text:
Are carbocations necessarily sp2 hybridized and trigonal planar?
My copy of Pearson's Organic Chemistry (7e), Morrison and Boyd, under the section "Reaction intermediates", accords a succinct description on the structure of carbocations:

The central $C$-atom (of carbocations) is in an $\mathrm{sp^{2}}$ hybridized state, for which the carbocations have planar geometry. The $\mathrm{p_{z}}$-AO (atomic orbital) remains empty.
 The stuff in parentheses was added by me

Aided by this description, I conjured up the following "general" structure of carbocations:

 Though I pulled out the above image from Google Images, it was pretty much the same structure I've been visualizing this whole time...drawing my own would be messy
And as you can see, I've equated the "planar structure" mentioned in the book to "trigonal planar structure" (with an axial vacant $p$ orbital). This image of a carbocation's structure in mind proved rather handy, and didn't seem to be incorrect at all.
Wikipedia, on the other hand, doesn't sound so confident about the central $C$-atom's  $\mathrm{sp^{2}}$ hybridized state.

One could reasonably assume a carbocation to have $\mathrm{sp^{3}}$ hybridization with an empty $\mathrm{sp^{3}}$ orbital giving positive charge. However, the reactivity of a carbocation more closely resembles $\mathrm{sp^{2}}$ hybridization with a trigonal planar molecular geometry.
 (Emphasis, mine)

As you can see, Wikipedia doesn't appear to (completely) endorse the $\mathrm{sp^{2}}$ structure of the central $C$-atom.

I continued to keep the "trigonal planar" structure of carbocations in mind while studying them. This posed no hindrance until, I came across these carbocations (in a book not really worth mentioning):

 Created using PubChem Sketcher V2.4
I've faced multiple problems while trying to ascertain the hybridization cum geometry/structure of the central, positive $C$-atoms in those carbocations. I shall list them separately,
1) Issue with the Aryl carbocation

I visualized this as a particular Kekule structure of benzene having lost one hydrogen anion, thereby leaving a positively charged carbon atom in the ring. Considering the bonds involving the positive $C$-atom (in the particular Kekule structure I put up), I see two $σ$ bonds and one $π$ bond. Also, the $\mathrm{C=C^{+}-C}$ bond angle appears to be $\mathrm{120^{o}}$ (just like the normal benzene molecule. I honestly can't figure out the hybridization or structure/geometry of the positive $C$-atom here. I guess I should factor in the "delocalization of the positive charge" across the ring, but that hasn't borne fruit (for me).
2) Issue with the vinyl carbocation

I visualized this as an ethene molecule, having lost one hydrogen anion, thereby leaving a positively charged carbon atom (seen on the right end in the image). Here again, I see two $σ$ bonds and one $π$ bond. From my knowledge of the VSEPR theory, I suppose the $\mathrm{C=C^{+}-H}$ bond angle is $\mathrm{180^{o}}$ (i.e- linear). But I can't for the world figure out what the hybridization of the positive $C$-atom here, is. Heck, I'm not entirely sure if I predicted the geometry (linear) correctly in the first place...well, this case is alien to me.
3) Issue with the ethynyl carbocation

I visualized this as an ethyne molecule, having lost one hydrogen anion, thereby leaving a positively charge carbon atom (seen on the right end). Considering the bonds involving the positive $C$-atom, I see one $σ$ bond and two $π$ bonds. Hybridization? No clue. Geometry about the positive $C$-atom? Um...kinda looks like a ball at the end of a stick...not sure if there's any "angle" present ._.

Could someone please address these "issues" I've encountered for the above mentioned (aryl, vinyl, ethynyl) carbocations? I'm not sure if assuming "planar" structure" necessarily means "trigonal planar structure"... or if there's something about "hybridization" that I've grossly overlooked.
[Note- What I've been taught, is that a particular hybridization state, ensures a particular geometry/structure....the result of trying to combine "Hybridization" with the VSEPR theory]
My question(s), more explicitly put:
1) What is the hybridization state of the carbon atom carrying positive charges in the three examples I've used above? How is it determined?
2) What is the geometry/structure of the said hybridized carbon atoms? {If that isn't clear: I meant along the lines of "If it's $\mathrm{sp^{3}}$ it's tetrahedral, if it's $\mathrm{sp^{2}}$ it is trigonal planar, if it's $sp$ it's linear"}
I'm still in High-school, so I feel a bit overwhelmed at the moment (trying to wrap my head around this...hopelessly)

@Sawarnik Yeah, and the same goes for the ethynyl carbocation. Wanted to draw it using bond-line notation (meaning $C-H$ are implied)... https://www.google.co.in/search?q=vinyl+carbocation&rlz=1C1CHBF_enIN710IN710&oq=vinyl&aqs=chrome.1.69i57j69i59j0l4.3371j0j7&sourceid=chrome&ie=UTF-8

Don't forget the 1-adamantyl cation: http://pubs.acs.org/doi/abs/10.1021/ja00515a002

http://pubs.acs.org/doi/pdf/10.1021/jo990724x

Is this supposed to be carbenium ions?  (https://en.m.wikipedia.org/wiki/Carbenium_ion).  Carbocations are a much broader class.

@Oscar Ouch, "carbenium ions" and "carbonium ions" are new terms to me.  I've always used "carbocation" (oblivious of it's broader implications), and I guess that's because it's only as far as org. chem goes at my school. Now, I've tried to make comparisons between Wikipedia pages on "Carbocations", as well as "Carbenium" and "Carbonium" ions...however that leads me to believe use of "Carbocation" is more appropriate {Continued..}

{Continued..} The Wikipedia entry for "Carbenium ion", refers to a trivalent carbon with a positive charge. Whereas the entry for "Carbonium ion" refers to a pentavalent carbon with a positive charge. However, my examples (Aryl, vinyl, ethynyl carbocations) don't seem to fit either description. So I guess my use of "Carbocation" was, correct. I shall try to edit the question to make my stance clearer O:)

@paracetamol I think the -enium suffix is used for electron deficient (as in lesser than 8 electrons in the valence shell) while -onium is used for there being too many bonds. e.g. carbenium,  carbonium - both are carbocations, of which carbeniums are more common. Then, oxenium and oxonium, of which oxoniums are more common.

I actually have a (or many) big issue(s) with the quote:

The central C-atom is in an sp2 hybridized state, for which the carbocations have planar geometry. The p$z$-AO remains empty.

The authors here have clearly scrambled up their reasoning, making carbocations seem as something they are definitely not. Suffice to say (tl;dr) the above statement cannot be true. Let's get a few points straight before moving on to a more complex examples.

The p orbital remains empty.
We know that s orbitals ($\ell=0$) of the same principle quantum number $n$ have a lower energy than the corresponding p orbitals ($\ell=1$). It is therefore (almost) always energetically more favourable to occupy orbitals with as much s character as possible.
The coordination is planar.
Ideally one (any) of the p orbitals will remain completely unoccupied. Because of symmetry considerations, a planar arrangement of ligands around the central atom virtually ensures that. The planar coordination is a result of a favourable electronic state. Obviously there will be other interactions at play, but in a first approximation the above is always true.
(Also note that I am avoiding the word geometry, because that should rather be reserved for the whole molecule.)
Orbitals are hybridised, not atoms.
There is no such thing as an "hybridised state". There might be an atom of which the wave function can be described with hybrid orbitals. The colloquial phrase "the carbon is sp3 hybridised", which is especially popular with organic chemists, is a garbage simplification.
Valence Bond Theory is not a simplification; a.k.a. Bent's rule.
The description with sp$n$ orbitals is a relic of the very, very first days of VB theory. Nowadays this theory has well evolved past these rigid kinds of descriptions. Essentially, allowing $n\in\mathbb{R}$ produces better descriptions and a better agreement with experimental data. (Read more: What is Bent's rule? Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?)
Hybridisation is a mathematical description.
We would be completely fine without hybridisation. We choose to use hybrid orbitals, because they (in most cases) represent the geometry of molecules in a much easier view than the very generic canonical orbitals.
Unfortunately, hybrid orbitals became a tool of prediction in organic chemistry textbooks because they are so temptingly easy to understand. As a result many things get explained in this way where it would not the least be necessary. Often leading to wrong conclusions, other times being right only by coincidence (right for the wrong reasons).
Carbocations are nothing trivial.
It took a couple of years for the theory to be accepted and then confirmed by experiments, showing that there is nothing easy to fathom. In terms of electronic stability, only occupied orbitals count. Molecular entities will always adopt the lowest lying electronic state in the the optimal geometry.

Just because of Bent's rule it is only logical to assume that carbocations in general may differ significantly from the often taught 3×sp2 + p hybridisation scheme. In principle, only carbocations of the form $\ce{^+CR3}$ are symmetric enough to have this scheme. This already starts to break down with $\ce{R{ = }CH3}$ because of hyperconjugation. In first approximation, however, the convenient model holds. Just keep the limitations in mind.
With all of that we can go to your specific questions. All of your examples are what we often refer to non-classical carbocations. You may now ask yourself: What is a non-classical carbocation? I therefore recommend reading the linked Q&A before continuing. (Importance of such cations. Shameless self-promotion.)
I personally dislike the terminology and the definition in the gold book, as I find it a little reactionary, but we're stuck with it, there is no use in complaining.

nonclassical carbocation
  A carbocation the ground state of which has delocalized (bridged) bonding π- or σ-electrons. (N.B. Allylic and benzylic carbocations are not considered nonclassical.)

Note for the remaining part of the answer I am keeping things short as I am just summarising stuff from two sources on our network: (1)Do vinyl cations adopt a classical or non-classical structure? (2) Is the phenyl cation or ethynylium more stable?

Phenyl cation/ Aryl carbocation
In this case we have a cationic carbon which is already planar. Therefore the necessary change would be to adopt a linear coordination. This obviously is restricted by the cyclic backbone.

Technically this is not a non-classical carbocation according to the definition (or is it?), which is one of the reasons why I don't like this definition in the first place.
A true non-classical version with a bridging proton is not a stable stationary point on DF-BP86/def2-SVP.
While the bridging $C_\mathrm{5v}$ symmetric $\ce{^+C(CH)5}$ is a stationary point, it is about $\pu{145 kJ mol-1}$ higher in energy.
Vinyl cation
tl;TL;DR;dr: More recent work indicates that the bridged form of the vinyl cation with is slightly more stable (by about 1-3 kcal/mol).

Ethynylene Carbocation
tl;dr: The linear $\ce{HCC+}$ is not a stationary point at DF-BP86/def2-SVP. The stable structure is an almost three-membered ring, which is best thought of as a protonated dicarbon.

Conclusion (?!)
Throw out the restrictive thinking of hybridisation. It is almost always useless when it comes to carbocations (best case scenario) or even gives you the completely wrong ideas. Always remember that orbitals can be described hybridised, but not atoms, and that hybridisation itself is never a fixed deal.
Always keep in mind that the smallest molecular entities do the weirdest things, with the most complicated bonding situations.
Stay open-minded.

The structure you show for the phenyl cation is similar to that in the isoelectronic borinine, discussed here. The position of the carbocation center or boron atom is such that the sigma bonding may bridge around the electron-deficient atom, passing through the "empty" outward-pointing orbital. This structural feature would seem to qualify the carbocation as a "nonclassical" ion. (Neutral borinine might also be considered nonclassical.)

@Oscar Agreed. This is very interesting and quite similar. I don't have the time (or probably access) to read the publication, and I have not saved those calculations to compare them to this, but your claim is reasonable. However, I think we have reached the point where terms like "classical" or "nonclassical" loose their helpful meaning.

That notion is far from true.  There are many examples of carbocations where, through the use of delocalized bonds, carbon can be bonded to five or more atoms.  See, for example, https://en.m.wikipedia.org/wiki/Carbocation.  Among other things, this shows that even methane can be protonated to give not $\ce{CH3+}$ but $\ce{CH5+}$!

These are separate class (carbonium ions).

Carbonium ions are a type of carbocation.  And the question uses "carbocation".

Well, I think @para thought about carbenium ions, looking at his examples, nice catch though.

@Oscar Sorry I was late in responding to this >_<. Your answer was useful, but I'd be grateful if you could expand on it a bit more. Being the idiot schoolboy I am, I'm faced with... "difficulties"... in accurately comprehending the subtleties present in most sources on this topic [My confusion with "Carbocation", "Carbenium ion" and "Carbonium ion" is an example]. More specifically, I'd love it if you could elaborate on "...through the use of delocalized bonds, carbon can have a valence of five or more...".

In addition to the above; could you also explicitly address why I was unable to determine the hybridization and structure of the "carbocations" that I used as example in my post?


text:
What exactly is "B-strain"?
On factors that influence the stability of carbocations, I came across this peculiar one:

Bulky groups attached to the positively charged $C$-atom (of the carbocation), stabilize it by B-strain, which prevents the C-atom for returning to the $sp^{3}$ state.
 Source: Pearson's Organic Chemistry (7e), Morrison and Boyd

The issue? I've never heard of "B-strain" before. I mean, it's possible that bulky groups stabilize the positive $C$-atom via inductive (and other) effects...but what's with "B-strain"?
I ran a cursory Google search, but I couldn't find any reliable Chemistry-literature that deals with this. In fact, it raised more questions than answers, most notably, the presence of another effect called the "F-strain".
So I must ask:
What is the "B-strain", and how does it contribute to stabilization of carbocations?
Additionally, is there such a thing as the "F-strain"? If so, what's it all about?

B stands for back or behind (Carey, F. A.; Sundberg, R. J. Advanced organic chemistry Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007., p. 417). F stands for front (Houghton. Metal Complexes-Organic Chemistry, 1 edition.; Cambridge University Press: Cambridge ; New York, 1979. p. 66). If you have troubles accessing the sources, I can try to provide an answer including the book parts.

@andselisk I see! I know for certain that my library doesn't have the books you mentioned, so I'd really appreciate an answer from you. Meanwhile, I'll try looking for the PDF versions ...

B-strain
Both B- and F-strains are the terms I found were originally used to describe steric hindrance in complexes [1, pp. 66-67]:

The concept that steric hindrance in a monodentate ligand can affect the affinity of that ligand for a metal was first introduced by Brown and Bartholomay (1944) who suggested that co-ordination of a tertiary amine with bulky substituents would create strain within the amine. This strain, termed B-strain on account of it occurring behind the ligand atom with respect to the metal, would arise because although in the unco-ordinated amine any steric interaction between the bulky substituents could be relieved to some extent by an increase in the $\ce{C-N-C}$ angles.

Interestingly, further it is pointed out that at the same time there is no other solid evidences that B strain exists. Although modern organic chemistry textbooks are widely using this term for tertiary compounds.

From [2, p. 417]:

Steric effects of another kind become important in highly branched substrates,
  and ionization can be facilitated by relief of steric crowding in going from the tetrahedral ground state to the TS for ionization. The relative hydrolysis rates in 80% aqueous acetone of $t$-butyl $p$-nitrobenzoate and 2,3,3-trimethyl-2-butyl $p$-nitrobenzoate are 1:4.4.

This effect has been called B-strain (back strain), and in this example only a modest rate enhancement is observed. As the size of the groups is increased, the effect on rate becomes larger. When all three of the groups in the above example are t-butyl, the solvolysis occurs 13,500 times faster than in t-butyl p-nitrobenzoate.

According to [3, p. 350]:

Not all steric effects decrease reaction rates. In the hydrolysis of $\ce{RCl}$ by an $\mathrm{S_N1}$ mechanism ..., the ﬁrst step, which is rate determining, involves ionization of the alkyl chloride to a carbocation:

The central carbon in the alkyl chloride is $\mathrm{sp^3}$ hybridized, with angles of $\sim 109.5^\circ$, but when it is converted to the carbocation, the hybridization becomes $\mathrm{sp^2}$ and the preferred angle is $120^\circ$. If the halide is tertiary and the three alkyl groups are large enough, they will
  be pushed together by the enforced tetrahedral angle, resulting in strain. This type of strain is called B strain (for back strain), and can be relieved by ionization to the carbocation.
[...]
Substitution of ethyl groups for the methyl groups of tert-butyl chloride
  does not cause B strain; the increase in rate is relatively small, and the rate smoothly rises with the increasing number of ethyl groups. The increase is caused by normal ﬁeld and resonance (hyperconjugation) effects. Substitution by one isopropyl group is not greatly different. But with the second isopropyl group the crowding is now great enough to cause B strain, and the rate is increased 10-fold. Substitution of a third isopropyl group increases the rate still more.

Further in [3, p. 419]:

For some tertiary substrates, the rate of $S_N1$ reactions is greatly increased by the relief of B strain in the formation of the carbocation. Except where B strain is involved, $\beta$ branching has little effect on the $S_N1$ mechanism, except that carbocations with $\beta$ branching undergo rearrangements readily.

There is also a Baeyer strain aka angle strain most noticeable for three-membered rings ($60^\circ$ is far from expected tetrahedral angles), which can also probably be abbreviated alike. [3, p. 193]
It also seems like earlier sources used B-strain as a reference to H. C. Brown [4, p. 291]:

Hammond has proposed a steric "B-strain" (see H. C. Brown, J. Chem. Soc. (London) p. 1248 (1956)) argument to explain the greater stability of primary over secondary or tertiary carbanions. From the viewpoint of hybridization, structure A with $90^\circ$ geometry is preferred in carbanions. But this conformation brings the substituents too close together, and a compromise structure with larger bond angles results. Methide ion, $\ce{^-CH3}$, with the smallest substiuent groups, should be the most stable simple carbanion, since it can more nearly adopt the small angled conformation A preferred on electronic grounds. The more alkyl groups are added, the less able sterically are
  the anions to adopt the favorable conformations with smaller angles, and
  the less stable and more reactive the intermediates become.

F-strain
[1, p. 68] illustrates the origins of the term F-strain:

The boron trifluoride and boron trichloride complexes of diethyl ether are less stable than those of dimethyl ether, and the same order of stability is observed for the complexes of diethyl and dimethyl sulphides. As expected, the steric interaction is decreased as the distance between the metal and ligand atom is increased. Thus it is decreased when the metal atom is changed from boron to aluminium, and when the ligand atom is changed from oxygen to sulphur. It is largely eliminated, of course, if the substituents on the ligand atom are held back away from the metal, and thus very stable complexes are formed by the tertiary amine quinuclidine, while complexes of tetrahydrofuran and tetrahydropyran (and also the analogous sulphur heterocycles) are far more stable than those of their acyclic analogues.
Strain which arises as the result of this type of steric interaction has been termed F-strain by H. C. Brown -- who has carried out extensive investigations in this area -- on the grounds that the interactions responsible for the strain occurs mainly in front of the ligand atom with respect to the metal. For a critical but now rather dated review of H. C. Brown’s work, see Hammond (1956).

References

Houghton, R. P. Metal Complexes-Organic Chemistry, 1st ed.; Cambridge University Press: Cambridge; New York, 1979.
Carey, F. A.; Sundberg, R. J. Advanced organic chemistry Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013.
Hartley, H.; Karabatsos, G. J. Advances in alicyclic chemistry; 1966; Vol. 1.

The term "B-strain" was coined in the article Acid-Base Studies in Gaseous Systems. II. The Anomalous Base Strength of the Methylamines; A New Manifestation of Steric Strain J. Am. Chem. Soc., 1944, 66 (3), pp 435–442.

The term “B-strain” is proposed for this effect.
(The letter B is used to indicate that the
interference which causes the strain is localized
at the “back” of the molecule-trimethylamine
in the case under discussion-away from the entering
group; it is thus contrasted to F-strain
which results from interference at the “front” of
the molecule, at the interface between the two
components of the addition compound.)

Now consider an acid (Say Boron trifluoride) forms acid-base adduct with a base  with bulky substituents (Say tripropylamine), then there will be steric hinderance between the bulky substituents which results in opening bond angle (Propyl-Nitrogen-Propyl) of the alamine. This opening in bond angle causes more s-character in the N-R (Propyl-Nitrogen) bond more p-character left in the lone pair of nitrogen due to which lone pair is poorly situated for donation. This phenomenon is called B-Strain.
Note: F-strain decreases the basicity of the base.
       More s-character results larger bond angle
 For more detail Refer Inorganic Chemistry By James E. Huheey, Ellen A. Keiter, Richard L. Keiter and Okhil K. Medhi

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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

sort: votes


text:
Calcium-containing acidic solution
I want to prepare an acidic solution containing Ca ions. I thought about dissolving calcium sulphate, acetate or phosphate in $\pu{2 M}$ $\ce{H2SO4}$ or $\ce{H3PO4}$ solutions, but it seems that $\ce{Ca}$ cations react with sulphate anions.
Is it true? And is there any way to make a $\ce{Ca}$-containing acidic solution?
(I forgot to say I need $\ce{Cl}$  & $\ce{F}$  free solution...)

You could dissolve calcium chloride in water and add a small amount of HCl.

Then go for nitrate. And if you don't like it for some reason, there are quite a few more acids out there.

The best way to produce a calcium containing acidic solution is using calcium nitrate in HNO3 containing acidic solution. And please avoid sulfate or phosphate ions, which produce precipitates with calcium ions. Why do you want to avoid chloride ions ?

no chloride allowed

Probably your most flexible option is to make a buffer solution of a weak acid with its calcium salt. For instance, calcium acetate plus acetic acid with equal molarities of acetate ion and acetic acid will have a pH of about 4.8. If you want a more strongly acidic solution, you can use sulfamic acid ($pK_a\approx 1$) as the acid. Calcium sulfamate, unlike calcium sulfate, is readily soluble in water. Sulfamic acid is in fact used commercially in lime-scale cleaners.

If nitric acid is not suitable sulfamic as mentioned or possibly methanesulfonic or benzenesulfonic acids supposedly have soluble calcium salts and should be strong enough. A good idea would be to start with CaCO3, chalk, and dissolving it. then there would be only the acid anion present after the CO2 is evolved.

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


text:
Is there a systematic naming system to name carbonyl derivatives?
I had asked a question earlier about how the “$\ce{-COOCl}$” functional group would be named.
I had received a satisfactory answer that helped me completely understand how that compound was named.
However, I have a much broader question in mind now.
Consider the "$\ce{-COOH}$" functional group. We know that it's called a carboxylic acid group, however I used to think, why wasn't ethanoic acid simply called as a 1-hydroxyethanal? I now know that "1-hydroxyethanal" has properties vastly different from both aldehydes and alcohols, and as a result, was named differently to acknowledge this difference. Many acid derivatives have been named (and behave) differently, as a result of the $\ce{C=O}$ group nearby.

However, these may not be the only molecules that are affected by the nearby $\ce{C=O}$ group. My earlier question had dealt with the $\ce{-COOCl}$ functional group (not to be confused with acyl chloride $\ce{-COCl}$). However, there are a lot more:

Of course, I did try to name these compounds myself.

Cyanide is sometimes treated as a pseudohalide, so my guess is ethanoyl cyanide.
It may be an anhydride of hydrogen sulfide and ethanoic acid, but I'm not very sure.
Loong gave me an interesting way to name certain "anhydride" compounds in my previous question. This looks like an anhydride of hydroxylamine and ethanoic acid. So according what Loong told me, my guess is ethanoic hydroxylamine anhydride.
This one look's rather crazy to me. Had this been posessing a $\ce{C-O-S}$ instead of a $\ce{C-S}$, this would have been an anhydride of sulfuric acid and ethanoic acid. But it's not, so I have no clue on how to name it.
This one looks very familiar to an amide, infact a phosphorus analog of amide. For some reason, I feel like calling it ethanephosphide, although I don't know why.
Like cyanide, even azide is considered a pseudohalide, so my guess is ethanoyl azide.

Please keep in mind, I am not asking for the explanations to the naming of all the six compounds I showed above, although it would be great if I could get the names alone. Rather, what I mean to ask is is there any simplified nomenclature that can deal with all these types of compounds in a simple way? Although I only showed six, I can go on as long as my imagination permits.

I think it is a great question. As a suggestion from my side, could someone also comment on how bright is the idea to use an algorithmic name generation bundled with ChemBioDraw/MarvinSketch for these structures?

I don't know if it's really worthwhile including all your thoughts on the matter.

@orthocresol Well, I asked quite a big question, and it wouldn't be nice if I didn't show a proportional amount of effort that I took before asking someone else to help me.

If I were to write an answer, I wouldn't really want to pay attention to what you've written. I think that writing the correct answer for 6 compounds, complete with justification, is enough work. I've already spoken out, on multiple occasions, about my stance that too much effort is not useful. Maybe I'm the only one who thinks that way, but I just wanted to put it out there for consideration.

These are very different compounds. It's not like you can get whole knowledge of carbonyl compounds' nomenclature in one go.

Yes @Mithoron, but they all share something in common, the attached $\ce{C=O}$ drastically changes their usual properties. However you're probably right that I simply can't learn the entire carbonyl nomenclature in a single go.

https://en.wikipedia.org/wiki/Sulfonic_acid https://en.wikipedia.org/wiki/Thiocarboxylic_acid

@Mithoron thank you for your first link, although I was already aware of sulfonic acids. However, are you trying to imply that, in addition to naming it as "alkyl sulfonic acid", we may also name it as "acyl sulfonic acid"?

@Mithoron, Thank you very much for your second link, that helped me name the second compound completely.

@orthocresol If I were to write an answer, I wouldn't really want to pay attention to what you've written. That is perfectly correct and I feel that's rather a human tendency. However, we do close down simple homework questions because we don't feel like wasting time helping a person who can't help themselves. Also, right now, my main motive is to get an answer. I don't want by any means to discourage any potential answerer by making him think that I don't show as much interest into getting my own answer as the answerer is in trying to answer my question. I may be silly right now, so...

... I'd just want to clarify if my thoughts are valid.

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

$\ce{CH3CONH2}$ is not an anhydride.

@mykhal Whoops sorry, that must've been a typing error.

The names (2 of 6) are rather correct (but note that acetyl and  acetic names are preferred over ethanoyl and ethanoic), except following:

$\ce{Ac-SH}$: ethanethioic acid, more specifically ethanethioc S-acid (since there are two possibilities of “monothiation”)

P-15.5.3.4.1
(…)
$\ce{CH3-CO-SH}$
  thioacetic S-acid
  ethanethioic S-acid (PIN)
$\ce{CH3-CH2-CS-OH}$
  thiopropionic O-acid
  propanethioic O-acid (PIN)

$\ce{Ac-O-NH2}$: Anhydride name cannot be used as hydroxylamine is not a name of acid. It's also not a recognized acid functional group, so the name probably should be regarded a substituted ketone, with preselected prefix aminooxy (P-68.3.1.1.1.2), 1-(aminooxy)ethan-1-one. (Although pubchem entry lists wrong name: amino acetate; O-acetyl hydroxylamine is acceptable I think.)
$\ce{Ac-SO3H}$: “α-oxosulfonic acid“, 1-oxoethane-1-sulfonic acid
$\ce{Ac-PH2}$: phosphide as alleged analogy to amide (with different meanings in inorg. and org. chem.) is not used, the name should be 1-phosphanylethan-1-one (or acetylphosphane).

P-64.1.2.1 Pseudoketones
(…)
$\ce{H2P-\overset{1}{C}O-\overset{2}{C}H2-\overset{1}{C}H3}$
  1-phosphanylpropan-1-one (PIN)
  propanoylphosphane

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is the activation energy of iron rust?
Iron rusts to Geothite (FeO(OH)). What is the activation energy for this reaction? Where can I find this information?

Try google scholar.   Iron can rust to many things under many conditions.  This question is broad.

It is goethite. As for the activation energy, I don't believe it is even defined here. The reaction is too complex.

@Agriculturist Any single activation energy for any product.

In that case google "Activation energy for iron oxide"


text:
Why does the reaction between Na (s) and Cl2 (g) occur spontaneously if the entropy of the system decreases?
1- The reaction is exothermic.
2- My book says that it is because the total entropy change is very positive.
Here is my book's explanation and it is the highlighted part which i dont understand :
"Despite the fact that there is a decrease in the entropy of the system, the energy leaving the system produces a substantial increase in the entropy of the surroundings, because there are more ways of arranging the energy quanta in the rest of the universe than there are of arranging them in the system alone.

The book is correct. Spontaneity is governed by the change in entropy of the universe which is the sum of the change in entropy of the system and the change in entropy of the surroundings. The change in entropy of the surroundings is derived from the enthalpy change of the system.

@Zhe  okay what i didnt understand is what i wrote in bold. So what if there are more ways of arranging the energy quanta in the rest of the universe than there are of arranging them in the system alone?

That's the definition of entropy...

@Zhe  i think i got it. It is because there will be a smaller chance of having the energy quanta divided equally over the molecules in the surroundings right ?

"It is because there will be a smaller chance of having the energy quanta divided equally over the molecules in the surroundings" Not sure what that means... The number of microstates available in the surroundings per unit increase in energy is larger. That is the definition of entropy.

@Zhe  yeah never mind what i said in that comment but here is what i am thinking of now : energy is gained by the surroundings. Now the entropy of the surroundings is greater because there are more ways of sharing the energy quanta between the particles in the surroundings than there was which means that there is a less chance of the particles in the surroundings having equal amounts of energy. But why did the book compare the number of ways in which the energy quanta could arrange themselves in the universe , with the number of possible arrangements in the system ?

The count in the system decreases. The count in the surrounding is increases. The total for the universe increases because the surroundings' increase is larger. Also, be careful, but entropy works on density of states, not raw state count...

@Zhe  Sorry for not being clear when i explained my idea. I will try to make it more simple now. Lets imagine that all the energy is transferred to the molecules in the air (not to anything else). The surroundings now contain more quanta of energy than before, and all these quanta(s) are contained within the molecules in the air. An ordered arrangement would mean that these molecules are stacked together and not moving and all these molecules possess the same amount of quanta. Now what i understood is that there are more quanta of energy in the surroundings

@Zhe  (continued) , so that there are more possible ways of sharing these quanta(s) of energy between the molecules. More ways of sharing these quanta(s) means that there is a less chance that each molecule will have the exact same amount of quanta. But why did they compare the number of ways of arranging the energy quanta in the surroundings with the number of ways the energy quanta can be arranged in the system ?

Max white: While writing an answer to that analogous question, I suddenly had the idea how to answer your question easily and without obvious paradoxes. Perhaps you are interested to read the radically  new edit of my answer.

As you point out there are two things to consider, the heat released in the reaction and the change in entropy. In forming NaCl from chlorine gas and solid sodium the entropy decreases because the gas is converted into solid NaCl which has a far far smaller total volume. The energy released has the effect of heating the solid NaCl just formed and then the surroundings as energy flows away to eventually reach equilibrium again but now at a higher temperature than that at equilibrium before reaction occurred.
When a quantity of heat $dq$ from a body at $T_2$ transfers to one at $T_1$ the overall entropy change is $dS = dS_1 + dS_2 = dq/T_1 - dq/T_2$. As $T_2 \gt T_1$ the entropy change is positive.
In the reaction the entropy change overall depends on comparing that decreased due to volume decrease vs. that increased due to the energy released on reaction.
Entropy is a measure of the number of ways that the particles can be placed into their energy levels, this is often called the number of configurations $\Omega$, and Boltzmann found that $S=k\ln(\Omega)$ so that increasing the distribution of energy into the energy levels increases $\Omega$ and hence S. As the temperature increases more vibrational and rotational levels are populated and the kinetic energy distribution widens, all this means that there is an increased number of configurations and hence entropy just because energy can be placed into  energy levels and in more ways.
If there was just one level then all 'particles' would have to be in this level and so entropy would be zero, $\Omega=1$.  When there are two levels some can be in the first and some in the second level and now the entropy is increased ($\Omega \gt 1$), and so on for many levels.

So the released energy increases the entropy of the surroundings because there are more possible arrangements of energy quanta which means that there is a lesser chance of the particles being in the same energy level, right ?

yes, just so, entropy interpreted at a molecular level is all about the number of ways a molecule's energy can be distributed among its energy levels.

"...there are more ways of arranging the energy quanta in the rest of the universe than there are of arranging them in the system alone."

This sentence is poetical gibberish if one does not first study the
chemical reaction at hand by the fundamental equation.
1 Prelininary remarks
Because of his question about statistical mechanics, I suppose that the questioner has (at least superficial) knowledge of the fundamental equation of (chemical) thermodynamics $\ce{\Delta G} = \Delta H – {T\Delta S}$, otherwise it would be nearly impossible to explain the touched entanglement without gibberish.
In the formation reaction of $\ce{NaCl}$ from the elements, $\ce{\Delta G}$ is negative if the amount of (negative) $\Delta H$ (exothermic reaction) exceeds the amount of (positive) $\ce{–T\Delta S}$; this is the indication that this reaction may proceed spontaneously. If during the reaction, the punctual temperature near the piece of sodium in the porcelain boat is “relatively low” (perhaps 1000°C, when chlorine is channeled slowly over the sodium), the positive value of $\ce{–T\Delta S}$ does not exceed $\Delta H$.

2 The core argument
While it is otherwise a good practical jingle, this antagonistic view of $\Delta H$ and ${T\Delta S}$ is confusing if one tries to apply it (as in the question) from the point of view of the law of increasing entropy. Entropy is defined as $\Delta
S$ = $\frac{\Delta q}{T}$ (where q is transferred “heat” $\Delta H$). The real meaning of $\ce{\Delta G}$ may only be revealed if it is transformed into its entropy representation, by dividing the fundamental equation by (-T) (note that $\Delta H$ and $\Delta S$ are by definition $\Delta H_{sys}$ and ${\Delta S}_{sys}$):
$$\ce{-\frac{\Delta G}{T}} = -\frac{\Delta H_{sys}}{T} + {\Delta S}_{sys}$$
In this equation $\frac{\Delta H_{sys}}{T}$ is the entropy change of the system by “heat” exchange with the surroundings, hence $\frac{\Delta H_{sys}}{T} = -\Delta S_{surroundings}$. On this basis, the above equation may be identified as:
$$\Delta S_{total} = {\Delta S_{surroundings} + \Delta S}_{sys}$$
$\Delta S_{sys}$ and $\Delta S_{surroundings}$ strive mutually to a maximum of $\Delta S_{total}$.

Based on the comments, it looks like you are struggling with the base concept of spontaneity.
Spontaneity is characterized by having the total entropy change in a process be non-negative:
$$\Delta S_{\mathrm{universe}} \geq 0$$
This we can decompose into two pieces:
$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{surroundings}} + \Delta S_{\mathrm{system}}\geq 0$$
So, the system or the surroundings can experience a decrease in entropy as long as the other component more than compensates for the decrease with an increase in entropy that is at least as large as the decrease.
The definition of entropy is:
$$S = k \log \Omega$$ where $\Omega$ is the density of states, that is the number of states available for some small increase in energy:
$$\Omega = \frac{\delta \Phi}{\delta E}$$
The bold statement is trying to say that because the "number of states" (even though they mean density of states) is decreasing in the system, the entropy change is negative. On the other hand, because of the high exothermicity of reaction, the density of states in the surroundings is increasing by an amount that more than compensates for this (i.e., the entropy of the surroundings goes up a lot). This maintains the overall relationship that $\Delta S_{\mathrm{universe}}\geq 0$ even though the system's entropy change is negative.
Specifically, in your reaction, the increase in entropy of the surroundings is derived from the massive quantity of heat that you're dissipating from the system.

Okay i understood what you just said in this answer but what i dont understand is what i wrote in bold. Did i write something wrong in the last two comments ?

@Maxwhite No, the comments were getting long, and I thought I understood your problem, but maybe not. I'm having trouble understanding what about the bold statement you don't understand...

The thing is that i dont see why they even mentioned the number of possible arrangements of quanta in the system. They are explaining why the energy leaving the system produces an increase in the entropy of the surroundings right ? If i had to explain why that energy increases the entropy of the surroundings i would say : After the energy has been transferred to the surroundings, there are more possible arrangements of quanta in the universe. If we thought of the "universe" as the gas molecules in the air, this transfer of energy to the surroundings means that now there is a smaller chance of

(Continued) the molecules sharing their quanta equally since there are more possible arrangements and only one of these arrangements has all the quanta shared equally between the molecules. Now, is my explanation correct ?

@Maxwhite Updated answer. Hope this helps.

Okay now i think we are talking about two different points because i understood what you said but it is not what i am talking about. Look, they said "the energy leaving the system produces a substantial increase in entropy of the surroundings because,". They want to explain how the energy released increases the entropy of the surroundings. Not why the reaction is spontaneous even though the entropy of the system decreases (i understand why it is spontaneous). So we are only discussing how this released energy increases the entropy of the surroundings.

And what i dont understand is : what does the entropy change of the system have to do with the way the released energy increases the entropy of the system. Is my explanation for the way in which the released energy increases the entropy of the surroundings (in the previous comment) correct ?

The released energy does not have anything to do with the entropy of the system, only the entropy of the surroudings.

Dude please i just need to know if the explanation that i wrote in response to your answer is correct (about how the energy released increases the entropy of the surroundings).

Ugh. You've written so much now that I can't tell what is what... :(

There is a little inaccuracy in your post, the $\Omega$ you use is the number of configurations not the density of states, which is just the number of levels/unit energy (and so has dimensions  1/energy). The $\Omega$ is the number of ways of placing N molecules  into its energy levels so that $n_1$ are in level 1, $ n_2$ in level 2 and so on then $\Omega = N!/(n_1!n_2!\cdots)$

Ok lol ... I was talking about the explanation which i wrote right before your comment "Updated answer. Hope this helps."

@porphyrin Yes, wrong term. It's the number of states within $\delta E$.

@Maxwhite You're looking for the derivation of $\Delta S$ for entropy of surroundings. We're using $dS = \frac{q}{T}$ to compute this value from the enthalpy change of the system.

@Zhe I still think that its wrong, (1) $\Omega$ its the total number of configurations  and (2) your density of states its wrong dimensionally as the log has to be dimensionless. The density of states is just how many energy levels there are /energy gap,  not the number of ways  particles are placed into these levels which is $\Omega$

@porphyrin No, we're saying the same thing. $\Omega$ is not the density. It is the number of states within $\delta E$ of $E$, i.e. $\omega(E)\delta E$.

@zhe still don't agree; as in my previous comment the number of states in any energy interval is not important because if the temperature is zero then only one state is populated out of all the thousands of them  and entropy is zero . It is the number of ways of arranging the energy among the states that counts. If you want to calculate the extra entropy then  start with $\Omega =N!/(n_1!\cdots n_i!)$ , remove one from $n_1$ and add it to state $n_i$  and recalculate $\Omega'=N!/((n_1-1)\cdots  (!n_i+1)\cdots)$ , take logs of both and find $\delta S= -k\ln(n_i/n_1)$ when $n\gg 1$

@porphyrin For example, see: http://www.harding.edu/lmurray/themo_files/notes/ch09.pdf

@Zhe, yes its interesting but very unintuitive. His $\Omega(E)$ is the number of microstates which he then maximises (page 9 and eqn 44), this is just the same as my $\Omega$. He defines his $\Omega(E)=w (E)\delta(E)$ with $w(E)$ as density of states and then finds from this the number of microstates that are most probable; again just the same as the $\Omega$ I used. (Your $\Phi$ is the number of accessible states and the number of states at E is $\Omega=(d\Phi/dE)\delta(E)$, then from this the max number accessible microstates is found. You skip too many steps when you define your $\Omega$.

I think I've just gotten used to the statement right after equation 9.41...


text:
Is there regular decrease in atomic radii of 3d series?
The following fragment of the table of atomic radii from CrystalMaker — Elements, Atomic Radii and the Periodic Table shows a regular decrease in the atomic radius of 3d series:

Table of Atomic Radii
The following table contains some of the atomic radius data used by CrystalMaker.
This is a brief summary of a far more extensive body of work — please see the notes at the end of this page for more information.
$$
\begin{array}{ccc}
\hline
\text{Atomic} & \text{Element} & \text{Atomic} \\
\text{Number} & \text{Symbol} & \text{Radius [Å]} \\
\hline
\vdots & \vdots & \vdots \\
21 & \ce{Sc} & 1.84 \\
22 & \ce{Ti} & 1.76 \\
23 & \ce{V}  & 1.71 \\
24 & \ce{Cr} & 1.66 \\
25 & \ce{Mn} & 1.61 \\
26 & \ce{Fe} & 1.56 \\
27 & \ce{Co} & 1.52 \\
28 & \ce{Ni} & 1.49 \\
29 & \ce{Cu} & 1.45 \\
30 & \ce{Zn} & 1.42 \\
\vdots & \vdots & \vdots \\
\hline
\end{array}
$$
Notes

Atomic Radii: values are calculated from:
E Clementi, D L Raimondi, W P Reinhardt (1963) J Chem Phys. 38:2686.

While comparing atomic radii, two factors are important:

Decrease in size due to increase in effective nuclear charge.
Increase in size due to increase in shielding effect.

I was surprised to find a regular decrease in the atomic radius of 3d series elements on this site because that would mean that effective nuclear charge is dominating in case of each element more than shielding factor.
On some posts like Variation in atomic sizes in the transition elements I have found that from Ni to Zn there is a regular increase in size.
Are the values shown on this site wrong? If yes, what are the correct values?
If not, why shielding wouldn't be as effective as increase in nuclear charge?

I looked up the numbers in Greenwood & Earnshaw 2ed. I don't know about the exact values, (and it probably varies from source to source depending on how the data was obtained), but there is definitely a decrease. I really have no clue why that question that you linked says that there is an increase.

there is gradual decrese in the atomic radius ,due toeffective nuclear charge but tehey are almost the same due to shielding effect

Many books do show an increase from Ni to Zn. The Elements by John Emsley (Clarendon Press 1998) actually has an increase from Fe (124) to Co, NI (125), Cu(128), Zn (133). I suspect the increasing values are largely the result of different measurement methods/precision etc., but the apparent exception to the periodic trend was then accepted as truth and post-hoc rationalized and then cemented as a source of questions for students along with other periodic trend exceptions. I've seen questions about this on a number of exams.

Periodic Table Website confirms these atomic radii.
So why does the atomic radius decrease? The nucleus of the atom gains protons moving from left to right, which increases the positive charge of the nucleus which attracts electrons. Although electrons are also added as the elements move from left to right across a period, these electrons reside in the same energy shell and therefore do not offer increased shielding.
Source: Periodic Trends

According to Work Out Chemistry ‘A’ Level[1], there should be some mistake in the data.
In $\mathrm{3d}$ series, radius of $\ce{Zn}$ is greater than that of $\ce{Cu}$. The thing that brings difference in atomic size is effective nuclear charge. When the increase in nuclear charge and shielding is different, atomic radius changes.

When increase in nuclear charge is greater than increase in shielding, atomic radius decreases. This is the case form $\ce{Sc}$ to $\ce{Mn}$.

When increase in nuclear charge is equal to increase in shielding, atomic radius remains almost constant. This is the case for $\ce{Fe}$, $\ce{Co}$ and $\ce{Ni}$.

When increase in nuclear charge is less than increase in shielding, atomic radius increases. This is the case for $\ce{Cu}$ and $\ce{Zn}$.

Figure 23.1 The Metallic Radii of the First-, Second-, and Third-Row Transition Metals

Reference:
[1]: Burgess, D. A. The D-Block Elements. In Work Out Chemistry ‘A’ Level; Macmillan Education UK: London, 1987; pp 150–157.


text:
How do chemical bonds increase molecular refractivity?
I see that the molecular refractivity of a compound consists of the refractivity indexes of its composite atoms and chemical bonds, and that the refractive index $n$ of a substance increases with its polarizability $\alpha$, as described with the Lorentz-Lorenz equation:
$$
{n^2-1\over n^2+2} = {4\pi N \alpha \over 3},
$$
where $N$ is the number of molecules per unit volume.
I think that the formation of a covalent bond restricts the electron cloud of its composite atoms, delaying its release of an electromagnetic wave after being stimulated by a wave of the same frequency, and decreasing the wave length of the resulting created by the interference between the two electromagnetic waves affected and unaffected by the electron cloud. In the same way, I would explain the comparatively high refractivity index of a double bond between the same atoms as a stronger restriction of the shared electron cloud, delaying the release of an electromagnetic wave later than the covalent bond.
So I want to know, is there anything fundamentally mistaken with my interpretation?

The polarisability is the ability of any atom to have its electron 'cloud' displaced by an electric field. Polarisability is defined as the ratio of the induced dipole, formed by electron displacement, divided by the electric filed strength. It has units proportional to volume usually; values are in units $4\pi\epsilon_0\times 10^{-30} \mathrm{m^3}$. Larger atoms/molecules  naturally have larger values, in general, than smaller ones, e.g. $\ce{CH4}\; 2.6,\; \ce{CCl4}\; 10.5$. In a general sense also the number of electrons and how tightly they are held affects the magnitude of the polarisability.
In the case of light it is its oscillating electric field that affects the atoms in the material, a solvent medium for example. In doing so the speed of light in the medium is reduced and the ratio of this to that in vacuum is the refractive index. The energy of the photon remains the same inside as it is outside the material (assuming no absorption or scattering ) so the frequency remains the same, but inside the material the wavelength changes because $c/\lambda =\nu$ is constant and c changes. In forming the refractive index, the tails of electronic absorption bands  are important and are usually far into the ultraviolet.
In terms of working out the refractive index of some arbitrary molecule usually a semi-empirical method is used with the very many measured values available; values for different bonds can be extracted and parameterised. From these values then the refractive index of unknown compounds can be calculated. I found the pages www.pirika.com which has some explanation of how this is done.
In bringing atoms together to form molecules the atom/bond polarisability is used to form the refractive index and  should be additive provided that in forming the bond a strong absorption band is not brought into the visible part of the spectrum.  This is just what happens when a double bond or aromatic is formed so in these cases some extra empirical adjustment is needed.

I see that I need to study electron absorption bands and the properties of light in depth. So the strength of the restriction of the electrons in a covalent bond would affect the polarizabiliy of the molecule(which is connected to its molar refraction), rather than just directly delaying the release of the electromagnetic wave of the same frequency as the incident light. Much clearer now. Thank you for your response.


text:
Entropy of Activation and Temperature Dependence
I am trying to understand the extrapolation of enthalpy $\Delta H^{\ddagger}$ and entropy of activation $\Delta S^{\ddagger}$ from the Eyring equation. It's typically cast as:
$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$
where $k$ is the rate, $T$ is the temperature, $R$ is the ideal gas constant, $k_\mathrm{B}$ is the Bolztmann constant, and $h$ is the Planck constant. However, the entropy itself also has a temperature dependence in principle (i.e. harmonic oscillator entropy is a function of temperature). Is there a way to determine the temperature dependence of the entropy of activation from the Eyring equation for a real experiment?

That would depend on the temperature range; the traditional experiment for this sort of thing fits the left side to a linear function of $1/T$ which works out on a small enough temperature range that the enthalpy and entropy changes can be assumed constant (or close enough to constant that other errors dominate). Without that assumption it seems that you will need much finer scale information to get anywhere.

The equation
$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$
does not assume that $\Delta S^{\ddagger}$ is temperature independent. To evaluate its T dependence you might proceed as follows:
$$\left(\frac{\partial \Delta G^{\ddagger} }{\partial T}\right)_p = -\Delta S^{\ddagger}$$
$$\rightarrow\left(\frac{\partial [RT\ln K^{\ddagger}] }{\partial T}\right)_p =  \Delta S^{\ddagger}$$
$$\rightarrow \Delta S^{\ddagger}=\left(\frac{ \partial [RT \ln (hk/k_\mathrm{B}T)] }{\partial T}\right)_p $$
It follows that you can evaluate $\Delta S^{\ddagger}(T)$, $\Delta S^{\ddagger}$  at different temperatures, from the change of $[RT \ln (hk/k_\mathrm{B}T)]$ with respect to temperature.


text:
Does freezing water do work?
My brother and I were talking about how he needed to take his beers out of the freezer because they would burst open. This made me think of something no professor ever addressed... Obviously work is occurring on the beer can, but oxymoronically it's due to the removal of heat energy. The secret I imagine is in the phase change itself, which likely takes its own energy (which would normally be given off as heat) and uses it to expand (do work).
My questions extend on this idea...

What if we replace the beer can with a 5 inch thick steel capsule. Does the capsule prevent the water from freezing because there isn't enough room to expand?
What about if you take even more energy away? Would absolute zero finally break the capsule?
Can a strong enough container simply prevent water from phase changing to a solid altogether, regardless of temperature?
Normally increasing the pressure of a substance will cause it to freeze (in chemistry class my teacher made water boil at room temperature with a vacuum pump removing pressure). With water though, it expands while freezing... So paradoxically if you have water in a strong capsule and keep increasing the pressure (instead of temperature), will the water eventually freeze or simply remain very cold liquid?

Thanks!

In scenario 1, it turns out that the water will expand. Steel may not be strong enough to prevent this. See the 50 second excerpt from this video: https://youtu.be/31pLJyReFXw?t=6m49s (which is a modern rehashing of a Faraday lecture)...

Always interested in beer-related posts. That said, it's a bit broad as written and you might benefit by breaking it up into separate queries.

True Todd , but any of them answered would be interesting! They are also fairly related, basically the behavior of water on strong containers with varying temperature and pressure. And cool paracetemol, I couldn't find a similar one but would appreciate if you can locate it! Thanks Zhe will watch when I get home

https://chemistry.stackexchange.com/questions/20009/hypothetical-what-happens-to-water-as-pressure-increases-to-infinity https://chemistry.stackexchange.com/questions/9803/what-happens-to-water-under-high-pressures-without-possibility-of-escape

Long story short, it goes as follows:

Yes, the freezing water does perform some work. You put a load on top of a bucketful of water, and as it freezes, it will lift the load all right.
Yes, a capsule would prevent the water from freezing, at least for a while (provided that the capsule is immensely strong). Then some of the beer (water?) will freeze. Then... see the next item.
No, the absolute zero would not finally break the capsule. Rather, it would turn water into some alternative kind of ice which is more dense than water. BTW, you won't have to get precisely to absolute zero to achieve that.
True, normally increasing the pressure of a substance will cause it to crystallize ("freeze"). Water is different in this regard, which is quite rare, but by no means unique.


text:
Toxicity of vanadium in pottery
I mix and develop glaze recipes for midfire stoneware (fired to 1220°C). The colorants in pottery are mostly iron, copper, cobalt, tin, titanium, chrome, and a few other transition metals. Many of these materials are somewhat toxic in the raw form but safe in the finished work. These colors usually make up less than 3% of a glaze recipe.
Vanadium is a colorant that potters seem to fear without good reason. I find resources such as MSDS sheets to be very unhelpful on this subject.
The heart of my question is: in your humble-I-am-not-going-to-sue-you opinion, is vanadium (specifically tin-vanadium) safe enough? How about relative to cobalt?

See this abstract on Pubmed for information concerning the toxicity of vanadium and its salts. The net of it is don't inhale: "Most of the toxic effects of vanadium compounds result from local irritation of the eyes and upper respiratory tract rather than systemic toxicity."

I'd point out that there are two different toxicity problems, first to the potter and then to the user of the fired pottery.

Yes, vanadium is safe enough relative to cobalt.

Thank you.
Everything with mixing dry glazes is a respiratory hazard so that isn't a new worry.

I'd be more worried about what sort of toxic species might be released when the glazed items are fired. You'll breath all of that in.


text:
What is "Woods Metal"
I have a piece of what I have been told is "Woods Metal" it melts at 168F.  I am "curious" about how they make it, how low can the melting temperature be?

Wikipedia entry.

Wood's alloy is but a certain mixture of metals, and that' just how they make it: they mix the metals and melt them together. Its composition is adjusted so as to reach the lowest possible melting point. You can get beyond that if you use mercury, but then you'll get poisoned and die. Other than that, you would be able to reach way lower.

Here is some additional info from web
Wood's metal, also known as Lipowitz's alloy or by the commercial names Cerrobend, Bendalloy, Pewtalloy and MCP 158, is a eutectic, fusible alloy with a melting point of approximately 70 °C (158 °F). It is a eutectic alloy of 50% bismuth, 26.7% lead, 13.3% tin, and 10% cadmium by weight.
I have experience using the material for fixturing of difficult to secure /  shaped parts for machining. It can also help in fixturing/holding - then later melted away.
The eutectic composition will privide lowest Tm and get the info from a phase diagram.
Do not add mercury for safety reasons.


text:
Proof of Brillouin's theorem using the diagonality of the Fock matrix
I'm working my way through Szabo and Ostlund's Modern Quantum Chemistry. I am having trouble understanding the proof and significance of Brillouin's theorem. According to the book:

Now, by definition, solving the Hartree-Fock eigenvalue problem requires the off-diagonal elements to satisfy $\langle \chi_i|f|\chi_j\rangle = 0,  (i \neq j)$.[1]

However, isn't the Fock matrix only diagonal because we used "nice enough" spin orbitals that we constructed by unitary rotations? That is, for any other choice of spin orbitals, the Fock matrix is not diagonal but is still a perfectly valid matrix with a mostly-eigenvalue statement, right?
What am I missing here?
[1] A. Szabó, N. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover Publ., 1989, p. 129.

Your question actually contains two linked inquiries. The proof itself depends on the nature of the Fock-operator to which I'll have a few words afterwards. For the proof itself it is only important, that the Fock matrix is diagonal. Let's restate Brillouin's theorem:
In case of the Hartree-Fock (HF) reference determinant $\Phi_0$, the Hamiltonian matrix element between it and any singly excited determinant $\Phi_a^r$ reduces to zero.
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$
$$\langle\Phi_0|\op{H}_\mathrm{el}|\Phi_a^r\rangle = 0$$
The proof of this then rather simple:
\begin{align}
&& E &= \ll \Phi_0 \rb \op{H}_\mathrm{el} \lb \Phi_a^r\rr\\
&& E &= \ll \phi_a \rb \op{H}^\mathrm{c} \lb \phi_r \rr +
     \sum_j^{N/2} \ll \phi_a\rb \left(2\op{J}_j -\op{K}_j \right) \lb \phi_r \rr\\
&& E &= \ll \phi_a \rb \op{F}_j \lb \phi_r\rr\\
\text{with}&& \op{F}_i\phi_i &= \varepsilon_i\phi_i\\
\text{and}&& \ll \phi_j \rb \op{F}_i \lb \phi_i \rr
          &= \varepsilon_i \ll \phi_j | \phi_i \rr = \varepsilon_i\delta_{ij} \\
&& E &=\varepsilon_j\delta_{ar} = 0
\end{align}
The direct consequence and therefore significance of this is that for any post-HF method the singly excited determinants only mix indirectly with the ground state. This simplifies a couple of the integral evaluations.
The remaining question to solve is why the Fock matrix needs to be diagonal. Even though the equation
$$\op{F}_i\phi_i = \varepsilon_i\phi_i
     \tag{1}\label{fock-pseudo}$$
suggests and eigenvalue problem, it is not. Remember the definition of the Fock operator and the operators contained
\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}
As you can see, the "one-electron" Fock operator depends on the solution of all "one-electron" Fock operators (ref. Szabó-Ostlund p. 115). The only way to achieve self-consistency and therefore $\eqref{fock-pseudo}$ is to use unitary transformation and diagonalise the Lagrange multipliers from
$$\sum_j \lambda_{ij} \lb \varphi_j \rr =
  \op{F}_i \lb \varphi_i \rr.\tag{2}\label{fock-lagrange}$$
In other words, the Fock operator is not actually associated with the total energy, but only with the variation of it (due to the Lagrange minimisation). To know one orbital you have to know all orbitals, and to know one orbital energy, you have to know all orbital energies. The Hamilton operator is not the sum of all Fock operators, and the total HF energy is not the sum of all orbital energies.
Therefore the Fock operator is also dependent only on the total wave function and any unitary transformations of the canonical orbitals will retain the total energy. However, the canonical orbitals are the only ones for which $\eqref{fock-pseudo}$ holds.

The Hartree-Fock method minimizes the energy by diagonalizing the Fock matrix, therefore by definition we have
\begin{equation}
  \langle \chi_i|f|\chi_j\rangle = 0,  i \neq j
\end{equation}
where $\chi_i$ are called canonical orbitals (whether you use spin or spatial orbitals is not important here). In other words the "nice enough" orbitals are constructed that way.
But of course we can do unitary transformations of these orbitals, which leave the total Hartree-Fock energy unchanged, but have a non-diagonal Fock matrix. For example orbital localizations do such things.
I didn't check that, but I would expect that Brillouin's theorem would not be true for these cases anymore.


text:
What is the meaning of writing a catalyst name with two compounds split by a hyphen?
In literature, sometimes I see catalysts written with "-" between elements/compounds. Does this mean the first element/compound is supported on the second, or is the second a promoter? (e.g. $\ce{Fe}$-$\ce{KCl}$).

Can you give an example from literature? I only know it with a /. Pd/C is palladium on charcoal, Pd/MgO is palladium on MgO and so on.

If we consider palladium on carbon or alumina then it means that the palladium particles are dispersed onto the carbon or alumina. The carbon or alumina acts as a support for the catayltic particles of the palladium.
If we consider Lindlar's cataylst this can be written as Pd/CaCO3/Pb
The Pd (palladium) is the hydrogenation cataylist
The CaCO3 is the support
The Pb is a poison which is used to modify the chemistry of the palladium
Sometimes one of the things separated by a / line is either a promotor or a poison.

Could you cite some nomenclature sources? That would make your answer much more complete. Thanks!


text:
For chemical reactions occurring in the interstellar medium, which involve metal-ions, will the metal ions be solvated or not?
Usually, metal-ions in terrestrial chemistry will be solvated - either by gas-phase solvent molecules or by liquid-phase solvent molecules. But the interstellar medium (ISM) is never in local thermodynamic equilibrium. Moreover, the solvation of any system is driven by the system's proclivity to attain thermodynamic stability. But given the lack of local thermodynamic equilibrium in the ISM, solvation should not at all be an important factor in the ISM. Is it right or wrong? Any reference from the Astrochemistry/Astrophysics community will also be highly appreciated.

I'm not exactly seeing what you'd imagine as solvation in sth like this.

Per Wikipedia, the upper bound on the particle density in the interstellar medium is approximately $\pu{E6}$ molecules per cubic centimeter, in "cool, dense regions."
In the context of terrestrial aqueous solvation, where the concentration of water is approximately $\pu{55M}$, the particle density is on the order of $\pu{3.3E22}$.
So, even if every single particle in the vicinity of a metal ion in the ISM were a water molecule, the ISM would be roughly $\pu{E18}$-fold more dilute in neighboring water molecules as compared to aqueous solution.
No, metal ions in the interstellar medium will never be solvated.
Note that this does not imply that it is impossible for aquo-complexes of transition metal ions to occur in the ISM, given their significant affinity for ligands in the first coordination shell and for the appreciable organizing influence (thanks, MaxW!) such ions have on more distant coordination shells. It's just highly improbable that one would find a metal ion associated with a sufficiently statistically large number of water molecules to be considered solvated.

Further, to address the portion of the question about thermodynamic equilibrium: I don't think that's a particularly significant factor, no. On Earth, it's frequently the case that ions are solvated but not in thermodynamic equilibrium: the proton gradient that is sustained in the aqueous milieu of biological organisms to drive ATP synthase is one of the most significant examples that comes to mind. Thus, an absence of thermodynamic equilibrium does not necessarily entail a lack of solvation.

It is even more improbable. Solvation isn't just about ligands. Because of the dipole moment of water there will be a number of "spheres" (layers)of oriented water molecules around the complex before the water molecules are oriented randomly.

your answer is enlightening. I fully get your point. Can you please tell if the qualitative thermodynamic argument I listed - ISM not being in thermodynamic equilibrium; can also be a suitable reason for the lack of solvation in the ISM?

Interstellar medium is in equilibrium all right (when they say it isn't, they mean an entirely different thing, mostly related to radiation). But this has nothing to do with solvatation. With densities that low, the equilibrium state would not be solvated either.


text:
Thought experiment for evaluating total energy
Atomization Energy ΔHat is the energy required to disintegrate a molecule into isolated atoms that are infinitely away from each other. The question is about atomization energy and its relation to the total energy (E) in Schrodinger equation (Eψ=Hψ).
If we disintegrate a molecule one step further to isolated electrons and nuclei, will the overall energy spent be equal to negative of the total energy of Schrodinger equation? I am assuming an energy reference will come into play as well as thermal energy. So let's say we are conducting the thought experiment of disintegration at 0 K (zero Kelvin) and we correct the values by assuming a reference energy point. Then, is the total Energy E equal to disintegration energy of the molecule into isolated electrons and nuclei?

You can easily test that when looking at a H atom. Calculate the energy on different distances between proton and electron. You can then extrapolate your findings to bigger atoms and molecules. Of course, you need to use an approximated Hamiltonian, but eh, for a basic discussion it's alright.

The interaction between the isolated particles would vanish, therefore $\langle \hat V_{ij}\rangle=0$, but you still have the contributions from the kinetic energy $\langle \hat T_i\rangle$ of each individual particle.

@Feodoran Makes sense. I also think that all identical nuclei when isolated have the same kinetic energy and all electrons when isolated have the same kinetic energy as well. So at least for two isomers it should be taken care of, should it not? In other words, the total disintegration energy difference between two isomers is equal to total (SE) energy difference. Is that about right?

@Fl.pf. Right this can be tested through actual calculations. I was trying to get an answer through conceptual thinking without doing the actual calculations.

@Kinformationist proof of concept assisted by calculation?

If we disintegrate a molecule one step further to isolated electrons and nuclei, will the overall energy spent be equal to negative of total energy of schrodinger equation?

By convention, yes.
Energy as a absolute quantity is not well-defined, energy must be defined with respect to an energy zero.  By convention, when we perform quantum chemical calculations our energy zero is free electrons and nuclei infinitely far apart, which we define to have zero potential energy as they cannot interact.
Also, quantum mechanically, we consider only the case at $0$ K.  We can however add on corrections to the free energy due to entropy and internal motion at finite temperatures, as well as vibration zero point energies.
How is this different from the atomisation energy?
We can see that to get from the products of atomisation to the products of "disintegration" (as you refered to it), is to remove all the electrons from the neutral atoms.  That energy is given by the sums of all the ionisation energies for each atom in the system, removing each electron one by one.  This allows us to construct a Hess cycle.
Finite temperature corrections are added to the quantum mechanical energies as the potential energy and free energy are different at finite temperature.

What do you mean by finite T corrections. Also E (SE) plus Nuclei and electrons leads to the Molecule. E(SE) includes individual kinetic contributions of Nuclei and electrons. What you wrote on the scheme is probably only the potential energy part of the E (SE).

When the electrons are bound to nuclei they have finite kinetic energy that is accounted for within the Electronic Schrodinger equation, and presuming you include nuclear kinetic energy it will include the vibrational zero point energy of your nuclei.  So that kinetic energy is included in an ordinary Schrodinger equation energy.  However that is all calculated to $0$ K.

As such that doesn't include the effects of entropy.  At finite temperature, multiple vibrational states will be accessible to the molecules, some of which will be occupied.  This introduces vibrational entropy; rotational, translational and in extreme cases electronic entropy can also be introduced.  This changes the free energy of the molecule as $G = U + PV - TS = H - TS$.  However as you can see above my diagram used enthalpy $H$ rather than free energy $G$ so this corrections aren't required to complete the cycle as you point out.

Finally, free energy due to entropy and (temperature in general) is not relevant to the case of a single molecule whose vibrational and rotational state we can measure/know.  It is only relevant to and ensemble of many molecules at a given temperature where different numbers are in different rovibrational states.  I'll try and clear up the main answer when I have a bit more time.


text:
What is the role of the sulfuric acid in a titration between potassium permaganate and iron ammonium sulphate solution?
So basically my titration was of finding the concentration of potassium permaganate against an iron ammonium solution. In the iron ammonium sulphate solution it consisted of water, iron ammonium sulfate and sulfuric acid. Im not sure about the role of sulfuric acid in this case so if someone could explain it to me that'd be great!

related https://chemistry.stackexchange.com/questions/67272/potassium-permanganate-iron-sulfate-titration-didnt-work-and-i-cant-work-out

https://chemistry.stackexchange.com/questions/67753/potassium-permanganate-to-analyze-iron-compounds

And this: https://chemistry.stackexchange.com/questions/32750/minimising-the-formation-of-manganese-dioxide-in-the-titration-of-potassium-perm more or less answers it


text:
Is [7]сirculene aromatic or anti-aromatic?
This compound is called [7]circulene and is one of the examples of non-planar aromatic systems (1). From the first overview of this compound, it is observed that there are 7 benzene rings so one would conclude that it is an aromatic compound:

There are $7 \times 4 = 28$ $\pi$-electrons, so this compound satisfies the criteria of anti-aromacity. Then how come this compound is aromatic?
(1) Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013, p. 48, compound 40.

TL;DR: We don't count π electrons here, rather the π conjugated circuits [1]. [7]-сirculene has all conjugated circles of size $4n + 2,$ and none of $2n,$ what makes it an aromatic compound.
Strictly speaking, the aromaticity $4n + 2$ and antiaromaticity $4n$ rules in their original interpretation are very limited and cannot be universally applied.
According to the original Hückel $4n + 2$ rule, if a monocyclic, planar molecule has $4n + 2$ π electrons, it is considered aromatic.
But the count of π electrons is not what matters for polycyclic compounds, therefore a generalized $4n + 2$ rule exists: if a conjugated molecule has only $4n + 2$ π conjugated circuits, it is considered aromatic.
Non-benzenoid compounds can have both $4n + 2$ as well as $4n$ number of conjugated circuits counted by taking into account all Kekule structures. Among non-benzenoid non-alternant polycyclic hydrocarbons [5]-сirculene and [7]-сirculene both have only $4n + 2$ $(6, 10, 14, \dots)$ numbers of circuits.
Further reading: Is there any special rules for checking the aromaticity of polycyclic compounds?
Reference

Gutman, I.; Cyvin, S. J. Conjugated Circuits in Benzenoid Hydrocarbons. J. Mol. Struct. THEOCHEM 1989, 184 (1), 159–163. DOI: 10.1016/0166-1280(89)85141-3.


text:
Why does having identical groups on either side of the pi bond lead to no geometrical isomerism?
As far as I was taught, cis and trans isomers were categorised under stereoisomers and sub category of geometric isomers.  Qualifying properties like, rigid and restricted double bond rotation (due to pi bond electron cloud), no terminal alkenes (those with double bond at the end of carbon chain).
But, here was a quote from a LibreTexts article:

If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH3 etc. there cannot be any geometric isomers.

What is the basis of above claim? Why cis and trans with identical groups can't be qualified as a geometric isomer? If not what is this? Am I missing something here?

"If one carbon of the double bond has two identical groups …"

@Loong I am still not getting there :/ so identical groups means, there's symmetry, it's isn't in geometry? so what is this then?...

Hmm, you seem to be having a problem visualizing the structures in the case of 2 identical groups on one carbon. But never fear, 3D molecules are here.
Check out chloroethene:

It has two identical groups (hydrogens) on the left side. If you tried to make a geometrical isomer for this, you'd get:

But hey! That's just the previous image flipped upside down! That's not an isomer, it's the same thing!
Indeed it's the same molecule. These molecules can't have a geometrical isomer because it possess an plane of symmetry:

So there you go. Go ahead and visit this link if you want an interactive 3D model that you can spin around.

Added based on comments from @Zhe, there's a plane of symmetry in following angle,

Gee! why didn't I think about it? I wasn't looking at this molecule (or the type). I was trying out this : 2,3-dimethyl-2-pentene and its isomers and 2-methyl-1-butene... Basically, there can't be geometric isomers (even with non-identical groups attached), when a double or a triple bond isn't present. 3D vitalisation link is indeed big aid - but is it possible to  use this Jsmol molecular dynamics app online to try out on my own for any other molecule/compound?

In the last figure, we should not refer to this is an "axis." It is a plane.

My bad @Zhe , I'll fix it soon.

Actually, reading this again, you should call it an axis, but it's just not an axis of symmetry for the single molecule. The two structures are related by an axis of rotation.

@Zhe Alright then, I will fix it soon, although not right now, no time currently.

@Zhe is that because the existence of an axis of symmetry disqualifies the molecule of being an isomer?

This molecule does not have an axis of symmetry. It has an plane of symmetry. The lack of isomerism comes from the two substituents on the left side being identical. @bonCodigo

@Zhe With reference to the molecule in the answer $\ce{C2H3Cl}$, I understand the identical substituents contributes to the lack of isomerism. And the axis of symmetry (for rotation) doesn't concern it. However I am concerned about your comment of axis of symmetry vs plane of symmetry (it's opening another door a pages of digging conceptual details...). Speaking of plane of symmetry, this molecule still doesn't have a plane of symmetry as the imaginary plane that bisects this molecule doesn't give mirror images of each other... (cut me in to half now : sagital plane, pun intended). Does it?

@Zhe Actually now that I bisected the molecule horizontally (transverse...)... I see the light you mentioned. It has a plane of symmetry in that angle, but not from vertical though.

@PrittBalagopal I think we should remove that axis of symmetry picture, what do you think? or label is as , plane of symmetry. My classmate said it's confusing. LOL

I don't find the first plane of symmetry relevant to the matter being discussed. Even $\ce{HClC=CClH}$ has a plane of symmetry - that too in the same manner as you described for $\ce{H2C=CClH}$ - however the former does show GI. So, please do you mind explaining this? :) (sorry if I missed something! also I skipped all these back and forth comment discussions because they pertained to an earlier revision of your answer)

@GaurangTandon The plane of symmetry was just mentioned to explain why chloroethene does not form geometrical isomers. It's due the symmetry that even exchange a chlorine and hydrogen atom still produces the same molecule.

@PrittBalagopal (I hope you to are referring to the first plane of symmetry)  Doesn't $\ce{HClC=CClH}$ also have the same plane of symmetry? Why does it show GI then?

I take issue with the way a plane of symmetry is used in the currently accepted answer (see comments). So, here is my explanation of the question:
Have a look at some example molecules: $\ce{H2C=C(Cl)(CH3)}$, $\ce{H2C=C(Cl)(H)}$, $\ce{H2C=C(CH3)(CH3)}$. All of these do not show any geometrical isomerism. The first two have two identical groups (hydrogen) on the left side, while the last example has identical groups (hydrogen and methyl) on both sides.
Why does having identical groups on either side of the pi bond lead to no geometrical isomerism? (In fact, it is the only restrictive condition in pi-bond geometrical isomerism.)
We know that, to obtain the other geometrical isomer of a given compound, we just need to flip the groups on one side of the molecule. For example, here are the GIs of 2-butene:

You can also go in the reverse direction, and flip either side, with the same result. This rule of flipping-and-fixing will hold true for any pi-bonded molecule.
Now, let's apply this on 2-chloropropene:

Turns out, it's not a geometrical isomer! In fact, you can take the final molecule formed and rotate it 180deg (about an axis passing through a double bond) to obtain the original compound in the question. Recall that any molecules that become identical with each other on complete rotation are actually identical. Hence, in the above diagram, there's not two new isomers. It's still the same molecule, just shown rotated.
Note that if you had from the side where both groups are identical, you'd immediately seen both the molecules being identical:

Hence, this is how having identical groups on either side of the pi bond leads to no geometrical isomerism. Now, you should go back to the original example of 2-butene, and notice how rotating it through the pi bond, or any other axis does not lead to the same representation.


text:
Nomenclature of acid anhydride
I have been struggling hard for writing IUPAC naming of this compound.

Anhydride with benzene having five membered ring can be named as benzene dicarboxylic anhydride and five membered ring without benzene can be named as pentane -1,5-dioic anhydride. But what can I do for this compound?

It looks like a isochroman-1,3-dione to me.
For the reference there is a similar compound (published in French): Kakou-Yao, R.; Saba, A.; Ebby, N.; Pierrot, M.; Aycard, J. P. Tautomérie de la 4-(hydroxyphénylméthylène) isochroman-1,3 dione à l'état solide, Acta Cryst C 1999, 55 (9), 1591–1593. DOI: 10.1107/S0108270199006782

Are you sure that "isochroman" is an appropriate IUPAC prefix or root? It looks rather "trivial name"ey to me.

@PrittBalagopal I'm never 100% sure. I checked CCDC database, suppliers and did a little bit of googling around, and it looks like this is the only name used and it is IUPAC-compliant.

@PrittBalagopal ChemDraw also suggest a similar name

@EashaanGodbole ChemBioDraw sometimes suggests utter nonsence, I wouldn't just blindly trust this thing.

@andselisk I was using ChemDraw Direct. Anyways, thanks for informing me that it might not always be correct.

For preferred IUPAC names according to Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), cyclic anhydrides are generally named as pseudoketones. For example, the preferred name for succinic anhydride is oxolane-2,5-dione.
For the parent structure of the compound that is given in the question, the retained name isochroman, which was used in previous editions of the IUPAC recommendations, is still acceptable in general nomenclature – even with full substitution,
including characteristic groups expressed as suffixes (see Subsection P-31.2.3.3.1). Thus, the name isochroman-1,3-dione is acceptable in general nomenclature for the complete compound that is given in the question; however, it is not the preferred IUPAC name.
The preferred IUPAC name for unsubstituted isochroman is the systematic benzo name 3,4-dihydro-1H-2-benzopyran, which can be directly derived from 1H-2-benzopyran.

Likewise, the preferred IUPAC name for the complete compound that is given in the question can be derived from 1H-2-benzopyran by introducing the two principal characteristic groups of the pseudoketone as ‘1,3-dione’. However, this change requires the addition of another hydrogen atom, which must be cited in the name.

P-58.2.2.1 General methodology
The presence of at least one hydrogen atom on a ring atom that is attached to adjacent ring atoms by single bonds only that results from the introduction of a principal characteristic group, a free valence, radical, or an ionic center into a mancude polycyclic system in the absence of, or lack of, sufficient hydrogen atoms, is cited by using the capital italic letter H following the locant of the ring atom for each such position. This ‘added indicated hydrogen’ designation is enclosed in parentheses and inserted into the name immediately following the locant(s) for the free valences, radical or ionic centers, or principal characteristic groups.

Therefore, the complete name for the compound that is given in the question is 1H-2-benzopyran-1,3(4H)-dione.


text:
Finding the kp with density, temperature, and pressure

Consider the following reaction
$$\ce{3O2(g) <=> 2O3(g)}$$
At $\pu{175°C}$ and a pressure of $\pu{128Torr}$, and equilibrium mixture of $\ce{O2}$ and $\ce{O3}$ has a density of $\pu{0.168g/L}$. Calculate $K_p$ for the above reaction at $\pu{175°C}$.

I used the ideal gas law $pV=nRT,$ $K_p=K_c(RT)^n$, and $\rho=M \times c$ but I am not certain what to do after finding the total $c$ of the reaction. I was thinking of using the ICE chart to find the concentration of reactant and products which will lead to the $K_c$ and then $K_p$ value but since I don't know the initial concentration of $\ce{O2}$ I end up with an equation with two variables.
I was thinking of using the density to find the molar mass of the reaction and then somehow the  weight of $\ce{O2}$ and $\ce{O3}$ in grams but it seems like a stretch.
$$pV=nRT$$
$$T=175\ \mathrm{^\circ C}=448\ \mathrm K$$
$$p=128\ \mathrm{Torr}=0.1684\ \mathrm{atm}$$
$$0.0821=\frac nV\times\left(0.0821\ \mathrm{\frac{atm\ l}{K\ mol}}\times448\ \mathrm K\right)$$
$$0.0821=\frac nV\times36.78\ \mathrm{\frac{atm\ l}{mol}}$$
$$\frac nV=0.00457\ \mathrm{\frac{mol}l}=c$$
$$ \begin{bmatrix}
  & \ce{3O2} & \ce{2O3} \\
  \mathrm I & y & 0 \\
  \mathrm C & -3x  & 2x \\
  \mathrm E & y-3x & 2x
 \end{bmatrix}0.00457\ \mathrm{mol/l}=(y-3x)+2x$$

Do you know how to get the partial pressures of the two gases in the mixture?

If I had the Kp, yes kp=products/reactants and use the ICE chart

If x is the mole fraction of O2 and (1-x) is the mole fraction of O3, what is the mole fraction of O2 in the mixture if the molar density of the mixture is 0.00457 moles/liter and the mass density is 0.168 g/liter?  Based on this, what are the partial pressures?

Explanation and example about mole fraction to go along Chester's comment.

@ChesterMiller O2 =x and O3=(1-x) makes sense for the mole fraction percentage (1=100%) but I can't find the molar mass of O2 in the reaction if I don't know the percentage of is in the reaction and vice Versa. Every equation I come out with consist of3 variables which I'm sure isn't suppose to be the case. I know A full 100% of O2 molar mass is (2x15.99) O3=(3x15.99) and the total mass of the reaction is 35.75

The molar mass of oxygen is 32 and the molar mass of ozone is 48.

@ChesterMiller yes like I said. O2=2x15.99=32 and O3=3x15.99=48

So $$32x+48(1-x)=35.75$$  What is the value of x?

@ChesterMiller I'm glad you said that because I thought of that equation yesterday as well but it doesn't make sense to me. The answer is around 70% . This equation makes it seems like you are asking for what's 70% of O2 and not what's 70% of the reaction. That's where my two other variables came. If O2=x percent of rxn and O3=(1-x) percent of rxn than what factor of O2 makes it x percent of the reaction? That's why I was creating for O2 32y=x36.76 and for O3 48z=(1-x)36.76

I have no idea what you are talking about here.  But, if the mole fraction of oxygen is 0.70 and the mole fraction of ozone is 0.3 (and the total pressure is 128 torr), what are the partial pressures of oxygen and ozone?

That raises another question I had. Now .70 is supposedly the mole fraction but we used the grams of O2 (32)  and O3 to get it. So would we use .7019=O2 g/36.76g  then convet it to molarity or .7019=O2M/.00457? I'm assuming it's the latter since the former creates a M higher than .00457. Using the latter equation I get O2=[.003207] O3=[.00136]. Also, I know O2 and O3 are gases but we have been using grams and L so wouldn't it be the Molarity and not the partial pressure? Anyway [.00136]^2/[.003207]^3=56.07(which is a very high Kc). Plug that number into kp=kc (RT)^-1 and get 1.52

If you are trying to calculate Kp, all you need to know are the partial pressures.  Do you know how to calculate the partial pressure of a species once you know the total pressure and the mole fraction of the species?

@ChesterMiller feel free to update my answer. I am running out of wits to explain further to OP's comments.

I don't know what else you can do.  You've pretty much said it all.

You can use the formula for the ideal gas
$$\rho=\frac{p\cdot M_\mathrm{mean}}{R\cdot T}$$
As pressure, temperature and density are given, you get the mean molar mass.
$$M_\mathrm{mean}=\frac{\rho RT}{p}$$
From that, you get molar fractions.
$$x_{\ce{O2}}=\frac{ M_{\ce{O3}}-M_\mathrm{mean}}{ M_{\ce{O3}}- M_{\ce{O2}}}$$
$$x_{\ce{O3}}=1-x_{\ce{O2}}$$
From that, you get partial pressures.
$$p_{\ce{O2}}=p\cdot x_{\ce{O2}}$$
$$p_{\ce{O3}}=p-p_{\ce{O2}}$$
From that, you get $$K_\mathrm{p}=\frac{p_{\ce{O3}}^2}{p_{\ce{O2}}^3}$$

Looking at your struggle and @ChesterMiller's support, I am sharing how I would solve it. Anyone else please feel free to point out any mistakes in this answer.
Short Answer
here. (combining Dalton's law, Avagadro's law, ideal gas law, reaction quotient at equilibrium $Q_p = K_p$ in terms of partial pressures of gas mixture) Molecular weight of $\ce{O2} = \pu{32gmol-1},~\ce{O3} = \pu{48gmol-1}$.
$$\text{density}, \: d = \frac{\text{mass}, \: m}{\text{volume}, \: v}$$
$\text{num of moles,} \: n = \frac{\text{mass}, \: m}{\text{Molecular Weight}, \: M}$
Therefore we consider molar density,
$$\frac{d}{M} = \frac{\text{mass}}{\text{volume}}$$
$$\text{total density } T = \frac{\text{mass}( \ce{O2})~+~\text{mass} (\ce{O3})}{V \text{ volume}}$$

Using ideal gas law, the density of a gas is directly proportional to its molecular mass (M).
$d_\mathrm{i} = \frac{P_\mathrm{i}\times M_\mathrm{i}}{RT}$
The density of a gas mixture is obtained using total mass (m) of the gases in volume V at temperature T.
$P_\mathrm{T} = \frac{d_\mathrm{T}\times RT}{m_\ce{O2} + m_\ce{O3}}$

Mole fractions, $\chi_\ce{O2} + \chi_\ce{O3} = 1$.
How?
Total number of moles in gas mix = N
If $\ce{O2}$ has n moles, $\ce{O3}$ has = $N-n$ moles
Therefore, $N = n + (N-n)$, get the mole fraction of each gas to total moles,
$\frac{N}{N} = \frac{n}{N} + \frac{N-n}{N}$
$ 1 = \frac{n}{N} + \left(\frac{N}{N}- \frac{n}{N}\right)$
The mole fraction of $\ce{O2}$,
$\frac{n}{N} = \chi_\ce{O2}$
$\frac{N-n}{N} = \chi_\ce{O3}= (1 - \chi_\ce{O2})$

Now apply this to the ideal gas law equation that we derived earlier,
$\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}} = \chi_\ce{O2} * M_\ce{O2} + \chi_\ce{O2} * M_\ce{O3}$
$\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}} = 32x + 48(1-x)$
$$\chi = \frac{48 - \left(\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}}\right)}{16}$$
Partial pressures of gas mix in the system at equilibrium, (I wouldn't simplify these calculations as total pressure to density cancels out each other at the end)
$p_\ce{O2} = P_\mathrm{T} \times \chi_\ce{O2}$
$p_\ce{O3} = P_\mathrm{T} \times \chi_\ce{O2}$
$K_p = \frac{[p_\ce{O3}]^2}{[p_\ce{O2}]^3}$
$K_p = \frac{[P_\mathrm{T} \times \chi_\ce{O3}]^2}{[P_\mathrm{T} \times \chi_\ce{O2}]^3}$
$$K_p = \frac{\left[P_\mathrm{T} \times \left(1-\left(\frac{48 -\left(\frac{d_\mathrm{T}\times RT}{P_\mathrm{T}}\right)}{16}\right)\right)\right]^2}{\left[P_\mathrm{T} \times \left(\frac{48 - \left(\frac{d_\mathrm{T}\times*RT}{P_\mathrm{T}}\right)}{16}\right)\right]^3}$$
Please plug in the numbers, bear in mind the conversion factors.

I honest most of your steps but I am still confuse on how the molecular mass of rxn equal to 32x+48(1-x). As you written. XO2+xO3=1 which is just x% + X%=100% of the rxn mass. Let's say x is 60%. The equation then tells me that O2 is 60% of the rxn total mass and 03 is 40%. Which makes sense. This is where 32x+48(1-x) confuses me because it's asking for 60% of O2 mass and 30% of O3 mass instead of 60% and 30% of rxn mass. If the total rxn mass was 70, 70(x)+70(1-x)=70 makes 32(x)+48(1-x)=70 doesn't. Also, I got x=[(dt x RT/PT)/-16 ]-48. My 16 and 48 are negative

@S.Polk Let's consider this: What is the difference between mass of a molecule/compound, molar mass and molecular mass? What does this mean to you? Have you learnt about mole fraction? Isn't it in your gen-chem text book? Because if you can grab this concept, you are through.

Yes I do and that still help me understand that specific part of the equation or why you don't have a negative 16 or 48 when 32x+48(1-x) will be 32x+48-48x-->48-16x

@S.Polk have you resolved your doubts? Do you know how to measure the mass of a molecule in a molecule mixture? There's a mole ratio and there's partial pressure ratio, and you also know the molar mass of that molecule. Does this ring any bell to you?

A little. You can use the M and density to get the mass of a molecule. I'm still confuse on how they are getting their mole ratio formula.

Take a look at this. Exactly the same concept.

for that example are we suppose to assume that the reaction is N2O4-->NO2 and not N2O4-->2NO2? It's the only way I can get kc=2. It will make kp=(2/3)/(1/3)=2 and delta n=0 which would cause kp=kc(RxT)^n to become kp=kc so 2=kc


text:
D/L and +/− labels of glucose
I had just started to read about carbohydrates when I came across D-(+)-glucose. My book says that the '+' sign refers to its optical rotation measured through the polarimeter, and that it could either be + or −, but throughout the book D-glucose is always designated with the + sign.
Is it possible for there to be D-(−)-glucose?

Related: What is the difference between D and L configuration, and + and −?

For a given configuration there is a corresponding and fixed rotatory effect. However, the actual rotatory effect - that seen by a polarimeter - can either be + or −, independently of the D- or L-configuration of the stereogenic centre that shows up in the compound name (in the case of sugars, it refers to the carbon furthest away from the =O group).
In brief, D-glucose is always D-(+)-glucose.

Yes, infact in the case of glucose it's always D(+) and L(-), but for some other sugars like fructose it's D(-) and L(+).

The naming along D (or L) is convention, based only about this peculiar $\ce{OH}$ group adjacent to the $\ce{CH2OH}$ in the Fischer projection.  The rotatory effect however depends on the whole molecule.

In Emil Fischer's day, there was no D,L terminology. Configuration was designated as d,l and rotation as +,-. Today they are synonymous. The D,L nomenclature evolved from the Delta,Lambda configurations of Rosanoff (1906). See, http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/history/Fischer/rosanoff.html

This is general not necessary limited to sugars. The label is given by priority rules. The verse must be measured (perhaps calculated with some quantomechanical method).