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Does entropy contribute work?
I've seen that Gibbs free energy is defined as the maximum amount of work that a system can do. From this I gather, this is how much energy it will release - right?
But, I've also learnt that the heat released when I burnt ethanol was due to the enthalpy of the reaction.
Given that the formula for Gibbs free energy is $\Delta G=\Delta H - T\Delta S$, it is quite clear that a change in enthalpy is not a change in Gibbs free energy. Rather, the entropy seems to also be able to contribute work.
If the total energy released by a reaction is greater than just the heat energy released as enthalpy, what form does the energy released by entropy take? Is there an intuitive way to understand where this energy is/goes?

Heat. You feel heat in the form of kinetic energy of air around you.

I've seen that Gibbs free energy is defined as the maximum amount of work that a system can do. From this I gather, this is how much energy it will release - right?

No, this conclusion is incorrect, the Gibbs energy is not how much energy a process releases. A system can exchange energy with the surroundings in the form of heat or work. The Gibbs energy is a measure of the maximum non-PV work a system can do (if the Gibbs energy is negative), or the minimum non-PV work that has to be done on the system to make the process go forward (if the Gibbs energy is positive). The Gibbs energy is practical to use for constant pressure processes (where we know what the PV-work is for a given process).

But, I've also learnt that the heat released when I burnt ethanol was due to the enthalpy of the reaction.

That is true for a constant-pressure process in the absence of non-PV work. In the presence of non-PV work, you would have to add that to the heat exchanged (with the proper conventions for the +/- signs of heat and work).

Given that the formula for Gibbs free energy is $\Delta G=\Delta H - T\Delta S$, it is quite clear that a change in enthalpy is not a change in Gibbs free energy. Rather, the entropy seems to also be able to contribute work.

Both enthalpy and entropy contributions could have positive or negative values. So it might contribute to the work the system can do, or the work you have to do on the system to make the process go in the forward direction.

If the total energy released by a reaction is greater than just the heat energy released as enthalpy, what form does the energy released by entropy take?

There is no energy "released by entropy". The entropy portion determines how much choice there is of running a process with respect to work vs heat. In some processes, the work you can extract is larger than the enthalpy provided by the process (the additional energy comes from cooling down the surroundings).
One simple example of having a choice how to run a process is discharging a battery (i.e. an electrochemical reaction). You can short the battery (in the system), giving off heat to the surrounding. Or you can have some wires into the surrounding, running a motor. In this case you let the reaction do some work (there will probably also be some heat released).

Is there an intuitive way to understand where this energy is/goes?

I have not found an intuitive way to understand any of thermodynamics. Its beauty is the high level of abstraction, which might make it one of the less intuitive topics. I get surprised everytime I set it aside and then think about it again. One way to deal with this is to have a couple of examples you are comfortable with and where you know the proper description of what is going on. Then, you can return to these as you are looking at a situation that is new to you (or one you have forgotten about).

[from the comments] Can you give a concrete example of this for a non-electrochemical cell? I've only been able to understand this in terms of electrochemical reactions. How about say, combustion?

For a typical combustion reaction (e.g. burning some olive oil in an oil lamp), there is no work, just heat. However, our bodies manage to capture the free energy (e.g. heart using oxidation of fats as source of Gibbs energy to beat) by running the same reaction in steps near equilibrium. So again, in one case the chemical reaction produces heat only, and in the other case it generates a mix of work and heat. For a combustion reaction, it is possible to capture some of the thermal energy in a steam engine; however, you need some cool water that gets heated up as well, so you can't transform 100% of the thermal energy into work.

See also: https://chemistry.stackexchange.com/q/49145/72973 and https://chemistry.stackexchange.com/q/112900/72973

that's interesting. Thank you. My question is, when you say that we can "extract" the energy, what form is this energy in. Can you give a concrete example of this for a non-electrochemical cell? I've only been able to understand this in terms of electrochemical reactions. How about say, combustion?

@JohnHon For a typical combustion reaction (e.g. burning some olive oil in an oil lamp), there is no work, just heat. However, our bodies manage to capture the free energy (e.g. heart using oxidation of fats as source of Gibbs energy to beat). So again, in one case the chemical reaction produces heat only, and in the other case it generates a mix of work and heat. For a combustion reaction, it is possible to capture some of the thermal energy in a steam engine; however, you need some cool water that gets heated up as well, so you can't transform all the thermal energy into work.

your last comment on the cells absorbing the work into the other chemicals was brilliant - totally helped me understand. Thank you!

@KarstenTheis Oops, sorry, I meant to open up my answer for editing and accidentally opened yours--sorry for the two trivial edits!

@theorist Those are good edits - once I post the answer, it is not really mine anymore, so you are welcome to edit, even accidentally.

Suppose you have an ideal gas in a container in mechanical and thermal equilibrium with its surroundings (same $T$ and $p$—say, 298 K and 1 atm).  It can't do any work on the surroundings.
Now compress it isothermally to, say, 100 atm.  Now it certainly can do expansion work on the surroundings.  What change in thermodynamic state function accounts for its increased ability to do work?
Well, since the internal energy, $U$, of a sample of ideal gas depends solely on its temperature, its internal energy hasn't changed.  So the additional capacity to do work has not resulted from an increase in $U$.
But its free energy is much higher. Since $\Delta U = 0$  for that isothermal compression, the increase in the system's free energy—and the corresponding increase in the its ability to do work—has resulted solely from a decrease in its entropy.
So yes, a change in entropy can be used to do work.
It's tricky to talk formally about free energies for the above system, since while it's constant $T$, it's neither constant $p$ nor constant $V$.  Rather it's constant $p \times V$.
So let's consider some more easily characterized systems (ideal gases at constant $T$ and $V$ or constant $T$ and $p$):
In a closed system at constant $T$ and $V$, the maximum total work the system can do, which is the reversible work, is given by $\Delta A$, the Helmholtz free energy.  Under those conditions, $\Delta A=\Delta U - T\Delta S$.  If $\Delta U = 0$, the magnitude of the maximum work is given by $T\Delta S$.
And in a closed system at constant $T$ and $p$, the maximum total non-$pV$ work the system can do, which is the reversible non-$pV$ work, is given by $\Delta G$, the Gibbs free energy.  Under those conditions, $\Delta G = \Delta H - T\Delta S$.  If $\Delta H= 0$, the magnitude of the maximum non-$pV$ work is given by $T\Delta S$.
So you can see we can formally connect the work a system can do with the entropy change. This shouldn't be surprising, because nothing happens to a macroscopic system unless it's associated with an increase in the entropy of the universe.  That's the 2nd law of thermodynamics, which provides an arrow of time for the universe.  Thus anytime work is done anywhere, it's always ultimately driven by an increase in entropy (not of the system, but of the universe).
An increase in the entropy of the universe is the driving force behind every macroscopic process.
At constant $T$ and $p$, the only reason a decrease in enthalpy  (release of thermal energy) corresponds to the ability of a system to undergo a change is because it causes an increase in the entropy of the surroundings: $\Delta H_{sys} = -T\Delta S_{surr}$.  Thus ultimately the abilty to undergo any process, including those in which the system does work, isn't about the change in the energy of the system.  It's about the change in the entropy of the universe.
Let's make this more general (i.e., not restrict our discussion to ideal gases):
In a closed system at at constant $T$, $V$, both of the following hold (I'm taking positive $đw$ to be work done on the system):
$$dA = -T(dS_{sys}+dS_{surr}) = -T dS_{univ}$$
$$dA = đw_{max}$$
Thus:
$$đw_{max} = -T dS_{univ}$$
And in a closed system at at constant $T$,  $p$:
$$dG = -T(dS_{sys}+dS_{surr}) = -T dS_{univ}$$
$$dG = đw_{non-pV, max}$$
[G only gives the non-$pV$-work because, at constant $p$, the $pV$-work is subtracted from G because it's subtracted from the enthalpy: $H = U +pV$; at constant $p$: $dH = dU + pdV = dU - đw_{pV}$.]
Thus:
$$đw_{non-pV,max} = -T dS_{univ}$$
In sum, it's all about the entropy, baby!  The change in the energy or enthalpy of the system is only a bookkeeping surrogate for what really matters—the change in the entropy of the universe (unless you're doing industrial chemistry, and need to know the heat flow for practical reasons, in which case you would want to know $\Delta H$ or $\Delta U$, for processes at constant $p$ or $V$, respectively).

That's a very cool point. However, isn't Gibbs free energy the non expansive work done? If so, then considering say burning fuel, what form does this "entropy energy" take?

If you consider free energy in terms of Helmholtz free energy, then it is okay since the change in pressure doesn't matter.

@JohnHon You're correct, I was being too sloppy.  At constant T and p, the decrease in the Gibbs free energy of a closed system is the theoretical maximum non-pV work that can be done.

@M.L  Yes but then you need to constrain the system to constant volume for a decrease in Helmholtz free energy of the system to correspond to an increase in the entropy of the universe.

@theorist "because nothing happens to a macroscopic system unless it's associated with a decrease in the entropy of the universe." Do you mean an increase in the entropy of the universe?

@GRANZER  You're absolutely right, and thanks for catching that typo!  I've corrected the sentence.

Constant volume is not a constraint on the Helmholtz free energy change being equal to maximum work for a closed system at constant temperature.


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What is a non-organic analog of a metabolic pathway?
The wikipedia page on the Belousov-Zhabotinsky reaction states the following:

The discovery of the phenomenon is credited to Boris Belousov. In 1951, while trying to find the non-organic analog to the Krebs cycle, he noted that in a mix of potassium bromate, cerium(IV) sulfate, malonic acid, and citric acid in dilute sulfuric acid, the ratio of concentration of the cerium(IV) and cerium(III) ions oscillated, causing the colour of the solution to oscillate between a yellow solution and a colorless solution.

Perhaps Boris Belousov didn't really know what he was looking for, but perhaps he did. My thoughts go to network analysis and graph theory to identify equivalence classes induced by graph isomorphism or other 'structure-preserving' maps between two representations of a collection of chemical reactions. Perhaps the structure would include information about which steps were oxidative/reductive or SN1/SN2/E1/etc.
What would have been true of such a non-organic system if Boris Belousov had succeeded?

And my thoughts are he neither knew or cared about "equivalence classes induced by graph isomorphism", but just wanted to make something remotely similar using the stuff he had in the lab ;> BTW too broad questions are closed not moved to meta, where mostly internal site stuff is discussed.

It simply means replacing enzymes with inorganic catalysts. In this case, the goal is to oxidize citrate to CO2, which is sort of what happens in the Krebs cycle. Finding inorganic catalysts for metabolic reactions was a way to approach understanding how the enzymatic catalysis might work

Let me first acknowledge that Belusov published very little, and what he did publish is hard to get a hold of in English, so this answer is not based on any specific writing of his, but rather is speculation based on what other scientists were doing at the time.
In those days, it was quite difficult to study enzyme-catalyzed reactions, because the structures of the active sites and even the identity of most of the relevant residues were generally unknown. Obviously, techniques such as recombinant overproduction systems and targeted mutagenesis were not available.
In this environment, cofactor-dependent enzymes (including metalloenzymes) were of particular interest because it was presumed that in those enzymes, the catalytic function might be primarily provided by the cofacter, and the role of the protein might be as simple as just providing a scaffold for the cofactor and a binding pocket to bring the substrate to it. If that were true, then it might be possible to mimic the enzyme and effect catalysis with a much simpler system consisting of the cofactor bound to a small scaffold or even free in solution. As an example, pyridoxal was found to catalyze a wide variety of reactions by itself in solution, albeit with slower rates and much less specificity than enzyme-bound pyridoxal phosphate.
The benefit of these simplified systems is that it is much easier to probe the reaction mechanisms, though with the caveat that the mechanism might not be the same as when the reaction happens on an enzyme.
In the case of Belusov, it had been observed that a number of the enzymes in the Krebs cycle are metalloenzymes, including the first two reactions from citrate, ie those catalyzed by aconitase (with an iron-sulfur cluster) and isocitrate dehydrogenase (with a Mn(II) ion). It is likely that Belusov was mixing citrate with different metals in the hopes of reproducing one or both of these reactions in a non-enzymatic system with the goal of furthering the understanding of the reaction mechanisms and the role of the metals.

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Optimal voltage for electrolysis
Reading Wikipedia for electrolysis, it provides the information that a potential of 1.23V is required to electrolyze water. How was that calculated? My intuition tells me that it is the sum of the potential of the hydrogen and oxygen ions, but I am afraid that it is wrong. There's the other problem of electrolyzing a solution other than a molten electrolyte, when the potential is definitely different. I also know that a overpotential must be provided to prevent the solution from reaching an equilibrium, but how much exactly is that overpotential?

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Purification of Buchwald-Hartwig amination
I am conducting a synthesis involving Buchwald-Hartwig amination. I use $\ce{Pd2(dba)3}$, $\ce{P(o-tol)3}$, and $\ce{Et3N}$ as base. In many proposed mechanisms, they often omit the precatalyst. I wondered how $\ce{Pd2(dba)3}$ and $\ce{P(o-tol)3}$ are converted to catalyst. I couldn't find exact mechanism, but I found one for $\ce{dppf}$.

In this mechanism, $\ce{dba}$ is replaced by $\ce{dppf}$. Assuming that this mechanism applies equally for my case, $\ce{P(o-tol)3}$, we can assume $\ce{Pd[P(o-tol)3]}$ or something is formed. I did MS analysis for my reaction mixture, and it had a major peak at $m/z=714$, which can be thought to be $\ce{Pd[P(o-tol)3]2}$. If this assumption is correct, then is $\ce{dba}$ mixed in my reaction mixture? Should I consider additional purification for $\ce{dba}$?

There will be free dba in your reaction mixture. What percentage catalyst loading are you using and what purification are you doing? If you are doing column chromatography it will run close to the solvent front in most systems

@Waylander I use toluene as solvent, 3 mol% Pd2(dba)3, 6 mol% P(o-tol)3 (relative to main reagent). I filtrated the reaction mixture through Celite pad, concentrated by rotary evaporator, and dried in vacuum oven at 130 ℃, ~11 Torr. According to Wikipedia, b.p. of dba is 130 ℃. Is dba evaporated? Column chromatography is not performed yet. I have trouble selecting solvent system when I test the TLC.

The dba may have been removed with the Toluene but you still have residues from the phosphine - most likely the oxide.

@KrangLee Please consider using appropriate formatting (mhchem) for chemical formulas (e.g. use your previous edited question as an example). Also, "MS analysis" in general doesn't provide sufficient information. Ideally you need to provide supplementary data regarding type of mass-spectrometer, ion source, positive/negative mode, mass analyzer, type of scanning etc. This might not be crucial for this question, but this is a good practice.

@andselisk Thank you for comment. Actually I am a new graduate school student, still I lack knowledge and experiences about spectroscopy and practical synthesis procedures. Your advice is of help.


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How can 2 molecular orbitals form simultaneously?
I understand (for example in the case of H2) that two s orbitals produce through constructive interference a bonding orbital and that the out of phase combination of two s orbitals leads to destructive interference, followed by the apparition of an antibonding orbital, but, as far as I know, these two orbitals form at the same time (the number of orbitals is conserved, although only the lower energy one is occupied in the first instance). How can this be possible? Are specific regions of the two s orbitals in-phase and others out of phase so as to produce two different outcomes at the same time? Do same-phase regions interact selectively with other same-phase regions or does the s orbital coexist in two phases at the same time, thereby simultaneously producing two different bonding orbitals (having to do with superpositions). Please help me, I am really confused and similar questions contain answers that rely on mathematical proofs - I would like to grasp it conceptually: are the two phases of the s orbital simultaneously occurring?

Hello. I'm afraid that deleting and reposting your question doesn't make it any less of a duplicate of this. You want something "conceptual", but in fact if you read it, the previous OP already asked for something similar. If the answers there don't satisfy you, then you should seek a new answer to that question, not a new question.

I (quite substantially) expanded on my answer there, with an analogy which you might hopefully find more helpful. You might want to check it out again. That's as far as I can help, though; this doesn't affect the fact that this is still a duplicate, so I'm not inclined to reopen, unless there is a convincing argument (and I don't see one now).

I will try to answer without using any mathematical formula, as your question is just made of words, without any mathematical symbols.
You should know that the wave function of an electron in the $\ce{H}$ atom is normalized. It means that the integral of the wave function over the whole space is equal to $1$. And it must be equal to $1$.
If you add two of these wave functions from two $\ce{H}$ atoms, the integral of this sum will be $2$. This is forbidden. It must be $1$.
So you cannot simply add two normalized wave functions, like the $\mathrm{1s}$ in the $\ce{H}$ atom. You have to take one-half of the wave function of the first atom and to add it to one-half of the wave function of the second atom. This would produce a new molecular wave function whose integral is equal to $1$. It is nice. This new molecular wave function describes a sigma bond, designed as $\ce{H-H}$ in the Lewis model.
Now you have to use and combine the other half of these wave functions, without adding them. The only thing to do is to subtract them. But such a subtraction produces a point between the two nuclei where the obtained wave function is equal to zero. As a consequence, this molecular orbital is a antibonding orbital. And the integral of this antibonding is also equal to $1$.
Hopefully the pure theorists will pardon me for having simplified the description of the problem. I should have spoken of the factor $√2/2$ instead of $1/2$.

You basically answered exactly the same question as this. You might wish to consider answering there, instead of, or in addition to, this question.


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Do covalent bonds fit in with the orbital theory of atomic structure?
Which atomic model do covalent bonds adhere to?
In the the covalent bond between $2$ oxygen atoms, we consider each oxygen atom to have $6$ valence electrons, thus resulting in a double covalent bond.
However, the electronic configuration as per the orbital theory is $\mathrm{1s^2 2s^2 2p^4}$, i.e. $4$ valence electrons. In such a case, how can an oxygen atom bond with another atom to form $\ce{O2}$?
I feel there is something simple that I haven't understood yet that explains this.

For compounds one does not use atomic orbitals, one uses molecular orbitals.

The total number of valence electrons does not determine the structure of the molecule $\ce{O2}$. The number of non paired electrons is more important. The Oxygen atom has $4$ electrons $2p$, but $2$ of these $4$ electrons are paired, and so are unavailable to make a bond. It remains $4 - 2 = 2$ free and unpaired $2p$ electrons (unpaired electrons are indicated as points in the Lewis model). These $2$ electrons can produce a valence bond. In the Lewis model, the oxygen atom may be drawn |O:, with one more bar under the O atom (Sorry : I am unable to underline). If two such Oxygen atoms are in contact, the $2$ free electrons from each atoms are producing $2$ valence bonds between them, like in the formula |O=O| (with underlined bar under each O). And this gives each atom a configuration respecting the octet rule.

That is a bit confusing. There are bonds, which we call dative or coordinate bonds, in which the electrons come from one partner. In the end all Valence electrons will be in molecular orbitals, no matter of whether they were paired before or not.

Covalent bonds may be dative or not. They are always made of paired electrons, also called doublets

I find it very odd that you chose dioxygen as an example in your answer, as it is proof that the statement you just made is completely wrong. Even the simplest molecule, with which thousands of people started to learn about quantum chemistry only has one electron.

@Martin. You are absolutely right. The example of $\ce{O2}$ is specially poor. Because $\ce{O2}$ is a rather special case. It has no double bond at all. It is a diradical. I was only trying to answer the initial question asked by Aditi, about the nature of the covalence. But you are right. it would have been much better to choose another molecule, like $\ce{H2O}$ or $\ce{CO2}$


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Where is my non-expansive work in combustion?
Consider if I burn some octane.
The enthalpy released is $5430 \;\text{kJ/mol}$. The Gibbs free energy released is $5537 \;\text{kJ/mol}$.
From my understanding, enthalpy is the heat change in the reaction - it is the heat released in the reaction. Gibbs Free energy is the amount of non-expansive work done. I can accept that out of the $5537 \;\text{kJ}$, there will be $5430 \;\text{kJ/mol}$ of thermal energy output, which is "non-expansive". But how about the other $107 \;\text{kJ/mol}$? Where did this energy go? Where can I see it?
Typically, the example of non-expansive work done is electrical or magnetic work. I can appreciate that, but in this context, since there are no magnets or electric charges moving, where is that non-expansive work?

I think a question underlying this is "how can a combustion reaction be carried out reversibly?"  Does that make sense to you?

It is the maximal work possible. Now if you had an octane fuel cell, you could capture some of this, https://www.sciencedirect.com/science/article/abs/pii/S2211285512000729

Related: https://chemistry.stackexchange.com/questions/154236/does-entropy-contribute-work


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Does the vapor pressure of a substance depend on the presence of other gases?
The Clausius-Clapeyron equation (CC equation) can be used to find the (saturated) vapor pressure of a substance ie. the gas pressure at which the two phases (vapor + liquid or vapor + solid) reach equilibrium. However, what happens when the vapor is mixed together with other gases, such as water vapor mixed together with air? The reason I am asking this is because the derivation of the CC equation in my book relies on the assumption that the two phases are at equal temperature and pressure, whereas in the presence of other gases, the temperature of the vapor and liquid/solid will obviously be equal at equilibrium, but the pressure is not (since the other gases contribute with some pressure as well, and in general the partial pressure of the vapor will be smaller than the total pressure). This same problem arises if you consider the the equilibrium between, say, a solid and a liquid, so long as one of the phases is in a mix (like if the liquid is mixed with other liquids). If the CC equation changes as a result of this, then how does, say, the vapor pressure as a function of temperature change too?

The net pressure changes, but the individual partial pressure of the vapour being looked at remains same, doesn't it?

@TRC Could you check out the comments I made under Chet Miller's answer? The issue is that in the derivation of the CC equation, you get a term involving the change in the pressure of the liquid (more specifically, a term like vdp, where v is the volume per particle of the liquid and dp is the change in pressure), which for most liquids around us should be zero (since the pressure of the liquid should remain at the ordinary atmospheric pressure). This would give a different result for the equation.

The vdp term for the liquid leads to the Poynting correction I referred to in my comment.  For the vapor, even the pure vapor, the ideal gas law leading to the CC equation is not valid, and you need to use VdP for that too.

I'm not familiar with the derivation as of now - I will look it up and try to solve your question.

If the system is at lower pressures where the ideal gas law is close to valid, then the partial pressure of the volatile species (mole fraction times total pressure) can be used in place of the vapor pressure with the Clausius Clapeyron equation.  At higher pressures, the free energy of the liquid is affected by the higher overall pressure, as is the partial molar free energy of the volatile species in the vapor.  So the CC equation cannot be used for these situations.  To quantify such cases, one needs to study and apply the thermodynamics of mixtures.  See Chapters 10 and beyond in Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.

Thanks for the answer, but I was thinking more in terms of how the derivation of the CC equation/relation would change in the presence of other gases. For instance, if you use the derivation that uses the fact that the changes in the chemical potentials of the liquid and vapor are the same, then the change in the pressure of the liquid will appear, which for constant atmospheric pressure would be zero. The overall result is that the relation will only have the volume of the vapor in it instead of the difference in volumes of the vapor and liquid. For a liquid-vapor transition, then maybe there

isn't a huge difference (since the volume of the liquid is so small), but for a liquid-solid transition, that would be a huge difference. Maybe I should have included this in my question, as I was wondering whether the relation would change like this or something.

Like I said, the chemical potential of the species in both phases changes.  In the liquid phase, it changes according to the Poynting relationship, and in the gas phase, the change is much more complicated because of molecular interactions between species in question with its own molecules, but also between its molecules and the other species.  It is not worthwhile for you to speculate about the possibility of minor modifications to the CC relationship until you study multicomponent mixtures.

Yeah, I've thought more about it, and perhaps as a first order approximation I don't really need to worry about the presence of other gases in the air to compute the vapor pressure. I guess I can ignore these gases first to derive the CC equation, and only later do I imagine that these gases come in and "fix" the pressure of the liquid to 1 atm (without changing the partial pressure of the vapor). The corrections you describe seem to be way past my current understanding. Do you think this would be a fine enough way to think about it for now?

Of course.  This is how a mixture of gases behaves in the ideal gas limit.  The chemical potential of a species in the  gas mixture is equal to that of the pure species at the same temperature and partial pressure of the species in the mixture.


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Ethyne reaction with HCN (red hot Fe) and CH2N2 (cold ether solution)
I have tried this and according to me
$\ce{C2H2 + HCN -> C5H5N (pyridine)}$
and
$\ce{C2H2 + CH2N2 -> C3H3N2H (pyrazole)}$
But according to my teacher the case should be vice versa, i.e.
$\ce{C2H2 + CH2N2->pyridine}$
I am not able to understand the reasoning.

These are reactions of acetylene not ethylene

Ethylene is the alkene, $\ce{C2H4}$. Ethyne or acetylene as @Waylander said is the correct name for $\ce{C2H2}$. Also, for formatting chemical equations, refer this - https://chemistry.stackexchange.com/help/notation - it's understandable if you didn't know of it in the first place, but the very least you could do was to capitalize the carbon and hydrogen atoms in your equations. Please be careful in the future. Better framed questions are seen more favourably.

It appears that your teacher is wrong!
The synthesis of pyridine from acetylene and hydrogen cyanide over red hot iron was originally reported by William Ramsay in 1876 wikipedia.
Further examples have been reported JACS paper here, patent with Rhodium and Cobalt catalysis  here
The second reaction is the Pechmann Pyrazole synthesis  article here.

The preparation of pyrazole derivatives with the participation of a 1,3-dipolar cycloaddition between diazomethane (or diazonium salt in some cases) and a molecule with carbon-carbon double or triple bonds with subsequent oxidation (if they are double bonds) of the Pyrazoline is generally known as Pechmann's pyrazole synthesis.


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g-tensor and hyperfine tensor for nitroxide spin label with 15N
Does anyone know where I can find numerical data for the $g$ tensor and hyperfine tensor for a nitroxide spin label where the nitrogen-14 atom has been replaced by a nitrogen-15 atom?
I'm really getting lost in the literature. I come from outside the field of biological EPR so it's very difficult to find the information I need from the plots and the jargon.

Some data here might be of use - it's actually a good paper with a large theoretical component and you can at least get ballpark numbers for $g$, A, and D.

This could certainly be useful for getting familiar with the nomenclature, thank you. However, I don't think it discusses the case of 15N, unless I'm mistaken. Maybe the numbers are not too different from 14N, but I don't know.

I finally found this book by Derek Marsh:
Derek Marsh, "Spin-Label Electron Paramagnetic Resonance" (CRC Press, Boca Raton, 2019). DOI: 10.1201/9780429194634.
It has many tables and the basic theory going with them. I would strongly recommend it.


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the ratio between high spin state and low spin state when they coexist
I am considering the spin state's effect on the charge carrier's lifetime, I was told that high spin and low spin state could coexist when the crystal field splitting energy and the pairing energy are close, but I am not sure how to estimate the ratio between high spin state and low spin state, is there any theory I could adopt to analyze this kind of situation?
Here is some background I find:

whether a complex is high spin or low spin depends on two main factors: the crystal field splitting energy and the pairing energy. The electrons will take the path of least resistance--the path that requires the least amount of energy. If the paring energy is greater than  Δ , then electrons will move to a higher energy orbital because it takes less energy. If the pairing energy is less than  Δ , then the electrons will pair up rather than moving singly to a higher energy orbital.

https://en.wikipedia.org/wiki/Boltzmann_distribution is your friend. By the way, you should use > for quoting instead of \...``.

So, the spin state distribution fits the Boltzmann distribution? could you give more hints?

If you've not seen the Boltzmann distribution before, I'd suggest looking it up in a textbook on physical chemistry, or something similar. You won't get an answer on the Internet that's better than that. (Once you understand it, it's a fairly straightforward application of the formula $n_i / n_j = \exp[-(E_i - E_j)/kT]$. After all, the ratio you're after is exactly given by $n_i / n_j$, right?)


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What are some of the lowest melting point glasses available out there?
I'm wondering about lead-free "glass solder" or any glass frit powder or paste that has a very low melting point (under 400 c).  What products are available to the layman, where can I find them?

Standard suppliers include places like Schott.

Thanks, Im assuming this is the one: https://www.schott.com/en-gb/products/sealing-and-solder-glass/product-variants?tab=solder-glasses

Though there are low softening-point (s.p.) silica-based glasses, such as Anywhere Powder D235 and D240, with claimed s.p. of 350 °C and 400                       °C, respectively.
However, it's unlikely a glass would serve as solder. Unlike malleable metals, glass is strong, but exceedingly brittle. If the coefficients of linear thermal expansion of the two glasses and of the "solder" are not closely matched, then the work will crack or shatter as it cools.
A standard test of the closeness of match of two types of glass is to pull a combined thread from two glass rods. As the glass cools, if the two do not have matching coefficients, the combined thread will curl. Before you could meld glass, with or without a "solder", you'd need to confirm that all types expand and contract correspondingly.

Thanks for the answer. What I was wondering about was if it was possible to add low melting point glass frit to a higher melting point light sensitive material, like a vat dye.  To make a photoceramic coating. Yes, to coat onto glass.

This need should, perhaps, be another question, with some more specifics, e.g., do you want photosensitive glass, e.g., https://en.wikipedia.org/wiki/Photosensitive_glass , or photochromic glass, e.g., https://www.sciencedirect.com/topics/chemistry/photochromic-glass, or some other effect, e.g., reproducing images , temporarily or permanently, in or on glass?

A non toxic, non dichromate, photoceramic method, whereby I could coat something on glass, expose, develop and fix, then fire: (it was you who helped me ;) https://chemistry.stackexchange.com/questions/150291/what-non-toxic-light-sensitive-materials-do-not-burn-away-at-high-heat  Not photosensitive glass, and not photochromic (unless there is a way to fix photochromic pigments). Maybe a better question is, what light sensitive materials have a high boiling point/melting point. Should I ask this question seperately?

@Curiousmarble, why not edit your question to fully explain your needs? Asking repeated simillar questions is likely to get them ignored or rejected. As for melting and boiling points, another important question might be stability at high temperature.

I have re-answered you on the original question.


text:
What are the references in multi-reference methods?
There is this question: What exactly is meant by 'multi-configurational' and 'multireference'?
But I still don't understand what the multiple "references" are. I understand that you can for example use a HF Ansatz and exited states for multi-configurational methods. And that multi reference methods use more than one configuration to build exited states. But what are these configurations? There is only one HF Slater determinant, how do the other references look like?
I would appreciate if you could answer the questions with some additional formulas instead of just describing it by words :)

The references are just other determinants that differ in their occupied/unoccupied orbitals. Multi-reference CISD would then generate all singly and doubly excited determinants starting from these references. It also allows you to include specific determinants without having to create all other higher excited determinants. This might seem strange when you think about a stationary calculation for one geometry but an example where it is useful is excited state molecular dynamics with surface hopping. There you need to describe many states adequately at very different geometries.

This question is more suited for https://mattermodeling.stackexchange.com/ imo.

@HansWurst: But if they differ in their occupied/unoccupied orbitals, these are simply excited states, right? So they should be already included in CISD.

An electronic excited state is in the general case a linear combination of determinants. The determinants themselves form a "basis" into which we expand the electronic state. For singlet groundstates it is most of the time sufficient to use only a single determinant, but for excited states many determinants can contribute. And the basis set of determinants that are available for the expansion are build up by reference to an initial determinant. Further the determinants are generated by "exciting" this initial determinant. By adding a second reference, you will generate a larger basis.

I’m voting to close this question because it has been asked by the same user on Matter Modeling

@SRMaiti If you continue to suggest other sites, please make sure to also mention that cross-posting is generally frowned upon in the SE network. Be more helpful and avoid getting these questions closed for no good reason.

@SRMaiti I also completely, utterly disagree with you. And your comment makes it sound that the question is off-topic here, which it is absolutely not.

@Martin-マーチン I continue to suggest matter modelling, because computational chemistry questions keep getting closed here. I usually mention that cross-posting is frowned upon when an user responds to my comment. Also, you are the mod here, and you voted to close the question, I didn't, so accusing me of getting it closed is a bit unusual. There is nothing in my comment which would suggest it is off-topic here.

@srmaiti I have closed the question as a direct consequence of it being cross-posted. I would not have had to do that if it wasn't. I could have migrated it, if that were the wish of the OP, but I do not think this would have been necessary at all. Your message, as well intentioned as it might have been, was quoted by the OP on matter modelling as the reason. So yes, in my view, you are directly responsible for it being closed here.

@Martin-マーチン The question remained unanswered here for 15 days after being asked. On matter modelling it received an answer within 3 hours. That in itself is justification enough that computational chemistry questions are better asked on that site, and this has been demonstrated repeatedly in the past on other questions. The fact of the matter is that there are more modelling people on that site than on this site, so I will continue to suggest that site to computational questions. You are free to disagree but you won't change my mind.

@SRMaiti You are confusing two things here: 1. Since MM is public I have not once questioned its validity; I have on multiple occasions chosen to recommend it myself. So while this would obviously have been a valid course of action we come to the real issue. 2. I am criticising the way you communicated your recommendation. It is rather obvious that it was conceived as a request rather than a recommendation. So I am asking you once again: When you suggest asking on another site, include a remark that cross-posting is discouraged, and be more friendly. Anything else: [chat]

@Martin-マーチン Ok, I will keep that in mind, thank you.


text:
Stannous chloride is not dissolving into water
I am trying pcb tin plating. I am following a formula that uses

50ml water
2gm sulfamic acid
0.5gm stannous chloride
3gm thiouria
50ml water

The problem is, my Tin(II) chloride is not dissolving. It is precipitating. I have bought tin (ii) chloride dihydrate. Is there any issue?
Later, I tried to dissolve 0.5gm stannous chloride in only 50ml water. It also precipitated. Also, the tin chloride and water mixture does not attack aluminium or iron (i think it should as per the reactivity series of metal).
Can anyone figure out what am I missing? Thanks in advance.

Your tin(II) chloride may have oxidized.

How to double check? The bottle they gave me looks old though. :-/

Update, I just tried 20ml water, 30ml HCl, 3gm SnCl2. It makes a turbid solution and also the tin plating is not working. Shouldn't it be a clear transparent solution?

Tin chloride $\ce{SnCl2}$ (hydrated or not) is not soluble in pure water. It is quickly hydrolyzed according to : $$\ce{SnCl2 + 2 H2O -> Sn(OH)2 + 2 HCl}$$ and $\ce{Sn(OH)2}$ is insoluble in water.  To dissolve $\ce{SnCl2}$, this reaction has to be reversed by adding $\ce{HCl}$ in $5$ M solution. In this case, a clear solution is obtained.
Unfortunately $\ce{SnCl2}$ solutions are easily oxidized by oxygen from the air, and transformed into $\ce{SnCl4}$ plus a basic and not very soluble salt $\ce{Sn(OH)Cl}$, according to : $$\ce{6 SnCl2 + O2 + 2 H2O -> SnCl4 + 4 Sn(OH)Cl}$$ If a pure solution of $\ce{SnCl2}$ is wanted, this reaction can be avoided by adding some metallic tin grains ($\ce{Sn}$), which react and reduce $\ce{SnCl4}$ according to : $$\ce{SnCl4 + Sn -> 2 SnCl2}$$ It should also be mentioned that the solutions of $\ce{SnCl4}$ are also slowly hydrolyzed at high dilution.
Ref.: F. P. Treadwell, Analyse qualitative, Dunod Ed. Paris, 1924.
F. A. Cotton, G. Wilkinson, Advanced Inorganic Chemistry, Interscience Pub., J. Wiley, New York, 1972, p. 330 - 331


text:
How is the equilibrium constant for the water-gas shift reaction calculated after it is disturbed?
Consider the following equilibrium process at $\pu{686 ^\circ C}$:
$$\ce{CO2(g) + H2(g) <=> CO(g) + H2O(g)}$$
The equilibrium concentrations of the reacting species are $[\ce{CO}] = \pu{0.050 M}$, $[\ce{H2}] = \pu{0.045 M}$, $[\ce{CO2}] = \pu{0.086 M}$, and $[\ce{H2O}] = \pu{0.040 M}$. (a) Calculate $K_c$ for the reaction at $\pu{686 ^\circ C}$. (b) If we add $\ce{CO2}$ to increase its concentration to $\ce{0.50 mol/L}$, what will the concentrations of all the gases be when equilibrium is reestablished?
I've answered (a) already, and I got $K_p$ and $K_c$ both equal to $0.52$; in other words, I'm certain that the system is in equilibrium. However, I'm having trouble answering (b). Do just add $\pu{0.50 mol/L}$ to all the other concentrations?

So at equilibrium, we can list the concentration of each species:
\begin{align}
[\ce{CO2}] &= 0.50 - x = 0.500 - x,\\
[\ce{H2}] &= 0.045 - x,\\
[\ce{CO}] &= 0.050 + x,\\
[\ce{H2O}] &= 0.040 + x,
\end{align}
where $x$ is the change in concentration that is given to all species.
Therefore, we get the following expression at the new equilibrium:
$$\frac{[\ce{H2O}][\ce{CO}]}{[\ce{CO2}][\ce{H2}]} = \frac{(0.050+x)(0.040 + x)}{(0.045 - x)(0.500 - x)} = 0.5168$$
We could plug the equation into WolframAlpha and solve for $x$ but if you can't do that, you can make some reasonable approximations.
First, since $0.500-x$ is likely a small change, we can approximate it to be only $0.500$, giving us a new expression of
$$\frac{(0.050 + x)(0.040 + x)}{(0.045 - x)(0.500)} \approx 0.5168.$$
At this point, you can't really approximate anymore. Honestly, at this point, either brute force it by solving the quadratic or plug it into an online calculator. You eventually get a value of $x = \pu{0.0251 M}$. That means the concentration of each species at equilibrium are:
\begin{align}
[\ce{CO2}] &= \pu{0.4749 M},\\
[\ce{H2}] &= \pu{0.0199 M},\\
[\ce{CO}] &= \pu{0.0751 M},\\
[\ce{H2O}] &= \pu{0.0651 M}.
\end{align}
It's good practice to plug these values back into the equilibrium expression to see if we made bad approximations, but in our case, it matches.

I've fixed your MathJax, please use \ce{...} only for actual chemical expressions. Here's some boilerplate with links: On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

The question says "add CO2 to increase its concentration to 0.50mol/L," so I interpret that to mean the new CO2 is .50. Change the denominator a bit to have (.500 -x). My old graphing calculator (TI83 Plus) gives x=.02516.


text:
Is there a simple way to distinguish grades of purity for copper metal?
Is there a simple way to distinguish different grades of fine copper purity (on a trace metal basis) that can be carried out in a small lab?
Specifically, I would like to be able to differentiate 4N, 5N, and 6N (the commercial nomenclature for 99.99/99.999/99.9999% pure copper, respectively) grades.
The samples are in the form of pellets, and significant trace impurities include S, Zn, Ag, As, Cr, Ni, Fe, Pb, Sb, Cd, Hg, Se, Sn, Bi, Te, Ge, In, and Tl (more or less in that order).

You don't state if the sample is a bar, or shot / powder of Cu (lower part here). Perhaps a gravimetric analysis of a dissolved sample may discern 4N from 5N Cu if the gravimetric factor is just high enough .and. you are willing to spend enough material. The abstract of e.g., ASTM E53 looks more like electrogravimetry (electrolysis-like, e.g. from Cu foil to Pt net) than proceeding via dissolution, filtration of the precipitate, and incineration of the dedicated ash-free filter paper.

The sample is in the form of pellets. Since copper has low reactivity and low solubility in some ways I was hoping I could either apply a reagent (that doesn't react with copper) and look at color change or dissolve a bigger sample, precipitate out either a copper salt or impurities, dry and measure weight difference, or something the like.

Did you consider using a handheld XRF analyser?

@Loong Excellent idea! I did a quick search and both purchasing and renting of portable XRF analysers is viable, depending on one’s budget. My implicit assumption was that only simple chemical testing was of interest.

@ Loong: If I informed myself correctly, a handheld XRF analyzer, even if accurate enough for the purpose at hand, would cost several thousand bucks, which is way too expensive for the purpose unless tons of copper would need to be handled.

Not my area ;but don't you need light spec for trace elements. ? In the stone age I worked on XRF of copper alloys and we only did the alloy elements  and did not look for Sb and Bi

If my answer was helpful, please consider accepting it! Thanks!

It is unlikely that there is a simple method, not involving relatively expensive chemical analysis, for distinguishing 4N, 5N, and 6N copper specimens. Shown below is a photograph of some copper specimens I have:

The evaporating dish contains copper shot, the large plate (4.76 mm thick) is OFHC (oxygen free high conductivity) copper and the third specimen is a copper bar that is 25.4 mm by 31.5 mm cross section. The standard banana is for scale. All of these copper specimens are supposedly 'pure', but how pure is 'pure'?
The problem with determining copper purity, when the specimens are already what might be considered 'pure enough,' is that the impurities are present at low levels and their identities matter a great deal. This figure, from Edward R. Tufte, The Visual Display of Quantitative Information, Graphics Press, Cheshire, CT, USA, ©1983, p. 49, shows published copper thermal conductivities on a log-log scale. The original publication reference is given as Ref.1 (page I-244).

This beautiful figure collects results from hundreds of studies in the published literature and shows, quite dramatically, that the large majority of those studies were incorrect, sometimes wildly so. What tripped up so many researchers is that they assumed they were measuring the thermal conductivity of copper specimens that were pure enough. But they were not pure enough. And oxygen turns out to be a major culprit in reducing the thermal conductivity of copper metal. Given that we live in an oxygen atmosphere, this is a problem. As well, trace oxygen determination is not an easy thing without appropriate instrumentation and methodology.
Bottom line: you need to know what the major impurities are (or might be) and then do trace determinations of those.

References:

C. Y. Ho, R. W. Powell, and P. E. Liley, “Thermal Conductivity of the Elements: A Comprehensive Review,” Journal of Physical and Chemical Reference Data 1974, 3 (Supp.  No. 1), pp. I-244 (ISBN-13: 978-0-88318-216-1).

In section «Journal of Physical and Chemical Reference Data Monographs or Supplements» NIST offers direct access to the plot (as figure 60, page I-244) Tufte cites. An abridged version (including a doi and citation report) is provided by AIP.

This is interesting information. If I understand well the discrepancy is predominantly attributable to oxygen content. I didn't think about this. I'm concerned with purity more on trace metal basis than oxygen content (I made an edit to the question to add this information).

For the figure, oxygen was the main problem. The example illustrates what happens if researchers get sloppy and assume what is not in evidence. In general, though, other trace species may be (and almost certainly are) problematic, e.g., S, Se, As, etc. So you need to know something about the significant trace impurities. No sane person, who needs 6N copper, would purchase and accept it without getting a serious chemical analysis certificate. The analysis would be done using, e.g., ICP-OES or ICP-MS. I see no real way around this.

@Ed V: Say the significant trace impurities are S, Zn, Ag, As, Ni, Fe, Pb, Sb, Hg, Se, Sn, Bi, Te, Ge, In, Tl (more or less in that order).

*I forgot Cr in the list, above Ni. And Cd under Sb.

I don’t think there is a good chemical option to solve your problem, unless you do tests for particular contaminants, which is too much work.
A relatively simple option is magnetic sputtering, or accelerator enrichment. It may sound complex, but there are YouTube videos of people doing it with microwave parts, glass bottle, vacuum pump, silicone and some wire.
Idea is to use a magnetic field and a electric field to concentrate and deflect ions. And ions deflect differently depending on their mass. Initial evaporation of ions is usually achieved by microwave as heating when hobbyists are doing it.
It is still simpler than doing chemistry to distinguish 5N from 6N without knowing contaminants.

Thank you for these innovative suggestions. I'll watch some youtube videos to learn more!

A more familiar term for this technique, perhaps, is "mass spectrometry".

One thing is identification, but what about the quantitative aspect of it?

@PeterMortensen cheap way is to measure how much residual is accumulated on a particular target spot through opacity. More precise is many electrodes and current measurments for each, on a target. But anyway, he only wants to distinguish very different examples, like 4N, 5N, 6N, with such huge difference in amount of contaminants, it is likely sensitive enough as is, just to say 'this is more likely to be X type of purity'


text:
Which electron should be considered the last one?
Today I came across a problem which asked some quantum number informations regarding the last electron of Zinc.
Now as far as I have studied 4s subshell is filled in preference to 3d subshell bcos of lower energyof 4s subshell. But 4s subshell is located at greater radius than 3d due to greater size of 4s subshell .
Which of the the electrons should be considered the last electron 4s one for the 3d one.

See https://youtu.be/7hDmtUomFu0

I'll get to zinc, but first, the ground state electron configuration of ground state gaseous neutral nickel is  [Ar]3d$^8$4s$^2$. Ref 1. Nickel has 28 nuclear protons, two less than zinc. If atomic structures build up linearly, this says that if you put two more protons into the nucleus, you have to fill the 3d shell of electrons, because that's what's empty.
This could also be the electron structure of Zn$^+$$^+$, but isn't! (Ref 2.) "The dication Zn$^+$$^+$ has 30 protons and 28 electrons. The electron configuration of Zn$^+$$^+$ is 1s$^2$2s$^2$2p$^6$3s$^2$3p$^6$3d$^{10}$. (The 4s shell is empty!) Zinc is a d-block transition element. For the d-block elements, the outermost s-sublevel has higher energy than the d-sublevel, which is contrary to what the Aufbau diagram

indicates. When d-block elements lose electrons, they lose the highest energy s electrons first, which in the case of zinc are the two 4s electrons. Having eight 3d electrons and two 4s electrons is much less energetically stable than ten 3d electrons and no 4s electrons." So you fill the 4s shell, because that's what's empty in zinc dication.
"Contrary". The bane of students everywhere. What difference does it make, except on tests? And is there any effect on zinc properties?
Let's ask what electrons are the highest energy in zinc metal, rather than trying to discover what electron label we need to neutralize gaseous Zn$^+$ monocation. Metals have atoms swimming in a sea of electrons: "Transition metals tend to have particularly high melting points and boiling points. The reason is that they can involve the 3d electrons in the delocalization as well as the 4s. The more electrons you can involve, the stronger the attractions tend to be." (Ref 3.). What are the labels on the electrons in the sea of zinc electrons? It is said that you can't tell one electron from another because you can't label an electron, but you can count electrons!
The boiling point of zinc is unusually low when compared to its neighbors in the periodic table:
This suggests that the number of electrons in the zinc sea is low, like 2 per atom, rather than high, like 10 or 12 per atom. So the additional 2 protons in zinc compared to nickel switches the energies of the 3d and 4s shells - with all those nodes and electron amplitudes interacting, it's a wonder that chemistry is as organized as it is!
The data point to the 4s level as the last to be filled.
Ref 1. https://www.webelements.com/nickel/atoms.html
Ref 2. https://socratic.org/questions/what-is-the-electron-configuration-for-zn2
Ref 3. https://chem.libretexts.org/Courses/Bellarmine_University/BU%3A_Chem_103_(Christianson)/Phase_3%3A_Atoms_and_Molecules_-_the_Underlying_Reality/9%3A_Chemical_Bonding/9.5%3A_Metallic_Bonding


text:
Brownian motion of asymmetric particles
The classic Stokes-Einstein relation ($D=k_BT/(6\pi\eta r)$) it is assumed that the particle undergoing Brownian motion is spherical, and experiences ordinary Stokes flow between (or during) consecutive impulses.  In the nonspherical case, however, translational motion (driven by an external field for example) is typically correlated with chiral motion; consider, for example, the helical motion of a maple seed pod that develops under gravity.  This suggests that there should be some correlation between angular and translational Brownian increments, unless angular increments are expected to be considerably larger or smaller than translational ones.  This also suggests, as angular diffusion can likely arise also through anisotropic forces exerted on the boundary of an asymmetric particle, that angular diffusion (or the ratio of angular to translational diffusivities) is non-universal and depends on the shapes of particles and interfacial properties of constituent materials.  Is this in fact the case?


text:
How is the vibration of multiple atoms treated in vibrational spectroscopy?
Most of the advanced undergraduate textbooks on physical chemistry/ instrumental analysis show the case of a diatomic harmonic oscillator and derive formulae for predicting the frequencies. The concept of reduced mass and force constant is also illustrated for simple atoms. The concept of reduced mass is limited to a two-body problem, although Wikipedia says that it can be generalized for multi-body problems. Reduced mass,
Similarly, the force constant is limited to two bodies; e.g., it is a typical teaching exercise to calculate a force constant from experimental vibrational frequencies.
I was wondering about more complex examples, such as aromatic ring breathing picture from here. The number 1 and 15 is the notation of Wilson for normal modes.

For example, how do physical chemists or theoreticians predict the vibrational frequencies where six atoms are executing harmonic motion (i.e., what methods are used to solve such problems?). I wanted to see an elementary exposition of such treatment, perhaps as a solved example in a good reference which does not need computer programs or fancy tools.
Is the generalized reduced mass still useful in polyatomic vibrations, and how is the concept of force constant generalized to more than two atoms? Thanks.

Are you asking if the G-matrix is relevant?

@ToddMinehardt, I haven't encountered G-matrix. My background is analytical chem.

Well, you're in luck. I happen to have done my doctoral dissertation on IVR (intramolecular vibrational relaxation) for benzene. I will happily send you reprints and take this off-line or into chat to discuss. Please Google Scholar me and ping me. I also know the experimental folks and am happy to discuss.

To be honest I would not be able to follow your hardcore QM papers. Is there a way to explain it here i.e. a qualitative picture of how this is done would be sufficient. Actually, I am asking out of curioisity and just for historical purposes. A colleague suggested that a C-H, and C-D bond distances (e.g., C6H6, C6D6) are the same because there is no technique to directly measure C-H vs. C-D distance (too small for X-rays) and this is where IR came in. Of course C-D, C-H vibration frequencies are different.

Let me put something coherent together. I'll also shoot a bunch of stuff up to github, which I should have done already. Wilson, Decius, and Cross is the canonical reference. There are matrix transformations (such as $\bf{R = LQ}$) where normal modes ($\bf{Q}$)  are transformed to rectilinear combinations (local modes) which include $\ce{C-H}$ stretches, $\ce{H-C-C}$ bends (in-plane and out-of-plane), and so on. More later.

For a much simpler intro, these systems can be treated semi-quantitatively as standard systems of coupled harmonic oscillators as summarized here: https://www.colby.edu/chemistry/PChem/notes/NormalModesText.pdf

@M.Farooq Why would the C-H distance be too small to measure by X-ray crystallography?

@Karsten does hydrogen atom produce any useful signal in X-ray crystallography? My experience is limited to X-ray photoelectron spectroscopy.

As X rays scatter off electrons, and there are not many electrons around H, the folk lore certainly used to be it was difficult to see H by XRD, especially in large molecules such as proteins. However this, and all experiments, are not my area, so I don't know the current state of the art.

Hm, I think the question is already answered here https://chemistry.stackexchange.com/questions/119941/the-concept-of-normal-coordinates-of-a-molecule
although the wording is somewhat different. Is there a particular point that should be explained in more detail ?

@HansWurst - I am under the (potentially mistaken) impression that the OP wants to know what takes us from normal modes to local modes, the latter of which include things like $\ce{CCC}$ bends, $\ce{CCCC}$ torsions, and so on, which are actually linear combinations of normal modes. While I am rereading my dissertation to refresh my memory about all of this, the answers you point to do address the question asked there but stop short of the transformation from normal modes to local modes. Also omitted is fitting the PES through beyond second-order.

@ToddMinehardt Then the question title could use some improvement. I certainly would not expect a discussion of localized modes given the current title; also considering the lack of the term "localized modes" in the question body. It certainly would be interesting to see an explanation about localized modes though. I am also aware that there are a lot more details that could be discussed about normal modes, which are not given in the answer that I linked, but covering all details goes beyond the scope of one question, which is why I asked for clarification.

@ToddMinehardt, You are right, please continue. My previous old question on normal  does not address polyatomic vibrations. including ring breathing.

@M.Farooq The answer to the old question does actually explain how you would obtain for example the ring breathing mode in benzene. I guess it isn't clear since that isn't the focus of that question/answer, but if you applied the described procedure to benzene, you would obtain the ring breathing mode. One of the $Q_i$ vectors would describe the motion that you have drawn in your figure.  Localizing modes does something different. It's purpose is to convert a set of delocalized modes, like the ring breathing mode for example, to a set of localized modes that aren't spread out over the molecule

Vibrational spectroscopy and normal modes are based on the Born-Oppenheimer approximation and the harmonic approximation.
The BO approximation yields the adiabatic electronic states and the corresponding adiabatic energy eigenvalues as functions of the nuclear coordinates. The adiabatic energy eigenvalues of the electronic states are also known as potential energy surfaces $E_\alpha(R)$. Associated with this electronic adiabatic state is a vibronic Hamiltonian for the nuclear wave function,
$$
H_\alpha(R) = \frac{1}{2}\nabla^T_RM^{-1}\nabla_R + E_\alpha(R)
$$
From now on I will fix the electronic state to be the groundstate and  drop the index $\alpha=0$, that defines the electronic state. In pure vibrational spectroscopy the electronic state remains the same and is thus of no further interest.
If we knew the exact potential energy surface $E(R)$, then we could try to solve the problem numerically without the harmonic approximation. But doing so is very expensive for a large amount of degrees of freedom.
To avoid this, we can make the harmonic approximation and expand the potential energy surface to second order around a equilibrium structure, i.e. the nuclear coordinates where the potential energy surface has a minimum. Let this structure be $R_{EQ}$,
$$
E(R)\approx E(R_{EQ}) + \sum^{3N}_{i=1}\frac{\partial E} {\partial R_i}|_{R_{EQ}} (R-R_{EQ})_i +\frac{1}{2}\sum^{3N,3N}_{i=1,j=1}(R-R_{EQ})_i \frac{\partial^2 E} {\partial^2 R_i R_j}|_{R_{EQ}} (R-R_{EQ})_j
$$
We can chose our energy axis such that $E(R_{EQ})=0$. We also are at a minimum of the potential energy surface, which means that the gradient vanishes $\frac{\partial E} {\partial R_i}|_{R_{EQ}}=0 \ \forall \ i$. All that remains in the harmonic approximation at the equilbrium geometry is the second order term with the Hessian evaluated at the equilibrium geometry $H_{ij}=\frac{\partial^2 E_\alpha} {\partial^2 R_i R_j}|_{R_{EQ}} $,
$$
E^{Harmonic}(R) = \frac{1}{2}\sum^{3N,3N}_{i=1,j=1}(R-R_{EQ})_i H_{ij} (R-R_{EQ})_j
$$
With this we can define the vibronic Hamiltonian in the harmonic approximation,
$$\begin{aligned}
H^{Harmonic}(R) &= \frac{1}{2}\nabla^T_RM^{-1}\nabla_R + E^{Harmonic}_\alpha(R)\\
H^{Harmonic}(R)&= \frac{1}{2}\nabla^T_RM^{-1}\nabla_R + \frac{1}{2}\sum^{3N,3N}_{i=1,j=1}(R-R_{EQ})_i H_{ij} (R-R_{EQ})_j
\end{aligned}$$
A short remark on what I have not done is also in order. I have not separated  the nuclear coordinates into translation, rotation and vibration thus far. Six(five for linear molecules) of the normal coordinates that we will obtain are going to describe translation and rotation. They are typically easy to recognize by their associated angular frequencies that are almost zero.
Now back to the problem. I will now drop the superscript harmonic. Our harmonic vibronic Hamilton operator is not a sum of 1D Harmonic oscillator Hamilton operators, but we can achieve that by doing some coordinate transformations.
First we introduce mass-weighted displacements
$$
x_i=\sqrt m_i (R-R_{EQ})_i \rightarrow \frac{\partial}{\partial x_i} = \frac{1}{\sqrt m_i}\frac{\partial}{\partial R_i}\rightarrow \nabla_x = M^{-1/2}\nabla_R
$$
In these coordinates we obtain
$$
H(x) = \frac{1}{2}\nabla_x^T\nabla_x + \frac{1}{2}\sum^{3N,3N}_{i=1,j=1}x_i \frac{H_{ij}}{\sqrt{m_i m_j}} x_j
$$
We define now the mass weighted Hessian $F_{ij} = \frac{H_{ij}}{\sqrt{m_i m_j}}$
which leads us to
$$
H(x) = \frac{1}{2}\nabla_x^T\nabla_x + \frac{1}{2}\sum^{3N,3N}_{i=1,j=1}x_i F_{ij} x_j
$$
In this equation we still have coupling between different $x_{i/j}$ due to off-diagonal matrix elements in the mass weighted Hessian. But we can eliminate these by transformation to a coordinate system in which the matrix is diagonal. To find these coordinates we solve the eigenvalue problem
$$
FV = V\Lambda
$$
where the matrix $V$ contains the eigenvectors of $F$ as columns. $\Lambda$ is the corresponding diagonal eigenvalue matrix.
The eigenvector matrix allows us to define the transformation to normal coordinates $Q_i$,
$$
\sum^{3N}_{k=1} V_{ik}Q_k = x_i \quad  \rightarrow  VQ=x \quad  \leftrightarrow Q = V^{-1}x
$$
The eigenvectors of the mass weighted Hessematrix are orthogonal and if normalized form a orthonormal set. This means that $VV^T=1=VV^{-1}$. Which means that $F$ can be diagonalized via
$$
\Lambda = V^TFV
$$
Plugging in the transformation into the Hamiltonian yields
$$\begin{aligned}
H(Q) &= \nabla_Q^T({V^{-1}})^TV^{-1}\nabla_Q + \frac{1}{2} Q^TV^TFVQ \\
H(Q) &= \nabla_Q^T\nabla_Q +\frac{1}{2} Q^T \Lambda Q
\end{aligned}$$
Expanding this equation shows that we now have a simple sum of harmonic 1D Hamiltonians since the Eigenvalues matrix is diagonal $\Lambda_{ij}=\delta_{ij}\Lambda_i$ and no longer couples different coordinates.The last step is recognizing that each 1D equation is the same as a harmonic oscillator equation with $\omega_i^2=\Lambda_i.$
$$\begin{aligned}
H(Q) &= \sum^{3N}_{i=1}\frac{1}{2}\frac{\partial^2 }{\partial Q^2_i} + \sum^{3N}_{i=1}\frac{1}{2}Q_i^2\Lambda_i\\
&=\sum_{i=1}^{3N}\frac{1}{2}\frac{\partial^2}{\partial Q^2_i} + \frac{1}{2}Q_i^2\omega_i^2\\
H(Q) &= \sum_{i=1}^{3N} H^{1D}(Q_i)
\end{aligned}$$
Meaning that our large coupled problem finally has taken the shape of a sum of 3N one dimensional harmonic oscillators, which we know how to solve analytically.
At this point I'd like to repeat that we have not separated translation and rotation. Due to this, the first six frequencies should be zero (or almost zero). The six normal modes corresponding to these zero eigenvalues are not oscillations and have to be omitted when calculating vibrational transitions.
It is possible to remove them before transforming to normal coordinates. How this is done is described in the book of Wilson, Decius and Cross, Molecular vibrations but discussing these points goes beyond the scope of my answer.
Let me now address how all of this connects to the figure that you show in your question. In particular how we know how the atoms are displaced relative to each other for a given normal mode.
By inverting the transformations, we see
$$
\Delta R = M^{-1/2}VQ
$$
with $\Delta R = R-R_{EQ}$.
This means that the columns of the matrix
$$\begin{aligned}
K &= M^{-1/2}V
\end{aligned}$$
describe the relative cartesian displacement amplitudes of the normal modes, easily seen by plugging in a unit vector for Q.
If we had calculated benzene then one of the columns of $K$ would contain the relative displacements as shown in your figure. Some quantum chemistry programs also print these column vectors after some sort of normalization, when they output Cartesian normal modes and use them for visualization.
Now let me say a bit about reduced masses and force constants. First of all reduced mass and force constants are no observables. They are helpful quantities that we can define to make the analogy to a 1D system with a mass attached via spring to a wall. Depending on the particular conventions that we use, we will obtain different values. A common convention is to define a reduced mass based on the Cartesian displacement vector norm. The normal mode vectors as I have defined above are assumed to be normalized to one with regard to mass weighted coordinates. A mass weighted normal coordinate vector of norm one corresponds to a displacement of atoms such that all displacements squared, multiplied with their atomic mass then summed and the root taken yield 1. Take as an example one mode with  many heavy atoms contributing to the motion and a mode where mostly light atoms contribute to motion. If both modes have now a normal coordinate value of 1, you would observe spatially when looking at the atom displacements, that the mode with the heavy atoms is only a little displaced while the mode with light atoms shows larger displacement.
We can however introduce a new coordinate that is constructed in a way that a value of 1 means always the same net spatial displacement of the atoms contributing to the mode irregardless of the mass of the contributing atoms.
Let us look at the norm of the Cartesian displacement vector associated with a mass weighted normal mode vector. A unit normal mode vector is of the form
$$
(\hat Q_\alpha)_j = \delta_{j\alpha}
$$
The transformation of this vector to Cartesian displacement coordinates yields
$$
R_\alpha = K\hat Q_\alpha
$$
The Cartesian norm squared that describes the actual displacement of the atoms in space is then,
$$
R_\alpha \cdot R_\alpha = K \hat Q_\alpha \cdot K\hat Q_\alpha
= \sum_{nml} K_{nm} (\hat Q_\alpha)_m K_{nl} (\hat Q_\alpha)_l \\
= \sum_{nml} K_{nm} \delta_{\alpha m} K_{nl} \delta_{\alpha l} \\
= \sum_{nml} K_{n\alpha} K_{n\alpha} \\
|R_\alpha|^2 = \sum_{nml} (K_{n\alpha})^2 \\
$$
We define now the reduced mass for mode $\alpha$ as
$$
\mu_\alpha = \frac{1}{|R_\alpha|^2}
$$
and the new set of coordinates
$$
y_\alpha = \frac{1}{\sqrt \mu_\alpha}Q_\alpha
$$
In this coordinates we obtain the Hamiltonians
$$
H(y_\alpha) = \sum_\alpha \frac{1}{2\mu_\alpha}\frac{\partial^2 }{\partial y_\alpha^2} + \frac{1}{2}\mu_\alpha \omega^2_\alpha y_\alpha^2
$$
and can define the force constants for each mode
$$
k_\alpha = \mu_\alpha \omega^2_\alpha.
$$
I am afraid you need a computer to do all of this or at least a very long breath and patience. I think for some systems the diagonalization can be done manually if you use symmetry, breaking down the matrices that have to be diagonalized to managable size, but generally you are going to let a computer do the linear algebra parts. I am pretty sure that is also the reason why almost all texts only discuss the trivial linear case where everything can be done by hand, although that is problematic since the simple case doesn't give you much intuition for the more general case.
All of this is also explained in the white paper https://gaussian.com/wp-content/uploads/dl/vib.pdf
I would not recommend the book from Wilson, Delcius and Cross as first reference. The book is good if you already have some understanding of the  matter but as first exposition it might be difficult. It is however a good source if your are interested in internal coordinates and a treatment based on internal coordinates and the problem of separating vibrational and rotational degrees of freedom.


text:
Order of Shielding Effect for orbitals
From what I know Shielding effect is the ability of inner electrons to repel outer electrons and reduce the Nuclear charge felt by the outer electrons and this is caused by electron-electron repulsion.
When I read the explanation from a site it said that

Shielding refers to the core electrons repelling the outer rings and thus lowering the 1:1 ratio. Hence, the nucleus has "less grip" on the outer electrons and are shielded from them. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity.
Because the order of electron penetration from greatest to least is s, p, d, f; the order of the amount of shielding done is also in the order s, p, d, f.

But I don't understand the fact that inner orbitals would have greater shielding effect than the outer orbitals as the closer the electrons are the more the repulsion they will have as it follows the inverse squared law. So, as the outer electrons are closer to other electrons hence they will show more repulsive force than the inner electrons as the outer electrons are closer.
One explanation which I thought was that as the electrons are not stationary and moving at great speeds, the charge could be considered as symmetrically distributed over the volume and as the charge density would be low for outer orbitals they would have less shielding effect.
Is this explanation correct or am I going wrong somewhere?
Source:https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Multi-Electron_Atoms/Penetration_and_Shielding

I suggest to look at the data instead of such explanation that you expect, they dont say much on quantity of the effect.
https://en.m.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page)
Here you can see that hydrogen and francium only differ about 3.4 times in energy needed to remove one outer electron. Now lets look at the atoms sizes:
https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1201/Unit_2._Periodic_Properties_of_the_Elements/2.08%3A_Sizes_of_Atoms_and_Ions
About 70% scroll, figure 3.7
So, if we assume that electron is just orbiting charge +1 but at different distance, result is quite similar to what we expect naively, assuming charge of lower electrons and protons in the nucleus just cancels out. As if outer electron orbits around one proton, and the rest of protons and electrons cancel out. Another position to look at this problem from is this trick:
https://en.m.wikipedia.org/wiki/Shell_theorem
Not sure it applies exactly to give solution exactly as if inner electrons and n-1 protons are cancelled, but I assume it is a good starting point as a path to dig deeper into this effect.
I dont think the way effect is described in your link is a good way to describe it, a bit too vague.


text:
Lother Mayer curve
While studying about atomic trends, I encountered the Lothar Meyer curve:

If the covalent radius decreases across a period, why is the graph of atomic volume reaching local minima halfway in the period?
And how did he measure the atomic volume by the way?

It is an interesting question, and I am not completely sure on everything I write below.
First of all, atomic volume is defined as the "quotient obtained by dividing the atomic weight of an element by its specific gravity". (Merriam-Webster) It is interesting to note that I could find only a few publications that have the term "atomic volume", and most of the references are from 1980's. So, it seems that the use of atomic volume has almost disappeared from mainstream chemistry, except perhaps a few niche areas.
The definition essentially means that you measure the volume of one mole of the substance, and then divide it by Avogadro's number to get the atomic volume. So, that means measuring atomic volume just requires a way to measure the volume of a fixed weight of substance. Perhaps Lothar Meyer did something like that. Atomic volumes can also be calculated from the packing density, if you have a crystalline element.
But anyway, let's look at the trend you showed in the question. The atomic volume is the largest in the alkali metals and smallest in the middle of the period. From Chemistry of the Elements1:

Fig 2.1 shows a modern version of Lothar Meyer's atomic volume curve: the alkali metals appear at peaks and elements near the centre of each period (B,C; Al, Si; Mn, Fe, Co; Ru, and Os) appear in the troughs. This finds a ready interpretation on the electronic theory since the alkali metals have only 1 electron per atom to contribute to the binding of the 8 nearest neighbour atoms, whereas elements near the centre of the period have maximum number of electrons available for bonding. ... Closer inspection reveals that a much more detailed interpretation would be required to encompass all the features of the curve which includes data on solids held  by very diverse types of bonding. ...

So basically the simplistic explanantion boils down to the number of unpaired valence electrons each element has. Remember that the atomic volume is the volume of one mole of element divided by a constant. The volume of one mole of element means that we are considering the volume of a fixed number of atoms i.e. it depends only on how close the atoms can pack in the solid state, and does not depend on mass.
As the number of unpaired valence electrons is the highest in the middle of a period, the packing is strongest there and the atomic volume is lowest. It's likely that the trend is atomic radius becomes negligible compared to this effect in solid state.
References:-

Chemistry of the Elements, N. N. Greenwood, A. Earnshaw, Butterworth-Heinemann, 1998, p 23-24

Thanks for the answer! It also explains that why atomic volume increase along a group.


text:
Can phosphoric acid be used to substitute sulfuric acid in the synthesis of thymolphthalein?
I am planning to synthesize thymolphthalein, a pH indicator. I am wondering if phosphoric acid can be used to substitute sulfuric acid in the preparation. I know it can catalyze Friedel-Crafts reactions, which are a part of the synthesis. Any advice would be appreciated.

It would be appreciated if you could attach an image of the normal synthesis involving sulphuric acid. Would help those who are not aware of the compound. Thank you :)

If you use polyphosphoric acid (PPA) I think it will work. PTSA will also probably work

What is your reasoning for, what is your reasoning against the substitution of the strong Bronsted acid $\ce{H2SO4}$ by phosphoric acid (i.e., review of the literature)?  By the way, the waxy polyphosphoric acid may be considered, too (an elder review, a current application).

I do chemistry in a workspace with wooden benches, and sulfuric acid would eat through them quickly. Thanks for suggesting those, I can make or get both of those.

But the question is: would straight phsophoric acid work?

@Jacob Are you asking about tthe commercially avialable 85% phosphoric acid solution or the 99% crystalline material? The 99% material might do it, I doubt the 85% solution would.

The phenol will be more reactive than a normal benzene, by being a phenol + the alkyl groups in addition; this with using an acid anhydride may suggest less need (quantity / strength) of catalyst / less harsh reaction conditions here.  However, consider the  concurrent esterification (phenol-OH + carboxylic acid), too.

The reaction is a friedel crafts reaction, and it forms water. The reaction to form the thymolphthalein is irreversible, but esterification is. Also, phenols are sterically bulky, so this is not favored.

Thanks, all! Polyphosphoric acid does indeed work to catalyze this reaction. I was successful in making it.

@Jacob If you find your answer, kindly self-answer it. It completes your post.

I used polyphosphoric acid and it worked.


text:
Fields of Study in Chemistry
Is there a standard classification scheme for different areas of study and research for chemistry like mathematics has Mathematics Classification Scheme (MSC)?

I think there is none. See the six major and traditional classifications by the American Chemical Society  here https://www.acs.org/content/acs/en/careers/chemical-sciences.html

Neither one of the following two might be rigorous enough from the perspective of a mathematician because of some overlap of the fields within chemistry, or by initial inspiration / interaction with fields outside chemistry.  Thus I think the answer is no.
However, curricula at universities typically divide chemistry into general, inorganic, organic, physical, analytical chemistry, biochemistry, etc. This is reflected in study guides like chemistry.se's compilation of resources for self-study, or the organization e.g., of bookshelfs Chemistry LibreTexts.  As an example of topical overlap however, there are spectroscopic or electrochemical methods which may attributed to physical chemistry as well as of analytical chemistry.  A topic like bioinorganic chemistry may be recognized as an offspring in common of inorganic chemistry and biochemistry.  There equally are subjects where chemistry may be very relevant, but "chemistry" is not part of the name, too, like toxicology, geology, materials science, to mention a few.
Complementary to this, for long, there have been attempts to classify knowledge into divisions.   This is reflected by the labels (not only scientific) libraries put on their media (books, journals, etc.).  One of them is the system established by the Library of Congress with starting letters QD as root for chemistry within science (cf. entry wikipedia, or entry LOC).  An example for this may be the record for Chemistry of the Elements by Greenwood and Earnshaw here and other topically related entries of shelf QD466.

An other attempt are IPC as the international and CPC as the cooperative patent classification (USPTO and EPO) assigning patents related to chemistry (and metallurgy) to class C (entry EPO/espacenet). This classification however may be less helpful than a library catalogue à la LOC since patents may be filed in more than one (sub-class) at once.  Equally, you may find multiple patents related e.g., to keyword thiophene in classes other than C, too.

I just upvoted this fine answer. One thing I might add is that many chemists nowadays (I know many of them) focus on solving problems and eschew traditional divisional lines as no longer relevant (if they ever really were).


text:
Can suspended solid particulate particles adsorb on surface?
Gaseous and liquid particles adsorb on solid surfaces. Can solid particles suspended in a fluid medium also adsorb on these surfaces in the same manner?

Suspended particles don't adsorb, molecules do.

@Mithoron, Bacterial cells are said to adsorb on glass surfaces. Are bacterial cells molecules?

@M.Farooq Maybe biologist would consider binding of bacteria to surface an adsorption, but for a chemist there's a whole phase inside a bacteria. How big particles are supposed to be to not be "adsorbate"? When calling something adsorption is gonna have a different meaning?

Terminology is a grey area and modern science is all interdisciplinary. Adsorption of very small "particles" is a well known phenomenon as illustrated below and in the field of biochemistry/biology.

Yes, even very small particles can "stick" permanently to solid surfaces in solution just like gases or liquids. Just note that the particles have to be nanometer sized. One classical discovery, which made million dollar business in the field of chromatography is that of so-called agglomerated ion-exchange phases (Dionex).
Basically you have negatively charged particles, such as a sulfonated resins. The sulfonate group on the surface has a negative charge. Now expose this resin to a nanometer sized suspension of polymeric particles bearing quarternary amine groups (so called latex). The positive charge on this colloid helps to stick to the negatively charged resin. In the scanning electron microscope, the solids bead look this this. I cannot find the original SEM at this moment (This work was done in 80s)
This "adsorption" is almost permanent. Nothing can get rid of this latex coating on the particles.

So we can say that any adsorbent doesn’t require a recognition of its state because of its particulate size, right?

Welcome, it depends on particles dimensions. You cannot have a 5 micron particle adsorp on a 10 micron particle.


text:
Does (BeCl2)n have 3-center-4-electron-bonding or not?
The title really explains what the issue is here. I would think that it has multicenter bonding akin to $\ce{(BeH2)_n}$ polymer, because they are pretty much iso-structural (although it shouldn't have 3-center-2-electron bonds, but rather 3-center-4-electron-bonds unlike $\ce{(BeH2)_n}$.
J. D. Lee is a bit vague, but it seems like it says there is no multicenter bonding in this polymer:

The anhydrous [Beryllium] halides are polymeric. Beryllium chloride vapour contains $\ce{BeCl2}$ and $\ce{(BeCl2)2}$, but the solid is polymerized. Though the structure of $\ce{(BeCl2)_n}$ is similar to that for $\ce{(BeH2)_n}$, the bonding is different. Both show clustering, but the hydride has three-center bonding, whereas the halides have halogen bridges, in which a halogen atom bonded to one atom uses a lone pair of electrons to form a coordinate bond to another beryllium atom.

I found another reference in Miessler & Tarr that is similarly vague about this issue:

Although beryllium halides $\ce{(BeX2)}$ may be monomeric and linear in gas phase, in solid phase the molecules polymerize to form halogen-bridged chains with tetrahedral coordination around beryllium. Beryllium hydride ($\ce{BeH2}$) is also polymeric in solid with bridging hydrogens. The three-center bonding is involved in bridging by halogens, hydrogen and other atoms.

Again here it is not explicitly mentioned here that halogens are involved in multicenter bonding for beryllium but are involved in general and it looks contradicting to what J.D.Lee says

Wikipedia cites this structure without any further comment. I would concur from this structure that it is involved in two seperate 2c-2e-bonds. My instructor had noted that this is 3c-4e-bond, too, which makes this matter rather confusing.
I would prefer if someone shows a primary citation with evidence for the type of bonding in this polymer (mentioned explicitly)
If there is no multicenter bonding like in the case of $\ce{(BeH2)_n}$ then is there any qualitative reasoning as to why? (Any quantitative reasoning is also welcome albeit not preferred.)
Sources:

Concise Inorganic Chemistry by J.D.Lee (5th ed.)
Inorganic Chemistry by Gary L. Miessler and Donald A. Tarr (3rd ed.)

IIRC, the Be-Cl bonds are equivalent. So, they are better considered as a 3c-4e bond (i.e. delocalised in a sense). The covalent Be-Cl and coordinate Cl->Be is just a Lewis model of the compound. If you are counting electrons, both give the same result, so it's a matter of the way of looking at things. IF one bond was covalent and the other was coordinate, you would be able to differentiate them, woudn't you?

@SRMaiti . But are 3 orbitals (2 of be and 1 of cl) really overlapping? I would like to see some explicit reference. Also by delocalisation you mean resonance? If so why would J.D.Lee detail the bondings separately? If you have any sources please direct them to me. Thanks.

Hmm I am not sure how the orbitals are overlapping. No, I didn't mean resonance, I was trying to say that the electrons are spread across two bonds, I couldn't find any other words except delocalisation. I think J D Lee wrote them separately because the Be-H-Be bonds are electron deficient, whereas Be-Cl-Be bonds are not. I don't have a reference right now, I am recalling something I read somewhere, so it might be wrong.

I don't think there's any contradiction here between "two 2c/2e" and "one 3c/4e" descriptions. I believe the use of "three-center" is just to emphasize that the halogen has two bonds rather than the typical one.  The 3c description also emphasizes the equivalence of the two bonds, even though the drawn Lewis structure necessarily makes one dative and one covalent. We could draw contributing resonance structures with each bond taking on covalent or dative character depending on the structure.

@Andrew when I say 3c-4e I imagine that there are (atleast 3 orbitals overlapping (sort of the situation where in a venn diagram we see AnBnC and if AnBnC=0 then it is not 3 centre overlapping) I am still in high school so I don't know if currently this is a valid definition though. Also can you site me a source that refers to delocalisation and that 2 2c/2e and 3c/4e are really the same (Although it does seem plausible just to be sure). My main confusion is with what J.D.Lee was implying really.

Lee appears to be using "multi-center bond" to refer exclusively to 3c/2e bonding (found only in BeH2), thus distinguishing it from the 3c/4e bonding of BeCl2. The difference is because H has no lone pairs that can participate in bonding. With regard to delocalization/orbital overlap, read any introductory text on MO theory and you will learn that electrons are always delocalized across a molecule. We would not expect significant overlap between atomic orbitals on the two Be atoms, since there is no bonding between them, but both will have overlap with AOs on the H/Cl atoms.


text:
Can different components show different deviations from Raoult's law?
Introduction:
Raoult's law is used to calculate the partial pressures of the vapour phases of volatile components of a solution
Given that there are two components, A and B, in the liquid-liquid solution, there are three types of intermolecular forces of attraction, A-A, A-B and B-B
A non-ideal solution disobeys Raoult's law
If A-B is lesser than A-A and B-B, the molecules of the components find it easier to escape the liquid phase. Thus, there is higher partial pressure than that predicted by Raoult's law. The solution is said to exhibit positive deviation
If A-B is higher than A-A and B-B, the solution exhibits negative deviation
My question:
Are there solutions in which A-B is greater than A-A but less than B-B? Would component A exhibit negative deviation, and component B exhibit positive deviation, in the same solution?
Can the magnitudes of these deviations cancel out each other such that the total partial pressure of the solution is still linear (with respect to change in more fractions of the components)? Can the overall solution obey Raoult's law even when the components don't?
Reference:
https://ncert.nic.in/textbook/pdf/lech102.pdf

It's hard to see how you might make this work. For instance, perhaps a complex AB would observe Raoult's law while the dissociated components A and B don't. But if you stray from an associated or stoichiometric balance then you would have A or B in solution, and Raoult's law would not be obeyed.

IMHO, there is no need for A-B to be either higher either lower than both A-A and B-B. Important is relating of A-B to the average of A-A and B-B.

When considering Gibbs free energy comparisons across a (binary) phase diagram, you get to pick the reference states at the extremes. In other words, one can set A-A and B-B to be zero, and A-B is then calculated relative to those reference states. Or you could set B-B to a billion J/mol and ... nothing changes in the analysis at all.

@JonCuster My textbook has shown the cases where A-B is less than both A-A and B-B or greater than both A-A and B-B. My doubt is whether A-B can be in between them instead


text:
Comparing geometries and substrates calculated with DFT for different pathways
As DFT calculation is an excellent tool for confirming and understanding experimental results, I got deep into this field as organic chemist. I successfully isolated interesting transition states (TS) for my system and performed frequencies, IRC, and optimization calculations.
However, I have a practical question for the interpretation of the relative energies obtained from the calculations. My system can be understood as a simple two-step reaction:
$$\ce{A + B -> C -> D + E}$$
I did all calculations for the first step toward the formation of $\ce{C}$ and I was able to isolate several geometries for the TS for the second step. From all possible geometries tested, two have shown to be more favorable as the difference of energy between the reactants complex (optimization of IRC in reverse direction) and the TS were the most little (around $\pu{200 kJ mol^-1}).$ The reaction pathway with relative energies is shown on the attached figure.

I have the following questions:

When having the choice between these two pathways, which could be the best choice as the difference of energy between reactant complex and TS for this step are the same? Can the orange pathway be preferred for the reaction for its more little TS absolute energy?

I want to compare this reaction to the same reaction with a different substrate. With a first substrate, I get a TS at an energy of the same order than the blue pathway, an alternative geometry like the orange giving the same energy. The difference of energy between complex and TS is also of $\pu{200 kJ mol^-1}.$
When I use my new substrate, I expect to get an activation barrier that is lower, which could fit with the orange pathway. May this new orange geometry with a lower TS energy explain why the reaction is faster with this new substrate (saw experimentally)?


text:
How to do a capacitively coupled electrode?
I have an AUTOLAB potentiostat: working electrode (WE, Pt wire), counter electrode (CE, Pt wire), reference electrode (RE, Ag/AgCl solid state).
I would like to capacitively couple the potentiostat and the three electrode cell system through the working electrode by connect a capacitor in series with the WE.
Am I doing it right? I read in a paper by van Dongen and Serdijn [1] that capacitive coupling could increase the cell potential during stimulation I would like to do. How should I do it to prevent this increase, or minimize it? Even the OCP is not the same, if I connect the capacitor into the cell.
Reference

van Dongen, M. N.; Serdijn, W. A. Does a Coupling Capacitor Enhance the Charge Balance during Neural Stimulation? An Empirical Study. Med Biol Eng Comput 2016, 54 (1), 93–101. DOI: 10.1007/s11517-015-1312-9.

Nice linked paper, but perhaps look at my answer here and the figure. It seems you want to insert a capacitor between the working electrode and the op amp’s inverting input. That yields a differentiator configuration: it blocks DC offset voltage, but differentiators are intrinsically noisy. So then you would likely need a somewhat more complicated configuration, e.g., some low pass filtering. So this is really an electronics problem and we have some electrical engineers here. Maybe one will suggest something or post an answer.

I do not think so, that low pass filter would be efficient in my case as these Pt microelectrodes (WE) will be used for cochlear implants. Could anyone describe what low pass filtering could be of use here, what are the details?

A differentiator configuration is a high pass filter, so you get the derivative of the response and you want to roll off the excess high frequencies. Have you considered not altering the front end and simply digitizing the response (with, e.g.,a 16 bit ADC) so you can use software processing (signal processing) to deal with the DC voltage offset? I will not comment further since we have users here who are also high rep in the electrical engineering stack exchange.


text:
Ortho effect and the relative acidity in aromatic acids

What is the order of the acidity of 2-methyl benzoic acid and 2-isopropyl benzoic acid?
I know about ortho effect which states that the acidity of ortho- substituted aromatic carboxylic acids is greater due to stearic influences of the ortho substituent.Thus the two are probably more acidic than benzoic acid
Coming to the relative comparison , both groups show hyperconjugative effect as well as +I effect . The +I effect of -CH3 is less than that of isopropyl group whereas the hyperconjugative effect of isopropyl group is less than -CH3 group.As hyperconjugation is stronger, 2-isopropyl benzoic acid will be stronger than 2-methyl benzoic acid.
Is my reasoning correct? Or am I going wrong somewhere?


text:
Regarding the reaction of xenon with fluorine gas
In my textbook, under the topic about xenon-fluorine compounds, the following three equations are mentioned:

$\ce{Xe(g) + F_2(g) -> XeF2(s)}$ (Xenon in excess / 2:1 ratio)
$\ce{Xe(g) + 2F_2(g) -> XeF4(s)}$ (1:5 ratio)
$\ce{Xe(g) + 3F_2(g) -> XeF6(s)}$ (1:20 ratio)

I am having trouble understanding how these ratios are obtained. For example, if we consider the second equation, 1:5 is definitely not the molar ratio, and it doesn't seem to be the mass ratio either. Similarly, for the other two equations too, I am not able to find a way to get the corresponding ratios.
Could anyone please explain on what basis are these ratios obtained?

They are assumed to be experimentally determined molar ratios to get the best (or good at least) yield of the particular fluoride.


text:
What is the basicity order of peroxide, superoxide and hydroxide anions?
My teacher told me superoxide and  peroxide anion are more basic than hydroxide but according to Wikipedia

Given that the hydroperoxyl radical has a pKa of around 4.8

and and pKa of H2O is

Acidity (pKa) 13.995[9][10][b]

Therefore, basicity of superoxide anion should be lesser than hydroxide as hydroperoxyl radical is more acidic than water.
For hydrogen peroxide I am not able to find pKa for both the hydrogen as it has two hydrogen so I don't know how to compare it, and I am not able to find the data of pKb of radical directly.

Water pKa is not the same thing as pKw.

Its given on wikepedia , i have provided the link you can check also

OK, I have seen the reasoning for both  14.0 and 15.7. Your Q is OK. See also https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/What_is_the_pKa_of_water%3F