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How is plane polarized light rotated by chiral compounds?
There are plenty of questions related to this topic on this site but no proper answer.
Can anyone please explain to me how plane polarized light is rotated by chiral compounds, and why it cannot be rotated by achiral compounds?

Good question. Anyone willing to answer should also share the "why" part of optical rotation. The standard explanation of slowing down of one component of the electromagnetic radiation by a chiral molecule is rampant everywhere.

This can be helpful.https://en.wikibooks.org/wiki/Organic_Chemistry/Chirality/Optical_activity#Why_Polarized_Light_Is_Affected

@Osal Thuduwage thanks for suggestion.  But it doesn't have proper info but it is quite useful. See if more details are available.

Does this help?

@Safdar Faisal thank you very much.  Nice and simple explanation. Yet if anyone wants to add something new feel free to do so.

@M.Farooq do you mean to say that the 'why' part of the answer is what you have shared or that it isn't correct?

@SafdarFaisal, I mean someone should explain in-depth. I don't know the in-depth answer. The links you and others have provided explain superficially. As I wrote in the above comment, textbooks have a surface treatment of the subject. A good answer may explain why do we only consider the electric field? What happens to the magnetic field-most textbooks are silent about it! How does the electric field interact with the chiral molecule and so on and how does the electric field "know" that it is in a chiral environment.

Of course,@M.Farooq has oriented this question  to a proper path.

@M.Farooq A more rigorous explanation is given link. The complexity of this would be the reason for the oversimplification, however this takes into consideration the turbidity of the solution.

I agree with @M.Farooq, these points should be in the question itself, to make it complete.

@M.Farooq The magnetic field is not relevant because it has nothing to interact with, in ordinary matter with paired electrons. A translucent body with a lot of unpaired electrons (in a magnetic field, so the magnetic dipoles orient) should give a weak "magnetic" birefringence, but I have never heard of an example for that.

Actually, it is easy to find papers on the various magnetic dichroism and birefringence effects, including linear, circular and elliptical.

@M.Farooq Actually, magnetic effects are relevant and there many papers on things such as magnetic circular dichroism (MCD). Here is an MCD paper from 40 years ago with my name in the author debris field: “Measurement of Magnetic Circular Dichroism of Matrix-Isolated High Temperature Molecules”, R. Brittain, D. Powell, E. Voigtman, M. Vala, Review of Scientific Instruments, 51, 905-910 (1980). As well, Zeeman background correction in AAS uses magnetic dichroism to do the job. See here.

https://chemistry.stackexchange.com/questions/83990/what-is-occuring-on-the-quantum-level-when-a-molecule-rotates-plane-polarized-li?noredirect=1&lq=1

And another one: https://physics.stackexchange.com/a/16402

Circularly polarized light is like a helix that twists through space. The two components are mirror images of each other.
Now, every molecule interacts with both the left-handed twisting light and the right handed twisting light. The interactions differ. Every molecule, in different orientations, interact differently with the left-handed and the right-handed circularly polarized light.

Animation Reference: https://en.wikipedia.org/wiki/Circular_polarization
Now, if the molecules in solution have any mirror symmetry themselves, then if averaged over all the molecule interactions the left-handed and right-handed circularly polarized light interact in the same way, and so the polarization does not change.
But this does not hold for chiral molecules: for a particular interaction between a molecule in a certain orientation and the left-handed circularly polarized light there is no corresponding identical interaction with the right-handed circularly polarized light. There is no such molecule of that mirrored shape around.
This results in a net difference in interactions of the molecules with the left-handed and right-handed circularly polarized light, which can be described as a difference in refractive index for the two light waves. And this difference in refractive index can be detected as a change in the direction of the polarization for the sum of the two light waves.

Now, if we want to consider this using the concept of the speed of light varying in media, light goes slower in matter than in vacuum. This slowing down of light in matter is measured using the refractive index n. A higher refractive index means that the light goes slower and so keeps the light longer in the medium. Now if the medium is chiral, it gives the light two different speeds, one for the light that rotates its polarization clockwise and the other one for rotating polarization counter-clockwise. Any polarized light has only two parts (clockwise and counter-clockwise).
The two parts are combined and so the light shows a direction of polarization. When these two parts of light pass through chiral matter, one goes faster and the other slower. The result is that the polarization of light is rotated.

Response to @ Michael Seifert comment;
The way of splitting the light really depends on the matter, rather not decided by the light itself. . If matter has two directions (like a calcite crystal), light is split linearly (parallel and perpendicular). If matter is chiral, (like a solution of bio-substance) light is split circularly (clockwise and counter-clockwise). If matter is both, then all four polarizations are possible, which makes the out-going light hard to nail down.

References:
https://en.wikibooks.org/wiki/Organic_Chemistry/Chirality/Optical_activity#Why_Polarized_Light_Is_Affected
https://en.wikipedia.org/wiki/Optical_rotation
https://www.quora.com/How-do-chiral-molecules-rotate-the-plane-of-polarised-light

Good answer (& +1), but it might be helpful to explain the relationship between circularly polarized light and linearly polarized light in a bit more detail.

@Michael Seifert: Thanks for the opinion and glad to hear your kind words. I have edited my answer response to your comment. If you want more details I like to suggest you to follow http://www.intelsat.com/wp-content/uploads/2013/02/Polarization.pdf

What is an example of a substance that is both chiral and birefringent?

@Ruslan Birefringence is a property of  macroscopically anisotropic matter, optical activity of microscopic chirality of e.g. molecules. I assume that a single crystal of basically any pure enantiomer would also be birefringent.


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$Q_3$ is symmetric with respect to the $\sigma_v$ operation?
I am currently studying the textbook Infrared and Raman Spectroscopy: Principles and Spectral Interpretation, second edition, by Peter J. Larkin. Section 8. Symmetry: Infrared and Raman Active Vibrations, says the following:

Fig. 2.15 shows the Cartesian displacement vectors (arrows) of the vibrational modes $Q_1$, $Q_2$, and $Q_3$ of the bent triatomic $\ce{XY2}$ molecule (such as water), and shows how they are modified by the symmetry operations $C_2$, $\sigma_v$, and $\sigma^\prime_v$. For nondegenerate modes of vibration such as these, the displacement vectors in the first column (the identity column, $I$) are multiplied by either (+1) or (-1) as shown to give the forms in the other three columns. Multiplication by (+1) does not change the original form so the resulting form is said to be symmetrical with respect to that symmetry operation. Multiplication by (-1) reverses all the vectors of the original form and the resulting form is said to be antisymmetrical with respect to that symmetry element. As seen in Fig. 2.15, $Q_1$ and $Q_2$ are both totally symmetric modes (i.e., symmetric to all symmetry operations), whereas $Q_3$ is symmetric with respect to the $\sigma^\prime_v$ operation but anti-symmetric with respect to the $C_2$ and $\sigma_v$ operations.

It is this part that concerns me:

As seen in Fig. 2.15, $Q_1$ and $Q_2$ are both totally symmetric modes (i.e., symmetric to all symmetry operations), whereas $Q_3$ is symmetric with respect to the $\sigma^\prime_v$ operation but anti-symmetric with respect to the $C_2$ and $\sigma_v$ operations.

Shouldn't it say that $Q_3$ is symmetric with respect to the $\sigma_v$ operation, since this is the same form, and anti-symmetric with respect to the $C_2$ and $\sigma_v$ operations? It seems like a typo to me.
I would greatly appreciate it if people would please take the time to clarify this.

$\sigma'$  is in the plane of the image and molecule so Q3 is symmetric wrt this but not with respect to the other operations. (Note that some point group tables may define $\sigma,\sigma'$ the other way round.)

@porphyrin I don't understand. The author says the following: "Multiplication by (+1) does not change the original form so the resulting form is said to be symmetrical with respect to that symmetry operation. Multiplication by (-1) reverses all the vectors of the original form and the resulting form is said to be antisymmetrical with respect to that symmetry element." For $Q_3$, multiplication by (+1) results in the same form, and this happens under the $\sigma_v$ symmetry operation.

the $\sigma_v'$ rectangle on the Q1 image (top tight) has the wrong perspective for C2v it should be in the plane of the paper, I did not notice this before (: It should be drawn as in image  D.

@porphyrin Do you mean $C_2$ or $\sigma^\prime_v$? You wrote $Q_1$, but then wrote both $\sigma^\prime_v$ and $C_2v$, so it's not clear what you are referring to. Also, this misses my point: For $Q_3$, it is only the symmetry operation $\sigma_v$ that results in multiplication by (+1) and therefore the same form, so shouldn't the author be saying that $Q_3$ is symmetric with respect to only the $\sigma_v$ operation, and anti-symmetric with respect to the $C_2$ and $\sigma^\prime_v$ operations?

Q3 is antisym wrt C$_2$ and $\sigma_v$ and symmetric wrt $\sigma'_v$  as the atoms remain in the plane of the page (as it were) and this is also the plane of $\sigma_v'$ even though the image at the top right of Q1 indicates otherwise. The $\sigma_v$ plane projects out of the page towards the viewer. Q3 should have B2 sym species so the last two labels are wrong $\sigma_v (-1) ;\sigma_v' (+1)$

@porphyrin Ok, if the labels are wrong then that what the author said would make sense. Thanks for the clarification.

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Maxwell's distribution function for mixture of ideal gases
I came across a question demanding relations between the individual distribution functions $f_n$ for the $n$th gas, and the overall $f$ for the entire mixture.
I believe there is no simple way to express this, and that the mixture will behave as one entity, with a nearly time-independent function inexpressible in terms of the original individual functions and that the individual data of each gas, (however herculeanly obtained) will be time-dependent. Please correct me if I am wrong, any useful reading material is welcomed.
To be absolutely clear, my question is: Do the individual components obey Maxwell's function separately and/or do they obey the function as a mixture with it's own properties such as average molecular mass etc?

On your comment "I believe there is no simple way to express this,...", my review of this educational text, for example,  to quote:

Additionally, the function can be written in terms of the scalar quantity speed c instead of the vector quantity velocity. This form of the function defines the distribution of the gas molecules moving at different speeds, between  c1 and c2, thus...

where the alternate use of a scalar function approach, as defined by Equation (2) in the referenced text, appears tractable. It also can be integrated to arrive at an average speed, for example, as given by Equation (5) in the cited source.
As such, modifying the average value integration to address a weighted average of say an ideal gas mixture may be relatedly facile.

AJKOER, I am not sure that quite answers my question?

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Conformers with extremely high energy barriers?
What are some conformers with very high energy barriers? When we think of e.g. the butane conformers, at room temperature, the conformational change plays almost no role. I was wondering how molecules look like that are almost exclusively in one energy valley or the other (so something like 50-50, not 99-1) and hence might even have different physical properties, depending on the conformation it is in.

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

Exactly what I was looking for, thanks. Googling for "conformer with high energy barrier" and similar didn't lead me to the atropisomers. Only thing is that the examples on Wikipedia don't have spectacularly different forms/physical properties but are extremely similar.


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How do I remove some of the glycerin when making soap?
In some older descriptions of soap making, I read that when the base is added to the fat the glycerin separates and collects at the top where it is drained off. Fine, but there are no details.
In modern recipes I have been using for soap, however, the instructions invariably say to constantly stir after adding the base. Of course, if I do this, then the glycerin gets mixed into the soap, making it too soft. If I don't mix, then there is the problem that the base might not react with all of the oil.
So, there is kind of a disconnect between theoretical descriptions of soap making and glycerin removal, and actual real modern recipes.
The basic question is: how do I get the glycerin out when making soap?


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Could blue flames arise within the hull of a burning ship?
In Joseph Conrad's Youth: A Narrative, set in 1876, a 400-ton barque ship, the Judea, is sent to Java Head to transport 600 tons of coal. Although various delays occur during its journey, the coal it is carrying spontaneously combusts while off the coast of Western Australia. The crew attempt to smother the fire, unsuccessfully. Then attempt to flood the fire, but this is also unsuccessful. Finally, the gases in the hull explode and blow up the deck, and the crew hail a steamer to tow them to an intervening location. However, during their towing, the speed of the steamer's ship cause the fire currently present on the Judea to smoulder into flames, which the narrator describes thusly:

The speed of the towing had fanned the smoldering destruction. A blue gleam appeared forward, shining below the wreck of the deck. It wavered in patches, it seemed to stir and creep like the light of a glowworm

My questions:

Given the amount of fuel and material of the ship available, is it possible that a blue fire of 2300-3000 degrees Celsius (from a quick Google search), could be produced within the hull of the Judea?

Or is it the case that due to the nature of coal gas, a lower temperature than aforementioned need only be met for flames to appear blue, similar to how the flames of a gas stove appear blue?

Additional information, if helpful:
At its last port to Java Head, the Judea had been recalked and copper sheathed before embarking on the last stretch toward Java Head. The Judea explodes and is towed at night.

Copper could add blue/green color to the flame..

Certainly case 2. A forge (coal with air actively blown in) barely reaches 1300°C.

The temperature of the flame will not be the primary cause of its colour. Many gases will burn with a blue-ish flame under the right conditions (methane, hydrogen and carbon monoxide, for example, all potentially associated with decomposing coal).

I will upvote your question as the extent of background and commentary will likely reveal and support the right answer. The actual underlying chemistry, whoever presents it, will hopefully enlighten and possibly serve as valuable lesson for others. Thank you.

In a Bunsen burner, the gas (methane or propane) may burn under two practical regimes : reductive or oxidizing conditions. In the reductive conditions, the flame is yellow and unstable, because the amount of oxygen is low with respect to the stoichiometric ratio. In the oxidative conditions, the flame is noisy, strong and stable, and it is due to an excess of oxygen in the mixed gases before the combustion. The oxidative flame is nearly colorless, but it has a characteristic blue cone at its base. This blue cone has a temperature of about $900°$C, and its spectrum has been analyzed by Hertzfeld in $1935$ : it shows a line spectrum typical of the diatomic "molecule" $\ce{C_2}$, with plenty of vibrational and even rotational sublevels. Maybe that was the origin of the blue flame described by Joseph Conrad. Why not ?

See also 'coal gas' - https://en.wikipedia.org/wiki/Coal_gas - produced by heating coal.

Maurice: I also enjoy reading your answer. Unfortunately, my expanded answer (see comments after [EDIT]) indicate the failure of your response on addressing the likely chemistry, and especially on explaining the blue reported flame coloration. Further, my rendition does account for the contributing factor to the ship's disaster, an important point for future ship fires, relating to the interaction of hot iron and steam (from the unsuccessfully attempt at flooding the fire ).

What is even more troubling with this incorrect 'answer' is that it does not even explain the major reason for the reported flame, the very obvious EXPLOSION!

Maurice: Your answer also does not explain the suggested presence of a very high-temperature flame!

Jon Custer: Thanks for being the first to politely point to the obvious failings of Maurice's answer. I just upvoted your comment.

@AJKOER Is it so necessary to push your answer to this extant? feels excessive.

Safdar:  Thanks for your comment, and now, for balance, please review Maurice's  answer, and present your honest opinion in a comment. Thanks again.

@ Ajkoer and Safdar. Two remarks. 1. 900°C in a flame is not considered as a high temperature. 2. Blue flames can be obtained by burning practically any coal gas, or any mixture of hydrocarbons with or without CO. My point is that this color may be due to the presence of C2 in the flame. That's all.

Thanks Maurice for that as you noted in your  answer as the  'why not' speculation. Interestingly,  how do you address the issue of the explosion or have you abandoned all other explanations? If you no longer believe your answer, just say so.

@ Ajkoer, I maintain my speculation. I simply don't see any better explanation for getting a blue flame. Blue is due to C2, that's all. And I have not tried to explain the explosion.

@AJKOER I am not talking about the correctness or validity of both answers. I am in no position to judge this. Rather the approach that you have taken to declaim Maurice's answer just felt like it was over the top and unnecessary.

Thanks Safdar: I am normally polite but downvoting to negative, my even possibly correct answer, which could be a matter of maritime safety and lives, is where I will voice loudly my counter-argument. Alas, these are rare occasions (well, at least,  were in the old world).

The observed blue flame and explosion (the latter following per the quote, "Finally, the gases in the hull explode and blow up the deck") are likely the products of a hydrogen and air explosion.
There are two potential sources for the hydrogen gas here. First, the coal fire heats the mass of coal releasing a gas mix, to quote Wikipedia on Coal Gas on the particulars:

Coal gas is a flammable gaseous fuel made from coal and supplied to the user via a piped distribution system. It is produced when coal is heated strongly in the absence of air.

And further:

Coal gas contains a variety of calorific gases including hydrogen, carbon monoxide, methane, ethylene and volatile hydrocarbons together with small quantities of non-calorific gases such as carbon dioxide and nitrogen.

The second possible source of $\ce{H2}$ is from the failed attempt to flood the fire producing a large quantity of steam in the presence of hot iron, to quote an educational source:

If you pass steam over hot iron the steam reacts with the iron to produce a black, magnetic oxide of iron called triiron tetroxide, $\ce{(Fe_3O_4)}$.

$\ce{3 Fe(s) + 4H2O (g) -> Fe3O4(s) + 4 H2 (g)(1)}$

The hydrogen produced in the reaction is swept away by the stream of steam.

So, the hydrogen is formed and is trapped in pockets in the hull where it may eventually mix with oxygen from air and finally, a hot splitter causes an explosive ignition forming a striking blue flame.
[EDIT] On the explanation of the 'reported' (albeit, not always actual) color of a hydrogen flame, I quote a source:

There are also several text-books on chemistry which assert that hydrogen burns with a characteristic faint blue flame. It is easy to prove, however, that the flame of pure hydrogen has no blue tinge whatever. The blueness so frequently associated with the flame of hydrogen is really due to the presence of sulphur as is shown in a little paper I published in the Philosophical Magazine for November 1865.*

Now, in the current context, in the presence of burning coal, likely containing a significant sulfur impurity, a pronounced more unusual blue appearance apparently was reported, which would be in agreement with the hydrogen gas impurity argument, as is, the mention of a possible very high-temperature flame.
Others believing that this was a more normal occurrence (associated with, for example, a flame  "with characteristic blue cone at its base") are likely incorrect. Further, not accounting for the chemistry as to how such a quantity of hydrogen gas was likely amassed, and its potential implications for future ship's fire, is a significant, perhaps even egregious, omission. More profoundly, it lies at the very explanation for the occurrence of the reported explosion and any resulting loss of human lives.

Why should the flame of H2 be blue ?

Actually, sorry you brought up this point, as it exposes unfavorably your otherwise well-presented answer above.

(+1) I will upvote immediately after posting this comment: your answer is plausible to me.  Funny thing about the way the stack exchanges work: in several cases, I have posted detailed answers that I consider to be clear and demonstrably correct, yet they get minimal upvotes. Then other answers, that are clearly incorrect, get numerous upvotes and get accepted by the OP. Frustrating, but not much to be done about it: it is a voting model, for better or worse, and we volunteer our efforts.

I also upvoted. But beside H2 or H, answer such as that of Maurice and point 2 in a comment were what I thought at first. CO burns blu and there should be sufficient oxygen at see. Explosion can originate from every flammable gas pocket and even fine solid particles previously formed in the fire. Personally I upvoted both answers and that comment.

I disagree on the idea that it was a hydrogen explosion. High grade coal would be prone to methane release, which can build up in the hold, one reason why modern regulations require online gas measurement for shipping coal. For those of you who have seen old style refinery flue gas burning at night, if it's "burning blue" that means methane and/or other light gasses are being fully oxidised, so Maurice has my vote on the blue flame. Pockets of trapped methane will also explode when the concentrations are right (~7-14% in air) when it comes into contact with open flame - ask any coal miner.

Perhaps Gwyn, but you realize you are discounting the action of steam on hot iron. Further, why is even the unsuccessful attempt on drowning the fire even mentioned? Perhaps an insurance investigation suggested that steam formation led to hydrogen gas which precipitated an explosion. The latter would be the cause of hull damage, and as it is due to human error, that part of the damage would not be covered under a 'fire' policy. If the fire formed explosive gases, the fire insurance would cover the total loss.


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Reaction between ammonia and dichloromethane
Could ammonia displace a chloride in dichloromethane to form a methylammonia? If so, given enough reagents and time, could this be used to make a 3D structure comprised of nitrogen bonded carbons, and carbons bonded to nitrogen and hidrogen?

You mean https://en.wikipedia.org/wiki/Hexamethylenetetramine ?

Kind of. I guess a exaustive alkylation could occur though, extending the structure.

I was not aware of that synthesis though. I guess if that stops there, the same would be observed in my experiment. Why does it stop there though? Why doesn't a nitrogen attack another aldehyde?

There apparently exists CH2Cl2.NH3, which suggests a possible answer of no.

CH2Cl2 is not nearly as reactive as CH2O - that's why it's used as solvent.

If the reaction with methanal does not yield your product, DCM, being less reactive, won't as well.
I cannot point out why the reaction stops there, but from Wikipedia it seems that methenamine can be easily converted to a quaternary ammonium salt by simple alkylation.
Furthermore, your product has a problem: if the nitrogens did become quaternary ammonia**, your structure would collapse due to Coulomb repulsion caused by the positive charges.
** I'm not a native speaker, but if equilibrium plural is equilibria, I think it is reasonable for ammonium plural to be ammonia.


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How to determine the product from my attempt of HOCl synthesis
This is a follow up to this question. Here is my short procedure:

start with 200mL distilled water
add 7mL of 8.25% sodium hypochlorite (.11M) to the solution. The resulting solution should be 3.7*10$^{-3}$M NaOCl
add bromocresol green indicator.
add acetic acid dropwise, titrate until the solution hits a green color (pH~5).

I think at this point, the solution should contain HOCl and a small amount of dissolved sodium acetate.
$$NaOCl + CH_3COOH \rightarrow HOCl + CH_3COONa$$
I took a pH test with litmus paper and it shows between 5-6. I took a free chlorine test to determine the concentration of HOCl, and it resulted in ~1000ppm, which was what I was aiming for. The chlorine concentration stayed comfortably within its max solubility in water. I used an NaCl meter to determine the concentration in the solution and it returned 0. However, I tested for salts with silver nitrate and it precipitated both white and dark brown/black.
White precipitate indicates presence of chlorine, and dark brown precipitate indicates presence of OH. How did I get AgOH in an acidic solution? Is it simply that the silver rips apart the HOCl to form AgCl and AgOH?
My boss is concerned that I have made dilute hydrochloric acid, not hypochlorous acid. How can I prove that the solution is one product or the other?

Ryan, sorry your approach to generate the desired compound is not the best one. This post also shows the problems with the concept of an "accepted" answer on StackExchange. The answer is also wrong and misleading. Please consult your academic supervisor to help you do a proper literature search. I am not sure why are you doing all this, you did not explain in your two posts. First of all, there is nothing such as pure 8% NaOCl soln. This bleach is a mixture sodium hydroxide and NaOCl. What you are apparently titrating is the free alkali with acetic acid (which you should not be doing). Simultaneously, the hypochlorite ion is decomposing into other products.
The proper place to do inorganic synthesis research is the library, not Q&A site. A good reference is Kirk Othmer Encyclopedia of Chemical Technology to look up bleach and its properties. Any good library must have it.
In your first question you linked a paper "Hypochlorous Acid as a Potential Wound Care Agent" in the Journal of Burns and Wounds. This journal has a zero impact factor which means this is of dubious quality. There was a reason to avoid organic acid such as acetic acid in that paper. Have you read about disinfection by-products in water? Search this topic this too, because disinfection byproducts can be cancer-causing.

Here is the reason why I’m doing this. My boss at work uses copious amounts of cleaning fluid, but since COVID, he hasn’t been able to buy any. He asked me to look into synthesizing HOCl so he could use it in a fogging machine to disinfect the workplace. HOCl meets epa criteria for use against COVID-19. What do you think of this literature? https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7315945/ Is electrolysis the better approach to making HOCl?

Vapor-phase hydrogen peroxide is a standard method of microbial decontamination/sterilization for labs, hospitals, etc. Acidified chlorine bleach, which you tried out, or just 5% unacidified bleach, are effective but will leave a solid residue on surfaces. In really challenging situations, such as decontamination of an anthrax-contaminated postal facility, chlorine dioxide gas has been shown to be highly effective, but that is probably overkill in this case.

Every university should have a Environment & Health Safety Department. They are best persons to contact for this problem.

The explanation is actually a reaction I once attempted, the reaction of a soluble Silver salt (here Silver nitrate, I experimented with Silver acetate) with HOCl. The reaction apparently per Mellor (see this reference) proceeds, assuming some chlorine per the reactions:
$\ce{ 2 HOCl (aq) ->  2 HCl (aq) + O2 (g)}$
$\ce{ HCl + HOCl  <=>  Cl2  + H2O}$
further, as follows commencing with the creation of the unstable Silver hypochlorite in an equilibrium reaction:
$\ce{AgNO3 (aq) + HOCl (aq) <=> AgOCl + H+ + NO3-}$
Followed by the disproportionation of the unstable hypochlorite forming a white precipitate of Silver chloride, which moves the above equilibrium to the right:
$\ce{3 AgOCl (aq) -> 2 AgCl (s) + AgClO3 (s)}$
Also, in the presence of HCl:
$\ce{ AgNO3 + HCl  -> AgCl (s) + H+ + NO3-}$
The dark precipitate you reported is likely Ag2O formed, I suspect, from the subsequent breakdown of the Silver chlorate (which is a strong oxidizing agent). Less likely, but possible, is Ag2O2, which is also described as a dark brown solid per Wikipedia.

Thank you! I guess I kind of split my question into two parts. Does this also prove the presence of HOCl and not HCl?

Anytime your have Ag+ and a source of Cl-, including from HCl, one quickly gets AgCl(s). So, as long as the Silver ion is in excess, no HCl.

so the presence of the dark precipitate means that there was excess silver ions, thus no HCl- thanks!

@Ryan, Sorry to say all these silver related equations are imaginary, except the HCl one. A misleading answer.

Farooq: Do you known that the action of NaOCl on soluble Pb salts forms a transient Pb(OCl)2? There are pictures on this Lead hypochlorite and its subsequent decomposition producing red PbO2 on warming. I have also performed this reaction.  Silver chemistry obviously related as AgClO3 is cited as  unstable except in the presence of excess Ag2O,

@AJKOER: Yes I can only see this in 19th century books. Any modern evidence of the existence of lead hypochlorite?


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How is Bent's rule consistent with LCAO MO theory?
Bent's Rule states that $s$ orbitals on a central atom participate contribute more to molecular orbitals directed towards more electropositive ligands. (For a longer explanation, see this question and answers: What is Bent's rule?). Lone pairs are considered to be the extreme limit of an electropositive "ligand".
Many sources claim that this Rule is consistent with LCAO MO theory representations of electron density. It seems, however, that this claim is easily disproved by considering the example of simple molecules of the type $\ce{AH2}$.
We can start with $\ce{H2O}$, whose LCAO MO diagram is well known:

Source: H2O-MO-Diagram.svg by Officer781 on Wikimedia Commons
The orbitals contributing to the O-H bonds are the lower two in energy. Ignoring orbital mixing for now, these are formed from linear combinations of the hydrogen $s$ orbitals with the oxygen $2s$ (giving $\mathrm{2a_1}$) and $2p_y$ orbitals (giving $\mathrm{1b_2}$). The lone pairs are a pure $p_x$ orbital ($\mathrm{1b_1}$) and a combination of $p_z$ with a small contribution from the hydrogen $s$ orbitals ($\mathrm{3a_1}$). Based on the contributing oxygen atomic orbitals, that gives 100% p lone pairs and $50\%$ $p$/$50\%$ $s$ for the bonds, the opposite of the result predicted by Bent's rule, which is that the lone pairs should have more $s$ character.
If we properly account for orbital mixing, there is no effect on the $\mathrm{1b_1}$ lone pair orbital, while the $\mathrm{3a_1}$ lone pair does in fact get more $s$ character, but this increased $s$ character is not expected to exceed $50\%$, since that would just mean a switch of label between the $\mathrm{2a_1}$ and $\mathrm{3a_1}$ orbitals. $50\%$ $s$ in the $\mathrm{3a_1}$ represents the maximum mixed case. That gives a final result of a maximum of $25\%$ $s$ in the lone pairs ($50\%$ in the $\mathrm{3a_1}$ and $0\%$ in $\mathrm{1b_2}$) and $25\%$ in the $\ce{O-H}$ bonds ($\mathrm{2a_1}$ and $\mathrm{1b_2}$), still contradicting Bent's Rule, which predicts that the lone pairs should have more $s$ than the bonds, not the same amount. [One a side note, we know that $\ce{H2O}$ does not represent the maximum mixing case, as $\ce{SH2}$ has a smaller bond angle consistent with even greater orbital mixing, but that's beside the point here.]
The paper by Clauss et. al.[1] reconciles this result by claiming that the 1b1 orbital is "always excluded from the Bent's rule competition for in-plane p-character", but I've never seen any formulation of Bent's Rule that indicates such an exception. If there is such a qualification to Bent's Rule, how do we know what lone pairs to include or exclude?
I conclude from this analysis that Bent's Rule is much like VSEPR - useful in some cases as a pedagogical tool, but ultimately inconsistent with LCAO MO theory. Or perhaps it's only consistent when applied to actual bonds rather than to bonds and lone pairs? I'm not concerned with why people still use it, but I am confused as to why so many people argue that it is consistent with LCAO MO theory. Am I missing something here?
References:

Clauss, A. D.; Nelsen, S. F.; Ayoub, M.; Moore, J. W.; Landis, C. R.; Weinhold, F. Rabbit-ears hybrids, VSEPR sterics, and other orbital anachronisms. Chem. Educ. Res. Pract. 2014, 15 (4), 417–434. DOI: 10.1039/C4RP00057A. Or as a free preprint

Beyond the educational value, Bent's rule has had significant impact on how we understand wave functions of molecules. It's simplicity is almost unique in the world of quantum chemistry; almost everybody can immediately understand it and apply it. Together with the VSEPR model it can lead to quite accurate predictions on the back of an envelope.
Molecular Orbital Theory vs. Valence Bond Theory (MO vs. VB)
In their respective infinite treatment these theories are congruent. For more on this topic please read What is actually the difference between valence bond theory and molecular orbital theory?
Important for this post is that both theories are approximations for the wave function of a molecule, hence they do not describe bonding beyond the stationary point they are applied to, and they operate within the clamped nuclei (Born-Oppenheimer) approximation.
In MO theory it is common to express the molecular orbitals as linear combinations of atomic orbitals. Those MO are usually chosen to be orthonormal. The MO themselves are delocalised, but there are algorithms to transform them into more intuitive VB type orbitals. This will provide us with a Lewis-like bonding picture. Famously used in this regard is Natural Bond Orbital (NBO) theory.
Bent's rule
Unfortunately I cannot find my copy of Bent's paper, so this answer will be based purely on the definition of Bent's rule.
The IUPAC gold book defines Bent's rule (DOI: 10.1351/goldbook.BT07000):

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

Water is a good example to demonstrate Bent's rule and its interplay with the VSEPR model. In $\ce{H2O}$ you expect oxygen to follow the octet rule, hence there will be four electron pairs around it, and it predicts a generally tetrahedral structure of the molecule. The angle for these shapes is about $\newcommand{\degree}{^\circ}109\degree$. Since hydrogen is more electronegative than no ligand, Bent's rule predicts a smaller angle than that.
Probably due to its simplicity, Bent's rule can be applied within the MO and VB bonding pictures.
Coulson's Theorem
Coulson's theorem is the formal theory which applies to Bent's rule (see Wikipedia).
We choose a set of orthonormal AO, as would be consistent with the LCAO approximation.
$$
\langle\chi_i|\chi_j\rangle = S_{ij} = \delta_{ij}
\begin{cases}
0, & i \neq j\\
1, & i =    j
\end{cases}\tag{1}\label{orthonormalAO}
$$
We can transform these orbitals into sets of orthonormal hybrid orbitals.
\begin{align}
\varphi_a &= \sum_i \lambda_i \chi_i &
\langle\varphi_a|\varphi_b\rangle = S'_{ab} &= \delta_{ab}
\begin{cases}
0, & a \neq b\\
1, & b =    b
\end{cases}\tag{2}\label{orthonormalHybridAO}
\end{align}
Let's look at a special case, where we choose atomic orbitals to be $\chi_i$ with $i \in \mathrm{s}, \mathrm{p}_x, \mathrm{p}_y, \mathrm{p}_z$. We can construct following hybrid orbitals (also compare Mathematical form of four hybrid orbitals):
\begin{align}\tag{3}\label{orthonormalSP3}
  \varphi_{xyz}
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_x
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_y
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_z
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}
\end{align}
If all $\lambda_i$ are chosen to be $\frac{1}{2}$, then we would come to the famous $\mathrm{sp}^3$ hybrid orbitals.
Let's generalise this a bit and drop the normalisation:
\begin{align}\tag{4}\label{hybrid}
  \varphi_a
    &= N\cdot(\chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi_{\mathrm{p}_k}) &
  \leadsto
  \varphi_a
    &= \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k}
\end{align}
From \eqref{orthonormalHybridAO} and \eqref{hybrid}, and ignoring the trivial case:
\begin{align}\tag5
  \delta_{ab}
    &= \langle\varphi_a|\varphi_b\rangle\\
  0
    &= \langle
      \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k} |
      \chi_\mathrm{s} + \lambda_{\mathrm{p}_l} \chi'_{\mathrm{p}_l}
    \rangle\\
    &= \langle\chi_\mathrm{s}|\chi_\mathrm{s}\rangle
    +  \lambda_{\mathrm{p}_k} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_k}\rangle
    +  \lambda_{\mathrm{p}_l} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_l}\rangle
    +  \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l}
       \langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle
\end{align}
Given \eqref{orthonormalAO} we can simplify this. Note that $\chi'$ is actually a linear combination of $\mathrm{p}$ orbitals, and therefore they don't necessarily have to be orthogonal. The angle between two such orbitals is given by the inner product, i.e. $\langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle = \cos \theta_{kl}$.
\begin{align}\tag6
  0 &= 1 + 0 + 0 + \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l} \cos \theta_{kl}
\end{align}
Therefore Coulson's Theorem is
\begin{align}\tag{7}\label{Coulson}
  \cos \theta_{kl} &= \frac{-1}{\lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l}}
\end{align}
If you call $\lambda$ the hybridisation index, then you see that the angle between two hybrid orbitals is dependent on the this index. In other words, the angle between two ligands around a central atom determines the $\mathrm{p}$ character of the hybrid orbitals that ideally describe this bond.
What you can also derive from this formula is, given a constant angle $\theta_{kl}$, if you use a hybrid-orbital with increased $\mathrm{p}$ character, the other hybrid-orbital must have less $\mathrm{p}$ character, i.e. $\lambda_{\mathrm{p}_k} < \lambda_{\mathrm{p}_l}$.
This is in principle all that Bent's rule requires in the framework of LCAO-MO theory to be consistent; the argument about the electronegativity of a ligand is based on observation of molecular structure.

I hope the above answers the titular question sufficiently. In the second part I'll try to show where the original arguments fail.
First of all, you are basing your argumentation on even more handwavy mathematics then I did above. You are also not considering the actual molecular structure and the implications of that in terms of the LCAO-MO approximation of the wave function.

The orbitals contributing to the O-H bonds are the lower two in energy.

This is not quite correct. In LCAO-MO (or MO in general) theory all orbitals contribute to bonding.

Ignoring orbital mixing for now, these are formed from linear combinations of the hydrogen s orbitals with the oxygen 2s (giving 2a1) and 2py orbitals (giving 1b2). The lone pairs are a pure px orbital (1b1) and a combination of pz with a small contribution from the hydrogen s orbitals (3a1).

The shown MO scheme does not ignore orbital mixing, and neither do you in your argumentation. If the oxygen's $\mathrm{s}$ and $\mathrm{p}$ orbitals were far enough apart to not mix, then only one sort of these orbitals could interact with the hydrogen orbitals to lead to bonding. This would either lead to a situation where there is no oxygen $\mathrm{s}$-character in the bonds, or all of it.
This basically follows the argument that for the heavier homologues of water the oxygen valence $\mathrm{s}$-orbital becomes the lone pair.

Based on the contributing oxygen atomic orbitals, that gives 100% p lone pairs and 50% p/50% s for the bonds, the opposite of the result predicted by Bent's rule, which is that the lone pairs should have more s character.

I don't have an optimal LCAO-MO solution at hand, but I have the next best thing: a HF/STO-3G wave function with $C_\mathrm{2v}$ symmetry, $\angle(\ce{HOH}) = 100.0\degree$, and $d(\ce{OH}) = \pu{98.9 pm}$. That is off by plenty, but it's Hartree-Fock, so that was expected. The general tenor wouldn't change, if you were to run it on the experimental structure. This is the output of the wave function:
 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-20.252 is O1-s=1.0006
 Alpha occ 2 OE=-1.258 is O1-s=0.7899 H2-s=0.0928 H3-s=0.0928 O1-p=0.0245
 Alpha occ 3 OE=-0.594 is O1-p=0.5365 H3-s=0.2317 H2-s=0.2317
 Alpha occ 4 OE=-0.460 is O1-p=0.6808 O1-s=0.1328 H2-s=0.0932 H3-s=0.0932
 Alpha occ 5 OE=-0.393 is O1-p=1.0000
 Alpha vir 6 OE=0.582 is H3-s=0.3144 H2-s=0.3144 O1-p=0.2947 O1-s=0.0766
 Alpha vir 7 OE=0.693 is O1-p=0.4635 H2-s=0.2683 H3-s=0.2683

Unfortunately this doesn't tell us much, apart from that there is obviously one $\mathrm{p}$ lone pair, which was expected due to symmetry. (In MO theory though, this would still count as a bonding π orbital).
From the above table, one would derive that the majority of bonding comes from MO 3, which is a nearly pure oxygen $\mathrm{p}$ orbital.
I've run an NBO6 analysis on this calculation, which is basically a unitary transformation of the MO, and arrived at the following description:
     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. (2.00000) CR ( 1) O  1            s(100.00%)
                                         1.0000  0.0000  0.0000  0.0000  0.0000
   2. (2.00000) LP ( 1) O  1            s( 71.29%)p 0.40( 28.71%)
                                         0.0000  0.8443  0.0000  0.0000  0.5358
   3. (2.00000) LP ( 2) O  1            s(  0.00%)p 1.00(100.00%)
                                         0.0000  0.0000  1.0000  0.0000  0.0000
   4. (1.99926) BD ( 1) O  1- H  2
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000 -0.7071 -0.5970
               ( 40.76%)   0.6384* H  2 s(100.00%)
                                         1.0000
   5. (1.99926) BD ( 1) O  1- H  3
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000  0.7071 -0.5970
               ( 40.76%)   0.6384* H  3 s(100.00%)
                                         1.0000
 ---------------- non-Lewis ----------------------------------------------------
   6. (0.00074) BD*( 1) O  1- H  2
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000  0.7071  0.5970
               ( 59.24%)  -0.7697* H  2 s(100.00%)
                                        -1.0000
   7. (0.00074) BD*( 1) O  1- H  3
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000 -0.7071  0.5970
               ( 59.24%)  -0.7697* H  3 s(100.00%)
                                        -1.0000

This results in two $\mathrm{sp}^6$ bonding orbitals, one pure $\mathrm{p}$ lone pair and a $\mathrm{sp}^{0.4}$ lone pair. From this we would conclude that more $\mathrm{p}$ character is directed towards the hydrogen ligands.

If we properly account for orbital mixing, there is no effect on the 1b1 lone pair orbital, [...]

That is correct, this is due to symmetry.

[...] while the 3a1 lone pair does in fact get more s character, [...]

In your argumentation 3a1 would have been a pure $\mathrm{p}$ lone pair, with mixing it does gain $\mathrm{s}$ character, which is obviously more than none.

[...] but this increased s character is not expected to exceed 50%, since that would just mean a switch of label between the 2a1 and 3a1 orbitals.  50% s in the 3a1 represents the maximum mixed case.

The part with the switch is only true if the energy of these orbitals also switches and the contribution of the hydrogen were equal in both orbitals.
Does this argument also imply that an $\mathrm{sp}$ orbital was the maximum mixed orbital?
I'm very sorry, but I cannot follow the logic as I do not understand what is meant with maximum mixed case. In LCAO you are able to continuously mix $\mathrm{s}$ and $\mathrm{p}$ orbitals, if that actually produces a sensible approximation to the wave function is a completely different topic.

That gives a final result of a maximum of 25% s in the lone pairs (50% in the 3a1 and 0% in 2b1) and 25% in the O-H bonds (2a1 and 1b2), still contradicting Bent's Rule, which predicts that the lone pairs should have more s than the bonds, not the same amount.

This would be exactly the result which you were to expect from tetrahedral coordination: two $\mathrm{sp}^3$ bonds, one $\mathrm{sp}$ and one $\mathrm{p}$ lone pair.

[One a side note, we know that H2O does not represent the maximum mixing case, as SH2 has a smaller bond angle consistent with even greater orbital mixing, but that's beside the point here.]

I am really interested in that side note. I still want to understand what is meant by the maximum mixing case, and especially how a smaller bond angle is consistent with more mixing.

Thank you. This is what I was looking for. I think one issue is that I am using short-hand "s-p" mixing when it is of course MO's mixing, not AO's, so I need to think through the effect of the relative contribution of the H AO's. But if you are correct, would you say then that Albright et al. are wrong when they say that describing the bonding orbitals in SH2 as nearly pure p is nonsensical? (Their exact words are "It is clear from the shape of 1a1 and 2a1 that this is far removed from reality.")

Thinking more, I don't see any reason why I can't pre-mix the s and p orbitals and combine those pre-mixed orbitals with the H orbitals. In that case, the contribution of the H orbitals to the MO should be a function of the energy of the s-p mixed orbital, so it should make no difference whether I start with an unmixed 2a1 orbital and increase p from 0 to 50 or if I start with an unmixed 3a1 and decrease p from 100% to 50%. The contribution of the H orbitals will adjust accordingly, and I will end up in the same place. Mixing any further than 50% sends me back along the path towards the start

Orbitals aren't real, but besides that, in $\ce{SH2}$ 1a1 and 2a1 should be core orbitals of sulfur, so I'm not sure where the argument was heading. Apart from that, in MO theory you can't really differentiate between bond orbitals and lone pairs. In that line of reasoning, yes, the bonding does need all valence orbitals from sulfur. If you follow the localised bond picture, i.e. with NBO, then the lone pairs will be (1) pure p and (2) 80% s character, leading to the conclusion that the majority of bonding is from the p orbitals of sulfur (about sp⁸ iirc).

It doesn't matter how you set up the basis functions you put in the MO formalism, if they can be transformed into each other, the solution will be the same. However, AO typically will be eigenfunctions of the hamiltonian of an atom, while linear combinations thereof won't. Only eigenfunctions will have eigenvectors, or energies. In MO theory, the MO we are looking at are eigenfunctions, they have eigenvalues; any linear combination of those loses this quality. The overall expectation value for the energy will stay the same. The rest is interpretation, which is challenging.

I don't if you saw that I updated my answer with explanation of reconciliation and where I believe I was going wrong. The key for me was finding in Levine a representation of the localized bonding orbitals and lone pairs as linear combinations of the delocalized orbitals, which helped make the connection between the two clear. The second case of H2S would be an interesting further discussion, but not necessary for the explanation here.And last, it does seem that Albright et al have made the same error I did in extending the bond and nonbond nature of the delocalized MOs to localized MOs

@Andrew I have not yet had the time to read your revision, maybe tomorrow. Another thought came to mind though. Nowhere in Bent's rule does it say that the total s character of the lone pairs does increase to being more than in the bonds, but that's something to think about more. Maybe another time.

tl;dr Bent's Rule should be relegated to the dustbin alongside Pauling's claim that $\ce{H2S}$ and $\ce{PH3}$ bond with nearly pure p orbitals.
Corrected My analysis erred in assuming that the nonbonding 3a1 orbital contributes primarily to the lone pairs in a bonding orbital analysis and that the bonding 2a1 orbital contributes primarily to O-H bond orbitals.
The above statement is deliberately provocative in the hope that someone with a different perspective will chime in. In the meantime, here's my understanding.
see update at end
My first thought was that Bent's Rule was never intended to apply to lone pairs, but in Bent's original paper (Bent, H.A. (1961) Chem. Rev. 61:3, 275-311), the very first sentence after he states his rule is "Lone pair electrons are regarded as electrons in bonds to very electropositive atoms," so there is no doubt that he felt the Rule applied equally well to lone pairs.
A thorough reading of the paper revealed that his reasoning was flawed in essentially the same way as Pauling's reasoning regarding $\ce{H2S}$, which is not surprising given that Bent drew heavily on Pauling's ideas regarding hybrid orbitals. The fallacy in both their reasoning can be summed up in two related errors:

They failed to account for the energy of stabilization when a bond is formed and
They did not factor in the electrons contributed by the hydrogen atoms.

Using $\ce{H2O}$ as an example since that is discussed in the original question, Bent and Pauling would start with a neutral oxygen atom with valence shell electron configuration $(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1$. [I am using numbers to index the p orbitals so as not to suggest any bias towards a specific Cartesian axis.]
In order to form $\ce{H2O}$, they would describe the oxygen atom as rearranging to four hybrid $sp^3$ orbitals, two of which are completely filled:
$(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1\rightarrow (sp^3_1)^2(sp^3_2)^2(sp^3_3)^1(sp^3_4)^1$
Looking at these two electron configurations, they make the following logical argument.
Premise: The energy levels of the $sp^3$ orbitals are such that the total energy of four filled $sp^3$ orbitals is the same as a filled s orbital and three filled p orbitals.
$\implies$ Because two electrons have increased in energy from $s$ to $sp^3$ but only four have come down from $p$ to $sp^3$, the state $(sp^3_1)^2(sp^3_2)^2(sp^3_3)^1(sp^3_4)^1$ is higher in energy than the state $(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1$.
$\implies$ The lowest energy state for the $\ce{H2O}$ molecule is when the oxygen adopts a configuration as close as possible to $(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1$, i.e. with lone pairs in a pure $s$ and a pure $p$ orbital and with bonding to hydrogen involving only $p$ orbitals (creating an H-O-H bond angle of 90 degrees).
They argue that the only factor preventing $\ce{H2O}$ from achieving this configuration and having a 90 degree bond angle is repulsion between the partially positively charged H atoms. As a result, it adopts a configuration in which the lone pairs only have an increased s character rather than completely s:
$(s^{1+\lambda}p^{3-\lambda})^2(s^{1+\lambda}p^{3-\lambda})^2(s^{1-\lambda}p^{3+\lambda})^1(s^{1-\lambda}p^{3+\lambda})^1$
On the larger $\ce{H2S}$ molecule, there is more space between the H atoms so they can get much closer to the ideal 90 degree angle which (to Pauling and Bent) represented bonding with pure p orbitals.
You will hopefully notice right away that the above analysis does not allow for any effect of bonding on the electron energy levels, nor does it consider the electrons provided by the H atoms at all. Therein lies the problem.
Let us first consider the effect of bond formation. As every introductory presentation of qualitative MO diagrams informs us, a bonding orbital is lower in energy than either of the participating atomic orbitals, even in a valence bond treatment.
Looking again at the hybridized O atom, we need to form $\sigma$ bonding orbitals with the H atom 1s orbitals, leaving the lone pairs as non-bonding orbitals (nb):
$(sp^3_1)^2(sp^3_2)^2(sp^3_3)^1(sp^3_4)^1 + (1s)^1 + (1s)^1 \rightarrow (\sigma_{OH})^2(\sigma_{OH})^2(sp^3_{nb})^2(sp^3_{nb})^2$
Furthermore, because of the stabilization energy of bonds, we can conclude that the $\sigma_{OH}$ orbitals are lower in energy than the $sp^3_{nb}$ orbitals. Already, we have a contradiction with the Bent/Pauling model, since our lone pairs are now in the higher energy orbitals rather than the lower energy. Additionally, we have four electrons that have been stabilized in lower energy orbitals, because we are including the electrons from the H atoms.
Thus, there is no longer any basis for arguing that the s electrons are more likely to be found in lone pair orbitals or, by extension, in bonding orbitals to more electropositive elements.
Extending from this valence bond analysis to a delocalized LCAO treatment, we see again that the lone pairs are found in preferentially p orbitals, rather than s, and that the s contribution to the lone pair orbital only comes about because of the orbital mixing, which cannot occur to such an extent that the bonding orbitals have more p character than the lone pairs.
UPDATE: RECONCILIATION
Thanks to some prompting from Martin, I dug a little deeper, and I believe I have found my error, which is possibly a source of confusion to others. All of the quantitative results are taken from Chapter 15 of Levine's Quantum Chemistry text and are typically from computational work in the 1970's. As such, the numbers are not as accurate as what one would get from a modern computation, but seem to be close enough to be conceptually useful.
First, to address the questions about about degree of mixing. Levine provides coefficients for the contributions of each atomic orbital to the occupied delocalized MO's of $\ce{H2O}$. If we look only at the contribution from oxygen and convert the coefficients into percentage, we find that the 2a1 orbital is 97% s and 3% p, while the 3a1 is 30% s and 70 % p, confirming that there is not substantial loss of s from the bonding 2a1 orbital nor a more than 50% gain in s in the nonbonding 3a1 orbital. That the values do not add to 100% s and 100% p between them is due to differing contributions of the H orbitals and to the fact that the s and p orbitals are also mixed in the unoccupied 4a1 orbital.
The problem, as Martin suggested, is in the description of 3a1 as "non-bonding". This statement is based on the fact that there is very little orbital overlap between the H-centered orbitals and the O-centered portions of the orbital, which is supported by approaches such as Mulliken population analysis, which gives overlap populations of 0.53 for 2a1, 0.50 for 1b2 and -0.03 for 3a1.
Because of the described nonbonding nature of 3a1, I mistakenly assumed that it would contributed almost exclusively to a lone pair orbital in a bond orbital analysis, and likewise that 1a1 would contribute primarily to a bond orbital. Therein lies the error.
Although 3a1 is a non-bonding orbital, it achieves that by mixing a somewhat strongly bonding interaction (Opz with H1s) with a fairly strongly anti-bonding interaction (O2s with oppositely phased H1s).
In the bond orbital analysis, these two competing contributions are separated, and the pz bonding interaction is allowed to contribute significantly to the O-H bond orbitals. The s antibonding interaction acts to negate the bonding contribution of the s bonding interaction that comprised 2a1, such that the net result (as reported by Levine) is that the O-H bonding orbital is 85% p/15% s and the lone pairs are 62%p/38%s, supporting the Bent's rule statement that the lone pairs should have more s character.
A helpful analysis is provided in the construction of the bond orbitals by linear combination of the delocalized MOs, which shows that 3a1 contributes 18% of each O-H bonding orbital and 32% to each lone pair orbital, while 2a1 contributes 32% of each bonding orbital and 17% of each lone pair. Note that the distribution in both cases matches closely to the s/p distribution of 3a1, as we expect if the p interaction of 3a1 is bonding and the s interaction is anti-bonding (ie contributing to lone pairs).

I'm not sure I can follow this, and I cannot find my copy of Bent's review to read it again. In any case, here is what I remember: Bent's observation (later: rule) was rooted in or derived from (ab-initio) Valence Bond Theory. This is one possibility of constructing the wave function. Like in Molecular Orbital Theory, energy gain due to bonding cannot be included in this description (it can be approximated from considering different states though, which you can also do for VB). Since MOT is congruent to VBT at their respective full treatment, Bent's rule must be consistent with LCAO-MO.

@Martin-マーチン- I'm afraid your memory might be letting you down here. First, Bent himself used the word "rule" and formulated the exact one-sentence wording still used today. Second, he did not use ab initio VBT at all, nor did he derive his rule from theory. In his words, "Mathematical models of molecular structure. . .are not considered in this review." He used VBT only at the descriptive level, eg single bonds means sp3, and a double bond means sp2. Otherwise, he simply conjectured based on experimental data on bond angles, lengths, dipole moments and electronegativities.


text:
If we add surfactant in soft and hard water, which will require more surfactant and why?
I think it is hard water.But my doubt is even if we add more surfactant to hard water there will be no use because soap doesn't dissolve.then what is its use....

It depends on the type of surfactant. There are basically four types of surfactants: anionic, cationic, amphoteric and nonionic.
Anionic surfactants are probably the most common in everyday products (laundry powder, dish wash, shampoo etc.) because they are cheap and effective. By these types has the disadvantage that they will bind with the calcium-ions present in hard water, and thus loose their surfactant properties. In the end they precipitate and settle as soap-stone in pipes or the washing machine.
The ways to overcome this problem is either to add more surfactant to ensure a surplus, or add some chealant/binder in the product that binds the calcium stronger than the surfactant. Or you can off course choose another type of surfactant that are less sensitive to hard water.

No ! FrankS ! Anionic surfactants are not always reacting with Calcium ions to produce a precipitate in hard water. This reaction happens if the surfactant is the usual and well-known soap, produced by the action of soda on vegetable oils. This soap is made of oleate or other derivates from fatty acids, and it has the disadvantage that it binds with calcium in hard water, and settle as soap-stone in washing machines.
But this only happens with soap. With modern anionic detergents, this disadvantage disappears.

can you expalin the reason why carboxylate of soap have more affinity for cations than the sulfonates of detergents , it is also mentioned in wikepedia but i cant find the reason?

Hi Samardeep. No, not in detail. Maybe it relates to solubilities of inorganic carbonates vs. sulfates?


text:
Electrolysis using electrodes of metals whose ions are in solution
When we perform electrolysis of say an aqueous solution of silver nitrate with silver electrodes, silver ions, having a greater reduction potential than water, accept electrons and are deposited at the cathode. At the anode however, silver just leaves the electrode and enters into its solution.
The anode must just be a metal rod whose sea of electrons is much poorer than what it must be. Why does the silver get oxidized and leave the rod as ions? Water could get oxidized instead and that appears to be a more logical expectation.
Secondly, why does this not happen if we do the same reaction using platinum electrodes? What is preventing the metal from leaving the anode as an ion, moving to the cathode causing gradual thinning of one electrode and thickening of another?

The reason why silver dissolves (or water is oxidized) is well known. The reason why platinum does not dissolve is more elusive. Thanks to Maurice's comment for pointing this out.
Let's assume your solution is 1M in $\ce{AgNO3}$ and 1M in acid ($\ce{H+}$). The potential required to oxidize silver is 0.80 V; the voltage required to split water (oxidize it) is 1.23 V, but there are complications of overvoltage, mostly with hydrogen, but also with oxygen. To sum up: it is easier to grab an electron from silver than to pump another hydrogen ion into solution.
AHA! How about reducing the acid concentration, or not adding any at all? (Of course, the reason for adding the acid was to make the silver dissolve easier.) If the pH is increased so that hydrogen ion concentration is reduced to give a lower oxidation potential than silver, then water will be oxidized, hydrogen ion will be produced, and eventually, silver will become more easily oxidized. This electrolysis will go through a phase where hydrogen and silver are co-deposited, probably making a porous, weak deposit of silver.
For platinum to yield 2 electrons requires 1.188 V. Although this is less than required for oxidizing water, overvoltage effects due to oxygen and hydrogen are apparently important. Although platinum can be electroplated with a platinum anode, apparently the process fails to work well; older methods of platinum plating were kept refreshed with additions of platinum salts rather than relying on dissolution of the anode (Ref 1). More modern methods use a pulsed electropotential, but their success seems to be from a solving difficulties at the cathode rather than getting the anode to dissolve (Ref 2). They deposit a monolayer of platinum attached to a monolayer of hydrogen, then oxidize off the hydrogen and repeat. The electrode potentials strongly suggest that increasing the cell potential, if you had a platinum anode, would result in dissolving platinum metal, but the catalytic reactivity of platinum with respect to $\ce{H2O2}$ suggests that maybe the oxidation reaction occurs in the water layer, and the platinum acts merely as an electronic conductor.
Ref 1. https://www.technology.matthey.com/article/25/1/32-41/
Ref 2. https://www.chemistryworld.com/news/platinum-plating-at-the-flick-of-a-switch-/5720.article

hmmm... I didn't realise that silver oxidation was the thermodynamically favoured option. I was asking the question expecting that there would be a particular effect that occurred only when the same metal was used in the electrolyte as well as the electrode.If I were to choose a different solution, say, Copper Sulphate, would we still expect a silver electrode to get oxidised instead of water?

@Lucifer: Two silver electrodes in a copper sulfate solution (without applied potential) should be stable, because silver is less active than copper. When you apply a potential, copper is the only cation available, and so plates out at the cathode. The anode has the same situation as previously, except sulfate ion carries the current, not nitrate (which is a complication I did not discuss in the original answer). Under very acidic conditions, silver will dissolve; without acid, silver will dissolve - then go to the cathode and plate out preferentially.

When current flows in an electrolytic cell, the bulk solution has to remain electrically neutral. If one positive "ion" is deposited at the cathode as a neutral atom, another positive ion has to be generated. What is the source of the positive ion? The metallic anode.
Coming to your other question: What if the anode is a non-metal instead of any metal (an extreme case)? Say, it is made of graphite. In that case, water begins to oxidize, and a hydrogen ion is generated.

even in the silver nitrate case, water could get oxidized to maintain the balance. In the case of graphite, i understand that water has to get oxidized, but my question is concerning why the metal ion leaves here instead of that. and what is stopping such behavior on say a platinum electrode.

If I understand your question, you are saying that if we have Pt anode and a silver cathode. The electrolyte is silver nitrate. Why does Pt not dissolve?

yeah.why does Pt not dissolve while silver does

Probably nobody knows why Platinum does not dissolve at the anode.


text:
Why should duplicates compounds be removed from compound libraries before screening/activity prediction?
This is one of the first steps to be done while analysing chemical data or applying training models to predict chemical activity. However, I am unclear as to why is this done. Doe sit have a major effect on the result if duplicates are not removed? What is the main reason why we remove duplicate compounds?
Thanks

Let's consider a data set with three attributes for each compound: a chemical name, a CAS number (unique identifier for a compound) and a single empirically measured numerical interest, such as efficacy in lowering blood pressure in rats. If you had duplicate entries for a single experimental result, that could potentially bias the analysis. If you had two results for the same compound from two different experiments or measurements, you would surely want to include both in the analysis. If you had two results for the same compound (same CAS number) but named differently, you would want to keep them both in your data set, but use only the CAS number as the nominal variable for structure.

In any case it is well worth going over the data set with a fine-toothed comb to ensure that it is internally consistent, before running an analysis.


text:
Oxidation of ethanedioic acid with potassium permanganate

What are the products of the reaction of 1,3-butadiene with hot $\ce{KMnO4}$ (excess)?

I know that $\ce{KMnO4}$ will first cleave the double bonds and we should get
$$\ce{CH2=CH-CH=CH2 ->[KMnO4] 2CH2=O + O=CH-CH=O}$$
Now the aldehydes will further oxidise to
$$\ce{H2C=O <<=> H2C(OH)2 ->[KMnO4] O=C=O}$$
$$\ce{O=CH-CH=O ->[KMnO4] HOOC-COOH}$$
I don't understand how and why $\ce{HO2C-CO2H}$ would further oxidise to $2 \ \ce{CO2}$?

https://chemistry.stackexchange.com/questions/4310/what-is-the-reaction-between-oxalic-acid-and-potassium-permanganate Related

@Safdar, the link just provides the balanced equation. But the question is why does it react in the first place? Also what is the mechanism?

So, acc. to book, answer is 2 + 2 = 4 CO2?

@RahulVerma, yes that is correct. This question is from Solomon's Organic Chemistry 11e, Problem 13.20 (f), (Chapter 13, pg no 619).

The KMnO4 oxidation of oxalic acid mechanism is highly complex and is under debate.
This was the most recent paper I could find with possible mechanisms

@Safdar, is this reaction true for all general dicarboxylic acids with $\ce{KMnO4}$, or only specific to oxalic acid?

KMnO4 oxidations are complex in general.


text:
Caustic and Toxicological Determination
I need to determine whether or not a product that I am importing is "toxic" and or "corrosive" as defined below. But I am a chemistry n00b — that is to say I have virtually zero experience with and or knowledge of chemistry — which is why I am here. I am hoping that one of you guys can help me out.
Corrosive is defined as being

capable of inducing necrosis or ulceration of epithelial
tissue;
capable of causing an erythema or edema of the skin,
corneal or iris damage or conjunctival swelling or redness; or
let us not worry about this right now.

Toxic is defined as being

capable of causing a lethal effect on a human;
capable of causing a serious and irreversible but non-lethal effect on a human, such as a depressed level of consciousness, muscular weakness or paralysis, acute renal or hepatic failure, arrhythmia, hypotension, dyspnea, respiratory depression, pulmonary edema or optic neuritis; or
let us not worry about this right now.

Product Ingredients

Water
Water-soluble vinyl acetate polymer
Water-soluble natural wax
Water-soluble silicone
Natural glue

Please let me know if any information is missing or any other information is required. Also, please feel free to tag or re-tag the question however you see fit.
Thanks in advance for any help whatsoever.

@NilayGhosh no, i am not, but thanks anyways. i am trying to figure out whether or not the mixture is considered toxic and or corrosive according to the definitions that are provided.

@NilayGhosh i am too worried about the literal definition of corrosiveness, but rather the definition that i have provided in my question

As far as I can judge, your mixture is neither corrosive nor toxic. Water soluble vinyl acetate polymer is the scientific description of the popular paste, or white glue, that every pupil is using at school for  pasting images on sheets of paper. I don't know what is "Water-soluble natural wax", but it may be a sort of wax to make waxed paper or sealing. A silicone is a matter used to make water-proof joint or seal. The sum of all these stuffs should make a sort of cream or glue or anything similar which should not be corrosive or toxic.

@Maurice ohhh, wow, white glue. that was the one ingredient, that and the silicone, that worried me a little bit. thanks so much for your input. what i find weird is that, although water-soluble vinyl acetate polymer is white glue, the manufacturer also listed "natural glue" in the table of ingredients

Natural glue is probably a starch + water mixture. It may be used for stamps or for pasting labels on bottles.

@Maurice thanks. btw, you were indeed right that it is a liquid paste :)

The way to get the information you seek is to obtain a safety data sheet (SDS) for either your product or each of its ingredients.
Safety data sheets have slightly different content by jurisdiction, but most follow a common format, especially those that follow the Globally Harmonized system (GHS).
In a SDS that follows the GHS, Section II identifies the hazards of the substance using tightly defined common phrases. If the product or one of its compounds is toxic in some way or corrosive, Section II will have that information.
Section II contains "Hazard Statements" and "Precautionary Statements". Here is a SDS for acetic acid. There are three hazard statements:

H226 - Flammable liquid and vapor
H314 - Causes severe skin burns and eye damage
H402 - Harmful to aquatic life

There are numerous precautionary statements. The Hazard statements provide the information you want. Here is a list of GHS hazard statements.

i have the HS number of the good. i also have an SDS, although I am not sure how reliable it is since the specific regulations that are applicable in the country of origin are likely not applicable here. but its structure does seem to resemble that which is recommended by the GHS. however, the section 'Hazards Identification' simply states, "no special hazards", which seems especially vague lol. it certainly does not contain any GHS statements, so is that sufficient to conclude that it is neither corrosive or toxic according to the definitions above?

additionally, other than where it states "N/A" in the 'Toxicological Information' section, it states "Skin irritation acute: Direct contact may cause irritation" and "Eye irritation acute: Direct contact will cause irritation"


text:
PE expression for vibrational spectra of molecules using Newton's second law
I am currently studying the textbook Infrared and Raman Spectroscopy: Principles and Spectral Interpretation, second edition, by Peter J. Larkin. Section 10. Calculating the Vibrational Spectra of Molecules, says the following:

In the case of vibrational spectroscopy, the polyatomic molecule is considered to oscillate with a small amplitude about the equilibrium position and the PE expression is expanded in a Taylor series and takes the form:
$$V = V_0 + \sum_{i = 1}^{3N} \left( \dfrac{\partial{V}}{\partial{q_i}} \right)_e d q_i + \dfrac{1}{2} \sum_{i = 1}^{3N}\sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e d q_i d q_j + \dots$$
The above expression is expressed in internal coordinates, $q_i$ and $q_j$ which are directly connected to the internal bond lengths and angles. The above expression is simplified since:

The first term $V_0 = 0$ since the vibrational energy is chosen as vibrating atoms about the equilibrium position.
At the minimum energy configuration the first derivative is zero of definition.
Since the harmonic approximation is used all terms in the Taylor expansion greater than 2 can be neglected.
This leaves only the second term in the PE expression for $V$. Using Newton's second law the above is expressed as
$$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$

Even after taking the points above into account, I don't see how we get from the first equation to the second. I'm assuming that $V$ is the potential energy (PE), right? And Newton's second law in differential equation form is $F = m \dfrac{dV}{dt}$. So how does this and points 1., 2., and 3. then lead to $$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$?
I would greatly appreciate it if people would please take the time to explain this.

It's probably simplest to start with the familiar example of the 1D harmonic oscillator. Here
$$F=-\frac{dV}{dx}=-\frac{d\left(\frac{1}{2}kx^2\right)}{dx}=-kx\tag{1}$$
which is often called Hooke's law, $F=-kx$, where x is the displacement from the equilibrium position, and the potential is assumed quadratic (parabolic) in the position coordinate. The derivation above requires the substitution $V(x) = \frac12 kx^2$, which can be obtained by trimming a Taylor expansion of the potential about the energy minimum (assumed = 0) at the quadratic term, and the force constant is defined as
$$k=\left(\frac{d^2V}{dx^2} \right)_e\tag{2}$$
Also,
$$F=ma=m\frac{d^2x}{dt^2}\tag{3}$$
If we transform to mass-weighted coordinates (or equivalently set m=1), we therefore can write that
$$\frac{d^2x}{dt^2} = -\frac{dV}{dx}=-\left(\frac{d^2V}{dx^2} \right)_e x\tag{4}$$
The derivation in your book does essentially the same thing. Note that since the potential is assumed quadratic and the other listed conditions are imposed, the potential can be written as
$$V = \frac12 \sum_{i = 1}^{3N}\sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j \tag{5}$$
The double summation can be collapsed into a single one when you take the derivative with respect to a spatial coordinate, say $q_k$:
$$\begin{align} \dfrac{d^2 q_k}{d t^2}  &=-\left( \dfrac{\partial{V}}{\partial{q_k}} \right)  \\&=-\left(\dfrac{\partial{}}{\partial{q_k}} \left[  \frac12 q_k^2  \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e   + \frac12 q_k\sum_{j \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e  q_j  + \frac12 q_k \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i  + \frac12 \sum_{i \neq k }^{3N}\sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_j}} \right)_e q_i q_j  \right] \right)  \\ &= -\left( \left[ \left( \dfrac{\partial^2{V}}{\partial{q_k^2}} \right)_e  q_k  + \frac12 \sum_{j \neq k}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k}\partial{q_j}} \right)_e  q_j  + \frac12 \sum_{i \neq k }^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i}\partial{q_k}} \right)_e q_i    \right] \right)    \\ &= - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k} \partial{q_j}} \right)_e q_j \end{align}$$

Thanks for the answer. What are "mass-weighted coordinates"? And where did the $\dfrac{d^2 x}{dt^2}$ come from?

From equation 3 by setting the mass=1 (more correctly by transforming to mass-weighted coordinates). The latter combine mass with the displacement coordinate (normal mode). The substitution removes explicit mass terms from the equations and makes things look a little tidier. See for instance also http://gaussian.com/wp-content/uploads/dl/vib.pdf

Hmm, I see. What happened to the sum for $i$ in $$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$? It only has the sum for $j$.

I agree it's a tad unclear, I think it would be better to take the derivative wrt say $q_k$, but in the end if you do that you can swap labels $k\rightarrow i$ and you obtain the same result.

In other words: $$\dfrac{d^2 q_k}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_k}} \right) = - \sum_{l = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_k} \partial{q_l}} \right)_e q_l$$ where I've used new labels to make things clearer (I hope)

I'm very confused. Are you saying that the reason the other summation disappears is somehow because of the derivative? I don't see how this explains why the other summation disappears. In (5), we are summing over both internal coordinates $i$ and $j$; but in $$\dfrac{d^2 q_i}{d t^2} = - \left( \dfrac{\partial{V}}{\partial{q_i}} \right) = - \sum_{j = 1}^{3N} \left( \dfrac{\partial^2{V}}{\partial{q_i} \partial{q_j}} \right)_e q_j$$, we are summing over only one of the internal coordinate (in this case, denoted $j$).

My confusion is with what happened that the author went from two summations, over all of the internal coordinates, to a single summation, over just one of the internal coordinates.

I have included an explicit derivation that shows how the double summation collapses.

Ahh, yes, it is now all clear to me. Thank you very much for taking the time to explain this to me.


text:
Cis-trans isomerism with electron pairs
Is it possible to name compounds with lone pairs on the double bonded species using the cis-trans nomenclature? For instance, can't you name $\ce{Ph-N=N-Ph}$ by the relative position of the nitrogen atoms' electron pairs?

Phenyl groups are cis or trans and that's it.

Related: https://chemistry.stackexchange.com/questions/88240

yes it is. See the Azobenzene wikipedia page for more details.

Yes.
Azobenzene displays cis-trans isomerism. The two isomers can be interconverted using UV/Visible-irradiation. The less stable cis-isomer will relax back to the more stable trans-isomer, but the lifetime of the cis-isomer is long enough to allow observation and characterization.

is the fact that electron pairs and alkyl groups are the high-priority groups only because they're just...defined that way? Also, since you can't really use CIP to classify isomers with electron pairs, can you drop a link to a resource specifying the relative priorities of electron pairs and other such groups, if there's any list of that sort?

Actually, most of linear imines and exocyclic imines display $E/Z$-isomerism (commonly known as $cis/trans$-isomerism) with a lone pair on the nitrogen, which is a part of the double bond. The isomerism follows various mechanisms. For instances, $cis/trans$-isomerism of azobenzene a noted elsewhere is operated by UV/vis-irradiation is a good example. The mechanism of this transformation has been depicted as follows (Ref.1):

Similar to cis/trans-isomerism in alkenes, extensive studies has been done on azo-compounds and imines. For example, acid-catalyzed $E/Z$-isomerization of imines has been reported frequently (Ref.2,3). The abstract of Ref.2 states that:

The kinetics and mechanism of acid-catalyzed Z/E isomerization of O-methylbenzohydroximoyl chloride (1Za and 1Ea), methyl O-methylbenzohydroximate (1Zb and 1Eb), ethyl O-methylbenzohydroximate (1Zc and 1Ec and five para and meta substituted derivatives), O-methylcinnamohydroximoyl chloride (2Za and 2Ea), and methyl O-methylcinnamohydroximate (2Zb and 2Zb) have been investigated. The kinetics of Z/E isomerization of these imines have been studied in glacial acetic acid (1Ea and 1Zc) and in dioxane solutions containing $\ce{HCl}$, trifluoromethanesulfonic acid, or tetrafluoroboric acid (1Ea, 1Zb, 2Ea, and 2Zb). The isomerization takes place by either (a) rotation about the carbon−nitrogen double bond of the protonated imine (iminium ion rotation) or (b) nucleophilic attack on the protonated imine to form a tetrahedral intermediate that undergoes stereomutation and loss of the nucleophile (nucleophilic catalysis). The hydroximoyl chlorides 1Ea and 2Ea only isomerize by the nucleophilic catalysis mechanism. The hydroximate 1Zb appears to be capable of isomerizing by either mechanism. The hydroximate 2Zb may be isomerizing only by iminium ion rotation. Theoretical calculations support the notion that increased conjugation in the protonated imine increases the rate of iminium ion rotation.

Photochromic properties of a series of novel pyrrolidene imines bearing functionalized aryl or naphthyl moieties was studied by UV spectroscopy (Ref.4):

UV irradiation of these heterocyclic Schiff bases at room temperature promotes the trans–cis photoisomerization of the $\ce{C=N}$ double bond with formation of a variable amount of the cis-isomer that, in the absence of light, returns thermally to the original form in few seconds. The thermal cis–trans re-isomerization of these molecular switches is much more slower than for the common benzylidene aniline-type imines, allowing the observation of the photochromic phenomena at room temperature.

Another interesting studies have been done with indigo derivatives (Ref.5) noting that:

Substituted indigo derivatives undergo photoisomerization of the central double bond if both nitrogen atoms are functionalized. Indigo itself however does not photoisomerize because of a competing and highly efficient excited‐state proton transfer...

References:

H. M. Dhammika Bandara, Shawn C. Burdette, "Photoisomerization in different classes of azobenzene," Chem. Soc. Rev. 2012, 41(5), 1809-1825 (https://doi.org/10.1039/C1CS15179G).
James E. Johnson, Nora M. Morales, Andrea M. Gorczyca, Debra D. Dolliver, Michael A. McAllister, "Mechanisms of Acid-Catalyzed Z/E Isomerization of Imines," J. Org. Chem. 2001, 66(24), 7979–7985 (https://doi.org/10.1021/jo010067k).
James E. Johnson, Nancy M. Silk, Elizabeth Ann Nalley, Mohammed Arfan, "Mechanism of an acid-catalyzed geometric isomerization about a carbon-nitrogen double bond," J. Org. Chem. 1981, 46(3), 546–552 (https://doi.org/10.1021/jo00316a013).
Paulo J. Coelho, M. Cidália R. Castro, M. Manuela M. Raposo, "Reversible trans–cis photoisomerization of new pyrrolidene heterocyclic imines," Journal of Photochemistry and Photobiology A: Chemistry 2013, 259, 59-65 (https://doi.org/10.1016/j.jphotochem.2013.03.004).
Ludwig A. Huber, Peter Mayer, Henry Dube, "Photoisomerization of Mono‐Arylated Indigo and Water‐Induced Acceleration of Thermal cis‐to‐trans Isomerization," ChemPhotoChem 2018, 2(6), 458-464 (https://doi.org/10.1002/cptc.201700228).

is the fact that electron pairs and alkyl groups are the high-priority groups only because they're just...defined that way? Also, since you can't really use CIP to classify isomers with electron pairs, can you drop a link to a resource specifying the relative priorities of electron pairs and other such groups, if there's any list of that sort?

if available, please respond.


text:
Calculation of free energy from potential and kinetic energy
I am having trouble understanding how (and even whether it's possible) to get free energy from kinetic and potential energy.
I have run molecular dynamics simulations (NPT ensemble, CP2K software), but this is a basic thermodynamics question.  There's a results file called "energy."  The file contains energy-related output: temperature, kinetic and potential energy, and a "conserved" quantity.
(The "conserved quantity" is the sum of the kinetic and potential energies and also the thermostat energies).
From courses a long time ago and reading I have done online, I know that
kinetic energy + potential energy = total energy

I think "total energy" would be the internal energy.  It would be great if someone could please confirm/debunk that--I don't know if the situation is distinct for distinct ensembles, for example.  Whatever the case, I don't know how it relates to the free energy.  I would like to know if I have enough information to calculate the free energy.
If anyone could please point me in the right direction, I would appreciate it.  I can't find anything online and maybe that's because I don't even know where to start looking.

This question might fit better on Matter Modeling SE, though you should include what program you are using.

@Tyberius, thanks.  I'll look for that forum.  If it is not OK to leave this question up, please feel free to let me know.

The question is fine here, I just think you might have better luck with this sort of question there (here is the link). If you do wind up moving your question there, make sure to delete it here. The SE network generally discourages having the same question posted multiple places.

Thanks.  I did try posting it there and it won't post for some reason.  There are no errors.  So I guess I'll leave it here.

Thats strange, it wasn't just a review comment (like needing to add tags or the title containing some blocked words)? I've occasionally had issues on SE with a mobile browser, but otherwise I don't know what the issue would be.

No, I only added a sentence that I had first posted it here.  I'm not sure where the thermodynamicists like to hang out, so it's fine to leave it here.  If no one is interested, I guess I'll ask the moderators for help at Matter Modeling.  Thanks.

I'm actually one of the mods there, so if you do feel like moving, it let me know. If you leave off that sentence, we actually have a tracking tag we are using to mark question moved from other sites. The site is still kind of in its early days, so it always helps to get more questions. Plus, for whatever reason, Chem SE has tended to attract more electronic structure people than molecular dynamics. Matter Modeling, at least so far, has had a little more luck attracting MD researchers.

Thanks very much, Tyberius.  That's helpful.  I often notice that the people who really understand thermodynamics are not necessarily modeling people, so it might not hurt to leave it here for a bit.  I have a lot of trouble finding resources that relate atomistic simulation to basic thermodynamics.

Since the Gibbs free energy is G=H-TS, you would need to know the entropy of the system. Other than that it looks like you should be able to compute the enthalpy as H=E+pV, so that G=E+pV-TS. I agree, internal energy is the total energy here. As @Tyberius suggested it might be a good idea to include the software you are using.

@BuckThorn, thank you very much.  Unfortunately, I don't think I have the entropy but I will look more carefully at my results in case I have missed something.  The software is "CP2K."  I'll edit the question.

The sum of the potential and kinetic energies is the internal energy, E (or U).
However, you don't necessarily need to know the entropy to calculate the free energy.  If you perform an equilibrium molecular dynamics simulation, you can derive the free energy from the equilibrium constant.
See, for instance:
https://onlinelibrary.wiley.com/doi/full/10.1002/jcc.21776
If you are modeling a constant T and V system, where there is no work done, the equilibrium constant would give you the Helmholtz free energy:
$$\Delta A = - R T ln K^{'}_{eq}$$
where A = E - TS
If you are modeling a constant T and p system, where you have no non-pV-work, the equilibrium constant would give you the Gibbs free energy:
$$\Delta G = - R T ln K^{''}_{eq}$$
where G = H - TS = E + pV - TS
The reason for this is that, at constant T and V, with no work term, minimizing A_sys corresponds to a maximization of S_univ, and thus determines the equilibrium condition. By contrast, at constant T and p, with pV-work only, minimizing G_sys corresponds to a maximization of S_univ, and thus determines the equilibrium condition.
I've marked the equilibrium constants as prime and double-prime to indicate that these would be numerically different.

This is very interesting and helpful.  Thanks.  I thought about it today and now think I understand what's going on. I asked my question because information on the software I'm using indicates that I can't calculate the free energy (Gibbs or Helmholtz) without providing a further constraint (on the species present and their reactions) and re-running my simulations. I couldn't figure out why such a constraint would be necessary. From what you said, it's because I have multiple Keq associated with multiple species and need to determine the Keqs accurately to get accurate free energy.


text:
What is the origin of rotational energy barrier in ethane?
What is the origin of rotational barrier in ethane?
In most organic chemistry textbooks, we learned that it's originated from steric repulsion between two methyl groups. But recently I heard that there also exists other perspective which explains rotation barrier using hyperconjugation model.
So I searched and read many papers about the origin of rotational barrier in ethane and I notice that there are two competing models for explaining this rotation barrier.
But there are no significant conclusions and it still remains controversy "what is the primary energy term in rotation barrier?".
Because many authors suggested many calculations to prove what is mainly contributed to rotation barrier and there results for supporting there argument, but each calculation is based on different approximations and theories. (e.g. NBOs are not optimal --> hyperconjugation energy is overestimated)
So my question is that "Is there no conclusion? What factor mainly contributes to ethane rotation barrier?"

As you say, this issue appears to be unsettled. I find the argument in favor of hyperconjugation compelling, but there is theoretical evidence that steric strain contributes as well.
This debate is about 20 years old. Here are two papers summarizing the opposing views:

Pophristic, V.; Goodman, L. (2001). "Hyperconjugation not steric repulsion leads to the staggered structure of ethane". Nature. 411 (6837): 565–8. doi:10.1038/35079036.
Bickelhaupt, F.M.; Baerends (2003). "The case for steric repulsion causing the staggered conformation of ethane". Angew. Chem. Int. Ed. 42 (35): 4183–4188. doi:10.1002/anie.200350947.

A more recent paper that has an interesting take:

Baranac-Stojanović, M. (2015). "Theoretical analysis of the rotational barrier in ethane: cause and consequences". Structural Chemistry. 26 (4): 989–996. doi:10.1007/s11224-014-0557-5.

I cannot read the entire article, but the abstract offers the conclusion that hyperconjugation is responsible for the initial energy rise during rotation, but that steric factors are needed to fully explain the high energy of the eclipsed conformation.

However, the same Wikipedia article also states that the situation is not so clear cut with others suggesting that more involved MO calculations show that steric effects are most probably the cause. se Bickelhaupt, et al. Angew. Chem. Int. Ed. 42 (35): 4183–4188.

@porphyrin I edited my answer to properly portray the controversy, and provided a summary to a more recent article that suggests the correct answer is both hyperconjugation and sterics.

Another paper on this subject: https://www.dtc.umn.edu/publications/reports/2007_28.pdf


text:
Beckmann rearrangement v/s dehydration of Oximes

In the above question, A is an oxime, which is then reacted with phosphorus pentoxide. Now, phosphorus pentoxide is both a dehydrating reagent and a reagent used for Beckmenn rearrangement, but which reaction will form the major product mostly, ie, which reaction will dominate? My teacher told to consider beckmenn rearrangement with Phosphorus pentoxide, but the answer of this question is A, that is, dehydration is favored . What I think from my experience is that aldoximes will favor dehydration, and ketoximes beckmenn rearrangement( dehydration obviously cannot  occur in them), but I am not sure. Please tell the conditions under which any reaction will dominate over other. Also what  would be the answer if phosphorus pentoxide is replaced by phosphorus pentachloride?

The aldoxime will tend to dehydrate with P2O5 or PCl5. No Beckmann.

Or, A is the Beckmann product with aldoximes.

See beckmann fragmentation.

Do you think that the mechanism of "Beckmann" and "dehydration" are different?
Answer is No. They're essentially the same reaction. Aldoximes undergoes Beckmann rearrangement to form nitriles, or you can say "they dehydrate" (as hinted by Ben Norris in comments)
The mechanism for Beckmann rearrangement is as follows,

(Source: Wikipedia, Beckmann rearrangement)
Note: Just replace $\ce{R1}$ with $\ce{H}$ and then, in the intermediate (upper one) after second step, the arrow will be from $\ce{N-R1}$ bond to $\ce{N}$. That's dehydration!
Also, if you replace $\ce{P2O5}$ by $\ce{PCl5}$ or $\ce{H2SO4}$, the answer would be same for all.

The middle arrow in the intermediate is going in the wrong direction. 2) If you want phosphate to deprotonate intramolecularly, then draw the anti-oxime. 3) This is a dehydration, not a rearrangement. Atoms are not being reconnected as in the Beckmann rearrangement.

@user55119: due to limited search via Google, I can't find a standard mechanism for dehydration of aldoximes. Hence, I've edited it with Beckmann rearrangement.

What about Beckmann Fragmentation/abnormal Beckmann?

The top two arrows in the second  structure need work. R1 with its electron pair migrates to nitrogen as water is lost. Then the electron pair on nitrogen can stabilize the carbocation.

@user55119: "...Then the electron pair on nitrogen can stabilize the carbocation." - the intermediate which you're talking about is the lower one in square brackets. No?

@Safdar: you can see here for fragmentation. I think that's a different reaction.

No, the structure preceding the brackets. What is in the brackets is fine.