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Aluminium isopropoxide electronic structure
From Wikipedia:

A tetrameric structure of the crystalline material was verified by NMR spectroscopy and X-ray crystallography. The species is described by the formula $\ce{Al[(μ-O-i-Pr)2Al(O-i-Pr)2]3}$. The unique central $\ce{Al}$ is octahedral, and three other $\ce{Al}$ centers adopt tetrahedral geometry. The idealised point group symmetry is $D_3$.

I can understand that the 3 external $\ce{Al}$ atoms can be (inaccurately?) described as assuming an ${sp^3}$ hybridization. With 4 bonds per $\ce{Al}$ atom, the valence shell is complete, following the octet rule. This too may deserve some additional attention, since $\ce{Al}$ formally has only 3 electrons available for bonding, yet here it forms 4 bonds. At least one of these is dative, I assume?
I'm less clear on the central $\ce{Al}$ atom. My initial instinct was to invoke the ${sp^3d^2}$ hybridization. But this answer re $\ce{SF6}$ suggests that may not be the case. There are clearly similarities to the $\ce{SF6}$ case, but surely, also some differences? After all, $\ce{Al}$ is not $\ce{S}$.
In short, what is the electronic structure of the four $\ce{Al}$ centers?

It might be easier to consider all of the Al atoms as Al3+ actions and all the propoxide groups as anions. Each O has three lone pairs that can interact with the metal

Yes, that's true. Each $\ce{O}$ atom has one valence electron used up in the $\ce{O-C}$ bond, and each $\ce{Al}$ donates one electron to each of 3 $\ce{O}$ atoms, completing their valence shells. The oxygens should all be equivalent, so we can probably think of all of the $\ce{O-Al}$ bonds as dative.

More accurately, the $\ce{O}$ should be almost equivalent. For example, some of them are bridging, while others are not.


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Why is boiling point of Ammonia greater than Arsine?
I have previously read that the boiling point of Stibane(SbH3) is greater than Ammonia(NH3) as ammonia is gas at room temperature and no hydrogen bonding exists in the gaseous form of NH3.
Hence, as SbH3 is greater in size, has greater molecular weight; its boiling pint is higher than NH3.
why does Stibane have a higher boiling point than Ammonia
In the case, of AsH3, even this molecule is gas at room temperature and its boiling point should be greater than NH3, but this is not the case and NH3 has a higher boiling point than AsH3.
I want to know what is the reason behind this trend. (NH3 having a higher boiling point than AsH3)

My question is regarding group 15 of the periodic table. Hydrogen bonds are prevalent in a liquid state and both NH3 and AsH3 are gas at room temperature.

Have a look: https://chemistry.stackexchange.com/questions/64191/why-is-the-boiling-point-of-stibane-higher-than-that-of-ammonia

There is the general trend in the groups 15, 16, 17 of raising of boiling points for the binary compounds with hydrogen, going down the groups.
But the first members of each group - $\ce{NH3, H2O, HF}$ - have anomally with their boiling points being exceptionally high, due hydrogen bonds. The strength of these bonds and the boiling point decreases in the order $\ce{H2O > HF > NH3}$.
See the figure 2 on wordpress page for an illustrative picture:

Hydrogen bonds play role in (not limited to):

evaporation and boiling, as molecules need to gain an extra energy to overcome these bonds and to escape liquid.
condensation, acting as cohesive force for molecules to stick to each other or to a condensation center.
gas pressure, as the inter-molecular cohesive force decreases the gas pressure, compared to the pressure of an ideal gas at the same conditions.

Note that the fact that both NH3 and AsH3 are gases at room temperature is irrelevant for the above.

Not having in hand means for quantitative analysis, so just thoughts:

$\ce{NH3}$ has the least polar bonds and their dipole moments in large extent mutually cancel each other due sterical orientation.
$\ce{HF}$ is the most polar with the biggest electronegativity difference, but this seems about beaten by 2 polar bonds of $\ce{H2O}$, both having very similar dipole moments. What may play role is better sterical availability ( 2 versus 1 ) of water hydrogen for mutual hydrogen bonding.

Is there a reason why "The strength of these bonds and the boiling point decreases in the order H2O>HF>NH"?

@AdilMohammed That is a wrong question. If there was no reason, the trend would not be there.

Fair point, I'll rephrase my question. Is there a good reason why the trend occurs? If so what is it? If Hydrogen bonding depends on the EN of the other atom, the order would have obviously been different. I am suspecting the small size of Fluorine is the culprit here.

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Why doesn't base/acid reacting with water create a salt?
An acid/base nwutralization will create a salt + water. If one uses water itself as the acid or base, and have an acid/base neutralization, how come that doesn't create a salt?
For example:$$ \ce{HCN(aq) + H2O(aq) <<=> H3O+ + CN- ->[??] H3OCN}$$
Why doesn't that happen? What is keeping those conjugate bases and acids from forming a salt? And if there is something keeping these conjugate acids and bases from forming a salt, why doesn't it stop other acids and bases from forming salts? Like $$\ce{NH3 + HCl  -> NH4+ + Cl- + H3O+ + OH- -> NH4Cl + 2H2O}$$

The second part of this answer may have some useful information.

Salts with H3O+ cation aren't called salts.

This is a nice example where a semantic problem signals that everything should be redone almost from scratch. The answer is even surprisingly harder to be written. Beside that, as remarked above, those "salts" aren't called like that.

Remember that H3O+ is much stronger acid than HCN and CN- is much stronger base than H2O. Only the strongest acids form solid hydronium salts, like perchloric acid. [H3O+][ClO4]. Even anions like HSO4- are protonated by hydronium back to H2SO4.

@Alchimista Agreed, I feel like chemical terminology and the models they've made make simple concepts so confusing. It annoys the hell out of me. In reality, it is all as simple as electrostatic forces interacting in different ways. At least for me, the quantum physical descriptions of what's going on is easier to understand than the conceptual organization of these concepts.

Sorry but I was referring to your level of acquaintance not to the way chemistry is nor to nomenclature. Perhaps I may agree on the school chemistry curriculae, in part.

An acid/base neutralization will create a salt + water. is rather a secondary/junior high school teaching.
It is rather
$$\ce{AcidA + BaseB <=> BaseA + AcidB}$$
E.g.:
$$\ce{NH3(base) + H2O(acid) <=> NH4+(acid) +  OH-(base)}$$

When strong bases or acid dissolve in water, they completely dissociate:
$$\ce{HCl(g) + H2O ->[H2O] H3O+(aq) + Cl-(aq)}$$
$$\ce{NaOH(s) ->[H2O] Na+(aq) + OH-(aq)}$$
When they neutralize each other, it is reaction of hydronium and hydroxide ions. They do not form salt at all. The ions forming future salt are just "spectator ions", existing in solution before and after reaction:
$$\ce{H3O+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) <=>> 2 H2O + Cl-(aq) + Na+(aq)}$$
what is usually written just as:
$$\ce{H3O+(aq) + OH-(aq) <=>> 2 H2O}$$
When concentrations of ions $\ce{Na+(aq)}$ and $\ce{Cl-(aq)}$ reaches high enough values, there start to partially form fully separated, then "solvent sharing", then "contact" ion pairs $\ce{[Na+Cl-](aq)}$. They finally precipitate, forming the solid salt $\ce{NaCl(s)}$, consisting of alternating ions $\ce{Na+}$ and $\ce{Cl-}$.

Very weak acids like HCN have the dissociation equilibrium strongly shifted to the left:
$$\ce{HCN(aq) + H2O <<=> H3O+(aq) + CN-(aq)}$$
as there are weak acid and base on the left, while strong acid and relatively strong base on the right.
The ions on the right are present just in traces. If an acid like $\ce{HCl}$ is mixed with solution of $\ce{KCN}$ and the ions meet in high numbers, they mutually neutralize each other on the spot:
$$\ce{CN-(base) + H3O+(acid) <=>> HCN(acid) + H2O(base)}$$
Most of $\ce{HCN}$ will then probably escape as a gas, as it has boiling point $\pu{26 ^{\circ}C}$.

Neutralization vs Solvation vs Dilution:
Semantically, the reaction of acid or base with water is (usually) solvation and dilution, not neutralization. Pure water $\ce{pH}$ (at $\pu{25^{\circ}C}$) is always 7, i.e. neutral.
When you add an acid to water, you are dissolving the acid. The formation of $\ce{H3O+}$ ions necessarily lowers the $\ce{pH}$, away from neutral:
$\ce{pH \equiv -log10([H+]) = -log10([H3O+])}$
(Technically, $\ce{pH}$ is defined with activity of $\ce{H+}$, but concentration is a good enough approximation for most cases.)
As you can see, higher concentration of $\ce{H3O+}$ ions leads to lower $\ce{pH}$.
Similarly, when you add a base to water, you are increasing the concentration of $\ce{OH-}$ ions (which also destroy some of the naturally occurring $\ce{H3O+}$ ions, lowering their concentration), so $\ce{pH}$ rises away from neutral.
$\ce{pH \approx 14 - pOH = 14 + log10([OH-])}$
Higher $\ce{OH-}$ concentration leads to higher $\ce{pH}$.
When you then add more water, you are reducing the concentration of both $\ce{H3O+}$ and $\ce{OH-}$ ions (as well as that of all other dissolved species), thus bringing the $\ce{pH}$ closer to neutral. This is called dilution.
Neutralization is when you add a base to an already acidic solution ($\ce{pH < 7}$), or an acid to an already alkaline solution ($\ce{pH > 7}$) (one can also react acids and bases out of solution, but that's beyond the scope of this post). In the former case, new $\ce{OH-}$ ions react with pre-existing $\ce{H3O+}$ ions; in the latter case, new $\ce{H3O+}$ ions react with pre-existing $\ce{OH-}$ ions. Either way, the following neutralization reaction occurs:
$\ce{H3O+ + OH- -> 2H2O}$
Thus, the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions is reduced, and the $\ce{pH}$ is again brought closer towards neutral.
Note that when you dilute a solution with water, the above reaction doesn't occur.

Ions, Salts, and Hydrates:
A salt, by definition, is a chemical compound consisting of an ionic assembly of cations and anions. In solution, these ions dissociate from each other, and are surrounded by solvent molecules (solvation shells). So in solution, what you have isn't exactly a salt, but rather a collection of isolated ions. Some of these can and do form solid salts. This is determined by the salt's solubility product.
The $\ce{H3O+}$ ion will generally not form solid salts, because its affinity to water is too high. It's very similar to plain water molecules. However, if we look at solid samples of various hydrophilic substances, we'll often find that they're hygroscopic. Meaning, they attract water molecules even from air, and form hydrates. This is when a substance's molecules are intermixed with water molecules even in the solid phase.
If we look for example at a solid acid such as benzenesulfonic acid, we will find that it readily forms a hydrate. For very strong solid acids, some of the water molecules in the hydrate will in fact be in the $\ce{H3O+}$ form. But for weaker acids, most of the water molecules will be in the neutral $\ce{H2O}$ form. More info here.
If you add too much water to such acids, the hydrate will dissolve, and you'll get an acidic solution.
On the other hand, $\ce{OH-}$ ions form stable salts much more readily. For example, if you add a base to a solution of $\ce{CaCl2}$, the $\ce{Ca(OH)2}$ will precipitate:
$\ce{CaCl2(s) + H2O -> Ca^2+(aq) + 2Cl-(aq)}$
$\ce{CaCl2(aq) + 2NaOH -> 2Na+(aq) + 2Cl-(aq) + Ca(OH2)(s) v}$
The $\ce{Ca(OH)2}$ is itself a base, but poorly soluble in water. And $\ce{NaOH}$, a common base, is also technically a salt.

Regarding your two examples:
As Poutnik noted, $\ce{HCN}$ is a weak acid and is a gas at very close to room temperature. But if you could isolate your suggested "$\ce{H3OCN}$", you would find this is just the normal hydrate: $\ce{HCN.H2O}$. The water would be in the neutral form. Even in solution, the $\ce{H3O+}$ and $\ce{CN-}$ ions would only appear in trace amounts (since this is a weak acid).
For your 2nd example, $\ce{NH4Cl}$ does indeed exist, but it is a very soluble salt. So if you react say aqueous ammonia with $\ce{HCl}$, you will only get the separate ions:
$\ce{NH3(aq) + HCl(aq) <=> NH4+(aq) + Cl-(aq)}$
You would need to remove water for the salt to precipitate (very carefully, so that it doesn't decompose back to $\ce{NH3}$ and $\ce{HCl}$, which are both gases).

Even as a solvent, water is still an acid and a base, by both Broensted-Lawry and Lewis theories of acids and bases.

In the sense that it participates in acid-base reactions, yes, it is both. But in the sense of its $\ce{pH}$, pure water is neither (neutral, $\ce{pH} = 7$), as stated. But you're right, that sentence is inaccurate at best, so I'll remove it.

Being an acid  (acid(1)) and forming acidic solution (acid(2)) are 2 different things. E.g. HPO4- is an acid(1) as well, even if Na2HPO4 solutions have pH > 7 ( so it is not acid(2)), because it is stronger base than acid(1).

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HCl Concentration vs HDPE Permeation
I've learned that HCl escapes HDPE. I looked up why and it turns out it's because of the small size of the molecule. That factor is unlikely to be affected by the concentration of HCl in water, but I was thinking: does water exert an "attractive" effect on HCl such that HCl would escape an HDPE container from a solution of 30% HCl (for example) over an order of magnitude more than from a 20% solution?
Does this phenomenon take place? What order of difference in HCl permeating an HDPE vessel could I expect between solutions of 10%, 20%, 30% concentration? (just to have a general idea of how big the phenomenon is, if at all)

Check Fick's first law on diffusion : $J \propto - dc/dx$.

So the relationship is exponential (squared)?

? I don't think so.


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Aromatic Character in Percentage?
I was just searching whether Uracil is aromatic or not and in this link$^1$ it is said that (from the abstract):

High-level computational methods were tested for the enthalpy of hydrogenation of uracil, but only with G3 was possible to obtain results in agreement with the experimental ones. It was found that uracil possesses 30.0% of aromatic character in the gaseous phase.

What is the meaning of this sentence? I have only basic knowledge on Aromaticity limited to Huckel's rule and the names of a few aromatic compounds. I didn't expect Aromaticity to be expressed in percentages. Can someone explain why?
Reference

Tiago L. P. Galvão, Inês M. Rocha, Maria D. M. C. Ribeiro da Silva*, and Manuel A. V. Ribeiro da Silva. J. Phys. Chem. A 2013, 117, 28, 5826–5836.
https://doi.org/10.1021/jp404938u

Nothing is truly black or white. Consider pure ionic and pure covalent bonds to be unobtainable extremes. Every real bond is somewhere in between.

I found this article but I do not have  access to it.

Technically, a good point is made here. What is the meaning of "100% aromatic?"

@Zhe I'm sure the paper explains the meaning of the quoted percentage. ;) Ratio of prefactors in a linear combination of orbital models?

Assuming it is more or less clear to all of us what aromaticy means, and the ways it can be expressed or assessed, then one can construct many comparative scale. One can think of termochemistry, another of chemical shifts, and relate these to the corresponding in benzene, just to give an example. Or, it can come directly by MO calculations at various levels. The most straightforward would be that the phi of aromatic orbital has coefficient of 0.3; or as a comparison is likely made, a ratio between prefactors as @Karl said. This is to say that you can get the message without overthinking.

I took the liberty of expanding the quoted text for clarity. Feel free to rollback the change if you disagree this improves your post.

@MilanPaul if you'd like to access that paper you linked to in the comments, you could use Sci-Hub. Disclaimer: it's legality is questionable

G3 is generally a term used to describe a particular computational model used in calculating molecular bonding in quantum chemistry.  The G refers to Gaussian and 3 is just a third iteration of including additionally complex terms to the molecular orbital calculations being used.  There's a brief discussion on wikipedia, but you'll find more extensive discussion of these types of calculations in Quantum Chemistry text books or directly from the researcher's works just an example, a G3 Theory site.
Aromaticity can actually be measured quantitatively via Aromatic ring current and calculated with [NICS] nucleus-independent chemical shift.  The wikipedia article included a link to a paper describing the quantitative scale.  It's just a way of saying how aromatic is the molecule compared to another (call it a standard / perfect example molecule.)  The problem with such relative standards is there may always be one that exceeds the standard, so you could potentially end up with results larger than 100%.

Looks like I need to learn a lot....

Can you also explain the need for quantitative measurement of Aromaticity(in specific percentages). When we say that a compound X is y% aromatic w.r.t a particular method of quantitative measurement, what does it actually say about the molecule(other than the experimental/computational result itself)? If a compound X  has a higher 'percentage of aromaticity' than a compound Y, is X more stabilised by aromaticity than Y?


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How to compute Franck-Condon Factors for spin-orbit states?
I am computing Franck-Condon Factors between the neutral ground state and cationic states to simulate the photoelectron spectra (using Gaussian 16, freq=FCHT). The target cationic state is a pi state and has 3/2 and 1/2 spin-orbit components. I have done optimization and frequency calculation for the pi state in TDDFT. Thus I have FCF for the pi state before spin-orbit splitting. How do I compute FCF two spin-orbit components ? In another word, how can one optimize the two different components of a spin-orbit state. I used the same FCFs for both the components separated by the SOC. It provided reasonable agreement with experiment, however, I wonder if there is more accurate way of doing it.
Thank you for your help.

Is the photo-electron spectrum that you are comparing it to well resolved enough to distinguish effects from spin-orbit coupling?

Yes, the spin-orbit coupling is quite high and the components are well separated. Each component has associated vibronic lines.


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Decarboxylation of 2‐cyano‐2‐cyclohexylideneacetic acid
2‐Cyano‐2‐cyclohexylideneacetic acid (1) is heated and $\ce{CO2}$ is lost:
[
Why is this not formed?

In other words, why do π-electrons move into the ring?

As drawn, the reaction will not work. Conceptually, if you lose H+ from -CO2H and then lose CO2 you are left with the anion of the nitrile. Add the proton on the anion and you have CO2 and 2-cyclohexylacetonitrile. There is no double bond.

Oh I see what has happened. I think someone editing the question has copied one of the molecules incorrectly. I am changing that now.

You need to ping @user55119, or else they won't get a notification.

This reaction is an Organic Syntheses prep;http://orgsyn.org/Result.aspx?ga=pa KOH was not used. Now you can work on a mechanism. Notice that ammonium acetate is still present during the decarboxylation. BTW, I believe this very question was asked previously on ChemSE.


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The quantum mechanics behind periodicity of elements
Especially in high school/first-year undergraduate chemistry courses, we learn with great dedication the periodicities along groups and periods. There are various useful and interesting trends.
I wanted, however, to find the theoretical framework behind it which explains the trends through mathematical analyses. I skimmed through the book Physical Chemistry by Peter Atkins, but couldn't find much information.
So I was also considering the possibility that the trends were mainly discovered and investigated through experiments, and quantum mechanics was there only to provide some rough theoretical yet non-quantitative interpretations. (For example, explaining the fact that the $4s$ orbital has a lower energy level than $3p$ orbital using what's called the penetration effect, which says that there is a comparably significant value of $|\psi|^2$ for $4s$ than $3p$ for smaller radius despite the probability densities at higher radius or expectation values of radius)
Is there any qualitative analysis of such periodic trends using quantum theory? If so, where may I find it?

As you surmise, quantum mechanics was developed to help explain observations such as that periodicity, spectral lines and the photoelectric effect.
However, the solution of the Schrödinger equation is difficult, and not complete beyond hydrogen. For example Wikipedia states, "Unlike for hydrogen, a closed-form solution to the Schrödinger equation for the helium atom has not been found." Matrix mechanics can also predict some properties, such as spectral emissions. Neither, approach, though, is likely to give a clear insight into chemical behavior for heavier elements, particularly where relativity affects inner electron behavior for the actinides.
So you've raised a good question, and as mathematics and computational speed increase, exact solutions for other elements will be found.
[To me, it seems a bit odd (no pun intended) that there is one pair of s electrons, three pairs of p, five pairs of d and seven of f, as if it's numerology. But that's my limitation. Sigh... or ψ.]

Nice answer but I disagree with the final reference to oddity. Well, it is true that everything looks odd at the atomic level. But the fact that the atomic orbital are, and are filled, like that does not go to change with computing power. A more accurate description can be surely envisioned, but unless of dramatic change, the quantum description leads already to the "numerology" we observe. That Pauli exclusion pr. must be taken as a property of nature it is also true, in such an aspect Chemistry (the Periodic table) can be seen still open.

ok for me is quite the same. Then insert a line between the last paragraph and the bracketed sentence. It will be more clear. Like it is, one can read that computing power might send away this magic, which I think is an unforeseen situation unless changes happen at the core theory.


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What are the currently most accurate sources for lists of the values for elemental properties
I have a hopefully simple question. I am a tutor working with a student that wants to know where to find the most recent accurate measurements for elemental properties. He's only in Chemistry I, so he's not looking for every property out there calculated multiple ways. I think he (and I) just like looking at the best sources. So we're looking for the following:

Latent heat of fusion
Latent heat of vaporization
Melting point
Boiling point
Density
Specific heat
Ionization energy
Atomic radius (metallic and covalent)
Atomic mass
Electronegativity
Commonality and uncommonality of charges (like oxygen is commonly -2 but is also -1 in peroxide)

I can point him to wikipedia or https://ptable.com/#Properties/Series for some of the information, but I myself don't know where the most legitimate sources for these values can be found. My CRC handbook (97th edition) lists ionization energies, but for atomic weight it doesn't list the weight of the most common isotope of elements like Ac. Also, it's a few years old.
My question is, what are the current best sources of accurate values for each of these properties?

I don't you will find everything in one place. Have you explored the NIST website? They are supposed to have the most accurate/ or accepted values. Wikipedia for serious researchers should be their last resort but it is great starting place. The Landolt-Bornstein tables (partly in German) used to have tables of properties but they are not online completely.

For #9, google for 'atomic mass evaluation 2016'.

@Farooq: You know I had trouble finding anything on NIST and then you're comment made me think to good NIST instead of searching their website. That works better.

@Custer. That's a pretty cool reference.

For the radioactive elements greater than uranium there are a couple of considerations. First all elements fermium and beyond the half life is so short that bulk measurements are probably not available. Also there is no natural source for anything greater than uranium so the isotope mix of any element with a greater atomic number would depend on how that element was synthesized and how long it had "cooled off" since its synthesis.

Here's the link to the NIST chemistry webbook: https://webbook.nist.gov/chemistry/

I find the CRC handbook useful but access may be difficult.

You should teach your student not to use numerical values with as many significant figures as possible. Knowing how to make sense to decimals is part of the scientific culture. The distance between two cities cannot be given to the least micrometer : it has no meaning. The weight of an object has no use if it is given such as $123,45678910111213 g$. The least figures which can be given for sure is the milligram (here 6). Put on a balance, such an object will display fluctuating next figures. Already the figure $7 (0.1 mg$) will fluctuate at the least draft. To be sure of the next decimals you must put your balance in an insulated and closed room without nobody entering during the whole day, in order to avoid drafts. And even, the least dust grain falling from the air will change one or two of the following figures. So please do learn to use numbers with maybe $2, 3, 4$ or $5$ significant figures. But not with ten or twenty significant figures. It is not the way science works. So why looking for the newest published values which have one or two more significant figures ? The given weight $123.45678910111213 g$ is not better than $123.456 g$. It only gives an irrational feeling of superiority. That's all. It has no meaning for usual applications. Nobody will believe you if you say that you have calculated a weight of $123.4567891011121314 g$, or measured a temperature of $12.3456789 °C$.

I'd quibble a bit. Physicists try to measure many values such as the fundamental constants with great precision. However in chemistry 4 significant figures would be more than sufficient for 99.99% of the work, and it would take really precise lab work to get 4 significant figures.

@MaxW It is not always what is sufficient, but also available. Physicists  are often able to measure many things with high accuracy, or even predict them so ( like QED prediction of electron magnetic momentum with the accuracy of 11 digits ). OTOH, they are not so lucky with the gravitational constant. We chemists are often glad if we are able to get some data with 4 digits accuracy. E.g. the mean molar mass of elements with variable isotopic composition, or activities of ions in solutions.

@Poutnik - Agree totally, but Maurice said "It is not the way science works." Physicists know mass of each isotope with astonishing accuracy. The limiting factor for atomic masses of course is the variation in isotope fractions across the planet.

@Maxw Sure. E.g. $\ce{M_\ce{Zn}}=\pu{65.38(2) g/mol}$ . Once I have read a quote of a famous chemist in some textbook in a sense like: "Nothing is better sign of chemical ignorance but untamed precission of calculations." All is about known or estimated accuracy of the input data and constants and about rules of error propagation.

There's some misinformation and hyperbole in this post.  Those who understand science won't be misled, but my concern is those who don't might be:  (1)  "The least figures which can be given for sure is the milligram".  No, there are higher-precision enclosed balances.
(2) "To be sure of the next decimals you must put your balance in an insulated and closed room without nobody entering during the whole day, in order to avoid drafts."  No, you can certainly get beyond 1 mg without having to prevent anyone from entering the room "for a whole day" (!).


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Confusion on the Finkelstein reaction(AKA halogen exhange reaction)
So in the Finkelstein reaction, the iodide ion is supposed to replace another halogen(say chlorine) through SN2 mechanism.
But here, Acetone is used, which is a polar aprotic solvent. And in acetone, iodide ion is the weakest nucleophile. So how can it replace the chlorine?
Aren't we better off if we use water or ethanol(which are polar protic solvents) since (I-)
acts as the strongest nucleophile?

https://chemistry.stackexchange.com/questions/68081/why-isnt-a-polar-protic-solvent-used-in-the-finkelstein-reaction ... https://chemistry.stackexchange.com/questions/133984/why-is-finkelstein-reaction-not-conducted-using-naf ... https://chemistry.stackexchange.com/questions/55304/role-of-acetone-in-the-finkelstein-reaction .. https://chemistry.stackexchange.com/questions/6947/why-are-nacl-and-nabr-not-soluble-whereas-nai-is-soluble-in-acetone-in-the-fin

the driving force of Finkelstein is the insolubility of the product halogen salt (KCl for ex) so that it is removed from the reaction.

Iodide would be also be a worse nucleophile in a protic solvent.


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How exactly do Copper salts help in Sandmeyer Reaction (Halogenation)?
I have searched a quite few sources on mechanism for Sandmeyer's reaction, but couldn't find (and understand) how CuCl for example helps in replacement of N2 from Benzene diazonium chloride and gives Chlorobenzene? In Solomon-Fryhle-Snyder's regarding the mechanism its been given that "the mechanisms of these replacement reactions are not full undertstood", is this true and the reason most sources don't have mechanisms for Sandmeyers Halogenation reaction?

See: https://chemistry.stackexchange.com/questions/34948/why-use-potassium-iodide-instead-of-copper-iodide-in-sandmeyer-reaction


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19F NMR line spectra
I am trying to figure out the NMR spectra of F3P-BH3
I am solving this out using 2nI+1 rule.  IP = 1/2 and IB = 3/2
Hence for this compound,

(2nIP+1)(2nIB+1) = (2x1X1/2+1)(2X1X3/2+1) =
(2)(4) = 8 lines (quadraplet of doublet)

This is what I came up with, it would be great if you could correct me if I'm wrong. Trying to solve few problems before the exam =)

I think you have got the calculation right. However, I would also consider the coupling between F and H because coupling through 3 bonds is also usually visible. So, each of your 8 lines will be split into four by the H-atoms.
That will be the major part of the spectrum. You would also get weak signals with a different splitting pattern due to the other isotope of boron ($\ce{^10B, I=3}$, abundance=20%).


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Nomenclature of branched alkane
I have a compound named 7-(2,4-dimethyl)hexyl-3-ethyl-5,9,11-trimethyltridecane. I have a few questions:

why is it (2,4-dimethyl)hexyl and not (2,4-dimethylhexyl)? I asked this question because I saw a compound named 2,5-dimethyl-4-(2-methylpropyl)heptane, and not 2,5-dimethyl-4-(2-methyl)propylheptane.
why is the 7-(2,4-dimethyl)hexyl put before 3-ethyl? Shouldn't hexyl be put after ethyl, because e comes first than h?

Can you draw out the structure? I'm pretty sure re. point 1, you are correct and the parentheses should go round the whole thing. Re. point 2, the substituent's full name is dimethylhexyl, which starts with d, and d comes before e. Note that this is different from 5,9,11-trimethyl... where the substituent is methyl and the tri just indicates that there's three of it.

@orthocresol Thanks! Then what about 2,5-dimethyl-4-(2-methylpropyl)heptane? Should it be 2,5-dimethyl-4-(2-methylpropyl)heptane or 4-(2-methylpropyl)-2,5-dimethyl-heptane? Both the substituents start with m?

That's quite a different question; you should ask a new question.


text:
Confusion in barometer experiment

Now at A, there is pressure by mercury column which is nothing but it's weight. Then, at C there is pressure which is atmospheric pressure. Also, the density at A is more than at C. What they say according to the law is that the pressure at C and A are same. I am not able to understand why the pressures are same actually because of the difference I listed in my text. Also, the pressure at C is inside the barometer and at A is outside. So, how can they ever be same and if they are, what is the condition for it.
Pressure at A = Pressure at B +pgh. Now how is this possible? Pressure at A should be more than pressure at B since the height for pressure at is more. So, why is that? Acceding to me Z it should be Pressure at A + pgh= pressure at B.
Then, they also say that pgh is atmospheric pressure. Where they used denstiy of mercury. Now, This should be mercury pressure and not atmospheric right. Density for atmosphere is very low right

Said by other words, you do not believe that $3 + 2 = 3 + 2$

There are 2 $\ce{Hg}$ columns, sharing the same pressure at the bottom.
Both have the same height, so the pressure difference is the same.
Yet you have doubts the pressures at their tops are the same.

Pressure at A = Pressure at B +pgh

It should be $p_\mathrm{A} = p_\mathrm{B} + \rho gh$ where $\rho$ is the density of mercury.

The vertical gradient of the hydrostatic pressure in liquid of the density $\rho$ at gravitational accelerationg $g$ is:
$$\mathrm{d}p = - g \cdot \rho \cdot \mathrm{d}h$$
As we can consider $\rho$ and $g$ as constants, the difference of the pressures is then :
$$p_C - p_A = - g \cdot \rho \cdot \int_{h_A}^{h_C} {\mathrm{d}h}$$
$$h_C=h_A \implies p_C=p_A$$

If pressure at the same altitude within the connected liquid of constant density was different, there would be the liquid flow until it is equal.
The pressures at the same altitude would be slightly different, if e.g. mercury at both columns had different temperature and density.

Feedback to comments:
In fact, it does not matter if it is mercury or water, the important thing is the density is constant across the liquid. What other than atmospheric pressure would you expect at C ( regardless of the liquid below ) ? Your scenario description is not exactly clear. Now at A, there is pressure by mercury column which is nothing but it's weight.
If the density along the thought path from A down and then up to C is constant, then pressure at A and C are the same, if the liquid is in the rest. Any pressure difference would cause motion of the liquid.

Let assume the liquid is mercury with density $\rho$.
Let assume the atmospheric pressure at C is $p_0$.
Let assign to C the relative height $h=0$.
Then pressure anywhere below C ( with negative $h$ ) is
$$ p = p_0 - \rho \cdot g \cdot h \gt p_0$$
The same is anywhere below A. At A is the same pressure as at C, i.e. $p_0$.
Between A and B, at height $h$, is the pressure:
$$ p = p_0 - \rho \cdot g \cdot h \lt p_0,$$
but this time $h$ is positive.
B is then at the height
$$ h_\mathrm{B}= \frac {p_0}{\rho \cdot g}$$
with $p=0$
( if we neglect very low vapour pressure of mercury.)

Give me some recommendation of book , site pls so that I can improve my theory in this chapter till high school level.

That may be hard, as I did not study that from English books. A great site, written in the form of "cheat sheets", is http://hyperphysics.phy-astr.gsu.edu/hbase/index.html  Try searching for educational stuff, there is plenty on internet.

Ok.@Poutnik  sir . .

the moderators are very strict in physics stack exchange. They suspend if the question is of a homework but has a conceptual doubt in it.

Can I either talk with you in chat room on CSE regarding physics and chemistry questions.I can’t ask on PSE but if you could help me. I would be grateful.Even if you give me hints.I am more than happy. I don’t have a teacher with me . I do self study.

what do you think is missing in my question ? Moreover, what is quota site.

you can’t ask questions the way I do here.

Neither you can post pictures there on Quora.@Poutnik

You can upload it to SE, you can upload it anywhere. Note that comments under questions or answers are not intended for chatting, but  for improvement suggestions or asking for clarifications of Q or A.

Think about your questions before asking.

I will surely do that.

I will answer whatever I know in that question and then ask.@Poutnik . Thank you for being such a help

Your terminology makes seeing the correct picture harder.
Part of the problem is that the pressure at A is could be at any point in the column of mercury and that is not constant. The pressure in the column of mercury varies with height (depending on the weight of mercury above that point). When you fill the tube with mercury and invert it there is no pressure above the tube of mercury (neglecting the tiny value of mercury vapour). So all that can support the mercury in the tube is external pressure from the surrounding atmosphere acting on the external surface of the mercury.
The mercury in the tube will fill up to the point where the total weight of mercury produces just enough pressure at the bottom of the tube to balance the external atmospheric pressure. The density is irrelevant as liquids are essentially incompressible.
So the pressure at the top of the mercury column or B will be zero as there is nothing in the space above the mercury to exert any pressure. The pressure will rise at points down the mercury column until the point where it balances the external pressure on the external surface of the mercury. (if your intention was to define A as the point in the tube equivalent to the external level of mercury in the bowl, then pressure at this point is atmospheric but this is not true at other points in the column).
The point is that the pressures are equal and the densities are irrelevant. If you used a liquid other than mercury the same effect would be seen but the height of column would be different (water is about 14 time less dense than mercury so the total column would be ~14 times higher for the weight of water to balance the pressure of the atmosphere). This is the only point where density matters: it doesn't vary in the liquid column.


text:
Finding the atomic packing factor (APF) of Sodium Chloride and other FCC ionic compounds
For a piece of coursework I am doing, I need to calculate the atomic packing factor of some ionic compounds. I have had no formal teaching in this area, so what I know comes from information I have found on the internet.
I take that the atomic packing factor is given by:
$$
APF = \frac{V_\text{atoms}}{V_\text{unit cell}}
$$
Taking a unit cell of $\ce{NaCl}$, we get:
$$
APF = \frac{2\pi(r_\ce{Cl}^3 + r_\ce{Na}^3)}{3(r_\ce{Cl} + r_\ce{Na})^3}
$$
Is this right? I took the side length of the unit cell to be $r_\ce{Cl} + r_\ce{Na}$, and reasoned that 1/8 of each ion is actually "inside" the unit cell, so there is essentially 1/2 of a chloride ion and 1/2 of a sodium ion "within" the cell. $r_\ce{Cl}$ is the ionic radius of chloride, and $r_\ce{Na}$ is that of sodium.
Would this also hold for all alkali metal-halides with a FCC crystal structure? For instance, for a compound with anion $\ce{A}$ and cation $\ce{C}$:
$$
APF = \frac{2\pi(r_\ce{A}^3 + r_\ce{C}^3)}{3(r_\ce{A} + r_\ce{C})^3}
$$
Finally, is this a good measure of packing efficiency? As I mentioned, I have little experience in this topic, so I would be grateful if anyone knew of a better measure.

What you call a unit cell is actually 1/8 of it. (Not that it matters much, though.)

@IvanNeretin I wasn't sure, thanks for clarifying that. Does my formula look right otherwise? And will it hold for all alkali metal halides with a FCC structure?

Yeah, well, looks about right.

Coming to question 1, that is, is your calculation correct?
Quoting from Wikipedia,

In solid sodium chloride, each ion is surrounded by six ions of the opposite charge as expected on electrostatic grounds. The surrounding ions are located at the vertices of a regular octahedron. In the language of close-packing, the larger chloride ions (167 pm in size) are arranged in a cubic array whereas the smaller sodium ions (116 pm) fill all the cubic gaps (octahedral voids) between them. This same basic structure is found in many other compounds and is commonly known as the $\ce{NaCl}$ structure or rock salt crystal structure. It can be represented as a face-centered cubic (fcc) lattice with a two-atom basis or as two interpenetrating face centered cubic lattices. The first atom is located at each lattice point, and the second atom is located halfway between lattice points along the fcc unit cell edge.

Also refer this LibreTexts article.

What does this mean?
For starters, $\ce{NaCl}$ is an FCC structure, with $\ce{Na+}$ occupying the octahedral voids, and $\ce{Cl-}$ occupying the FCC lattice.
Edge length in FCC unit cell = $2\sqrt{2}r$.
You get unit cell volume from here.
There are 4 octahedral voids (OV) in a unit cell, and thus 4 $\ce{Na+}$ ions.
Take them as whole spheres, and find their volumes.
$V_{\ce{Na+}}={4 \times \frac{4 \pi {r_{\ce{Na+}}}^3}{3}}$
Number of chloride ions = 8 corners + 6 face centres = $8\times\frac{1}{8}+6\times \frac{1}{2} = 4$ (we are multiplying the various positions with their individual contributions.)
$V_{\ce{Cl-}}={4 \times \frac{4 \pi {r_{\ce{Cl-}}}^3}{3}}$
Now, volume occupied by the ions = $V_{\ce{ions}}=V_{\ce{Na+}}+V_{\ce{Cl-}}=\frac{16\pi}{3}\times ({{r_{\ce{Na+}}}^3}+{r_{\ce{Cl-}}}^3)$
Volume of unit cell = $V_{\text{uc}} = (2\sqrt{2}r)^3=16\sqrt{2}r^3$,
Assuming radius of the ions is the same (that's how I was taught, it's a simplification for the calculation),
The correct result is:
$$\text{Packing fraction}=\frac{V_{\ce{ions}}}{V_{\text{uc}}}=74\text{%}$$
The $74\text{%}$ number is standard for all FCC unit cells.
So you can refer to this for the correct calculation.

Question 2, will this hold true for all alkali metal halides with an FCC structure?
Of course! It holds true for all alkali metal halides with an FCC structure.
The only exception is $\ce{CsCl}$ which has a non-closed packed structure type, but is commonly mistaken to be body-centered cubic (BCC structure).

You can read this article to know more.

Citations:

Sodium Chloride, Wikipedia, The Free Encyclopedia, Wikimedia Foundation, 18 April 2004, en.wikipedia.org/wiki/Sodium_chloride#Solid_sodium_chloride
R. D. Shannon (1976). "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides". Acta Crystallogr A. 32 (5): 751–767. doi:10.1107/S0567739476001551.


text:
Corrosion rate of aluminium
In which conditions the corrosion rate of aluminium is faster?
a) acidic conditions
b) neutral
c) alkaline conditions
I know that corrosion can be in a and c cases but I dont know when it'll be faster. Also please describe why corrosion rate in one case is faster than in other one.

Not your fault, but this problem you were given is incompetently formulated and simply lacks details to be answered adequately. Well, at least they wrote aluminium, not aluminum.

In general, corrosion rate depends on too many factors. Which acid or base, which concentration, which temperature? Which other chemical species are present? Is there galvanic corrosion present? Is there a voltage applied, and which voltage? And so on.
How do you even you measure the "rate" of corrosion: depth per unit time, mass per unit time, other? Both may depend on surface geometry, defects, etc. If you measure as mass per unit time, that depends on how finely the material is divided, since corrosion occurs at the surface. More finely divided -> more surface. Depth per unit time may not even be uniform across the whole sample.
This applies not only to aluminum.
Assuming you keep all other factors the same, you still need to specify at least the type of acid or base, and their concentration. For example, aluminum won't dissolve in concentrated nitric acid.


text:
Why does charcoal not produce smoke when heated?
As I understand; the reason we have smoke in a fire is because the heat causes wood (or whatever material is burning) to dis-integrate, the disintegrated components react with oxygen to create more heat; causing a chain reaction.
Much of the smoke we see is the dis-integrated carbon in the air from incomplete combustion.
However, when we heat charcoal in a vacuum, there is no smoke. But if heat is what causes carbon to dis-integrate, then why does heating carbon (regardless of oxygen presence) not cause smoke?

Carbon doesn't disintegrate; wood does.

But then where does the carbon in the smoke come from when the wood burns?

From pyrolyzed wood and resulting vapours and gases, containing carbon. Charcoal as carbon is the result of wood pyrolysis. As all has already been done, there is nothing to make a smoke.

I see, can you confirm if this understanding is correct?: basically, when fire burns; all of the smoke comes from the components gassing off the wood, which separates into carbon from the heat. Some of the carbon reacts with oxygen to form light but not all of it. ---- However, the actual carbon in the wood only ever reacts with oxygen and therefore doesn't ever contribute to smoke.

In short, smoke is drops and very little pieces. In vacuum nothing can remove particles from a surface, so even if carbon would behave as it does in air, there is no convection. Instead what happen to C in these vacuum conditions is interesting and I don't know much.

You say that wood decomposes under heating and produces gases(1) which then burn. That's correct. It would be wrong, though, to assume that all combustible materials are like that. Indeed, many - arguably, most of them are. Wax, tar, and all kinds of plastic come to mind. But carbon is a different story.
See, carbon is very heat-resistant, about as much so as the most heat-resistant thing you know, only more. You heat it to $1000\;^\circ\rm C$, and nothing happens. Well, it may start burning provided there is air around, but short of that, it is nothing. You heat it a few hundred degrees more, it starts glowing red, then white, then steel melts, then you pass $2000\;^\circ\rm C$ and continue to push on, then everything else melts, by now your carbon glows so bright that you can't see a thing, but it is still intact.
Burning of carbon, unlike burning of many other materials, is a true heterophase reaction. You chip atoms off that piece one by one, hence no smoke is produced. Burning of wood or plastic is another story. Combustible gases are produced and fly away, and when some of them do not burn completely(2), they leave a solid residue (which, BTW, may or may not consist mostly of carbon). That's what we call smoke.

P.S. Thought I'd rather add two clarification notes:
(1) When heated, plastic turns into gases all but completely. Wood doesn't. Much of it remains in the form of charcoal, which is mostly carbon and burns like carbon, that is, producing no smoke.
(2) Incomplete combustion of carbon is also possible, and it results in carbon monoxide, which killed quite a number of people over the past few millenia. But it isn't smoke; it's a gas. You can't see it. You just see people dropping dead.
So it goes.

I think you missed the point that when something is being burned that products other than just carbon dioxide and water are created. Burning wood creates hundreds if not thousands of different compounds. Some of what burn completely other which do not. You often also get particulate matter when burning something. // Also if you heat something in a vacuum then there is no oxygen to promote burning. Rather what you are seeing in that circumstances is thermal decomposition and out-gassing.


text:
Steady state approximation rate equation for an E1cB reaction
Is anyone able to explain why species 'D' is not included in the steady state approximation derivation for this E1cB reaction?
The correct rate equation in the image has been derived first and then I have added in the line I am confused about below
Thanks!

Because it is the reaction solvent and is present in vast excess

The base is ethoxide in a solvent of ethanol. The concentration of ethanol is then essential constant and can be absorbed into rate constant $k_{-1}$.


text:
Quantum Espresso, phonon band dispersion
I am having a trouble calculating the phonon band dispersion of graphene at $K$-point using Quantum-Espresso, using normconserving+LDA and normconserving+PBE, ecutwfs=60, Ecutrho=400.0, forc_conv_thr=1.0d-6,
as you can see in the figure, the calculated acoustic and optical eigenvalues are ok at $\Gamma$ and $M$-points, however the calculated value deviates from what it should be at $K$-point:
TA and ZA branches are not overlapping at $K$-point,  and a strange bump can be seen when ZO approaches $K$-point,
I have checked the convergence for K-mesh, Ecutwfs, and Ecutrho,
I have also tried increasing the K and q mesh for scf-calculation,
using tr2_ph = 1E-14 and alpha_mix(niter)= .1 also does not improve the results,
()
Any help is appreciated,
Cheers

step 1, pw.x

&CONTROL
 calculation='scf'
 title='ml_gr'
 prefix='ml_gr'
 verbosity='high'
 restart_mode= 'from_scratch'
 nstep=100
 tprnfor=.true.
 outdir='./tmp'
 disk_io='default'
 pseudo_dir='../pp_NC_LDA/'  ! LDA+NC
 forc_conv_thr=1.0d-6,
/

&SYSTEM
ibrav=0,
nat = 2,
ntyp = 1,
ecutwfc = 60.0 ,
ecutrho = 400.0, ! 480
nosym=.true.,
occupations='smearing', smearing='cold', degauss=0.02

la2F = .true.,

/
&ELECTRONS
startingwfc='random'
diagonalization='cg'
conv_thr = 1.0e-12
mixing_beta = 0.2
electron_maxstep=200

ATOMIC_SPECIES
C  12.011   C.pz-vbc.UPF

CELL_PARAMETERS (angstrom)
   2.428479517  -0.000000605   0.000000000
  -1.214240282   2.103124989   0.000000000
   0.000000000   0.000000000  20.000000000

ATOMIC_POSITIONS (crystal)
C            -0.0002315717        0.0002344099        0.4996978700
C             0.6664267343        0.3335704066        0.4996978902

K_POINTS automatic
64 64  1 0 0 0

%--------------------
%      step 2 ,pw.x
%--------------------

same as above except

la2F = .false.,

K_POINTS automatic
32 32 1 0 0 0

%--------------------
%      step 3 , ph.x
%--------------------

 &inputph
  tr2_ph=1.0d-14,
  prefix='ml_gr',
  fildvscf='ml_gredv',
  amass(1)=12.011,
  outdir='./tmp',
  fildyn='ml_gr.dyn',
  electron_phonon='interpolated',
  !el_ph_sigma=0.005,
  !el_ph_nsigma=10,
  alpha_mix=0.1
  diagonalization='cg',
  trans=.true.,
  ldisp=.true.,
  nq1=4 , nq2=4, nq3=1
  ! nq1=8 , nq2=8, nq3=1   makes no difference
 /

%--------------------
%      step 4  , q2r.x
%--------------------

 &input
  zasr='simple',
  fildyn='ml_gr.dyn',
  flfrc='gr333.fc',
  la2F=.true.
 /

%--------------------
%      step 5 , matdyn.x
%--------------------

  &input
    asr='simple',
    amass(1)=12.011,
    flfrc='gr333.fc',
    flfrq='gr333.freq',
    la2F=.true.,
    dos=.false.
 /
58 ! number of q-points
0.          0.          0.
0.          0.02631579  0.
0.          0.05263158  0.
0.          0.07894737  0.
0.          0.10526316  0.
0.          0.13157895  0.
0.          0.15789474  0.
0.          0.18421053  0.
0.          0.21052632  0.
0.          0.23684211  0.
0.          0.26315789  0.
0.          0.28947368  0.
0.          0.31578947  0.
0.          0.34210526  0.
0.          0.36842105  0.
0.          0.39473684  0.
0.          0.42105263  0.
0.          0.44736842  0.
0.          0.47368421  0.
0.          0.5         0.
-0.01754386  0.50877193  0.
-0.03508772  0.51754386  0.
-0.05263158  0.52631579  0.
-0.07017544  0.53508772  0.
-0.0877193   0.54385965  0.
-0.10526316  0.55263158  0.
-0.12280702  0.56140351  0.
-0.14035088  0.57017544  0.
-0.15789474  0.57894737  0.
-0.1754386   0.5877193   0.
-0.19298246  0.59649123  0.
-0.21052632  0.60526316  0.
-0.22807018  0.61403509  0.
-0.24561404  0.62280702  0.
-0.26315789  0.63157895  0.
-0.28070175  0.64035088  0.
-0.29824561  0.64912281  0.
-0.31578947  0.65789474  0.
-0.33333333  0.66666667  0.
-0.31578947  0.63157895  0.
-0.29824561  0.59649123  0.
-0.28070175  0.56140351  0.
-0.26315789  0.52631579  0.
-0.24561404  0.49122807  0.
-0.22807018  0.45614035  0.
-0.21052632  0.42105263  0.
-0.19298246  0.38596491  0.
-0.1754386   0.35087719  0.
-0.15789474  0.31578947  0.
-0.14035088  0.28070175  0.
-0.12280702  0.24561404  0.
-0.10526316  0.21052632  0.
-0.0877193   0.1754386   0.
-0.07017544  0.14035088  0.
-0.05263158  0.10526316  0.
-0.03508772  0.07017544  0.
-0.01754386  0.03508772  0.
0.          0.          0.

This would probably better fit at [mattermodeling.se].

Adding q_in_cryst_coord=.true to the matdyn.x input file resolves the problem otherwise the code interprets the q-coordinates to be in Cartesian !


text:
The covalent bond in HCl and the Lewis model
Can the formation of the covalent bond in HCl be explained through the hybridization model?
We know for sure that it can be explained by the theory of molecular orbitals. In this way, it would be normal for the atomic orbital $s$ of the hydrogen to mix with one of the chlorine's atomic orbitals which has the same form or energy. Neither the symmetry of the form nor that of the energy can be preserved (totally), because the energy levels of the atomic orbitals of chlorine are much lower than that of the $s$ orbital of the hydrogen. Thus, the $s$ orbital of the hydrogen mixes with the $pz$ orbital of the chlorine and forms $2$ molecular orbitals (a bonding one and an antibonding one). The atomic orbitals $s$, $px$ and $py$ of the chlorine are nonbonding and remain at the initial energy.
According to the Lewis structure, between H and Cl we have a simple sigma bond and around the chlorine atom we are left with $3$ pairs of nonbonding electrons. The charges on H and Cl are null. With this in mind, we notice that $4$ hybrid orbitals would be needed to host the sigma bond and the $3$ pairs of nonbonding electrons of the chlorine, thus getting an $sp^3$ hybridization (the real geometry of the molecule is linear and the one in which are also involved the orbitals with pairs of nonbonding electrons is tetrahedral.
Considering the energy diffrence between the $s$ orbital of $H$ and the $s$ and $p$ orbitals of Cl (as we saw in the theory of molecular orbitals), is it possible to create hybrid orbitals in the HCl molecule?

Yes, you need 4 hybrid orbitals, but why should you assume that they must all be the same (hence sp3)? You could have sp2-hybridised chlorine (sp2 + sp2 + sp2 + p), which is still four orbitals, or sp-hybridised (sp + sp + p + p), or indeed completely unhybridised (s + p + p + p). Indeed, any combination of four atomic orbitals must give four hybrid orbitals, so knowing that we need four hybrid orbitals doesn't actually tell us anything about the resulting hybridisation.

@orthocresol Oh, I understand. I didn't know that we can split them i.e. that sp+sp+p+p is the same as sp3, but it totally makes sense. Thank you!


text:
How does strong acids conduct electricity in water?
How do strong acids conduct electricity in water?
I have just studied ionization of water and I'm confused about some points.
First: I know that water ionizes  in a very small amount ($10^{-14}$) so it doesn't conduct electricity and if we add strong acid (or base) to it , it becomes a good conductor of electricity. My question here is how it conduct electricity although if a strong acid is in water its $\ce{H+}$ ions concentration  in water would range from $10^{-1}$ to $10^{-6}$ (which is avery small amount) to keep $K_w$ constant.
I'm confused so if you understood me please explain and tell me if I have something wrong

Acids (and other electrolytes) in water solutions conduct electricity because ions. Ions come from dissociation. Water by itself doesn't dissociate very well. Acids (and other electrolytes) do. What's unclear about that?

@Eman, The "how" part of your query is pretty difficult and perhaps still not understood well. The fact that you should keep in mind is that the proton in water has the highest known conductivity, this is followed by hydroxide ion. So both H+ and OH- are strong conductor. Search the keyword Grotthuss mechanism.

Acidic solutions with $\ce{H+}$ concentrations around $10^{-5} - 10^{-6} M$ are not good conductors. They are better than water, but not that much.

@ivanneretin i am confused about how this small amount 10^-2 can conduct electricity in strong acid, while we say that sulphourous acid is a weak acid that doesnt conduct electricity  because its equilibrium constant equals 1.7*10^-2 tell me if I'm wrong

$10^{-2}$ is not a small amount. Then again, everything conducts electricity. Some things do it well, and some not so well, and some do it poorly, and some yet worse, and some are ten orders of magnitude worse than that, and there are levels upon levels upon levels even below that. What do you really mean when you say "doesn't conduct electricity"?

@ivanneretin Thanks. I know it, but I have misread it. I revoke my comment.

In water at 25 ºC $[H+]=[OH-]=10^{-7}M$, so a concentration of $[H+]=10^{-6}M$ is very low. With a strong acid like HCl you have to dissolve only around 35 micrograms in a liter of water to get this concentration.
Sea water on the other hand has a high conductivity and, in average, 35 grams per liter of salts dissolved in it. If you dissolve 35 grams of HCl en 1 liter of water you have $[H+]=1M$ and surely you can expect a high conductivity.
These are gross numbers, but I think they help to get the idea.
Conductivity of a solution depends on the free ions present on the mixture. With a strong acid you can get a high concentration of ions using a little quantity of acid. With a weak acid you need much more.


text:
Sodium Deoxycholate titration assay
Purchasing chemicals online often I see "Sodium Deoxycholate, ≥97% (titration)" offers. Is there any publicly available titration protocol I can use to confirm that statement in my very basic lab which does not have HPLC?

Probably not much of a help, but for what it's worth, Merck's spec sheet states "TITRATION (NT) HCLO4 0.1M".

@Alex For future reference: the addition of context will increase the value of your question for other readers of ChemSE.  In the present case, this could be / could have been where, or / and  when you see this.  For example «Purchasing chemicals, I often see their purity described in the pattern of e.g., "Sodium Deoxycholate, ≥97% (titration)."».  This may increase the likelihood for an answer, too.

Buttonwood, of course purchasing chemicals! What other kind of offers could I mean? Offers to spend a night on Tinder?
Buttonwood you are power-tripping

@Alex Buttonwood's suggestion is on-point and adds more clarity to your question. Please try to be more patient and don't bite the hand that feeds. Unfortunately, your question has been closed due to lack of clarity, but I think you can vote to re-open it after the edit—at this point I'm also interested in finding the answer:)

Alex I hope you find @Buttonwood and andselisk notes to be positive and helpful. For what it's worth, I too felt there wasn't enough context to expand on an answer.

Public (as in published, not as open access) are e.g., indirect assessments by titration calorimetry (e.g., 1995JPhysChem11742, or 2000Langmuir5267). Not a HPLC, but not this type of research is more likely seen in groups about enzyme binding, transition metal complexes, and physical chemistry.


text:
Why is the side chain of lysine more basic?
According to the images below, the NH3+ on the side chain of lysine is less acidic than the NH3+ closer to the COO-. Why is this? Wouldn't the NH3+ have an intramolecular H-bond with a negatively charged oxygen on the carboxylic acid group, stabilizing it?
One possible explanation I thought of is the carbons on the side chain donating electron density through hyperconjugation, but this seems unlikely because both amines are primary.
My instructor said that the inductive effect of the carboxylic acid makes the NH3+ close to the oxygens unstable, but in solution it would be a negatively charged carboxyl group, making induction also seem unlikely.
Any help would be appreciated.

What is the pKa for Propionic acid? How does that compare to lysine? Why the difference?

The alpha ammonium group can't make hydrogen bonds with the carboxylate group - they are pointing in the wrong direction and the connection is too tight.

@MaxW The pKa on Wikipedia is 4.88, which is less acidic than the COOH on lysine. Is this because of the alpha NH3+ stabilizing the conjugate base? What does this have to do with the acidity of the two amines?


text:
Balancing complex redox reaction of gold treated with sodium cyanide
How would you balance this reaction:
$$\ce{Au(s) + NaCN(aq) + O2(g) + H2O(l)-> Na[Au(CN)2](aq) + NaOH(aq)}?$$
Here’s what I’m came up with. First we can balance any atoms besides hydrogen and oxygen in the reaction (like $\ce{Na}$ and $\ce{CN}):$
$$\ce{Au(s) + 2 NaCN(aq) + O2(g) + H2O(l) -> Na[Au(CN)2](aq) + NaOH(aq)}$$
Then we can write out the half reactions:
$$
\begin{align}
\text{Anode:} &\quad &\ce{Au(s) &-> Au+} \\
\text{Cathode:} &\quad &\ce{O2(g) &-> H2O}
\end{align}
$$
Since the oxygen is going from an oxidation state of $0$ to $-2$ in the cathode half reaction, I figured you can rewrite $\ce{OH-}$ as $\ce{H2O}$ on the right hand side of the equation.
For the half reactions, I figured we can eliminate the ions whose oxidation number don’t change but I’m not sure whether you should include $\ce{NaOH}$ in the anode or cathode half reaction.

In principle, the oxidation of gold is the half-reaction $$\ce{Au ->  Au^+ + e^-}$$But as this oxidation is done in a cyanide medium, $\ce{Au^+}$ is included in a complex ion : $$\ce{Au  +  2 CN^- -> Au(CN)_2^- + e^-}$$ The second half-equation is $$\ce{2H2O + O2 + 4 e^- -> 4 OH^-}$$ So the final equation describing the dissolution of gold in an aerated cyanide solution is the sum of the two previous equations, after multiplying the first by 4 : $$\ce{4 Au + 8 CN^- + 2 H2O + O2 -> 4 Au(CN)2^- + 4 OH^-}$$ And if you prefer equations with neutral substances, you may write it this way $$\ce{4 Au + 8 NaCN + 2 H2O + O2 -> 4 NaAu(CN)2 + 4 NaOH}$$


text:
What is the 3D shape of an ethane molecule?
I am currently teaching a course on groups, and want to give my students "real life" examples of symmetry groups to think about. The ethane molecule would be one such.  But, the on-line sources don't all agree on the exact shape. I'm fairly sure ethane looks like either this

or this.

I oriented the molecules in such a way that the carbon atoms are on top of each other. The difference between the two choices is how the groups of three hydrogen atoms relate to each other. In the latter alternative the molecule is its own mirror image with respect to the plane bisecting the line segment joining the carbon atoms. Meaning that the hydrogen atoms also pair up in a way that they are on top of each other. However, in the former version this is not the case, and you would need to rotate one triple of hydrogen atoms by 60 degrees to move them on top of each other.
To turn this into a homework problem I would like to know which is correct?
Wikipedia suggests that the one on the top is correct. However, the picture below also appears in some places. Like here.
My theory: Whatever little I know about physics/chemistry suggests to me that the top one is more likely. The reason being that the similarly charged hydrogen atoms repel each other (possibly this is only a dipole or a higher order effect?). Therefore it stands to reason that one triple of hydrogen atoms is aligned in a way that they use the "in-between" angles.
Nitpick: The "correct" answer probably is that neither of these two images is quite right. You see, I generated them using the tetrahedron (=methane) angle of 109.5 degrees everywhere (simpler for me to do it that way). But Wikipedia says that the angle between the carbon-carbon bond and a carbon-hydrogen bond is slightly larger, 111.17 degrees. So the hydrogen atoms at the top should be pushed a teensy bit further up and similarly the ones at the bottom a little further down. A bigger similar effect is that in the actual molecule the distance between the carbon atoms is larger than the carbon-hydrogen distance. Neither of these has an effect on the collection of symmetries the molecule has, but the answer to my main question does have an impact. That's why I am asking.
If somebody is inclined to elaborate on the reason(s) why we don't have that methane angle everywhere, I will be educated further :-)

Sorry about not knowing which tags would be most appropriate. Or whether this question is a good fit for Chemistry.SE at all :-(

This question about the confirmations of butane might be of interest to you: https://chemistry.stackexchange.com/q/11015/4945

@Martin-マーチン  Conformations :-)

I'm sorry about the low level of this question. When I was in high school (40 years ago) I read a book (in Finnish) that could be called Organic chemistry 101+102. It did explain the existence of different conformations in more complex molecules such as cyclohexane. I was blissfully unaware of this phenomenon also appearing in such a simple molecule as ethane.

And I'm aware of the fact that the actual groups of rotational symmetries of the two conformations are isomorphic. In a sense the same group acting on different sets of points of the 3D space.

You need to know the terms "staggered" and "eclipsed", because that's what we call these two conformations. Now to the point: the real ethane molecule rotates and changes shape all the time, so both geometries are possible, albeit the top one is more likely, because it is the energy minimum. To make the molecule stop in the staggered conformation, you have to freeze it to nearly absolute zero, or maybe put some bulky groups in place of hydrogens.
This, BTW, is the reason why I dislike to use ethane molecule as an example when I teach space groups to kids. Better use something rigid.
So it goes.

Thank you. Particularly for the last comment of ethane not being rigid!

Plus 1 especially for the final advice.


text:
Reaction of SeCl4 with alkene and possible consecutive reaction?
There is a reaction that confused me a bit:

According to my research on $\ce{SeCl4}$, it can be used as a dichlorinating agent for alkenes. However, I am totally confused about the mechanism. I've read about several mechanisms using $\ce{PhSe-SePh}$ as a precatalyst, but in this reaction, there is no other information, apart from the reagent $\ce{SeCl4}$.
If I assume that there is a dichlorination with $\ce{SeCl4}$, I think, the syn-specific dichlorination happens on "top" as the methylene group is sterically less demanding.

Now, if this happens, could there be a subsequent $\mathrm{S_N2}$ reaction with the $\ce{OH}$-group? It could form a 5-membered cycle through a back-side attack on the carbon with the chlorine substituent.

Can you confirm or correct my thoughts about this reaction? Thank you a lot in advance!

I did a brief literature search, which turned up no hits for your particular reaction; basically, it'd be hard to conclusively say what would happen.

We humans are still at best mediocre at predicting the outcome of chemical reactions. Even computers are at best a little bit better. If there is no literature on this compound (and Orthocresol commented that he found none), no experiment has likely been performed and we cannot be certain.
You could find a computational chemist of your choice with free time and the appropriate software and let them simulate this reaction. This would allow a slightly better prediction as it would answer the question ‘is there a reasonable reaction pathway that this compound could follow?’ However, even when simulating ‘obvious’ reactions we chemists might overlook problems with a simple-looking pathway that make this pathway non-accessible in the first place. Thus, even if the simulation looks promising, the only true answer is to go into the lab and try it out.
On paper, your scheme seems promising. You may require additional base (maybe a weak base like carbonate will do the trick, maybe you need selective oxygen deprotonation e.g. by sodium hydride) or it may happen during the chlorination reaction.
There may be other side reactions. Without knowing whether it is applicable in your case, I would attempt to reseach potential non-classical cations and how they might interfer, in case this reaction would follow a more $\mathrm{S_N1}$-style pathway.
In addition, I wonder if an elimination reaction to a chloro-norbonene might be a competing side-reaction.
I don’t think that a nucleophilic substitution on the other chloride would pose much of a problem as five-membered rings are typically kinetically preferred over six-membered rings.

Well, I found a reference for the reaction between $\ce{SeCl4}$ and alkenes (Ref.1). The authors have discussed the preparation of organic selenides and their dihalides by addition of selenium halides to an unsaturated linkage. They were interested in the reaction because it produces compounds having reactive halogen on the $\beta$-carbon atom. Because I don't have access to J. Org. Chem. at home (working at home due to COVID-19), I didn't see the whole picture of the reaction. Yet, by reading the first page available in the journal website, I assume that, although it is possible to have dihalide products, the major product might be the helo-selenide (similar to a chloro-hydrin) by electrophilic addition to an alkene starting compound.
My assumption was supported by the another article, which is in German! (Ref.2) The abstract reads:

Selentetrachlorid wirkt bei vollständigem Feuchtigkeitsausschluß auf organische Substanzen chlorierend ein. Bei Anwesenheit von Spuren Feuchtigkeit werden selenorganische Verbindungen gebildet, wahrscheinlich durch die Entstehung von Selenoxychlorid aus Selentetrachlorid und Wasser. Aus salz‐ bzw. bromwasserstoffsaurer Selendioxyd‐Lösung und einer organischen Komponente kann man leicht und sauber selenorganische Verbindungen in wäßrigem Medium herstellen. Schwefeldioxyd‐ bzw. Tellurdioxyd‐Lösungen geben unter gleichen Bedingungen keine schwefel‐ bzw. tellurorganische Verbindungen. Ferner findet keine Bildung von selenorganischen Verbindungen statt, wenn man $\ce{SeO2}$ in anderen als den oben genannten Säuren löst. Tellurtetrachlorid bildet mit organischen Substanzen nur unter Feuchtigkeitsausschluß tellurorganische Verbindungen.

The Google translation: With complete exclusion of moisture, selenium tetrachloride has a chlorinating effect on organic substances. In the presence of traces of moisture, organic selenium compounds are formed, probably due to the formation of selenium oxychloride from selenium tetrachloride and water. From hydrochloric or hydrobromic selenium dioxide solution and an organic component, organic selenium compounds can be easily and cleanly prepared in an aqueous medium. Sulfur dioxide or tellurium dioxide solutions do not produce any organic sulfur or tellurium compounds under the same conditions. Furthermore, there is no formation of organic selenium compounds when $\ce{SeO2}$ dissolves in acids other than those mentioned above. Tellurium tetrachloride forms organic tellurium compounds with organic substances only when moisture is excluded.

The translation explains clearly why both products (di hal compound and halo-selenide compound) are possible. Funk and Papenroth have later described the dissolution of primly saturated aliphatic hydrocarbons with selenium dioxide and hydrochloric or hydrobromic acid in an aqueous medium, which is believed to be an oxidation condition (Ref.3).
Further, if the mechanism of the reaction of $\ce{SeX4}$ and an alkene follow the same mechanism as the reaction of $\ce{SeX2}$ and an alkene (Ref.4 and 5), then the intra-molecular cyclization with an appropriate group is possible, even if it is a five-membered ring (e.g., Ref.5 and 6). However, cyclic compound is not a halo compound, but a selenide.

It is note worthy that the products similar to major product from Ref.1 (that's what I believe) have been achieved by the reaction of $\ce{SeX2}$ ($\ce{X = Br, Cl}$) and alkenes (e.g., Ref.4 and 5). An example is depicted below (From Ref.4):

References:

Reed F. Riley, J. Flato, D. Bengels, "Addition of Selenium and Sulfur Tetrachlorides to Alkenes and Alkynes," J. Org. Chem. 1962, 27(7), 2651–2653 (DOI: https://doi.org/10.1021/jo01054a513).
H. Funk,  W. Weiss, "Über die Umsetzungen der Tetrachloride des Selens und Tellurs mit organischen Stoffen und Über ein einfaches Verfahren zur Darstellung selenorganischer Verbindungen in wäßrigem Medium (The Google translation: About the reactions of the tetrachlorides of selenium and tellurium with organic substances and about a simple process for the preparation of organic selenium compounds in an aqueous medium)," Journal für Praktische Chemie 1954, 1(1-2), 33-40 (DOI: https://doi.org/10.1002/prac.19540010104).
Herbert Funk, Wolfgang Papenroth, "Darstellung Selenorganischer Verbindungen in wäßrigem Medium (The Google translation: Representation of organic selenium compounds in an aqueous medium)," Journal für Praktische Chemie 1959, 8(5-6), 256-263 (DOI: https://doi.org/10.1002/prac.19590080502).
Vladimir A. Potapov, Maxim V. Musalov, Evgeny O. Kurkutov, Vladimir A. Yakimov, Alfiya G. Khabibulina, Maria V. Musalova, Svetlana V. Amosova, Tatyana N. Borodina, Alexander I. Albanov, "Remarkable Alkene-to-Alkene and Alkene-to-Alkyne Transfer Reactions of Selenium Dibromide and $\ce{PhSeBr}$. Stereoselective Addition of Selenium Dihalides to Cycloalkenes," Molecules 2020, 25(1), 194 (17 pages) (DOI: https://doi.org/10.3390/molecules25010194).
Maxim V. Musalov, Vladimir A. Potapov, Evgeny O. Kurkutov, Maria V. Musalova, Alfiya G. Khabibulina, Svetlana V. Amosova, "Regio-selective syntheses of bis(2-haloalkyl) selenides and dihalo[bis(2-haloalkyl)]-$\lambda^4$-selanes from selenium dihalides and 1-alkenes and the methoxyselenenylation reaction," ARKIVOC 2017, (iii), 365-376 (DOI: https://doi.org/10.24820/ark.5550190.p010.351).
Scott E. Denmark, Michael G. Edwards, "On the Mechanism of the Selenolactonization Reaction with Selenenyl Halides," J. Org. Chem. 2006, 71(19), 7293–7306 (DOI: https://doi.org/10.1021/jo0610457).


text:
How to make foggy and cloudy plastic fully transparent?
I used sandpaper of different grits to sand the surface of plastic. The surface is smooth but it looks dull after sanding. I read that one could use toothpaste to remove the foggy look but it did not work. I also tried vinegar with baking soda. When the plastic was inside the vinegar+baking soda bath, it looked transparent and shinny. However, when I took it out to wash it with water to clean the smell, it looked shinny and transparent when being washed but it became foggy and cloudy again after washing. I put it back to vinegar+baking soda bath. It became transparent again but when I used a clean microfiber cloth to clean it, it became foggy again. Looks like some kind of thin film on it. Putting it in vinegar+backing soda bath for about 8 hours did not help. What caused the foggy/cloudy look? Would you please tell how to I make the plastic fully transparent?

It is still porous and therfore diffuses the light. When wet, the solution fills gaps and roughness. With somewhat matching refraction indexes, the transparent looking is recovered. I am afraid it will stay opaque. Perhaps grinding with special creams and powders of very fine size might help. But those are expensive niche products. Likely new foils are much cheaper.

Thanks. Do you have any good recommendation for special creams and powders? I tried "Flitz Green Metal, Plastic and Fiberglass Polish" but it made things even cloudier and dull looking. What sandpaper grade do you recommend to use to start sanding? I started from 120 grits. I wonder if this created more porous and made it more difficult to achieve good transparency.

Which plastic? There are many and their properties are very different. As are the reasons why they appear cloudy.

See also Matt Black comment below. Also, it is possible that depending on the plastic mechanical properties you can't recover a smooth surface. In the sense that no matter how fine you go, the coming out particle are always such. To give you an exaggerate picture: you can polish quartz but not a polystyrene foam block.

Sorry, I have no idea what kind of plastic is used to make those boxes bought in store. I am trying to clean up scratched surfaces of transparent boxes to make them look brand new again.  I read that one could use sandpapers to remove the scratches but they are still there.  Does that mean I need to use some kind of compound or wax to fill in the scratches while sanding to make the scratches disappear. What kind of compound or wax do you recommend?

After doing the sanding, hit it with a couple coats of clear high gloss spray paint that has UV protection. Works like a charm.

The tiny scratches on the plastic surface reflect and diffract light in different directions, so edges of objects behind the plastic become distorted, and the clear areas of the plastic are less clear.
There are several ways to make the plastic more transparent.

Use a finer grit sandpaper - really really fine: 1500 grit or finer, if you can get it. Plastic abrasive sponges are more gentle and can be used wet, so may create smaller, less deep scratches.

Use toothpaste to "sand" the plastic in the first place. You have to get the paste to be pasty thick, not watery thin. It will reduce scratches, but if you don't put deep scratches on in the first place, it will be easier to shine up a plastic surface. Toothpaste will reduce the shininess of a perfect surface, but usually will improve an ordinary surface. There are commercial pastes for cleaning and shining stovetops that are more abrasive, and you can make your own with sodium bicarbonate (no vinegar) - but you have to give it some energy, some elbow grease, and it takes a while. You could use a buffer wheel on a power drill.

Filling in the tiny scratches with a liquid helps even out the surface, but water evaporates and the effect does not last. An oil (e.g., silicone) or petroleum jelly (very thin layer) might last long enough for you - days, maybe weeks.

Depending on the plastic film, you might be able to heat it slightly to allow the scratches to deform slightly and become slightly wavy instead of sharp valleys and projections. This will require a lot of experimentation to get enough heating without distorting the plastic beyond your requirements. Hot air or hot smooth metal, or perhaps hot water or steam might work.

The least likely method to work is using a solvent to smooth out the surface. You might get some immediate effect that looks like an improvement, but any solvent that dissolves the very top surface will also diffuse into the plastic and when it evaporates, you will have a plastic that is disturbed even deeper, and will be foggier. Working this approach further suggests that you could use what would be better described as a plasticizer, a high molecular weight liquid that is compatible with the plastic, but does not migrate (out, usually, but in here), buff the surface and quickly clean the plastic to remove all the solvent/plasticizer. Now, since you probably don't have access to different kinds of plasticizers, you could try some vegetable oil plus a little of a more powerful solvent, like a few percent of alcohol, or mineral spirits, or even acetone, and you could add some sodium bicarbonate for a little abrasion, and use a buffer wheel.

Thanks James. 1. I sanded from 120 to 5000 grits. Is it better to start from 1500 grit? 2. Do you mean sand with paste first and then 1500 grit and up sandpapers? Any recommendation on which paste to use? I used Sensodyne Pro-Email. 3. When washing, at what temperature should the water be? I wonder if too cold or too hot would affect the transparency.

The problem is that the plastic might be cloudy because it is porous not just because the surface is rough. This might explain why it isn't cloudy when immersed in some liquids as they "fill in" the holes with something other than air, reducing the scattering.

@matt_black: Good thought. If the plastic is cloudy to begin with, polishing might be useless. sunflower is asking specific questions, but providing no information to work with! What kind of plastic are you working with, what dimensions, for what purpose, what have you tried, and WHY?

Use a headlight restoration kit or if you feel confident that you can use a clear gloss spray paint well this will also restore plastic clarity. You've done all of the prep that would normally be taken to restore a headlight or plastic pieces on a car but you need the final coat.


text:
Which enol form is more stable?
I came across an example today while I was studying tautomerism.

It is an example of tautomerism exhibited by acetylactone/1,3-diketone; given under the keto-enol systems under 1-3 migration/triad systems. (G.R.B. Organic Chemistry - O.P. Tandon 4.7.2(i) pg.122)

Soon I wondered - Why out of all the possible enols only this particular one - as shown in the example,
is the most stable form? ( 76% of the total content is only that particular enol, while the total enol content is 80.4%
testifies that it is much stable than the other possible enols)
As per my knowledge, the stability of enol increases in structures exhibiting:

Aromaticity
Intramolecular Hydrogen Bonding
Conjugation
Hyperconjugation

Where, Aromaticity stabilizes the most and hyperconjugation relatively the least with others in between just as they are arranged. (although I am not sure about it because it was just mentioned by my Chemistry teacher and I could not find sources mentioning it in such a hierarchy - so any edit suggestion in the hierarchy is welcome)
I drew all enol structures possible by 1,3-hydrogen migration:

Legend for the workings - 3C,5C,6C represent the number of atoms involved in conjugation while the numbers with alpha after them represent the number of alpha hydrogen present to stabilize a double bond.
(Here it is without any workings/markups)

Now, I'll be stating my assumptions about the stability of each of the 5 structures that I've drawn as per my knowledge.

Has conjugation extending only to 3 atoms & only 2 alpha hydrogens for hyperconjugation.
Has conjugation extending only to 5 atoms & only 3 alpha hydrogens for hyperconjugation.
Has conjugation extending only to 6 atoms & only 4 alpha hydrogens for hyperconjugation.
Consecutive double bonds are often unstable so I decided not to bother about this, anyway it is least conjugated although has 6 alpha hydrogen hyperconjugation shared by two double bonds.
6 atom bad conjugation since it is not spread throughout the chain & 2 alpha hydrogens.

So with all this, I concluded that option 3 is more stable but the book and the teacher said it is option 2.
I asked this question to my teacher about why option 3 cannot be counted in to which I got a rather unconvincing answer - the double bond is usually stable if it is placed more towards the center of a chain. Double bonds at the ends are usually unstable.
Having already accounted for hyperconjugation this is such an unconvincing answer!
It would be really helpful if someones can tell me the hierarchy of various stabilizing factors that make the enol more stable and how to determine which structure is more stable.

Firstly, and most importantly, the stability of certain tautomeric forms is very dependent on the solvent. Secondly, counting alpha hydrogens and carbons involved in conjugation is a very, very rough way of guessing. Don't forget about electronegativity and other effects. Third, (intramolecular) hydrogen bonding is usually the most important factor. Everything else pales in comparison. Ketone forms are usually more stable that enol forms (0th order approximation).

Structure 2 is a vinylogous carboxylic acid, which is supported by its 1H NMR spectrum. https://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi

Is there a source where I could read more about this and understand how & how much these stabilizing effects have an influence on the structures? I'm in 12th grade so NMR spectroscopy is like alien to me :(

related: https://chemistry.stackexchange.com/q/18904/102629


text:
Does urine make concrete hygroscopic?
TLDR: Does urine make concrete hygroscopic enough to absorb moisture from the air so much that a spot of humidity is visible? Are uric acid crystals moisture-absorbent? (Urea apparently is according to Wikipedia)
Here is the story:
In my parents' house, there is a mystery that we were unable to solve.
During the renovation, certain spots on the cement floor appeared wet (used to be covered by glued carpet). Even worse, some spots appeared to evolve and sometimes grow.
We assumed the worst. Since this was the ground floor (there is no basement) we thought groundwater might be infiltrating the house. One particularly intriguing spot was near the hinge side of an interior door. The door is in a wall of cored brick next to an interior concrete pillar. The spot on the floor was next to this pillar.
Several investigations have been made:

A dig outside the house to check the depth at which water could be found. It was deep enough.
Checking if there is hydro-isolation on the foundation. There was.
Checking on the outside of the base of the house for humidity. It was dry.
Checking the heating pipes buried in the floor. They were nowhere near and eventually the in-floor pipes were abandoned.

All of the people who investigated (engineers, constructors) were mystified as the spot kept evolving even though we properly ventilated the area for a significant time.
Even more puzzling, the concrete pillar was dry.
Eventually, believing there might be a spring under the house (and that the pillar was ok due to the hydro-isolation on the foundation) we broke the first layers of the cement floor.
Underneath the first layer, there was a layer of extruded polystyrene thermal insulation. Underneath that, a hydro-isolation layer. And underneath that the concrete slab of the floor. All were dry.
At this point, all other people just gave up and some even suggested we call an exorcist.
However, this meant a return to one of the first theories about these spots.
In the house used to live several cats and unfortunately, there have been plenty of times when some of them peed on the carpet floor. The spot near the pillar was one of the preferred places. This was one of the reasons for completely removing the glued carpet. This theory was initially rejected because the house was ventilated for more than 2 months with the carpet removed and the cats long gone. During this time and more the spots evolved and grew. We believed that since the spot grew noticeably (we traced the edge) and since it didn't smell it could not be because of the cats and it had to be from below (of above, but the roof and the ceiling were ok so we eliminated that possibility).
Now with layers below also dry, the only possible source of humidity left was the air.
So, does (cat) urine make concrete hygroscopic enough to produce a visible wet spot?
I have searched Google and Wikipedia about the moisture absorbant properties of uric acid but I have not found anything relevant. According to Wikipedia, urea is hygroscopic and highly soluble in water. I decided to ask this here since it is about the physical properties of a chemical compound.
The closest I have found is this:
https://pubs.acs.org/doi/10.1021/ja00836a027
What makes these particles hygroscopic?
Salt + Sodium Bicarbonate releasing water in carpet

Urine is not only urea and water, but there are some colored compounds present, too, e.g. from bile which sometimes is recorded as billirubin. Concrete is porous, so by capillary action water may ascend and travel; water may partially evaporate (thus increasing the concentration of the dyed compounds) till it is visible to the eye.   If aiming for repair, perhaps SE DIY/homeimprovement (https://diy.stackexchange.com/) is more suitable, then ChemSE.

Very interesting question - and thorough. But I would like to get proof that the spot is due to moisture. #1 Does the spot grow and shrink dependent on the relative humidity at that position? #2 If the spot is misted with a water sprayer, is it sharply defined? #3 If the spot is gently warmed with a space heater, does it disappear? #4 What color is the concrete? Very light (low iron), medium gray (medium iron), or dark, or brown?
If multiple cat pees have been located there, it's possible that attempts at cleaning or deodorizing have also been applied to the carpet, and have soaked thru and into the concrete. At this point, so much could have happened to the cement compounds that you and your parents should consider yourself fortunate that the deviation from perfection is only visual and not oderiferous! I have a spot on an outside patio that is dribbled upon from an aluminum gutter, only occasionally, when it rains (the end cap on the gutter has a tiny leak). The spot is about 4 inches in diameter, and is only visible at certain times (usually high humidity, like when rain has almost completely evaporated). Without a microscopic examination, I can't see anything unusual, but I suspect that some of the $Ca(OH)_2$ in the concrete has been dissolved, or perhaps carbonated more, or both, and the water in the pores at that spot is slower to evaporate.
If the discoloration is obnoxious, you might try painting (or dusting) that area with a non-paint paste of calcium carbonate in water. It would be harmless to the concrete, but provide some lightening. But then the area might be too white...

Concrete exposed to air has tendency to catch humidity, regardless of additional exposure (e.g., https://www.youtube.com/watch?v=17wnf7ehcZw).

The Youtube video is woeful. I learned that concrete is acidic and corrosive to steel, and a battleship in water is supported  indefinitely, whereas a battleship placed on concrete (dry land) will fall down ("pancake") because its sides are not supported. I do believe concrete exposed to water will be somewhat wet, but I did not learn this from the video.


text:
What is the correct numbering & order for this molecule?

I'm having a hard time understanding why the numbering for this molecule is: 4-bromo-1-tert-butyl-2-methylcycloheptane, instead of: 1-bromo-4-tert-butyl-3methylcycloheptane. Is it just because the overall numbers are lower? I just figured the numbering out by going in alphabetical order, which I now know is wrong. But is that the reason the first numbering is correct?

Note that the parentheses are not needed here.

The parent hydrocarbon chain for this molecule will be the 7 membered ring.

2.There are no functional groups with suffixes.
3.A halogen substituent(Bromine) and two side chains are present. ( methyl and t-butyl or (1,1-dimethylethyl) )
Having identified all these we can proceed to numbering the parent hydrocarbon chain.
After numbering the Parent hydrocarbon chain in all possible ways(ideally) we choose that numbering that satisfies the following criteria:
Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence.
1.Has the lowest-numbered locant (or locants) for the suffix functional group. Locants are the numbers on the carbons to which the substituent is directly attached.
2.Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).
3.Has the lowest-numbered locants for prefixes.
In the above molecule since there are no suffix functional groups or multiple bonds we can apply criterion 3.
After numbering the parent chain all possible sets of locants are written down in the ascending order.(in this case)
eg: If the set of locants are (2,4,3) and (1,4,3) we write them as (2,3,4) and (1,3,4)
Then choose the one with the lowest possible locants. First compare the first locant, choose the ones with the lowest locant, among these sets of locants compare the second locant, choose the ones with the lowest locant and so on...
eg: If the sets of locants are (1,3,8) and (1,4,5) we choose (1,3,8)
It is only when we get two different numbering for the same compound following all the above mentioned rules that we should choose the set of locants which assign the lowest locant to the substituent or side chain on the basis of alphabetical precedence.


text:
Relationship between molecularity and the reaction order
Should the molecularity be equal or less than the reaction order?
Consider the reaction:
$$\ce{H2(g) + I2(g) → 2 HI(g)}$$
If we consider the mechanism of this multi-step reaction, the molecularity is 3 (termolecular reaction):
$$
\begin{align}
\ce{I2(g) &<=> 2I(g)} &\quad&\text{(equilibrium reaction)} \\
\ce{2I(g) + H2(g) &-> 2HI(g)} &\quad&\text{(rate-determining step)}
\end{align}
$$
The rate equation for the reaction is
$$\mathrm{rate} = k[\ce{H2(g)}][\ce{I2(g)}].$$
The order of the reaction is 2 here. The molecularity is greater than the reaction order. Is it possible?
Or is the actual mechanism for the above reaction is different?

For eventual writing and formatting of chemical/mathematical formulas or equations, see  MathJax usage guide   It is not mandatory, but highly recommended and it gives huge advantage for writers and readers.

It is not $2 \ce{ I}$ that reacts with $\ce{H2}$, but plain $\ce{I2}$. The collision  $\ce{H2 + I2}$ produces an intermediate excited complex $\ce{H_2I_2}$ which can get decomposed into $2 \ce{HI}$ or back to $\ce{H_2 + I_2}$ Here the order an the molecularity are both 2.

@Maurice Are you sure ? As synthesis of HBr has more complicated mechanism with non integer reaction order.

@Poutnik. I know that the law governing the synthesis of $\ce{HBr}$ is rather complicated, and has no order. I also know that the synthesis of $\ce{HCl}$ is a chain mechanism. So all halogens behave differently when reacting with $\ce{H2}$. But the existence of a transition state$\ce{H2I2}$ has been developed and proved by Eyring, which has shown that the rate constant of $\ce{HI}$ synthesis is $\ce{2.0·10^9 √T e^{- 42500/RT}}$, and that the transition state $\ce{H2I2}^*$ is decomposed with a vibration constant equal to $\ce{210 cm^{-1}}$

Molecularity applies to single elementary reaction steps and is seldom higher than 2.
The reaction order is the "effective molecularity" of the whole set of linked reactions, if it had been a single step reaction. It need not to be integer and has just indirect relation to molecularities of elementary steps.
The reaction order is either determined experimentally, either derived from the reaction schema via determined conditions for steady states of intermediate products.

Does an elementary reaction $\ce{A->B}$ remains elementary for $\ce{2A->2B}$ ? If no, then the order (which is equal to molecularity) will not change…

@Apurvium If there is mechanism of collision 2 A, forming 2 B, then it is elementary. OTOH, A->B would not be.

Sorry, I did not get it. It is given that A forming B is elementary. On that basis I was asking…

@Apurvium Why then 2A -> 2B? Just a formal multiple for matching a bigger reaction schema, or for state function change evaluation? If so, then obviously the elementary reaction, molecularity and order remain, as the multiple is just formal.

Reaction rate and rate constant changes as we manipulate stoichiometry, hence the confusion…

@Apurvium They do while the above do not.


text:
Distance of subshells from nucleus
Which subshell is more closer to the nucleus.4f or 6s?
6s gets filled first so does it have to be more closer to the nucleus?

Distance of sub shells to nucleus cannot be determined as subshells are not points. They are extending around the whole atom. Their energies can be measured. But not their distances to the nucleus. Forget about Bohr's model !

Then how do you explain lanthanide contraction?

The question is misguided by the obsolete idea of shells and subshells as separated exclusive regions of the space around the nucleus.
All orbitals largely overlaps and statistical distribution of electron density, (aside of being in the particular orbital) depends on the nucleus charge and overall electron configuration.
It can be said, that distribution of electrons density for 4f is in average closer to the nucleus than for 6s. But due weak screening of f electrons, the effective nucleus charge for outermost ekectrons is growing across the lanthanoid serie, what leads to the contraction.
OTOH, higher s orbitals have very significant density near nucleus as the local maximum, what leads to relativistic effects due high speed., especially for the near right end of the period 6. And, this is reportedly responsible for about 10% of lanthanoid contraction.


text:
Why can't we use 30 mL of water and 70 mL of alcohol to make a 70% (v/v) solution of Isopropyl Alcohol?

Volume percentage (V/V): Solutions that are formulated from two liquids are often expressed as volume percent
with respect to the solute. The volume percent is the volume of a liquid in 100 mL of solution. The label on a bottle
of ordinary rubbing alcohol reads “isopropyl alcohol, 70% by volume”. Such a solution could be made by mixing
70 mL of alcohol with water to make a total volume of 100 mL, but we cannot use 30 mL of water, because the
two volumes are not necessarily additive:

What exactly does "not neccessarily additive" mean?

You can use 30 + 70 mL to get 70% v/v solution, but the resulting liquid need not have 100 mL,  E.g. 1 L of water and 1 L of ethanol give ( after cooling of selfheated mixture) about 1.96 L of 50 % V/V solution.  In chemistry, the volume fraction φi is defined as the volume of a constituent Vi divided by the volume of all constituents of the mixture V prior to mixing. https://en.wikipedia.org/wiki/Volume_fraction

Assuming you are talking about either isopropyl alcohol or ethyl alcohol (ethanol) and not both. They are different beasts, they are both used as disinfectants and solvents, but isopropyl is way more toxic to humans (who more or less regularily drink etanol solutions).
70ml ethanol and 30ml water, whem mixed, do not make 100ml mixture, but somewhat less.
That's what "non additive" volume means.
I don't have the numbers at hand, but the "contraction" is pretty much measurable and in particular important in the mass production of desinfectants and drinks.

I'm very sorry, I made a typo, I meant "70% alcohol" in the question.

You are free to edit your question to make it self-consistent. BTW my answer holds for both isopropyl and ethyl alcohol. You just can't trivially make one from the another.