text:
Why is the distribution of electrons of calcium in K,L,M,N shells 2,8,8,2 instead of 2,8,9,1?
I'm a beginner to this topic, so this would likely sound dumb. As far as I know, when distributing electrons in energy shells, the last energy shell can't have more than 8 electrons. So for calcium, it can't be 2,8,10. However, why isn't it 2,8,9,1?

This was determined by chemical properties of calcium, the way it combined with other chemicals. Bohr did the early work on electron configurations.

8 electron rule does not count with d nor f orbitals.

The answer of your problem lies in the existence of sub-shells, that you seem not to know. Let me explain.
If you give a number $n$ to the successive shells, $K$ shell gets $1$, $L$ gets $2$, $M$ gets $3$, etc. The maximum amount of electrons per shell is $2n^2$ , which is $2$ for $K$ shell, $8$ for $L$ shell, $18$ for $M$ shell, etc. $n$ is called first quantum number or principal quantum number.
But whatever the $n$ value, the $2$ first electrons of any shell are called with the letter $s$ after the first quantum number $n$. Their energy is always a little bit lower than the other electrons of the same shell. They form what we call the sub-shell $s$. Examples of sub-shells $s$ : $1s, 2s, 3s, 4s$, etc.
If a shell has more than $2$ electrons, the supplementary ones are called $n$ plus the letter $p$. They form a sub shell $p$. Example of sub-shells $p$ : $2p, 3p, 4p$, etc.  There may be up to six such $p$ electrons, and their energy is a little bit higher than the two $s$ electrons of the same shell.
If a shell has more than $8$ electrons, the ten supplementary ones are called $n$ plus the letter $d$. They form a sub-shell $d$ and their energy is higher than the electrons $s$ and $p$ of the same shell. Examples of sub-shells $d : 3d, 4d$. etc.
But surprisingly, electrons of the sub-shell $4s$ have a lower energy than $3d$. So when filling up the electronic configuration of Calcium ($Z = 20$), the sub-shell $4s$ will be filled before the $3d$. This is why Calcium contains $2$ electrons in the $K$ shell, $8$ in the $L$ shell, and not more then $8$ electrons in the $M$ shell. After finishing the $3p$ sub-shell, the two last electrons are not going to the $3d$ sub-shell of the $M$ shell. They will prefer going to the $4s$ sub-shell of the $N$ shell. This is why Calcium is not $2,8,9,1$ or $2,8,10,$ as you wrote. It is $2, 8, 8, 2$


text:
Matrix elements evaluation in Hartree-Fock method
In Roothaan's matrix formulation of the Hartree-Fock method, we eventually must evaluate the elements in the Fock matrix in the form:
$$F_{\mu\nu} = \langle \mu | {-\frac{1}{2}}\nabla^2 | \nu \rangle
             - \sum_m^{\text{nuclei}} Z_m \langle \mu | \frac{1}{r_m} | \nu \rangle
             + \sum_{\lambda\sigma} P_{\lambda\sigma}[\langle \mu\nu|\lambda\sigma\rangle - \frac{1}{2}\langle \mu\lambda | \nu\sigma\rangle]$$
I'm having difficulty "descending" from the realm of abstract symbols to the actual number crunching that a computer must do to evaluate each term. At the end of the day, each term in the equation above is a number. My question is in particular about the second term, which represents the coulomb attraction between electrons and nuclei.
How is it computed, exactly? Is $1/r_m$ an operator? I know it represents the distance from the nucleus (for each nucleus, since it's inside a sum over all nuclei), but what is it? How do I start with a basis function and end up with a number that depends on the distance from the nucleus? The distance of what, by the way? The basis function is not a point in space . . .
I realize I'm not asking a clear question, but that's because I'm confused, so please bear with me and try to address some of the stuff I asked. Thanks guys.

Yes, it's an operator and the integral - which is commonly is referred to as $\hat J$ - is then solved in polar coordinates or in Fourier space.

Electronic structure theory at the end of the day is "advanced quantum mechanics". IMO, one should understand the symbols from a QM perspective before applying it here.

Is $1/r_m$ an operator?

Recall that in QM, observables are represented by operators. Even the humble $x$ in classical mechanics is represented by an operator $\hat{x}$ in quantum mechanics. When acting on a wavefunction, the effect of the $\hat{x}$ operator is to multiply the wavefunction by $x$:
$$\hat{x}[\psi(x)] = x\cdot\psi(x)$$
Now, that's kind of boring, but it doesn't stop it from being an operator. Formally, an operator is just something which takes a wavefunction (aka a state, or a ket) and transforms it into another wavefunction. The action of $\hat{x}$ did exactly that: it transforms $\psi$ into $x\psi$.
Generally speaking, the same is true of $1/r$. The only difference is that your wavefunction now is expressed as a function of three spherical coordinates $(r, \theta, \phi)$. In the equation you wrote, there aren't any hats involved; this is mostly for notational convenience, and also because it barely makes any difference (notice how similar the LHS and RHS are on the above equation). You can imagine that there's a hat there, though, if you're more comfortable with that.

How do I start with a basis function and end up with a number that depends on the distance from the nucleus?

You integrate over all space, which reduces the function into a number. Presumably you know what $\mu$ and $\nu$ are, because this is your basis set which you have chosen.* These are both functions of $(r, \theta, \phi)$. Then:
$$\langle \mu | \frac{1}{r} | \nu\rangle = \int \left(\mu^* \cdot \frac{1}{r} \cdot \nu\right) \,\mathrm{d}V = \int_0^\pi \int_0^{2\pi} \int_0^\infty \left(\mu^* \cdot \frac{1}{r} \cdot \nu\right) r^2 \cos\theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$
Actually evaluating this integral with a computer is a totally different challenge, and not one that I'm particularly familiar with, but fundamentally speaking, the above integral tells you exactly what to do.

The distance of what, by the way?

The off-diagonal elements of QM operators ($\mu \neq \nu$) are quite hard to interpret classically, so I'd suggest to not think too much about it. At best one may view it as some kind of "interaction" term between the two states $\mu$ and $\nu$, but that generally does not yield much insight.
For the diagonal elements (i.e. $\mu = \nu$), and for a generic operator $\hat{A}$, it's possible to interpret the integral $\langle \mu | \hat{A} | \mu \rangle$ as representing the average value of the observable $A$ when the system is in the state $\mu$. (This is a fundamental postulate of QM, often one of the first things that is taught, so it's worth revising that if you don't remember it.) In this case the system is an electron, the state is $\mu$, and the observable is $1/r_m$ (distance to the $m$-th nucleus). So the term above may be interpreted as one over the average distance of an electron in the state $\mu$ from the $m$-th nucleus.

The basis function is not a point in space . . .

Indeed, and that's the whole point of performing the integration over all space (i.e., all parts of the functions $\mu$ and/or $\nu$ contribute to the integral).

* It's been a long time since I studied this; please correct me if this is wrong. It doesn't change the point, though. $\mu$ and $\nu$ should be functions that are fully known.

It's actually quite easy to think what the off-diagonal terms is once you realise the product of two basis functions describes a charge distribution, so the term is simply the electrostatic interaction  between the (typically point charge) nucleus m and the charge distribution described by that product.

@IanBush Hmm, I never thought of it that way before. To me, it still makes more sense for the diagonal term, because then $\mu^\mu$ may be associated with the charge distribution of orbital $\mu$; but for $\mu^\nu$, there isn't an object (or an orbital) which has that charge distribution...? Not too sure myself.

Writing the total charge density as expanded in the basis and you get mu nu overlaps. Then write down the line interaction energy between that and the nuclei. Then you get the appropriate Hamiltonian element by taking the derivative of the energy wrt the corresponding density matrix element

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text:
Why does almost every drug that causes dependence have this "-ine" suffix?
I noticed that many drugs that cause dependence have the suffix word "-ine". For example:
Caffeine, nicotine, benzodiazepine, methamphetamine, cocaine, and morphine.
My questions:

Why do many of these drugs that cause dependence have the word "-ine" in them?
What does "-ine" mean in a chemical?

It might be worth checking out https://en.wikipedia.org/wiki/Alkaloid#Naming.

Also check out things that end in "-ol."

Here are twelve substances, all starting from the letter A, that are not drugs : alanine, aniline, acridine, adenosine, alizarine, amine, anthocyanine, arginine, asparagine, aspirin, atropine, aspirine.

Different but (somewhat) related and not at all straightforward to answer: What is the definition of an opioid, beyond that it's something that stimulates opioid receptors?

Put succinctly, -ine is just a common termination for chemical substances in general. You're suffering from selection bias.

@Maurice: To be pedantic, adenosine, aspirin, and atropine are drugs.

Related: https://chemistry.stackexchange.com/questions/88299/why-do-most-drugs-eg-oxycodone-have-seemingly-arbitrary-names

@Vikki Yep, but in many languages there are different terms to describe a "mind-affecting drug, used for recreational purposes or for stunning/torturing people" and "drug used to cure people (i.e. medicine), maybe not mind-affecting at all". For example, in Italian, "droga" mainly means "drug" as what is sold by drug-dealers (funnily enough it can also mean "spice", as in "spices used for cooking") or something used to alter mind perceptions (e.g. morphine as a pain-killer). You would never call "aspirin" a "droga" in Italian. In English the two meanings are conflated.

@NicolauSakerNeto are you sure is just a common termination? Because I noticed many of "-ine" dependence drugs have simple benzene ring in their molecular structure that have minimal one nitrogen atom. So i think "-ine" suffix must have something related with that molecular structure.

This is silly. Tennessine and saline, among many others, use this common ending. There is no ‘there’ there.

I don't think one can get addicted to Strychnine. :-)

The suffix "-ine" is not necessarily a bad thing causing dependence, proline is just an amino acid and chlorine is a gas. So there must be another reason!
As per the Elsevier's Dictionary of Chemetymology

-in(e): ultimately derived from -inus (Latin suffix meaning belonging to or being similar to)

One can then immediately guess the ancient organic names of natural products, with a little knowledge of plant names. What is the botanical name of tobacco? Nicotiana tabacum, so a chemist or whoever extracted nicotine named it after deriving from tobacco plant.
The word coffee and hence caffeine are derived from Arabic. Its botanical name is Coffea arabica so one can guess that caffeine was derived from this plant.
Note the halogens form an exception and plenty of other common organic chemistry names in your list. Not all natural products are named after plants, it just could be a fictitious deity like morpheus. Get hold of a good dictionary and look up the word origins of each.
Addedndum related to the comments: This is not the reason that "-ine" was chosen to indicate nitrogen in medicines. The suffix "-ine" has a wider meaning. As stated, fluorine, bromine, iodine etc. have no nitrogen or compounds such as phosphine, arsine have no nitrogen at all. If you have access to the unabridged Oxford dictionary (>20 volumes) they have full page on "-ine" and the original meaning of "-ine" is derived from, related to, therefore amine is from "ammonia" and "-ine". Divine is belonging to or related to God. In short, "-ine" is well beyond chemistry.

Makes sense. Cocaine is derived from coca plant. Except Methamphetamine is not a natural alkaloid and achieved by reduction of ephedrine.

This answer feels incomplete to me without addressing that many of the -ine names one is likely to see in the context of drugs result from the substances being amines/alkaloids, which is indeed related to the molecular structure as the OP suspected (-ine for this reason implies there is a nitrogen atom).

@electronpusher, I addressed your point.

Magazine must be a very addictive drug, as many scientists are abusing it very frequently.

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text:
Inorganic question regarding Phosphorus-32

I am learning Inorganic Chemistry recently, and doing some pass year paper, and found this question. i am not sure how to identify  A3+.
Pls Help for question 1 ,2 ,& 3

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For question ii, phosphorus-32 has the same electronic configuration as the usual phosphorus atom which is phosphorus-31.

Noted, will improve for the next question

Isoelectronic means it has the same electronic structure and number of electrons, for example a negatively charged hydrogen ion (hydride) is isoelectronic with a helium atom.
Now phosphorus forms a simple 3- ion which appears in solids such as aluminium phosphide, the phosphide anion is isoelectronic with an argon atom. If we count three atomic numbers forwards from argon we go
Argon, Potassium, Calcium, Scandium. So A is scandium. If we remove three electrons from a scandium atom then we end up with a trication which is isoelectronic with argon.

Thx for that explanation, i really appreciate !


text:
Can uv light cause heterolytic cleavage?
All sources I've read stated that UV-light can only cause homolytic cleavage (homolytic fission), producing radicals while not mentioning whether it was possible for it to cause heterolytic cleavage instead. Hence is it possible for UV-light to result in heterolytic bond cleavage?

It is just a matter of energy of a single photon. Heterolytic cleavage of a symmetric bond requires more energy and can be less brobable, but see the totalitarian principle.
E.g. the extreme UV (tens of eV per photon) can indiscriminately cleave any chemical bond both homolytically and heterolytically.

I agree (+1). However I would note in the arena of photolysis, there appears to be an element of selectivity which may be a little more complex than just an energy argument.


text:
Dimerization of aniline to biphenyl
I've been looking for ways to synthesize biphenyl - $\ce{(C6H5)2}$.
The Ullmann coupling is inaccessible, as bromobenzene is inaccessible in my country due to the manufacture of PCP.
I had the idea to use a version of the Sandmeyer coupling between 2 molecules of aniline - $\ce{C6H5NH2}$.
My assumptions that this is possible are because of the synthesis of diphenic acid, wherein anthranilic acid is diazonated and coupled using a copper(I) ammonia hydroxylamine complex, as per this paper: http://orgsyn.org/demo.aspx?prep=cv1p0222.
I believe it would involve dimerization of the phenyl radical, similar to how the Sandmeyer reaction involves radical addition to copper(II), where the phenyl radical comes from copper(I) induced expulsion of nitrogen from diazonium salts. Is this mechanism correct?
Is is possible to do something similar with anilin? Is there literature for this method?
Please be mindful that I'm not too experienced in the mechanisms of these types of reactions.

Looks like you have a good idea. If the route via the diazonium salt fails, you can prepare bromobenzene from it via the Sandmeyer reaction. If you have access to benzene and bromine, you can make bromobenzene.

I see...thanks for the feedback. Would you mind leaving an answer?

As OP commented that they don't want benzene in the synthesis due to its carcinogenicity, here is an alternative route involving halobenzene:

Sandmeyer reaction to form halobenzene

Wurtz reaction or Ullmann reaction

The coupling reaction can be greatly improve (in terms of selectivity and % yield) by using high-end catalyst and improved reaction conditions. Here are some methods:

Fluorobenzene and phenyllithium

Using palladium based catalyst

Palladium on carbon. Here is a paper describing the reaction

Suzuki-Miyaura coupling

Kharasch coupling

Reaction sources:

Continuous Processes for the Synthesis and Isolation of
Functionalized Biphenyls via Suzuki-Miyaura Cross-Coupling
Reactions by Georg Johannes Lichtenegger (pdf)
Zenish J. Jain, Paraag S. Gide, Rani S. Kankate, Biphenyls and their derivatives as synthetically and pharmacologically important aromatic structural moieties, Arabian Journal of Chemistry, Volume 10, Supplement 2,
2017, DOI: 10.1016/j.arabjc.2013.07.035

OP's route through Diphenic acid can also work. Diphenic acid can be decomposed thermally as well as catalytically to form various PAH and its derivative, one of the product being biphenyl. [Beware: PAH are carcinogenic]

Ref.: Oxidative Ring-Opening of Aromatics: Decomposition of Biphenyl
Carboxylic Acids and Zinc Biphenyl Carboxylates by Natalia Montoya Sanchez and Arno de Klerk. (pdf)

Thanks, but I don't want to work with benzene. It's a carcinogen, and can't be stored in a nonvolatile state.

@Jacob Tbh, biphenyl is also carcinogenic and it can cause kidney stone formation and liver tumors. That bein said, do you want an alternative route involving making a biphenyl derivative and then reduce it to biphenyl?

Sure. Benzene is more volatile than biphenyl, hence I don't want to use it.

The thermolysis of diphenic acid works for me.


text:
Why are group 13 compounds in the +1 oxidation state more ionic than those in the +3 oxidation state?
I know that for group 13 metals in the p block, the stability of +1 OS is generally more than the stability of +3 OS as it is energetically not favourable to attain the higher OS of +3 and such ions act as strong oxidising agents.
$$\ce{\overset{+3}{Tl}+2e^-->\overset{+1}{Tl}}$$
This is called the inert-pair effect. Today, however, I learned in group 13 of p-block metals that the metals in +1 OS are more ionic than those in +3 OS for the same metal. Why is metal in its lower OS of +1 more ionic but in its higher OS of +3 more covalent? For example, Thallium(I) chloride $\ce{TlCl}$ (in its +1 OS) is more ionic or less covalent, and Thallium(III) chloride $\ce{TlCl3}$ (in its +3 OS) is more covalent or less ionic. Does this also suggest that ionic is more stable than covalent in this situation?

https://chemistry.stackexchange.com/questions/30854/why-does-thallium-form-tlf-and-tlf3-but-aluminium-only-forms-alf3 ... https://chemistry.stackexchange.com/questions/146851/in-what-form-does-the-respective-halides-of-thallium-exist-as ... https://chemistry.stackexchange.com/questions/49477/why-isnt-thallium-triiodide-stable

It can be explained by fajans' rules. As the positive charge on an atom increases its size decreases as a result polarising power increases. If polarising power increases then it distorts the electron cloud towards itself and hence ionic nature of the bond decreases.   This is the reason why Thallium(I) chloride is more ionic than Thallium(III) chloride.

Does this also suggest that ionic is more stable than covalent in this situation?

It may not be. Because you can't say that ionic is more stable than covalent. For example, consider diamond which is the one of the hardest substance known to us which is a giant covalent polymer. The actual reason for the stability of Tl(I) over Tl(III) is inert pair effect and it is what you have texted first.

There is another effect : relativity. In the 6th line of the periodic table (containing $\ce{Tl}$), relativity effect becomes important. Even though electrons have no velocity, their behavior does change as if outer electrons were moving at nearly the velocity of light. This was shown by Pekka Pyyko, Accounts of Chemical Research, Vol.$12$, No. $8, 1979$, p. $276 - 281$.
As a consequence, electrons of the outer shells are so fast that their dimensions decrease, so that they are "absorbed" by the other inner shells. Their chemical properties look like atoms having "lost" $26$ electrons and protons. This is why Thallium ($Z = 81$) is chemically similar to $\ce{Cs}$ $(Z = 55 = 81 - 26)$: both form soluble and stable $\ce{M^{+}}$ ions.
For the same reason, its neighbours have properties which are also changed in the same way. On its right, lead ($Z = 82$) looks like baryum ($Z = 56 = 82 - 26$), as both $\ce{PbSO4}$ and $\ce{BaSO4}$ are insoluble compounds. On its left, mercury ($Z = 80$} is "nearly" a noble gas like xenon ($Z = 54 = 80 - 26$) : it is liquid, boiling at low temperature ($356°C)$ : so it is "nearly" a gas, at least with respect to its neighbors, boiling at $1457°C˙ (\ce{Tl})$, and  $2600°C˙(\ce{Au})$. On the other hand, mercury is chemically unique : it is the only metal producing an ionic dimer in aqueous solution : the cation $\ce{Hg2^{2+}}$. This property is also known for noble gases, which may produce ions like $\ce{He2^{+}}$ in special circumstances. And left to left, gold ($Z = 79$) is "nearly" a halogen : it looks a little bit like iodine ($Z = 53 = 79 - 26$), as it produces an ionic compound with cesium : $\ce{CsAu}$.

Mercury is not the only metal to produce a dimeric ion in the +1 oxidation state. Zinc, cadmium, and even the alkaline earth metals beryllium, magnesium and calcium have been caught doing the same thing. Mercurous ion is, however, the most common case.


text:
Orientation of amide bonds in a simplified polypeptide drawing
The syllabus for my course requires I draw protein polymers like this, where the patterned boxes represent different bodies (note that this is an extremely simplified drawing stripped down to its bare minimum for high-school students):

Is there any particular reason why the monomers alternate in direction — i.e. up, down and then up?

In this simplified way to draw the primary structure of a protein, there is no reason to enforce flipping orientation of the peptide bonds with each amino acid.
The image you show in your question was probably simply made by taking a more complete drawing of a polypeptide chain similar to the one shown below and then replacing the amino acid center parts with boxes.

Image source: https://en.m.wikipedia.org/wiki/Peptide#/media/File%3ATetrapeptide_structural_formulae_v.1.png

In a Lewis structure, you are free to draw any bond angles and sketch any conformation, no matter whether it matches real bond angles and common conformations.
In the way the polypeptide is drawn in the figure posted by the OP, however, it resembles the conformation found in beta strands forming beta sheets. This type of secondary structure has the polypeptide maximally extended, with consecutive $\ce{CO}$ groups pointing in alternate directions, allowing for hydrogen bonding with an adjacent strand. The two relevant torsion angles, phi and psi, are approximately 180 degrees.
The animation shows an example of a strand that is part of a twisted beta sheet in the structure of the protein LysC:

Source: https://commons.wikimedia.org/wiki/File:Beta_strand.gif
The point where the animation pauses corresponds roughly to the 2D image, with CO-groups (gray/red lines) pointing in alternate direction and in the opposite direction of the NH-groups attached (blue/white lines).
The conformation where CO-groups are opposite of NH-groups of the same peptide bond is conventionally (and slightly inaccurately) called trans peptide. Trans peptides are about a hundred times more common than cis peptides. So it makes sense to draw the Lewis structure hinting at the trans conformation.
There are multiple low energy conformations of peptides (including that of a beta strand) specified in the so-called Ramachandran plot. The other very common comformation features consecutive CO-groups pointing in the same direction, as found in the secondary structure called alpha helix. Drawing an alpha helix in 2D, however, is much less straight forward then drawing the extended conformation of a beta strand.

It has 2 reasons, but not really critical ones:

Visualization, clearly separating aminoacids within the peptidic structure.
Orientation of atomic bonds is not generally rectangular. If one insists then alternating of $\ce{O}$ and $\ce{H}$ is closer to reality than keeping them on the same side. But it is even more often written e.g. as $\ce{R1-CO-NH-R2}$


text:
Do ketdione molecules have diol tautomers?
In my chemistry textbook there's a keto-enol tautomerism example:

But will it also be in equilibrium with pent-1,3-dien-2,4-diol in a reaction like this or will it just be a functional isomer?

No, such tautomer doesn't exist in practice, or at least is too unstable to be anything more then intermediate

@Mithoron So it's just a functional isomer and can't exist at equilibrium?


text:
How does the Standard Cell potential of the Nernst equation work?
$$
E = E^\circ - \dfrac{RT}{nF} \ln Q
$$
For calculations, the standard cell potential is typically given with a reference page, using elements at standard conditions. While doing an experiment with varied concentrations of one of the electrodes, there will be an effect on the voltage due to the reaction quotient from this equation.
But, would you also have to account for a change in the standard cell potential of that electrode due to concentration, or is that value kept at the same standard condition cell potential?
For instance, the copper-zinc reaction is around $\pu{1.1 V}$ but if you conducted an experiment with copper at $\pu{0.0 1M}$ instead, would you still use the value of $\pu{1.1 V}$?
How about with an experiment of varying temperatures?
If experimenting with a system at $\pu{50 ^\circ C}$, would you need to use a standard potential with respect to the elements at this new temperature of $\pu{50 ^\circ C}$, or do you still use the standard potentials of $\pu{25 ^\circ C}$?
Effectively just wondering if the $E^\circ$ value for this equation is dependent on solely the elements and to look at their data values for STP or do you have to take into account the elements and their conditions to form an $E^\circ$ value.

If the concentration $\ce{[Cu^{2+}]}$ is $0.01 M$, the potential of the copper electrode becomes $\pu{E_{Cu/Cu^{2+}} = 0.34 V - 2· 0.0295 V = 0.28 V}$. If this copper electrode is coupled with a zinc electrode where $\ce{[Zn^{2+}] = 1 M}$, the potential of the cell becomes $\pu{0.76 V + 0.28 V = 1.04 V}$
Now the effect of temperature change can be calculated from the formula $2.30\frac{RT}{zF}$. At $25°C$, this factor is $0.059/z$. At $50°C$, it becomes $0.0636/z$

Okay, so I want to provide another answer just to elaborate on what was said.
$\ce{E^\circ}$ refers to the cell potential at standard conditions (By the way, it is not STP. STP is for gases primarily). So if we were doing an experiment at 50°C instead of 25°C, then we would use the Nernst to find the cell potential at the specific temperature (E). The standard cell potential ($\ce{E^\circ}$) value for something will always be the same.
Again standard conditions are $\pu{1M}$ of solutions, 1 atm of pressure, and 25°C.


text:
Order of acidity among fluoroacetic acid and chloroacetic acid
I've heard that fluoroacetic acid is more acidic than chloroacetic acid. But isn't the carboxylic acid a better electron withdrawing group than both fluorine and chlorine? If so, how's it that fluorine withdraws electrons towards itself in the compound? (and be more acidic)

First off, based on acid strength carboxyl is not more electron-withdrawing in this setting than fluorine. From the respective WP articles on the compounds below, we find that carboxyl is actually about as electron-withdrawing as chlorine, with fluorine beating both.

Name
Formula
$\mathrm{p}K_\mathrm{a}$

Chloroacetic acid
$\ce{Cl{-}CH2COOH}$
2.86

Malonic acid (1st dissociation)
$\ce{HOOC{-}CH2COOH}$
2.83

Fluoroacetic acid
$\ce{F{-}CH2COOH}$
2.59

Carboxyl would be more electron-withdrawing when the carbonyl component is conjugated to a π-bonded system such as an aromatic ring.
Regardless of how the halogens compare with carboxyl, a more strongly electron-withdrawing group would always stabilize the anion more and thus make the acid stronger. However, the difference in acid strength is relatively small because the carboxylate ion is already fairly well-stabilized by its own charge and bond delocalization.


text:
Mechanism of successively melting ice with table salt
Hi I was reading about how regular sodium chloride in itself can melt ice outside, for example ice on roads during low temperatures. However, I am unsure about the exact details, as I have found some conflicting information, which I would like to clarify here.
When we neglect any temperature change and assume the air as well as the ice are at the same temperature, maybe -2 degrees celsius, what would happen is the following:
When pouring the sodium chloride crystals on the ice surface, because of the room temperature crystals, heat is transferred to the ice, inducing a phase change to water, so that the NaCl dissolves and forms a concentrated solution, additionally the emerging solution should also cool down due to the positive solvation enthalpy of NaCl.
(On this website, it is stated that ice always has a water layer on top of it; to my knowledge, that equilibrium H₂O(s)⇌H₂O(l) only exists at 0 degrees celsius so should not apply here. https://www.scientificamerican.com/article/why-do-we-put-salt-on-icy/)
This concentrated solution has experiences freezing point depression (a 1M solution would freeze at around -4 degrees celsius), so that any ice surface in contact with the solution should dissolve, as the ice in equilibrium with the undercooled (negative degrees celsius) solution should not be able to mantain its solid form and also experience a phase transition to water.
(Would this actually also mean, that solid ice would also dissolve by itself in a concentrated solution of sodium chloride at negative degrees celsius?)
This process continues until the sodium chloride solution is too diluted to dissolve any more ice in it.
Would this be the correct explanation of the individual steps of this process? Would other salts execute this process more quickly? (For example salts with positive solvation enthalpy)
Any help or input would really be appreciated, as there is some conflicting information out there.
Cheers!

https://chemistry.stackexchange.com/questions/87310/melting-ice-with-salt-how-does-it-start https://chemistry.stackexchange.com/questions/5023/why-calcium-chloride-is-used-to-melt-ice-over-sodium-chloride?rq=1

It is well-known that “Solids do not react together at room temperature over normal time scales and it is necessary to heat them to much higher temperatures, often to 1000 to 1500 °C, in order for the reaction to occur at an appreciable rate.” Ref 1. One reason this is true is that even when molecules of one solid diffuse into another solid, the resultant is still a solid; the only thing observed is a diffusion, and this is so slow that it seldom qualifies as a “reaction”.
But “Thus, we study several reactions that have been recently reported to proceed “in the solid phase” and clearly show that, in most cases, grinding the two solid reactants together results in the formation of a liquid phase.” Ref 2.
To be very accurate, the salt doesn’t melt the ice; a salt solution causes ice to melt. If you don’t have a liquid phase to begin with, you won’t get any ice melted.
Consider ice and water in dynamic thermal equilibrium at 0ºC. Water molecules pass between the solid ice and liquid water at equal rates in opposite directions. The ice crystals may rearrange slowly, but the amount of ice and water remain constant. This balance is maintained at 0ºC, but can be upset, for instance, by adding salt.
Combining salt and ice (at the same temperature, below 0ºC), cause two reactions to occur simultaneously: #1. A solution is formed from traces of water, or from grinding. The temperature is reduced slightly because energy is required to dissolve the salt: about 1 kcal/mole of salt. This does not involve melting of the ice, just dissolution of salt by a (tiny bit of) liquid H2O. #2. Water molecules in the ice pass into the salt solution, but fewer pass back to the ice. This means that the ice is melting. The heat required to melt the ice (68.3 kcal/mole of H2O) comes from the salt solution and reduces its temperature much more.
These two reactions result in ice being drawn into the salt solution, increasing its volume at the expense of ice volume, until the ice and solution temperature are reduced and the water concentration is increased to a point where the water molecule exchange rates become equal again.
This process is used in two ways: #1: In the lab, for cooling baths, with excess salt added so that the mixture reaches its lowest temperature (-21ºC Ref 3.) #2: On roads, salt is spread onto a large amount of ice/snow, liquifying it by creating a combination of salt and water which has a melting/freezing point below the ambient temperature. Salt is spread as a fine granule or as a solution for faster action. Other salts, e.g., CaCl2, usually hydrated, and therefore somewhat wet, are preferred for melting ice at lower temperatures.
Ref 1. https://en.wikipedia.org/wiki/Solid-state_reaction_route
Ref 2. Gadi Rothenberg, Andrew P. Downie, Colin L. Raston, and Janet L. Scott, J. Am. Chem. Soc., 2001, 123, 36, 8701–8708, https://doi.org/10.1021/ja0034388
Ref 3. https://sciencenotes.org/why-salt-makes-ice-colder-how-cold-ice-gets/

So basically, even though the salt and the ice are both below 0 degrees celsius, through trace water, and subsequent dissolution of ice in the salt solution, a significant amount of ice can be dissolved?

And, based on your explanation, it would be correct that ice by itself would melt in a conc. solution of sodium chloride at temperatures even lower than 0C right?

@Mäßige: Yes to both comments/questions. The key concept is that the ice-water system is in equilibrium, and is dynamic: H2O molecules go both ways. When you disturb the equilibrium (with NaCl), the H2O movement adjusts, which changes the amounts of ice and water (usually set up with enough NaCl to melt all the ice, which is possible because the ambient temperature is not too low).


text:
What does n(Na2CO3) mean?
In my book, $n(\ce{Na2CO3})$ was used to indicate the number of moles of $\ce{Na2CO3}$ without any clarification. Is it common knowledge among chemists that $n(\ce{Na2CO3})$ indicates the number of moles of $\ce{Na2CO3}$?

Yes, the standardized quantity symbol according to ISO 80000-9:2009 Quantities and units — Part 9: Physical chemistry and molecular physics as well as the recommended quantity symbol according to IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book) for amount of substance is $n$.
The quantity “amount of substance” shall not be called “number of moles”, just as the quantity “mass” shall not be called “number of kilograms”. In the name “amount of substance”, the words “of substance” can for simplicity be replaced by words to specify the substance concerned in any particular application, so that one may, for example, talk of “amount of hydrogen chloride”, or “amount of benzene”. Although the word “amount” has a more general dictionary definition, this abbreviation of the full name “amount of substance” may be used for brevity.
Thus, $n(\ce{Na2CO3})$ would be called amount of sodium carbonate.

You are unclear if you are confused by $n$ itself, or by the particular notation $n(\ce{Na2CO3})$
$n$ is widely accepted to denote amount of substance, expressed in moles, as @Loong , IUPAC and Wikipedia say.
If $n$ means the amount of the particular substance like $\ce{He}$ or $\ce{H2O}$, it is often written in index form like $n_{\ce{He}}$ or $n_{\ce{H2O}}$.
Multiple levels of sub/superscripts  often bring typographic difficulties. For these reasons, instead of typing $n_{\ce{Na2CO3}}$, some may prefer rather $n(\ce{Na2CO3})$ instead.
It is not official recommendation and less controversial may be substitution by some placeholder.

I often think the term amount of substance has been chosen rather unluckily, as at overlap with less scientific domains, the term can be confused with primary understanding of amount in common language with very  vaque meaning.

Who recommends writing amount of sodium carbonate as $n({\ce{Na2CO3}})$? Is it IUPAC? I don't recall seeing this recommendation in textbooks?

It is not official recommendation, rather practical one, applicable generally on complex indexes, not just on amount of substance. I remember one of mods suggested it.

This is I am wondering, who is making this practical recommendation?

Personally, I think less controversial way would be substitution method, like let A is , then using $X_\mathrm{A}$, where X is general quantity. As $n_{\ce{^{235}_{92}U}}$ or similar may not always look good.

Yes, I have seen subscripted versions but rarely with in-line brackets.

Na2CO3 is fine as subscript yet, but displaying complicated sub+superscripted expressions, when the major quantity has its own superscript, is challenging. AFAIK, the original recommendation was made in context of standard potentials.


text:
How does soap help in cleaning clothes and rinsing off oil and grease?
When we say that we agitate clothes so that grease particles get separate out from clothes as ionic part of hydrocarbon chains of soap is attached to water and non polar part of chain is dissolved in oil and grease on clothes surface.
So it is said that if we shake clothes than these oil molecules get dispersed in water.Because ionic part keep attached to water thus snatching oil molecules from clothes
But how? As if we shake clothes can't the non polar part of hydrocarbon carbon chain keep attached to clothes grease and oil and keeping along the ionic part of chain with it?
I mean Does the ionic attraction between water and hydrophilic part of chain is stronger(thus keeping the chain stable and fixed in water help in snatching out grease particle when cloth shaked)? and that's why not keep attached to grease molecules even after rinse with water?
Basically , I am asking
"Why hydrocarbon chain snached off the oil molecules from clothes when shake and rinse with water instead of keep attached  with clothes , can't ionic part of chain moved along with shaking of clothes and therefore keep oil molecules on stable  clothes?"

The attraction between oil and cloth must be rather weak, and not to be compared with attraction between water and hydrophilic end of the soap chain

Consider the law of mass action -- if there is a lot of water, and an excess of detergent, then the odds are the detergent will wind up in water, not attached to the clothes, and the oily material will wind up in the detergent. If not, repeat and rinse -- that's using mass action, again.

Let's look at the system: clothes + grease + water + soap. To be more specific:
a) We prefer to wear hydrophilic materials to manage moisture removal (sweat). The clothes could be cotton, wool or linen, which are naturally hydrophilic, or synthetic, like polyester or acrylic, which are naturally hydrophobic, but are treated with a finish to make them hydrophilic, because we don’t like the feel of wearing plastic. Ref 1.
A material, whether hydrophilic or hydrophobic, can be wet by either kind of liquid, but the degree of attachment, or the ease of removal, will vary. Hydrophilic materials tend to be more polar and find ways to bind together by polar attractions, while hydrophobic (oleophilic) materials tend to bind with lower energy.
b) The grease is a long-chain hydrocarbon, a viscous, hydrophobic, oily material. Even tho the grease is hydrophobic, it will wet hydrophilic materials, and it still takes energy to get it off.
c) The soap is an alkali salt of a fatty acid: somewhat soluble in water; it ionizes at very low concentrations to a cation (+) and a long chain hydrocarbon with a negative end (carboxyl). At higher concentrations micelles form as hydrocarbon ends get close and extrude any water between them (the chain is hydrophobic after all). So a bunch of soap molecules agglomerate in little spheres or rods, or other shapes, with all the carboxyl groups on the outside affiliated more closely with water molecules.
Synthetic surfactants are not soaps and are very different chemically, but still have a hydrophilic end and a hydrophobic end. In the picture, the micelle on the far right is a mixed micelle, having grabbed some oil from some dirty clothes.

It's important to note that the flakes of soap, or the liquid soap, and the micelles are a reservoir of active material, and the first reaction of cleaning is done by a very low concentration of individual soap molecules.
d) The system also includes stirring or agitation. Chemical reactions between separate phases, like two liquids, or a solid and a liquid, almost always take place easier when there is plenty of agitation, which allows more opportunity for tiny molecular motions to do their thing.
Let's observe the process:
Case 1: Macro scale; visual observation.
If you take the oily/greasy clothes and wet them, the clean spots will get wet, but greasy spots will tend to repel water. If you wet the clothes with soap solution, everything gets wet, because the surface tension of the water is reduced. Shake the clothes in the soap solution, drain and rinse to carry away the emulsified grease, and your clothes are now clean. Like magic!
We explain the cleaning by saying that the surface tension of water is reduced by the soap so the solution can wet the grease, emulsify it and carry it away. Surface tension is like a skin formed from the outermost layer of water molecules; the skin on water can support a paper clip or a bug or resist wetting some low surface energy material like grease.

Surface tension does not explain everything: agitation is important to break up the grease; soap is important to encapsulate the grease into mixed micelles (which look like soap micelles, not grease droplets). Rinsing is important because it removes the excess soap (as micelles) and grease (as mixed micelles).
Case 2. Micro scale; visualize with imagination. When you put the greasy clothes into the soap solution, not much happens unless you agitate, which brings the soap molecules into better contact with the grease. Hydrocarbon tails find the grease, adhere and mix in. Now the grease becomes covered with soap that has its hydrocarbon end sticking in, and its carboxyl end sticking out into the water. The clothes are even dirtier! The OP’s question seems to address this very early stage of the cleaning process. But we don’t stop there - shake, shake, shake, and mechanical activity splits off a layer of grease containing the soap into the solution, where it rearranges into a mixed micelle. The reservoir of soap micelles supplies more individual soap molecules to the solution, some get onto the surface again; continual shaking knocks off more grease/oil/soap. The clothes get cleaner; the water gets greasier.
The process of mechanical agitation will continue to knock soaped up grease off more efficiently than grease can get out of a mixed micelle and get back onto the clothes. If the soap is inefficient in knocking off and trapping the grease, it will never make it to the big time.
Now you are halfway finished. You must remove the greasy water emulsion by rinsing the clothes. The clothes may be clean enough; if not, you could rinse some more, or wash again and rinse.
Ref 1. https://www.fsw.cc/type-finish-fabric-need/


text:
Navigating through the various characterisation technique?
We synthesis a lot of ceramic catalysts. For a long time now, we are facing problems in accurately and efficiently characterising the synthesized powders. We have identified the main problems as follows -

We have a doubt whether we are doing the sample treatment correctly or not. A lot of vendors tell us that, “providing a dry, fine powder is sufficient”.  But is there any comprehensive guide or something that can help me help in deciding the appropriate sample treatment for a particular characterisation method?

For a lot of techniques/methods, depending on the instruments the operator has to choose between a lot of parameters depending on the inherent characterisation properties of the powders. We feel the operators that we are dealing with nonchalantly input the default settings and this makes the output coming out not so accurate. Again, how do I navigate efficiently through this space?
Anyone who has experience in operating various machines and analysing the output, any help will be appreciated.

PS – We are doing the following techniques –

Powder XRD

ICP-AES

TEM/SEM

XRF

Chemisorption


text:
Does lithium grease damage plastics such as ABS & PBT?
I have read that petroleum-based greases damage plastics due to the petroleum-derived nature of most plastics. However, I have seen lithium grease being used in electronics such as the gears in CD players. What is determines the resistibility of certain plastics to petroleum-based grease?

Usually, no chemistry is involved. (Some) plastics may (partly) dissolve or swell in contact with organic solvents and also grease. This alters material properties, such as mechanical strenghth, optical appearance etc. This does not occur for all combinations of plastics and grease.


text:
IR Stretching in Hexacarbonyl Iron

With reference to this graph, my professor stated the following: "Since the IR band for the bridging carbonyls is single, it implies that the bridging carbonyls are oriented linearly. Since we have two bands for the terminal bonds, they are not collinear."
Precisely how and why?

This has to do with symmetry. If two CO groups are symmetric (i.e. interchangeable with a symmetry operation) then the vibrations of those stretches will couple to each other, and form a symmetric and antisymmetric band. Now the symmetric band would be invisible in the IR spectrum, if the CO are aligned perfectly opposite to each other, because there would be no change in the dipole moment in the symm. stretch...

...however, the asymmetric stretch is usually always visible because there is a dipole moment change. If you can see both symm. and asymm. peaks that means your CO ligands are not perfectly opposite to each other i.e. not collinear. This is a kind of simplified analysis that works for pairs of CO ligands, for higher symmetry (i.e. three fold) you need to use more rigorous group theory.

@SRMaiti, why not turn your comments into an answer?


text:
What's the factor governing the order of wavenumbers of CO in metal carbonyls?

As a higher wavenumber means a lower wavelength, which in turn means higher energy, the complex with the most effective backbonding would have the least wavenumber. The backbonding, as I know, is proportional to the number of electrons in the d-orbital, so all this says that as we go rightwards on the periodic table, wavenumbers should decrease, which clearly is not the case.
Where am I going wrong?

The backbonding, as I know, is proportional to the number of electrons in the d-orbital

That's indeed part of it — and that explains why the early d-block metals, Sc and Ti, don't form (neutral) carbonyls at all. However, the backbonding also depends on how willing the metal is to share those d-electrons. $Z_\mathrm{eff}$ increases going across the 3d block, which leads to poorer overlap between metal 3d orbitals and the CO π* orbital, and a poorer match in terms of their energies.
Around the middle there is a sweet spot where there is enough electron density to satisfy the CO ligands (or, equivalently, to satisfy the 18 electron rule), and the metal $Z_\mathrm{eff}$ is not so high.


text:
What is the general formula for tertiary alcohol?
According to my book, the general formula is $\ce{R-OH}$.
But according to the internet, it is $\ce{R3COH}$.
Are both the general formulas correct?

First, please use the proper term: tertiary alcohol. "3°" is a shorthand notation, fine for taking your own notes, but not appropriate in formal writing. Then, look up the definition of a tertiary alcohol, and you can figure out which one is correct from that. If your book really said that it's ROH, I strongly suggest dumping it and getting a new book.

I agree. You need a better textbook. The internet definition is much more accurate.

The formula $\pu{R-OH}$ could be used for all sorts of alcohols, and even for other substances like phenols.

I wouldn't be so quick to say dump the book. Whether the formula was used correctly depends on the context. For example, "Consider the following reaction involving ROH (a tertiary alcohol) and..."

The formulae are not inconsistent. "R" can represent anything including "R3C" so the second formula is merely a more specific subset of the first.

A tertiary ($\mathrm{3^\circ}$) alcohol is compound in which the hydroxy group ($\ce{-OH}$) is attached to a tertiary carbon; which again indicates a saturated carbon that is attached to three carbons.
So, the general structure would be (stereochemistry not shown, only connectivity):

Your confusion arises from the fact that in organic chemistry, the symbol $\ce{R}$ can be used to represent any carbon chain, and so the precise meaning depends on the context.
The most general formula for tertiary alcohol should be $\ce{R^1R^2R^3COH}$, as shown in the picture, to indicate that all three carbon chains can be different. However, $\ce{R3COH}$ is a reasonable way to represent it, as long as you keep in mind the caveat.
$\ce{R-OH}$ is probably not the best way to represent a tertiary alcohol, because it hides the tertiary nature, but it might be acceptable to draw in a reaction scheme in case there was no change in the carbon chain, so it is okay to hide that complexity.
At any rate, it is better to know what it means, than rely on a specific formula, because representations in chemistry are flexible, and can change depending on context.


text:
When pigments absorb light only around a single particular wavelength, why aren't they still white?
I'm trying to understand how organic molecules get their colour.

One major factor are conjugated double bonds which create delocalized pi-orbitals (e.g. https://en.wikipedia.org/wiki/Conjugated_system). When a photon hits an electron in the HOMO, and the photon carries the right amount of energy, the electron gets lifted to the LUMO. Other photons don't get absorbed, they are reflected/transmitted/whatever and create the colour of the pigment/object.Since everything down there is a wobbly probability distribution, the energy of the proton doesn't have to be at exactly a single    value, but can the close to it. But, as far as I understood it, it is a very narrow energy band that gets absorbed.

When light hits the cones in human eyes, we can model the activation of the different cones by scalar-multiplying the spectral power distribution of the incoming light with the sensitivity functions of the cones (e.g. https://en.wikipedia.org/wiki/LMS_color_space and https://en.wikipedia.org/wiki/CIE_1931_color_space). If we assume a uniform power distribution of the light and shine it on a colour pigment, only a small band of wavelengths should be absorbed.

Now... My problem is, that if the last sentence is true, the pigment should still appear white/grey or, at best, have a very desaturated colour. The incoming light appears white/grey because the scalar products of the incoming power spectrum with the CIE 1931 colour matching functions all produce the same value. Introducing a narrow, deep dip in the incoming power distribution via photon absorption won't change these values enough to give a strong, saturated colour. (For the actual human cone sensitivity functions, white/grey light might not be represented by three equal values, but the argument is the same.)
To get a strong colour, the pigment would need to absorb a large portion of the wavelength spectrum. And looking at the absorption spectrum of, for example, beta-carotene, this seems to be the case: https://www.photochemcad.com/compound-detail.php?name=Beta-carotene
If the $\lambda_{max} = 451\mathrm{nm}$ stated there represents the energy required to cross the HOMO-LUMO gap, why are so many other wavelengths, which are "far away" from $451\mathrm{nm}$ also absorbed?

Nice interactive link for beta-carotene! So it absorbs pretty much all light below about 510 nm, which means it transmits, from incident white light, red, orange, yellow and green. Then the color we perceive depends on our eyes: most of us are trichromats, some people have one of the color blindnesses, etc.

By the way, beta-carotene has broad absorbance, so it does not really support your query about what would happen if a narrow absorbance removed just a narrow band of wavelengths from a white light source. An example of narrow band absorbances in the visible is here. And the color of the holmium chloride aqueous solution dramatically changes, depending on the source of the illuminating light: white LED lamp, white fluorescent lamp or sunlight. So, not so simple.

Beta-carotine having broad absorbtion is exactly the point that confused me! It has to be that way to appear orange and the linked measurements confirm it. But looking at explanations how pigments such as beta-carotene get their colour, I only ever find the argument that the single HOMO-LUMO jump is responsible.

In introductory chemistry courses, students may be introduced to a tiny quantum mechanics section, which includes the particle in a box solution of the Schrodinger equation. Then beta-carotene, with its conjugated double bonds, is modeled as a particle in a box. What could be sadly misleading later on? ;-)

Being a pigment and absorbing just around single wavelength mutually excludes each other. You would need atomized matter in gaseous state for that.  Molecules and/or condensed phases have quite broad absorption bands or even wavelength regions.

You are right that an individual transition from one state to another by absorption of a photon is connected to a very well defined energy. However, an absorption band in UV/VIS results from transitions between various energy states.
For example, you might start from the electronic ground state before absorption. However, your molecules are in various different vibrational states. So your starting levels are different. The excited state you reach after absorption again has different vibrational levels, which have a different spacing than in the ground state. Thus, your absorption band is a superposition of many individual transitions between different states with different energy differences, and consequently is rather broad.
The situation can be much more complex than this short description. For example, also different electronic states can participate.
You might want to search for a Jablonski diagram for a visualization, e.g. here.

Thank you! That makes a lot of sense, and I suspected something similar.
The only thing that bothers me is that in my mind, the different vibrational levels lead to a small number of narrow energy bands. I have a hard time imagining that they are numerous enough and close enough to each other to give a broad, "smooth" absorption spectrum overall. But I guess they do?

Yes, in solution, there are so many molecules, with slightly different energies that there are so many transitions, that it looks like a continuous curve.

Yes, you could expect some fine structure, and look at the iodine gas phase spectrum in @M.Farooq's answer, there you have it! It is a matter of resolution of the spectrometer if you can see it. However, in condensed phase (liquid) you have additional interactions between the molecules that influence the energy states of the molecules. Thus, so many different transitions occur that no spectrometer can resolve this, and you get your typical smooth absorption bands. You can observe similar effects e.g. in infrared spectroscopy.

Color perception is a complex topic and one cannot make generalized statements.
If you are seriously investigating and want to have a deeper understanding, then get hold of a beautiful book called The Physics and Chemistry of Color: Fifteen Causes of Color by Kurt Nassau. It could be available on Internet Archive. It is a readible book without involving too much math.
The sky is blue for one reason, a Bunsen burner flame is blue for another reason, and the carrot juice is orange for another reason. The key criterion for a colored solid/ liquid or gas is to interact with visible light (~ 400 to 700 nm) in some way. There is no heavy restriction on how many wavelengths must be absorbed in for something to appear colored. There are a very few exception which something appears colored to us because of interaction with high energy particles, which are not visible us, but the interaction of those particles with matter leads to emission of light in the visible region.
When chemists talk about colors, they are usually talking about molecules in solution or in gas phase. Unlike gaseous atoms where only discrete wavelengths can be absorbed, like sodium vapor, molecules have vibrational and rotational excited energy levels besides electronic energy levels. If you excite molecules with visible light a broad range of wavelengths can be absorbed just like your beta carotene example. Consequently, this assumption

But, as far as I understood it, it is a very narrow energy band that
gets absorbed.

is not right. Molecular absorption in solution or gas phase is pretty broad. I am avoiding solids. There you have to deal with their reflectance spectra. We are focussing on absorption/ transmission spectra.
One of the best examples of colored gas is that of molecular iodine vapor. It has a striking violet color. Photo credit here
.
Since the vapor is appearing colored to us, it must be interacting (absorbing in this case) with light wavelengths in the range 400-700 nm. Have a look at this absorption spectrum here. The iodine molecule in the gas phase absorbs green to orange portion of the visible spectrum. As the result, transmitted light from iodine vapor looks violet.

Thank you for your answer! You wrote "Molecular absorption in solution or gas phase is pretty broad", but can you explain why that is?

Because unlike atoms, molecules can also vibrate in the excited state. There are plenty of vibrational energy levels for a molecules. This is why it is not a single sharp line like atoms, but a series of lines, as you can see in the iodine spectrum. Each peak corresponds to a vibrational level.

Some types of light absorbance show very narrow lines while others show broad peaks.
Atomic spectra have sharp lines when the atoms are in the gas phase. One example is the absorption of sunlight by gases on the surface of the sun. The sunlight appears white, especially when observed in outer space. However, there a various species absorbing the light at specific wavelengths, allowing spectroscopists to infer the composition of the sun's surface.

Source: https://calgary.rasc.ca/solarspectrum.htm
On the other hand, visible and UV absorption of samples in the liquid or solid phase are often broad. Typically, these also arise from electronic transitions, but those of valence electrons. The energy of the electronic states of valence electrons depends on bond lengths and angles (i.e. couples with vibrational states) and intermolecular interactions (i.e. will be different from molecule to molecule).

To get a strong colour, the pigment would need to absorb a large portion of the wavelength spectrum.

Yes, so atomic absorption does not change the perceived color much. On the other hand, atomic emission results in clear bright colors like in fireworks or flame tests.

Beautiful figure. A small note for readers. There is no instrument/spectrograph which can generate a raw solar spectrum like this. This marvellous solar spectrum is computer generated to look like a echelle spectrographs output, which also does not look like this at all. As novices, I and a good friend here were confused for a couple of years by this beautiful spectra thinking that this could be the raw image of a fancy spectrograph.

I should mention that while many of the images available for the solar spectrum might suggest that these gaps are something that need fancy instrumentation to measure indirectly and then later "reconstruct", this is not necessarily the case - the gaps are very much tangible, and can be captured even with very rudimentary tools. See for example these photos of several light spectra (including the solar spectrum) by using a CD in a cardboard box as a monochromator.

Yes such experiments with CD spectroscopes are worth the time.

As others have brought up, indeed the absorption features of most compounds with electronic transitions in the visible light range are rather broad, and this breadth is important in generating a perceived colour.
But what would happen if you somehow had a substance which acted as an "anti-laser", where somehow a very tiny sliver of a white light spectrum were absorbed? What would that look like?
As it turns out, the solar spectrum is already full of "holes", called spectral lines. These are caused by isolated atoms in the gas phase in the solar atmosphere. Because these absorption features are caused by electronic transitions in single isolated atoms, effects capable of broadening the absorption of the electronic transition lines are much more limited. For example, the solar spectrum has dark slivers at 589.158 nm and 589.756 nm, corresponding to electronic transitions of isolated sodium atoms. These "holes" in the solar spectrum are very thin, covering a wavelength region on the order of 0.001-0.01 nm wide. There are hundreds of these holes in the spectrum of sunlight, even when measured without the additional complicating effects of Earth's atmosphere. Of course, this doesn't stop sunlight from looking white, as you suspected.
To make this all seem a little less esoteric, there is a rough but rather interesting experiment which can be performed, as shown in this Youtube video. A crude gas of sodium atoms can be created simply by adding table salt to a flame. When illuminated with white light, there is nothing out of the ordinary. However, if illuminated with a sodium lamp, suddenly the fire becomes dark, almost black. What's happening is that the sodium gas in the flame is absorbing photons of exactly 589.158 nm and 589.756 nm wavelength, and if your light source contains those photons almost exclusively, then you can clearly notice their absence. But importantly, the flame was always absorbing these photons - it's just that with a white light you couldn't really see the effect, it made virtually no difference to the overall white light spectrum.
You might now be wondering whether "white" sunlight can truly be considered "white", or whether there is a "whiter" light source. In principle, you could generate an almost perfect Planck-distributed blackbody spectrum with the same colour temperature as the surface of the Sun (~5700 K) and you could imagine comparing the perceived colour of this blackbody spectrum and actual sunlight. These will almost certainly be indistinguishable to the human eye.

@EdV Seems like we had the same idea, I was just slightly slower at answering! But I agree, the answers are complementary.

@NicolauSakerNeto, Beautiful experiment.

Thank you for your answer! I understand the "narrow gaps" in the solar spectrum and other spectra produced by atomic absorption. But it is still a bit unclear to me, WHY molecular absorption has a broad absorption spectrum.


text:
How can you explain the pKa of the first protolysis of phosphoric acid without using resonance
The pKa1 of phosphoric acid is 2.14.
As far as I am concerned, phosphoric acid does not have any double bounds and, as a main-group element, obliges to the octet-rule. This does create formal charges but these are preferred over breaking the octet-rule. This is further explained in this article.
This begs the question though: How can one explain the low pKa.
Resonance seems to be off the table due to the fact that there cannot be any resonance if there is no conjucated system.
As far as I am concerned, expanded octets are incorrect and cannot be applied here - if they can even be applied anywhere since I believe this theory is somewhat outdated.
Inductive effects seem like they could be of use here since there are indeed 3 OH-groups and one O (the one with the negative-formal charge). But I believe that this should not be able to explain this pKa for two reasons.
First of all, all of these groups should theoretically draw electron-density to themselves. This would not make the bond between the H and O more unstable and thus would not result in a higher chance of dissosiation of the proton.
Furthermore, when comparing the pKa from the first protolysis of phosphoric acid to the one from ethanol (pKa=15.9) the differnece seems to be far too great to be due to an inductive effect.
Continuing this thought though, I have had the idea that maybe this relativly strong pull of electon density - due to the number of groups with a negative inductive effect -could stabilise a negative charge, thus making the first protolysis more appealing. A similar thought, but this time with the polarizability of the phopshorusatom in mind, is explored in the following.
Assume one or more protons have already been taken away, leaving behind a negative charge. The elektostatic repulsion of the bonding elektrons (and maybe others aswell) from this negative charge would push the electron-density towards the comparibly high polarisable phosphorusatom. This in turn could lead to a somewhat stabilised negative charge. Now I know that there is no "real" delocalisation to apply here - this is just a thought I have had.
I have also thought about something similar to the erlenmyer rule. My theory was that due to the multiple OH-groups on the same phosphorus there might be a similar reactivity as with carbonatoms with multiple OH-groups. But as far as the feedback I have recieved from discussing this posibility, this seems like a far fetched assumption and I have to agree.
The last idea has something to do with constitutional isomers (more specific:tautomers). H-Atoms are know for their behaviour to swap places. If, for example, one proton has already been taken away, the remaining H-Atoms could swap their place to the negative charge because of the electrostatic attraction. This would lead to a somewhat delocalised proton (or multiple protons) leaving the the negative charge to be also delocalised making it harder for it to accept a different proton (for example from water). (I believe the term "delocalisation" is misplaced here since there is no conjucated system and thus no resonance but I could not find a better way to describe it)
As you can see I am very confused but intrigued by this topic. Most of the sources I have looked at and professors + colleagues I have talked to, draw the phosphoric acid with a double-bounded O-Atom (oxo) and explain its pKa with resonance. This makes it incredibly frustrating to figure anything out about this. This is why I am searching for help here and sharing all of my thoughts - no matter how wrong some of them sound.
I hope somebody can help me - and others with a similar question -out.
Thank you

If you write a single or double bond there doesn't matter. Both oxygens are equivalent and charge is delocalised.

related https://chemistry.stackexchange.com/questions/13796/nitric-acid-vs-chloric-acid-strength

But what about the octet? Shouldn't the ocetet-rule be valued over creating fromal charges? There are only four available AOs for phosphorus so it cannot have 5 bonds. The phosphoric acid is also not hypervalent as you can see here. I may be missing something but because of the non existent 5th bond I cannot comprehend there being any resonance. I am well aware that even the octet is only a law made up my us and thus doesn't depict the nature perfectly.

Think if dative bond to oxygen is any different then P-O- created in dissociation. You don't even need resonance.

@Mithoron I am unsure if I unterstand you correctly but I think you want to mention the difference between the coordinate bond (P-O) to the product of dissociation (P-O^-)? Because of the negative charge at the oxygen the dissociation of the next H is less likely. I do get the differing pKa's between phosphoric acid/dihyrogen-phosphate/hydrogen-phosphate. The problem I am having is explaining the first pKa in comparison to something like ethanol - the pka is way to low for a OH-group so there has to be something that explains it. If you meant something different please comment...Thanks

The answer there^ shows my point.


text:
Multibranched fluorinated thiol IUPAC name
I've been reading some papers about using self-assembled monolayers as hydrophobic coatings and came across this molecule:

Throughout the entire article its simply referred to as "multibranched fluorinated thiol" and I was wondering if anyone knew what the proper IUPAC name for it might be? Here is the DOI to the journal article in case anyone is interested in it:
10.1021/acssuschemeng.8b00777

What IUPAC name, PIN?

You have already correctly identified the principal characteristic group of the compound as thiol. Thus, the parent structure is propane-1-thiol.
Next you can add the neopentyl ether group and get 3-(2,2-dimethylpropoxy)propane-1-thiol.
The fluorinated tert-butyl groups are named 1,1,1,3,3,3-hexafluoro-2-(trifluoromethyl)propan-2-yl. Including the ether group, these substituents are [1,1,1,3,3,3-hexafluoro-2-(trifluoromethyl)propan-2-yl]oxy.
Adding three of these substituents gives the complete name 3-[3-{[1,1,1,3,3,3-hexafluoro-2-(trifluoromethyl)propan-2-yl]oxy}-2,2-bis({[1,1,1,3,3,3-hexafluoro-2-(trifluoromethyl)propan-2-yl]oxy}methyl)propoxy]propane-1-thiol.

That's a hell of a mouthful as a systematic name. Surely with something this symmetric there must be some neat simple but non-systematic name? It could be seen as a derivative of pentaerythritol, surely? Please. ;-)

My organic tutor would have said something like "Just call it Fred's compound". Change Fred as appropriate.

Thank you @Loong, that really is quite the mouthful! I can see why they decided to go with multibranched fluorinated thiol instead of giving the full name.


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How to visualize an endothermic/exothermic reaction?
Let’s say X and Y bond together to create XY, where this reaction is exothermic. What I’m visualizing in my head is X and Y floating about with certain speeds, which contributes to their kinetic energy. But when they bind together, the kinetic energy of the X-Y system decreases, because their bonding is a perfectly inelastic collision.
But where does the missing energy go? Does it go to surrounding atoms, making them speed up faster? And how does it do that?—-I want an explanation on the level of moving subatomic particles and subatomic forces like Coloumb’s law. What forces cause the surrounding particles to move at a faster rate? And where does this force originate from?
According to this logic, all synthesis reactions should be exothermic, but this is obviously not the case. How can I visualize a synthesis reaction that is endothermic?
Or is this the wrong way to think about it?

Looks like a duplicate of https://chemistry.stackexchange.com/questions/13536/is-bond-formation-strictly-exothermic https://chemistry.stackexchange.com/questions/119093/how-does-the-energy-released-during-a-bond-formation-typically-manifest-itself-o

And possibly a few other - too broad.

@Mithoron not exactly, my primary request is a visualization of how a bond can affect the speed of neighboring particles, my discussion about inelastic collisions is merely to elucidate what type of answer i want.

In physics, the concept of energy can be derived from the concept of force, and vice versa. So we should be able to explain everything we can explain in terms of energy in terms of force. Thus there should be a “force” explanation of how bonds can affect the speed of neighboring particles

Seems you didn't check the second link... also links which are there https://chemistry.stackexchange.com/questions/143335/why-is-a-third-body-needed-in-the-recombination-of-two-hydrogen-atoms?noredirect=1&lq=1 https://chemistry.stackexchange.com/questions/55914/how-is-heat-physically-released-in-an-exothermic-process

With interest for subatomic particles, energies/forces, your question likely is better suited for physics.se than chemistry.se.


text:
What is the product when alcohol is added to carbonyl compounds?
How do we know if, on adding alcohol to carbonyl compound:

nucleophilic addition reaction will take place

or

alcohol will take alpha hydrogen to form a carbanion and consequent
aldol condensation will take place. Here B is alcohol (e.g. $\ce{EtOH}$)

Note the conditions and what other reagents are added - they are very important.

"Here B is alcohol (e.g. $\ce{EtOH}$)" Not exactly. What is it instead? The reaction mechanisms are totally different.

B is an alcohol. see the first step of aldol condensation.

@aaravm  Incorrect! B is an alkoxide - big difference.

oh yeah i hadnt realised that reagents were different. sorry!

You should have been more concerned with comparing ketal formation (1) with an acid-catalyzed aldol condensation.

@aaravm if you got your answer, feel free to make a self answer.

As already pointed out in the comments, the second reaction reagent is not exactly an alcohol but an alkoxide. For example, $\ce{EtONa}$ and not $\ce{EtOH}$.
This makes the reaction condition for the second reaction basic whereas the first reaction is taking place in acidic conditions.


text:
Confusion regarding continuity of state
I was reading about Andrew's isotherms and about the continuity of state.
What confused me was the continuous conversion of gaseous state to liquid state.
Now my textbook says that by keeping volume constant and increasing pressure and then keeping pressure constant and letting the gas cool we are able to convert the substance from gaseous to liquid state.
Now what confused me is this sentence in my textbook - "Thus  in the absence of the surface of discontinuity, there is no way of distinguishing between liquid and gas".
What does this mean?
Also, how is this conversion continuous, as along line LD wouldn't the gas convert to liquid in such a way that both states coexist? How is it possible that this conversion is continuous?

Ok so in ALD there is a point (critical point) after which the entire phase is liquid and the quote from my textbook basically means that we don't know where exactly this critical point lies. Is that what you are saying?

Do you consider air liquid?

The good initial step before asking is reading relevant textbooks and googling  keywords and studying the search results.
This prevents cases of asking for answers that have been already written in many ways on many places. The search scope can be eventually limited to the particular address domain by adding the additional term site:, where the  can be e.g.  stackexchange.com,  libretexts.org or wikipedia.org

It should be clear to you even just from the chart. Think of drawing the third p=f(V) curve above the upper one, for higher temperature. There would be no edges, pressure would continuously increase as volume would decrease. The parabolic-like curve tells you properties of liquid and gaseous phase converge to each other for raising temperature. See also Critical point.
Imagine a circular slope platform closing the loop by a single step. If you go upwards by stepping up, can you return without stepping down? A small child would tell you that you can. Being down=liquid, being up=gas.
When you look at a compound phase diagram, than the neighborhood of the critical point is like that slope and step platform. You can make liquid to evaporate many times and never condensate. Or vice versa, if you reverse the path.
When gas or liquid (under enough pressure) are heated up above critical temperature, they become gas-like or liquid-like supercritical fluids, than can gradually become each other with changing pressure, without phase change or phase boundary. It is like going along the slope instead of making a step.
To comment the ALD transition:
Along AL, when passing the upper curve, gas becomes gas-like supercritical fluid without a phase change.
Along LD, when passing the upper curve back, liquid-like supercritical fluid becomes liquid without a phase change.

Thanks i was really confused about the ALD transition. I did read some standard texts I found but none of them addressed the ALD transition.

It was rather just like applying the general principle of criticality to this particular scenario.


text:
Can endo and exo products occur when the diene is not cyclic?
I have always thought that the diene needs to be cyclic for there to be endo and exo products. Perhaps, I formed this impression because most textbook examples that are used to illustrate the concept of the endo preference in Diels-Alder reactions involve a cyclic diene. Recently, I came across this website and if you look at Point 7, Example 1:

Firstly, as pointed out by someone in the comments section on that webpage, the use of the terms "endo" and "exo" are incorrect, as the products are not bridged. Thus, according to IUPAC rules, the terms cannot be applied and instead, "endo-like" and "exo-like" may be used. However, this is not my point of contention.
Perhaps, I simply lack the ability to visualise this but can someone explain to me how the aldehyde substituent would be considered to be nearer to the double bond in the stated "endo" product, compared to the stated "exo" product.
In general, I would also like to ask if there exists an endo preference in Diels-Alder reactions where the diene is not cyclic.

Take a look here. Cyclic or acyclic diene, it doesn't matter.  In general, endo is kinetically favored.


text:
Naftiridine hydrogenation
I was reading an article about the synthesis of 1,2,3,4 tetrahydronaphtyridine and I found this image.
I don't understand why the major product of hydrogenation is the less substituted and not the most. I thought that with an alkyl chains there was more negative density on the ring and so hydrogenations was most likely to happen in that ring.

Welcome to ChemSE. How major is the major isomer? What is "negative density"? The unsubstituted ring is less hindered favoring absorption on the catalyst surface. The alkyl group provides a steric factor and likely some hyperconjugation stabilization.

Another interesting question is the distribution of 3 and its isomer different in acid vs base.? I'd expect more 3 in base than acid.

It would be helpful if you provided a reference to the synthesis and a description of the conditions used for the hydrogenation step


text:
Will strong base corrode aluminium in absence of oxygen?
From what i understand, putting iron in a vacuum chamber (airtight container) will stops further oxidation take place.
However in aluminium, with only oxygen, aluminium will form thin oxide film that will stops further oxidation take place.
I wonder if aluminium can be put into vacuum chamber to prevent corrosion take place, even if submerged in strong acid/base solution?
My assumption it won't, because of oxygen and acid/base is different oxidation agent
But i could be wrong

Vacuum and water solutions do not like each other.

Oxygen is irrelevant to the reaction between aluminium and strong bases. It is relevant to passive protection of aluminium but bases corrode away the passivating layer. Consider, for example, the process of making Raney Nickel. A NiAl alloy is reacted with a strong base like sodium hydroxide which dissolves the Al leaving very finely divided, high surface area nickel. In labs this is usually done with the exclusion of air because the product is pyrophoric. But the Al still dissolves.

Approximately 0.01 hour of searching, including the time taken to type in the search words, reveals this on Wikipedia:

In hot concentrated hydrochloric acid, aluminium reacts with water with evolution of hydrogen, and in aqueous sodium hydroxide or potassium hydroxide at room temperature to form aluminates—protective passivation under these conditions is negligible.[1]

So water with a strong nonoxidizing acid or base, but otherwise isolated, will corrode aluminum.
Cited Reference
1.
Beal, Roy E. (1999). Engine Coolant Testing : Fourth Volume. ASTM International. p. 90. ISBN 978-0-8031-2610-7. Archived from the original on 24 April 2016.

Do you mean the potassium hydroxide will still corrode aluminum, even in absence of oxygen? Pardon for my limited english, your sentence structure a bit difficult for me to understand clearly

Water with strong acid or base does not need oxygen to corrode aluminum.

If that's the case I wonder if there is any way to stop further corrosion of aluminum when it's dipped in strong acid/base solution?

@digda  the way to stop corrosion of aluminium by strong acid/base is to not dip it in the solution

OK, a reference noting: "protective passivation under these conditions is negligible.[1]", which supports  the term "dissolving", and not "corroding".

Actually, the question is correct in citing the seeming "corroding" and not dissolving of "Aluminum" in the context of a strong base, like Lime (Calcium hydroxide), which is apparently commonly attacking Aluminum alloy structures, for example, from the release of CaO dust from cement plants.
For example, see this question, to quote:

I have a question, when our aluminum tank coverings for upright storage tanks were exposed to Lime from a cement plant they all turned black. What is the reaction here and what if anything can we do to protect the tank coverings?

Apparently, the net reaction is given by:

$\ce{Ca(OH)2 + Al(s) + 8 H2O = Ca3Al2((OH)4)3 + 6 H2 }$

where $\ce{Ca3Al2((OH)4))3}$ is known as Katoite, which per a mineral source, to quote:

This progressive exchange can also be represented as Ca3Al2(SiO4)3-x(OH)4x. In end-member katoite, the replacement is complete (so x = 3).

Wikipedia further notes:

Hydrogrossular is translucent to opaque, and found in green to bluish green, pink, white, and gray. The cause of the green color is chromium, and possibly iron. Pink hydrogrossular is caused by the presence of manganese. Hydrogrossular may have dark gray to black small inclusions.[1].

The "Aluminum" as referenced here is likely an alloy with several elements, to quote Wikipedia:

The typical alloying elements are copper, magnesium, manganese, silicon, tin and zinc.

So, to answer the question a seeming corrosion with a dark coating is possible and chemically not dependent on the presence of oxygen.


text:
Is air made 100% from atoms?
I know air is made up of roughly 78% nitrogen, 20% oxygen, etc.. but I would like to know what percentage of air is just space and has no atoms at all.

You can try to estimate the "length" of empty space among molecules by a quantity called mean free path. Look at the table there, https://en.wikipedia.org/wiki/Mean_free_path#Kinetic_theory_of_gases
On the other hand this is a philosophical question as. Is the space between atoms and molecule empty? Is the space really empty?

I once watched a VSauce video where he mentioned that space has only 10 atoms per cubic cm and interstellar space has 1 atom per cubic cm. In comparison, the best vacuum achieved in laboratory is ultra high vacuum (UHV) where there are 100 atoms per cubic cm.

I see the clash of 2 meanings of air. Air(1) as a particular mixture of gaseous components, surrounding the Earth. Air(2) as the space occupied by Air(1).  Air(2) = vacuum + Air(1).  Comparing densities of liquid and gaseous air is trivial.

@M.Farooq atoms and molecules are quantum objects so defining a mean free path may be a little bit tricky.

@JunSeo-He, Yes, still mean free path is very well used concept.

I don't think this is a homework question. It is a naive question but not entirely trivial or inappropriate here. Not sure it should be closed.

I agree, this is not a homework question.

One way to answer this is to compare nitrogen gas at standard pressure and temperature to liquid nitrogen at 1 atm. The density ratio is about 1500. In the liquid, space is nearly filled by the N2 molecules, so it follows that the molecules in the gas only occupy 1/1500 of the total volume. Also, since the cube root of 1500 is about 12, it must be true that the typical distance between molecules is about 12 times the mean molecular diameter.

Good answer and a cool user name!

Air is mostly molecules not atoms and most of the space in a gas is empty
Most of the gaseous components of air are not individual atoms but molecules (oxygen, nitrogen are diatomic, carbon dioxide and water vapour are triatomic). The only common single atom is argon (about 1% of the atmosphere by mole ratio).
But this is mostly irrelevant to the underlying question. Gases occupy far more space than the molecular size of the components of the gas. To a good approximation, all gases occupy just over 22 L of space at standard temperature and pressure whatever the molecule is. This is far larger (by a factor of 1000-2000 times the volume taken up by the molecules or atoms making up the gas if we use the normal size of the molecules (as 10ppb points out in their answer).
So most of the space in a gas is not "occupied" by anything at a given snapshot in time. But, remember, gas molecules move around quickly which is why they "fill" a space so a view at one moment in time is a misleading way to think about the issue.

Molecules are made up of atoms, aren't they?

@Barmar Yes, but when working out the volume occupied by a gas, you count molecules not atoms. The distinction matters.


text:
Hydride shift or Methyl shift

For the carbocation (A) given in the reaction, first we can rearrange it by ring expansion (4 to 5) then we have two choices either H-shift or methyl-shift. My teacher told me that Hydrogen is the best migrator so according to me product (D) should form but in my textbook (C) is given as final product ?? Where am I wrong ? (or is my textbook wrong ?)
Also I have been taught that whenever there is a choice among H-shift, alkyl-shift or phenyl-shift we prefer H-shift since H is the best migrator. So why methyl-shift is happening here ? Please explain.
What's the concept I am missing ??

C is tertiary carbocation whereas D is a secondary carbocation.

Your question is an example of Cationic Rearrangements. When a reaction is set to progress under correct condition for thermodynamic control, the final product is based on the thermodynamic stability of intermediates. The migratory aptitude matters if the intermediate stability increased by the migration (see the examples given in here).
For instance, in $\bf{A} \rightarrow \bf{B}$ rearrangement, there are two migratory choices: hydride shift or alkide shift. Although hydride is superior to alkide in migratory aptitude, why is alkide shift dominating here to give the ring expansion? Because it gives the most stable intermediate releasing ring strain energy (5-menbered ring versus 4-membered ring). Now, in intermediate $\bf{B}$ rearrangement to $\bf{C}$ (by methide shift) or $\bf{D}$ (by hydride shift), it has two choices as indicated. Again, although hydride is much superior to methide in migratory aptitude, it prefers methide shift because it would give more stable tertiary carbocation instead of energetically similar secondary carbocation by hydride shift.

This is essentially a question of thermodynamic control versus kinetic control. The statement "my teacher told me that Hydrogen is the best migrator" has to do with the migration barrier, which is related to kinetic control. Since H atom is very light, its vibrational frequency can be high, which may facilitate the migration. Being a light atom, sometimes H can even display wave character and therefore can undergo quantum tunneling without needing to fully climb up the reaction barrier. However, the yield of final products is mainly determined by the stability of the products, which is related to thermodynamic control. As long as you let the reaction take place for sufficient length of time, you will get the more stable structure as the main product, unless the barrier connecting it with the reactant is prohibitively high at the given reaction temperature.

Image is taken from http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch10/ch10-3-3.html

As long as C and D are not interconvertible under the reaction conditions, then both C and D are formed under kinetic control. If C and D are in equilibrium with each other, perhaps through reprotonation of the olefin related to D, then the reaction is under thermodynamic control. A thermodynamic product may be formed under kinetic control even though it is not being formed under thermodynamic conditions. Base elimination of 2-bromobutane to give the Zaitsev product (2-butene)  is a case in point. 1-Butene and 2-butene are not in equilibrium under the reaction conditions.

Good point! But I noticed a difference b/w the base elimination of 2-bromobutane and the cationic rearrangements. Once HBr is eliminated, it is dissipated into the solvent, making the equilibrium between 1- and 2-buene unlikely. But cationic rearrangement is different in that the molecule does not lose any atoms. Not sure if a direct conversion between C and D by [1,3] H-shift is possible (probably too far). However, indirect conversion between C and D, e.g. through B, will lead to an equilibrium between C and D. Consequently, the reaction shown above most likely is thermodynamic control.

Under basic conditions there is no HBr because it is neutralized. In the solvolysis (SN1) of 2-bromobutane, the alkenes formed are under thermodynamic control owing to the presence of HBr. See this. @Vishal Anand

That makes sense (under basic conditions there is no HBr). It may be the reason that the less stable 1-butene can be formed by kinetic control. Good to know that the presence of HBr can lead to thermodynamic control, despite the fact that it leaves the molecule during elimination. (It may be a different situation in case H2 is formed but pumped away by vacuum.) In any case, the cationic rearrangement shown above is most likely thermodynamic control.

Yours textbook answer is correct. We can't just say that hydrogen is a best migrator. Shifting of groups depends on the stability of transition state and final product. In this example, you can observe that carbocation in C is a tertiary carbon with 6 hyperconjugative structures whereas in D carbocation is a secondary carbon with only 4 hyperconjugative structures. Therefore, carbocation in C is more stable than in D. Hence compound C is the major intermediate.

Carbonium ion rearrangement results in a more stable carbonium ion and probably happens in the actual formation of the carbonium ion not in a time delayed subsequent reaction. If this is the case there is little sense in first looking at the migrating moiety but to first look for the most stable carbonium ion then second to look for the better migrating group to give that carbonium ion. In your reaction the ring opening is probably driven by loss of ring strain with an aid from the formation of the tertiary carbonium ion by methyl migration. If there is formation of the secondary carbonium ion there is no energetic reason for rearrangement to another secondary but there is for rearrangement to the tertiary.  These are my impressions on these reactions based on some limited experience and some long almost forgotten book learning.
I found your reaction. It seems the secondary cation is formed blowing my one step proposal out of the water. https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/rearrang.htm
Time to check the original data.


text:
How to calculate a citric acid and trisodium citrate buffer?
I am trying to calculate how to prepare 500 mL of a citric acid (H3Cit) and trisodium citrate (Na3Cit) buffer with a pH = 4.5 and a total concentration of 0.1 M. I know that it is necessary to use the Henderson Hasselbalch equation but I don't quite understand how to perform the calculation, taking into account that at that pH there are several species of the citrate anion present and that citric acid first dissociates into citrate anion in the form H2Cit(1-). I found in this forum an equation, which I think I could use:
How do I calculate the amount of polyprotic acid and base needed to achieve a certain buffer pH?
However I don't understand how to get that equation. Can someone explain to me, please? Thank you

See here: https://chemistry.stackexchange.com/a/116226/79678. From the alpha fractions in my answer at the link, you see that at pH 4.5 you have the hydrogen citrate ion and the dihydrogen citrate ion. Citric acid and citrate ion are negligible at that pH. So one citric acid plus one citrate ion yields one each of the two other species.

Is it a computation exercise, or a practical task? If the latter, why not to use e.g  Sigma-Aldrich citric acid pH buffer preparations table ?


text:
Why is Gallium-69 stable, Gallium-70 unstable and Gallium-71 stable again?
Why is gallium 70 suddenly unstable, whereas it becomes stable again when you add another neutron?
Shouldn't isotopes become unstable when there are either too few or too many neutrons? Why is gallium then unstable again in the middle?

You may want to read https://en.wikipedia.org/wiki/Even_and_odd_atomic_nuclei.

Related: Why is tin-112 stable, but indium-112 radioactive?

@Poutnik - well, that "rule" is a result of nuclear physics as you note. It isn't like we imposed the rule, and 70Ga was a bit late to the party so was the odd isotope out once 69Ga and 71Ga took the spots...

@JonCuster Well, even if it were not odd-proton number element, 70Zn and 70Ge took the place already. // For most of chemists, the neighbors rule and the max 2 for odd ones rule are taken as they are, regardless of if they are results of nuclear physics reasoning or not. Like they were empirical rules that may or may not have theoretical background.

@Poutnik - indeed, as can  be seen from many questions here, chemistry "rules" range from pretty fundamental to suggestions to vague hints. Early chemistry courses should note that isotopes exist, and move on quickly.

There is the general neutron/proton ratio rule, regarding stability of nuclides, described by the Valley of stability.
Additionally, in beta decay context, there are 3 easily remembered rules, based on reasoning of nuclear physics:

Elements with the odd proton number have maximally 2 stable nuclides. (Some have just one, technetium and promethium have none.)
If there are two isobars (nuclides with the same nucleon number) of elements that differ by 1 proton (and 1 neutron) then maximally one of them is stable.
The least stable are nuclides with both odd numbers of protons and neutrons. More stable are odd-even nuclides and the most stable are even-even nuclides. The reason is the nucleon pairing stabilization, increasing bonding energy per nucleon.

The only stable odd-odd nuclides are $\ce{^2_1H}$, $\ce{^6_3Li}$, $\ce{^{10}_5B}$,
$\ce{^{14}_7N}$,$\ce{ ^{180m}_{73}Ta}$.

For very light elements, the n/p ratio based stability overrules parity based stability. As the valley of stability is initially very narrow. The deviation from the ideal ratio causes a higher nucleus energy raise than would be the energy release by reaching the parity.
For tantalum, there is the interesting fact the stable one is the naturally occurring meta-stable nuclear isomer, while the lower energy isomer is unstable.

Back to gallium-70:

Gallium is the odd-proton element and there are already stable gallium-69 and gallium-71.
Gallium-70 has an odd-odd nucleus, generally predicted to be unstable.
Both even-even neighbor isobars are stable - zinc-70 and germanium-70.

Gallium-70 has therefore 3 reasons why it is not stable, beta decaying to both zinc (electron capture) and germanium (electron emission).

Care to add a comment about the pathological stability of $N_{14}$?

@PrimeMover  Why pathological? By chance, just seconds ago I added the list of 5 known stable odd-odd nuclides. For very light elements, n/p ratio based stability overrules parity based stability.

P.S.: You must mean $\ce{^{14}N}$.

Wasn't there when I posted it. Whatever the notation is, nobody really cares. It's pathological because it's one of those exceptions.

@PrimeMover That does not make it pathological. The least stable does not mean unstable, especially if parity is not the only contribution to stability. // Chemists care, as $\ce{N_{14}}$ means a nonexistent nitrogen molecule with 14 atoms.

wotever........