text:
Why don't Alkyl Fluorides form Grignard Reagents
The reactivity of forming Grignards reagent were given as RI > RBr > RCl > RF and was noted that Grignard reagent with alkyl fluorides cannot be formed. What is the reason for this?

Consider the strength of the C-F bond compared to the other C-Hal bonds

As suggested by Waylander, the greater strength of the carbon-fluorine bond versus other carbon-halogen bonds raises the activation energy for obtaining a Grignard reagent with fluorine. A secondary issue is that if you do run a Grignard-reagent synthesis using a fluoride, you may complicate the workup by forming magnesium fluoride precipitate or (especially hazardous and glass-corrosive) hydrofluoric acid. For both reasons it is more feasible in most applications to use a halogen that is easier to work with.
Fluoride Grignard reagents can be made by first using dioxane and heating to promote the disproportionation of a heavier-halide Grignard reagent forming the dialkyl magnesium compound ($\ce{2RMgX <=> R2Mg + MgX2}$; see Cope[1]) followed by reaction of the dialkylmagnesium compound with appropriate fluorine compounds (see Ashby and Nackashi[2]).

Arthur C. Cope (1935). "The Preparation of Dialkylmagnesium Compounds from Grignard Reagents". J. Am. Chem. Soc. 57, 11, 2238–2240.
https://doi.org/10.1021/ja01314a059.

E.C. Ashby, John Nackashi (1974).
"The preparation of organomagnesium fluorides by organometallic exchange reactions".
Journal of Organometallic Chemistry,
72, 1, 11-20.
ISSN 0022-328X,
https://doi.org/10.1016/S0022-328X(00)82011-2.

A third reason - alkyl chlorides/bromides/iodides are generally easier to form/more widely available than alkyl fluorides

Grignard reagents where the halogen is fluorine was previously not taken into interest owing to the unreactive nature of the carbon-fluorine bond and the preparation of such species was very time and resource consuming. Until recently, chemists looked for ways to create this species:

Using specially activated magnesium (Rieke metals)
Organometallic exchange reactions (See @OscarLanzi's answer especially the 2nd paper)
Selective direct fluorination (Deyoung et.al., 1992)
Mechanochemical activation (Speight et. al, 2020)


text:
meaning of ingredient labelling `w/v`
A mouthwash states it has among other ingredients, menthol 0.042% w/v.  What is the correct interpretation?
This chemistry.stackexchange.com answer says it's weight per volume with unspecified units, so it is ambiguous. Kilograms per litre or grams per cubic light-year or whatever.
Alternatively, could it mean that 0.042% w/v is the concentration by weight and by volume, presumably because they are interchangeable?
The only reason to think they are interchangeable is that the density of menthol is 0.890 g·cm−3 which is close to water's density so perhaps it is also close to the density of this solution before the addition of menthol.

In chemistry we usually use grams for mass and for volume we use litres or millilitres. I don't think that they would put 0.042 gram menthol in 100 litres of water. So in your scenario w/v means $g/mL$ or $kg/L$ This is the what w/v mostly means. Always use comparable units. You cannot compare grams with cubic lightyear. Kg and L ; g and mL are comparable units.

Percentage (one of hundred), ppm (one of million), ppb (one of billion) and similar units are relative quantities, relating two values of the quantity of the same kind. Like two masses/weights, two volumes, two molar amounts.
As this is source of possible misinterpretation, it is often marked explicitly like w/w, v/v, n/n.  For these reasons, it is generally recommended to avoid these relative units, especially without explicit meaning, using explicit absolute quantities like e.g. $\pu{mg/L}$, $\pu{g/kg}$, $\pu{\mu mol/mol}$ etc.
The mixed percentage (w/v) is a trick, assuming $\pu{1 mL}$ of solution has mass $\approx \pu{1 g}$ and therefore 1 % w/v as $\pu{1 g}$ in $\pu{100 g}$ means also  $\pu{1 g}$ in $\pu{100 mL}$.
Particularly, $\pu{0.042 \% \text{w/v}}$ means $\pu{420 mg}$ of the substance in $\pu{1 L}$ of the product, as it is considered as $\pu{0.042 g}$ in $\ce{100 g}$(=$\pu{100 mL}$).

But it doesn't say w/V.  It says w/v.  Both lowercase.

It is the same, is was just capitalization glitch.  :-) As volume has as the symbol the capital V, but in this percentage context is used small v.

Please show your rationale for equating  0.042 % w / v  with  420 mg / L .   I can start by equating the original to 0.00042 w / v by reversing the implication of the percent sign. Please show the next step that you employed to equate it to 420 mg / L.

As previously noted,  1% w/v means 1 g/100 mL or 10 g/L, 0.042% w/v means 10 . 0.042 g/L = 0.42 g/L = 420 mg/L

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text:
Why does the Nernstian cell potential remain ideally constant during charge or discharge of an electrochemical cell?
I was reading a couple of papers by Prof. B E Conway, including his book "Electrochemical Supercapacitors", where he came across this statement that the Nernstian potential of an electrochemical cell remains ideally constant during its charge/discharge.
Can anyone please explain this statement? Please try not to use any analogies with water pumps and stuff. Instead, I would highly appreciate proper chemical and mathematical equations being used wherever necessary.

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

Supercapacitor is a different thing than a simple cell https://onlinelibrary.wiley.com/doi/10.1002/aenm.202003311

Mithoron: The Wiley article you suggested  does not even mention the word Nernst! The OP is reading a book and the book can talk about the Nernst equation!

@Mithoron I know that supercapacitors are different, but the book compared supercapacitors to batteries, where this particular statement was mentioned

@AChem By the way, what is 'OP'? I saw it in a couple of other places, too.

@JitinSathishKumar, OP is the original poster.

According to Wikipedia, the Nernst equation is given as follows.
$$
E = E^\ominus -\dfrac{\text{R}T}{z\text{F}}\ln{Q_\text{r}}
$$
Notice the dependence of cell potential $E$ on the reaction quotient $Q_\text{r}$. Thus, during charging or discharging, when $Q_\text{r}$ changes, so will the cell potential.

Nernst equation has nothing to do with AC or DC current.

@Achem AC current allows for the measurement of cell potential without changing the electrolytic concentrations because the reversal of current also reverses the chemical reactions that occur, preserving the reaction quotient.

No ananta. I think there is a misconception here. AC current is not used to measure cell potentials. You are confusing cell potential with conductivity measurements. The latter is indeed dependent on high frequency AC for the reasons you suggest above.

@AChem please find a source to support your statement. Meanwhile, I will do the same.

Check any university level physical chemistry textbook or Wikipedia. Search the terms conductivity measurements and potentiometry. In potentiometry the "potential" is measured with zero current drawn. Conductivity exclusively uses AC.

@AChem but the OP has specifically asked for the potential measurement during charging and discharging, not when there is zero current. Although not being an experimental chemist, I may be wrong about the method of measurement, there is no doubt that changing the reaction quotient when charging or discharging changes the cell potential.

Let us continue this discussion in chat.

Of course the cell potential changes while charging/discharging; the bulk cell electrolyte concentrations change and the reaction quotient in the logarithm changes until it reaches the equilibrium constant K, where the cell potential drops until zero.

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text:
Reason for large barrier of hydrogenation of ATP
Why does hydrogenation of ATP have high barrier (mostly doesn't happen spontaneous in typical biological conditions) compared to hydrogenation of OH- (does happen spontaneously). Both contain Oxygen atom with one covalent bond and -1 formal charge. One would naively expect the two reaction have similar properties.
Is the reaction product of hydrolysis of ATP somehow high in energy, while water is low in energy? Does the phosphates somehow "magically" make the reaction barrier higher?

Sorry. English is not my first language. I'm a physics grad student who is not familiar with nomenclature of chemistry. I meant to say hydrogenation or whatever adding a hydrogen in this context is called.

It's not the wrong word, definitely not hydrogenation. You compare hydrolysis of ATP with what really? Protonation of hydroxide anion? It's completely different thing!

"You compare hydrolysis of ATP with what really? Protonation of hydroxide anion?" I'm comparing protonation of ATP with protonation of hydroxide anion. I saw figures of ATP whose phosphates carry oxygen anion (I heard this charge repulsion is why ATP has higher energy than ADP). I saw a post talking about how the protonation of APT's oxygen anion has large barrier. I don't understand why, so I asked this question here. BTW, I sense some annoyance/agitation from you. You seem to be a generous person who's willing to share your knowledge to others. Having a positive attitude makes things better.

Adding a proton to ATP is no different than adding it to hydroxide anion. ATP is way less basic, but there isn't any high barrier. Both get protonated and than lose proton in a matter of nanoseconds. So... I still don't get what you're talking about.

Couldn't you just write the reactions?

Okay. I think I know where my confusion is. I was confused why is ATP often drawn with phosphates connecting negative anions ( P-O- ) instead of hydroxial group (P-OH). When I googled this, I saw the answer to "Why is ATP stable? Why doesn't it get hydrolyzed easily in the cell?", is that it has high barrier. I thought it means that the reaction from PO- + H+ -> P-OH has high barrier... but I now know that hydrolysis of ATP doesn't mean the reaction PO- + H+ -> P-OH.

The later reaction, like you said, happens all the time at nanosecond scale. It is less basic due to resonance between O=P-O and O-P=O, similar to Carboxylate ion.

Well, I did say it's a different thing. knowing right jargon is very important in making useful answer.


text:
How original is my general matrix formula for rate equations?
I was dissatisfied with the method my teacher taught me to solve rate equations, where you're given experimental values for the concentrations of reactants, and the rate of reaction, which given by
$$r = k\,[\ce{A}]^n\,[\ce{B}]^m\,\ldots,\tag{1}$$
as I thought the method wasn't very computational. Instead I rearranged the system of equations and was able to create a general formula to solve using matrices for any reaction
$$\ce{R_1 + R_2 + \ldots + R_n -> P[ ]}\tag{R1}$$
$$r = k\,[\ce{R_1}]^{N_1}\,[\ce{R_2}]^{N_2}\,\ldots\,[\ce{R_n}]^{N_n}\tag{2}$$
$$
\begin{pmatrix}
\ln k \\
N_1 \\
N_2 \\
\vdots \\
N_n
\end{pmatrix}
=
\begin{pmatrix}
1 & \ln_1R_1 & \ln_1R_2 & \cdots & \ln_nR_n \\
1 & \ln_2R_1 & \ln_2R_2 & \cdots & \ln_nR_n \\
1 & \ln_3R_1 & \ln_3R_2 & \cdots & \ln_nR_n \\
\vdots & & & \ddots &  \\
1 & \ln_nR_1 & \ln_nR_2 & \cdots & \ln_nR_n \\
\end{pmatrix}^{-1}
\begin{pmatrix}
r_1 \\
r_2 \\
r_3 \\
\vdots \\
r_n
\end{pmatrix}\tag{3}
$$
I'm wondering if this was an original formula for solving rate equations, and if the formula is obsolete as there must be different equations for rate of reaction dependent on temperature and pressure, which are just reduced to the constant $k$ in this equation.

What are $\Gamma _i$'s?

No, these are $r_i$.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

OK what are the $r_i$? You define the version without subscript, but not the one with. Also showing your derivation would help

I tried my best to convert text from the image, but please do feel free to check notations as well as define them.

Also $ln_n$, what is meant by that?

A few problems there. One of the first one that I notice is, your $ln R$ matrix is not a square matrix, and I am not sure that you are allowed to take "inverse" of a non square matrix.


text:
What does it mean if a calibration curve regression doesn't cross (0,0)
I have been taking many lab courses in my second year of university and I have come across the technique of creating a calibration curve quite a bit. One thing I noticed in a lot of my plots is that many of them do not intersect at the origin (0,0) when I make a linear regression.

This is one of my more recent plots from an MP-AES experiment we did, and the part that I cannot wrap my head around is that the blank solution for this calibration showed basically zero intensity. But we see a completely different spike in signal in the 2ppm trial than the normal sensitivity of the trial that the slope shows.
My questions are, what exactly does it mean that my calibration curve doesn't start at (0,0), and, why is it different from the sensitivity (normal slope of the points)?
I have additionally gotten solutions that get negative concentrations when I try to use external standards (plugging the solution intensity into the equation and solving backwards) and I am wondering  what kinds of errors could account for something like this happening in terms of external standards.

Well, you made error(s), most likely. Who knows what exactly.

I can understand that as the explanation for the negative concentrations, but what about the fact that many calibration curves don't start at 0,0 when doing external standards? Or did you mean both.

In the plot you've given, it looks like you might have saturated your detection method. If you do much lower concentrations (maybe 0-1 ppm), do you get a steeper slope?

We unfortunately only gathered data for a blank and 2 ppm+ solutions. But the part that was confusing is that the reagent blank showed, what is to me at least, an "expected" near 0 absorption at this wavelength.

Am I correct in assuming you are doing microwave plasma atomic emission spectroscopy and detecting potassium by its emission at, say, 766.5 or 769.9 nm? So a decent blank may well produce negligible emission there. Your graph does not include the blank. Hate to say it, but the data are pretty bad: big error bars, not linear response, and not good R squared. @Andrew may be right, but those big error bars make it harder to test his hypothesis.

By the way, it is almost always a bad idea to try to force a calibration curve to pass through the origin: https://chemistry.stackexchange.com/a/116224/79678. Doing so eliminates a possibility of learning the full story that the data are telling.

Yeah, I saw a couple of reports about how forcing it was bad but I was just wondering if there was any other explanation for why my absorbance might be so high. Also yes we did measure it at 769.9nm. I guess if it is mostly error it might be with making the solutions because the trial we did for copper was pretty good and had R2 of like 0.96.

I also get a calculated LOD of like 49 so I guess yeah something probrably went wrong

For given data, using linear approximation does not seem to be numerically justified. It seems the data series is curved toward the origin.

Thank you, after looking at it for a bit I also started to notice that it looks much more like a square root type of function.

Who is "we"? Are you the only one who ran this lab? Did others obtain similar results? Or did you aggregate the results and this is what is shown here? How did you determine the uncertainty in the individual measurements (what is the source of the error bars?)? What is your experimental protocol? Is this an instrument that is in good shape, well maintained? Who did the calibration, and when? Some of these qs might be difficult to answer, but more information will lead to a better answer.

I saw a couple of reports about how forcing it was bad but I was
just wondering if there was any other explanation for why my
absorbance might be so high.

You are doing emission spectroscopy (AES= atomic emission spectroscopy) not atomic absorbance. Most well prepared calibration curves, or and carefully prepared standards have a very small intercept.
What was the composition of the blank?

My question is what exactly does it mean that my calibration curve
doesnt start at 0,0 and why it is diffrent from the sensitivity
(normal slope of the points). I have additionally gotten solutions
that get negative concentrations when I try to use external standards
(plugging the solution intensity into the equation and solving
backwards) and I am wondering what kinds of errors could account for
something like this happening in terms of external standards.

Calibration curves and most of the experimental analytical chemistry follows the GIGO principle, i.e., if the standards are not prepared with utmost care, the sample analysis is a useless exercise. The only absolute way to test the accuracy is to use a reference material with a known concentration of potassium. If the experimental calibration curve produces the correct result, it means everything is working alright. If it produces wrong numbers and they do not match the reference, it means the experiment has serious errors. Negative concentrations simply imply the standard preparation was erroneous. Recheck all calculations, purity of the potassium salt, and of course, the blank.
I also suspect ionization issues, the calibration curve is nowhere linear. Potassium ionizes very easily.

Thank you for your comment, In that particular lab I think we were also comparing it to FAES and I suspected that it might have to do with the temperatures that an MP-AES can reach and how that may cause loss of signal because it might strip electrons. That was my first time using one of these machines so now I really have an appreciation for the precision that we need in sample preparation. Thank you again for the responce.


text:
Mulliken Labels: How do I tell apart E, doubly degenerate, symmetry apart from T, triply degenerate, symmetry for a molecular orbital?
With A and B, it's pretty simple. If you rotate the molecule about its principle axis to where the orbitals line up, and the signs change, it's Mulliken label has a B. otherwise, it's A.
I heard E symmetry means there are two configurations that are equal in energy, and T means there are three configurations that are equal in energy. My professor posted some slides that included examples of E symmetry and T symmetry, but it's still confusing:

Why is this E symmetry? It shows a configuration with one sign along the Z axis and the other sign for the other two axes, and then a configuration with one sign along the Y axis and the other sign for the other two axes. Why can't we have a third configuration, with one sign along the X axis and the other sign for the other two axes, making this have T symmetry? Could somebody explain what E and T symmetry are in their words?
EDIT: I just talked to my professor, and I forgot his exact explanation, but he said the above diagram has a mistake in it. He said two of the orbitals should not be there.

In point groups with three fold or higher axis of symmetry ($C_n$ where $n>2$) there is the possibility of degeneracy, in other words rotation about a horizontal axis and about a vertical one have the same moments of inertia, and hence angular momentum, and will have identical rotational energy levels. (You can see that the structures you draw differ only in how xyz are labelled). Similarly vibrational levels may have as a result of symmetry identical energy levels.  These degenerate motions have to be treated as a set and are labelled E for double degeneracy and T for triple. The symmetry species A, B E etc., are defined by their row of characters in the point group which describes how they behave to symmetry operations such as rotation, reflection, etc. Each row is unique and hence has a unique label, typically 1, -1, 2, 0 etc. this label is a representation of how the symmetry operation affects the molecule.
To be able to generate the characters in a point group and find what orbitals these belong to you need to put vectors on xyz and operate on them with matrices representing rotations, mirror planes etc. so it is not straightforwards but the book by Carter (see below) explains this.
(The Mulliken labels are actually quite complicated, if you want the original see Mulliken, J. Chem. Phys. v23, p1997, (1955), or a summary by R. Carter 'Molecular Symmetry & Group Theory' publ Wiley, or in detail by G. Herzberg "Molecular Spectra and Molecular Structure' vol II van Nostrand Rienhold)

Just a historical trivia: T is not for triple. Actually, E for Entartet is also dubious. Since Mulliken borrowed someone else's symbols from a German chapter in an obscure book called Handbuch der Radiologie. Someone had shared his unpublished book chapter on group theory with him and Mulliken got the full credit for the symbols he never invented. Actually none of Mulliken symbols are his.

Interesting, I should have used three fold instead of triple. Some author use F instead of T.

You are right, F is the older symbol for T.

First of all — it should be noted that this is not a unique representation of the $\mathrm{e_g}$ set. You can take any two linear combinations of these two orbitals (i.e. add and subtract them), and as long as they are linearly independent (i.e. the resulting combinations have no net overlap with each other), they are valid depictions.
But to address the question about why there isn't a third member: the three axes are not entirely equivalent. In fact, there is an example that may be closer to home — there is a similar 'asymmetry' in the set of atomic d-orbitals as well: there is a $\mathrm{d}_{z^2}$ but there isn't a $\mathrm{d}_{x^2}$ or $\mathrm{d}_{y^2}$.
The three axes are equal in the sense that we could have chosen any of the three axes to be 'special', but by convention (and for mathematical reasons) we chose $z$ (see: Why is there a z² label for d orbitals, but no x² and y² labels with corresponding shapes?).
In an octahedral metal complex, the metal d orbitals are split into $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ sets. The former contains the $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ orbitals; the latter the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$. These can overlap with symmetry-adapted linear combinations (SALCs) of ligand orbitals, as long as the SALCs have the same symmetry as the metal d orbitals.
The SALCs are precisely what are drawn in the diagrams in your question. You'll see that the first of the two SALCs that you drew overlaps with the $\mathrm{d}_{z^2}$ orbital, and the second overlaps with the $\mathrm{d}_{x^2-y^2}$ orbital. In other words, the $\mathrm{e_g}$ SALC overlaps with the $\mathrm{e_g}$ set of d-orbitals, so all is well and good.
I'm aware I'm not directly answering why the SALCs lack a third member. This is mainly because I don't want to dive into that maths. But I hope that the discussion above, and the link, explain why the d-orbitals aren't perfectly 'symmetrical'; and I hope that the symmetry overlap between the SALCs and the d-orbitals at least suggests that the same argument can be extended to the SALCs.


text:
Dewpoint of water in relation to composition of atmosphere
A couple of colleagues and I were talking about the nature of dewpoints, specifically the dewpoint of water in relation to the chemical composition of the atmosphere in a closed space.
If I remember my theory, then the chemical composition of the atmosphere should not have any impacts on the dewpoint of the water (assuming ideal gas law and that the atmosphere won't react with the water vapors)
My colleagues disagree with this, as they are of the opinion that different gasses can hold different concentrations of water vapor, e.g. $\ce{CO2}$ versus $\ce{N2}.$ As such, they argue that water would "crash out" of the gas phase at higher temperatures in atmospheres that could "hold less water".
Instead of wasting countless hours on research, I thought I might as well ask you brilliant people, if any of you could settle this dispute, or point me in a direction where an answer could be found.

Does this answer your question? How does composition of air affect vapor saturation limit?

Gases do not hold water vapor. Vapor is gas on its own. Neither oxygen nor nitrogen need other gases to hold them either. Gases are mutually toward each other just  bystanders. Vapor can be present in vacuum alone.
The dewpoint is the temperature, at which would liquid water have the same saturated vapor pressure as the partial pressure of vapor present in the existing gaseous phase. If such vapor comes to contact with surface of temperature at or below the vapor dewpoint, part of vapor starts to condense, until the surface temperature is above the new dewpoint of vapor with lower partial pressure.
At special conditions near the liquid critical point, the dewpoint does depend on other gases, but in sense of total pressure. Or if there is strong interaction between gases. But neither of it is the case of water vapor in air near ambient conditions.

Somewhat related to deviations from independent vapor pressures are your answers to Can precipitation occur in states of matter other than liquids? and Why exactly does high pressure (72 bar) nitrous oxide's pseudocritical¹ temperature drop so far when oxygen is added to make 50%v/v mixture? (Entonox)

@uhoh Yes, they are, I did not want to go to these details here

yep, but I have a hunch at least some readers of this answer will enjoy doing so in the future ;-)


text:
Non-octet structures of stable nitrogen compounds
A question in my exercise book demonstrates that nitrogen cannot form non-octet structures since the most common examples — nitrogen dioxide and nitrogen monoxide — are unstable/reactive. For example, $\ce{NO2}$ quickly dimerizes to $\ce{N2O4}.$ $\ce{NO}$ is quickly oxidized to $\ce{NO2}.$ Other elements, like phosphorus, on the other hand, can form stable phosphorus pentachloride.
I do not think the question is precisely set. Despite the instability of nitrogen dioxide and nitrogen monoxide, it does not mean nitrogen “cannot” form non-octet structures.
Could anyone give me examples of stable nitrogen compounds where the nitrogen atom does not achieve an octet? I just want to prove that the answer is wrong.

What do you mean when you say NO2/NO are unstable? Do you mean reactive?

"Unstable" and "Reactive" are two different things. 2. What's the basis of comparing nitrogen oxides with phosphorus halides?

Your comparison of the nitrogen oxides with phosphorus pentachloride is not apt. The former have unpaired electrons whereas the latter has multicener bonding with all electrons paired. These represent different departures from the octet rule.
Nitrogen forms trifluoramine oxide, $\ce{NOF3}$, whose structure is described here. The nitrogen is bonded tetrahedrally to the other atoms, and the normal octet rules "should" allow only single bonds between all the connected atom pairs. But empirically the nitrogen-oxygen linkage shows double-bond character as if the nitrogen were forming five bonds. "Outer $d$" valence orbitals can't explain this when the atoms are all in Period 2. Molecular-orbital theory can.


text:
Colour of coordination compounds due to d-d transition
Hexaaquanickel(II) is green but tetracyanidonickelate(II) is colourless.
On internet I found two explanations.
One  used valence bond theory that claimed there is no excitation of electron in tetracyanidonickelate(II).
Other used crystal field theory that claimed there is excitation in tetracyanidonickelate(II) but not in visible region.
So my question is that which one is more correct?
At university level both theories might be obsolete but which explanation is 'less wrong'.
Some people would say VBT is more correct but then question arises that if every d orbital was having paired elections in tetracyanidonickelate(II) then where does our excited electron went.

Your link goes to an answer that does not mention VBT, nor does it say that all of the d orbitals are full. It says only that all d electrons are paired. There is no explanation of why that would mean that an electron cannot be excited. Can you provide your reasoning for why VBT means an electron cannot be excited?

@Andrew if there is no vacant orbital left in case of cyanide then where does our excited electron went?

why do you think there is no vacant orbital?

@Andrew 4 d orbitals keep 8 electrons of nickel and 5th one is busy in dsp2

even if you use a hybrid orbital (which you can't for the purposes of determining energy gaps), the ligand electrons only occupy the bonding orbitals. The antibonding "dsp2" orbitals are vacant and those are the ones that would be higher in energy than "d" orbitals.  But the general answer to your question is that hybrid orbitals cannot be used for determining energy differences between orbitals.

@Andrew your explanation is nice. But this gave rise to dilemma. According to CFT electrons jump in split d orbitals but according VBT they go to antibonding dsp2. Which theory should prevail?

As I said before, you can never use hybrid orbitals to compare energy gaps between orbitals.

Here is a copy of the published explanation as printed in the first reference, that looks reasonable :
In $\ce{Ni(H2O)6]^{2+}}$, $\ce{H2O}$  is a weak field ligand. In this complex, the d electrons from the lower energy level can be excited to the higher energy level and the d−d transition may be present in the visible region. Hence, $\ce{Ni(H2O)6^{2+}}$ is coloured.
In $\ce{[Ni(CN)4]^{2−}}$, the electrons are all paired as CN- is a strong field ligand. Therefore, d-d transition is not possible in $\ce{[Ni(CN)4]^{2−}}$ in the visible region. The transition occurs in the UV region. Hence, it is colourless.

If there is no vacant d orbital left in case of cyanide then where does our excited electron went.

You are right. I had not thought of this situation.

So valence bond theory wrong?

@ Hariot : No idea about validity of VB theory !..


text:
dilution and pH and pOH
The question is: given a $10$ ml solution with pH = $12.3$, which is then diluted to a $100$ ml solution, find the pH of the new solution. My attempt was as follows:
[H$_3$O$^+$] = $10^{-12.3}$, and $c = \frac{n}{V}$ gives $n = 10^{-12.3} \frac{\text{mol}}{\text{dm}^3} \cdot 0.01 \text{dm}^3 = 10^{-14.3} \text{mol}$. After dilution, we have $V_{\text{new}} = 100 \text{ml} = 0.1 \text{dm}^3$ and hence $c_{\text{new}} = \frac{10^{-14.3}}{0.1} \implies \text{pH} = 13.3$. This answer, however, seems wrong since the answer in the book is $\text{pH} = 11.3$. What did I do wrong? When I however switch pH to pOH and to the same process (but doing $14.0 - \text{pOH}$ at the end to get pH) I get the correct answer $\text{pH} = 11.3$. Anybody knows what's going on?

Instead of writing $100$ ml ($100$ ml), write $\pu{100 ml}$ ($\pu{100 ml}$).

Dilution of water solutions causes convergence of pH values toward 7 at infinite dilution (at 25 deg C). // Remember  Kw = [H+][OH-]=10^-14 // Dilution of the above solution causes generally raise of H+ concentration, being produces by water autodissociation. // The target pH can be in interval approximately 11.3(strong base solution) - 12.3  (pH buffer). The question as is does not have a single right numerical answer.

As Poutnik writes in the comments, the question does not have a single answer unless you specify the components of the solution.
Given that the pH is fairly basic at pH = 12.3, you could guess or assume that this is a strong base dissolved in pure water, and ignore ionic strength effects, and get the numerical answer given by the answer key.
However, you can exclude your first answer even without knowing more information than given:

pH=13.3

If you add water to a solution, you are diluting it. If you dilute it infinitely, you approach pure water with a pH of 7. You don't expect the pH to get more extreme as you add water (maybe there is a case where this happens at very high concentrations, but not in a textbook scenario such as this).

If you started with a $\pu{10^{−12.3} M}$ solution of some inert material, then diluting it by a factor of 10 would indeed result in a $\pu{10^{−13.3} M}$ solution.
$\ce{H3O+}$ is not inert. It reacts with $\ce{OH-}$ and water (and of course with many other things too, if they are present). This means that the quantity of $\ce{H3O+}$ can, and does, change as the solution is diluted. Ignoring this reaction gives you the wrong answer.
The authors of the problem almost certainly intended you to solve it in the way you did (computing $\ce{[OH-]}$ and dividing by 10). But let's try to solve it from first principles to see what's actually happening; hopefully this makes things less mysterious.
Let's assume that the solution was formed by dissolving a strong base in water. This essential assumption is not stated in the problem, because the problems in elementary textbooks are often poorly written. Suppose the base is $\ce{NaOH}$. We will then have the following formulas:

$\ce{[Na+] + [H3O+]} = \ce{[OH-]}$ (charge balance)
$\ce{[H3O+][OH-]} = \pu{10^{-14} M}$ (equilibrium for the acid/base reaction)

In the system before dilution, $\ce{[H3O+]} = \pu{10^{-12.3} M}$ is given and so we can use the two equations to find the remaining two values. For all practical purposes, $\ce{[OH-]} = \ce{[Na+]} = \pu{10^{-1.7} M}$. Technically the two values are not exactly equal, but the difference is $\pu{10^{-12.3} M}$, about 10 parts per trillion difference, so it can be ignored. (If that makes you nervous, go ahead and do the calculation exactly and see how much it affects the answer!)
During the dilution, the quaitites of $\ce{H3O+}$ and $\ce{OH-}$ can change, because there is a reaction involving those species and water. So the post-dilution values of these are unknown But there is no reaction involving $\ce{Na+}$, so the quantity must remain unchanged. You can therefore deduce that the final concentration of $\ce{Na+}$ is $\pu{10^{-2.7} M}$. Thus we once again have two equations and two unknowns, and we can solve them for the final values of $\ce{[OH-]} = \pu{10^{-2.7} M}$ and $\ce{[H3O+]} = \pu{10^{-11.3} M}$


text:
Why is INDO suitable for covalent systems despite neglecting two-center two-electron integral?
I do not have a good basis understanding of basic chemistry so I am hoping to obtain an explanation that is as narrow as possible.
My current understanding of the semi-empirical method INDO is that

it inherits the traditional assumption of the Hartree-Fock method through
1.1 neglecting correlation effect
1.2 the Born - Oppenheim assumption
1.3 construct a wave function $|\psi\rangle_{\mathrm{HF}}$ using a Slater determinant chosen from one of $\bar{M}$ configurations that arose from $N$ electrons distributed over $K$ single-electron spatial orbitals

One can extend the Hartree-Fock method to a semi-empirical method of which a class is the INDO (Intermediate Neglect of Differential Overlap) method.
INDO assumes

inherits the assumption of ZDO which assumes that product of basis function $\chi_{m}$ (describing single - electron spatial orbitals) are ignored.
ignores all two-center two-electron integrals (this qualifies electron-electron interactions) except for Coulombic-type interactions.

In this way: INDO is suitable for systems exhibiting covalent and polar covalent bonding

My question: If electron-electron interactions are responsible for covalent and polar covalent bonding, why is INDO suitable for such systems?

After seeing this again, it might not gain much traction here. I think it's a good question and on topic here though. You might get an answer on [mattermodeling.se]. We could migrate it there. Please raise a custom flag on the post. Please don't just ask the question there again unless you delete it here first. Cross-posting is not recommended across the network.


text:
Aminium/Ammonium?

Chemdraw names this as 2-carboxyethan-1-aminium. Shouldn't it be ammonium? What's an aminium ion?

https://goldbook.iupac.org/terms/view/A00277, https://iupac.qmul.ac.uk/BlueBook/P6.html#62

@andselisk I have seen this, but unfortunately it provides no information on aminium ions for me to understand.

"Aminium" is the cationic suffix for the parent neutral amine group $\ce{R-NH2}$. There is no ammonium/ammonia present in your structure.

But it's a substituted ammonium ion. https://en.m.wikipedia.org/wiki/Ammonium This wiki page says that substituted ammonium ions in IUPAC nomenclature are called aminium ions. Ir then gives an example forming dimethylammonium ion by protonation of dimethylamine.

"Substituted ammonium ion" is amine. The cationic form of β‐alanine you have drawn is an amino acid, not ammono acid.

For example, we have H3CNH3+. It's called methanaminium. But why is (CH3CH3)NH2+ called dimethylammonium while it's from a secondary amine?

No, $\ce{Me2NH2+}$ is dimethylazanium. According to IUPAC, the names stemming from ‘ammonium’ are only for quaternary salts.

Then what is (CH3)3NH+ and (CH3)4N+ called respectively. The second one tetramethylammonium? And the first one?

PINs for $\ce{Me3NH+}$ and $\ce{Me4N+}$ are trimethylazanium and, yes, tetramethylammonium, respectively. Casually you can see and informally use trimethylammonium for $\ce{Me3NH+}$ too, but that name wouldn't be the preferred IUPAC one which ChemDraw usually tries to produce.


text:
Calculating the pH after adding hydrochloric acid to the potassium hydrogen phosphate solution
I am trying to calculate the $\mathrm{pH}$ of a solution obtained by mixing $\pu{20.00 mL}$ of $\pu{0.2 M}$ potassium hydrogen phosphate solution and $\pu{10.00 mL}$ of $\pu{0.25 M}$ hydrochloric acid solution $(K_{a2} = \pu{6.34 \times 10^{-8} mol/L})$.
My solution:
So, hydrogen phosphate reacts with oxonium to form dihydrogen phosphate: $$\ce{HPO4^2- + H3O+ <=> H2PO4- + H2O}$$ The equilibrium concentrations of hydrogen phosphate and dihydrogen phosphate are: $$[\ce{HPO4^2-}] = c_0(\ce{HPO4^2-}) - [\ce{H3O+}] \quad \text{and} \quad [\ce{H2PO4-}]=[\ce{H3O+}]$$
where $[\ce{H3O+}] = \frac{V(\ce{HCl})}{V(\ce{HCl)} + V(\ce{HPO4^2-)}}c_0(\ce{HCl})$.
Finally, I plugged everything in the Henderson–Hasselbalch equation $$\text{pH} = \text{p}K_\text{a2} + \log{\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}}.$$
I got $\mathrm{pH}$ of 7.34, but the answer sheet says 6.97. What did I do wrong?

What concentrations did you calculate? You have a good explanation of the correct answer from Poutnik, but I'm not sure it is clear how you got 7.34, and where the error was.

There is initially $\pu{0.02 L} \cdot \pu{0.2 mol/L} = \pu{0.004 mol}\ \ce{HPO4^2-}$
There is added $\pu{0.01 L} \cdot \pu{0.25 mol/L} = \pu{0.0025 mol}\ \ce{H+}$
This leads to approximately $\pu{0.0025 mol}\ \ce{H2PO4-}$ and $\pu{0.0015 mol}\ \ce{HPO4^2-}$.
As $\text{p}K_\text{a} = -\log {(\pu{6.34E-8})} \approx 7.20$,
then $\text{pH} \approx \text{p}K_\text{a} + \log {\left(\frac{n(\ce{HPO4^2-})}{n(\ce{H2PO4-})}\right)}=7.20 + \log{\frac{0.0015}{0.0025}} = 6.976$

@Poutnik. Sorry ! One zero is missing in the calculation of the ions after mixing. Mixing $4$ millimol $\ce{HPO4^{2-}}$ + $2.5$ millimol $\ce{H+}$ does not produce $25$ millimol $\ce{H2PO4^-}$ and $15$ millimol $\ce{HPO4^-}$ as written, but it produces $10$ times less. This mixture produces $2.5$  millimol $\ce{H2PO4^-}$ and $1.5$ millimol $\ce{HPO4^-}$.

@Maurice Thanks. Fixed.

I got pH of 7.34, but the answer sheet says 6.97. What did I do wrong?

I'm showing here where you make the mistake as an answer to your above question (If you have already compared your answer to Poutnik's correct answer, I'm sure you'd have find where you make the mistake).
You have given following to solve the question:

The equilibrium concentrations of hydrogen phosphate and dihydrogen phosphate are
$$[\ce{HPO4^2-}] = c_0(\ce{HPO4^2-}) - [\ce{H3O+}] \quad \text{and} \quad [\ce{H2PO4-}]=[\ce{H3O+}]$$ where $[\ce{H3O+}] = \frac{V(\ce{HCl})}{V(\ce{HCl)} + V(\ce{HPO4^2-)}}c_0(\ce{HCl})$.

Even though you haven't define what are the $c_0(\ce{HPO4^2-})$ and $c_0(\ce{HCl})$, I assume correctly that they are $\pu{0.20 M}$ and $\pu{0.25 M}$, respectively.
Since correct reaction is:
$$\ce{HPO4^2−+H3O+ -> H2PO4^− + H2O}$$
If initial concentrations of $\ce{HPO4^2−}$ and $\ce{HCl}$ are $\alpha$ and $\beta$, and they are added in volumes $V_1$ and $V_2$, then final volume of the mixture is $(V_1 + V_2)$. Since initial $n_\ce{HPO4^2-} \gt n_\ce{HCl}$, all $\ce{HCl}$ will be converted to $\ce{H2PO4^-}$ as reaction indicated. Thus your calculation of $[\ce{H2PO4-}] = \frac{V(\ce{HCl})}{V(\ce{HCl)} + V(\ce{HPO4^2-)}} [\ce{HCl}]$ is correct before equilibrium.
However, your assumption of $[\ce{HPO4^2-}] = c_0(\ce{HPO4^2-}) - [\ce{H3O+}]$ is incorrect, simply because you have forgotten to include the fact that volume of $\ce{HPO4^2-}$ has also subjected to be diluted to $(V_1 + V_2)$ from initial volume $V_1$. Thus, your calculations gave:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a2} + \log{\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}} = 7.198 + \log{\frac{0.2 - \left(\frac{0.01}{(0.01 + 0.02)}\times 0.25\right)}{\frac{0.01}{(0.01 + 0.02)}\times 0.25}} = 7.198 + 0.146 = 7.344$$
This is obviously the wrong answer. To correct it, you should have dilution factor to $\ce{HPO4^2-}$ as well:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a2} + \log{\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}} = 7.198 + \log{\frac{\frac{0.02}{(0.01 + 0.02)}\times 0.20 - \left(\frac{0.01}{(0.01 + 0.02)}\times 0.25\right)}{\frac{0.01}{(0.01 + 0.02)}\times 0.25}} = 7.198 - 0.222 = 6.976$$


text:
What is in the output of VeloxChem python based program?

After running SCF optimization with VeloxChem code the output is a lengthy document at the end of which is this section.
I am wondering what these numbers mean, more specifically in the parentheses?

I would guess they are the coefficients with largest magnitude for the MOs expressed in the AO basis. But without digging through the manual I can't be sure.

There should be documentation shipped with the program, either physical copy, or/and via their web site.  Is it https://veloxchem.org/docs/intro.html to point e.g. to literature references (https://veloxchem.org/docs/refs.html)?  Does this workshop (https://enccs.github.io/veloxchem-workshop/) cover somewhat computations you launched?

I have read through the manual and various references. Nowhere does it say what that part of the output is.

For future Reference: Please include citations and references. Any text, that is text, like in the case here, should not be included as an image.

Based on the output you can search the source code, which leads to molecularorbitals.py, function _MolecularOrbitals_print_coefficients:
 """
 Prints molecular orbital coefficients to output stream.

    :param eigval:
        The eigenvalue (orbital energy).
    :param focc:
        The occupation number of the orbital.
    :param iorb:
        The index (0-based) of the orbital.
    :param coeffs:
        The AO coefficients of the orbital.
    :param ao_map:
        The string representation map of basis functions.
    :param thresh:
        The threshold for priting AO coefficients.
    :param ostream:
        The output stream.
    """

  ostream.print_blank()

    valstr = "Molecular Orbital No.{:4d}:".format(iorb + 1)
    ostream.print_header(valstr.ljust(92))
    valstr = 26 * "-"
    ostream.print_header(valstr.ljust(92))

    valstr = "Occupation: {:.3f} Energy: {:10.5f} a.u.".format(focc, eigval)
    ostream.print_header(valstr.ljust(92))

    tuplist = []

    for i in range(coeffs.shape[0]):

        if math.fabs(coeffs[i]) > thresh:
            atomidx = int(ao_map[i].split()[0])
            anglmom = to_angular_momentum(ao_map[i][-3].upper())
            valstr = "(" + ao_map[i] + ": {:8.2f}".format(coeffs[i]) + ") "
            tuplist.append((atomidx, anglmom, valstr))

So the stuff in parentheses is a string representation map of a basis function (param ao_map) followed by the corresponding AO coefficient of the orbital (param coeffs).

It's always a good sign when you have to turn to the source code to understand what the software is trying to say…

@Martin-マーチン could be worse :) The code contains some annotations, at least, and is open source. But you need to read closely, because it's like Ian Bush says: "if math.fabs(coeffs[i]) > thresh:"

Yes, try things like that with Gaussian. I think there's not a single comment in the source code. Any other choice than what they have done would also have been silly. Once you use real life molecules with real life basis sets, you get a whole bunch of naught zeros. I've also ad a brief look at the docs and found it surprisingly hard to find the source code repo. Anyway, in my next hacking session, I might give this a go.


text:
Why is more work done for a reversible process?
It is said that reversible processes do more work than irreversible ones. A graph as the one below showing an irreversible and reversible isothermal expansion is used to justify this.

However, I don't see how this is a proof or an intuitive explanation. Why wouldn't it be possible for the gas to, say, expand to $\text{V}_2$ at constant $\text{P}_1$, then drop to $\text{P}_2$. Since the gas in this case begins and ends at the same temperature, shouldn't it also be a valid irreversible isothermal expansion? In this case, though, the work done would be clearly greater than the reversible path, since the area below it would be greater.
What am I missing?

For processes between the same initial and final thermodynamic equilibrium states, there can be reversible paths that give less work. than an irreversible path.

How then should I interpret the statement that reversible processes do more work than irreversible ones? I hear this a lot.

Imagine. a gas expanding against p_ext < p

How you interpret and judge statements like those are up to you.  I would say that there would have to be more constraints placed on the processes used in such a comparison.  Here is a reference to a thread in another forum giving specific examples of reversible paths between the same two end states in which, for some of the reversible paths considered, the reversible work is less than the work for the irreversible path: https://www.physicsforums.com/threads/focus-problem-for-entropy-change-in-irreversible-adiabatic-process.1054165/

You're right to be confused. The diagram you show is neither a proof nor an intuitive explanation. It can't be a proof because the statement "reversible processes do more work than irreversible ones" is not true without significant additional, unstated assumptions.
It's this sort of thing that makes thermodynamics incredibly confusing to students.
It's hard to say what assumptions your teacher/textbook/whatever wanted to add here. But I'll give you a pair of assumptions that do the job, that might be close to what they had in mind:

We aren't considering processes that require putting work into the system, even temporarily.
The only heat source available is the environment, which has the same temperature $T$ as the system's initial and final temperature.

Together, these assumptions prevent you from increasing the temperature above $T$. That means your process is forced to follow a path that lies at or below the reversible isothermal expansion, and hopefully that makes it intuitive that the total work done by any process following these rules must be less than or equal to the reversible process.
There are certainly irreversible paths that do more work, but they are disallowed by the assumptions. You gave one example; it includes an isobaric expansion that requires adding heat, violating assumption 2.


text:
I'm writing a theory for a game called Timberborn, but I need help identifying a chemical from it
As the title says, I'm trying to write a theory for a game called Timberborn where I figure out the backstory of the game. But I need help identifying a chemical in the game. It's called Badwater, and according to the game, it's polluted water, but I'd like to know your guesses about what it could be. Here are the details I already know:

It's an Orangish-Reddish colour, about the colour of bloody lava.
If beavers get doused in it, they get contaminated, which means they're tired, can't work, and eventually die, but if they walk on land that its contaminated, carry containers of it, or stand close to large pools of it, they're perfectly fine, which makes me think it's not radioactive, despite the game using the radioactivity symbol.
You can use it to make "Extract", which you can use to make medicine for contaminated beavers and as an Ingredient to strengthen Dynamite.
It comes from "Badwater Sources" which are old human ruins found near mines.
As it flows, it soaks into the ground and spreads about 7 blocks, killing all plant life on top of it, and makes the ground look like it has cracks in it.
It dilutes in normal water.

What makes you think there's anything really matches it? That's just some arbitrary properties.

I just want to know if there's anything that could be similar.

By the way, thanks to whoever changed the tags. I'm really terrible at choosing good tags.

While this is a game, and I wouldn't like to assume the developers researched environmental issues which occur in the real world, what you describe sounds very similar to what I learned about in undergraduate chemistry concerning acid mine drainage (AMD).
To give you a specific quote, according to Akcil and Koldas:$^1$

Acid Mine Drainage (AMD) is produced when sulfide-bearing material is
exposed to oxygen and water. The production of AMD usually – but not
exclusively – occurs in iron sulfide-aggregated rocks. Although this
process occurs naturally, mining can promote AMD generation simply
through increasing the quantity of sulfides exposed.
Naturally-occurring bacteria can accelerate AMD production by
assisting in the breakdown of sulfide minerals.

The specific type we were taught about was in connection with pyrite contamination:
$$\ce{FeS2(s) + 7/2 O2(g) + H2O(l) -> Fe^{2+}(aq) + 2SO4^{2-}(aq) + 2H^+(aq)}$$
Microbial oxidisation leads to:
$$\ce{4Fe^{2+}(aq) + O2(aq) + 4H^+(aq) -> 4Fe^{3+}(aq) + 2H2O(l)}$$
$$\ce{FeS2(s) + 14Fe^{3+}(aq) + 8H2O(l) -> 15Fe^{2+}(aq) + 2SO4^{2-}(aq) + 16H^+(aq)}$$
This is what it looks like in real life:

And here are some links to further information about it (as well as some more impressive pictures):
Acid drainage: the global environmental crisis you’ve never heard of (The Conversation)
Acid mine drainage - a legacy of an industrial past (RSC)
Hopefully this helps!

A. Akcil and S. Koldas, Acid Mine Drainage (AMD): causes, treatment and case studies, J. Cleaner Prod., 2006, 14, 1139-1145.


text:
Ozone formation
I have read about the formation of natural atmospheric ozone on Wikipedia, where it is claiming that the ozone creation step requires an extra molecule in order to conserve momentum and energy:
$$\ce{O + O2 + A -> O3 + A}$$
I understand that the stated reason implies that it is impossible to have
$$\ce{O + O2 -> O3},$$
but I do not understand why it cannot be:
$$\ce{O2 + O2 -> O3 + O}.$$
Specifically, why does the source claim that we have complete cleavage of the $\ce{O2}$ double-bond before the $\ce{O^{..}}$ radical attacks an $\ce{O2}$ molecule? Why is it not possible that after one bond is cleaved we have an $\ce{^.O−O^.}$ radical reacts with an $\ce{O2}$ molecule directly to form $\ce{O3}$ and an $\ce{O^{..}}$ radical? In particular:
$$\ce{O=O + ^.O−O^. -> O=O^+−O^- + O^{..}}$$
This not only solves the momentum+energy conservation problem, but also seems simpler. The other proposal requires 3 molecules to meet at the same time and does not really explain how the momentum/energy gets transferred.

Hmm I just thought about another possibility: Maybe ( O + O2 → O3 ) can conserve momentum and energy, by having the resulting molecule spin, but I'm not too sure..

It would be removing a wedge by a wedge, as there are 3 fundamental conservation laws (energy/mass, momentum, angular momentum). Generally, 2 molecules forming one need some inert object to pass energy there, respecting all these laws.

@Poutnik: Why can't the bond lengths in a spinning molecule compensate for the extra energy? Anyway, that's a separate question from the one I'm asking haha..

Resulting O3 cannot pass extra energy to its spinning, it would break angular momentum conservation. // BTW, the molecule O2 is at the same time double bonded and biradical (counter intuitive state explained by Molecular Orbitals theory and manifested by oxygen paramagnetism. Substances with molecules without unpaired electrons are diamagnetic.

@Poutnik: I think you missed my point: If the bond length is changed, the energy it stores is also changed. So we have an extra degree of freedom to compensate for the extra energy, no? By the way, I am aware that the Lewis structures are merely approximates, and MO theory is more accurate. But can it answer my question?

Let us continue this discussion in chat.

See this answer: https://chemistry.stackexchange.com/a/133039/17368

@NilayGhosh: I saw that previous post but it does not answer my question at all and quotes from the same source that I cited in my question and have doubts about.

The $\ce{O + O2 + M \to O3 + M }$ does appear written as such in many texts and is unusual in that other atom-diatom reactions that have been studied do not seem to need any $M$, such as $\ce{OH + Cl \to HCL +O}$, $\ce{ H +NO2 \to OH +NO}$,  $\;\ce{F + H2 \to HF + H}$, $\;\ce{O +HCl\to OH + Cl}$ and many others. How a reaction proceeds depends both on the shape of the potential surface between the four species involved as the reaction evolves and on how much rotational and vibrational energy the diatomic has and the mutual approach velocity.
There could be two reasons for writing the reaction with $M$ given that a three body collision reaction is very unlikely to happen on statistical grounds and that is either that the nascent ozone has so much energy that it needs a collision with a third body to remove this energy to stabilise it, or that a transient complex is initially formed such as $\ce{O + M \to OM}$ or $\ce{O2 + M\to O2M}$ which then reacts with the remaining species to give products. The $\ce{I + I \to I2}$ reaction proceeds via an $IM$ intermediate, for example.
(Any linear momentum considerations can be accounted for  by different velocities after reaction and changes in rotational energy levels for angular momentum conservation. The $\ce{O3 + O \to O2 + O2 }$ reaction has quite a large activation energy $\sim 18$ kJ/mol so the reverse reaction probably has a larger one making it even slower that this one).
Update in response to questions
The energy the two species have for reaction is relative to their difference in velocity, this may be quite small relative to any dissociation or activation energy. At $400$ K the average thermal energy is only $\approx 265\;\mathrm{cm^{-1}}$. The ozone and oxygen vibrational (B=rotational) (all frequencies are in wavenumbers) O2, 2063 (B=1.4) and O3, 1110,705,1042 (B=3.5,0.4,0.39) so the molecules are vibrationally 'cool' and rotationally 'hot'. The O3 dissociates at $\approx$ 11700, i.e. a huge energy relative to thermal energy, but not to UV photon energy.
When then O+O2 species approach (or any A+B species) they experience one another's potential before collision (they are not hard spheres) this potential is, approximately, of a Lennard-Jones type so the species are attracted initially and move towards one another at a distance many times typical bond length. At close range repulsion can occur. A reaction may occur provided the energy is not too great or too small and the approach trajectory correct or the O and O2 will partially orbit and then separate. (see figure from https://chemistry-maths-book.com/chapter-11/). Typically this last event is the most likely.
If reaction can happen the transient species $\ce{OO2}$ forms. The presence of the extra O atom will weaken the O2 bond and energy is transferred into new vibrations (this will take only a few picoseconds) and reaction may be complete, most of the time however the transient species falls apart with a bit of energy having being transferred between the departing species. If an O3 molecule is formed it has all the energy of the collision but this is small compared to dissociation energy and so the molecule will be stable but initially be vibrationally and rotationally 'hot'. This will remain so until a non-reactive collision with an other species removes some energy and thermalises the new O3 molecule. How long this takes depends on the pressure, but at experimental pressures of a few torr,  will occur within tens of nanoseconds

The impact parameter is the distance away from the line of centres. This diagram is for spheres, but should give the general idea even for a slightly no spherical molecule such as O2.

So you are claiming that with ( O + M → OM ), the extra energy can somehow be absorbed by the complex formation? I don't get it at all. We still have too much extra energy. (Take the reference frame centred at the centre of mass of the ( O + M ), and after complex formation it has 0 velocity...)

The essential is one versus two product entities. The former needs M, the latter does not.  All other examples have two products. 2 H -> H2 needs some M too.

And by the way none of the reactions you mentioned in your first paragraph has the conservation problem, since there are enough products to balance the equations.

@Poutnik: Then explain the problem with ( O + M → OM ), which again has only one product.

@user21820 M is supposed to be inert, being needed for mechanical reasons to honor conservation laws. It is not some catalyst in chemical sense. It serves just as a ball to be pushed.

You're not getting the point! In the COM reference frame, ( O + M ) has zero momentum, and hence also after forming a complex. But it initially had positive energy, and where does that go?

It does go to M :-) . 2 objects can have any momentum (and therefore kinetic energy) each if they cancel each other,

@Poutnik: What... how?? If you can explain, please post an answer, thanks! I need to go now, but I'll look at it when I get back. But I think you're wrong.. I know quite enough (classical) mechanics..

Let us continue this discussion in chat.

@user21820, the OM complex can form quite easily, even if M is an atom OM has  rotational and vibrational energy as well as translational kinetic energy. Momentum (a vector) will also be conserved. Experiment shows that molecules are more effective than atoms in forming a complex, more vibrational modes and more polarisability for electronic interaction . Such a (short lived) complex is the only possible explanation for the $\ce{I + I}$ reaction as observed by experiment, i.e. experiment shows that $\ce{I + M =IM, IM + I\to I2 + M}$ is the mechanism. (Porter  Proc. Roy. Soc. v261, p28, 1961)

@porphyrin: Your first sentence does not explain anything, especially since you say "even if M is an atom OM has rotational and vibrational energy". I could by your own reasoning equally claim that ( O + O2 ) has rotational and vibrational energy that can account for the extra energy. Please clearly explain why ( O + O2 → O3 ) is significantly different from ( O + M → OM ). (I'm not satisfied with being told "It is like this so just believe it.".)

Perhaps you would say that O3 is not a complex, and admits much less variation in vibrational energy. Fine, but why? Even if we pick the reference frame with zero linear momentum and zero angular momentum, the O3 molecule still has other modes of vibration, such as bond angle fluctuation, bond twisting and bond stretching. In a complex, you won't have complete freedom either, so you can't just say "it works".

species OM if M is an atom or a molecule will have vibrational and rotational energy, i.e. both a diatomic and a polyatomic species has vibrational and rotational energy, there is no basic difference. I don't understand your last posted comment . O3 is a molecule not a complex,  OO2 would be the complex with vibrational and rotational energy. I'm not asking you to believe, but its an experimental fact that I+M=IM, leads to IM+I=I2 as a scheme (particularly when M=I), so its not unreasonable to expect the same with O+O2. You will have to search literature for this.

@porphyrin: I'm not saying I dispute your statements. And yes O3 is a molecule, but why does it not have sufficient freedom in its vibrational modes to store the excess energy? I know for a fact that O3 can have excess energy in its vibration, but nobody has explained why OO2 (or another complex) is necessary for the amount of excess energy involved here, nor how it can even work; why would a weakly bonded complex be stable enough? By the way, regarding ( I + M → IM ), I'm unable to access the cited article, but did they account for adsorption on the container wall?

have a look at this article  https://www.nobelprize.org/uploads/2018/06/porter-lecture-1.pdf  which describes a little of what is in the paper.

I have just read your updated answer. I am reasonably happy with it, thanks! =)

I am quite confused. The $\ce{O + O2 ->[M] O3}$ is one step in the ozone cycle, but I cannot remember whether this actually is considered to be an elementary step (I think so). With the examples in the first paragraph though, I am quite positive these deal with gas-phase chemistry and are overall reaction equations and studies. For example $\ce{OH + O -> O2 + H}$ will always go through $\ce{OOH}$, which is a stable, reactive intermediate.

@Martin - マーチン,  You are right about OH+O and this was my mistake, I have removed it to avoid any more confusion. These are experiments done in molecular beams at v low pressures where two beams collide and products measured, so not under thermalised conditions. These would be the average of many molec. beam experiments made under many different conditions, such as collision energy, collision angle , vibrational energy and orientation.

Evaluating $\ce{2 O2 -> O3 + O}$ singlet oxygen hypothesis:

The molecule $\ce{O2}$ – also called triplet oxygen – has in its standard state a double bond, but it already is a biradical CH SE: why-is-o2-a-biradical. It is paramagnetic and has 2 unpaired electrons.
The next state with higher energy is the singlet oxygen, is paramagnetic and fits the classical idea about double-bonded  biatomic molecule. It is formed by specific chemical reactions and is very reactive.
The opposite reaction $\ce{O3 + O -> 2 O2}$ is fast.
Singlet oxygen is not $\ce{\cdot O-O\cdot}$, but rather $\ce{O=O}$, while the triple oxygen is like $\ce{O÷O}$.
Single oxygen in upper atmosphere is not formed from triplet oxygen, but by photolysis of ozone.
Molecular entities can emit a photon, but it is a process much slower than the resonance break.
$\ce{O2->[UV]2 O}$ is supported by the spectroscopic evidence of oxygen UV spectrum, when absorption below about $\pu{240 nm}$ confirms the breaking of oxygen molecules to atoms, as the needed energy is equivalent to $\pu{241 nm}$. There are not shown any local maxima at higher wavelengths,  suggesting some lower energy photon excited oxygen molecule.

The third object mechanism:
Reminding the known:
In reference frame of zero total momentum, without $\ce{M}$, resulting $\ce{O3}$ from assumed $\ce{O2}$ + $\ce{O}$ collision has zero momentum and zero translational energy. Due angular momentum conservation, it has comparable rotational energy as original molecule $\ce{O2}$. Plus lower potential energy due bonding. Something does not fit.  It is like  when a skateboarder runs through an U-ramp and ends up on the other side free. Molecules do one vibrational period, bounce and part away.
If there is any inert entity $\ce{M}$, both $\ce{M}$ and $\ce{O3}$ are parting with the same but opposite sufficient momentum, according to conversion of energy released by bonding to translational energy. Like we when a skateboarder hits something and does not make it to the other side, being "bonded to the U ramp".
When there are two products, one plays simultaneously the role of $\ce{M}$, like in case $\ce{NO + O3 -> NO2 + O2}$.
Involving $\ce{M}$ does not mean that collisions of $\ce{O}$, $\ce{O2}$ and $\ce{M}$ occur at the same time point. It means one collides when the other two are still attached during resonance meeting before parting.
Resonance can be understood by 2 hydrogen atoms, approaching each other. They reach maximum kinetic and minimum potential energy at the distance equal nuclei distance of a hydrogen molecule. But  the motion continues, kinetic energy decreases to zero, reaching maximum potential energy. Then the motion continues like if reversed in time. In fact, in is one vibrational period on the maximum energy level with one side amplitude in infinity (and then some).
Without existence of resonance, there would be no carbon in universe and no life based on carbon compounds. In fact there are even 2 resonances in row:
$\ce{2 ^4He <<=> ^8Be^{*}<<=>[+^4He]^{12}C^{*}->^{12}C + \gamma }$. These resonances have many orders shorter life than resonances of molecular entities. See Beryllium-8 and Triple-alpha_process.
As a curiosity, see Atomic hydrogen welding.

I seems that you did not read what I wrote in my comments objecting to your claim. I did not ask about M and O3. I asked about M and O, which you had claimed could combine into a complex. I stated that it faces the same issue, ( M + O ) has positive energy but zero momentum after combining, so all that energy must go somewhere. You must explain why ( M + O → MO ) is significantly different from ( O + O2 → O3 ), otherwise you have explained nothing at all.

I did not claim M and O form complex. I need not to explain why something happens if I think it does not happen. IIRC it was Porphyrin's idea. BTW, there is also the difference forming product and forming a resonance that falls apart unless passes energy to other object  . Like $\ce{2 H <<=> H2^{*}->[M]H2}$

M forms a resonance with O or O2, that falls apart, unless encounters O2 or O, forming O3 which together with M part with momentum and kinetic energy from the reaction. Direct O and O2 form a resonance O3, that falls apart, unless colliding with M, passing it energy and then not falling apart.

Please re-read; I asked clearly about ( O + M ) and you answered incorrectly. Since you clearly misread, your answer here is not useful since it does not answer my question and also does not address my objection to the "OM complex formation" idea. Your last comment (claiming resonance rather than complex formation) may be viable but still does not answer my question (i.e. why not O2 + O2 → O3 + O).

And your last sentence ("Direct O and O2 form a resonance O3, that falls apart, unless colliding with M [...]") fails to explain where the extra energy is stored in the time between the initial collision and the subsequent collision with some M. What exactly is this resonance that can hold so much extra energy? I asked about vibrational energy in my last comment to porphyrin, but you seem to think it is too little. So it's unclear what you are claiming at all.

Do not you think you have slided to a confrontational instead of cooperative approach? / Try to put more thinking into the topic.

Let us continue this discussion in chat.

I've put a LOT thinking into the topic, but you have to understand that I hadn't seen any clear answers as of my earlier comment. However, you have since updated your answer, and thanks a lot for that! I will get back to you in chat for further inquiries.

Thanks for your update. But if you take away the peak at about 250 nm, you will get another peak. Look at the graph of ( y = exp(−10·x^2)+exp(−10·(x−0.4)^2)/2 ). Your graph in fact has a peak shifted to the right, and even has a bump on the right as if there is really a second peak. Remember, I'm not at all claiming that my proposed mechanism is right, but that I'm not satisfied with the other proposed mechanism and the lack of evidence against my proposal. If you have any better evidence, I would be grateful.

@user21820 You may confuse the spectrum of oxygen and ozone. The former has no peak at 250 mm, the latter has.

Yes, I did not realize the diagram you provided is for ozone, not oxygen. Your diagram does not show absorbance below 10^−25, so it is useless for analyzing absorbance of oxygen. Ultimately, please look at the graph I gave you to see why it isn't as simple as you think just because you see one peak.

@user21820 Look at the diagram once more. It clearly shows the majority if not all absorption of UV is total cleavage of O2 -> 2 O. I am very well aware that minor side peaks need not to have local maxima, but that is irrelevant to oxygen case. // Aside of that, conversion triplet to singlet oxygen is spin forbidden transition, so such absorption is not impossible, but very highly improbable.// Note that Hg UV lamps filtering wavelengths below notorious 254 nm do not produce ozone. See https://uvresources.com/the-ultraviolet-germicidal-irradiation-uv-c-wavelength/

Ok thank you for that. I'm not sure that an advertisement webpage can count as a scientific source though.. There are many fake-science advertisements from many air-purifier companies.

The point is, it is not provided as scientific evidence, just as applied technology confirming what I said, as result of quick search. You can search deeply if you want, I am not going to pay much effort to this ozone topic any more.


text:
Electrolysis with sodium acetate solution - pH stratified layers - what happened?
I electrolyzed 0.5M CH₃COONa in an H-Cell with a pH-sensitive dye using platinum electrodes. From the attached pic, you can see that an acidic layer formed at the top of the anode solution, while an alkaline layer formed at the bottom of the cathode solution. I understand why there is an acid and a base, but I don't understand why they stratified into these layers. What happened?

You should at least try to analyze the situation. what is being formed, where, how it could affect density and if there are any convective processes.

@Poutnik ok, there is the upwards convective process from the gas bubbles forming on both electrodes, which might explain the acidity at the top of the anode chamber (I also see the alkaline solution temporarily rise by the cathode). I was expecting the only products to be H2 and O2, and to see localized pH changes while the electrolysis was running. What I don't understand is why the alkaline solution sunk. I feel it must be due to density, but am unsure what would form that is more dense.

I also get that water with NaOH is more dense, but I'm not sure why it would not have mixed with the rest of the solution. The experiment has been sitting there 1 day now (without current applied), and while the acidic layer is gone, I still have a nice stratified alkaline layer.


text:
Why do buffer solution only maintain pH to a certain limit?
I know that buffers have a capacity after which they are not able to maintain a relatively stable pH. This capacity is exceeded when we add too much acid or base but I was wondering why does this happen?
Suppose we have the buffer CH3COOH + CH3COONa If I were to keep adding HCl, the solution would actually lose its pH maintaining ability. Is it because when we add HCl, its H+ ions will start to get consumed by CH3COO- produced by the dissociation of CH3COONa and after some time, eventually, no CH3COO- ions to maintain the pH?
If this is the case then it means that the buffer solution actually has two capacities?

The capacity to maintain pH after acid is added
Capacity to maintain pH after base is added

Is this true?

Review the search results for site:stackexchange.com OR site:libretexts.org OR site:hyperphysics.phy-astr.gsu.edu OR site:wikipedia.org pH buffer capacity

Well, yeah, pretty much that.

It is right that a buffer solution has the capacity of maintaining pH after a little acid or some base is added. But this capacity has a limit. The amount of acid or base added must be smaller than (about half) the amount of acid and base present in the buffer solution.

The Henderson-Hasselbalch equation provides numerical insight. The pH of the solution is roughly close to the pKa since the adjustment factor is modulated by the log of the ratio between the acid and conjugate base. However, once this ratio becomes very large or very small, even the logarithm cannot keep it small enough.


text:
How do two orbitals need the same symmetry to bond? Do they need the same mulliken label?
I heard that in order for two atomic orbitals to form a bonding molecular orbital, they need:
Similar energy
Similar symmetry
and Possibility of overlap with the same sign
but, for example in methane, the carbon's P orbitals, with T1u symmetry, form molecular bonds with the hydrogen's S orbital, with A1g symmetry. If the orbitals need the same symmetry to bond, how does this work? Those are different mulliken labels.

This statement is somewhat inaccurate. If you're looking at the tetrahedral point group, then you're not really looking at the atomic orbitals of hydrogen in isolation. You're looking for linear combinations of those 4 orbitals that map onto the irreducible representation of the group. After that, then yes, they need the same symmetry to interact.

If the H atoms are represented as 4 vectors at the corners to the centre of a tetrahedron they combined produce A1 & T2 symmetry species. The H atoms can combine (as LCAO's) with the 2s C orbital (A1) and the three 2p orbitals, which in Td belong to T2 sym. species.

(1) There are no g and u labels in tetrahedral geometry. (2) There are four H 1s orbitals, and collectively they are A1 + T2, and only the T2 bit of it overlaps with the carbon p orbitals.

I'm counting these as answers, thanks.

This post appeared to require clarification, which it obtained in comments. However this could have received a proper answer rather than comments.


text:
Gaussian16 optical rotation calculation using a deuterated species
I am running into issues preparing an input file for DFT calculations. Currently, I am trying to submit an optical rotation calculation for an alpha-deuterated benzyl bromide (1 deuterium on the benzyl position, not 2). I have the isotopically-substituted structure optimized using various DFAs but am unable to proceed past this point. I have tried this with and without referencing the old .chk file (and removed the geom=check for that submission). If possible, could anyone guide me in how to solve this problem? My assumption is that the Freq=(Readisotopes) is not compatible with the Polar=OptRot specification, but I am unsure how to fix this and still have it calculate the isotopically labeled compound.
The variations of files I have tried are:
%nprocshared=128
%mem=180GB
%chk=/work/F32/m06l-optrot-bnbr-6311pgdp.chk
#p M06L/6-311+G(d,p) Polar=OptRot int=ultrafine empiricaldispersion=gd3 nosymm freq=(ReadIsotopes) CPHF=RdFreq

m06l-optrot-bnbr-6311pgdp

0 1
C    -1.153420000000      1.246497000000     -0.027676000000
C     0.218484000000      1.219778000000      0.206372000000
C     0.941392000000      2.405538000000      0.243455000000
C     0.302536000000      3.633029000000      0.048686000000
C    -1.074755000000      3.649411000000     -0.187339000000
C    -1.798641000000      2.464113000000     -0.224083000000
H    -1.718772000000      0.320541000000     -0.055616000000
H     0.725550000000      0.272906000000      0.361967000000
H     2.013053000000      2.384773000000      0.423440000000
H    -1.577463000000      4.600225000000     -0.344160000000
H    -2.868499000000      2.490001000000     -0.404738000000
C     1.075908000000      4.900664000000      0.094575000000
H     1.957993000000      4.831819000000      0.729465000000
H(Iso=2.014)     0.471732000000      5.751722000000      0.405436000000
Br    1.781396000000      5.410164000000     -1.691602000000

589nm

and
%nprocshared=128
%mem=180GB
%oldchk=/work/F32/m06l-isot-bnbr-6311pgdp.chk
%chk=/work/F32/m06l-optrot-bnbr-6311pgdp.chk
#p M06L/6-311+G(d,p) Polar=OptRot int=ultrafine empiricaldispersion=gd3 nosymm CPHF=RdFreq Geom=Check

m06l-optrot-bnbr-6311pgdp

0 1

589nm

Unfortunately this is impossible to debug. We do not have the checkpoint file and we do not know the actual error. Please include an input file that could be run on a computer. On another note: Are you sure you want to request 128 processes and 180 GB of memory? That seems excessive when we're talking about https://en.wikipedia.org/wiki/Benzyl_bromide. Does your compute node even have that much juice? And furthermore: The 6-311G basis set is not a good choice. And what about spin and multiplicity?

@Martin-マーチン I updated the post to better reflect the issue. To be transparent, I did not consider the request sizes, as I have been working with a template for a palladium catalyzed reaction. I will change the request sizes accordingly though. Please let me know if any other information would be helpful.

I don't currently have access to an installation of g16 to test this, so please also include the error with which the calculation fails. I guess though it is a syntax error, because it doesn't make sense to have a non-integer isotope specified. For the calculation resources requested I guess you are not having budget issues. But even with large platinum complexes that seems rather a lot to request. If you're running on more than one node, you might loose quite a bit through overhead. And you'll burn through calculation time in a jiffy.

If the second calculation run off of the first, the chk statements need to be reversed. Also, are you not running the first calculation from an optimisation? The result would be rather meaningless if you don't. And again, 6-311G is not a good basis set choice. Try something modern, like def2-tzvpp.

Related: https://chemistry.stackexchange.com/q/27302/4945


text:
Does a chiral allene have stereogenic centres?
This a picture of what my professor taught in class.

According the the definition of a stereogenic centre, allene should have two of them. So I don't understand what I am interpreting wrong.

https://chemistry.stackexchange.com/questions/6477/why-are-allenes-chiral

The definition of "stereogenic centre" is unhelpful and can be confusing
The large majority of chiral molecules are chiral because a carbon in them has 4 different things attached to it (atoms or more complicated groups). The tetrahedral arrangement around carbon guarantees that the result will be a chiral molecules (with some complications if there are multiple such centres).
But the underlying cause of the chirality is the overall symmetry of the molecule (or, better, lack of certain types of symmetry like mirror planes). Carbon atoms with 4 different substituents are the commonest way to guarantee that type of symmetry. As a result those carbon atoms are often described as "stereogenic centres". But there are many ways to get chiral molecules (and no mirror planes) without the need for that type of stereogenic centre.
Allenes are an example where the chirality does not result from the arrangement around a single carbon. The Wikipedia definition is confusing and unhelpful and most chemists would not use the term "stereogenic centre" for any atom in the molecule.
A more obvious example of a chiral molecule with no stereogenic centre is hexahelicene (which is made from 6 ortho-fused benzene rings). The terminal benzene units can't exist in the same plane so the whole molecule twists into a spiral to minimise spatial interactions. The 3D structure is shown below.

The molecule is forced by spatial interactions to be a spiral which leads to a strongly chiral spatial structure but one where no single atom cause the chirality and no definition of stereogenic centre makes sense.s
Allene is just a simpler example of the general point that, to understand chirality, the overall symmetry of the molecule needs to be considered and not some count of stereogenic centres.

Thank you very much!

Chirality is a hypernym for the observation that an object can not be superimposed to its mirror image.
Most frequently, examples of chirality in organic chemistry are caused by the spatial arrangement around a specific atom (hence stereogenic centre). An example are four different substituents around carbon (as in mandelic acid); sulfur (as in methyl phenyl sulfoxide), or conformationally locked amines (as in sparteine)*:

Note that a molecule can be achiral despite presence of stereogenic centers (meso compounds).
The idea of a tetrahedral arrangement of atoms around an other equally was the starting point for chemists to consider asymmetry and stereochemistry in a modern way.  In the 1870s, this allowed Jacobus Henricus van't Hoff and Joseph-Achille Le Bel to relate Louis Pasteur's observations on optical activity (macroscopic scale) with underlying structure (microscopic scale) with the now famous models, decades prior e.g., to the introduction of single crystal X-ray diffraction analysis.

An additional type is the one of axial chirality as seen with allenes.  If you orient the molecule model to look along its $\ce{C=C=C}$ axis, the sequence of substituents can be either clockwise (assigned by R), or anticlockwise (S):

(figure from the English Wikipedia)
An other example are biphenyl structures.  Large substitutents in 2,2',6,6' position can hinder the free rotation around the C-C single bond between the two arenes and is the axis to consider.  Consequently, the two structures below differ from each other (if you can, build a model of both):

Next is helical chirality about molecules in a shape similar to the one of a screw, or spiral stairs.  Beside the popular helix of DNA, helicenes are examples typically shown to demonstrate this type.  If six benzene rings are fused together to yield a hexahelicene this way:

(source)
there is steric hindrance where the two ends come close to each other.  To get away with this uncomfortable situation, one end gets under the other, yielding a screw turning either clockwise P (for plus), or anticlockwise M (for minus):

(source about heptahelicene, i.e. 7 benzene units)

Chirality is not limited to organic molecules, let molecules alone.  The spacial arrangement of molecules in the solid state can be chiral (chiral space groups, e.g. $P6_1$ and $P6_5$), as well as the habit of crystals (e.g. L- and D-quartz, sodium ammonium ($2R$,$3R$)-tartrate Pasteur worked on [see illustrations in  Flack's article]) can be chiral, too.
* The CIP rules assign an electron pair the absolute lowest priority (i.e. even lower than a hydrogen atom).
Flack, H. D. Louis Pasteur’s Discovery of Molecular Chirality and Spontaneous Resolution in 1848, Together with a Complete Review of His Crystallographic and Chemical Work. Acta Cryst. A 2009, 65, 371–389. doi 10.1107/S0108767309024088 and author's copy.


text:
Is polarizing power related to solubility, melting point of a compound?
It is stated in this websites that
"It is found that the greater the possibility of polarization, the lower is the melting point and heat of sublimation and the greater is the solubility in non-polar solvent"
Is it true for all circumstances? As the only application of polarizing power that I've known is to define the ionic or covalent character of a bond.

There is the general rule about rules in chemistry that all rules have their exceptions. Except those they do not, as this rules applies on itself too.
Most of the rules are partly relying on rough models, partly being empirical. As they frequently do not consider minor effects, or are not even aware of them, there usually have exceptions.
Raising of covalent character can lead to molecular structures with low melting/boiling points like hydrogen halogenides, or long range structures and high melting points like silicon carbide.


text:
Production of sulphuric acid
To my knowledge, sulphuric acid is produced by complex industrial processes like contact process.
The following structure shows a sulphate ion.

I wonder why sulphuric acid cannot be produced through the protonaction of sulphate ions, for example, by mixing HCl and Na2SO4.
After mixing the salt and acid, concentrated sulphuric acid can be produced by heating up the liquid.
However, this method is not commonly used, or not even used, why?

What are the costs of hydrochloric acid and sulfuric acid, per mole of protons?

What are you going to use to doubly protonate sulfate ion? It needs to be a stronger acid than sulfuric, there aren't many and HCl is not one of them, they are all more expensive than bulk sulfuric.

Why? Because the reaction you mention works in the opposite way. It's $$\ce{2NaCl + H2SO4 -> HCl(g) + Na2SO4}$$

Most sulphate salts are produced from sulfuric acid, so converting them back is hardly an efficient process.

Sulfuric acid is too strongly dissociating to be extracted from water solution in this way.
With sulfuric acid you have complete dissociation for the first stage (in strongly acidic solution) forming solvated $\ce{H^+}$ and $\ce{HSO4^-}$. You will have these in combination with the sodium and chloride ions.
With this solute composition, the boiling-down stage will likely produce sodium bisulfate ($\ce{NaHSO4}$) plus $\ce{NaCl}$ and fumes containing unreacted (or re-formed) $\ce{HCl}$.

But than you could keep heating the sodium bisulphate and eventually produce sulphur trioxide which you could hydrolyse to sulphuric acid :)


text:
Doubt regarding isothermal reversible process- Thermodynamics
Since its a reversible process, I can say
$P_{ext}=P_{int}\pm dP$  so now
$W_{int}=\int_{V_1}^{V_2} P_{int}dV$
My question is why can't I assume $P_{ext}=P_{int}$ here and take it as a constant outside the integral then finally get $ W = P_{ext}(V_2-V_1)$?
(I am aware of the equation $W = -nRTln({V_2}/{V_1})$ but I really want to know what was done above doesn't work

For isothermal reversible process.  p=f(V), (for ideal gas p=nRT/V),  so integral p.dV is not p . ΔV.

I'm not sure that your first statement is correct. Why is the external pressure so close in value to the internal pressure?

@Zhe if it was not, it would be irreversible process, out of scope of the question.

Oops. Thanks. I missed that.

Welcome to Chemistry SE Ainesh.
This is a common doubt that students all over the world have. In order to dispel it, it pays to have a glimpse of exactly what process each equation is describing. The irreversible expansion can be illustrated by the following example.

In this example, the external pressure is the same through out and is lower than the internal pressure. The only thing preventing the achievement of mechanical equilibrium between the two sides is the grey lock. Once the lock is removed, in the second step, the internal pressure is allowed to do work and the gas expands. Notice that from the second to the third step, the external pressure is constant at all times, and it only becomes equal to the internal pressure when equilibrium is reach. hence, when we compute the work exerted by the gas throughout the entire process, like so:
$$
W = - \int_{V_i}^{V_f} P_{ext} \mathrm{d} V
$$
The external pressure, as we have just established, is constant and can leave the integral, leaving us with the familiar expression for irreversible expansion:
$$
W = - P_{ext} \int_{V_i}^{V_f} \mathrm{d} V = - P_{ext} \Delta V
$$
In the reversible expansion we have more like the following situation:

Now here, things are different. The gas is not allowed to expand with its total power against a constant external pressure. Rather, there is an external gas, whose pressure is allowed to lower slowly, allowing the gast to undergo several differential steps of expansion against a constantly dwindling external pressure. At each step in the expansion, the internal and external pressure are in equilibrium, so at each differential step of expansion we have that $P_{ext} = P_{int}$.
If the expanding gas is an ideal gas, then $P_{int} = \frac{n R T}{V}$, which means that $P_{ext} = \frac{n R T}{V}$. Under these conditions we have that:
$$
W = - \int_{V_i}^{V_f} P_{ext} \mathrm{d} V = - \int_{V_i}^{V_f} \frac{n R T}{V} \mathrm{d} V
$$
Now, during this expansion we are assuming that there is no leakage in the internal gas. If this expansion can be made slowly enough and the temperature of the gas is maintained at a constant value throughout, then we can simplify the above expression to:
$$
W = -  n R T \int_{V_i}^{V_f}\frac{1}{V}\mathrm{d} V =
- n R T \ln \left(
\frac{V_f}{V_i}
 \right)
$$

If I do a force balance on a massless frictionless piston (or partition), I find, using Newton's 2nd law, that what you call $P_{Int}$ is equal to $P_{Ext}$ throughout the irreversible expansion.  How do you reconcile that with your diagram showing that they are not equal?  Do you think that, in an irreversible expansion, an ideal gas continues to exhibit gas behavior throughout the expansion, or is some other feature of gas behavior coming into play?

This clarifies it, thanks alot. I see now that $P_{ext}$ isn't necessarily constant. Also if P was constant then V would have to be too right? Then it would also imply work done is $0$ as $ln1=0$.

For a reversible process, you are right in saying that $P_\text{int}= P_\text{ext}$; however, this pressure is not constant. As the gas compresses or expands, the pressure changes, and as a result it is variable and cannot be taken out of the integration as a constant.

When the reversible process is also isothermal, Boyle's law applies and $P_\text{int} = P_\text{ext} = \dfrac{n\text{R}T}{V}$ where the numerator is constant and can be taken out of the integration:
$$
\begin{align}
\int_{V_\text{i}}^{V_\text{f}}P_\text{int}dV &= \int_{V_\text{i}}^{V_\text{f}}P_\text{ext}dV \\
&= \int_{V_\text{i}}^{V_\text{f}}\dfrac{n\text{R}T}{V}dV\\
&= n\text{R}T\int_{V_\text{i}}^{V_\text{f}}\dfrac{1}{V}dV\\
&= n\text{R}T\ln{\dfrac{V_\text{f}}{V_\text{i}}}
\end{align}
$$
Your approach works for a reversible isobaric process, the pressures are constant and can be taken out of the integration. Say $P_\text{int}=P_\text{int}=P$:
$$
\begin{align}
\int_{V_\text{i}}^{V_\text{f}}P_\text{int}dV &= \int_{V_\text{i}}^{V_\text{f}}P_\text{ext}dV \\
&= \int_{V_\text{i}}^{V_\text{f}}PdV\\
&= P\int_{V_\text{i}}^{V_\text{f}}dV\\
&= P(V_\text{f}-V_\text{i})
\end{align}
$$

For a massless-frictionless piston, applying Newton's 2nd law gives $$F_g=P_{ext}A$$where $F_g$ is the force that the gas exerts on the piston.  So the work the gas does on the piston is $$W=\int{F_gds}=\int{P_{ext}Ads}=\int{P_{ext}dV}$$
Do you think that, in a rapid irreversible expansion or compression, the average pressure of the gas on the piston is given by the ideal gas law?  Or does the ideal gas law apply only in a reversible process.

If I'm not wrong then ideal gas equation is applicable throughout a reversible process but only applicable on the initial and final states for an irreversible one


text:
When relaxing the strong field interaction, the (t2g)^2 configuration splits into 4 states. What is the symbolic configuration of the Eg state?

When relaxing the infinitely strong field, the electrons start to feel one another's presence. They will, therefore, give rise to sets of states such as T1g, T2g... for the (t2g)^2 initial configuration. Given that Eg is a doubly degenerated state, which symbolic configurations (arrangements of electrons in dxy, dxz, dyz--> t2g) would fit this state?


text:
Amount of PEG to add to copper plating solution?
I found a paper about making a copper plating solution and thought I'd give it a shot. Please keep in mind I have no formal education in chemistry. I have a few questions about the preparation of the solution.
Information:
The paper lists the following ingredients:

In the conclusion, the authors recommend the following additives:

500 ppm PEG
5 ppm SPS
30 ppm JGB

PEG:
The PEG I have is labeled as PEG 400. The paper doesn't say anything about this. I assume the paper calculates as if it was pure PEG, where mine is 400 g / mol? Could someone guide me on how to calculate the quantity needed to achieve 500 ppm?
Cl-:
For chlorine, I am considering hydrochloric acid. The concentration of the acid I have is 30%, and I would need 1.4 mM of chlorine. Since I need only a very small amount, It might be best to dilute the concentrated acid. I do have a very precise scale, but weighing out 0.05 gram seems a bit ridiculous.
Results:
Here is what I've calculated so far:

Chemical
Molar mass
Required
Result

CuSO4 5H2O
159.609 g/mol
0.79 M
126 grams

H2SO4
98.079 g/mol
1.02 M
100 grams

HCl
36.46 g/mol
1.40 mM
0.05 grams

SPS
-
5 ppm
5 milligrams

JGB
-
30 ppm
30 milligrams

PEG400
-
500 ppm
(Not sure)

This is for a 1-liter solution, with the remainder being deionized water. I'm aware of the acid-to-water addition protocol.
Prepare scaled down solutions:
Make 1% SPS solution:
1000 mg SPS + 9000 mg water = 10000mg of 1% solution.
Make 1% JGB solution:
1000 mg JGB + 9000 mg water = 10000mg of 1% solution.
Make 1% PEG solution:
1000 mg PEG + 9000 mg water = 10000mg of 1% solution.
Make 1% HCl solution:
100 mg HCl @ 36% + 3500 mg water = 3600 mg of 1% solution.
1% HCl => 36.46 / 0.01 = 3646 g/mol
Prepare electrolyte 1 kilo (more or less 1 liter)
Add CuSO4 5H2O:
CuSO4 5H2O = 159.609 g/mol
0.79M * 159.609 g/mol = 126 gram
Add H2SO4:
H2SO4 = 98.079 g/mol (Mine is 98%)
1.02M * 98.079 g/mol = 100 gram
Add HCl:
HCl @ 1% = 3646 g/mol
0.0014M * 3646 g/mol = 5.1044 gram
Add PEG:
500ppm of PEG = 500 mg total
500mg @ 1% = 50 gram
Add SPS:
5ppm of SPS = 5 mg total
5mg @ 1% = 0.5 gram
Add JGB:
30ppm of JGB = 30 mg total
30mg @ 1% = 3 gram
Power supply:
The paper specifies a particular waveform. For now I would like to ignore this and just use a current limited DC power supply.
References:
Shin S-H, Kim T-Y, Park J-H, Suh S-J. Optimization of Additive and Current Conditions for Void-Free Filled Through-Silicon Via. Applied Sciences. 2018; 8(11):2135. https://doi.org/10.3390/app8112135

Not sure about the significance of the choice of PEG MW, but for calculating ppm that would be irrelevant. Also, you have a lot of latitude in the choice of amount of PEG to add: $$\pu{100-500 ppm }= \pu{100-500 x 10^-6 g~PEG/g} ~\textrm{solution} \ \approx \pu{100-500 x 10^-3 g~ PEG/kg~\textrm{solution}}$$  If you choose to make 1 kg (approx 1 L) of solution at a concentration of 500 ppm PEG that means adding 500 mg of PEG. Given the wide suggested range you probably don't need to be particularly accurate.

The choice of waveform is probably important, but this is not my specialty.

@BuckThorn, So then, what's the difference between PEG and PEG400? I know you also have PEG 3000 for example. From what I've read, it's the molar mass of the peg. But I doubt my bottle contains pure PEG, since it's a liquid.

You can read more about PEG on the Wikipedia: https://en.wikipedia.org/wiki/Polyethylene_glycol It's a polymer, and the number of monomers in a chain determines whether its a solid or liquid at room temperature. The low MW stuff is liquid

I have found some, not so reliable, claims on the internet that the PEG should be between 1500 and 3500. Since it's not too expensive, I ordered some PEG 1500.

In the description of the polymer, 400 is the probable molecular mass of the polymer PEG. As the monomer has a molecular mass equal to $62$, it means that the polymer is a mixture of hexamers and heptameters, which may be liquid.

Some practical remarks. 1) $\ce{HCl 1}$% does not have a molecular weight $3646$ as given in the text. Solutions like $\ce{HCl}$ $1$% do not have a molecular weight. Only pure substance have a molecular weight. $\ce{HCl}$ has a molecular weight of 36.46 g/mol.

Some practical remarks. 1) $\ce{HCl 1}$% does not have a molecular weight $3646$ as given in the text. Solutions like $\ce{HCl}$ $1$% do not have a molecular weight. Only pure substance have a molecular weight. Pure $\ce{HCl}$ has a molecular weight of 36.46 g/mol. 2) See under ...

In the text, the $\ce{Cl-}$ concentration is obtained by adding $\ce{HCl}$. This is not a good idea, as first $\ce{HCl}$ is a gas. Gases are difficult to weigh, and to dissolve. And if you use commercial $\ce{HCl}$ solutions instead of the gas, it is not easy to weigh with precision small amounts of liquids, like less than one gram. You should rather replace $\ce{HCl}$ by $\ce{NaCl}$ which is easy to weigh with precision, whatever the quantity chosen.

If you want to make a solution of PEG $200$ ppm, it is not important to know which sort of PEG you use. $1$ liter PEG $200$ ppm, means $200$ mg or $0.2$ g PEG, whatever the choice of PEG. You may choose PEG $300$, or PEG $1500$, or any other sort off PEG. The only important thing is the necessity of adding $0.2$ g of this PEG in one liter water.

The same problem happens for $\ce{HCl}$ and $\ce{H2SO4}$. You may be sure of weighing $100$ g $\ce{H2SO4}$. But how will you be sure that this viscous liquid sample will be correctly transferred into water solution ? It is possible, but not easy. Think in advance about how to handle this corrosive liquid !