answer. You need to use the chain rule $(fg)' = fg' + gf'$:
$$\begin{align}
\frac{\mathrm dP}{\mathrm dv} &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\frac{\mathrm d}{\mathrm dv}(\mathrm e^{-mv^2/2kT}) + \mathrm e^{-mv^2/2kT}\frac{\mathrm d}{\mathrm dv}(v^2)\right] \\
&= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\left(\frac{-2mv}{2kT}\mathrm e^{-mv^2/2kT}\right) + \mathrm e^{-mv^2/2kT}(2v)\right] \\
&= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi\mathrm e^{-mv^2/2kT}\left(\frac{-mv^3}{kT} + 2v\right) = 0
\end{align}$$
Since an exponential cannot be zero, the polynomial factor must be equal to zero:
$$\begin{align}
\frac{-mv^3}{kT} + 2v &= 0 \\
v\left(2 - \frac{mv^2}{kT}\right) &= 0 \\
v &= 0, \pm \sqrt{\frac{2kT}{m}}
\end{align}$$
You're not interested in the $v = 0$ stationary point, and the negative square root is unphysical (we're talking about a distribution of speeds, which must be non-negative). So the maximum occurs at $v = \sqrt{2kT/m}$.
From a chemist's point of view this is the physically important quantity, as it represents the most probable speed of a gas molecule. You could substitute it back into $P(v)$ and find out the value of $P$ for this value of $v$, but I don't know what circumstances this would be of interest in.


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Post-synthetic Anisotropic Wet-Chemical Etching of Colloidal Sodalite ZIF Crystals
It's well known that Metal-organic Frameworks can be made by attaching inorganic metal ions to organic linkers via strong chemical bonds, ZIF-67 is one of this big family which could be prepared by using Cobalt source with some kind of organic linkers.
my question is how could we control the structure of ZIF-67 by using wet chemical etching in order to break the links and in particular get sheet like structure?


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Can the eigenvalue for a quantum mechanical operator be zero?
Can the eigenvalue for a quantum mechanical operator be zero?

yes it could be, here you could find a satisfying answer link

The question is way broader than you seem to think; you might want to be more specific. As it stands now, the answer is "yes", but it is a really useless "yes", about the same level as if you were asking "can a difference of two numbers be zero". Sure it can; so what?

$S_z$ (spin along the z-axis) is a QM operator, and it frequently is zero - such as in closed-shell molecules.
Also note that the zero energy point of quantum mechanics can be defined in more than one way: some quantum chemistry programs use "all electrons and nuclei at infinite separation", whereas some solid state programs used "neutral atoms at infinite separation". Nevertheless, they evaluate the total energy.


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Mechanism for interconversion of spin isomers of hydrogen
What is the mechanism by which the ortho- and para- spin isomers of hydrogen interconvert?
If such a mechanism exists, does this mean that ortho-hydrogen increases in concentration on increasing temperature?

Related: Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance?, Stability of ortho- and para-hydrogen

The first question is:

What is the mechanism by which spin isomers of hydrogen switch between the ortho and para forms?

There is some explanation in ChemPhysChem 2006, 7 (3), 551–554 (non-paywall version here):

One can define three situations. In the first, a magnetic conversion occurs without bond elongation or breaking. For example, in solid dihydrogen dipolar magnetic nuclear interactions are responsible for an extremely slow spin conversion.[4] In the presence of unpaired electrons the conversion is accelerated by the magnetic hyperfine interaction.[5] The second situation also involves a magnetic conversion mechanism, but is assisted by an intermediate H–H bond elongation, for example, by binding to a transition metal center. Finally, in the third situation, a H–H bond splitting and re-formation with other hydrogen atoms occurs, which corresponds to a chemical spin conversion.

The second question is:

If such a mechanism exists, why does ortho-hydrogen not increase in concentration on increasing temperature?

The equilibrium fraction of ortho-hydrogen does increase from 0% at absolute zero to 75% at high temperature.
para-Hydrogen is the lowest energy, most stable form, since it is capable of accessing the $J=0$ rotational level (whereas ortho-hydrogen cannot). There are three microstates corresponding to ortho-hydrogen:
$$|\!\uparrow\uparrow\rangle, |\!\downarrow\downarrow\rangle, |\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle,$$
but only one microstate for para-hydrogen:
$$|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle.$$
So, according to the Boltzmann distribution, the equilibrium fraction of ortho-hydrogen goes from 0% at 0 K to 75% at infinite temperature (in practice, a temperature at which the effects of rotational quantisation are not seen).


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How can the hybridisation schemes of transition metal complexes be determined?
Lets say I have to find hybridisation of $\ce{[Ni(CO)_4]}$ and $\ce{[Ni(CN)_4]^{2-}}$.
The metal atom/ion in these compounds are $\rm Ni$ and $\rm Ni^{2+}$ respectively.
So the outer shell configuration for $\rm Ni$ in $\ce{[Ni(CO)_4]}$ would be $$\rm 4s^2  \ \ 3d_{xy}^2\ 3d_{yz}^1\ 3d_{zx}^1\ 3d_{z^2}^2\ 3d_{x^2 - y^2}^2 \ \ 4p^0 \ \ 4d^0$$
The electrons from s oribital will jump to d orbital and so I expect $\rm CO$ to donate electron pairs in $\rm 4p$ and $\rm 4s$ orbitals and form $\rm sp^3$ hybridisation.
The outer shell configuration for $\rm Ni^{2+}$ in $\ce{[Ni(CN)_4]}^{2-}$ would be $$\rm 4s^2  \ \ 3d_{xy}^0\ 3d_{yz}^1\ 3d_{zx}^1\ 3d_{z^2}^2\ 3d_{x^2 - y^2}^2 \ \ 4p^0 \ \ 4d^0$$
Here also I expect the electrons to jump from s to d and hybridsation to be $\rm sp^3$, but the answer given here is $\rm dsp^2$. Why ?
I am heavily confused with this hybridsation stuff, is there a systematic and logical way of knowing hybridsation of coordination compounds ?

I tried to search this topic on internet and I came across  $H = \dfrac12(V + M - I)$ where $V$ is number of valence electron, $M$ is number of single bonds and $I$ is charge on the atom. Here if $H= 2$ then it is $\rm sp$ , $H = 3$ then $\rm sp^2$ and so on.
For $\ce{[Ni(CO)_4]}$, here $V = 10$, $M = 4$ and $I = 0$, so $H = 7$ hence $\rm sp^3d^3$.
Also for $\ce{[Ni(CN)_4]}^{2-}$, here $V = 8$, $M = 4$ and $I = -2$, so $H = 7$ hence $\rm sp^3d^3$.
As it can be seen, this formula is complete failure.
Why this formula does not work for coordination compounds whereas it works for compounds like $\rm PCl_5,PCl_3$ and $\rm SF_6$ ?

In this answer the author makes the claim that $\ce{[NiCl_4]^{2-}}$ has hybridsation $\rm sp^3$ and $\ce{[Ni(H2O)6]^2+}$ has hybridsation $\rm sp^3d^2$. How did he know that ? This is my question not why hybridsation is not a valid theory to study coordination compounds.

I know there is a debate about whether this is a duplicate of Why is it wrong to use the concept of “hybridization” while studying complexes?. This question has been closed, reopened, and was about to be closed again. Let me state it very clearly: this is not a duplicate. The linked question asks why an incorrect theory is incorrect; this question asks about the incorrect theory itself. While the linked question is obviously relevant, it is not a duplicate. I say this not only as a moderator, but also as the answerer of both questions.

Firstly, note that hybridisation theory as applied to transition metal complexes is an incorrect, flawed theory. It is an attempt to rationalise experimental observation (e.g. geometries), but in the process it invokes models of bonding which are highly unrealistic and demonstrably untrue. See: Why is it wrong to use the concept of hybridization for transition metal complexes?
Now, with that disclaimer out of the way, let's examine how that flawed theory works. In traditional hybridisation theory, tetrahedral and square planar 4-coordinate complexes are considered to be $\mathrm{sp^3}$ and $\mathrm{dsp^2}$ respectively. Since $\ce{[Ni(CO)4]}$ is tetrahedral and $\ce{[Ni(CN)4]^2-}$ is square planar, one can assign the hybridisations accordingly. If you have a textbook that still mentions this outdated theory, it should have a table telling you what hybridisation each geometry corresponds to. Here is an example from Housecroft's Inorganic Chemistry 4th ed. (p 667). To their credit, there is a subsequent section debunking this theory.

These are generally based on post hoc justifications. Essentially, somebody looked at the geometry and then tried to piece together which orbitals could be combined to form such a geometry. So, I do not think there is a good way to predict them a priori. This formula you have dug up does not work very well, even for main group elements. For transition metals the factors deciding geometries are much more complex, and simply cannot be captured in one line of mathematics.

The real problem with the approach above is that you need to know the geometry before you can assign a hybridisation. (Don't even try to do it the other way round - it simply does not work because hybridisation is an incomplete theory. It is like trying to predict wave-particle duality using classical mechanics.)
If you cannot predict the geometry of the complex, then you cannot predict the hybridisation. You could argue that cyanide is a strong field ligand and in conjunction with a $\mathrm{d^8}$ metal, a square planar geometry is preferred. On the other hand, in $\ce{[Ni(CO)4]}$, the central atom is $\mathrm{d^{10}}$ and hence even with a strong-field ligand like carbon monoxide, a tetrahedral geometry is favoured, simply based on steric grounds.
However, this does nothing but introduce more questions. The term "strong-field" has no meaning outside of molecular orbital theory. Hybridisation theory has no inkling of what strong- or weak-field ligands are, and consequently cannot explain why $\mathrm{d^8}$ Ni(II) only prefers to go square planar with strong-field ligands. On top of that, it is not at all obvious from hybridisation theory how the electronic configuration of the central metal should favour a particular geometry. It's simply not possible for your formula to explain any of this.

So, to answer your questions directly:

Is there a systematic and logical way of knowing hybridisation of coordination compounds?

Yes. You use a more advanced theory (MOT) to predict the geometry, and then you use the table above to find the hybridisation. That's a step backwards though, you already used MOT and there's no reason why you should go back to hybridisation.

Why this formula does not work for coordination compounds?

Because hybridisation theory is an incomplete theory and cannot capture the subtle factors influencing geometry of coordination compounds.

Main group compounds
I guess I will elaborate a bit more on main group compounds. The formula you quoted seems to work for main group compounds, but let's get two things out of the way.

It gives the correct hybridisations as espoused by old theory. $\ce{SF6}$ is not $\mathrm{sp^3d^2}$ hybridised because d-orbital involvement in bonding is negligible. So, this formula does not actually work.
It does not predict the hybridisation without predicting the geometry.

Point 1 is important, but let's not go into that now. Point 2 may seem confusing to you. Clearly it is predicting the hybridisation, simply from the number of valence electrons, etc. etc.?
The truth is that this formula only "works" because these quantities (number of valence electrons, number of bond pairs, ...) allow one to predict the geometry, using VSEPR theory. For example, the structure of $\ce{CH4}$ is tetrahedral, and VSEPR theory is sufficient to predict that. Once the geometry has been established as tetrahedral, the hybridisation can then be correctly assigned as $\mathrm{sp^3}$.
So, this formula is skipping a step in between. In fact, the correct sequence is to predict the geometry using VSEPR theory, and then obtain the hybridisation. Now, with transition metal complexes, VSEPR does not work. Thus, it is no surprise that that formula also does not work.

If the geometry is trigonal prismatic, then how can we decide between $\mathrm{sd^5}$ and $\mathrm{sp^3d^2}$ ?

To be honest I don't actually know. I suspect this is (as I wrote) not entirely possible to deduce, because these are post hoc justifications - meaning that instead of developing a hypothesis and testing it by experiment, people observed the geometry and tried to come up with a hypothesis that explains it - and in this case it just so happens that there is more than one scheme that explains it.

Is sd5 hybridisation the one involving the s and d orbitals of the same principal energy level? I couldn't find a clear explanation online.

Also, going with the above comment, you can't really use any theoretical arguments (why even more electrons from the inner-d orbitals don't excite so that the coordinating orbitals are lower in energy, or why s electrons don't themselves excite up to upper orbitals so that that s orbital can be used for the same reasons) and try to figure out hybridisation, since it's a hand-wavy property made up by us that's determined solely from geometry. Have I got it right?


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Magnetic anisotropy of triple bonds
Is the rule for shielding/deshielding areas in NMR bond dependent? The acetylene C triple bond C  has deshielding zone perpendicular to the triple bond. Will the same deshielding zone hold true for C triple bond O in metal coordinated carbonyl ligands.

In coordinated CO it's rather less then triple bond.

Let me rephrase it. In a metal carbonyl complex with a ligand that has proton located "above" CO bond, will I observe similar downfield shift for the proton as for the analogue situation with acetylene due to deshielding. Effect as illustrated for  the Anisotropy of Acetylenes  https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-02-delta.htm


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Brillouin zones of a compound after structural relaxation
My background in solid-state chemistry isn't much, so I apologize in advance if there arises the need to raise a basic fact while answering my question. I've been learning mostly from the web, with my main text resource  'Band Theory of Solids: An Introduction from the point of view of symmetry.' by Simon L. Altmann.
From what I understand, Brillouin zones are areas (or volumes in 3-D) in reciprocal space where you use to describe the energy levels with respect to the k-vectors. Thus typically in band structure plots the k-vectors range from the origin Γ along some straight path which contains points like Δ, K etc.
So far the common examples I've seen for a Brillouin zone are for 'pure' elements (Si, Cu etc.) where you have the atoms in a crystal formation, which you obtain direct lattice pattern, and then the reciprocal space, and finally the Brillouin zone polyhedron.
What I'm really curious about is, if I have a material that I know the Brillouin zone (like a graphite sheet), and this material reacts with another molecule (like a dopant), how would the Brillouin zone change? A scenario would be performing structural relaxation of a carbon nanotube with some dopants, and the nanotube deforms slightly (with the dopants moving to an equilibrium configuration). In this case even if I know the Brillouin zone of a pure (undoped) carbon nanotube, with the deformation, how should I determine which are the k-vector points then?

The answer is neither general nor easy. First, doped structures are non-periodic, therefore you need an assumption how you treat it. Second, the dopant will introduce new energy levels, and interact with the existing ones. How will they look like? You often have to make calculations to have a clue, no trivial answer for that.

If the doped structures are periodic (say I arrange them in a periodic configuration and the relaxed model turns out to be periodic as well), would that be still possible to do that? I was able to obtain relaxation calculations of the doped structures, but to have the software (I use SIESTA) obtain the band structure values I would need to input the band lines or band points, which is what I'm stuck at, since I don't exactly know if the Brillouin zone is the same (as the carbon nanotube), and if the X, K points still match..


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Estimation of pressure and kinetic energy density of stellar interior using kinetic theory of gases
In Physical Chemistry we are working a problem that is confusing. We started by calculating the pressure half way to the centre of the sun assuming that the interior consisted of ionized hydrogen atoms. We determined this to be $\pu{3.59e6 kPa}$. However, the third part of the question asks:

What is the kinetic-energy density half way to the centre of the Sun? Compare with the (translational) kinetic energy density of Earth's atmosphere.

Can someone expand on what kinetic energy density is, if it is different from kinetic energy, and what the formula is. I might just be overthinking the problem.

I would guess that this is asking you to use the Maxwell-Boltzmann distribution, but with the kinetic energy substitution. Namely,  make the substitution $v=\sqrt{\frac{2E}{m}}$ and then take the derivative to substitute $dE$ for $dv$.
So, that gives you the energy distribution and then the energy density is just this divided by the pressure you determined (maybe multiplied by a constant). Notice that [N]/[m^2] is dimensionally the same as [J]/[m^3]. Probably the simpler approach is to use a formula for average velocity then get KE that way.

I recognize that this is a crude approximation to use all these ideal gas equations, but without more information I would guess this is what you're supposed to do.

I'm voting to close this question as off-topic because it is dealing with physics, not chemistry.

This is a question from Atkins' Physical Chemistry. This question deals equally with physics and chemistry and I'm against closing it. The textbook defines kinetic energy density as $\rho_\text{k}=E_\text{k}/V$,  and in combination of kinetic energies of the particles $E_\text{k}=1/2NMc^2$ with the mass density  $\rho_\text{k}=3/2p$ ($p$ -- pressure in the interior of the Sun). The textbook also gives $\rho_\text{k}(1/2r_\odot)=\pu{0.11 GJ cm-3}$, $\rho_\text{k}(\text{atm})=\pu{0.15 J cm-3}$.

OP also missed as essential part of the question (Q1.39c from Elements of physical chemistry): "[...] Compare your result with the (translational) kinetic energy density of the Earth’s atmosphere on a warm day (25°C): $\pu{1.5e5 J m−3}$ (corresponding to $\pu{0.15 J cm−3}$)." Probably you want to add this into the question because it is rather important.

I would like to first point out that the pressure you've determined is too low, most likely because you used only molar mass of protons $\ce{H+}$ instead of average mass for star matter in order to fulfill the condition of plasma quasineutrality:
$$\bar{M} = \frac{M(\ce{H+}) + M(\ce{e-})}{2} = \frac{\pu{(1 + 0) g mol-1}}{2} = \pu{0.5 g mol-1} = \pu{5e-4 kg mol-1} \tag{1.1}$$
Referring to the question from Atkins' Elements of Physical Chemistry [1, p. 40], where this question has been taken from, we obtain necessary data to complete the solution:

In the Sun [...] the density is 150 times that of liquid water at its centre and comparable to that of water about half-way to its surface

and

(a) Calculate the pressure half-way to the centre of the Sun, assuming that the interior consists of ionized hydrogen atoms, the temperature is $\pu{3.6 MK}$, and the mass density is $\pu{1.20 g cm−3}$ (slightly higher than the density of water)

the pressure of star plasma (that can be treated as an ideal gas) can be found as following:
$$p = \frac{\rho RT}{\bar{M}} = \frac{\pu{1.2e3 kg m-3} \cdot \pu{8.314 J K-1 mol-1} \cdot \pu{3.6e6 K}}{\pu{5e-4 kg mol-1}} = \pu{7.18e13 Pa} \label{eqn:1.2}\tag{1.2}$$

Now for the second part, kinetic energy density is defined as kinetic energy per volume:
$$\rho_\text{k} = \frac{E_\text{k}}{V} \label{eqn:2.1}\tag{2.1}$$
Kinetic energy for a set of entities (ion, atom, molecule) is
$$E_\text{k} = \frac{Nm\bar{v^2}}{2} \label{eqn:2.2}\tag{2.2}$$
where $N$ is the number of particles, $m$ is the particle's mass, $\bar{v}$ is an average (root mean square) particle velocity. From the kinetic model of an ideal gas, the pressure obtained from \eqref{eqn:1.2} is also
$$p = \frac{Nm\bar{v^2}}{3V} \label{eqn:2.3}\tag{2.3}$$
Equation \eqref{eqn:2.3} implies $Nm\bar{v^2} = 3pV$, so kinetic energy from \eqref{eqn:2.2} can be rewritten as $E_\text{k} = \displaystyle\frac{3pV}{2}$. Finally, kinetic energy density, taking \eqref{eqn:2.1} into account, is
$$\rho_\text{k} = \frac{3p}{2} = \frac{3 \cdot \pu{7.18e13 Pa}}{2} = \pu{1.08e14 J m-3} \quad (\pu{1 J} = \pu{1 Pa}\cdot \pu{1 m3})$$
Comparison between kinetic energy densities $\rho_\text{k}(1/2r_\odot)$ (halfway to the Sun's center) and $\rho_\text{k}(\text{atm})$ (Earth's atmosphere) is really a matter of comparing the pressures:
$$\frac{\rho_\text{k}(1/2r_\odot)}{\rho_\text{k}(\text{atm})} = \frac{p(1/2r_\odot)}{p(\text{atm})} = \frac{\pu{7.18e13 Pa}}{\pu{101325 Pa}} = 7.09\cdot10^8$$
References

Atkins, P. W.; De Paula, J. Elements of physical chemistry, 5th ed.; Oxford University Press: Oxford; New York, 2009. ISBN 978-1-4292-1813-9.


text:
Carbocation Rearrangement in SNi
In Peter Sykes' A Guidebook to Mechanism in Organic Chemistry  regarding the retention of configuration of reaction of alcohols with $\ce{SOCl2}$:

The rate at which the alkyl chlorosulphite intermediate breaks down to $\ce{RCl}$ is found to increase with increasing polarity of the solvent and also with the increasing stability of the carbocation $\ce{R+}$: an ion pair $\ce{R+SOClO-}$ is almost certainly involved.

But it is nowhere mentioned whether a carbocation rearrangement is possible or not.
In "NCERT Chemsitry Class 12" in the section of Preparation of Alkyl halides from Alcohols:

The hydroxyl group of an alcohol is replace by halogen on reaction with concentrated halogen acids, phosphorous halides or thionyl chloride. Thionyl chloride is preferred because the other two products are escapable gases.

If that method is preferred, then I assume rearrangement does not occur.
So, why does not a carbocation rearrangement (like Wagner–Meerwein rearrangement) occur in the $\mathrm{S_Ni}$ pathway eventhough the intermediate is involved? Is it because the intermediate breaks down too quickly before any rearrangement is possible?

Have a look at http://www.masterorganicchemistry.com/2014/02/10/socl2-and-the-sni-mechanism/

@pritt thanks but nothing is mentioned regarding this in that post but someone has asked this question in the comments and there was no answer to that.

It was just a link I found relevant, since it explain the mechanism of thionyl chloride substitution. Oh, and asking questions in the comments isn't really what comments are for. If you find that comment again, be sure to flag it as "No longer needed".

https://www.youtube.com/watch?v=gqTQX4rQyfs
Here is a link to the answer to your question. Please jump to 19:22 minute of the video. If someone could also fact-check this it would be helpful.

The Mechanism of the reaction
The reaction of an alcohol with thionyl chloride under base-free conditions is one of the most significant examples of the SNi mechanism (DN+ANDe to give it it’s correct IUPAC designation, which is perhaps more instructive in this case).
There are two steps to the SNi mechanism, as shown below (figure taken from Bruckner, Organic Mechanisms [1]):

Step 1: The alcohol reacts with thionyl chloride to afford an alkyl chlorosulfite, which can often be isolated.[2] At this stage, the oxymethine stereocentre hasn’t been touched– it’s the second step that will determine the stereochemistry.
Step 2: The chlorine is delivered intra-molecularly, with extrusion of sulfur dioxide gas. Since the sulfur is being delivered from the same face that the hydroxyl was on, the reaction occurs with retention overall.  This can be seen in the dotted lines in the figure above.
Two things are worth noting at this stage:

If pyridine is added to the reaction mixture, we see overall inversion via an SN2 mechanism.[2] This occurs because the pyridine interacts with the intermediate alkyl chlorosulfite formed, liberating a free chloride (nucleophilic) and hence allowing the substitution to take place  inter-molecularly. This can be seen in the solid lines in the figure above.

The second step in which I described as happening intra-molecularly doesn’t quite happen intra-molecularly. In reality, the C-O bond begins to fragment in an SN1 fashion, before the chloride attacks. The reason why we still see retention (rather than racemisation as is common with  SN1) is due to a phenomenon called contact ion pairs.[#]

Carbocation rearrangement
The simple answer to your question is that at no stage of the reaction is it thought that there is any free carbocation concentration, making rearrangement less likely on kinetic grounds (along with elimination, which could equally be argued by your logic).
The ion pair theory is such that the carbocation remains closely associated with the leaving group, allowing the nucleophilic attack to occur rapidly without the need for an additional collision to take place.
[1]: Bruckner, R. Organic Mechanisms- Reactions, Stereochemistry and Synthesis; Springer: Berlin, 2007
[2]: J. Am. Chem. Soc. 1952, 74, 308
[#]: You can read about this in any advanced organic chemistry text such as Carey or March, and is a bit too long of an explanation for me to give satisfactorily here


text:
Chemical reaction between dichloromethane and solid carbon dioxide
I studied heat transfer liquids with low melting point (coolants), and how fast their temperature decreases when they are contacting with solid carbon dioxide. Nearly all liquids (brines, glycols etc.) that I studied have similar behavior, except dichloromethane.
When I put solid carbon dioxide grains into dichloromethane, the temperature of the liquid dropped nearly immediately! At first, the temperature was $\pu{20^\circ C}$, and after just $\pu{5 s}$ the temperature of the liquid was $\pu{-70^\circ C}$! Do you know why is it possible? When I freezed $\ce{KCl}$-brine, the temperature had been decreasing from $\pu{20^\circ C}$ to $\pu{-40^\circ C}$ for more than $\pu{5 min}$!
This property of dichloromethane may be very useful. I think it is due to its low viscosity.
Can you help me and explain this phenomenon?

In brine cooling happens only on the surface of the solid carbon dioxide and causes the water to freeze there, forming an insulating layer (the same happens with ethylene glycol). In dichloromethane, which freezes below $-97$ degrees, no freezing occurs. Moreover, carbon dioxide is well soluble in dichloromethane, which makes the heat transfer more efficient. The latter property may cause you problems if you want to use your coolant in closed installations, because after warming the solution may release the excess of carbon dioxide.

Heat capacity might also be a factor.


text:
Questions on concentration
Currently I am attempting two questions on concentration but I am not sure if I am correct. Can someone help me with these questions? These are the questions:

1) Calculate the volume of pure water added to dilute $\pu{0.2 M}$ $\ce{HCl}$ solution in $\pu{250 cm3}$ to $\pu{0.12 M}$ $\ce{HCl}$ solution.

$\pu{0.2 M}$ solution contains $\pu{0.05 mol}$. $\pu{0.12 M}$ solution contains $\pu{0.03 mol}$. The difference between the two solutions is $\pu{0.02 mol}$ (Which is how much water was added), so
$$V(\ce{H2O}) = 0.02 \times (2 \times 1 + 1 \times 16) = \pu{0.36 L}$$

2) Calculate the mass of $\ce{Na}$ that will form $\pu{0.15 M}$ solution of $\ce{NaOH}$ by reacting with $\pu{400 cm3}$ of water.

I know that this method is totally wrong but I'll show you:
$$\pu{0.15 M} = \frac{n}{0.4} \implies n = 0.06$$
$$m(\ce{Na}) = 0.06 \times 23 = \pu{1.38 g}$$
where $n$ just refers to the concentration ($C = n/V$).
Edit:
Thanks to @George Tian I have made another attempt for the second question.
I have calculated the number of moles of NaOH for a 0.15M solution:
c = n/v
n = 0.4 x 0.15
  = 0.06

And I calculated how much Na is needed for this reaction:
The balanced equation is: 2 Na + 2 H20 -> 2 NaOH + H2
So, 0.06 mol of Na is needed (Due to the molar ratio)

Then I converted moles to grams
Mass = 0.06 x 23
     = 1.38 g

To clarify, I've reworded your questions. Please comment if I changed the meaning:

Calculate the volume of pure water that needs to be added to dilute a  0.2M HCl solution with a volume of 250cm3 to a 0.12M HCl solution

The first step would be to work out how many moles of HCl is in the original solution. As you have stated
c = n / v
c = 0.2M
v = 250cm3 or 0.250L.
Therefore n = 0.05m
Then work out how much water is needed to make a 0.12M solution.
c = 0.12M
n = 0.05m
Therefore v = 0.42L
To work out how much water needs to be added, subtract the original 0.25L from 0.42L, which equals 0.17L

Calculate the mass of Na needed to react with 400cm3 of water to form a 0.15M NaOH solution.

As you haven't progressed much on this question, I'll give a few hints.
Write a balanced equation
Work out how many moles of NaOH is needed to make 0.15M solution
Work out how many moles of Na is needed to form that many moles of NaOH
Convert from moles to mass.
Hope this helps.

Thank you. Could you please check my edited question for the second question?

@kimchiboy03 Well done, that is correct. However, I'd use 4 significant figures for the molar mass of Na (22.99), and ensure the amount of significant figures in the final answer is correct.

There is no way to take a 0.2M solution of anything and get a 0.12M solution. Although it is trivial to take a 0.20M or even a 0.200M solution and dilute it to 0.12M. Have you ever heard of significant digits?
You state that a 0.12M solution contains 0.03 moles...so if I get a million tons of 0.12M solution or a millionth of a microgram of 0.12M solution they both contain 0.03 moles of solute?? I. don't think. so. In a dilution the number of moles of the solute do not change (of course, I'm talking about a one pot dilution, in serial dilutions, you only use part of the solutions, so the amount of solute does change). If you start with .05 moles of solute, then you will end with 0.05 moles of solute (ignoring losses).
So, you calculate how much HCl you have (moles): 0.20 x 0.250 = 0.050 moles
Then you want to convert that to a concentration of 0.12 moles/L: 0.050/x = 0.12 or x = 0.05/0.12 = 416.67 mL (during calculations, you should bring thru more digits that you need in the result). You started with 250 which means you need 416.67-250.00 = 166.67 ml or 167 ml after adjusting for significance.
When solving these kinds of problems, I generally try to convert to things which are conserved. Atoms are conserved, density and concentration and volume are not, so converting from concentration to moles means you're less likely to make an obvious error. At least, that's my experience.


text:
Can a solid be both crystalline and amorphous?
The first line in the Wikipedia article for "Glass" reads:

Glass is a non-crystalline amorphous solid that is often transparent and has widespread practical, technological, and decorative usage in, for example, window panes, tableware, and optoelectronics.
 Emphasis, mine.

From what I've been taught at school, if a solid is amorphous then it is necessarily non-crystalline and vice versa. (Of course, you could convert a given sample of an amorphous solid to some crystalline form...but it would no longer be amorphous now would it?).
The "non-crystalline amorphous" pleonasm adopted by Wikipedia seems to suggest the existence of (yes, I know this sounds stupid) crystalline amorphous solids.
Can something (I'm using glass as an example) be both "crystalline" and "amorphous?"
If not (as I suspect is the case), then why does Wikipedia use the pleonasm "non-crystalline amorphous" when describing glass? Is it purely for emphasis?

Yes, I've already seen this question here, and no, my question's not a dupe of that.
Sure, it's possible to convert an amorphous solid to a crystalline solid (proper melting and cooling should do it), I'm not questioning that.
What I am questioning, is whether a crystalline amorphous solid can exist. For example, glass has both (non-crystalline) amorphous  and crystalline (non-amorphous) phases.... but does it have crystalline amorphous phases? Yes? No? Why?

Here is an amorphous crystal https://en.m.wikipedia.org/wiki/Crystal (look in the first images)

Possible duplicate of Does crystalline glass exist?

Maybe you can say " amorphous crystal" because it has some periodic arrangements but they are lost in the distance. You can find this on Wikipedia also: amorphous (from the Greek a, without, morphé, shape, form) or non-crystalline solid is a solid that lacks the long-range order that is characteristic of a crystal.

This seems to be a question on English usage more then Chemistry. I prefer the Wikipedia phrase since it emphasizes the difference.

I'm voting to close this question as off-topic because it's about semantics

@Jon Eh? No... "Can something be both crystalline and amorphous?" doesn't sound like an English usage question. People seem to  be missing this point, which is why I paraphrased my question (see under EDIT) ;P

So what do you call a quasi-crystal? It would be a non-crystalline, non-amorphous solid, perhaps. Again, at best you are questioning somebody else's phrasing.

@Mith No, it isn't about semantics. Re-read the question again please (The "terminolgy" tag must be blacklisted then if you're right) O:)

@Jon I want to know: Do amorphous crystalline solids exist? (an example would be nice too)

Well, if you'd knew more about semantics, you wouldn't ask it in the way you did. Putting 2 similar description together, may or may not be redundant and even if is still may be useful. I'm quite against turning the site into "translate this passage for me 'cause I'm not getting the point" as happens on other sites.

@Mithoron But "non-crystalline amorphous" is redundant and useless (another pleonasm) :/

How do you feel about 'non-glassy amorphous solid'? If you think that is redundant and useless you would be incorrect. So, yes, the phrasing is overly precise - not exactly a crime in science.

@paracetamol I edited to add clarity. I think this question could be reopened, but I want to make sure I understand what is being asked.


text:
Copper plating, is sulphuric acid necessary? Alternative to H2SO4?
Many links and tutorials on copper plating use a solution that contains copper sulphate / sulphuric acid and hydrocloric acid.
In Sweden, Sulphuric acid is almost impossible to get a hold off. You need a license to buy it, and it costs 100 USD to get one...
What does the H2SO4 do in the electrolyte? Is there an alternative to the acid?
I want to plate PCB copper clads.

Try plating without an acid, and you'll see why it was important. I bet copper will deposit anyway, but will be all spongy and soft.

Usually you can buy 36% $\ce{H2SO4}$ where you would get lead acid batteries. Isn't it the case in Sweden? As a last resort, buy the lead acid battery and pour the acid out of the cells (be careful though).

Copper acetate can be used, at least for a thin layer, and it can be made with chemicals found at home.
To get a bright copper coat, extra acid (vinegar, 5% acetic acid, in this case) or gelatin can be added.
That said, if the plating is for through-hole continuity, it might be simpler and cheaper to use a solderable conductive ink.


text:
Effect of Zn dust reduction of phenolic -OH group on other groups
We use distillation with zinc dust to remove -OH group from phenol.
$$\ce{Ph-OH + Zn \rightarrow Ph-H + ZnO }$$
What I want to know is that whether this reaction has any effect on other groups attached to the benzene ring, like $\ce{-CH_3,-NH_2,-NO_2,-CN,-CONH2}$ etc. or not?

Nitro and cyano will react

@Waylander Please explain, in detail about the reactions. Answer with examples if possible.

Zinc plus proton source and heating is a standard way of reducing aromatic nitro. I have run this with ammonium chloride as the proton source in methanol

@Waylander, I understand. But can you give the whole reaction? I mean, a NO2 group can be reduced to NHOH or NH2 group, and CN group can be reduced to CH=NH, or CH2NH2 groups.

I think at this point you should do you own research into the reductive properties of zinc.

@Waylander: On your Oct 02, 2017 comment: Yes, it is a one way of receiving aniline form nitrobenzene. I have used AcOH as my proton source. A drawback is can't use in large scale. I got ~95% yoeld with using $\pu{1.0 g} $ of nitrobenzene, but less than 50% when increase the scale to $\pu{5.0 g}$,

ZnO is a photocatalyst.  Any introduced electrons and electron holes are not likely inert here.

To whoever is trying to repeatedly edit my question: please stop. I understand that C6H5 is a synonym for Ph but in this case, I feel that Ph is better because it explicitly mentions a phenyl ring instead of C6H5 which might be a number of other things.

I did some research for this question many years ago. Nisarg's answer prompted me to post an answer.

Nisarg's answer showed how nitrobenzene can be reduced to form different compounds but my answer focuses on some different compounds. Doing some literature surveys gave me results from very old books. All the reaction involves only zinc dust. I tried quoting but the majority of the terms/compound names are obsolete, so I have wrote them as points:

formanilide and its derivatives is reduced benzenenitrile and aniline (and its derivatives) as major products  and other side products such as N-Methyldiphenylamine. Yield of nitrile 10-20% weight of initial formanilide.
N-alkyl-o-toluidine and N-alkyl-p-toluidine yields corresponding nitrile: yield 15-20%
methyl-α- and methyl-β-naphthalamine yield α- and β-naphthonitrile
xylidine and its isomers yields xylonitrile (prone to hydrolysis to yield xylic acid(isomeric carboxylic acid of xylene)) : yield 12%
Phenyl isothiocyanate and thiocarboanilide forms benzonitrile and some aniline.

References:

K.Gasiorowski and V.Merz, Journal of the Chemical Society, 1884

Do these reactions occur in pure zinc dust distillation or do they require NaOH, NH4Cl etc.?

@SRMaiti only zinc dust is used. No other catalyst/reagent used.

Out of the groups you mention $\ce{-NO2}$, $\ce{-CN}$ will most certainly react, $\ce{-CH3}$ and $\ce{-NH2}$ not so much and I am ambiguous about $\ce{-CONH2}$.
Although I said that $\ce{-CN}$ will react, I can't pin point the product which will form in a certain condition. But in case of $\ce{-NO2}$, I certainly can predict the product based on a list I complied a couple of years ago.
The "List":
$$
\begin{array}{|ccc|}
\hline
\ce{Ph-NO2}&\xrightarrow[\ce{NH4Cl}]{\ce{Zn}}&\ce{Ph-NHOH}\\
\hline
\ce{Ph-NO2}&\xrightarrow[\ce{NaOH}]{\ce{Zn}}&\ce{Ph-NH-NH-Ph}\\
\hline
\ce{Ph-NO2}&\xrightarrow[\ce{NaOH}]{\ce{Zn+alcohol}}&\ce{Ph-N=N-Ph}\\
\hline
\end{array}
$$

References:
(1)   Kamm, O. β-Phenylhydroxylamine". Organic Syntheses. ; Collective 1941, 1, 445.
(2)   Bigelow, H. E.; Robinson, D. B. Azobenzene". Organic Syntheses. ; Collective 1955, 3, 103.

Why doesalcohol cause it to become double bond

Nitrobenzene can also get reduced to amines through zinc reduction. Only the reaction conditions and reagents are slightly altered. (Here, Here, Here).

@Buraian Possibly condensation. My guess: The reaction creates nitroso and hydroxylamine products which then condense to azoxy and then ultimately to azo product.

@Nisarg As a side note: Although this doesn't count as reduction but cyanobenzene does react with zinc dust and bromoacetic acid to form a red crystalline compound. This reaction is somewhat similar to Blaise reaction.

The original zn dust reduction of phenol is done by distilling with zinc dust. The reactions you have mentioned use other compounds as reactants/catalysts. I don't think this answers the question, sorry.


text:
Harnessing phase separation to easily make Li film
I want to make a thin, smooth film of pure lithium in a glovebox.
In general, I think that I can use a phase separation approach. I will heat a bath of mineral oil above Li's melting point, then pour liquid Li inside and let it cool slowly. The Li should cool on top and the only curvature should be due to miniscus effect.
I would think that if I used a third, lighter liquid than the Li, I could put a large column of it on top of the Li. However, this requires having a lighter liquid. Li, at melting point, has a density around $\pu{500 kg m-3}$.
Can anyone suggest a suitable lighter liquid?

I can't think of any. Then again, what's the point? The meniscus would still be there.

Top liquid would supply more pressure and reduce meniscus. At least thats my thinking in theory, doesnt mean itll work in practice

I don't think there's any liquid that could be used here, pressurised gas may work.

I have a suspicion it won't work. Molten metals usually have strong surface tension, so a thin film might fragment into rather large drops.

Do you have any better suggestions? I'm just trying to make an Li foil/film with a very, very smooth surface.

You are thinking in the wrong direction. Meniscus is not about pressure at all.
You seem to be imagining external hydrostatic pressure as a huge invisible flat lid that we put on top of your molten lithium and kinda force it to stay flat. This is very much not so. The pressure is the same all around, thanks to the Pascal's law. It does not affect the shape. It presses on the curved sides of the meniscus with the same force as it does on top. The very notion of "flat" is simply not there.
If anything, it is gravity force that reduces the meniscus. In effect, adding a liquid on top will partially negate the gravity and thus make things worse. Keep in mind that such liquid, if it is found (I still can't think of any, and I'm not a novice in chemistry, mind you) can't be much less dense than lithium, which almost brings us to the conditions of Plateau's oil drop experiment, with lithium playing the role of oil. Not quite the outcome you wanted, is it?
It is the surface tension that matters, and adding liquid might change that in an unpredictable way. Then again, you can achieve the same result without any extra liquid. Just use different material for the container. I guess using metal instead of glass would make the meniscus bend in the opposite direction. With some tweaking, it might be possible to find the material with wetting angle of $90^\circ$ that would make lithium surface perfectly flat, but I don't think it is worth the effort.
So it goes.

Solid answer. I learn a little every day. Thanks.


text:
I need to compare GaussView output files (same reaction, one HF method and the second PBEh1PBE)
I added 2 TS molecules. I need to compare them. The left picture was done via PBEh1PBE and the one on the right via HF.
Where can I find information about each level of theory that would be relevant for comparing the structure differences between the two?

At the moment your question is a little unclear. What are you trying to compare between the two structures? What sort of information do you want about the level of theory? What is the goal of this comparison? It would help to know all of this just to make sure the comparison you are trying to make is reasonable, let alone possible.

Assuming this is a technical question and that the output files are text files, maybe start with manual comparison with any of the diff-tools (say, kdiff3). Also, that Windows Taskbar looks so disturbing to me:)

I'm looking for information about the HF and PBEh1PBE levels of theory. Preferably if that information can help me explain the difference between the molecular structures of each level of theory.


text:
How is solid titanium(II) oxide an electrical conductor?
Most salts in the solid state are electrical insulators, since the ions are not mobile (e.g. sodium chloride). However, solid titanium(II) oxide, $\ce{TiO (s)}$, is a conductor. How is this so?

The electron configuration of $\ce{Ti^{2+}}$ is $\mathrm{[Ar]~3d^2~4s^0}$.
$\ce{TiO}$, like e.g. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$, adopts the rock salt ($\ce{NaCl}$) structure.
Due to the lower nuclear charge of $\ce{Ti}$ its 3d orbitals are less contracted than that of $\ce{Fe}$, $\ce{Co}$ and $\ce{Ni}$ which allows for overlapping of the 3d orbitals and the formation of a metal d band in $\ce{TiO}$. The d electrons of $\ce{Ti^{2+}}$ partially fill this band and lead to electrical conductivity. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$ on the other hand are electrical insulators.
Note that in this case it is electrons that act as the mobile charge carriers, not the ions.

I always thought it was related to defect structures.  $\ce{FeO}$ is a fairly good conductor on that account IIRC.

As far as I know conductivity of TiO is not related to defect structures. However, $\ce{FeO_{1-x}}$ might not be the best example for an insulator since it rather has the conductivity of a semiconductor than an insulator at normal temperature. Its conductivity is indeed due to $\ce{Fe^{3+}}$ holes hopping between Fe sites in the lattice.

$\ce{Fe_{1-x}O}$ maybe? Much of my job involves dealing with oxide scales, hence I see the q through wustite-colored glasses.

TiO has a defect NaCl structure (1/6 lattice points are vacant); I've read that this permits a more contracted structure and hence better overlap between 3d orbitals on neighbouring Ti. Disclaimer, I have only very superficial knowledge of solid state chem.


text:
Can you experimentally determine activation energy if the rate expression and rate constant are unknown?
When drawing a graph to find the activation energy of a reaction, is it possible to use ln(1/time taken to reach certain point) instead of ln(k), as k is proportional to 1/time?
For example, if an experiment I'm doing has a specific end point, can I use ln(1/time taken to reach that endpoint) instead of ln(k), and do this for a range of temperatures in order to work out the activation energy? Or would I have to use the rate equation and determine the rate constant for each temperature, as I am not sure of the rate equation for the reaction.
Thank you:)

Well, of course you may plug random combinations of anything into the equations instead of anything else, just don't expect the result to be very useful. Do you know how the concentration changes with time in a (say) 2nd order rate law? Try treating that expression as you suggest. Will the resulting graph turn into a straight line? See where this gets you.

This is valid only provided that the reaction is strictly first-order and provided that the “specific end-point” is a fixed concentration ratio. Note that the slope of $\ln(t_\text{end})$ vs. $\frac{1}{T}$ is $\frac{E_a}{R}$ but the intercept is not $\ln(A)$.
If the rection is first order ( $X {\xrightarrow{k}} Y$) then the product concentration at time t is:
$[Y]_t = [X]_0 .e^{k. t}$
ie:
$\ln\left( \frac{[Y]_t}{[X]_0}\right) = k. t $
If the end point is some fixed concentration ratio ($\frac{[Y]_{end}}{[X]_0}$), the time to reach the endpoint is:
$ t_{end} = \frac{\ln\left( \frac{[Y]_{end}}{[X]_0}\right)}{k}$
If we substitute this into the Arrhenius equation ($\ln(k) = \ln(A) - \frac{E_{a}}{R.T}$), we get:
$\ln(t_{end}) = \ln\left( \ln\left( \frac{[Y]_{end}}{[X]_0}\right)\right) - \ln A + \frac{E_{a}}{R.T} $
so a plot of $ln(t_{end})$ vs. $\frac{1}{T}$ will have a slope of $\frac{E_{a}}{R}$

This seems more like a comment than an answer. Could you expand, please?

What you need for calculating $E_a$ or $\Delta H^\ddagger$ and $\Delta S^\ddagger$ are rate constants at different temperatures. So the real question here is how to get the rate constant from a, what you call "specific end point". I am assuming here that you are talking about reactant/product concentrations at a specific time point and you also know the the starting concentrations.
This is in principle, at least for simple lower order reactions, possible, but only if you know the order of the reaction. How to do it depends on the order of reaction and the setup.

For example for zero order:
$ A \rightarrow B $
$[A]_t=[A]_0-kt$
If we got 10% left after 90 seconds and we started with a concentration of 1 M this means:
$0.1=1-90 k$
which means:
$k=0.01 M/s$

Or for a second order reaction following $2A \rightarrow B$:
$\frac{1}{[A]_t}-\frac{1}{[A]_0}=kt$
Let's say again that after 90 seconds 10% of A was left and we started with 1M:
$\frac{1}{0.1}-\frac{1}{1}=90k$
$k=0.1M^{-1}s^{s-1}$

Please note that this is very close related to half-lifes of reactions. You might want to look into how to calculate those, the reverse process is how you can calculate the rate constant knowing the conversion and the starting concentration.


text:
How do IR spectrometers produce light spanning the entire infrared spectrum?
An IR spectrum covers thousands of wavenumbers, typically $\sim 4000$ to $\pu{400 cm-1}$.
How are the spectrometers able to generate different wavelengths of light covering this entire range of wavelengths?

This could be accomplished by a scanning instrument. A broad-band light source could output IR radiation over the whole range. This incident radiation is passed through a monochromator to individually select wavelengths to send to the sample. This monochromator can work many ways, using mirrors, gratings, prisms, etc.
Alternatively, you could expose the sample to the broad-band radiation and use a detector that captures all of the transmitted radiation at once. UV/Visible spectroscopy can be done with a photodiode array detector capable of this, but I know of no IR spectrometers that use this method.
Most modern IR spectrometers use Fourier transform. An interferometer, converts an IR laser beam into broad-band IR pulses through interference generated with moving mirrors or by other methods. This action converts the coherent beam into an interference pattern in time. The interferogram below shows what this might look like (where this mirror position is also the time axis). The image has been released to the public domain by Wikipedia user Sbyrnes321.

Absorption of IR radiation by the sample perturbs the interference pattern before it hits the detector. Electronics or software perform the Fourier transform, converting time data into frequency data.

A couple of nit picking points. The standard FTIR instrument only uses a laser to measure  the position of the interferometer mirror, (by counting interference fringes crossing zero) and the source is continuous thermal ir radiation; no  'converting IR laser into broad band IR pulses' is necessary. The Fourier transform is performed directly by the interferometer in the machine, the software does the reverse transform to produce the spectrum

IR spectrometers detect the IR light via either by a monochrometer or an FT interferometer (as per Ben Norris's answer) They generate the light with a fairly simple heating element (normally an electrically heated ceramic block).  Being hot, it produces broadband black body radiation.