text:
Why is beryllium hydroxide amphoteric?
When we react beryllium hydroxide with acid we have an acid base reaction resulting in a salt and a water molecule. Similarly when we react it with base it results in formation of a  salt. Most sources cannot provide me with the actual reason but instead state its amphoteric because it tends to react with acids and bases both. thats not a satisfying answer.

Don't get hung up on the distinction between "basic" and "amphoteric" metal hydroxides. The difference between beryllium hydroxide and "basic" metal hydroxides like magnesium hydroxide is less than meets the eye; under the right conditions just about any metal hydroxide can act as as acid, even sodium hydroxide[1]:

Sodium amide, NaNH2, has recently been shown to be a useful catalyst to decompose NH3 into H2 and N2, however, sodium hydroxide is omnipresent and commercially available NaNH2 usually contains impurities of NaOH (<2%). The thermal decomposition of NaNH2 and NaNH2–NaOH composites is systematically investigated and discussed.  ... We report that 0.36 mol of NH3 per mol of NaNH2 is released below 400 °C during heating in an argon atmosphere, initiated at its melting point, T = 200 °C, possibly due to the formation of the mixed sodium amide imide solid solution. Furthermore, NaOH reacts with NaNH2 at elevated temperatures and provides the release of additional NH3. [emphasis added]

Reference
1.
Lars H. Jepsen, Peikun Wang, Guotao Wu, Zhitao Xiong, Flemming Besenbacher, Ping Chen and Torben R. Jense; "Thermal decomposition of sodium amide, NaNH2, and sodium amide hydroxide composites, NaNH2–NaOH", Phys. Chem. Chem. Phys., 2016, 18, 25257-25264. https://doi.org/10.1039/C6CP01604A

Beryllium ion in solution exist as $\ce{Be(H2O)4^2+}$. The small positively charged beryllium ion at the centre of the complex pulls electrons in the water molecules towards itself (strong polarizing effect on the water molecules).
Now, when we say "beryllium hydroxide", it is not structurally a cationic-anionic salt. It is actually a neutral complex which is covalently bonded. It is formulated as $\ce{Be(H2O)2(OH)2}$.
Adding sodium hydroxide initiates proton loss from the water molecules in the complex and you get the sodium salt product, sodium tetraberyllate.
$$\ce{Be(H2O)2(OH)2 + OH- -> [Be(OH)4]^2- + 2H2O}$$
Now, conversely when you add acid (say hydrogen chloride), you get a beryllium salt which contains the beryllium ion $\ce{Be(H2O)4^2+}$.
$$\ce{Be(H2O)2(OH)2 ->[H+] Be(H2O)4^2+}$$
This is how beryllium hydroxide exhibit amphoterism. Now the question arises, why can't other group 2 hydroxide exhibit amphoterism? The reason is that the other hydroxides are purely ionic and contains only hydroxide ions, $\ce{OH-}$. These react with hydrogen ions from an acid to form water and so the hydroxide reacts with acids. There isn't any equivalent to the neutral complex. Adding more hydroxide ions from a base has no effect because they haven't got anything to react with.

What you tell may be a convincing "lie for students" but $\ce{Be(OH)2}$ isn't an aquo complex or even hydrate and is almost insoluble in water. It has significant ionic component. In general it's not much different then Mg(OH)2. It's just that $\ce{[Be(OH)4]^2-}$ isn't basic enough to make its creation in aqueous base impossible.


text:
In cation analysis, can reagents of higher group precipitate lower group cations?
My teacher told me that Al$^{3+}$ and Zn$^{2+}$ hydroxides are frequently precipitated together and that is why we have to be careful about the concentration of base (NH4OH) added.
Does this mean that the group reagent of a higher group (say, group 5) can precipitate a lower group cation (say, group 3)?

No, group reagent of a particular group can't react with other groups. However, if we talk about aluminum and zinc precipitating as hydroxide, yes it is correct and that the conc. of amm. hydroxide is of utmost importance. In schools, most of us finds that zinc and aluminum are kept in separate groups, but it is in fact present in same group(III) but in different subdivisions (aluminum -  grp IIIA ; zinc - IIIB)

The reason for the subdivision is due to the fact some cations (iron, aluminum and chromium) are precipitated as hydroxides (NH4OH + NH4Cl). However, other cations (zinc, cobalt, manganese, nickel) are precipitated as sulfides (H2S + NH4OH) because their hydroxides hydrolyzes. Since, there is a common reagent (NH4OH), they are kept in same group but have two subdivisions.

Yes, technically they can precipitate cations from lower groups.
However, while doing the analysis, due consideration is given to the concentration of the added reagent so that unwanted cations do not precipitate. For example, cations of group 2 and group 4 are both precipitated as sulfides.
On the other hand, since group 2 cations require lower concentration of $\ce{S^2-}$ to precipitate, $\ce{HCl}$ is added to group 2 reagent along with $\ce{H2S}$ to reduce $\ce{S^2-}$ (common ion effect). This provides $\ce{S^2-}$ just enough to precipitate group 2 cations and not group 4 cations.

source: Stack Exchange

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

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text:
I have tried to remove oxidation from a metal scraper using vinegar, now the metal looks and smells burnt, what went wrong?
I have recently tried to remove some oxidation from a metal scraper spatula by using vinegar. After leaving it there for a couple of hours I was left with a burnt smell (with a hint of burnt candle wax). Scraper has literal burns on it alongside vertical abrasions.
Since I have no background on chemistry besides an undergrad intro course and IB chemistry back in high school I'm trying to figure out what kind of a reaction I triggered?
Here are 2 photos that show the scraper while including a kitchen knife for color comparison. the scraper was lighter and more metallic looking than the knife. In the close up you can see the original color, the rust and the burn marks clearly.
Thanks in advance.

Medical issues are strictly off topic here, but it is all but impossible to see how there could be a health hazard. I suggest rinsing the scraper with plenty of water and then use another cleaning method, i.e., steel wool or similar, in future.

thank you. I am more interested in learning what might have caused that kind of a reaction though.

If the scraper is in ordinary steel (about 99% iron, ≤ 1% carbon), the iron will be dissolved in acidic solution, but the carbon atoms may react to produce some heavy hydrocarbon that has a burnt smell. This observation is not unusual with steel.

Ah, in that case, is it possible to edit to include a photo of the scraper? That might enable someone here to elaborate on Maurice’s comment or post a helpful and valid answer.

added some photos that I believe might be of help

If you want to clean it, use sand paper like 320 SiC . The discoloration is VERY thin. The knife is 13 % Cr and resistant to most household materials.

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text:
What should be the IUPAC name of [(NH3)5Co-NH2-Co(NH3)4(H2O)]?
What should be the IUPAC name of $\ce{(NH3)5Co-NH2-Co(NH3)4(H2O)]}$?
This is similar to $\ce{[(NH3)5Co-NH2-Co(NH3)5](NO3)5}$, i.e. μ-amido-bis[pentaamminecobalt(III)] nitrate. But the two parts are different.

Related : Inorganic IUPAC for Compounds with bridged ligands?

The name of the listed product is said as $\ce{Co_2(NH_3)_9(NH_4)(H_2O)}$

Nomenclature for Bridged Coordination Compounds

I shall respond to this question with all the rules to be followed when there is a compound in the given formats:

$\ce{[L_1M_1 < (1-2 ligands) > M_2L_2]^n+}$
$\ce{[L_1M_1 < (1-2 ligands) > M_1L_1]^n+}$

I hope you are familiar with the basic rules to be followed (ligands in alphabetical order, oxidation number written at the end in roman numerals, atom of the ligand bonding with metal to be written in the case of ambidentate ligands, etc...). You can look up here - I have linked a Chemistry LibreTexts document.

Rule 1: The ligands in the middle should be written with Greek letter $\mu_n$ prefixed to it - where $n$ indicates the number of atoms to which the ligand is bonded. In practice the subscript $n$ is omitted when $n=2$. For ease of reference, we'll call this part of the name as Part 1.

Rule 2: The metal atoms must be named separately - $\ce{L_1M_1}$ and $\ce{M_2L_2}$. Call this part of the name as Part 2 and Part 3 respectively. The OP has named an example - $\ce{[(NH3)5Co−NH2−Co(NH3)5](NO3)5}$ is called μ-amido-bis[pentaamminecobalt(III)] nitrate.

Rule 3: If metal + ligand part is repetitive, use prefix bis-, tris-, tetrakis- etc.

Oxidation numbers should be included separately, and for an individual metal it is calculated by taking net negative charged ligands around it.
The final name would be in the format $\mathbf{\text{Part 2 Part 1 Part 3}}$.

The compound in your contention, $\ce{[\color{red}{(NH3)5Co}-\color{blue}{NH2}-\color{magenta}{Co(NH3)4(H2O)}]}$ : I have divided it into 3 parts (refer the colour scheme).
The $\ce{\color{red}{red part}}$ is named pentaamminecobalt(III).
The $\ce{\color{blue}{blue part}}$ is named μ-amido.
The $\ce{\color{magenta}{pink part}}$ is named tetraammineaquacobalt(III).

Merging all the parts together:
pentaamminecobalt(III)-μ-amido-tetraammineaquacobalt(III).

Another example for your understanding:

is named as
Tetraamminechromium(III)-μ-amido-μ-hydroxidobis(ethylenediamine)cobalt(III) ion.

Sources:

LibreTexts - Inorganic Chemistry - Coordination Chemistry - Nomenclature


text:
What can i safely use to remove grease and dirt from plastic (Polycarbonate, Polyethylene, Polystyrene and ABS)?
I need advice of how to clean plastic without produce cracks and white haze on them in plastic, exactly in:
-Polycarbonate
-Polyethylene
-Polystyrene
-ABS
Basically the materials of optical media (CD, DVD, Bluray), their cases and videogame console plastics
I thought that Isopropyl alcohol mixed with distilled water could make the work, but i read that in labs, the polycarbonate parts of computers are clean with Ethanol due to can be damaged in some way on the long run (Source: https://labproinc.com/blogs/chemical...ics-in-the-lab)
But the compatibility charts of the materials reports that Ethanol alcohol is just "good" and Isopropyl alcohol is "excelent"
Some studies of optical media even  recommend the isopropyl alcohol to clean it (Source: https://www.clir.org/pubs/reports/pub121/sec6/)
But, in other hand, the optical media manufacturers, recommend put in discs only water
Right now, i am a little confuse, what do you recommend to use to clean plastics without damages like paint strip, cracks or white haze?
In case that alcohol do the work, ethanol and isopropyl dry fast? When it dry it becomes the plastic flammable?
Thanks in advance

I would renounce to ethanol, because commercial ethanol (spirit) is often mixed with 1% impurity that makes it undrinkable when mixed with water to produce liquors like vodka. These impurities have practically no effect on chemical reactions of ethanol, but they can remain adsorbed on plastics. Isopropanol is purer than usual ethanol.

If in doubt, a damp washcloth and some elbow grease. Plastics do not all behave the same, obviously, so there may not be a single ‘best’ cleaning agent for all of the plastics you mention.

It isn't possible to generalise as the variety of polymers and formulations is too large. But one solution is to test samples of each polymer before committing to a specific cleaning agent (CDs and DVDs are usually polycarbonate, so test an old useless one with the solvent you have and see if it deteriorates). But start with isopropanol as it is widely available as a pure solvent and often works.

The washcloth suggested by Ed V could be dampened with plain water, or with solvent, for example, diluted isopropanol.
The amount of dirt and grease you will be dealing with, in 95% of cases, IMHO, is something that can be wiped off, rather than requiring dissolution of the gob. If you have a gob of dirt/grease, remove most of it mechanically, then gently mechanical (the washcloth), then with a reduced surface tension liquid. Avoid thinking that dirt and grease must be dissolved - they need only to be removed.
Low surface tension materials include solvents, but it has been noted that many plastics are degraded by solvents. Pure solvents are usually flammable (and often toxic). Diluting the solvent with water can still result in a low surface tension liquid, with the added advantage that the water will reduce the tendency of the solvent to penetrate (i.e., dissolve) the plastic. Diluting the solvent will be safer, cheaper, less damaging, and almost as efficient, requiring only a little more elbow grease. The figure shows how surface tension stays low even when solvents are diluted with water:

There is no absolute guarantee that dilution will solve all your problems; some risk is always involved. For example, even using diluted acetone to clean polystyrene would be a no-no.
And there is one more approach: use a diluted surfactant for your cleaning liquid. The lowest surface tension achievable by surfactants is often reached at concentrations of less than 1000 parts per million - that is 0.1% by weight. At this point, it is the water that could damage sensitive electronic equipment, so dampen the washcloth, don't soak it.

Thanks for the answer, i think that i understand what you point in the tecnical text, basically, the dirst must be removed, not disolved, with water or isopropanol. So, i think that in a order, the best option would be: Only distilled water, followed of the option of a mix of distilled water and isopropanol at 50/50, and the most sturdy ones with a mix of distilled water and soap, due to water low the tension of the disolvent, and is more safe to apply, but the mechanical effect do more work that even the disolvents. I am wrong with the interpretation?

Well, your way is safest. But to be faster, I would try 25% isopropanol in water (tap water might be OK) or 0.5% dishwashing liquid in water. You might make up a bottle of each and see how they work for you. This way, if the isopropanol solution is good but not great, you could add more, up to ~50%; same with the detergent (up to ~1-2%, not 50%!). Experiment. Water alone is good, if it works, and you don't use too much - but probably not very effective.

I will mix them and I will check, the distilled water more than anything is to be as pure as possible, although... It said that distilled water is more acidic. Would it affect the mixture in something?  And about where to mix it ... I had thought of some plastic perfume-type spray bottles, to control the amount expelled from the container. But would there be a better option to mix it?  Some glass maybe? And again, thanks for the answer

Oh, and another point, ethanol could be a substitute for isopropyl? Isopropyl is not so easily found in my area, but ethanol is, hence the question. I mean, i understand that Isopropanol is just Isopropyl alcohol not?

See Maurice's comment about ethanol. Get isopropyl alcohol rubbing alcohol from your grocery store. Spray bottles are OK if you spray onto the washcloth, not the plastic.


text:
Can dimethyl sulphide be used for ozonolysis of benzene?
There is literature that suggests that ozonolysis of benzene occurs with ozone and $\ce{Zn/H2O}$. Can it take place with ozone and dimethyl sulphide?

Yes, various reducing agents can be used, dimethylsulphide in particular.

I agree, DMS is a standard reductive workup for ozonolysis

Converting the comments into an answer; Yes, DMS can be used in place of $\ce{Zn/H2O}$ during the reductive workup for ozonolysis. The mechanism involves DMS reducing a $\ce{O-O}$ fragment of the trioxolane intermediate produced from ozonolysis by one oxygen and itself oxidizing to DMSO (dimethyl sulfoxide).

(Image source)
I did some literature survey and found out that benzene ozonolysis to glyoxal under DMS reductive workup gave 73% yield based on absorbed ozone (Keaveney et.al. 1969).


text:
InChI issue with organic salts, inconsistent approach?
I am wondering why there is the following inconsistency in InChI:

penicillin G vs. penicillin G potassium salt uses the /p layer to drop a proton
but chloramine-T vs. tosylchloramide drops an H from the formula instead

Here is penicillin G vs. penicillin G potassium salt

Here are their two InChI's aligned:
InChI=1S/C16H18N2O4S.K/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9;/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22);/q;+1/p-1/t11-,12+,14-;/m1./s1
InChI=1S/C16H18N2O4S  /c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9 /h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)          /t11-,12+,14-/m1/s1

they are nicely aligned in the formula and /c and /h layers. The only difference to construct the salt is the /q;+1 ionizing the potassium and the /p-1 removing a proton from the total, thus balancing the charges.
Now chloramine-T vs. tosylchloramide should do it the same way.

but it doesn't. I show first (1) what its InChI actually is, then the InChI of the tosylchloramide base, and finally what the chloramine-T InChI should be in my (unlearned) opinion:
see here    v                                                     vv            vvvvvvvv
InChI=1S/C7H7ClNO2S.Na.3H2O/c1-6-2-4-7(5-3-6)12(10,11)9-8;;;;/h2-5  H,1H3;;3*1H2/q-1;+1;;;
InChI=1S/C7H8ClNO2S        /c1-6-2-4-7(5-3-6)12(10,11)9-8    /h2-5,9H,1H3
InChI=1S/C7H8ClNO2S.Na.3H2O/c1-6-2-4-7(5-3-6)12(10,11)9-8;;;;/h2-5,9H,1H3;;3*1H2/q  ;+1;;;/p-1
see here    ^                                                     ^^--------------^^------^^^^

never mind the trihydrate, just focus on the formula, why is in the penecillin G case the proton removed with a /p-1 while in the chloramine-T case it is removed from the formula and the /h layer, then the electron moved from K to the base with /q-1;+1?
Since no answer came yet, I wonder what's different between the two. O vs. N. O tending to give off a proton, and N tending to capture one. Is it this? Still I think it's sad that InChI cares about the difference, it's always nicer if you can actually perform transformation operations on the InChI, separating the moieties and after adjusting the /q and figuring out to which moiety the /p difference belongs, generate the two InChIs for the two moieties separately. Alas, it doesn't work that way.
And if you answer this question, could you also say why the /p layer is so unspecific? Why doesn't it say from where the proton is dropped exactly? If you have multiple moieties here wouldn't it be clearer if the /p layer was like the /q layers, a list of additions and subtractions for each moiety?

I guess in the latter, this can be regarded as a potentially mobile proton. I also guess that the amide functional group is programatically treated quite differently than the sulfonamide functional group. The technical documentation might clear that up; otherwise the mailing list might be a good starting point for asking these questions. Obviously there is the documented source code. The charge layer (/q,/p) is intentionally vague to capture (e.g.) tautomers/any protonation state. If needed, a fixed hydrogen layer may be added to clarify if a specific structure is meant.

For penicillin G, potassium salt: what is the particular reason for you that the carboxylic acid kept its proton, while the imidic acid (conceptually like keto-enol tautomerism to the amide, possibly to a small extent only) did not?  Is the imidic acid this much stronger by pKa?

This is an extended comment to support Martin's comment by illustration, i.e. to generate InChI with a fixed H layer is going to account for the differences about peripheral hydrogens in question.  By this, you leave the sub set of standard InChI which may affect subsequent processing of the strings in other programs.  Reading InChI trust's FAQ e.g., about tautomers (e.g. «The Fixed-H layer is useful if you wish to represent a particular tautomer of a given structure.», section 6.2) may help; there are some details (e.g., «It is not possible to use InChI syntax to compare molecules with different but similar connection tables.», section 12.5) to be aware when using InChI.
In the second part of your question, the structure drawing and the InChI written do not match each other; the presence/absence of the sodium cation and water molecules influences the generation of the InChI in question (and subsequent InChIKey) and the two are discernible by standard InChI/InChIKey.

Regarding the case of penicilline, I departed from a set of eight canonical SMILES
O=C(Cc1ccccc1)N[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)O)(C)C
O=C(Cc1ccccc1)N[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)[O-])(C)C
O=C(Cc1ccccc1)N[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)[O-])(C)C.[K+]
O/C(=N\[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)O)(C)C)/Cc1ccccc1
[O-]/C(=N\[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)O)(C)C)/Cc1ccccc1
[O-]/C(=N\[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)O)(C)C)/Cc1ccccc1.[K+]
[O-]/C(=N\[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)[O-])(C)C)/Cc1ccccc1
[O-]/C(=N\[C@@H]1C(=O)N2[C@@H]1SC([C@@H]2C(=O)[O-])(C)C)/Cc1ccccc1.[K+].[K+]

which cover neuter, anionic, and forms of a potassium salt. To share the same page, here a visual survey of them by OpenBabel:
obabel -ismi penicillin_G_can.smi -osvg -O test.svg -xl --addinindex

Subsequently, a .sdf was written
obabel penicillin_G_can.smi -O penicillin_G_can.sdf -h --gen3d

With InChI trust's reference binaries (version 1.06, December 2020, repository), this container file was processed once to generate standard InChI/InChIKey, and once InChI/InChIKey considering a fixed H layer as a tabulator separated list:
./inchi-1 -tabbed -AuxNone -Key penicillin_G_can.sdf  # standard assignment
./inchi-1 -tabbed -AuxNone -Key -FixedH penicillin_G_can.sdf  # non-standard

In the standard form, there indeed is pairwise match of the strings for entries (1, 4), (2, 5), and (3, 6):
Structure: 1    InChI=1S/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/t11-,12+,14-/m1/s1    InChIKey=JGSARLDLIJGVTE-MBNYWOFBSA-N
Structure: 2    InChI=1S/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/p-1/t11-,12+,14-/m1/s1    InChIKey=JGSARLDLIJGVTE-MBNYWOFBSA-M
Structure: 3    InChI=1S/C16H18N2O4S.K/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9;/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22);/q;+1/p-1/t11-,12+,14-;/m1./s1 InChIKey=IYNDLOXRXUOGIU-LQDWTQKMSA-M
Structure: 4    InChI=1S/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/t11-,12+,14-/m1/s1    InChIKey=JGSARLDLIJGVTE-MBNYWOFBSA-N
Structure: 5    InChI=1S/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/p-1/t11-,12+,14-/m1/s1    InChIKey=JGSARLDLIJGVTE-MBNYWOFBSA-M
Structure: 6    InChI=1S/C16H18N2O4S.K/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9;/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22);/q;+1/p-1/t11-,12+,14-;/m1./s1 InChIKey=IYNDLOXRXUOGIU-LQDWTQKMSA-M
Structure: 7    InChI=1S/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/p-2/t11-,12+,14-/m1/s1    InChIKey=JGSARLDLIJGVTE-MBNYWOFBSA-L
Structure: 8    InChI=1S/C16H18N2O4S.2K/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9;;/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22);;/q;2*+1/p-2/t11-,12+,14-;;/m1../s1  InChIKey=PEPWXSJDCDQNTL-UDPGNSCCSA-L

If the optional fixed H layer is used, each member of this set is assigned a unique string:
Structure: 1    InChI=1/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/t11-,12+,14-/m1/s1/f/h17,21H   InChIKey=JGSARLDLIJGVTE-JGBPDRTNNA-N
Structure: 2    InChI=1/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/p-1/t11-,12+,14-/m1/s1/fC16H17N2O4S/h17H/q-1   InChIKey=JGSARLDLIJGVTE-JSKXZDGENA-M
Structure: 3    InChI=1/C16H18N2O4S.K/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9;/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22);/q;+1/p-1/t11-,12+,14-;/m1./s1/fC16H17N2O4S.K/h17H;/q-1;m   InChIKey=IYNDLOXRXUOGIU-UBIYMBTGNA-M
Structure: 4    InChI=1/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/t11-,12+,14-/m1/s1/f/h19,21H/b17-10-   InChIKey=JGSARLDLIJGVTE-SCGMZQNYNA-N
Structure: 5    InChI=1/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/p-1/t11-,12+,14-/m1/s1/fC16H17N2O4S/h21H/q-1/b17-10-   InChIKey=JGSARLDLIJGVTE-HOPLNNLJNA-M
Structure: 6    InChI=1/C16H18N2O4S.K/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9;/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22);/q;+1/p-1/t11-,12+,14-;/m1./s1/fC16H17N2O4S.K/h21H;/q-1;m/b17-10-;  InChIKey=IYNDLOXRXUOGIU-XNDLOSBFNA-M
Structure: 7    InChI=1/C16H18N2O4S/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22)/p-2/t11-,12+,14-/m1/s1/fC16H16N2O4S/q-2/b17-10-    InChIKey=JGSARLDLIJGVTE-JRFSBVOJNA-L
Structure: 8    InChI=1/C16H18N2O4S.2K/c1-16(2)12(15(21)22)18-13(20)11(14(18)23-16)17-10(19)8-9-6-4-3-5-7-9;;/h3-7,11-12,14H,8H2,1-2H3,(H,17,19)(H,21,22);;/q;2*+1/p-2/t11-,12+,14-;;/m1../s1/fC16H16N2O4S.2K/q-2;2m/b17-10-;;  InChIKey=PEPWXSJDCDQNTL-KBRAHMAKNA-L

In the case of chloramine, a discern of the two entries already is possible with standard InChI/InChIKey.  With canonical SMILES by OpenBabel
ClNS(=O)(=O)c1ccc(cc1)C
Cl[N-]S(=O)(=O)c1ccc(cc1)C.[Na+]

processing as above eventually yields the standard strings
Structure: 1    InChI=1S/C7H8ClNO2S/c1-6-2-4-7(5-3-6)12(10,11)9-8/h2-5,9H,1H3   InChIKey=NXTVQNIVUKXOIL-UHFFFAOYSA-N
Structure: 2    InChI=1S/C7H7ClNO2S.Na/c1-6-2-4-7(5-3-6)12(10,11)9-8;/h2-5H,1H3;/q-1;+1 InChIKey=VDQQXEISLMTGAB-UHFFFAOYSA-N

Will have to digest that more. Seems to me that your answer states what is, but not sure somewhere in there is the answer for why this is?
Clearly we can use fixed layer to nail them down. But I never doubted that. I wanted to know why the amino group's N-H gets to be treated differently in the formula, while the penicillin's O-H gets to have a /p dropped. I didn't see the answer to that.
I think an answer expanding on Martin's comment should address this "I also guess that the amide functional group is programatically treated quite differently than the sulfonamide functional group."

By the way I am told "This bounty has ended. Answers that other users post to this question are eligible for a +50 reputation bounty. You have 4 hours to award this bounty to an eligible answer." -- but what if the only offered answer does not actually contain the answer to the question? (Not trying to be dismissive or combative, I just don't understand this pressure of the 4 hours.)

So is it about the two chemicals in the neuter form instead of e.g., the potassium salt of penicillin?  Like in a comparison of acetic acid (InChI=1S/C2H4O2/c1-2(3)4/h1H3,(H,3,4)) vs. e.g., DMF (InChI=1S/C3H7NO/c1-4(2)3-5/h3H,1-2H3) vs. e.g. methanesulfonamide (InChI=1S/CH5NO2S/c1-5(2,3)4/h1H3,(H2,2,3,4))?  And it wasn't for the 50 pts («This is an extended comment to support Martin's comment»), indeed hoping a second/third/additional perspective would be put forward, but to understand the InChI better.  Just too long for a comment.

I don't know what "neuter form" means, I mean, the base with proton removed, of course I don't care about the counter-ion, Na+, K+, Ca++ whatever, I care about the base. In one case a /p-1 is used, in the other case the formula loses an H and a /q-1 is added. I want to know why that is done so differently in InChI, because frankly, it's annoying, but also, perhaps it can give more ideas about the difference of treating O vs. N.

From https://www.inchi-trust.org/technical-faq-2/: section 10.1: «For most compounds the /q layer uses a positive or negative integer to represent the actual charge on the species; the formula represents the correct composition.» And later (10.2): «What does the /p layer mean? This is the number of protons that must be added to or removed from the formula to give the input composition.» And 4.1: «each layer holds a distinct and separable class of structural information, with the layers ordered to provide successive structural refinement.» Though related, a mutually independent assignment.


text:
Computational determination of the diagram of molecular orbitals of a polyatomic chemical species
A diagram of molecular orbitals for $\ce{H2O}$ would be as follows:

The above diagram has been constructed qualitatively following the steps that I have been instructed in various classes on this bonding theory.
Nevertheless, I have seen that the Guassian software allows to visualize the molecular orbitals of a species. For example, for the case of water one obtains:

The computational results agree with the qualitative diagram data, since all electrons are paired and there are also five atomic orbitals.
What I do not know is how to obtain the energy of the orbitals of the oxygen atom and the two hydrogen atoms. On the other hand, I would not know how to construct the diagram either, that is, I would not know that molecular orbital 5, the LUMO, comes only from the 2p orbital of the oxygen.

You can visualize isosurfaces of the molecular orbitals in gview. This allows you to identify the orbitals visually. That is a way to determine that the HOMO is a lone p-orbital of the oxygen atom. I think that you have to load the *.chk file with gview and not the *.log file, if you want to plot isosurfaces since the required information is saved there. Here is a figure that shows to get to the isosurface in gview,

An alternative to this qualitative/visual approach is looking at the atomic orbital coefficients of the molecular orbital. I don't know where Gaussian prints them but if you really needed a quantitative measure you would look those up. In the case of this molecular orbital, they should show that there is only 1 atomic orbital that contributes to the 5'th molecular orbital. I.e. the coefficient of the atomic oxygen p_z orbital should be 1 (or almost one if there are some numerical deviations).
For the oxygen-atom orbitals you would need to calculate a lone oxygen atom. For the $\ce{H2}$-orbitals you would need to calculate a lone $\ce{H2}$ molecule. You can also calculate with symmetry and Gaussian will assign the molecular orbitals to the irreducible representations. This helps to identify molecular orbitals. This information is given somewhere in the log file if the symmetry was properly determined by Gaussian  during the calculation.
I also think you are mistaking LUMO(lowest unoccupied molecular orbital) with HOMO(highest occupied molecular orbital). Orbital 5 in the Gaussian output is the HOMO; orbital 6 is the LUMO.

+1 But please use MathJax! Chemical formulae can be rendered with  $\ce{...}$.

@SRMaiti Thank you for pointing the \ce environment out. I am familiar with mathjax but wasn't aware of this package for typesetting chemical equations.


text:
Quantifying soapiness; there's pH, pKa and pO2, is there a p_soap or p_surfactance?
Yes, $\mathrm{pH}$ is a concentration, $\mathrm{p}K_\mathrm{a}$ is a dissociation constant, and $\mathrm{pO_2}$ is a partial pressure. These are (roughly speaking) ways to indicate how much of a key ingredient is in a mixture or how active it is.
When I go away for a long weekend and have unwashed dishes, I just soak them in extra-soapy water; arguably to do some pre-cleaning but mostly as a lame attempt at preventing life from taking hold and multiplying exponentially in the hot summer weather before I get back.
All soaps are not created equal
I want to ask in Biology SE about the level of soapiness necessary to prevent this from happening, but first I want to ask here if there is a way to quantify the soapiness of soapy water on some recognized or at least recognizable scale.
There is a wide range of soaps available in a household and when there's no dish detergent per se available and nobody is looking I've been known to use other products. A gram of laundry powder, window cleaner, bar soap, and dish detergent could potentially have very different levels of soapiness and therefore ability to strip living cells of their protective lipid membranes, either in bulk or just key, vulnerable constituents.
Question
For purposes of quantifying soapiness, is there a recognized, or at least recognizable parameter, something like p_soap or p_surfactancein like there's $\mathrm{pH},$ $\mathrm{p}K_\mathrm{a}$ and $\mathrm{pO_2}$ for other situations?

@BuckThorn Thanks! Ya "accepted" hopes for some workable scheme that's already out there, whereas "acceptable" means I'm certainly open to working out something new. And you've hit the real crux of the upcoming biology question, once I can quantify soapiness, then words like "dilute" take on a more concrete meaning.

@BuckThorn and re clever circumvention; soap molecules don't diffuse through and into bits (okay, chunks) of food the way water molecules do, even if they are the tiniest soap molecules

Does "p" in "p_soap" refers to the same operator as in $\mathrm{pH}$ $(-\log_{10}a(\ce{H+}))?$ What is "surfactancein"?

"mostly as a lame attempt at preventing life from taking hold and multiplying exponentially" - a lot of detergents are perfectly acceptable food for a wide variety of bacteria.

@fraxinus yikes!

@EdV yes I remember! 1, 2

@fraxinus good point. Many of those detergents contain preservatives for that reason,  which I guess wouldn't be effective once the detergent is dissolved.

If you want to just solve your stop life from proliferating, the easiest solution is to just do the dishes!  I guess if you don't mind wearing gloves when you get back, you could soak your  dishes in lye.  No aluminum though. Not sure what would happen to stainless steel if you keep doing this.

If you want to apply the operator p like pX as negative decadic logarithm of X then X must be well defined value in the first place. What is the value of "soap" or "surfactance'?

@Poutnik Yes I see what you mean; you can even write that up as a supplementary answer if you like.

@fraxinus I've just asked Can some bacteria eat soap molecules in soapy water rather than get killed by it? How do they hang on to their surface lipids? Evolutionary advantage?

A concept that captures how effective a detergent is at doings its job is aptly called "detergency." As might be expected this is a complex property and difficult to describe unambiguously with a single parameter. Quoting Ref. 1 :

Detergency is difficult to evaluate because it depends on  a  multitude  of  variables  that  in  most  cases  are elusive  to  monitor  and  measure.

Given its practical importance it should not be surprising that a lot of effort has been expended to characterize this property, which in technical lingo is referred to as "detersive efficiency". Ref. 1 explains that standardized detergency-testing methods have been developed by the American Society for Testing and Materials (ASTM) or the International Standardization Organization (ISO). The method codes for various tests are listed in Table 1 of that publication. The AISE (International Association for Soaps,Detergents and Maintenance Products) has also developed laundry detergent testing guidelines.
In Ref. 1 detergency is quantified as a parameter $De$, the ratio of mass of soil suspended in the bath after treatment to the total mass of soil in the system.
References

E. Jurado Alameda, V. Bravo Rodrıguez, R. Bailon Moreno, J. Nunez Olea, and D. Altmajer Vaz. Bath-Substrate-Flow Method for Evaluating the Detersive and Dispersant Performance of Hard-Surface Detergents. Ind. Eng. Chem. Res. 2003, 42,4303-4310.

Soapiness or anything like that cannot be represented by a single number. Hence no point in inventing such a quantity. Just like we cannot associate a plain number to odors, soapiness is scientifically meaningless because it will be an umbrella term. Just like the term polarity is misused, soapiness could be even worse. The only common property of surfactants is the critical micellar concentration (CMC) but that has nothing to do with how clean our dishes or clothes look after a wash cycle. CMC would be useful if there were a single component in dishwashing or laundry soaps. Alas, our synthetic laundry detergents are a mix of really fancy chemicals.
To give you an idea, I quote from a monograph on how the detergents are tested so one can feel the complexity and understand why quoting soapiness with a single number is not useful. Take an example of a laundry detergent. See where does soapiness fit in? You can extend the same ideas to a dishwashing liquid.

Single wash cycle performance (soil and stain removal and bleaching)

Multiple wash cycle performance, e.g., after 25 or 50 washes (soil antiredeposition properties, degree of whiteness, buildup of
undesirable deposits, fiber damage, stiffness, color change,
fluorescent whitening)
Special characteristics (powder characteristics such as density, free flowability, dispensing in a washing machine, homogeneity,
dusting properties, solubility, foaming, rinse behavior, and such
storage characteristics as chemical and physical stability,
hygroscopicity, color, odor, and tendency to form lumps)

The
literature describes numerous methods for testing according to the
above criteria, some of which are standardized. Standardization is a
concern not only of national bodies (e.g., ANSI, the American National
Standards Institute; JISC, the Japanese Industrial Standards
Committee; DIN, Deutsches Institut für Normung [559]; AFNOR,
Association Française de Normalisation; BSI, British Standards
Institution), but also of international groups (e.g., ISO,
International Organization for Standardization). The above national
organizations are all members of ISO and can, therefore, exercise
influence on questions of international standardization [560]. Another
particularly important organization concerned with international
standardization of test methods is the CID (Comité International des
Derivés Tensio-Actifs) with its subcommittee, the CIE (Commission
Internationale d'Essai). This organization was disbanded in 1978 , but
in the meantime, its activities have been taken over and carried
forward by the Working Group TMS (Test Methods for Surfactants) of the
CESIO (Comité Européen d'Agents de Surface et Intermédiaires
Organiques).

Reference: Laundry Detergents. E. Smulders Wiley-VCH Verlag GmbH & Co. KGaA,
If there detergent properties could be summarized by a single number, all these govt. and regulatory agencies would have come up with a single number. To the best of my knowledge, there is none! I would say that synthetic detergent chemists and those who analyze them (analytical chemists) are certainly very smart and creative people.
Come to Nature's detergents like soapwort seeds. These "fruits" when rubbed or boiled make a good "detergent" solution. Got a chance to use them a couple of times as they are pretty common in South Asia. Their detergent action comes from saponins and even that is not a pure compound but a class of compounds which foam in water naturally but there are not typical surfactants. Again soapiness is not useful even for natural detergents.
Caution: A small note of caution of soaking dishes long term for days in dishwashing liquid solution is not a good idea. Depending on the  dishware, small cracks, chips, poor coatings can lead to absorption of detergent in the dishes. Of course, this should exclude all wooden items and even plastic items.

I saw a question by uhoh over at Astronomy SE (pertaining to using multiple prisms and having multiple interfaces) and posted some temporary stuff, of possible interest to him, in the Sandbox III: https://chemistry.meta.stackexchange.com/a/4757/79678. Thanks again for the paper about von Littrow! I also posted my most recent CFL echellegram (you have already had it).

@uhoh I am not sure how much pratical echellography you have done. We discussed that last year or two years ago. Please do see link posted by Prof. EdV and his echellogram. We can chat about that.

@M.Farooq yes I'd love to! We'll need an appropriate chat space. My practical experience is one long night forty-ish years ago but I spend a lot of time immersed in optics and imaging systems, math, physics and simulations, and these are beautiful (and colorful!) instruments.

@uhoh, Then your experience and optics experience would be great for our echelle spectrograph project. How can we (EdV & I) contact you?

The laundry detergent example appears to be measuring lots of things that are definitely not soapiness. Buildup of undesirable deposits, fiber damage, stiffness, colour change - none of these have anything to do with "soapiness".

@M.Farooq I live within Stack Exchange. I love the question and answer format. If you'd like something calculated analytically or scripted and simulated in Python, or explained or trouble-shot see if you can fit it into the structure of a Stack Exchange question. Thanks!


text:
Balancing octane combustion reaction for internal combustion engine
I am trying to balance out the equation for combustion of octane in $120\,\%$ of excess air with products $\ce{CO2},$ $\ce{H2O},$ $\ce{N2},$ and $\ce{O2}$:
$$\ce{C8H18 + 1.2$a$(O2 + 3.76 N2) -> $b$ CO_2 + $c$ H2O + $d$ O2 + $e$ N2}$$
My approach was as follows:
$$
\begin{align}
&\ce{C}: & b &= 8 \\
&\ce{H}: & 2c &= 18 \\
&\ce{O}: & 2b + c + 2d &= 2\times 1.2a \\
&\ce{N}: & e &= 3.76\times 1.2a
\end{align}
$$
However, I think somewhere in my steps there's something I am not doing correctly as I am unable to determine $a$ with the system of equations I tried developing.

Octane combusted in air yields carbon dioxide plus water. The nitrogen in air does nothing, unless you are in, e.g., an internal combustion engine. So ignore nitrogen.

This is for an internal combustion engine.

I have found the already balanced equation online, it includes all those reactants and all those products, I am trying how to figure out the balancing aspect.

A chemical equation must not mention substances that are not transformed, like $\ce{N2}$and part of $\ce{O2}$ here.

Maurice I understand what you are trying to explain; however, these reaction is written including O2 and N2 and the balancing process was done including them. If you google the reaction you can see what I am talking about.

If this actually is in the context of chemical engineering, then this should be mentioned in the question. Ideally the whole exercise should be quoted verbatim with a citation. From the chemical point of view, there really is not much use in including these; and for the problem at hand the best solution would be to strip the exercise down to the actual reaction that happens then include whatever additional information what was given.

It is remarkable that many textbooks and lectures select the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion. Note that internal combustion engines actually do not run on octane. Typical gasoline contains less than 1 % octane. See also my answer here about common misconceptions related to octane.

It is not clear whether you are trying to represent an idealised reaction of octane or a realistic reaction from an engine. You show nitrogen but not any reaction products involving it (which just overcomplicates things unless you intended to include reactions producing nitrogen oxides). If your want the latter, you won't get a simple equation as the side reactions are highly sensitive to the detailed reaction conditions.

Nitrogen is included in the reaction as it is carried out with theoretical air which is (O2 3.76N2)

Since the reaction is done in excess of air, you can assume complete combustion and accordingly you should break down the reaction to only this.
Apparently the question demands an answer in the following form:
$$\ce{C8H18 + 1.2$a$ (O2 + 3.76 N2) -> b CO2 + c H2O + d O2 + e N2},$$
however, this is the same as
$$\begin{multline}
\ce{C8H18 + x O2 + 0.2x O2 + (1.2$x$\times3.76) N2}\\
\ce{-> y CO2 + z H2O + 0.2x O2 + (1.2$x$\times3.76) N2},
\end{multline}$$
which is the same as
$$\ce{C8H18 + x O2 -> y CO2 + z H2O},$$
just more complicated.
From there you can form
\begin{align}
\ce{C}: && y &= 8\\
\ce{H}: && z &= \frac{1}{2}\times18 = 9\\
\ce{O}: && 2 x &= 2y + z \Longleftrightarrow x = 12.5.
\end{align}
Now you can obviously go back to the requested form:
$$\ce{C8H18 + 15 (O2 + 3.76 N2) -> 8 CO2 + 9 H2O + 2.5 O2 + 56.4 N2}.$$

(+1) And, of course, no internal combustion engine was involved, even putatively: no nitrogen oxides were produced. As well, the problem did not involve considerations of thermodynamics, e.g., the heating up of the nitrogen in the air.

Key things I did not consider properly:
120% of excess air, which means 0.2 of oxygen goes free, this changes my equation to:
$$\ce{C8H18 + 1.2$a$ (O2 + 3.76 N2) -> b CO2 +c H2O + 0.2a O2 + e N2} $$
So,
\begin{align}
\ce{C}:&& b  &= 8\\
\ce{H}:&& 2c &= 18\\
\ce{O}:&& 2b + c + 2(0.2) &= 2\times1.2a \implies a = 12.5\\
\ce{N}:&& e &= 3.76\times1.2a = 56.4
\end{align}

This is exactly the same as $\ce{C8H18 + x O2 -> y CO2 + z H2O}$, just more complicated.

The point was for me to understand how to workout the problem using this kind of forms. I still need to write down the balanced from of the more complicated expression.


text:
Comparing rates of substitution reactions
The following question was asked in an exam I gave recently

Compare the overall rates of the following substitution reactions
$$\ce{CH3Cl ->[OH-][Weak P.A.S]
CH3OH}$$

$$\ce{CH3CH2Cl ->[OH-][ Weak P.A.S]
CH3CH2OH}$$
$$\ce{(CH3)2CHCl ->[OH-][ Weak P.A.S]
(CH3)2CHOH}$$
$$\ce{(CH3)3CCl ->[OH-][ Weak P.A.S]
(CH3)3COH}$$ (P.A.S means polar aprotic solvent).
I know that $\ce{CH3Cl , CH3CH2Cl}$ react mainly via $S_{N^2}$ mechanism, $\ce{(CH3)2CHCl}$ react via both $S_{N^2} , S_{N^1}$ mechanisms considerably and $\ce{(CH3)3CCl}$ via $S_{N^1}$ mechanism.
The rates of the reactions can be given by the following rate equations
$r_1 = k_1[\ce{CH3Cl}][\ce{OH-}] + k_1^☆[\ce{CH3Cl}]$,
$k_1^☆$ being negligible
$r_2 = k_2[\ce{CH3CH2Cl}][\ce{OH-}] + k_2^☆[\ce{CH3CH2Cl}]$
,$k_2^☆$ being negligible
$r_3 = k_3[\ce{(CH3)2CHCl}][\ce{OH-}] + k_3^☆[\ce{(CH3)2CHCl}]$
can't neglect anything here,
$r_4 = k_4[\ce{(CH3)3CCl}][\ce{OH-}] + k_4^☆[\ce{(CH3)3CCl}]$
$k_4$ can be neglected here.
So, I would assume the reaction-3 occurs at least rate as ${k_3}$ and ${k_3^☆}$ are both low, but I don't know how to compare between $k_1 , k_2 , k_4^☆$.
Is there any theoretical way to compare
them? If yes, then how?
Answer given:

 The answer is 4>1>2>3

What is P.A.S? Also it should be $\ce{S_N2}$ and $\ce{S_N1}$ and not $\ce{S_{N^2}}$ and $\ce{S_{N^1}}$.

By "exam I gave recently" you mean that you are the author of this question? Or do you mean that you wrote this exam? I think it should depend on whether the solvent is protic or aprotic, but in neither case I see why this order is correct

@Azamat No, I am not the author of the question. I've recently written a mock test for JEE conducted by a local institute.

$S_n2$ in general is faster reaction than $S_n1$ and rate of $S_n2$ directly depends on stability of transition state (here bulky transition states are less stable). Also you can neglect $S_n2$ here because polar aprotic solvents favor $S_n2$ (since as transition state proceeds charge density reduces which is better stabilized by aprotic solvents). I would think (theoretically) the order is $S_n2$ based unless relevant data is provided.

How do you measure the rate of substitution based on RCl concentration when elimination is appreciable in the secondary and tertiary halides? Isn't the  appearance of alcohol more relevant?

Actually one can't always find answers through logical formulations to these JEE problems on organic chemistry , but we can understand and compare logical constraints. I remember this question was asked in two different ways in my examinations - one considering only $\mathrm{S_N2}$ among the reactions and the other one is same as yours.
$$\ce{CH3Cl ->[OH-][Weak P.A.S]
CH3OH}$$
$$\ce{CH3CH2Cl ->[OH-][ Weak P.A.S]
CH3CH2OH}$$
$$\ce{(CH3)2CHCl ->[OH-][ Weak P.A.S]
(CH3)2CHOH}$$
$$\ce{(CH3)3CCl ->[OH-][ Weak P.A.S]
(CH3)3COH}$$
If we only consider $\mathrm{S_N2}$ among these reactions order will be $\mathrm{1>2>3>4}$.
But we have to consider how large the other atoms or functional groups bonded to the carbon are, compared to $\ce{OH^-}$ also known as steric inheritance or steric crowding, then
For this question we are considering a substitution reaction..
You said the answer yourself - $\ce{CH3Cl, CH3CH2Cl}$ react mainly via $\mathrm{S_N2}$ mechanism, $\ce{(CH3)2CHCl}$ react mainly via both $\mathrm{S_N2, S_N1}$ mechanisms considerably and $\ce{(CH3)3CCl}$ via $\mathrm{S_N1}$ mechanism.
Now you have to consider steric hindrance in $\ce{(CH3)3CCl}$ where the three $\ce{CH3}$ groups makes $\mathrm{S_N1}$ more favourable. Also the tert-butyl carbocation $\ce{(CH3)3C^+}$ is more stable beacause of 9 alpha hydrogens which involve in hyperconjugation. In most cases when steric factors and hyperconjugation both favour carbocation formation then $\mathrm{S_N1}$ is the major pathway preferred over $\mathrm{S_N2}$.
So for $\mathrm{4}$, strongly favoured $\mathrm{S_N1}$ pathway increases rate of reaction. For the other 3, $\mathrm{S_N2}$ is dominant which follows the order $\mathrm{1>2>3}$. Hence overall order is $\mathrm{4>1>2>3}$.

But can you say that $\ce{S_N1}$ is always faster than $\ce{S_N2}$? I don't understand what "Weak P. A. S." means but the rates of the two mechanisms and what mechanism is dominant depend on the solvent. Or am I missing something?

@Azamat No we can't say $S_N1$ is faster than $S_N2$ vice versa. without knowing the solvent .this may help you know about solvents also about  "Weak P. A. S." [link]https://www.quora.com/Why-are-SN1-reactions-favored-by-protic-solvents-whereas-SN2-reactions-are-favored-by-aprotic-solvents

The question is inane.  The conditions are unknown the mechanisms are various and what is a weak PAS.  Also the logic that SN1 is favored over SN2 makes the SN1 reaction faster than another SN2 reaction is fallacious.


text:
Which is the correct parent chain in this compound?
In this compound:
Will the parent chain be the one inside the loop, or the longer one including the single bonds to the left of the vinyl group?
I know that carboxylic acid is the highest-priority functional group, and that it must therefore be included in the parent chain. However, once we have selected a highest-priority group, don't the rest of the functional groups (in this case, a double bond) take on substituent priority? Therefore, I believe that the longer chain will be the parent chain in this case.
Is my attempt correct?

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

The illustration looks like being cropped from an illustration.  So, it is not clear if all of the structure currently is shown (then a hexanoic acid).

@Loong in that question, they talk exactly about unsaturation. Is it the same for carboxylic acid group though?

@Buttonwood I've cropped it shoddily but the entire molecule is shown. Thanks for answering.

You are quite right about the feeling that the longer chain must be the parent chain,
According to the newly revised (IUPAC Blue Book 2013), the longest chain containing the functional group regardless of presence unsaturation (here the double bond) will be the parent chain
Check this answer for details regarding this here
Hence the parent chain will be a 6-Carbon chain with the carbonic acid functional group. Therefore the correct Preferred IUPAC Name (PIN) will be 4‐methyl‐3‐methylidenehexanoic acid


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Is the correct parent chain the longer one or the one including two double bonds?
I want to know which is the correct parent chain in this compound:
Is it the one inside the loop, or the longer one that includes the part to the left of the allyl group?
I know that once a functional group of highest priority has been selected, lower-priority functional groups take on the same priority as substituents. How does this rule work with two instances of the same functional group? Do we select the parent chain including both, or select one of them and the longest chain possible thereafter?

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

@Loong Thanks, but my school still expects me to use the old naming rules! If unsaturation took priority over chain length the correct parent chain would be the one in the loop, right?

@RayBradbury according to old naming rules, yes

Does this answer your question? How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

This may be of help.

According to the IUPAC 1993 Blue Book the name of the compound would be : 3-Pentylsept-1,6-diene (The circled one being the parent chain)
According to the new IUPAC 2013 Blue Book the name of the compound would be : 5-(1-ethylidene)-dec-1-ene (The longest continuous chain being the parent chain)
The reason being the new Rule in the 2013 conventions that state that the longest chain irrespective of the degree of unsaturation will be the parent chain.

The substituent at C-5 is a vinyl group, not ethylidene. Ethylidene would have the double bond in the trisubstituted position and have an E/Z designation. 5-Vinylundec-1-ene.

Both suggested names are wrong. It's "hepta" not "sept", and it's "ethenyl" not "(1-ethylidene)".


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Calculating electrons emitted when quantum yield is given
In a problem, the wavelength $\lambda$ of a monochromatic light, power $P$ of the source, and work function $\phi$ of the metal on which the light is shown is given. Quantum yield is $n$. I have to find the rate of emission of electrons.
My attempt:
Energy in a single photon = $hc/\lambda$
Number of photons falling on metal per unit time = $\dfrac{P}{hc/\lambda}$
Quantum yield is the "number of times a specific event occurs per photon absorbed by the system." Then, the number of electrons ejected = $\dfrac{nP}{hc/\lambda}$
Is my approach here correct? I am mostly confused whether I've interpreted the meaning of "quantum yield" correctly.

Looks good: the quantum efficiency is between 0 and 1 and is generally wavelength dependent, so your n is for your particular wavelength.

Also do a dimensional analysis to convince yourself. Quantum yield of a process is "no. of particles emitted"/ "Number of particles absorbed".


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Finding solubility of salts of polyprotic acids
Problem

Given, for $\ce{H3PO4}$,
$\mathrm{pK_\mathrm{a_1}=2.2}$
$\mathrm{pK_\mathrm{a_2}=7.2}$
$\mathrm{pK_\mathrm{a_3}=12.4}$
For the salt $\ce{Mg(NH4)PO4}$, $\mathrm{pK_\mathrm{sp}=12.6}$ and the equilibrium concentration of $\ce{NH4^+}$ is $\mathrm{0.1 M}$. If the solubility of $\ce{Mg(NH4)PO4}$ in the solution, with $\mathrm{pH=10}$ is S, find the value of $\mathrm{-log(S)}$.

Answer

 $\mathrm{4.6}$

Question
I assumed that the contribution of $\ce{NH4^+}$ ions from the salt is to be neglected, so $\ce{[NH4^+]=10^\mathrm{-1}}$. The solubility S is then equal to the concentration of $\ce{Mg^2+}$ ions. Also, the initial amount of $\ce{PO4^3-}$ will also be S, out of which, I took some $x$ amount to have hydrolyzed to $\ce{H3PO4}$, $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$.
Then,
$$\ce{K_\mathrm{sp}=\ce{[Mg^2+][NH4^+][PO4^3-]}}$$
$$\ce{K_\mathrm{sp}=\mathrm{(S)(0.1)(S-x)}}$$
And using the values of $\mathrm{pK_a}$, I can get three more equations, with two more variables for the amounts of $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. Four equations, four variables - that's fine in general, if one has a computer. But this particular exam doesn't even allow a calculator, so I don't think that is what we are supposed to do here.
Another similar, but much simpler question - one is asked to find the solubility of $\ce{AgCl}$ in an $\ce{NH3}$ solution of given molarity, with the values of $\mathrm{pK_\mathrm{sp}}$ and formation constant of complex $\ce{[Ag(NH3)2]^+}$ given. In that case we assume that all of the $\ce{Ag^+}$ initially released, turns to $\ce{[Ag(NH3)2]^+}$ and solve accordingly.
But in this case there are three different ions and products possible by hydrolysis, so I am unable to find out which of them should be considered as the dominant species.
EDIT - M.L.'s answer clarified the calculations involved. My main confusion now is regarding why $\ce{HPO4^2-}$ (and not $\ce{H2PO4^-}$ or $\ce{H3PO4}$), will be the major ionic species in the solution. Is there a way to predict this based on values of $\mathrm{pK_a}$ and the $\mathrm{pH}$ of the solution?

Okay, so I managed to get a solution without too many equations by making the right approximations.
First with your expressions:
$\ce{K_{sp} = [Mg^2+][NH_4^+][PO_4^3-]}$
$\ce{K_{sp} = (S)(0.1)(S-x})$
Here we want to find S-x which is $\ce{[PO_4^3-]_{eq}}$. To do this, I did the following:
$\ce{\frac{[PO_4^3-][H+]}{[HPO_4^2-]}=K_{a3} = 10^{-12.4}}$
Assuming $\ce{[H^+] = 10^{-10}}$:
$\ce{\frac{[PO_4^3-] 10^{-10}}{[HPO_4^2-] 10^{-12.4}}=K_{a3} = 1}$
$\ce{\frac{[PO_4^3-] 10^{2.4}}{[HPO_4^2-]} = 1}$
Now we could continue on and find $\ce{[H_2PO_4]}$. But it turns out the value is quite small and can be essentially ignored. Here is what we'll get if we solve for it:
$\ce{\frac{[HPO_4^2-][H^+]}{[H_2PO_4^-]} = K_{a2} = 10^{-7.2}}$
$\ce{\frac{[HPO_4^2-] 10^{-2.8}}{[H_2PO_4^-]} = 1}$
Replacing $\ce{[HPO_4^2-]}$ with $\ce{[PO_4^3-]}$ in our earlier expression, we get
$\ce{\frac{[PO_4^3-] 10^{-0.4}}{[H_2PO_4^-]} = 1}$
Now, we could set the balance for $\ce{S = [PO_4^3-]_{eq} + [HPO_4^2-] + [H_2PO_4^-] + [H_3PO_4]}$ (Since S should equal the total concentration of any phosphate species) and then replace each concentration with the ratios we solved above. If we ignore $\ce{[H_3PO4]}$ (which makes because if we solve for it's relative concentration, $\ce{[H_3PO4] = 10^{-8.2}[PO_4^3-]}$), we get:
$\ce{S = [PO_4^3-]_{eq} + [HPO_4^2-] + [H_2PO_4^-] = [PO_4^3-]_{eq} + 10^{2.4}[PO_4^3-]_{eq} + 10^{-0.4}[PO_4^3-]_{eq}}$
Solving, we get $\ce{\frac{S}{1+ 10^{-0.4} + 10^{2.4}} = [PO_4^3-]_{eq}}$ and we can plug this into the original expression:
$\ce{K_{sp} = (S)(0.1)(S-x})$
$\ce{10^{-12.6} = (S)(0.1)(\frac{S}{1+10^{-0.4} 10^{2.4}})}$
Here, just because it is non-calculator, I will approximate the denominator of the fraction to be simply $10^{2.4}$ as the $1 + 10^{-0.4}$ term is very small compared to $10^{2.4}$ and is just complicating the calculations
$\ce{\frac{10^{-12.6} 10^{2.4}}{0.1} = 10^{-9.2} = S^2}$
So we can find that $S = 10^{-4.6}$ and $-\log(S) = 4.6$
If we didn't make the approximation we did for $1+10^{-0.4} + 10^{2.4}$ term, the value we would get is $-\log(S) = 4.59879$ which obviously shows that the approximation we made didn't alter the resulting value by too much.

How did you realize that the concentrations of $\ce{H2PO4^-}$ (and subsequently $\ce{H3PO4}$) will be small enough to be negligible? Is there a logic/way to predict the major ionic species in the solution depending on conditions such as pH? Thank you :)

Hi. I made a few edits to my answer to clarify that up. Essentially, you can try to solve for the concentration of the next phosphate species if you'd realize that it is so small that it could be ignored. For me, I understood what I was doing and my brain just spontaneously decided that it would be suitable to make the approximations I did so I didn't actually solve for $\ce{H_2PO_4^-}$ in terms of $\ce{PO_4^3-}$. But you could as it shouldn't take too long and you'd realize the value is so small that you can essentially ignore it.


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Clarification on standard hydrogen electrode diagram
When looking at standard hydrogen potential electrode diagrams in the case of a negative standard reduction potential for the other half cell, I came across one part that does not make sense to me: (image from chemistry libre texts)

Unless I'm reading it wrong, it appears that hydrogen gas is being added to the system at 1 atm. Why is hydrogen gas being added if it does not participate in the reduction reaction of protons in solution? It seems to me that adding hydrogen gas will hinder the process by pushing the chemical equilibrium to the left.

It is not specific for the hydrogen electrode. An electrode potential depends on activities of both oxidized and reduced forms, regardless of the direction of the net reaction.

There is a good explanation and diagram here, https://www.chemguide.co.uk/physical/redoxeqia/introduction.html

@porphyrin From what I interpret from the article, would the diagram I posted be misleading as to how the standard hydrogen electrode would be used? It appears to me that minimal amounts of current should be allowed to flow through the voltmeter in order to truly measure the standard reduction potential of the other half cell.


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Major product formed in Reimer-Tiemann reaction when Potassium Hydroxide is used as a base
Reimer-Tiemann reaction is known to give ortho major product when NaOH is used as a base. However there seems to be some confusion regarding the major product in case KOH is used as the base.
It is known that when KOH is used as a base in Kolbe-Schmitt reaction para product is favoured whereas ortho product is favoured in case when NaOH is used for the same.
Now the accepted answer in this post states that the same thing will happen even in reimer-tiemann i.e. para will be the major product when KOH is used. However it does not states any credible sources which back it up and hence this creates confusion since the Wikipedia page on Reimer Tiemann displays the reaction with KOH as its base and it prefers the ortho product. But again I don't see any specific talk about KOH on this page or the books I have.
Hence to clear the confusion, I'm posting this question once again. If anyone has any citable/reliable source which clearly and specifically examines the case of KOH in reimer-tiemann reaction please answer this question. Also reasons as to why it should or why it should not are welcome.

From here:

Reimer and Tiemann (Ber., 1876, 9, 824) state that, contrary to experience in the Kolbe-Schmidt reaction, in their synthesis of hydroxy-aldehydes the proportion of $ortho-$ and $para-$ derivatives  is unaffected by changing from  sodium  hydroxide to potassium  hydroxide

The improvement in technique afforded by the determination of aldehydes by dinitro phenylhydrazine has enabled  us to show the effect of change of alkali in the Reimer-Tiemann reaction  where the original authors failed (loc. cit.). In 15N-alkali the ortho-para-ratio  was $\ce{NaOH} \ 2.08$, $\ce{KOH}\ 1.24$, $\ce{CsOH}\ 0.98$, and $\ce{NMeEt3.OH}\ 0.52$. In 2N-alkali the difference almost disappeared, the ratio being $\ce{NaOH}$ $0.90$, $\ce{KOH} \ 0.91$, and $\ce{NMeEt3.OH}\  0.86$. In 15N-lithium hydroxide the ratio was $0.9$ but, as much lithium phenoxide crystallised out, the concentration in solution was low.

When the Reimer-Tiemann  reaction is carried out in less concentrated solutions, e.g., in 2N-alkali, the ortho-para ratio, when tetra-alkylammonium hydroxide is used, corresponds to the substitution of the phenoxide  ion  plus that of free  phenol  formed by hydrolysis. When 2N-sodium hydroxide  is  used,  only a small amount of co-ordination  compound is formed, so that the ortho-para ratio is  not appreciably altered.

When a concentrated $\ce{NaOH}$ or $\ce{KOH}$ solution is taken, clearly the ortho product is major (o/p ratio $2.08$ and $1.24$ respectively). In the 2N-alkali solution, the para product is major and the o/p ratio for $\ce{NaOH}$ and $\ce{KOH}$ are nearly the same ($0.90$, $0.91$); this can be explained by saying that the concentration of the $\ce{Na+}$ and $\ce{K+}$ cations in the solution is low enough that they do not effect the reaction, hence the para-product is major and the o/p ratio is same for both $\ce{NaOH}$ and $\ce{KOH}$.
In the Kolbe-Schmidt reaction, the formation of the para-product with $\ce{KOH}$ is due to the inability of the $\ce{K+}$ cation to sufficiently stabilize the transition state due to it's larger size when compared to $\ce{Na+}$. The same effect can be seen with the Reimer-Tiemann reaction, where the o/p ratio for $\ce{KOH}$ ($1.24$) is significantly lower than the ratio for $\ce{NaOH}$ ($2.08$). Nevertheless, the ortho product is still the major product with $\ce{KOH}$, and the linked answer is wrong.
Reference: O.L. Brady and  J. Jakobovits, J. Chem. Soc., 1950, 767-777

+1, interesting indeed. Had a doubt, what is the plausible reason for para product being major at low concentrations, is it due to steric reasons?

@Ashish Yes, you can explain it using both steric reasons and the negative inductive effect of the oxygen atom.

Ortho product is also stabilized by intramolecular h bonding


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Does a UVC transparent polymer exist or can it be made?
Does a UVC transparent polymer exist or can it be made?
I am trying to find a polymer that can transmit UVC light. Most materials absorb this 265 265 nm range. for example quartz passes UVC but glass does not. I have tried a UVC transparent silicone but the losses are to great.

What are other requirements, aside of UVC transparency? Generally, provide to any question enough of relevant info to avoid clarification ping-pong. What do you mean be "265 265 nm range" ?


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How Does Voltmeter Sense/Measure All Potential Differences Present in an Electrochemical Cell?
Voltmeter is an instrument which measures electric potential difference between two points.
When measuring electrode potential of some redox system (vs SHE for example), it is said that voltmeter reading contains sum of all potential differences present in a cell. This includes all electrode/electrolyte potential differences, contact potential differences between electrodes and probes of voltemter and possibly liquid junction potential. Probes of voltmeter are connected with two electrodes in a cell.
However, since voltmeter measures potential difference between two points, how can it sense sum of all potential differences in a cell if it is connected only between two electrodes and not all potential differences are created at these two points?

In addition to the answer by @Poutnik, you might also look into three and four wire potentiostats. They can be a bit hard to fathom, but they are part of electrochemistry’s common instrumentation.

@EdV These devices are indeed interesting, but they are still kind of like separation of two quarks ( leading to 2 quark doubles ). Advantage is some of the potential differences can be considered in this case constant and some  variable.

@Poutnik No problem: I already upvoted your answer.

Yes, I think I need to check how voltage devices work as I thought they can measure only potential differences between two points on which they are in direct electronic contact. I will check potenciostats as well.

What is strange to me is that as far as I know measuring potential difference between two points is done so that you take a voltmeter and connect these two points to voltmeter in parallel. I am not sure how can voltmeter measure all these potential differences in a cell with only two probes.

It does not measure "all these", it measures their sum, like if you would measure summary potential difference on e.g.  a serie of resistors, which were integrated in a single package ( for the sake of example ). The cell can be formally simulated by an integrated circuit, including variable voltage source, current dependent resistors and potential difference dependent capacitors.

Maybe I misunderstand, but are you familiar with Kirchhoff’s voltage law? There is an article in wikipedia on Kirchhoff’s circuit laws.

Yes, I am familiar with the 2nd Kirchoff's law which expresses the fact that electrostatic field is conservative. Circulation of electrostatic field is always zero. Voltmeter does measure a sum, yes. If I wanted to measure a voltage on series of 3 resistors for example, I would need to connect a voltmeter on start of 1st resistor and the end of 3rd resistor. If I wanted to measure every resistor independently, I would adjust it accordingly. In any case I need to connect voltmeter on two points between which I want to know what voltage is

In a cell, we connect voltmeter on two electrodes and how is it that it doesn't read only voltage between metal electrodes and not all other potential differences. It would not be of much help if I connected voltmeter on some resistor and it includes a voltage on some other resistor as well in the reading.

Quite frankly, I cannot see why this is puzzling you. If you put the voltmeter leads on the terminals of an ordinary automobile battery, just to pick an example, you get a measurement that well approximates the battery’s open circuit voltage. The chemistry is happening inside, but so what? It is a black box, as it were, and could be replaced with its Thevenin equivalent in that regard.

Yes, but saying that something is a black box doesn't help in understanding of what happens inside.

True, but also, in some sense, perfectly fine: the voltmeter doesn’t care what is in the black box. All it can do is what you stated at the beginning: measure the potential difference between the two places contacted by the voltmeter’s electrodes. I once saw someone get less than 12 V from a 12 V automobile battery by screwing a wood screw into the side of the battery! Astoundingly bad and dangerous idea on multiple levels. But they figured if they put the screw halfway between the ends, they would get about 6 V. Then their voltmeter gave a measurement of about 6 V. It did not care.

Let us continue this discussion in chat.

It is the topic for electronics, as measuring of electrochemistry related potential differences by a voltmeter is not different to measuring other potential differences. In both cases they are potentials of different pieces of conductor.
Potential of the wire connected to SHE is conventionally taken as the zero potential.
Even in pure electronics, all what a voltmeter measures is the sum of potential differences across all the circuit path from one measuring point to the other one. That includes all kinds of intended or parasitic components of the circuit. It is the same as for circuits involving electrochemical cell.
There is a close parallel, as you e.g. cannot in general directly measure potential differences inside enclosed integrated circuits.
Partial potential differences within the cell are not measurable this way, must be measured indirectly ( e.g.by analysis the cell behaviour under the load ) , or differently, or estimated theoretically.


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Using salt and ice vs cold brine to rapidly cool an object
This a question regarding the practicalities of using salt+ice vs just cold brine to cool something in the home (inspired by this recent post: https://cooking.stackexchange.com/questions/116437/cooling-coffee-rapidly-without-dilution). It's similar-ish to this one: Will ice cream freeze faster with frozen brine or adding salt to ice?, but the question does have a key difference from mine I think
A couple answers/comments in the coffee question point out that combining ice+salt to create a slurry can be used to rapidly cool a cup off coffee. The question is, since home freezers are recommended to be set at 0F/-17.8C, and potentially can be set even lower, will this have any practical benefit over just storing a container of brine in the freezer and using it for cooling the coffee? My guess is that at or near -21C, it just doesn't matter anymore, whereas if things were closer to 0C, the ice+salt would see a significant advantage. Assume a sufficient freezer temperature and brine concentration that the brine doesn't actually freeze solid and become inconvenient to use.

Trying to better understand what's going on by reading more, is the ice absorbing energy to 0C, melting, dissolving the salt, then dropping back down to -20-ish C?

It can be said, that ice effectively melts at lower temperature in NaCl presence, down to near -20 deg C, depending on ratio ( usually salt:ice 1:3 ). For other relatively available salts, CaCl2 . 6 H2O gives in ratio salt : ice 2 : 1  even near -50 deg C. But it is not very practicar for regular repeated use.

The freezing but liquid concentrated brine may have  better thermal conductivity, so cooling down could be initially faster. But brine would be warming itself up immediately. Ice in the salt and ice slurry would have ( more or less) melt before the resulting brine would start to warm up.
Ice+salt would be warming up slower. Energy needed to melt ice is equal to energy needed to warm water by about 80 Deg C. OTOH, practical manipulation with ice and salt would make total time impractical, compared to brine, or even cold water.
Brine has advantage of reusability and readiness to be used.
IMHO, the most practical is just water bath with cold water, possible continuously replaced. Or metal cup immersed into hot liquid. Brine would bring opportunity of accidental splashing into coffee or other cooled down liquids.
Another way could be cooling by cold water streem via tube/pipe spirale, the best if from a heat conducting metal chemically compatible with the liquid. Stainless steel would be on safe side for acidic fruit tea, where aluminium should not be used.

But as the ice melts, won't the salt dissolve and change the heat capacity?

Comment moved to the updated answer.


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Hessian matrix and mass-weighted Hessian eigenvalues
If I have the Hessian matrix and the mass-weighted Hessian matrix for a system, I can find the normal modes from the mass-weighted Hessian and the force constants from the Hessian.
I have a list of the eigenvalues and eigenvectors, but I'm trying to figure out how to map each force constant to each vibrational mode. How would I go about doing that?
I have two sets of eigenvalue/eigenvector pairs. The eigenpairs for the mass-weighted Hessian gives me the vibrational frequencies and the eigenvectors for those frequencies. The eigenpairs for the non-mass-weighted Hessian (so, just a normal Hessian) give the force constants and the associated eigenvectors. What I want is to match the eigenpairs from the Hessian to the eigenpairs for the mass-weighted Hessian so then I'd have the force constant for each vibrational mode.
Right now I have a list of force constants and vibrational modes, but I have no way to know which force constant goes to which vibrational mode (since I don't know the reduced mass for the vibrational modes). If I can figure out how to relate the eigenvectors between the Hessian and mass-weighted Hessian, then I'd be able to pair the vibrational frequencies and force constants.

How did you obtain the eigenpairs? It's normally pretty obvious which eval is connected to which evec. Or is it visualising what the evec is telling you that you are asking about?

@IanBush Oh, I wasn't very clear in my post.  I have two sets of eigenpairs, one set for the Hessian and one set for the Mass-weighted Hessian.  I got the eigenpairs using python's numpy.linalg.eig.  So I have:

A list of vibrational modes (eigenvalues of the Mass-weighted Hessian) and the corresponding eigenvectors
A list of force constants (eigenvalues of the Hessian) and the corresponding eigenvectors.

I'm trying to figure out which eigenpair from the Hessian matches which eigenpair from the Mass-weighted Hessian, so I then know the force constant for each vibrational mode

The eigenvectors between the Hessian and Mass-weighted Hessian should be related.  I'm just struggling to find what that relationship is.  For example, if I just had two C atoms, I figured the eigenvectors would be related by a factor of 12.011 or  sqrt(12.011).  If that were true then I could easily take an eigenvector from the mass-weighted hessian, multiply by this factor, and just solve for the eigenvalue of the non-mass-weighted hessian.  When I tried this, multiplying the eigenvector for the mw-hessian by 12.011 or sqrt(12.011) didn't yield an eigenvector from the non-mw-hessian.

+1 But you should mention that you used python in the question, and post a code snippet if possible. If you don't get any answers here, you may also want to check out Matter Modelling SE.

The eigenvectors of the Hessian and the mass weighted Hessian are not connected by a single scalar factor.
You might want to look at this white paper from the Gaussian Developers, https://gaussian.com/wp-content/uploads/dl/vib.pdf
It explains how mass weighted eigenvectors are connected to reduced masses and cartesian displacement eigenvectors.

To put it another way the two matrices are not related by a similarity transform (https://en.wikipedia.org/wiki/Matrix_similarity) so there is no simple reason why there should be a direct mapping of the eigenpairs.

If this question focuses on programming in any way, it should be migrated to [mattermodeling.se] or [compsci.se]. Otherwise you might get better answers from the theoretical physics community at [physics.se]. In the context of theoretical and physical chemistry it is obviously on-topic here. You can pick, but you should pick one as we discourage cross-posting. So maybe you would like to familiarise yourself with these sites first.

@HansWurst Okay, so I really just need the mass weighted hessian, according to that document.  If you look at second 2.6, this l_cart matrix is essentially what I was doing.  In my case, I don't have the "D" matrix that they use, since I don't need to pull out translation/rotational modes.  So matrix "L" is just the matrix containing the eigenvectors of the mass-weighted hessian.  Then L*M where M is a diagonal matrix with elements 1/sqrt(mass_i).  So now we have mass-weighted eigenvectors.  Then taking the inverse of the norm^2 for each column vector should be the reduced mass for that mode?

section 2.6, not "second 2.6"

@NicholasH. Normal modes and their eigenvalues and eigenfrequencies are based on the mass weighted Hessian. I think the only case where you can use directly the unweighted Hessian is, the case when all particles have the same mass, which isn't the case for molecules in general.  You can define a reduced mass and transform afterwards from the mass weighted modes to Cartesian displacement normal modes. These Cartesian displacement modes are what people typically indicate with vectors on atoms to illustrate normal modes.

@HansWurst Well, so I've done some random playing around with Hessians and mw-Hessians.  I got the force constant two ways.  First way is just like in the gaussian document, using the eigenvectors of the mw-hessian to calculate a reduced mass and then getting the force constant.  The other thing I did was just to sort of do a quick and dirty comparison between mw-hessian eigenvectors and hessian eigenvectors and just mapped out which eigenvectors from the mw-hessian corresponded to hessian eigenvectors.  Then I could map eigenvalues from the hessian and mw-hessian, so then... (next comment)

@HansWurst previous comment continued... so then, I would have frequencies from the mw-hess and the force constants from the hessian.  So, what was interesting to me here is that these methods gave different answers, but I thought they would be the same.  So method 1 was just using the mw-hessian, calculating a reduced mass, and then a force constant.  Method 2 was just mapping mw-hessian eigenvectors to hessian eigenvectors, which then also maps eigenvalues.  But I actually got quite a bit different force constants. For low freq modes, the force constants were similar, but for (next comment)

@HansWurst For the highest frequency modes, method 2 gave force constants that were almost exactly double the force constants from method 1.  For my research, it's only the lowest frequency modes that we care about anyways, and these methods provided very similar answers for those, but it seems strange to me that the methods gave different force constants for the higher frequency modes.

I'll avoid extending the comment section by posting this as an answer. But this is mostly intended to address your last comments.
The actual value of the reduced masses associated with normal modes depends on the convention that you are using. There is not a single "true" value for the reduced mass, it is a property that we define for convenience. It is not a physically measurable quantity in the case of vibrations of polyatomic molecules. The value of the force constant changes with your choice for the reduced mass, since we have
$$
k_i = \mu_i \omega_i^2
$$
This should always hold true and needs to be consistent. If this does not hold, your are doing something wrong. Per default, when we diagonalize the mass weighted Hessian, we more or less have all reduced masses equal to 1. The force constants are then simply the angular frequencies squared. If you introduce reduced masses, your force constant has to scale according to this equation.
You also shouldn't diagonalize directly the Hessian matrix. The resulting eigenvectors do not lead to decoupled equations of 1D harmonic oscillators. This is due to the fact that the mass matrix does in general not commute with the matrix of eigenvectors of the unweighted Hessian. There is a special case when this works, and that is the case when all atoms have the same mass. In this special case, the mass matrix commutes with any matrix since it is simply a multiple of the unity matrix. In all other cases you do not obtain the proper eigenvalues and eigenvectors when using the unweighted Hessian matrix.
Force constants are thus based on the eigenvalues of the mass weighted Hessian and the choice of reduced mass. You do not obtain the force constants from the unweighted Hessian matrix in the general case.
The underlying math of what I said can be found here,
How is the vibration of multiple atoms treated in vibrational spectroscopy?


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Why is the carbanion stability order the way it is? (3*<2*<1*)
According to the lecture notes I made, a 1* carbanion is more stable than a 3* carbanion. I can't grasp exactly why. (I had a look at the similar questions, the answers didn't help)
With my very limited knowledge, there can be possibly two effects at play here:

The Inductive Effect
Hyper-conjugation

The mesomeric effect can't take place here since there are no pi-bonds, as far as I know.
Now, for the inductive effect, I know that CH3 is considered a +I group (electron-donating), but I don't know why, since the inductive effect operates through electronegativity, and two carbon atoms have no relative electronegativity difference. Also, my textbook tells that the electronegativity difference between carbon and hydrogen can be considered to be zero, so I can't see why the inductive effect would do anything here. (If anything, it seems to me that since there's a negative charge on the central carbon, it should spread out to the methyls, since there's no electronegativity difference between carbons, stabilizing the tertiary carbanion the most)
For hyperconjugation, I've read that it is the interaction between sigma-electrons and pi-electrons, or sigma electrons and empty p-orbitals(as in the case with carbocations). Since there's no pi bond in these compounds, and the central carbon's octet is full (3 bonds, 1 negative charge), I can't see why there would be any hyperconjugation in these compounds.
What's happening?

Please typeset the material in the photo - it is not searchable.

Added a different picture @ToddMinehardt

Which textbook mentions electronegativity difference between carbon and hydrogen as zero? That's a fantastically wrong statement.

@TRC My coaching modules do, they're not standard textbooks, maybe that;s why the error. So is that the reason?

Electronegativity of H is 2.2, C is 2.5. So in a $\ce{CH3}$ group the electron density is pulled towards the carbon side - even if by a (non-negligible) small amount. This increased electron density on the methyl carbon in turns makes this more electron rich, and it tries to push its electrons towards the next carbon. Hence causing +I effect. In carbanion case, if the methyl carbon is already electron rich due to the three attached hydrogen, it wouldn't like to accept more negative charge from the neighbour, would it?

Okay, thanks :)  My age-old query answered sigh


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How come the methyl group (-CH3) is electron donating?
If the inductive effect operates purely through electronegativity differences, how come the methyl group is considered to donate electrons through the inductive effect? My textbook says that there is negligible electronegativity difference between carbon and hydrogen, and definitely none between two carbon atoms, then how come CH3 shows the +I effect?
Reading other similar questions on CSE, (Why is the methyl group electron-donating via the inductive effect?) the answers seem to say that hyperconjugation is the reason for the inductive effect, which I cannot grasp at all.
As far as I've read, hyperconjugation does not depend on electronegativity at all, then how come both the effects are related in any way?
Coming back to the question, why exactly is CH3 a +I group?

It makes some acids weaker then -H group, and not via resonance - that's about how much reason such rules of thumb need.

And if it actually does make them weaker depends largely on solvent effects (or lack of thereof).

Hi when there is a bond b/w H and C there is a slight charge speration like this

In methyl there are three H so it looks like this

Notice the delta negative charge on carbon this is enough for carbon to "donate electron"
When case is with ter. carbocation the delta negative charge is satisfying positive charge of carbocation stabilizing it so

So it is a +I group

@new_to_code would highly advise you to either give the images in proper handwriting / source images from the internet


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Does the Li-F cell possess the highest possible voltage?
My question is quite straightforward: would a theoretical cell with fluorine and lithium yield the highest possible standard potential of all?
The table of Standard electrode potential in Wikipedia lists several potentials that are "better" than those of lithium and fluorine but I am skeptical about those values. The more common tables often list lithium with a potential of -3.040 volts and fluorine with a potential of 3.053 volts. Thus the standard potential gives 6.093 volts. However, can this number be improved? Are there materials that have more negative potentials than lithium or a higher potential than fluorine?
Please ignore the cost of the cell or practically.

See https://chemistry.stackexchange.com/questions/102502/what-is-the-strongest-oxidising-agent which is all but a duplicate of this. It gives an example of a theoretical cell with a potential of a bit over 2,000,000 Volts.

Be aware that standard electrode potentials are thermodynamicaly equivalents of standard Gibbs energy of reaction a ox + b H2(g) <=> 2b H+(aq) +  a red as Delta G_r = -nF E_std.. Some of them are  just theoretical, with limited or hypothetical possibility to be directly measured, or to design a cell based on it. Note that above potentials are defined for water based solutions and hydrated Li+ and F-, what would definitely not be the case if such a cell were ever designed.


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Selection of p-orbital(s) when considering sp or sp2 hybridisation
I am considering $z$-axis as the internuclear axis in all cases.
When we consider overlap of s and p orbital to form a $\sigma$-bond, the chosen orbital must be p$_z$ orbital for a proper overlap.

Now, when we consider two hybridised sp orbitals, their properties are different from s or p orbitals. And I think they can reorient themselves in any direction to form bonds.

Is there any particular direction of this hybrid sp-orbital w.r.t. coordinate axes?
(i) First, take the case of $\ce{BeCl2}$. In the excited state of $\ce{Be}$, there is only one $\ce{e-}$ in the p-orbital. Can any p-orbital participate in hybridisation to form two sp hybridised orbitals?
But these two sp hybridised orbitals must overlap with two p$_z$ orbitals of two $\ce{Cl}$ atoms to form two sp-p$_z$ $\sigma$-bonds.
(ii) Secondly, take the case of $\ce{NO2-}$, in which $\ce{N}$ is sp$^2$ hybridised. In this case, the remaining p-orbital with $\ce{N}$ atom must be a p$_x$ or p$_y$ orbital which will form a $\pi$-bond with p$_x$ or p$_y$ orbital (respectively) of $\ce{O}$ atom. As two p$_z$ orbitals cannot form a $\pi$-bond.
Thus, in the three hybridised sp$^2$ orbitals, one must be a p$_z$ orbital. And one of the half-filled p-orbital of $\ce{O}$ must be a p$_z$ orbital to form sp$^2$-p$_z$ $\sigma$-bond?

This concept of the “internuclear axis” always being z (which is the basis of your claim that sigma bonds must be formed from pz) is fine for linear molecules, but breaks down completely for anything nonlinear like NO2. The bonds point along different axes; clearly, they can’t all be z at the same time.

You need $d_{x^2 - y^2}$ $d_{x^2 - y^2}$. And that comment is implicitly assuming that there's only one bond under discussion, i.e. a diatomic molecule, which is necessarily linear. As long as you only want to talk about one bond, it's perfectly fine to call that axis $z$. If you want to talk about two bonds at once, you can't call both of them $z$ at the same time.

@orthocresol Understood. But then could not understand your comment to this question: {d${xz}$, d${yz}$}→π , and {d${xy}$, d${x^2−y^2}$}→δ; Why these particular orbitals would form these bonds?

@orthocresol OK. But then one more confusion, why do we choose d$_{z^2}$ orbital for $\ce{sp_3d}$ hybridisation and likewise there are other examples... Does it anything to do with the orientation of lobes, which are along the axis?

Well, why not? In NO2, it’s perfectly permissible to choose one of the N-O bonds to be the z-axis, so you can indeed say that that bond is formed from s and p_z. You just can’t also claim that the other bond is also made from p_z. With an octahedral transition metal complex, it’s also perfectly permissible to choose one of the three bonding axes to be z, so that the d_z2 orbital points along that axis. The only restriction is that you can’t claim that the other axes are also z. Also, this sp3d thing is dubious at best, but I’ll let that slide for now.

Or in a square planar complex, it’s also permissible to choose the “out-of-plane” axis to be z (i.e. the axis in which there are no ligands), and we nearly always do so. Hence the d_z2 orbital doesn’t interact (much) with the ligands. That’s all perfectly fine to say, as long as you’re internally consistent. Calling both axes z in NO2 is not internally consistent, hence it’s not okay.

Note, that there is one more issue. You seem to suggest that an sp orbital is formed from one s and one p orbital. That is true in the simplest of cases, but you picked a difficult case with NO2-. The bonds don’t point at right angles to each other, so you cannot assign one p orbital per bond: the geometry doesn’t work out. Actually the truth is that the sp orbitals are a mixture of s and p orbitals, such that the total contribution of s orbital is equal to the total contribution of p orbitals. In other words, you could have something like s plus half px plus half py.

@orthocresol N is sp² hybridised. So, there will be contribution from all 3 p orbitals such that p orbitals contribute two-third part. Now the whole notion of selection of p orbitals seem to be collapsed!