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Stability comparison between tropylium, cyclopropylmethylium and dicyclopropylmethylium cations

Write the correct stability order of the following compounds:

I don't have any concern with estimating stability of (1), only (2), (3) and (4). Our teacher said that unless the carbocation has three isopropyl groups, it won't be as stable as the quasi-aromatic compound — in this case, the tropylium cation.
My friends claim the isopropylmethylium cation should be more stable than the tropylium cation.
Is there any experimental data saying which one is more stable? How many cyclopropyl rings are needed to “dominate” over a tropylium cation?
Previously, I reviewed the following questions about triisopropylmethylium carbocation:

Stability Comparison between the Tropylium and Tricyclopropylcarbinyl Carbocation
Why does cyclopropane react with bromine?

There are quite a few quantitative ways to measure the stability of carbocations (you could have a brief image of them if you look at March's Advanced Organic Chemistry or F.A.Carey's organic chemistry).
In this case, I'll take HIA (Hydrogen Ion affinity) as the criterion. The lower the HIA, the more stable it is.
Data below are from [1] if not mentioned. All of the values are in kcal/mol.
(2)'s HIA is 201, 200, 238 (Last one could be taken as an outlier, as [2] gives a value of 199, consistent with the former two results.)
(3)'s HIA is 241.
I failed to find the exact HIA for (4) in essays, but the HIA for 1,1‐dicyclopropylethylium is 207. Taking the stabilization of an extra methyl into consideration, (4) is less stable than (2). As (4) is obviously more stable than (3), the stability is (2)>(4)>(3).
[1] D.H.Aue and M.T.Bowers, Gas Phase Ion Chemistry, M.T.Bowers, ed., Academic Press, New York, 1979.
[2] Würthwein, E.-U., Lang, G., Schappele, L. H., & Mayr, H. (2002). Rate-Equilibrium Relationships in Hydride Transfer Reactions:  The Role of Intrinsic Barriers. Journal of the American Chemical Society, 124(15), 4084–4092. doi:10.1021/ja0121540


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Why the IUPAC naming of this compound is 6-hydroxycyclohex-2-en-1-one?

Should it not be 2hydroxycylohex-5-en-1-one?
Functional groups are usually given more priority than double/triple bonds, except in cases of Nitro, halides, etc.
Shouldn't the priority order be Ketone>alcohol>double bond for assigning locants. Ketone gets the least, then asigning rest in clockwise fashion with-OH at 2 and double bond at -5.
After Ketone, alcohol should be given the priority, not double bond,

eg  (CH3-CH=CH-CH(OH)-CH3) is pent-3-en-2-ol, not pent-2-en-3-ol.

Shouldn't the same logic apply to this compound too?

"Shouldn't the priority order be ketone>alcohol>double bond for assigning locants?" No.
"Shouldn't the same logic apply to this compound too?" Yes, and that's why 6-hydroxycyclohex-2-en-1-one and pent-3-en-2-ol are both correct.

"Functional groups are usually given more priority than double/triple bonds" - citation needed! You're probably talking about suffixes, not functional groups (hence your exclusion of nitro and halides which are not suffixes), but here 'hydroxy' is not a suffix.

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Which way is this coordination equilibrium likely to lie (entropic vs enthalpic factors)?
The question is asking where this aqueous equilibrium is likely to lie.
My thought process is that, because Li+ is a smaller cation than Na+, with a greater charge density, upon displacement it will order solvent (water) molecules around itself more than Na+. This would result in the forward reaction likely having a positive entropy change, making it unfavourable. This means that the equilibrium would likely lie to the left.
On the other hand, I can see the argument being made that Li+ is so small that it is unlikely to coordinate well to the O and N lone pairs, meaning the M-L bonds on the left hand side of the equilibrium equation will likely be weak and easily broken in favour of stronger Na-O and Na-N bonds. This would mean that the equilibrium would likely lie to the right.
I have a suspicion that the entropic considerations outlined in the first paragraph are more important in determining the direction of equilibrium (and outweigh the enthalpies factors in paragraph 2) but just wanted to confirm if this is the case?

There's no need for some lengthy analysis - Li is simply too small to fit well - that's like the main point of crown ethers and cryptands - they are size selective.

@Mithoron ok got it thank you very much. Would entropic considerations be of any significance here though, or is this purely driven by enthalpy (strength of M-L bond)?

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What can visual observations of pi-bonding MOs explain about LCAO expansion coefficients?
For examples of CO and N2, what can their MOs tell us about their expansion coefficients?
I've thought that in N2, its symmetric so the expansion coefficients would be equal from the pi-bonding MO due to the symmetry as shown in the image below.

And for CO, the coefficients could be different with the oxygen having less character due to it being more electronegative?

I'm thinking like this because I know each AO of an atom is multiplied by an expansion coefficient to build MOs. But I don't entirely know what the expansion coefficient is or where it really comes from so I can't answer the question fully. Any help is appreciated.

I've thought that in N2, its symmetric so the expansion coefficients would be equal from the pi-bonding MO

Yes, that's right; symmetric molecules have symmetric MOs which have symmetric coefficients of the constituent AOs.

And for CO, the coefficients could be different ...

And yes, conversely for unsymmetric molecules, the coefficients are going to be different.

... with the oxygen having less character due to it being more electronegative?

This is where it starts to get rather murky.
In this specific case, where the π MOs are formed from carbon and oxygen 2p orbitals only, then you can say that the bonding π MO has more oxygen character (not less!). This is because the oxygen 2p orbitals are lower in energy, so the bonding MO (which is lower in energy than antibonding) will take on more of the oxygen 2p than the carbon 2p. On the other hand, the π* antibonding MO will have more carbon character.
(You could say that this is because of electronegativity, but I think that effective nuclear charge is a slightly better rationalisation. That said, they're all interrelated, of course.)
However, this line of logic only works when you have the case of two AOs combining to form two MOs. The moment you have more than two, you need to be very careful about claiming which one has greater contribution. For example, the σ framework of the same molecule, CO, turns out to be pretty complicated, and it's not obvious at all which combination of carbon 2s, carbon 2p, oxygen 2s, and oxygen 2p forms any given MO.
Also, even in the 2 AOs/2 MOs case, the electronegativity argument must be applied with caution. We could say above that oxygen 2p is lower in energy than carbon 2p, but if you're not comparing the same orbitals, then there's no guarantee that the more electronegative element has orbitals that are lower in energy.
Generally, if you wanted to make definitive statements about which atom had a greater contribution, you'd have to actually use a computer programme to calculate these actual contributions. This is known as electronic structure theory, which is a fairly prominent subfield of computational chemistry. If you don't have something like this, then the conclusions you can draw in a qualitative fashion are fairly limited.

Orthocresol, this has nothing to do with the post, but I am unable to find it in Stack Exchange. How do you make those vertical lines, and the text citing the OP at the right, while writing your answer? Thanks in advance.

@MetalStorm They're blockquotes: https://chemistry.stackexchange.com/editing-help#simple-blockquotes If you're new to Markdown then the other pages on there will be useful too. Alternatively, you can click 'edit' on my post to see the original source code that I typed. (If you don't save it then nothing will be edited.)


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Why is AChE inhibited by triflates not susceptible to reactivation?
According to a report, certain esters of trifluoromethanesulfonic acid are capable of acting as irreversible inhibitors of acetylcholinesterase. Also, 3-PAM, while capable of reactivating AChE inhibited by methanesulfonates, will not reactivate AChE inhibited by a triflate. Why is this the case? Does the size of the fluorine atoms sterically hinder the approach of the reactivator? Is the adduct simply unreactive? Does it age, and if so, what is the aging mechanism?


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What results from the neutralization of copper citrate with Na2CO3
I found this question that describes PCB etching with citric acid. The answer to that question explains how hydrogen peroxide, salt, and citric acid can be used to ionize copper metal in water, which provides:
$$\ce{Cu + H2O2 + 2 H+ -> Cu^{2+} + 2 H2O}$$
The equation above excludes the citrate ion, so maybe it isn't party to the reaction.  This observation prompted the following:
Question:

Assuming the citric acid ion is $\ce{C6H7O7^-}$, what is the equation for the neutralization reaction (I kept $\ce{H^+}$ in this, but it's been a while since highschool/college chem, this may not be represented right):

$$\ce{Cu^{2+} + C6H7O7^- + H^+ + Na2CO3 -> } ??? $$

What remains ionized in solution after neutralizing with $\ce{Na2CO3}$?
Will there be a precipitate? If so, what?

(This will inform our choices for responsible disposal.  I'm not sure what other details to provide, as this seems like a pretty simple question.  Let me know if you need more information.)

You don’t dispose of any chemicals down the drain. Check your municipal hazardous waste disposal folks.

You would prepare a variant of Benedict reagent, copper kept in stable citrate complex. If you added glucose and boil, the blue solution would start to form yellow to brick read insoluble cuprous oxide.

Copper ion forms with citrate stable chalate, that is not disturbed even by addition of sodium carbonate, what is used in Benedict's reagent(*) for detection of reducing sacharides.
Copper etching:
$$\ce{Cu(s) + H2O2(aq) + H4Citr(aq) ->  2 H2O(l) + CuH2Citr(aq)}$$
Solution neutralization:
$$\ce{2 CuH2Citr(aq) + CO3^2-(aq) -> H2O(l) + CO2(g) + 2 [HCuCitr]-(aq)}$$
$$\ce{[CuHCitr]-(aq) + CO3^2-(aq) -> HCO3-(aq) + [CuCitr]^2-(aq)}$$
Copper can be eliminated by reducting agents, like glucose, or perhaps by sulphite/bisulphite, forming insoluble copper(I) oxide.

(*) Benedict's reagent  is a deep-blue aqueous solution. Each litre contains:
$\pu{17.3 g}$ copper sulfate
$\pu{173 g}$ sodium citrate
$\pu{100 g}$ anhydrous sodium carbonate or, equivalently, 270g sodium carbonate decahydrate

Does no positive ion exist in solution after neutralization?  I ask because (perhaps this is where my chem knowledge breaks down) I had always thought you needed both positive and negative ions in the solution.  You wrote $\ce{-> HCO3-(aq) + [CuCitr]^2-(aq)}$ and both ions are negative.  So is the result really a solution of only negative ions?

There are no solutions with only negative ions. Positive ions like Na+(aq) are called here bystanding/spectator ions, not taking part in reactions, just maintaining electroneutrality.

So the NaCl that was added to the original etching solution remains Na+ and Cl- through the entire process, from etching to neutralization?

True for Na+. Cl- takes part in parallel reactions of oxidation to Cl2 by H2O2, reaction of Cl2 with Cu and forming weak complex with Cu^2+.

So does Cl2 remain in solution, or can it come out as Cl2(g)?

I case of concentrated solutions of HCl  + H2O2, Cl2(g) would be formed. In our case it would stay as minor product of side reactions.


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Looking for terminology for when a hygroscopic chemical gets moisture in it
Scenario: A jar of zinc chloride (which is very hygroscopic) is left on a shelf in a chemistry classroom for several years.  You open it and it is hard.  What is the word to describe that?  It was ruined by moisture in the air leaking into the bottle.  So basically, what is the past tense of something hygroscopic that has been exposed to moisture? I've been searching for a bit, can't find it, and it is driving me crazy. Thanks in advance.

I am afraid that there is no single word to describe this process because it must a multistep process with a lot of complications. Hygroscopic substances can become deliquescent, which means these solids continue to absorb moisture from air till they become liquids or form a saturated solution. That saturated solution, if left undisturbed, can form large hydrated crystals from slow evaporation, or perhaps it hydrolyzed over time in a lab environment to other oxychloride.
NaOH pellets behave similarly. They become liquid, form a coating of sodium carbonate and then those pellets fuse and recrystallize. That is also a hard mass.
The verb is "deliquesce".

I assume this addresses just the extreme hygroscopicity that manifests as deliquescence, what is not the OP case. The Czech informal lab speech uses for such a moisture triggered sintered state the adjective "spečený", what means literally "baked up together"

@Poutnik, I think sintering involves heating. Let us see the dictionary meanings. sinter: A hard incrustation or deposit formed upon rocks, etc., by precipitation from mineral waters; esp. siliceous sinter. (2) Material which has been subjected to sintering; spec. iron ore prepared for smelting by sintering the powdered material, usually together with coke and other materials; So sintering does not apply.

@AChem  I know what sintering means. I should have used moisture triggered "sintering", for nonavailability of better word. // Hygroscopicity has a wide range and deliquescence is just at one of its end.

The word for powder that turned into a lump is "caked". The process is "caking". Anticaking agents are used to prevent this.

Caking may not involve deliquesce and it is usually used for surfaces. For example, Oxford English Dictionary defines "cake" as a verb "To form, harden, or coagulate into a (usually flattish) compact mass; to form into a thick crust or deposit, esp. on a surface. Also with together." If the OP shared a picture, that would have been the best!

@AChem The OP described rather "thick crust" than saturated solution for his jar of zinc chloride.


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Watercolor pigment conundrum - can a change in ionic profile induce flocculation?
A client of mine has installed an ion exchange water purifier to de-harden her tap water. After that, however, she noticed that it has negatively affected her watercolor painting. From what she describes, it seems that the pigments have flocculated. I have a theory of what may have caused it, but I wish to run it by the community here/inquire about any alternative theories.
The ion exchange filter captures Ca2+ and Mg2+ and exchanges them for Na+, while carbonate and sulfate ions slip through. The original hard water had in other words the presence of partially dissociated CaCO3/CaSO4/MgCO3/MgSO4 species. The ion exchange of Ca2+ and Mg2+ for Na+, however, leaves a different ionic profile consisting of Na+, carbonate, and sulfate ions. This resembles the ions present when N2SO4/N2CO3 dissociate completely, forming a strong electrolyte solution.
If that is correct and we assume that the new water behaves like a strong electrolyte solution compared to regular hard water, then the electrical double layer surrounding the pigment particles ought to be affected --> electrostatic repulsion decreases --> pigments flocculate. However, the DVLO theory states that it is the valency of the electrolytes present and their concentrations that affect particle dispersion, not whether the solution is considered a strong or weak electrolyte. Therefore I am not entirely sure whether I am on the right track.
Am I on the right track thinking that it is the change in electrostatic repulsion that causes my client's watercolor problems or is there something else I am missing? If it is the electrostatic repulsion that is affected, then how exactly?

What you state seems reasonable: salting-out can be used to precipitate some substances. If it were tempera paint, you could point out that proteins in egg albumin might precipitate, but I don't know about water color pigments.
It would be easy to test, though. Mix up batches of paint with a few choices of solvent, perhaps:

Distilled water (not just deionized water)
The client's treated tap water
The client's untreated tap water, if readily available, perhaps from a garden hose that bypasses the ion-exchange filter
Distilled water with specific amount(s) of the salt used in the purifier added.

She should be party to these tests, and would not only learn what solvents are acceptable for her use, but could help others if that were reported here as an answer!


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Converting Quartz data from one (P3221) space group to another (P3121)
For my studies, I’ve done a lot of work on coal fly ash QXRD QPA on TOPAS which has been a fun and interesting challenge. Apart from theory reviewing, a lot of my learning so far has come from a prior colleague of mine who had done an extensive amount of fly ash QPA.
One thing I’ve noticed in my prior colleague’s work is that he has occasionally gotten quartz .cif/.str files with a spacegroup of P3221 (left-handed quartz) and then, when loading it into a .inp file, seemingly changed the spacegroup to P3121 (right-handed quartz). Originally, I wasn’t sure why he would do this (and simplistically thought the simple right vs left-handed geometry would correct itself in a refinement) until I ran a refinement with the original .str file and had a big intensity error with my quartz. Since then, I’ve realised the massive difference alternate site positions can have which are seemingly mitigated when switching to the alternate space groups (which I presume are the quartz found in the original sample). From this, I’ve been trying to develop a model to easily convert between spacegroups as many quartz .cif/.str files I’ve seen have the former spacegroup and, obviously, I don’t want these refinement errors when I’m trying to get meaningful results out.
So, from the above, my questions are the following:
1.  Apart from significant intensity errors, how can the suitability of left vs right-handed space groups (such as with quartz) be noticed during a refinement?
2.  Is there a general rule when incorporating crystal enantiomorphs as to which ‘hand’ should be applied or should it be determined empirically when refining a given diffractogram?
3.  Are the amended site positions correct? If so, or not, can anyone provide a reference so that I can create a model to do this conversion on Python (or JEdit) before doing a refinement?
a)  The original .cif file (PDF database entry no. 000461045) had the following data:
Spacegroup: P3221

site Si_1   x 0.4723   y =0;   z =2/3;   occ Si+4 1.0   beq  0.59533
site O_1   x 0.416   y 0.2658   z  0.7881   occ O-2  1.0   beq  1.37938

b)  The final .inp file used for analysis had the following information:
Spacegroup: P3121

site Si num_posns  3 occ Si+4  1 beq  0.46 x  0.4698 y  0 z =1/3; :  0.33333
site O num_posns  6 occ O-2  1 beq  0.46 x  0.4151 y  0.2675 z  0.2139


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Would a "chain" of ring polymers be possible?
Basically what the title asks. By a "chain", I mean a structure where you have two or more ring polymers (polymers where the ends are joined to form a large circular/ring-shaped molecule; different from something like a benzene ring), which interlock like links in a chain.
I'm not able to find much information about ring polymers (seems DNA is the most documented instance of something similar, since it sometimes takes the form of a ring in bacteria/viruses). Would it be possible for this sort of chain to exist? I'm mostly just asking out of curiosity, since if it is possible, one could imagine fun things like rings of ring polymers and molecule-scale chainmail.

See this answer. Examples in the answer involve just two or three "links", but a comment refers to a case with seventy-one!

Graphene (if we are not considering interlocking)

Frazer Stoddart got his Nobel for partly for pioneering just such things.

Absolutely! Organic rings can interpenetrate. A complex, downright amazing, example is forming molecular Borromean rings. These rings interlock so that removing any single ring causes the other two to come apart.

In 2004, J. Fraser Stoddart et al created a "Borromeate" of an organic zinc complex.

Try making the far simpler Olympic ring symbol before synthesizing Zn Borromeate ;-)


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What would be the correct IUPAC name of this compound?

How many carbon atoms will be there in the parent carbon chain? I have proposed its few names, which one is more appropriate?

4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol
3-(1-Chloromethyl)-2-(1-methylethyl)-pentan-1-ol

Although the answer given in my book is option (1), I have a little confusion regarding why option (2) is incorrect. Since we know that halogens are treated as substituents(suffixes are never used for them), not functional groups in nomenclature, shouldn't number of carbons be given more priority than substituents like Cl? Then, in that case, option (2) must be the right answer. Could anyone who is familiar with latest IUPAC rules enlighten me regarding this? And if you think of any more possibilities of naming this compound other than these two, please share it with me.
This question has already been asked on Quora, but I couldn't find any of the answers satisfactory as some of them choose 5 carbons in PCC, some 4 and some even 6, which made me even more confused.(https://qr.ae/prlmhm)

Looks to me like your textbook is wrong: there are six carbons in the primary chain. Tell us which textbook so we can warn others about it.

NCERT book std 12. I doubt it is available outside India also, so no need to be worried about. Fun fact: It is considered the most standard book in India that even wrong things are considered to be true if it is mentioned in the book.

Possibly of relevance: https://v.cx/2010/04/feynman-brazil-education

@matt_black You are right that the textbook is wrong; however, there are five carbons in the primary chain.

@Loong I count 6 unless I'm behind on IUPAC rules, which is possible. It would be worth a clarification in an answer to explain why I'm wrong.

@matt_black See my answer. When selecting the parent chain, the suffix ("-ol") comes before the longest chain.

@Loong Good clarification. Thanks for the correction to my shonky IUPAC rule knowledge.


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Formal explanation for why ammonia prefers C3v over D3h
The 2 major possible geometries of ammonia are D3h and C3v, but it prefers C3v. My thought process is that this is simply Walsh's rule - the HOMO in C3v is bonding (2A1), whereas the HOMO in C3v is non-bonding (1A2''), so the molecule prefers the trigonal pyramidal geometry based on the HOMO being best-stabilised.
Is this the complete explanation or does the 2nd order Jahn-Teller distortion come into effect here too?
I included a Walsh diagram for PH3 (couldn't find one for ammonia) just for completeness' sake.

By the 'second-order JT distortion', are you referring to the mixing of $2a_1$ and $3a_1$ when symmetry is lowered? It's been a while since I studied this stuff, but I don't think your descriptions are mutually exclusive. It's precisely the mixing that makes $2a_1$ more stable in $C_{3v}$ (if it didn't mix, it'd just be a pure p orbital, same as $a''2$ in $D{3h}$).

Past me wrote about this here: https://chemistry.stackexchange.com/a/58633/16683

You write "the HOMO in C3v is bonding (2A1), whereas the HOMO in C3v is non-bonding (1A2'')," but that does not work…


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What's the relation between two symmetry groups, if one has all the symmetry of the other and some more?
Question:
Consider two molecular symmetry groups, for example $C_s$ and $C_{2v}$.

$C_s$ has one inversion plane, and two irreductible representations: the symmetric $A'$, and the antisymmetric $A''$.
$C_{2v}$ has one inversion plane and one 2-fold axis; and four irreductible representations: $A_1$, $B_1$, $A_2$, $B_2$.

It is clear that if a molecule has a $C_{2v}$ symmetry, it also has a $C_s$ symmetry. It also stands (in a selected coordinate system), that an orbital of a molecule with $C_{2v}$ symmetry that belong to $A_1$ or $B_1$ also belong to $A'$, and similarly, orbitals belonging to $A_2$ and $B_2$ also belong to $A''$.

What mathematical term describes the relationship between $C_s$ and $C_{2v}$? Is $C_{2v}$ a subgroup of $C_s$? Is it rather a subset of it? Should we rather say that $C_{2v}$ implies $C_s$?
What's the correct term for the relationship between $A_1$/$B_1$ and $A'$, and between $A_2$/$B_2$ and $A''$? Are they subspecies? Does $A_1$ imply $A'$?

Background:
I'm comparing some orbitals of two similar molecules. One is a root with $C_{2v}$ symmetry, and the other is an alkylated version of it with $C_s$ symmetry. For certain reasons, symmetry classifications of the molecules play an important part of the writeup.  I want to explicitly point out that the symmetry and the irreductable representations of the root correspond to the ones of the alkylated molecule, even though they're labeled differently.
While this sounds like a basic question, I found it surprisingly hard to find an answer. I've been struggling for almost a week at this point. All textbooks and online resources talk about how you can obtain $C_{2v}$ from $C_s$ with certain symmetry operations, and about what properties they have individually, but never mention the terms I'm looking for.
I don't really have a strong background in group theory, so if I used any other terms incorrectly, I'm glad to be corrected.

With groups, subgroup is indeed the correct word. Smaller group is a subgroup of the bigger one.

I think you mean "irreducible". I'm not aware of "irreductible" being used, but I could be unaware of its use. "orbitals belonging to A2 and B2 also belong to A′′"
My understanding is that elements of A2, B2, and A'' represent symmetries of the molecules. Aren't the symmetries of the molecules different from orbitals? "One is a root with C_{2v} symmetry, and the other is an alkylated version of it with C_s symmetry." Alkylating makes molecules more symmetric?

@Acccumulation 1) $C_{2v}$ is more symmetric than $C_s$ as it also has a 2-fold axis in addition to the inversion plane. 2) Irreps are not the same as symmetries. Vaguely (in this case) they classify whether the orbitals are symmetric or antisymmetric WRT to the specific symmetry operations of the group, but u/orthocresol is much more equipped to comment on this. 3) re spelling, I'll check, it may be a typo that sticks around in my team.

If the user who downvoted the question could gently provide an explanation, I'd be more than glad to attempt to improve it. Right now I'm unsure what's the issue.

@Neinstein I wouldn't worry about it too much; in my experience there's a slightly higher (but still low) level of spurious background downvotes without comments here than in some other STEM sites.

"It is clear that if a molecule has a $C_{2v}$
symmetry, it also has a $C_s$
symmetry." Doesn't that mean that $C_{2v}$ has fewer symmetries?

@Acccumulation If a molecule has a symmetry plane and a 2-fold axis, ergo has a $C_{2v}$ symmetry, then it has a symmetry plane ergo $C_v$ symmetry. I'm not sure why this would mean that $C_{2v}$ has less symmetries.

If I say "Everyone who has a TV also has a phone", does it not follow that there are more phones than TV? "If something has an X then it has a Y" implies that there are more Y than X. The disagreement may stem from the question of whether "symmetry" refers to an automorphism that preserves some quality, or a group of such automorphisms. You seem to be using in the latter sense (and if so, it might be better to say "$C_s$ symmetry" rather than "a $C_s$ symmetry".

If "a symmetry" refer to an automorphism rather the group of such, then the statement "If G has a symmetry, then H has that symmetry" means that H is more symmetric than G, because every automorphism in G is also in H. There's the further confusion engendered by you asking "Is $C_{2v}$ a subgroup of $C_s$?" If G is a subgroup of H, then H is more symmetric than  G.

@Acccumulation I'm not sure I'm following. Your example with the TVS and mobiles would translate to "there are more molecules with Cs symmetry than with C2v", which certainly stands.

@Acccumulation I also don't really follow the second point with G and H. C2v has all the symmetry elements of Cs and one more, therefore C2v is more symmetric (I.e. describes a higher degree of symmetry). You seem to state the opposite which feels disingenuous to me. The groups don't have the symmetries themselves (what would that even mean), they are groups consisting of symmetry operations. A molecule is then said to have e.g. Cs symmetry if it's invariant to all symmetry operations in Cs.

My example with TVs translates to your statement meaning that there are more $C_s$ symmetries, not more molecules with $C_s$ symmetry. Also, $C_{2v}$ has more than just one more symmetry than $C_s$. I am not saying that $C_s$ has more elements, I'm saying that your wording implies this. And calling me "disingenuous" is a rather uncivil accusation.

@Accumulation I didn't call you disingenuous, I called my interpretation of your comment seemingly disingenuous. I didn't mean to offend you.

@Neinstein "You seem to state the opposite which feels disingenuous to me" does mean that if Accumulation stated what you feel they "seem to state", then they were being "not candid or sincere, typically by pretending that one knows less about something than one really does." (from google). Perhaps it's not the word you were looking for? Or perhaps you meant to say "...feels disingenuous of me" instead of "to me"?

@uhoh Probably I used the word wrongly. English is not my native, and until now I believed "disingenious" is pretty much a slightly stronger but still formal and uncharged synonym of "unintuitive". I've seen it a couple times on SE in contexts which led me to this interpretation. Thanks for clearing that up.

What mathematical term describes the relationship between $C_s$ and $C_{2v}$? Is $C_{2v}$ a subgroup of $C_s$? Is it rather a subset of it? Should we rather say that $C_{2v}$ implies $C_s$?

$C_\mathrm{s}$ is a subgroup of $C_\mathrm{2v}$, yes.
I'm speaking very informally here, but a group is a set of elements, together with a binary operator $*$ that 'combines' the elements to form other elements. In the case of symmetry groups, these elements are symmetry operations. The binary operator is composition, i.e. $B * A$ means do symmetry operation $A$ first and then $B$ (or the other way round, it's just a matter of definition); and it turns out that $B * A$ is always also a symmetry element. So, a group has more structure than just a set. A set is just a collection of elements, it doesn't have to have a way to combine them.
It's true that the elements of $C_\mathrm{s}$ are a subset of the elements of $C_\mathrm{2v}$. However, when referring to the groups themselves, subgroup is the correct term.
I don't think 'implies' is appropriate here, that's typically used in a context like propositional logic.

What's the correct term for the relationship between $A_1$/$B_1$ and $A'$, and between $A_2$/$B_2$ and $A''$? Are they subspecies? Does $A_1$ imply $A'$?

$A$, $B$, etc. are representations of a group. Defining the word 'representation' is actually quite involved! I've written a bit about it here before, but it's a whirlwind tour, and I don't necessarily think it makes for good reading.
If you're just looking for the correct term, though, then my understanding is that $A'$ is the restriction of the representation $A_1$ to the subgroup $C_\mathrm{s}$: https://en.wikipedia.org/wiki/Restricted_representation
Writing this in a chemistry paper/report will probably throw most readers off. (I reckon most chemists probably don't understand the formal meaning of a subgroup, but would have an intuitive understanding of it. I don't think a restricted representation means anything to the typical chemist, though, except for the very mathematically inclined.) So if I were in your position, I'd probably write that there is a 'correspondence' or a 'correlation' between the two irreps. That's not super-precise language, in that it doesn't describe the nature of the 'correspondence', but it'll get the message across. (Of course, you could define it properly once at the beginning, and then stick to vague language afterwards.)
For a different perspective, these kinds of relationships between representations are tabulated in so-called descent in symmetry tables, which you can find online. I also spent some time typing them up on Chemistry Meta, and in the original source from Atkins et al., similarly vague language is used:

The following tables show the correlation between the irreducible representations of a group and those of some of its subgroups. In a number of cases more than one correlation exists between groups.

"I also spent some time..." seems like a bit of an understatement :-)

As the discussion of these issues in the comments has gotten rather extensive, I think that I should move it to an answer.
In math, it's common to use the word "symmetry" to refer to an automorphism that leaves an object essentially the same. For example, see this page that says that a cube has 48 symmetries. Thus, what you refer to as a "symmetry operation" would in math often be referred to as simply a "symmetry".
As another example, a benzene molecule, if we ignore the single/double bond issue, has 12 symmetries, six rotations and six reflections. This is known as the dihedral group (note the statement "A regular polygon with n sides has 2n different symmetries", again using the word "symmetry" to refer to an automorphism) $D_6$. There are 12 symmetries in $D_6$. $D_6$ is a group of symmetries.
If we alkylate one of the carbon atoms (don't worry if this doesn't make sense chemically, I'm just dealing with the math), then the molecule has only one non-trivial symmetry (a reflection about the alkylated carbon). This is the binary or $Z_2$ group. If we alkylate two carbon atoms at opposite side of the molecule, then there are four symmetries, all of which are members of $D_6$. Since all symmetries of alkylated molecules are members of $D_6$, it would be correct to say "If the alkylated molecule has a symmetry, then the root has that symmetry".
Note that there is only one way to have a non-alkylated benzene molecule, but many different ways to have an alkylated benzene molecule. So:

The root has more symmetries than the alkylated molecule.
For each symmetry, if the alkylated molecule has that symmetry, then the root does as well. For instance, if a molecule has a $Z_2$ symmetry, then it has a $D_6$ symmetry.
The root is more symmetric than the alkylated molecule.
There are more alkylated molecules than root molecules [at least, in terms of possible configurations, not necessarily instances].
The symmetry group of the alkylated molecule is a subgroup of the root's.

Note that there is an opposite relationship between the first and fourth of these points. If the symmetry group of things of type X is a subgroup of the symmetry group of things of type Y, then the set of things of type Y is a subset of the set of things of type X.
For your question "Is $C_{2v}$ a subgroup of $C_s$?" the answer is a definite "no"; $C_s$ is a subgroup of $C_{2v}$. Your statement "It is clear that if a molecule has a $C_{2v}$ symmetry, it also has a $C_s$ symmetry." is misleading at best. It would be more clear as "It is clear that if a molecule has is invariant under $C_{2v}$, then it is invariant under $C_s$."

I've been following the discussion and while I'm not a chemist I struggle with symmetries of unit cells in crystals (cf. Is the completeness of the 230 known space groups in three dimensions proven?) and wonder if a question something like "How closely related are molecular symmetry groups to space groups used in crystallography? Are there also 230 of them?" could might be askable here, in Matter Modeling SE or in Math SE?

Also, any thoughts on Straightforward formalism to get sets of lattice vectors of fcc (111), (110) and (100) facet surfaces that I can also cite? A comment there suggests "don’t you just have one set of unique lattice vectors and the rest could be computed by symmetry operations?" but no further insight on symmetry operations were forthcoming. My theory is that matter modelers just let their big commercial programs handle all things symmetry-relate and don't really want to wade into it explicitly :-)


text:
Why does H2O appear as spikes in IR of air?
In the IR of air shown below, the region for water at 1500 cm-1 is hard to individually assign peaks due to their "spikey" nature. Is there any specific reason for this or does it just happen? Is it potentially due to interactions from other hydrogen in the air sample maybe binding to the water or interfering with it in some way?

This may help: https://chemistry.stackexchange.com/a/116536/79678.

Instead of spikes, the proper spectroscopic term is rotational line structure. Those are actual transitions.

Can the rotational line structures ever be individually identified to a vibrational mode or are there too many that it's not possible?

yes. A standard undergraduate experiment is to measure at the vibrational -rotational spectrum of HCL. In the HCL as in the water the spectrum measures transitions from rotational levels associated with one vibrational level to rotational transitions in the next vibrational level.  (A spectrometer with good resolution is needed also, $1, \mathrm{cm}^{-1}$ or better)

Those "spikes" are actually the rotational transitions of water. With a high enough resolution spectrometer, you can see the rotational structure associated with a given vibrational energy level for most gas phase molecules.

Since the gas molecules are separated the lines are not collisionally broadened as theynare in the liquid spectrum

@jimchmst You'll still see collisional broadening in the gas phase at pressures greater than 1 Torr. You can start to assume doppler broadening at low pressures, but in most cases, collisions still affect the line shape


text:
how to quickly dissolve sea shells?
Our condos are by the bay and seagulls are ubiquitous,so is their detritus. The shells they drop have been drawn by gravity and rain water to the down spouts and gutters of our buildings for 40 years and the drainage pipes underground are so backedup that the down spouts hadto be disconnected. Is there an economic way to dissolve the seashells quickly?

Not very quickly, not very economic, not very environment friendly.

And not likely to work without dissolving a metal drain.

I suspect the only real way is with mechanical means, beefier than the standard sewer cleanout snake, more like a root cutter. One might think the local plumbing companies would know what to do.

There are high strength acid drain cleaners such as Rooto, which contain concentrated sulfuric acid, available at local hardware stores.  This would be for the underground pipes.  The downspouts and gutters can be manually cleaned.
Please be very careful with this approach, and first check the resistance of your underground pipes to acidic drain cleaners.
First cautious application of dilute acid is recommended, followed by a good water rinse.  If the pipes are good with this, keep at it until they flow well again. Be patient, and make sure you buy acid drain cleaner.  The basic ones, made with sodium hydroxide, will not work.

Sulfuric acid is not very good at the purpose in question.

@IvanNeretin I realize calcium sulfate has limited solubility, but it will break up calcium carbonate sea shells.  HCl may generate fumes.

My "not very good" was not a token of courtesy, but a realistic assessment of its capacity. It is not bad, no. Just not very good.

@IvanNeretin I welcome improvements or better ideas. Here is evidence that sulfuric acid may be a better solvent than water.  HCl might be even better. Depends on the pipes.

Think I must apologize for looking somewhat judgemental. Chances are there is no ideal solution. H2SO4 is of course better than water, but, er, well, CaSO4. HCl (which would be my first choice) might be better still, but like you said, fumes. Also, both may damage pipes. Acetic acid is slow. So it goes.

@IvanNeretin no apology needed!  Calcium acetate has much higher solubility.  Appreciate the input. It is worth a try, perhaps on a small scale first.  3 acids in 3 different containers before adding to the pipes.  Work continues ...


text:
How to synthesize lanthanum(III) chloride & Cerium(III) chloride from their oxides?
Interested in obtaining anhydrous lanthanum/cerium chlorides

$\ce{La2O3}$ / $\ce{СeO2}$ in a flow of carbon tetrachloride at 500 degrees
$\ce{La2O3}$ / $\ce{CeO2}$ with ammonium chloride when heated (As I understand it, this method has a problem with the formation of oxochlorides)

Does it practically work? Is there any information? Maybe there is an easier option?
UPD1:
The oxide reaction is not feasible with all oxides, it depends on their production.
I tried this with different concentrations of hydrochloric acid, my cerium oxide sample didn't dissolve.
Kinetics and mechanism of the reduction-dissolution of cerium oxide catalyzed by fluoride ions in hydrochloric acid solution.
The reduction-dissolution of CeO2 in hydrochloric media obeys Eq. (4):
The protocol used here to investigate this reaction has been described previously (Lemont et al., 2015). The experimental setup is shown in Fig. 2. Two types of CeO2 powders were considered, shown previously to react one completely (one supply: CeO2 Acros 99.995% batch n° A011584501) one not at all (three supplies: CeO2 Acros 99.995% batch n° A010849501, CeO2 Alfa 99.995% batch n° 97,077, CeO2

Buying it maybe?

Wikipedia mentions synthesis of $\ce{LaCl3}$ from $\ce{La2O3}$:

Anhydrous lanthanum(III) chloride can be produced by the ammonium chloride route. In the first step, lanthanum oxide is heated with ammonium chloride to produce the ammonium salt of the pentachloride:
$$\ce{La2O3 + 10 NH4Cl → 2 (NH4)2LaCl5 + 6 H2O + 6 NH3}$$
In the second step, the ammonium chloride salt is converted to the trichlorides by heating in a vacuum at 350-400 ºC:
$$\ce{(NH4)2LaCl5 → LaCl3 + 2 HCl + 2 NH3}$$

As for $\ce{CeCl3}$, refer to this prepchem article:

$$\ce{CeO2 + 4HCl -> CeCl4 + 2H2O}$$
$$\ce{2CeCl4 ->[\Delta] 2CeCl3 + Cl2}$$
85 ml of concentrated hydrochloric acid is mixed with 40 ml of water and poured into the 500 ml flask. 48 g of cerium(IV) oxide are added to the solution of hydrochloric acid. The solution is warmed gently until the reaction starts and heated for 5 minutes. When the reaction is complete the solution is filtered and evaporated to 70 ml. The obtained solution is poured into a small crystallizing dish, cover with a watch glass, cooled slowly. The obtained crystals are separated by filtration, and the additional amount could be obtained from the filtrate.

Thank you very much for the information. CeCl3: According to your link, I turned to the source: https://research-solution.com/uplode/book/book-33147.pdf there really is a good recommendation.
LaCl3: I would like to find experience from practice.

@1st32 if the answer helped you, please make sure to accept it.


text:
Is there any special meaning to the angle between two peptide planes?
Take a look at the following image:

Is there any special meaning to the angle between two peptide planes (marked with red arrow heads)?
If so, why isn't a specific name (like phi, psi, etc.) given?

Hmm, I'm pretty sure that angle is just $\phi - \psi$ (or $\psi - \phi$; I guess it depends which way round you define your angle, but my point is that it's not a 'new' value).

There is an angle defined by three consecutive alpha-carbon positions, and pseudo-torsion angles defined by four consecutive alpha-carbons (for course-grained modeling). Because this is a 3-dimensional structure, the angle shown in the picture is unrelated to the angle of intersection between the peptide planes (which is about 90 deg for a alpha-helical segment and maybe 50 deg for a beta strand).

@orthocresol It is more complicated. Plus, the angle shown is ill-defined (is it a proper angle, or a dihedral, and does it depend on the position of the amide proton - which is ill-defined itself).

I stared at it again, and to be honest, I still think it is some kind of combination of $(\psi, \phi)$; but I'm fully willing to accept that I'm missing something, so I'll defer to you. I don't have a proof, it's just a strong intuition, which may be wrong. FWIW, my reading of it is this: if the peptide units are indeed planar, then the angle between them is the angle between the two normal vectors (= $\arccos(\vec{n}_1 \cdot \vec{n}_2/|\vec{n}_1||\vec{n}_2|)$). There's an ambiguity in that $\cos(x)$ is symmetric about 180°, but that can be resolved through careful definition (I think).

@orthocresol If the angle at the alpha carbon were 180 deg, then I would agree. In the conformation shown, the peptide planes are almost coplanar. If you rotate both planes by 90 deg (so that there are far outside of the plane of the paper), they would have an angle closer to 60 deg. The figure can be found here and is ambiguous about the direction of rotation (what are we keeping constant and what are we rotating).


text:
Why is CrO3 an oxidising agent, but WO3 is not?
So apparently CrO3 can be used as an oxidising agent, but WO3 can not. I saw this on an assignment question which I will include below:

I think the reason why W6+ is 6-coordinate and Cr6+ is 4-coordinate is because W6+ has a larger ionic radius than Cr6+ based on being lower in the periodic table and therefore can accommodate more anions around itself, based on basic geometry.
I am not quite sure why WO3 is not an oxidising agent though because since they are both in the same group and have the same oxidation state, they should have the same number of d electrons in the ion (namely 0 because they both have a noble gas configuration - Cr6+ is [Ar] and W6+ is [Xe]). So in theory, their ability to act as oxidising agents (i.e. to accept electrons) should be the same - i.e. not very potent because they have a stable full shell noble gas configuration?
Does anyone have any insights into this?

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.

Here's a different question, but the same answer is responsible for most of this effect.

It could be generalized to the oxidation state +VI.

I wrote about CrF6 vs WF6 here: https://chemistry.stackexchange.com/q/48971/16683

Just a note: the quote says “not a strong oxidizing agent”. That is not the same as “not an oxidizing agent”. I would certainly expect, under appropriate conditions, that tungsten oxide oxidizes hydrogen gas to yield tungsten metal and water vapor.

We can look at both ionization energy and coordination aspects. Both factors differentiate the relative stability of both $\ce{MoO3}$ and $\ce{WO3}$ from the strong oxidizing tendency of $\ce{CrO3}$.
Ionization energy
Wikipedia's table of ionization energies reveals that the fourth through sixth ionization energies of chromium are significantly greater than those of its heavier congener molybdenum. In kJ/mol, the relevant values are:
$\ce{Cr}: IE_4=4743, IE_5=6702,IE_6=8745$
$\ce{Mo}: IE_4=4480, IE_5=5257,IE_6=6641$
Values for tungsten are not given, but they are expected to be similar to molybdenum based on typical periodic trends. Thus although bonds with these elements in high oxidation states (+4 or higher) are expected to be polar covalent instead of predominantly ionic, the electron transfer required to achieve the higher oxidation states is more easily accomplished with molybdenum, and thus probably tungsten, than with chromium; and so higher oxides of the heavier elements would expect to be more energetically favored.
Why is there the difference in ionization energy? In this discussion Nicolau Saker Neto notes that certain valence orbitals, such as $4f$ in the lanthanides and $3d$ in the lightest transition metals like chromium, are more difficult to ionize because they lack radial nodes. The interested reader can review the arguments in that answer.
Coordination
When we have a binary compound with a stoichiometric ratio far from $1:1$, lattice stability may be difficult to achieve because of the forced differences in coordination requirements. If the element with fewer atoms cannot achieve a sufficiently high coordination number, the bonding in a lattice will be limited and the material becomes both less stable and more reactive. Nitrides of alkali metals ($\ce{M3N}$) are a good example; only with lithium, whose relatively small ions enable the nitrogen to assume a coordination number of six or more (there are different allotropes of lithium nitride) without severe steric strain, is the nitride reasonably stable.
Similarly with the Group 6 $\ce{MO3}$ oxides, molybdenum as well as tungsten can form trioxides with six-coordinate metal, allowing more bonding than that in chromium oxide where (presumably) steric interference limits the metal coordination number to four. (Chromium does achieve a coordination number of six in lower oxidation states, where the metal center has more valence-electron density and is larger in size.)


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Gaussian simultenous scan with two variables
I am trying to rigid scan two bond distances simultaneously in Gaussian 09/16. For example, I have taken a water molecule and started a rigid scan. Below is my input.

# b3lyp/6-31+g(d,p) scan

Title Card Required

0 1
 O
 H                  1            B1
 H                  1            B2    2            A1

   A1           109.50000006

B1 0.96 S 20 0.1
B2 0.96 S 20 0.1

The output scan plot is this:

It is a 3D plot of 441 (21 steps for H1 atom * 21 steps for H3 atom) grid points. I want both the H atoms to move simultaneously, so I get a 2D plot of only 21 points. Below is the type of plot I want for scanning 2 H atoms. Any help with the input?

I'm not sure, but it's certainly easy enough to try: use a single variable for both bond lengths.

That seems fine, though. However, I would be thankful to you if you please share any sample input.

In your z-matrix substitute B2 with B1 and delete the last line before the mandatory blank line.

it works, thanks

From the obvious hints @Martin provided, it became easy. We can simultaneously scan only one coordinate. So, for this particular problem the input should be as follows,

# hf/3-21g scan

Title Card Required

0 1
 O
 H                  1            B1
 H                  1            B1    2            A1

   A1           109.50000006

B1 0.96 S 20 0.1

Explanation: Since bond the bond lengths of OH are same, one has to put only B1 parameter in the z-matrix and vary that parameter.
But, for a special case, if $OH_1\neq OH_2$ then this will not work.
I will look for more general case and post that as update here.

In my opinion don't bother looking for a more general case. Gaussian interface is not flexible enough for that. There is possibly an easier way to generate a series of xyz coordinates programmatically with python of so and run single points on these. For future questions like these, I recommend looking at [mattermodeling.se].

Oh... I forgot, you might have luck with GIC. But in all honesty, learning python is probably easier...


text:
Adding Bicarb to Sparkling Water
This may seem a trivial request of mine, but it is a major part of my health regime,
I add Bicarb (Sodium Bicarbonate - NaHCO3) to Sparkling Water (H2O + CO2 (bubbles) + a weak solution of Carbonic Acid - H2CO3)
There is an effervescent reaction.
But I cannot find an explanation of this - using chemical formulae - online.
I do this a lot of this for health reasons - with the approval of my GP.
It changes my pH levels from acidic to alkaline/neutral.
So I'd like to know the gas that is released, and the residual solution.
Thank you

Virtually any powder added to sparkling water, or seltzer, club soda or any other carbonated drink cause outgassing by nucleating bubbles of $\ce{CO2}$. See Chemistry and Physics Stack Exchange, and this talk on YouTube.

"talk on YouYube" and not some dozen questions about that here?

Thank you. So the gas released is Carbon Dioxide. Is the residual solution Sodium bicarbonate dissolved in water or is there a chemical reaction ?

It's a mixture of dissolved solids and dissolved gases, no reaction is likely. BTW, Your "formula" reminded me of T. Edison's boyhood exploit: he "once persuaded a boy employed by the family to swallow a large quantity of Seidlitz powders in the belief that the gases generated would enable him to fly. The awful agonies of the victim attracted attention..." https://www.heritage-history.com/index.php?c=read&author=meadowcroft&book=edison&story=boyhood It seems you're close to reinventing that powder ;-)


text:
Comparison of acidic strength in organic compounds by assigning charges to rings
Generally we compare acidic strengths by comparing relative stability of negative charge formed by removal of acidic hydrogen. However I can't understand the following comparison

The solution given was

Now I can't understand why the 7-member ring is getting a positive character and the 5-member ring a negative character. I can't get any resources related to it.
Is this related to aromaticity in some way?

Yes, it is because of Huckel's rule of aromaticity. This states that for a ring to be aromatic, it must have 4n+2 pi electrons, where n is an integer. For the conjugate bases to be stabilized they would need to be aromatic. For the five-membered ring, the -1 charge is stable as it has 4(1)+2=6 pi electrons. For the seven-membered ring, the +1 charge is stable as it has 4(1)+2=6 pi electrons.
https://en.wikipedia.org/wiki/H%C3%BCckel's_rule
Note: the compound azulene is quite similar and also exhibits dipoles due to aromatic stabilization: https://en.wikipedia.org/wiki/Azulene

The link you provided states that "Hückel's rule is not valid for many compounds containing more than one ring" in "Polycyclic Ring" section. However i think here different parts of the compound are trying to get partial aromatic character, so it seems correct

@hemantkumar I think for those compounds you have to analyse each ring individually. Usually, some electrons are shared between rings or do not participate in aromaticity so the system cannot be analysed as a whole.


text:
Henderson-Hasselbalch equation and degree of dissociation
Is it possible to evaluate the degree of dissociation in unbuffered solutions, through the Henderson-Hasselbalch equation?
For example: place acetic acid in a sodium hydroxide solution. After the reaction, the pH of the solution is lowered. Is it possible to exploit the Henderson-Hasselbalch to evaluate the degree of dissociation of acetic acid, using this final pH, even if we are not in the presence of a buffer solution?
Generally, Henderson-Hasselbalch is introduced in the presence of buffers, which is where my doubt arises. Infact, by placing the same acetic acid in a buffered solution at a given pH, it is very simple to obtain the ratio [A-] / [HA] using the H-H. Is this reasoning also applied to the previous example?
Thanks in advance.

Yes, of course.  NaOH is a strong base.  pH tells you starting concentration of NaOH.  Final pH tells you how much NaOH has reacted.  This should agree with the formula pH = pKa + log salt/acid.  There will be no buffering effect until pH gets close to pKa.  Otherwise it's just an acid-base reaction.

H-H equation is nothing else but equation for the acid dissociation constant, rewritten in the logarithmic form. Any mathematical pH evaluation must respect this equation in either of its forms.

What do you mean with: even if we are not in the presence of a buffer solution? What you have described is the way to create a buffer solution.

@RobertDiGiovanni and Poutnik. Therefore, what is found with the ICE method must coincide with the H-H: for this reason the H-H can be very well exploited to evaluate the [acetate] \ [acetic acid] ratio at the end of the reaction with NaOH, exploiting the final pH.
However, it would not be possible to derive the pH directly with the H-H, as we would not consider the possible contribution of the NaOH remaining after the reaction. Right?

@Martin-マーチン i mean: if we have a buffer where pH doesn't change, we can simply use H-H equation to find [A-] / [HA] ratio. If we put acetic acid in a NaOH solution (unbuffered solution), final pH Will be different from the starting one. So, although we have this situation, can we always find [A-] / [HA] with H-H, although pH Is different? Generally we talk about H-H when we have buffers.

@Luckenberg. Right.  The "equilibrium" between NaOH and acetic acid is 100% conversion to acetate.  Acetic acid will be consumed as any other acid will be until you get close to the PKa of acetic acid.  This why other buffers, such as phosphate, are made with KH2PO4 and K2HPO4 in a known ratio (conforming with H-H ) and are only adjusted slightly with H3PO4 or KOH using a pH meter.  It might be fun to just try acetic acid/NaOH titration in the lab and experimentally measure pH as more titrant is added.

A higher [A-]/[A] ratio will calculate a higher pH.  But an [A-]/[A] ratio of infinity means ... you're not in the buffering range.

The equilibrium constant expression for a weak acid is:
$$K_\mathrm{a} = \frac{\ce{[H3O+]}_\mathrm{eq} \ce{[A-]}_\mathrm{eq}}{\ce{[AH]}_\mathrm{eq}}\tag{1}$$
for the reaction
$$\ce{AH(aq) + H2O(l)<=> H3O+(aq) + A-(aq)}$$
In equation (1), the brackets either stand for activity, or for concentration divided by $\pu{1 mol L-1}$. If it is the latter, it is an approximation that works best at constant ionic strength and lower concentration of the weak acid/base conjugate pair. The subscripts "eq" specify that these are to be measured at equilibrium.
If you solve (1) for the hydronium ion activity and take the logarithm, you get:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left( \frac{\ce{[A-]}_\mathrm{eq}}{\ce{[AH]_\mathrm{eq}}}\right)\tag{2}$$
I would argue that this is not the Henderson-Hasselbalch equation, rather it is:
$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\left( \frac{\ce{[A-]}_\mathrm{initial}}{\ce{[AH]_\mathrm{initial}}}\right)\tag{3}$$
Equation (2) is universal and exact. If you know the pH of the solution and the pKa of the weak acid (this could be an indicator, a substance intended as buffer, any weak acid you add to the solution or the conjugate weak acid of a weak base you add to the solution), you can figure out the ratio of protonated to deprotonated species. Granted, the pH might change when you add your substance of interest to the solution, so you should measure it in the final solution.
Equation (3), on the other hand, is an approximation which obviously does not work for a weak acid in water (with nothing else added) because the initial concentration of conjugate base would be zero. It does work fairly well when you add similar concentrations of weak acid and conjugate base to water (one way of making a buffer). In this case, the equilibrium concentrations are not far off from the initial concentrations, so it is a useful approximation.

Is it possible to exploit the Henderson-Hasselbalch to evaluate the degree of dissociation of acetic acid, using this final pH, even if we are not in the presence of a buffer solution?

First, when you mix a sodium hydroxide solution with an excess of acetic acid, you do have a buffer solution (mixing weak acid with strong base while monitoring the pH is a common way of making a buffer). If you measure the pH, you can estimate the ratio of acetic acid to acetate equilibrium concentration using equation (2).
Technically, if you calculate concentrations by assuming that the reaction of sodium hydroxide with acetic acid goes to completion, and call these initial concentrations, you can also apply equation (3).
So it is a matter of opinion or perspective whether the answer to your question is yes or no.

"Technically, if you calculate concentrations by assuming that the reaction of sodium hydroxide with acetic acid goes to completion, and call these initial concentrations, you can also apply equation (3)". With this, do you mean that the concentrations we find after the reaction are the 'initial' concentrations of the buffer, so we can use (3), that gives us (about) the same result as (2)?

And using the same reasoning, if we have an excess of sodium hydroxide, than we can't find the final pH using H-H, because we must consider the effect of remaining NaOH, which isn't considered by H-H. However, we can still find the degree of dissociation of acetic acid using the final pH, determined by the remaining NaOH..

@Luckenberg Yes, exactly. With an excess of NaOH (e.g. pH higher than10), less than 1 in 100,000 of the acetate ions will be protonated at any time.


text:
Non-Constant Pressure and Temperature Processes - Gibbs Energy as Equilibrium Condition
I reviewed several questions already, but none address the same point except the following:
Is Gibbs energy minimized for processes at constant temperature are pressure only?
However, this question remains unanswered until today. The derivations and equations below are from Robin Smith's Chemical Process Design (Chapter 5) and Kolt's Chemical Thermodynamics (Chapter 6 and 7)

For the calculation of the equilibrium constant, it is the case the $\Delta G$ is set to zero at the equilibrium condition. However, doing so seem improper from what I know, and I am wondering what I am getting wrong.
I am considering a reaction of ideal gases, in which the total pressure is changing while the reaction is progressing. As the reaction progress, Gibbs energy can be described by the following correct equation:
$$\Delta G_{\mathrm{r}} = \Delta G - \Delta G^0=RT\ln{K}$$
However, it is claimed that at equilibrium $\Delta G = 0$. Therefore,
$$\Delta G^0=-RT\ln{K_{\mathrm{eq}}}$$
This latter equation is the one I am concerned with due to the imposition that $\Delta G = 0$, which does not seem proper as the pressure is changing while the reaction is progressing. My concern stems from the logic detailed below.
The standard thermodynamic condition for equilibrium is $\mathrm{d}S_{\mathrm{tot}} = 0$. This then can lead to an equilibrium condition involving Gibbs energy as follows:
$$\mathrm{d}S_{\mathrm{total}} = \mathrm{d}S_{\mathrm{env}} + \mathrm{d}S_{\mathrm{sys}}$$
Rewriting the equilibrium condition in terms of system variables only gives:
$$\mathrm{d}S_{\mathrm{sys}} = \frac{DQ_{\mathrm{sys}}}{T_{sys}}$$
This assumes $T_{\mathrm{env}} = T_{\mathrm{sys}}$.
Embedding the first law of thermodynamics and assuming only $PV$ work leaves us with:
$$\mathrm{d}U + P\mathrm{d}v - T\mathrm{d}S = 0$$
Imposing the assumption of constant temperature and pressure allows adding P and T differentials since they are zeros.
$$\mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P - T\mathrm{d}S - S\mathrm{d}T = 0$$
Once integrated, this gives:
$$\Delta U + \Delta (PV) - T\Delta S = 0 $$
or equivalently
$$\Delta H - T \Delta S = 0 \text{with} \Delta G = \Delta H - T \Delta S$$
However, this final result only came to be with the assumption that $\mathrm{d}P$ and $\mathrm{d}T$ are zero. If this assumption is removed, the equilibrium condition differential would be as follows:
$$\mathrm{d}U + P\mathrm{d}v + V\mathrm{d}P - T\mathrm{d}S - S\mathrm{d}T = -V\mathrm{d}P - S\mathrm{d}T$$
or
$$\mathrm{d}G = V\mathrm{d}P - S\mathrm{d}T$$
This latter equation does not produce $\Delta G = 0$. In fact, the $\Delta G$ and $K$ relation is derived from it. The end result is that we are applying an equation for constant pressure and another with non-constant pressure for the same system at the same time; this does not make sense to me, as pressure is either constant or not.
How is it then that $\Delta G^0=-RT\ln{K_{\mathrm{eq}}}$?

See also constant T,V and minimum Helmholtz free energy as equilibrium condition.

You are dealing with a problem of chemical equilibrium for a single reaction in a homogeneous phase. The Gibbs free energy, in this case, is a function of pressure, temperature, and moles of all the chemical species.
The only thing the following equation will tell you
$$ \Delta_r G = 0 \longrightarrow K_\pu{eq} = \prod_j a_j^{\nu_j } \tag{1} $$
is that there exist one and only one state, described by the macroscopic variables $(P,T,y_k)$ once you reach chemical equilibrum.
So I will clarify:

As the reaction progress, Gibbs energy can be described by the following correct equation.

When the reaction progresses, you are not in chemical equilibrium, this is why your system "moves". In this state, there are dissipative forces/driving forces that will modify the macroscopic variables in space and time of your system. If you want to describe the system in this state, you resort to transport phenomena in order to obtain the relationship between those macroscopic variables with space and time.

This latter equation is the one I am concerned with due to the imposition that $\Delta_r G=0$, which does not seem proper as the pressure is changing while the reaction is progressing.

Indeed you are right, since there is one and only one state described by $(P,T,y_k)$ achieved for a certain set of initial conditions. I understand your question, because in books they ask you things like:

If the pressure is increased, what happens to equilibrium?
If the temperature is increased, what happens to equilibrium?

So you are crushing your head to the wall, wondering why they are talking about changes and equilibrium at the same time. The fundamental idea is the following: it does not matter how you reach that final state, what interests you is that $ \Delta_r G = 0 $ will tell that new state described by a new set of $(P,T,y_k)$. So the question should be reformulated as:

If at equilibrium we have a system at $(P_1,T_1,y_{k1})$, what will happen to the new equilibrium established if now the temperature is $T_2$?
If at equilibrium we have a system at $(P_1,T_1,y_{k1})$, what will happen to the new equilibrium established if now the pressure is $P_2$?

For example, you may already know that if there is no change of moles in the chemical reaction, nothing will happen if the pressure is changed to $P_2$. You can prove it with Eq. (1), considering ideal gases, and noting that Eq. (1) will be satisfied regardless the value of the pressure.
The derivation you made is related to a closed system for a pure substance. In that case, the internal energy is only a function of $2$ variables $ U = U(P,T) $ if you like. This is not the case we are analyzing, dealing with a chemical reaction.

Metal Storm did provide a correct answer. My initial confusion indeed stemmed from talking about equilibrium and changes (to the equilibrium) simultaneously. I will re-write their answer to address my concerns specifically in the way I thought about them previously.
At Equilibrium, the reaction is no longer changing position. Therefore, both $T$ and $P$ are constant at the equilibrium (there is nothing changing them). Therefore, $\mathrm{d}G = 0$ and $\Delta G = 0$ per the derivation in the original posted question.
Now, if the pressure was getting changed somehow under constant temperature, what will become of the equilibrium condition?
The total differential for $\Delta G$ is:
\begin{align}
\mathrm{d}\Delta G &= \Delta V\mathrm{d}P - \Delta S\mathrm{d}T &
\text{with } \left.\frac{\delta \Delta G}{\delta P}\right|_{T} &= \Delta V
\end{align}
Thus, assuming an ideal gas:
$$G_{\Delta P_{\mathrm{final}}} - G_{\Delta P_{\mathrm{initial}}} = \int \Delta V\mathrm{d}P = \int \frac{nRT}{\Delta P}\mathrm{d}P = nRT \ln\left(\frac{\Delta P_{\mathrm{final}}}{\Delta P_{\mathrm{initial}}}\right)$$
Or, once we incorporate multiple components in a reaction (derivation skipped),
$$\Delta G_{\mathrm{components}, \Delta P_{\mathrm{final}}} - \Delta G_{\mathrm{components}, \Delta P_{\mathrm{initial}}} = RT \ln(K)$$
This latter equation is correct regardless of whether equilibrium is achieved on the final state or not, and is correct regardless of the chosen initial and final conditions are. It has no relation to spontaneity and it makes no claims regarding equilibrium. It did assume a reversible pathway connecting the two pressures though.
However, if we are concerned with a final state that is at equilibrium, the equation can be re-written as:
$$\Delta G_{\mathrm{components}, \mathrm{equil}} - \Delta G_{\mathrm{components}, \mathrm{standard}} = RT \ln(K_{\mathrm{eq}})$$
Per the initial discussion, at equilibrium, $\mathrm{d}G = 0$ and $\Delta G = 0$. Therefore,
$$- \Delta G_{\mathrm{components}, \mathrm{standard}} = RT \ln(K_{eq})$$

Your equation for $\mathrm{d}G$ is incorrect.  It should read
$$\mathrm{d}G = -S\mathrm{d}T + V\mathrm{d}P + \sum{\mu_jdn_j},$$
where the $\mu$s are the chemical potentials of the various species and $dn$'s are the changes in the amounts of substances.

This answer was deleted with no explanation. I am resubmitting trusting the moderator will explain the problem.
As you finally seem to conclude when equilibrium is attained T, P, and activities of all participating species are constant and these conditions satisfy an equilibrium constant at the given T and P. When T and P are standard Conditions the Gibbs [free] Energy between the condition of all activities equal to 1. and the equilibrium condition is the standard Delta G of the reaction, Delta G[0]. Delta G[0] can be measured or calculated from thermal or electrochemical data or from Hess's Law calculations or calculated from a measured equilibrium constant. Standard free energies are simply a way to express equilibrium constants in a compact, recognizable, form [energy] similar to pKa and pH in acidity. At equilibrium Delta G and Delta A = zero and the chemical activities satisfy the equilibrium constant. It took Gibbs, Helmholtz et al. to realize that both energy and entropy are involved in attaining equilibrium and determining the availability of work from a reaction.
Equilibrium at other than standard conditions results in a different Delta G[0] and Keq. To reduce to standard conditions energy is removed or added as appropriate. Kinetics may or not allow equilibrium under all conditions.

At constant temperature and pressure, the graph of dependence of the system Gibbs energy as function of the reaction quotient has minimum at the reaction equilibrium.
With constant volume instead of pressure, the same is valid for the Helmholtz energy.
If the above $G=\mathrm{f}(Q)_{T,P}$ or $A=\mathrm{g}(Q)_{T,V}$ are plotted for variable temperature, pressure, or volume, the respective graphs would be changed and the function minimum does not generally match the equilibrium any more.


text:
How to shift SO3/SO2 equilibrium?
I am struggling with a chemical equilibrium question and would appreciate some help. The question involves a closed system where the equilibrium
$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$
has been established. The forward reaction is exothermic.
My question is, how can I decrease the amount of $\ce{SO3}$ in the mixture? The options given are to change the temperature, volume, or amount of $\ce{O2}$ present.
So far, I have tried to recall the principles of Le Chatelier's principle, which states that when a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will shift its equilibrium position to counteract the change. In this case, if I want to decrease the amount of $\ce{SO3}$, I need to shift the equilibrium to the left, meaning that I need to decrease the concentration of $\ce{SO3}$.
Based on my understanding of Le Chatelier's principle, changing the temperature will have the opposite effect on the equilibrium because the reaction is exothermic. Decreasing the volume of the container will also have no effect since the reaction involves gases. Therefore, the only option left is to change the amount of $\ce{O2}$ present. If I decrease the amount of $\ce{O2}$, the equilibrium will shift to the left, reducing the concentration of $\ce{SO3}$.
Am I on the right track with my answer? I would appreciate any feedback or additional information that could help me better understand this concept.

Be sure to review available info about the principle in offline and online sources, including this site, to avoid creation yet another redundant info.

"Decreasing the volume of the container will also have no effect since the reaction involves gases." ?!?  If you change the volume of a closed system, you obviously change the pressure.

In an exothermic reaction like this one, heating the mixture will push the system to the left hand side of the equation. So it will decrease the amount of $\ce{SO3}$ and increase the proportion of $\ce{SO2}$ and $\ce{O2}$. This is what we are looking for.
Increasing the volume will have the same effect, as increasing the volume will produce the pressure to decrease. And in such a mixture, Le Chatelier's principle states that moving the equation to the left hand side will increase the pressure, so counteract the effect of changing the volume.
As a consequence, both increases (temperature and volume) will have the same effect : destroy $\ce{SO3}$ and produce more $\ce{SO2}$ and more $\ce{O2}$

What about the question the OP asks? Are they on the right track about $\ce{O2}$?

Of course decreasing the relative amount of $\ce{O2}$ will push the equation to the left-hand-side.


text:
What is the driving force of Bartoli indole synthesis?

The Bartoli indole synthesis mechanism begins with a series of attacks on the nitroarene regent by the Grignard reagents. But the atom $\ce{O}$ in $\ce{-NO2}$ should be negative, and Grignard reagent is also be negative, they should repel each other. So, why the first step happens?
I calculate the ESP and the LUMO of nitrobenzene by NBO.

If it is because HOMO(Grignard)-LUMO(nitrobenzene) is bigger than their electrostatic repulsion, why aromatic nitro compounds perform $\ce{S_{N}Ar}$ on benzene ring instead of $\ce{-NO2}$?

"why aromatic nitro compounds perform SNAr on benzene ring instead of −NO2" — I feel like this is the sort of thing that is traditionally explained with so-called HSAB theory, although there is a strong argument that HSAB theory is completely wrong, and should be replaced with transition state theory / Marcus theory which may be relevant here. https://onlinelibrary.wiley.com/doi/full/10.1002/anie.201007100 I wonder if it's enough to model the starting material; you may need to study the entire reaction (and also consider the potential reversibility of competing pathways).

Where did you find this mechanism? It's overly elaborate and, likely, simply wrong.

OK, I found the picture is from Commons - better link directly. Ref. for the mechanism seems to be https://pubs.rsc.org/en/content/articlelanding/1991/P2/p29910000657

@orthocresol, Thanks. According to HSAB, hard-hard preference  is electrostatic stabilization, soft-soft preference is HOMO-LUMO interactions, the LUMO is mainly on $\ce{−NO2}$. The $\pi$ orbital of $\ce{−NO2}$ may be harder than benzene ring, and Grignard reagent is kind of hard soft, so hard-hard should be on $\ce{−NO2}$? Then why on atom $\ce{O}$ instead of $\ce{N}$?


text:
Empirical formula and rounding
I am very confused with the procedure to find the empirical formula. In the part where we have to multiply by subscripts to make them integers, are there clear rules?
For example, if I have the formula (I don't know if it exists in reality)
$$C_{2}H_{3.75}O_{1}$$
In this case, it seems that I have to multiply by 4 and it would be:
$$C_{8}H_{15}O_{4}$$
But, if I work with more or fewer digits, I could end up with something like this:
$$C_{2}H_{3.60}O_{1}$$
and I should multiply by 5:
$$C_{10}H_{18}O_{5}$$
So which one is correct?
And another question is when can I just approximate and not multiply by an integer?
Are the empirical formulas we find only probable?
Is there a way to verify the empirical formula?

In the part where we have to multiply by subscripts to make them integers, are there clear rules?

No. The number of significant figures in the given data can give some guidance. If there is a large uncertainty in the underlying data, you should round more.

Are the empirical formulas we find only probable?

Anything based on measurement is only probable. In this case, you would go for the simplest empirical formula (lowest numbers) that is a reasonably good fit to the data.

Is there a way to verify the empirical formula?

Yes. You could either repeat the experiment or, even better, do a different experiment that gives you supplemental data. If you somehow measure the molar mass, this will put a limit of the magnitude of the subscripts in the empirical formula. For example, if the molar mass of hydrogen peroxide is measured as 35 g/mol, an empirical formula of $\ce{HO}$ would be a better fit than an empirical formula of $\ce{H134O135}$.

So which one is correct?

Without knowing the underlying data, there is no way to give an answer. $\ce{C8H15O4}$ would be suspicious to someone familiar with organic chemistry (it looks like a radical), but without knowing the pedagogical situation in which a question was asked, I would not reject $\ce{C8H15O4}$ on those grounds.

Not necessarily a radical - one possibility is https://pubchem.ncbi.nlm.nih.gov/compound/164754000

@ToddMinehardt Radical or ion or something with a hat (caret) that I am not familiar with :-)


text:
The 'bite size' effect in chelating ligands
My question is in relation to the following data, which describes the log K for the equilibrium of Mn+ + Ligand $\ce{<=>}$ [ML]n+

As far as my understanding goes, this is an example of the effect of changing the 'bite size' of the ligand. In TRIEN, the bite-size is greater than in 2,3,2-tet, as shown on the image below, which means that it prefers to bind to metals of larger radius. However, this data suggests that both ligands bind better to Cu2+, which is to be expected for 2,3,2-tet, but not for TRIEN.

Granted, the logK reflect this 'bite-size' effect somewhat as log K increases for
Cu2+ and decreases for Pb2+. However, my question is why this effect is not more pronounced in the TRIEN ligand because I would expect logK(Cu2+)>logK(Pb2+) based on the arguments above?

Can we get a citation for the data?

@JamesGaidis the data comes from one of the Oxford university undergraduate chemistry past papers (https://www.whatdotheyknow.com/request/chemistry_past_papers_3?unfold=1), so I doubt it's going to be wrong in case that's what you're thinking

For James Gaidis information, the OP is correct and the data given in OP's table is accurate, which is from a peer-reviewed paper (Ref.1). In that paper, binding affinity of $\ce{Ni^2+}$ was compared with various other metal ions with different diameters:

Here, the ligand $\ce{L^6}$ is TRIEN (10-member chain) and the ligand $\ce{L^7}$ is 2,3,2-tet (11-member chain). As explained in the above chart, TRIEN is preferred when by the ions when atomic radius increased over larger 2,3,2-tet (larger the metal ion, the more strongly it prefers the smaller ligand, TRIEN). Similarly, the smaller the metal ion, the more it prefers to complex with the 13-membered chain, 2,3,2-tet For example, for $\ce{Cu^2+}$ (smaller cation), $\Delta K = \log K_\ce{L^7} - \log K_\ce{L^6} = 3$. For same two ligands, for $\ce{Pb^2+}$ (significantly larger cation), $\Delta K = \log K_\ce{L^7} - \log K_\ce{L^6} = -2.5$ (assumed from the plot). When they have changed the 10-member chain of TRIEN to 12-member macrocycle $\ce{L^1}$, and 11-member chain of 2,3,2-tet to 14-member macrocycle $\ce{L^3}$, the same trend continued:

This behavior is simply explained by using macrocycles. It is the fact that the hole size in $\ce{L^1}$ is actually larger than that in $\ce{L^3}$. This rather surprising result comes about because the most stable form of $\ce{L^1}$ as indicated by molecular mechanics calculations is the trans-I(++++) form (as shown in following diagram; Ref.2), which has all its nitrogens oriented so that their hydrogens are on the same side of the macrocycle. The hole size in this conformer of $\ce{L^1}$ is such that metal ions showing $\ce{M-N}$ lengths of $\pu{2.11 Å}$ fit best. On the other hand, the hole size in $\ce{L^3}$ in its most stable conformer, the trans-III(++--) form, is such that metal ions with $\ce{M-N}$ bond lengths of $\pu{2.05 Å}$ fit best:

This contraction-expansion ability can  be shown by the drawing of the $\ce{[Cu(L^1)Br]+}$ cation, showing how the copper atom is raised above the plane of the donor atoms. The ligand has the trans-I(++++) configuration. In addition, the macrocyclic ring of $\ce{L^1}$ is much more flexible with respect to expansion and contraction, and is, in particular, able to accommodate large metal ions much more easily than can $\ce{L^3}$:

All of these arguments (changing confirmation, etc.) are applied to $\ce{L^6}$ (TRIEN) and $\ce{L^7}$ (2,3,2-tet) ligands as well such that changing the so-called "bite distance" to adjust the size of metal ions. A valuable discussion of other factors affecting chelating including solvent can be find in Ref.3.
References:

Vivienne J. Thöm and Robert D. Hancock, "The stability of nickel(II) complexes of tetra-aza macrocycles," J. Chem. Soc., Dalton Trans. 1985, (9), 1877-1880 (DOI: https://doi.org/10.1039/DT9850001877).
Vivienne J. Thöm, Christine C. Fox, Jan C. A. Boeyens, and Robert D. Hancock, "Molecular mechanics and crystallographic study of hole sizes in nitrogen-donor tetraaza macrocycles," J. Am. Chem. Soc. 1984, 106(20), 5947–5955 (DOI: https://doi.org/10.1021/ja00332a032).
Robert D. Hancock and Arthur E. Martell, "Ligand design for selective complexation of metal ions in aqueous solution," Chem. Rev. 1989, 89(8),  1875–1914 (DOI: https://doi.org/10.1021/cr00098a011).


text:
Are there xenon compounds with two xenon atoms bonded directly to each other?
I've seen so many compounds containing xenon, and xenon was mostly the central atom, and if we had 2 xenons they would never bond directly to each other. I was curious if it is possible to have for example 2 xenons in a compound that are bonded directly to each other?

As you said yourself, we haven't seen any examples yet.

From a quick search, https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7435756/ explains: "Noble gas (or aerogen) bond (NgB) can be outlined as the attractive interaction between an electron-rich atom or group of atoms and any element of Group-18 acting as an electron acceptor."

One example is the $\ce{Xe2+}$ cation. Quoting from the abstract of Formation of the dixenon cation, $\ce{Xe2+}$, in fluoroantimonate(V) media by oxidation–reduction methods: spectroscopic properties of the ion

A green species has been generated both by oxidation of gaseous xenon
with a dioxygenyl salt and by reduction in antimony pentafluoride
solution of the $\ce{XeF+}$ cation, e.g. with water or a suitable metal; the
Raman, u.v.-visible and e.s.r. spectra of the product are all
consistent with the identity $\ce{Xe2+}$, representing the first homonuclear
noble-gas cation to be characterized in the condensed phases.


text:
Transesterfication: Which is a suitable acid that does not contaminate the polar product?
In my PhD Project I am doing a transesterfication of a phosphonic ester with methanol. If you use HCl 37 % you introduce water necessarly which slows the reaction. Sulfuric acid is more or less dry, but in LCMS after a couple of days I observed methylbisulfate formation (111 m/z -TIC ESI). This is a problem, because my product is also very polar and can not easyly purified (DCM/water might work, but the previous reaction contained a lot of white precipitate and this made the workup problematic). Will methanolic HCl work better (premix without water)? It can only form methyl Chloride which is volatile, but, I am wondering if HCl gas will just evaporate quickly under reflux conditions. Or can you recommend others acids for transesterfication? Methanesulfonic acid for example will form Methyl methanesulfonate which is like chlormethan carciogenic. Phosphoric acid would show the same problem.

Consider using a resin-bound acid such as Amberlyst 15 (hydrogen form) which can be removed by simple filtration

Interesting idea. Also should withstand the temperature of 70 degrees regarding the datasheet. The materials I found are indicated as wet. Does wet mean that there is residual water? Maybe it can be conditioned with methanol beforehand....

It is available dry  https://www.sigmaaldrich.com/GB/en/substance/amberlyst15hydrogenform1234539389203   Could also consider using TFA

Wash the Amberlyst with MeOH and dry in vac oven overnight should dry it.

The thing with tfa: The reaction runs for a couple of days. So I would expect formation of TFA-Methanol ester as Well. This ester may not be volatile anymore and might go more to DCM Phase than it goes to water phase. Am I wrong?

Boiling point of methyl trifluoroacetate is 43C

Hmmm, TFA might be worth a shot indeed then. Might try this and amberlyst 15.

Chemistry is an experimental science. Run hte reactions, at the very least it gives you more to put in your thesis. If there isn't a study of transesterification of phosphonic esters in the literature then maybe you can publish one.

Maybe it would be worth testing some sort of solid acid catalyst, maybe a zeolite, as separation of the solution from the solid catalyst would be much easier

@Walander that is right. I did two experiments, base and acid catalyzed beforehand and in the next round I want perform them under optimized conditions.

@raptorlane  Zeolites have been used to catalyse transesterification in biodiesels so well worth a try  https://www.sciencedirect.com/science/article/pii/S246802572030193X

@Waylander When I get this publication right they produce a catalyst under hydrothermal conditions - Seems way to complicated for me to be reproduced by me. Also, in a previous experiment me observed deteriation of my product in sodium methanolate 25 w%, so I would prefere the acid route until now. I saw the product emerging in time, Problem was the contamimation with the product of MeOH + sulfuric acid reaction.

@raptorlane zeolites are commercially available for sure you can pick up a couple of grams for quite cheap.

@Cbornes. I know them mainly as mol sieve. Feel free to link to a reference...


text:
Do (some or all) phosphates thermally decompose?
I know that metal carbonates decompose to a metal oxide and carbon dioxide upon heating and I was curious if phosphates (eg K3PO4) displayed similar behaviour? Or are they like sulphates where some decompose, such as Iron (II) sulphate, while others such as sodium sulphate do not (this behaviour also confuses me). I realise that unlike carbon and sulphur oxides, phosphorus oxide is not a gas, so will this have an impact? Or will it just not decompose for some reason?
So far my online research hadn't resulted in anything substantial, with searching of
'decomposition of sodium phosphate' resulting in academic papers that are very hard to understand (and don't appear to directly answer my question) and a single result that it will decompose to Na + PO4 (AI generated nonsense, clearly wrong).
Thanks

Metal phosphates containing the $\ce{PO4^{3-}}$ ion are in general thermally stable at temperatures commonly achieved in laboratory and industrial processes. In principle the phosphorus oxide component, $\ce{P4O10}$, is volatile and can be evolved as a gas; but the entropy of this gas evolution is not great enough to balance the large enthalpy release of the acid-base reaction between metal oxides and $\ce{P4O10}$.
Hydrogen-phosphates, where the phosphate ion has more complex stoichiometry, can be decomposed by forming denser polyphosphate salts. Banach and Makara [1]  describe the decomposition of sodium hydrogen-phosphate mixtures, $\ce{Na5P3O10}$, involving several stages and ultimately leaving the metaphosphate salt $\ce{NaPO3}$ (in which the phosphate functions are fully polymerized). Decomposition all the way to the metaphosphate requires a temperature of only $320°\text{C}$.
Reference

Marcin Banach and Agnieszka Makara (211). "Thermal Decomposition of Sodium Phosphates". J. Chem. Eng. Data, 56, 7, 3095–3099. https://doi.org/10.1021/je200381z

In order for a salt to be decomposed by heating it to high temperature, it must produce a gaseous substance by heating. This is the case for carbonates who deliver CO2, and for nitrates who deliver nitrogen oxides at high temperature. But it is not the case for phosphates, as phosphoric oxide is not easily gasified.

Volatility of $\ce{P4O10}$ itself is not an issue, sublimating at $\pu{360 ^{\circ}C}$. Rather releasing of the oxide from the salt ionic matrix.


text:
Elegant and fast way without chromatography to remove fluoride ions from water phase without contamination of the polar water soluble product?
I have a challenging purification routine. I want to remove fluoride from an aqueous solution, my product is a phosphonic acid, which is readly water soluble and an anion.
The anion is about 200 Da in weight and carries two negative charges. It would be nice, if purification could work in the 100mg - 1000 mg range (Basically 100 mg of validated pure substance should be sufficient for most biological and physical testing, but its just nice to have some scaling capacity).
In the solution the anions -in an ideal scenario- are my product and the abstracted fluoride (and maybe OH-, depending of the base). The product is very water soluble
and fluoride abstraction is likely to make it even more polar. After the fluoride abstraction, the F- anions need to be removed from solution. I know that you can precipitate fluoride with Ca salts, but are there also other ways to achieve this? I am worried that Ca will also precipitate part of my product out of solution. In previous experiments I saw loss of product because of Ca addition. Some kind of selective catching would be nice. I am wondering if scavenger exist, that are selective for F- only.
If there is no easy approach I would use ionic chromatography, but usually this is kind of time consuming, because you need to mix buffers with specific pH, equilibration takes quite long and than evaporation of water is also a factor, so any hint for streamlining is much appreciated.
Edit: Added more information on product.
Edit2: If you downvote, please be so kind and provide some explanation, otherwise no clearification can be provided.

What are major or important anions there? // Always include enough relevant details of the question purpose, context or background to make the question clear,  preventing  misunderstanding and need of  clarification.

The product itself is an anion. Fluoride is an additional anion, when it is abstracted from the product. An additional anion would be OH-,  because you need a base for hydrolysation.

I have been previously going to suggest anion exchanging resin, but it not applicable. How big is the anion and what are the quantities? What about ion exchanging of gel exclusion chromatography?  // I understand the product may be under business secret restriction, but less you reveal, less satisfying answers you obtain. It is like asking us how the forest can be improved, if you lead there the advisor after nightfall with a band cross eyes. Consider what you can  afford to share and share it, otherwise I am not sure you will receive suitable answers.

@Poutnik The anion is about 200 Da in weight, carries two negative charges. It would be nice, if purification could Work in the 100mg - 1000 mg. (Basically 100 MG of pure substance should be sufficient for most validation, but its Just nice to have some scaling capacity) I assume you mean its not applicable, because it will Stick to the Resin and is not going off anymore?

Put additional info rather to the question. Essential is also to know at least the compound class. Questions should not be riddles.

@Poutinik Added additional information and adjusted some passages language wise to make things more clear.


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Is there any chemical that has a fire diamond (NFPA 704) of 3-4-2 COR OXY?
At Walmart the other day I spotted this fire diamond. It was randomly there on a wall on the side of the store, with seemingly no explanation. There were no chemical tanks or anything near the sign, but there was a door that was a little bit away.
I cannot find anything that has this fire diamond, let alone a corrosive oxidizer.
Any help in finding this mystery chemical is appreciated.
The fire diamond is shown below. It does use the informal "COR" symbol, which means corrosive.

There's probably many examples that would fit.

I can't find one.

Well, that's a matter of right method of searching, probably best to go through the original source like andselisk did: https://chemistry.stackexchange.com/questions/85342/what-is-the-chemical-compound-that-has-a-nfpa-704-fire-diamond-of-2-2-1-ox

I've tried finding a place that can search for chemicals by fire diamond. I can't seem to find one.

The final values in the diamond are a combination of all chemicals within the placarded area. Whether that is from one single chemical, or multiple, is not definable from just the placard.

I agree with Jon Custer's comment. This symbol (see below) is not for any specific chemical:

The ComplianceSins.com says following about this NFPA 704 Diamond with 3-4-2-COR_OXY Hazard Level Indicators:

Please NOTE: The symbol in the white Special Hazard area of this specific diamond is not considered an "authorized symbol" by the NFPA. Before using this symbol, consult your local emergency response team regarding the use, size and placement of non-authorized NFPA symbols with your NFPA diamond.


text:
Dynamic equilibrium of Girard T-derivatized oxysterols
This is a question relating to organic chemistry, moreso than chromatography.
I am running LC-MS on various hydroxycholesterols (HC) and dihydroxycholesterols (diHC). To increase ionization efficiency with the electrospray, I derivatize my oxysterols with Girard T reagent. The picture below shows the two forms, syn and anti, that the derivatized oxysterols take (here showing derivatized 7α,24S-dihydroxycholesterol).

It is common knowledge that these isomers often have different retention times through an LC column, leading to split peaks. Some of my oxysterols do show splitting, but my group believes this not to be caused by the syn/anti-situation, as they claim the stationary phase I am using (phenylhexyl from ACE) does not show this. Rather, they say it is probably  contamination of our standards (perhaps of some precursor).
I have observed that the split peaks converge when column temperature is increased. Below is a chromatogram of 7α,24S-diHC splitting at 40 C, but becomes a single broad peak at 70 C. This is consistently true for all my diHCs, but only one (22R-HC) of my HCs (the rest show no splitting at 40 C).

The mobile phase moving the oxysterols along is around 67% water, 26.7% acetonitrile and 6.3 % methanol. The pH is 2.8 via 0.1 % formic acid.
My question is this: Could there be a dynamic/active equilibrium ongoing between the syn and anti conformations, or is this unlikely given the presumably rigid hydrazone bond? This could, perhaps, explain the "bridge" between the two split peaks.

What were the reading of MS for two peaks at 40 C? That would easily tell you if they are geometric isomers.

It is also possible that peak at 20.08 min would be belongs to unreacted Girard T hydrazine. Temperature might affected its retention time more effectively than that of hydrazone.

@MathewMahindaratne The chromatogram shown is with selected reaction monitoring (SRM) for a specific fragment from the derivatized oxysterols (530.432->471.358). Both peaks are the same exact mass.
It shouldn't be the unreacted Girard T, both because of the SRM, and because I'm using a setup where I load the oxysterols onto an SPE, flush it with water to remove excess Girard T reagent to waste, then switch and elute the oxysterols to the column with organic solvent.

Also, regarding them being potential geometric isomers, usually I would agree, but it is still possible for the split peak to be some precursor used in the production of the oxysterol that for some reason happens to have the same m/z (using a triple quad. resolution isn't the highest)


text:
If NaOH is a Brönsted-Lowry base then, what is its conjugate acid?
Brönsted-Lowry theory doesn't go against but adds to the Arrhenius theory on bases and acids.
$\ce{NaOH}$ is an Arrhenius base because it releases hydroxide ions in water solution.
NaOH is also a Brönsted-Lowry base because it dissociates in water and its hydroxide ion reacts with water(which acts as an acid).
$\ce{NaOH = Na+ + OH-}$
So if the "actual" Brönsted-Lowry base in NaOH is the hydroxide ion then is the conjugate acid here water? Is the conjugate acid of NaOH, water?
My book on the other hand, says that "$\ce{Na+}$ is the conjugate weak acid to NaOH which is a strong base. So when NaCl is dissolved in water, $\ce{Na+}$ will not react with water at all" .  But how can $\ce{Na+}$ be an acid when it doesn't have any hydrogen to give?
So which one is the conjugate acid, $\ce{H2O}$ or $\ce{Na+}$ and how can $\ce{Na+}$ be an acid at all?
Can $\ce{Na+}$ be a conjugate lewis acid but not be a conjugate Brönsted-Lowry acid?

Duplicates https://chemistry.stackexchange.com/questions/35673/what-is-the-conjugate-acid-of-naoh-using-the-br%c3%b8nsted-lowry-definition-of-acids https://chemistry.stackexchange.com/questions/97341/is-sodium-hydroxide-a-bronsted-lowry-base https://chemistry.stackexchange.com/questions/98084/how-can-arrhenius-bases-also-be-br%c3%b8nsted-lowry-bases and low hanging ones at that.

The conjugate acid to the Brønsted-Lowry base $\ce{OH−}$
is $\ce{H2O}$, just added a proton to $\ce{OH−}$.
$$\ce{B(aq) + H+(aq) <=> BH+(aq)}$$
$$\ce{OH-(aq) + H+(aq) <=> H2O(l)}$$
where $\ce{B} = \ce{OH-}$
OTOH, $\ce{H2O}$ is also the conjugate base to $\ce{H3O+}$.
$\ce{Na+(aq)}$ is an extremely weak Lewis acid. But in a gaseous phase, as a naked ion, it reacts with Lewis base $\ce{Cl-}$, forming an ionic pair $\ce{Na+Cl-}$.
$\ce{Na+}$ could be formally considered as a conjugate Arrhenius acid to $\ce{NaOH}$, but I am not sure if the concept of conjugate pairs is ever used in the Arrhenius theory.

In Arrhenius there's no conjugate acid, but $\ce{Na(H2O)+}$ complex can be considered one in B-L.