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Hazardousness of Chlorothalonil & metabolites in shower water?
Recently it came out, that in some parts of Switzerland (where I live), there is Chlorothalonil & metabolites/breakdown-products of it in the drinking water. Yikes.
Problem is, chlorothalonil synthesis frequently result in contamination of it with small amounts of hexachlorobenzene (HCB), which is toxic. Worse, one of the metabolites/breakdown-products is 2,3,7,8-Tetrachlorodibenzodioxin, which is one of the most potent carcinogens known.
Now, according to official sources, the drinking water in my region, which exceeded the maximum permissible value by law (and those are not necessarely low, e.g. the nitrates permissible values is far too high), has been diluted with lake water from Lake Constance, and according to the official narrative, all is fine now. That's what the same official sources said before, too - until someone discovered that this wasn't the case. However, Syngenta, noble originater of the problem-causing fungicide(s), is moving in courts to successfully forbid obligatory measurements, because "it has not yet conclusively been proven that Chlorothalonil causes cancer". That's bullshit, but the scientific proof is indeed still owing, unfortunately, yet. Unfortunately, that little detail also means their court-work sabotage is successful. However, the correlation is clear, and I'm sure it will be proven in the not so distant future, sooner or later (but too late for the courts). Syngenta's behaviour equalling the reprehensible ethics of the tabacco industry.
Now, me being a distrustful person, I have bought a reverse-osmosis table-appliance, to be sure with my cooking and drinking water, just in case this is a lie. I have read that this filters the chemicals in question and more (or at least it should), and is also good for decalcification, which is an additional benefit.
However, assuming the reverse-osmosis-appliance itself is safe, not to mention it working as advertized, there remains the problem of showering and bathing water.
Is that even relevant for concentrations of Chlorothalonil breakdown products of <= 0.1 microgramms per liter, or is the main danger from oral consumption ?
For absorption by skin, can such concentrations be harmful ?
lab blog for reference

In January 2020, Syngenta Agro AG appealed to the Federal
Administrative Court against the ban on chlorothalonil. The
proceedings are still pending, but two interim rulings have already
been issued in which Syngenta Agro AG's applications for precautionary
measures have been approved. In the second interim ruling, the FSVO is
required not to designate four metabolites of chlorothalonil,
including R417888 and R471811, as toxicologically relevant for the
time being (this would once again result in a high maximum level of 10
µg / l). According to a media release from the Federal Administrative
Court of February 18, 2021, it will only be necessary to assess in the
main decision which classification with regard to carcinogenicity
(category 2 or 1B) is to be assumed and whether all metabolites are
automatically considered relevant with a possible classification in
category B1, and which maximum drinking water levels (10 µg / l or 0.1
µg / l) are applicable.
Due to the pending proceedings, Labor Veritas AG has decided to
suspend the conformity assessment of water samples in accordance with
the TBDV in connection with the chlorothalonil metabolites until the
legal situation has been clarified. The following note therefore
appears on our test reports: "Due to the currently unclear legal
situation with regard to the applicable maximum values for the
metabolites of chlorothalonil, the samples are no longer evaluated for
conformity". In our Spotlight No. 34 from May 2020 , all metabolites
of chlorothalonil are described as relevant due to the initial
situation at the time. This statement should be treated with caution
for the time being. This spotlight will not be updated for the time
being, especially because the statements made in it could become valid
again.

Note that medical/health advices are explicitly off-topic on this site //   There is  many funny but true evidence even strong  correlation does not automatically mean causality. //  There are still high level debates and you want us to say the verdict. You will not get it // Generally, questionable concentration levels for drinking/cooking are safe for showering/bathing, with exception of microbiology and radon // BTW what about microbiology and ion composition of the output of the  RO unit?

@Poutnik: As always, correlation is a necessary condition for causation, but not a sufficient condition. But it also couldn't be a sufficient condition if there were no correlation. Certainly exactly the point of one side of the debate. Generally, if you don't store the water, but use it immediately, the microbiology shouldn't be a problem, as long as you change the filters in the necesary interval as described in product maintenance. The water is deionized/demineralized, which is why it is being remineralized, requiring mineral cartridges in the RO unit. Fair point, nobody really knows, yet.

correlation is a necessary condition for causation, but not a sufficient condition. But it also couldn't be a sufficient condition if there were no correlation. Sure, that is the same by other words. The world has enough struggles already with correlation/causality in COVID vaccination. // It can still be replecement of evil by devil, as you do not have operational feedback about water quality, unless you pay regularly for it big money. // I would guess RO is not needed unless required for other reasons. Regularly replaced active-C based filterer in flow through devices should be enough,

@Poutnik: A TDS-detector is monitoring water quality at the inlet and outlet of the RO appliance. According to https://www.sciencedirect.com/science/article/pii/S0043135420306035
active-C filters have difficulty with TP R471811. Don't know
how relevant that is for small-time use. Also, active-C doesn't filter nitrates,
and after having seen the legally allowed values for nitrates,
i also think filtering them would be a good idea.
Because legally allowed or not, values that are well in the range
to cause cancer are pointless BS. Also, the RO has an active-C filter 4 VOCs: B0817DCXG5

TDS is too simple check to say all is OK, but it can serve as indicative parameter for maintenance or emergency shutdown. // Well, there is no lower limit under which alcohol does not cause cancer either. Many areas of life work with "acceptable risk level" for potential risks. Nitrates in low scale can be eliminated easily by RO or ionexes, but that is not always the case for other risks, all aspects considered.

@Poutnik: I know (both alcohol and simple check), which is why i don't (knowingly) drink alcohol, and don't use the covid disinfectants :) Yea, unfortunately, there is never 100% security, and life remains life-threatening-dangerous :)

Let us continue this discussion in chat.

In principle you can estimate the risk yourself. Estimate the maximum concentration of dioxin in water (from the solubility), and multiply by the volume of water that contacts your skin during a shower. A small fraction of that solution will result in dioxin binding. This would be an upper limit. In practice the fraction available for absorption through your skin might be much smaller than this, in part because the concentration of the compound in the water may be much lower.

In dermal studies on rats the compound was applied solubilized in acetone which might facilitate absorption. In that case ~50% of the stuff was absorbed into the body, but very slowly. See eg https://www.sciencedirect.com/science/article/abs/pii/0041008X91902106


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Is strength of pi bonds greater than sigma bond of nitrogen molecule?
I am referring to molecular orbital (MO) theory. In the nitrogen molecule (and elements of lower atomic number) $\textrm{sp}$ mixing occurs and the sigma orbital set is raised above the pi orbital set. Shouldn't that mean that pi bond is stronger than sigma bond in this case?

One occupied sigma orbital is raised above the pi orbitals. Another two are further lowered. (See for example How to rationalise with MO theory that CO is a two-electron donor through carbon?) And additionally it is not possible to conclude about bond strength from a single molecular orbital diagram.

You can actually calculate orbital "energies" with MO theory. The resulting "energies" for $\ce{N2}$ are the following (calculated with molpro and CASSCF(10,8)/aug-cc-pVTZ):
Orbital "energies" in atomic units

orbital
energy/a.u.

1σg
-1.11

2σg
-0.99

1σu
-0.77

πu
-0.59

πg
0.29

2σu
1.22

As you can see, the "energy" of the π orbitals is above the "energy" of the 2σg, which is even below the 1σu "energy". This disagrees with virtually every MO scheme of $\ce{N2}$ you can find. So why is that?
In the end, MO schemes are used in what is called "qualitative molecular orbital theory" (QMO theory). There are usually no underlying calculations (or the calculations were performed in the 60s) and the schemes are adapted in order to explain and understand the stability and reactivity of molecules.
On the other hand, there is computational MO theory, which could in principle be used to improve these diagrams. Why is this not done? In the end, electronic states of a molecule exist and have energies, while orbitals don't. They are not in any way 'real' or observable and are in general best understood as a tool for building approximate wave functions. The orbital "energies" are eigenvalues of the so-called Fock operator, but they should not be given too much attention. For the most accurate wave functions (the so-called full-CI wave functions), the orbital energies are actually arbitrary.
In the end, MO schemes are a tool used to understand and simplify the discussion of chemical bonding. If they cease to be simple, they should not be used.
As for the discussion of the π bond energy vs. the σ bond energy: again, there are not really separate π and σ bonds. However, if you want to calculate some energies aligning with these concepts, valence bond theory is better at that than MO theory.

On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

MO schemes are essentially a visualisation of a single determinante self-consistent field, or Hartree-Fock. Sorry for being nitpicky here, but the Fock-Operator, whether it is so-called or not, doesn't actually have eigenvalues as it is dependent on its own solutions. This obviously didn't stop generations of chemists to use it as such. Are for a FCI wave function the "energies" really arbitrary, or are the not defined? But I actually like where this answer is heading to.

@Martin-マーチン thanks for the syntax improvements! Regarding your other comment: you are right, that MO schemes are essentially a visualization of a single determinant, I just performed the CASSCF calculation instead of HF in order to have reasonable energies for the unoccupied orbitals, which are not optimized with HF. The Fock operator really has these eigenvalues: see here.

As for the FCI wave function: with FCI, you build all possible configurations with all orbitals, construct the wave function as a linear combination of those and optimize only the linear coefficients. The shape of the orbitals does not matter, if they span the same vector space. They can be constructed by any orthogonalization procedure performed on the atomic basis functions.

Not the first time Wikipedia is wrong. The equation they mention is a pseudo-eigenvalue equation, you'll see that when you derive it. I guess I skipped the important steps here: LCAO (Linear Combination of Atomic Orbitals) and Phases

The definitions: Fock operator invariance under unitary transformation

I think, that Wikipedia is not necessarily wrong, more a bit misleading. Given a set of orbitals (e.g. after the converged SCF cycles), F is a properly defined operator, and the orbitals are also its eigenfunctions with eigenvalues epsilon. Of course, the equation is pseudo-eigenvalue, however, this still allows for calling the eigenvalues eigenvalues. PS: If you think, that wikipedia its wrong, you should change it =)

There is a similar question on StackExchange from 6 years ago (Ref 1), from which I have taken Diagram #1:
DIAGRAM #1

And a quote from the answer: "Many of the diagrams (on the internet) are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance." Something like Diagram #2 below, which implies a very different degree of hybridization (none). Just to be clear, "N3" means the nitrogen molecule (which is N2).
DIAGRAM #2

The four orbitals (2 sigma atomic and 2 pi atomic) interact and separate into 4 sigma bonds, two bonding and two antibonding. The lowest three sigma MOs are filled in N2. The take-away from both diagrams is that the three lowest energy sigma MOs are  energetically more stable than the atomic s and p orbitals, even tho formally, the uppermost filled sigma level is antibonding.
The degree of splitting is dependent on the amount of hybridization entered into the calculation, and if that is small, the actual energy levels will resemble Diagram #2, and if the amount of hybridization calculated is large, the molecular orbital energies calculated will resemble Diagram #1. Calculations are an important part of the discussion.
Both diagrams predict an increase in ionization potential on going from atomic N to molecular N2 (actual ionization potential for N is 1402 kJ/mol; for N2 is 1503 kJ/mol), but the difference does not allow a definitive choice between the models.
So it comes down to estimating the hybridization and doing a calculation. I suspect that if you try to determine something about the first ionized state of N2, it would resemble Diagram #1 more than #2, because the nitrogen atoms would be drawn closer together.
Another differentiation would be to see how nitrogen molecules bind to other atoms. Ref 2 indicates that the bonding is linear, in agreement with both diagrams. Nitrogen sometimes (rarely) bonds with two atoms (a side bond). The question allows for a lot of pictorial discussion, but not much in a numerical sense.
It could be looked at as a comparison between a stabilizing, but formally sigma antibond against a pi bond and an interesting way to bring up hybridization and overlaps.

Ref 1. In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?
Ref 2. https://en.wikipedia.org/wiki/Transition_metal_dinitrogen_complex

This is kind of in a tangent to the question OP asked. Also, I would question that the two MO diagrams you drew above are actually different. They have the same ordering of the orbitals, just the dotted lines joining AO and MO are different.

@S R Maiti: The 2nd says that the N s orbitals overlap (sigma bond), and three p orbitals form a sigma and two pi bonds. The first  says an s and a p orbital on each N hybridize to sp orbitals: two lobed, one lobe much larger. That's the 4 sigma MOs in the third diagram. (The pi double bonds in N2 are the same in both diagrams.) Draw the 4 orbital shapes with a) no hybridization and b) with complete hybridization. Is it "fair" to compare a filled sigma anti-bond (#3) with a pi double bond? I don't think so.

Oh, so you are referring to the third diagram as the 2nd? I thought you meant the two diagrams at the top. It's a bit confusing to follow.

The fist diagram (from 6 years ago) is made with complete hybridization to sp-p2. The next one down (with blue and pink) shows MOs with no hybridization. The next one down (third; with blue lines) shows only sigma MOs, but with hybridization. Since sigma #3 is filled, and is below the p level (acc to diagram #2, but not to #3), it could be considered a sigma bond - but is actually a sigma anti-bond, but still better than the original p atomic orbitals. You're right - the diagrams meant to clarify are in contradiction. I will edit. Thank you!

Hybridisation and s-p-mixing are completely different things! You can always have an explanation with hybrid orbitals as these are just mathematical descriptions. It will tell you absolutely nothing about the order of the molecular orbitals. And on that note, the diagram 2 is almost completely wrong. And moving further, it is impossible to conclude on bond strengths from an MO diagram. On the other hand, you can calculate (predict) bond strengths (hypothetically of course) without ever needing to resort to hybridisation or drawing a MO diagram.

@Martin: I agree with you that #2 is not accurate! Good definition of mixing and hybridization is given at https://chemistry.stackexchange.com/questions/26445/is-s-p-mixing-referring-to-hybridization-or-is-it-the-mixing-of-one-atoms-s-orbi. The question was whether a sigma bond could be weaker than a pi bond. And the answer is yes, using EITHER MO diagram, which depend on different mathematical assumptions. Relative bond strengths can be estimated/compared if you draw the diagram accurately. Structural data suggest that a sigma MO has the highest energy, therefore weakest. Do you concur?

No, on all accounts. First, I've made my objections to Ron's answer already there. Second, the second MO diagram is not inaccurate, it's wrong. If it were assuming no s-p-mixing, then the second sigma(g) orbital must be lower in energy than the pi orbitals. Third, and I have no way of saying that differently, it's impossible to conclude to bond strength from a MO diagram. The orbital energies do not represent bond energies, as they also do not add up to the total energy of the wave function. A MO diagram cannot capture electron correlation, but that's crucial for bonding.

@Martin: I see your edit on Ron's answer, but no objections. Can you write a better answer? I would appreciate it!

source: Stack Exchange

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Can molecular systems display negative magnetic susceptibility of a higher magnitude than bismuth?
Bismuth has the biggest value when it comes to a negative magnetic susceptibility than all other elements, ignoring superconductors, as they would have a value of -1.
Yet, when tables displaying diamagnetic materials show these, none show those shown by molecular systems above Bismuth. Are there any molecules or compounds with a higher negative negative susceptibility than Bismuth?
Any examples?

Pyrolytic carbon supposedly has higher diamagnetism than bismuth metal. It is easy to demonstrate the diamagnetism of a graphite pencil lead (a refill lead for a mechanical pencil) by using two of the pencil leads and a strong Nd-B-Fe magnet. I used to do this demo in class. Lay one pencil lead on the table top, lay the other one across it, like a tiny see-saw, then move the magnet close to the top pencil lead: it will move away from the magnet. For the demo, I did this using a video overhead projector in the lecture hall, so everyone could see it.

Thank you for your response. Are there any examples for singular molecular systems with diamagnetism stronger than bismuth?

I have no idea. If I had to guess, I would say no simply because “combining” something diamagnetic with something less diamagnetic does not look like a winner. But it would be nice to be proven wrong, so maybe a good answer will happen.

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


text:
How to dry sludges?
I sometimes need to dry sludges or thick, sticky slurries. The way this is typically done in highly equipped labs is to use a spray dryer which liquifies the slurry, heats it up, then sprays it at high pressure into a open chamber, and finally collects it in a venturi. Obviously such machines are incredibly expensive and not really where I want to go.
So what are my options? I have tried vacuum chambers, but the problem is that the steam tends to be intrapped in the material so it puffs up like an angel food cake and does not get evacuated. Also, heating is kind of problematical because I am not sure about the wisdom of putting my vacuum chamber on top of a heater. The lid of the vacuum chamber is a thick plastic and the body is aluminum, so if I heat the chamber, it could potentially damage the lid.
I guess one idea would be to vacuum it, then take it out of the chamber, then shear it, then vacuum it again, then shear it again, etc. This seems kind of tedious.
Note that I typically use a GAST type vacuum pump which produces only a moderate vacuum. One option is that I could use a high power, two-stage oil pump which can generate a significantly stronger vacuum. Will this give me much better drying than the GAST, or not significant improvement?

I guess you are setting up a small scale detergent plant (based on your previous questions). Nobody can provide you good advice online. Talk to the people who actually do it at a small scale and perhaps consult some chemical engineers in the detergent business.

Fine grains of slurry may clog a filter, eventually, neither gravity, nor vacuum filtration with a Buchner will advance at reasonable rate.  Using a centrifuge then may be an alternative.  A side form those with beakers, there are ones with a sieve / filter cloth, too (basically a spin dryer in the kitchen for salads, or the tumble dryers) so your solid not only settles faster (like in the beaker centrifuge), but gets drier, too.


text:
Is it possible to run column chromatography in reversed phase?
I'm conducting a synthesis involving aza-crown-ether.
Whether aza-crown-ether acts as a base or not is unclear, but when I do TLC using dichloromethane/methanol, it is hardly drawn up with the mobile phase (see image below).

I assume it is very polar (or basic), and I have to run reversed phase chromatography.
Whenever I search for reversed phase chromatography, all texts consider reversed-phase 'HPLC'.
Is column chromatography in reversed phase impossible? I have never found sources about it.
I used a dichloromethane/methanol ratio of 3:1. Do I have to keep the ratio of methanol to not exceed 10%? Should I use reversed-phase TLC or try using dichloromathane/methanol in 10% ammonia and add some triethylamine?

AFAIK there are/were RP (HP)TLC alu sheets or glass plates, or free SF as well, but you could have the different problem being washed away.// You may need more polar MF as you mention. Or some conditioning of SF, e.g. in a chamber with  aquaous ammonia.

That TLC makes it seem DCM:MeOH is still doable with normal phase chromatography. Reverse phase is much more expensive, the silica costs 10-100x as much. It's not uncommon to go as far as adding a few percent water or aqueous ammonia to the eluent if necessary, for extremely polar compounds.

I second the suggestion of adding ammonia to your eluent. I have routinely used DCM/MeOH/NH3 systems for columning basic materials. Ammonia is MeOH is commercially available which is what I used. You could also consider chromatography on alumina.

What are eluents used for structures similar to the one you target?  What stationary phase is reported there?  Note, alumina Waylander suggested is available as acidic/neuter/basic. Is TLC intended as to monitor the progress of the reaction (because of what looks like a pale spot at Rf ~ 0.8, perhaps easier to see if the lane were stained with $\ce{KMnO4}$)?  If aiming for purification, could protonation and subsequent extraction represent an alternative (e.g., cyclam) to chromatography?

@Buttonwood TLC image I uploaded was visualized by ninhydrin. In other papers using aza-crown-ether, they use ethyl acetate or hexane/ethyl acetate (5:1 to 1:1). One strange thing is when I use these solvent systems, TLC results were still challenging.

@Poutnik Thanks for comment, but what does 'SF' and 'MF' stand for?

Ah, I see reason for confusion, I applied abbreviating of Czech words, not English ones :-D  - so it would be  SP and MP, or SPh and MPh, ( stationary/mobile phase), not quite sure which is common now.

You may have to consider whether you are looking at a metallated or non-metallated aza-crown. If there's a metal ion inside it, the affinity for the silica will increase enormously.

@KrangLee If possible, run a TLC with multiple lanes: starting material(s), (quenched) reaction mixture, co-spot starting material and quenched reaction mixture. Maybe the differences in the Rf values are already good enough (the pale perhaps-spot around 0.8?) and  a short column already provides sufficient separation / purification.  Or, that you elute with a gradient.  This assumes you stick to chromatography as purification, as (back) extraction / recrystallization may be alternatives.


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Understanding packing in different dimensions
In one dimension, there is only one way of packing, that is keeping the balls next to each other.

In two dimension, we can keep a line of spheres on top of another line directly or we can keep the second line in the cavities of the first line.

In my book, for three dimensional packing, only structure is shown when keeping the square packing layers on top of each other:

Would it be possible to get another form of 3D packing from square close packing in two dimensions by making the spheres go into cavities? (similar to the way we kept the line in a staggered way to get hexagonal close packing)
Similarly for packing 2D hexagonal packed spheres in 3D, could we keep the spheres on top of each layer on top of each other( in axis of 3-D packing)?

The simple cubic packing is one, but not the only approach to pack spheres.  There is a good chance you see e.g., close packing in the groceries.

See such a  caesium chloride packing  https://en.wikipedia.org/wiki/File:Caesium-chloride-3D-ionic.png,

The textbook missed a beautiful opportunity to show a relationship between simple cubic and close-packed cubic structures. As students learn, mental (spatial) operations need to be explained - once shown, they are there. if not explained, these operations grow in unclarity as more and more spatial operations and pictures are piled upon the original, unclear one.
Here is an explanation: Take the pile of spheres (4 x 4 x 4) and shift them sideways as in the picture below:

I did a shift in only two dimensions, but you see the change in what is obviously not closest packing to a tilted pile with a lot closer packing. It's still two dimensional, but the diagram says "hexagonal packing"; to get real closest packing, you have to shift the spheres in the third dimension (into the paper, or screen). Then the spheres will lie in the lowest valleys of a packed plane. Then the densest plane of spheres is not one of the original x,y,z axes, but a tilted one. You will have converted the simple cubic lattice into a closest packed lattice.
You can play with this idea in order to make it more concrete in your mind: imagine a closest packed lattice and unfold it to a simple cubic lattice.

I am not sure how this answers the question. The image you provide is already contained in the question, albeit a bit less accurate. In two dimensions the is only one way to pack the circles according to a square grid closest: square (close) packing. There is a more dense packing: hexagonal (close) packing. The point of the question is, in my opinion, what kind of structure you would get (in three dimensions) when you extend the hexagonal packing in the same way as you would go from square packing to simple cubic. Hence I believe the answer should focus on FCC and HCP structures.

@Martin: You have great vision and long experience. For those of us who take one simple step at a time, the answer above spreads out the movement of spheres into two steps, the first in 2D, and then the second in the third dimension, with one of two possibilities. Doing it all at once is a wonderful mental capability - not all of us have it. Another way to "get it" is to play with about 64 (or 125) magnetic spheres. Alas - we have here only a flat screen, typed words and some pictures.


text:
If rate of SN1 is more does that automatically mean rate of SN2 is less, and vice versa?
If I have a list of compounds and I compare their SN1 reactivity, for SN2 reactivity, is it correct to use the same list but order inverted? Are there any exceptions?

Short answer: no, it is not correct.
You can have compounds that are thermodynamically/kinetically impeded for a variety of reasons and present both slow SN1 and SN2 reactivities.
The exceptions would be carefully handcrafted “everything else being equal” lists.

Also it really depends on external factors. There is no absolute order to use.

No.
Example 1 - neopentyl halide: slow SN2 as it has considerable steric hindrance; slow SN1 as it is a primary carbon, so it does not form stable carbocation intermediate.
Example 2 - benzyl halide: fast SN2 being a primary carbon; fast SN1 due to a resonance-stabilized benzyl carbocation intermediate.


text:
How to thermally insulate beaker glass on hotplate?
I bought a magnetic stirrer with a hot plate almost identical to this one.
I am trying an experiment where I need to heat the beaker/mixture to about 100 degrees C. However the mixture/beaker is losing a lot of heat to the environment, especially when being stirred and it needs to be stirred.
I tried using the included thermometer but the algorithm seems to be imperfect because while trying to heat the mixture to my desired temperature the apparatus would heat the hotplate to such high temperatures that the bottom of the mixture starts boiling even when stirring at high speeds and this is suboptimal as the mixture becomes a gel at a certain not-to-high temperature and I need to avoid that.
Since I'm quite new to this, is there any way to insulate the beaker or in some other way prevent heat loss without heating the room to 100 deg C?

It sounds like you should take a page out of the culinary world and use a bain-marie. You can stay below or just at 100 °C pretty reliably with it.

Which is more important, not overheating or heating evenly? It is far from obvious that insulating the vessel will make either better. Use a water bath around the vessel to prevent both overheating and to give an even heat. heat loss isn't the same problem as getting an even heat distribution.

Is it to be open or can be closed ?
If the former, the major loses can be due evaporation, keep the opening as small as possible and safe. If there is a wide opening as beakers have, try to put aluminium foil cap freely sitting on the beaker top, if applicable. Or a "watches/clock glass", typically used to cover beakers.
Additionally, try to wrap the beaker by aluminium foil that reflects thermal radiation, leaving just a bottom to be observed. You may try also a foil sandwich with isolation inbetween, like polyurethane foam or some fabric( like wiping/washing fabric for kitchen or floor cleaning or similar ).
If the primary problem is rather undesired overheating, than a water bath is the solution as @nicolausakerneto suggests, eventually isolating all, if still desired.

It can be closed and I usually put something over the lid, I'll try the foil idea. Is this the usual approach in lab conditions in such cases? I thought there would be some sort of procedure or something else that would give a higher effectiveness.

It may not be the common need and the lab workers often use DIY approach. At least we did when I was there.

If the beaker «of interest» has a flat and dry outer bottom, than you can put it in a second beaker of dry and flat inner bottom, slightly larger in inner diameter than the outer diameter of the inserted beaker.  It isn't perfect, but this «coat» equally shields the inner beaker from air current (e.g., next to the open sash of a fume hood).  The contacting bottom surfaces of the two beakers must be dry, otherwise evaporating water may damage the setup.  Gentle heating, of course.

@Buttonwood I would afraid it may be tight trade off between heating enough the inner beaker and not overheating the bottom of the outer one or heater itself. Unless one has handy a usable metallic plate to put inside.

@Poutnik .True., immersion of a temperature probe into the inner beaker of the two and dialing up the heating plate to the maximum power indeed is a good approach to shatter the setup by uneven thermal expansion. The gentle increase of temperature mentioned is easier to obtain if you still have an old «flat iron»-type heating plate instead of the ones with a glass surface and proceed slowly (multiple plateaux) over time.

Getting good, even, contact across the bottom is as important as wrapping the side and covering the top. A thin glass fiber paper (asbestos is good, too), or a sandwich of aluminum against the hotplate plus a thin cotton fabric (like a handkerchief) plus an upper layer of aluminum touching the beaker bottom will help to even out the beaker bottom temp. Glass electrical tape is useful (high temp); big stir bar to sweep the bottom and/or boiling stones to whip the liquid up more.


text:
Are pseudo-zeroth order reactions possible?
Are pseudo zeroth-order reactions possible?
I have been taught that pseudo first-order reactions take place and they can be considered pseudo zeroth-order reactions because one reactant is in excess of the other(s).
What are the necessary conditions to label a reaction pseudo-zeroth order?

I've added this as a comment, below, but it warrants attention here: "Hence, even for elementary reactions, the molecularity of a reaction does
not have to be the same as the order of the reaction. When the molecularity is
not the same as the order of a reaction because the concentration of one or two
species is kept constant either due to the concentration of the species is high or
because the concentration of the species is buffered, the reaction order is also
referred to as pseudo-order."

I think to some extent, you can think about all zero-order reactions as "pseudo-zero-order" reactions. This is because it's not possible for a reaction to be true zero-order. How can it be that the rate of a reaction does not depend on any one of the reactants? So, anything that is zero-order means that there is some artifact in the system, or that something else is present.
In general, zero-order reactions happen when the reactant is not able to react at the theoretical maximum rate due to something preventing it. For example, for a catalytic process, the reaction happens at the surface of the catalyst, so if you have a large excess of reactant, all of the surface of the catalyst will be covered up with the reactant, and the rate will become constant. When the concentration of the reactant goes down so that all of the catalyst surface is not in use, the reaction will show a first/second or higher order rate equation.
An example of such reaction is the decomposition of nitrous oxide:
$$\ce{2N2O\overset{\Delta, Ni}{->}2N2 + O2}$$
When nickel catalyst is used, the rate is limited by the total surface area of the catalyst. Once the catalyst is working at the full capacity, the rate won't increase, even if the concentration of the reactants is increased further.
Without the catalyst, the reaction is slower, but it shows the standard second-order rate equation that we would expect by looking at the equation.
For reactions that are driven only by light (not just initiated by light!), the rate would be limited by the intensity of the light. Once there are enough species that can absorb all of the light, the reaction rate won't go up even if more reactants are added. Then there are biochemical reactions which are diffusion controlled.
So, in summary, all zero-order reactions are zero-order because there is some reactant which is in excess than something else the reaction depends upon (which can be catalyst, light, diffusion etc.). You can say they are probably all pseudo zero-order.

This is incorrect. There are plenty of reactions where the rate does not depend on one or more of the reactants, which is the defining part of "pseudo zeroth-order." Example: decomposition of $\ce{PH3}$ on hot tungsten at high pressure (taken from this).

@ToddMinehardt Yes, because the tungsten is the catalyst here, and as long as all the catalyst surface sites are engaged, the rate doesn't depend on the conc. of PH3. Nothing in my answer contradicts anything you say. Did you read my whole answer? Please do so before downvoting.

@ToddMinehardt I have added another word in the answer, hopefully that makes it clearer.

A standard example in biochemistry are enzyme-catalyzed reactions with a single substrate (e.g. ATPase, where we ignore the second substrate, water, because it is the solvent). At low substrate concentration (low reactant concentration), the reaction is first order (actually, pseudo-first order because of the water). At high susbtrate concentration (high reactant concentration) the reaction is pseudo-zero order. The theoretical expression was derived by Maude Leonora Menten and Leonor Michaelis, and the results are shown as a graph below:

Source: https://commons.wikimedia.org/wiki/File:Michaelis_Menten_curve_2.svg
As to whether it is pseudo or not, here is a textbook quote that shares the OPs view:

Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.

Source: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.10%3A_Zero-Order_Reactions
Of course, if you disagree with that view, you can always call the source a pseudo-textbook...
In the case of the enzyme-catalyzed reactions, the reaction would proceed at almost constant speed until the substrate concentration is depleted to a concentration in the range of the parameter K$_\mathrm{M}$, at which time it would slow down appreciably and continually until the substrate is depleted.

For more about the Michaelis-Menten derivation, see e.g. https://chemistry.stackexchange.com/q/97470/72973

Zero-order kinetics can also appear in certain industrial settings. In some steel strip annealing processes where improved bendability of the product is required, steam is applied to decarburize the steel near the surfaces of the strip according to the reaction
$\ce{C(s) + H2O(g) -> CO(g) +H2(g)}$
The decarburization process requires carbon to diffuse out of the steel in order to reach the water molecules at the surface, and this process has a limited rate no matter how much steam is introduced outside the strip. Thus at a high concentration of steam in the atmosphere the decarburization process becomes zero-order in the steam concentration, dependent only on the diffusion of the carbon.

In this case, wouldn't the diffusion rate of the carbon to the surface be directly proportional to its concentration, making the overall reaction (ideally) first order with respect to it?

Still zero-order with respect to steam.

Wouldn't the reaction rate still increase at very high steam concentrations as steam starts to diffuse into the steel?

Steam would not diffuse into the steel as water molecules, and if individual oxygen atoms were to enter there would be no way to form CO molecules in the solid matrix either. Also I am ignoring a side reaction where the steam oxidizes the iron, which the manufacturers avoid by having hydrogen in the atmosphere.

@OscarLanzi By that reasoning, every reaction is "zero-order with respect to" one of the reactants that's in excess, which is a fairly uninteresting fact

@253751 that is true of the eventual conversion, not the rate in all cases.

The best example of a zero-order reaction is the combustion of a candle. If the candle weighs $m_o$ grams at the beginning, and $m$ at any time afterwards, and if $a$ grams of it are burned per minute, the reaction rate $r$  is constant from the very beginning to the end of the candle, and it is : $r = dm/dt = a$ in grams per minute. The integrated rate law is : $m = m_o - at$. This is a zeroth-order kinetics.

Is this a real candle or a theoretical candle? Seems pretty contrived.

Contrived or not, a zeroth-order reaction is one where the rate is independent of the quantity of reactants/products. This is a good write-up. I've encountered these (and the defined pseudo-type, which I don't have time to write up now) in the context of geochemical kinetics.

@ToddMinehardt I am not saying zero-order reactions don't exist. But I have seen real wax candles burn and they don't burn at the same rate from the start to finish. At the beginning it will burn slowly, and after a while as more of the wick comes out, it will burn faster. It's okay if the candle example is given as a thought experiment, but actual candles definitely don't burn at the same rate from start to finish.

@SRMaiti A bit of calculus might help here. The reaction is 0th order with respect to mass of the candle but non linear with respect to temperature and wax shape. If i give you access to a candle of unknown length who's end you can't and see and its pushed upward such that the flame stays in the same place there is no measurement you can perform even when allowed to turn the candle off and on again that allows you measure the length of the candle without just burning it completely The non linearity with regard to candle warming yields no information about the total length of the candle. => 0th

@worldsmithhelper I am not talking about the length of the candle. I am saying that I believe the rate of mass loss would increase after more of the wick gets exposed. I don't understand all of the stuff you wrote about length and warming etc.

@SRMaiti The rate of mass loss (~length loss) of a candle that has been turned on for a while is independent of the remaining fuel. The wick has an influence, as has a lighting a cold vs a hot candle however the wick effect becomes irrelevant really fast. If we compare two identically unlight candles of different lenght and light them we see that the burn rate is independent of the total mass, so the effect of the wick is not order 1 or higher either. The burn rate at the wick therefor does not depend on mass remaining but mass burnt and quickly levels of once the constant thickness is reached

@worldsmithhelper Ok, I guess I understand what you are saying. But I can't convince myself without an experiment.

First, be aware that the order of a chemical reaction depends on its mechanism and a great deal of chemical reactions have more than one mechanism, depending on the conditions.
The traditional example for zeroth order reaction is a reaction fully dependent on a catalyst (e.g. enzymes in biochemistry).
When you have enough reactants and a scarce amount of the catalyst, you get a reaction rate that depends only on the concentration of the catalyst. It is zeroth order in regard to reactants.
Another example: topochemical reactions. The site(s) of the reaction may limit the reaction rate.
Related example: reactions limited by the mixing rate.


text:
Solubility of lead(II,IV) oxide
I was solving a question:

Which of the following liberate $\ce{O2}$ upon hydrolysis?

$\ce{Pb3O4}$
$\ce{KO2}$
$\ce{Na2O2}$
$\ce{Li2O2}$

The answer is no doubt $\ce{KO2}$ but if I am not wrong $\ce{Pb3O4}$ is insoluble so it doesn't undergo hydrolysis right and I would like to know the reason why $\ce{Pb3O4}$ is insoluble?

Better tell your reason to think it should be.

Most of metallic oxides are insoluble, as breaking of covalent bonds or the ionic lattice is not thermodynamicaly favorable.
Soluble exceptions mostly belong to electro-positive metal oxides ( together with peroxides and superoxides) from the first two groups of the periodic table, forming more or less soluble hydroxides. So it is not true dissolution.
Other exceptions are some oxides in high oxidation metal states, forming independent molecules, getting hydrated as acid anhydrides, like $\ce{Mn2O7, CrO3}$.


text:
IUPAC naming for a ring system containing different ring structures
Regarding the IUPAC naming convention for compounds in which one ring is more highly $\pi$-conjugated than the other ring but contains fewer carbon atoms:

I can name this compound as either:
(a) 1-[cyclohexenyl]-cyclopenta-1,3-diene, giving preference to the ring with the greater number of $\pi$ bonds; or
(b) 1-[cyclopenta-1,3-dienyl]-cyclohexenyl, giving preference to the ring with the greater number of carbon atoms.
What is the correct answer?

Before deciding the parent structure, there are two problems. Firstly, why square brackets? You only need square brackets as a second layer if you already have parentheses. If not, you should only use parentheses. Secondly, hyphens are not used to separate substituents from the main part of the name. You don’t say 2-methyl-butane, right?

If you use "cyclopenta-1,3-diene" in the end of the name (your first attempt), then this would be the stem system.  In case like the one presented in the question -- two rings connected via a single bond -- it is recommended to recognize the cyclohexene as the stem (like parent), and the cyclopentadienyl moiety as a substituent (like child).  Because counting the number of ring atoms has a higher priority, than the number of double / triple bonds in the cycles, your second attempt is in this perspective the better one of the two.
To detail out the substituent, use normal parentheses. Square brackets would be suitable e.g., to describe annulated bicyclohexanes' bridges (e.g., decalin as bicyclo[4.4.0]decane); however, this motif is not present in your structure.
This yields the name of: 1-(cyclopenta-1,3-dien-1-yl)cyclohex-1-ene
Addition: The answer by Safdar Faisal refers to the 2013 edition of IUPAC's Blue Book (online content here).  IUPAC published  a brief guide about this revision (Hellwich et al. 2020) with the following flow chart relevant to the original question:

and specific polychromatic examples e.g., about structures with two cycles joined by a single bond:

As SI to this publication, there equally is a four-page excerpt available, too.
Reference:  Hellwich K.-H.; Hartshorn R. M.; Yerin A.; Damhus T.;  Hutton A. T. Brief guide to the nomenclature of organic chemistry (IUPAC Technical Report). Pure Appl. Chem. 2020, 92, 527-539; doi 10.1515/pac-2019-0104 (open access).

You have two choices for choosing the parent hydride here in this case.
1. Choosing the cyclohexene ring as the parent hydride
If we consider the cyclohexene ring as the parent hydride, we get the cyclopentadiene ring to be a substituent leading the name to become 1-(cyclopenta-1,3-dien-1-yl)cyclohex-1-ene
2. Choosing the cyclopentadiene as the parent hydride
If we choose the cyclopentadiene as the parent hydride, the cyclohexene ring becomes the substituent and  we get the name of the compound to be 1-(cyclohex-1-en-1-yl)cyclopenta-1,3-diene
Now, what do we choose - the cyclopentadiene or the cyclohexene?
The preferred IUPAC naming
According to P-44.2.1 of the Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), we gain the rules to determine seniority in ring systems.

P-44.2.1 The senior ring or ring system:
(a) is a heterocycle;
(b) has at least one nitrogen atom;
(c) has at least one heteroatom (in the absence of nitrogen) that occurs earlier in the following sequence:
F > Cl > Br > I > O > S > Se > Te > P > As > Sb > Bi > Si > Ge > Sn > Pb > B > Al > Ga > In > Tl;
(d) has the greater number of rings;
(e) has the greater number of skeletal atoms;
(f) has the greater number of heteroatoms of any kind;
(g) has the greater number of heteroatoms occurring earlier in the sequence:
F > Cl > Br > I > O > S > Se > Te > N > P > As > Sb > Bi > Si > Ge > Sn > Pb > B > Al > Ga > In > Tl.

Therefore, via P-44.2.1 (e), we have the cyclohexene ring as the the main parent hydride. Therefore, the name found in choice 1 is the PIN of the compound. Thus, the PIN of the compound becomes 1-(cyclopenta-1,3-dien-1-yl)cyclohex-1-ene

Because our university library did not (yet) engage in purchasing the 2013 edition of the Blue Book (neither paper, nor electronic).  So it was a nice find (eventually) that there is a brief guide addressing at least some points (cf. vide infra).  Do you know if the current Blue Book equally contains such flow charts and examples where color is used to highlight the structure motifs?

@Buttonwood IIRC the blue book is black and white only and it isn't actually text searchable with the pdf. So you have to do an OCR of your own if you would like to use that...

@Martin-マーチン I thought a text of reference this relevant would retain the text layer.  So the content mirror from Queen Mary University and Brief Guide mentioned below may be more accessible from the get-go.

@Martin-マーチン As it happens, an HTML version of the Blue Book was just published last month. Sadly, search engines have not gotten around to indexing it yet.

@Buttonwood I've found that OCR works reasonably well on the Blue Book PDF; the main annoyance is that names can pick up extraneous spaces around their punctuation, and 1 is often mistaken for l. The online HTML version is probably the better option now that it exists, though.

Finally! Thanks, @LegionMammal978. Even though this is not for actual learning, it should probably included in the 'Resources' Q&A.


text:
Protein drug conjugate exercise
I am supposed to draw the structure of a derivative of the following structure which could be used to make a protein drug conjugate. And then I should draw the reaction for the protein modification.
So I thought it would be easiest to turn the carboxylic acid into an aldehyde and use this as the derivative.
Then the aldehyde and primary amine on the protein could form a imine. I am not a chemist or pharmacist. So it's been a while since I had to do organic chemistry and I don't have solutions for this exercise. So does this derivative and reaction make sense?

The reaction for formation of imine is correct, as far as I  (not an expert) can tell. Before that you'll need to reduce the acid to aldehyde, though. Whether this will work as a protein drug conjugate or not - I frankly haven't got the faintest idea

The solution suggested by the OP might work, but has a low chance of success.
Because the drug is linked directly to the protein, it might not bind to its target. There are multiple ways this problem is solved in the literature. You could have a long linker attached to a region of the drug that is unimportant for binding to the target (you would have to be lucky, try different sites, or have more information on how the drug interacts with the target). Or you could design a linker that opens up again when the drug reaches the compartment of the target (through a pH change, a slow hydrolysis reaction, or something fancy like photoactivation).
For a super-deluxe conjugate, here is an example involving a photocaged compound: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6857671/


text:
2-Iodoxybenzoic acid oxidation reaction
I was thinking about using 2-Iodoxybenzoic acid to selectively oxidize primary alcohols to aldehydes. I've found papers using various solvents like acetone/water mixture1, 2-methyl-2-propanol2, DMSO or bromobenzene3.
My question are:

How to choose the most suitable solvent for an IBX reaction?

IBX reduces to iodosobenzoic acid which can be filtered out together with leftover IBX and oxidized back to IBX using Oxone and reused. After filtering the reaction mixture and distilling the solvent, we get reasonably pure aldehyde for further use.

Is this workup procedure correct?

Could be IBX prepared in situ from 2-iodobenzoic acid and potassium persulfate?

References:

Mild Oxidation of Alcohols with o-Iodoxybenzoic Acid (IBX) in Water/Acetone Mixture in the Presence of β-Cyclodextrin, K. Surendra, N. Srilakshmi Krishnaveni, M. Arjun Reddy, Y. V. D. Nageswar, and K. Rama Rao
The Journal of Organic Chemistry 2003 68 (5), 2058-2059
DOI: 10.1021/jo026751w

Scott A. Van Arman, 2-Methyl-2-propanol as solvent for o-iodoxybenzoic acid (IBX) oxidation of 1˚ alcohols to aldehydes, Tetrahedron Letters,
Volume 50, Issue 33, 2009, DOI: 10.1016/j.tetlet.2009.06.045.

Iodine(V) Reagents in Organic Synthesis. Part 4. o-Iodoxybenzoic Acid as a Chemospecific Tool for Single Electron Transfer-Based Oxidation Processes, Nicolaou et.al, J. AM. CHEM. SOC. 9 VOL. 124, NO. 10, 2002, DOI: 10.1021/ja012127

A Simple and Advantageous Protocol for the Oxidation of Alcohols with o-Iodoxybenzoic Acid (IBX), Jesse D. More and Nathaniel S. Finney
Organic Letters 2002 4 (17), 3001-3003, DOI: 10.1021/ol026427n

@Waylander I'd like to avoid bromates as bromide impurities in the final product increase explosion hazard. I'd like to generate iodoxybenzoic acid in situ using potassium peroxymonosulfate (oxone) but more favourable would be potassium persulfate for me. I'm a bit worried if it could oxidize the starting alcohol to carboxylic acid.

I cannot find a ref for persulfate oxidation of 2-Iodobenzoic acid but Potassium persulfate and 3-methylimidazolinium triflate has been used to oxidise alcohols to aldehydes.  https://www.tandfonline.com/doi/abs/10.1080/00397910802374117?journalCode=lsyc20

@Waylander I meant for the preparation of IBX, not for actual oxidation. (2-iodobenzoic acid and potassium persulfate)

I realised that, hence my comment about not being able to find a reference for IBX prep using potassium persulfate.


text:
How practically to determine dispersion's Hamaker constant using CCC?
I know that Hamaker constants can be determined from CCC data.
$$ccc=\frac{9.85 \cdot 10^4 \epsilon^3 k_B^5 T^5 \gamma^4}{N_A e^6 A^2 Z^6}$$
Where, $\epsilon$ is solvent dielectric constant, $N_A$ is Avagadro constant, $k_B$ is Bolcman constant, $Z$ is electrolite charge, T is temperature of solution, A is Hamaker constant, e is electron charge.
$\gamma=\frac{exp(Z e \psi_0 )/2k_B T+1}{exp(Z e \psi_0 )/2k_B T-1}$.
I can get all constants to determine A except for $\gamma$. Where can I get $\gamma$ values?
Lets assume, that I have CuS stabilized by oleic acid nanoparticles dispersed in hexane, how do I practically find $\gamma$? I can grasp that $\psi_0$ is somehow related to zeta-potential.


text:
How are 2nd, 3rd, etc. ionization energies measured?
I am learning about ionization energies and I understand roughly how the first ionization energy can be measured. But how are the subsequent energies measured?
For the 1st ionization energy the neutral atoms are excited by photons or electrons of known energies. When an atom is ionized the negative electron and the positive ion are attracted to a detector giving a readout voltage.
After the first electron is removed the atom is charged and moves towards the detector plate. How then can we measure the second energy if the ion ''goes away'' after the first ionization? Is there some way of holding the ion in place until a second ionization occurs?
Our notes used potassium as an example listing off the 17 ionization values. So how was say the 17th ionization energy for potassium measured?

Moving ions can still be subjected to incident radiation to trigger the higher ionization.


text:
How does volume change as a piston separating two compartments of a thermally insulated container moves to reach thermodynamic equilibrium?
The situation is a thermally insulated container divided into two compartments, which are separated by a diathermic piston. Initially, the left compartment has a volume (V1) of 2L, and a temperature of 400K, and the right has a volume (V2) of 4L and a temperature of 200K. Both contain one mole of ideal gas each. The piston is first clamped, and then released. The question states that work associated with the movement of the piston may be neglected. The question asks to describe the graph of volume and pressure versus time for this spontaneous change, and these were the correct solutions (pictures below).
I calculated that the final volumes are 3L, and the final pressures are 100 atm in both compartments. (using the fact that that temperatures equalise to 300K while the piston cannot move, final pressures of both compartments must be equal, and number of moles in each compartment must remain the same)
What I do not understand is the manner in which this final 3L is reached. Specifically, I do not understand why V1 rises to a peak and then reduces and why V2 decreases to a minimum and then increases to the final value.
I've researched any other similar piston questions, but could not find an explanation for it. Could anyone please help with this? (I am in 12th grade, and this question is from an exam meant to be given by 12th graders)

That seems an unnecessary complication added to the problem. It looks like the response of a damped oscillator (see for instance https://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator). What happens is that the hotter and compressed compartment initially expands while losing heat to the second one. The piston moves into the second compartment with some inertia which results in the volume of the first compartment expanding past the equilibrium (halfway) point. The details are less important than the final condition of $V_1=V_2$.

@BuckThorn thank you so much for taking the time to reply, I understand what's happening now! these were only 2 of the options, the other one was a graph with a shape similar to the pressure one but for volume, with the same condition of V1 = V2, hence my confusion. I have a follow up question, if that's okay: why does't the pressure oscillate like the volume? and what causes the damping here? as I understand it, damping usually involves some form of energy loss, but I can't figure out where energy might be lost here.

They must also give you data on the thermal conductivity of the piston, right?

@ChetMiller it’s a diathermic piston! Sorry, I didn’t have the space to include that in the main question

They just say it is diathermic, or do they give you more information about the piston?

@ChetMiller the question has described it exactly as ‘rigid, movable, frictionless, and diathermal’

In my judgment, more information is necessary to calculate the temporal variation.


text:
Comparing intensity of colour in coordination compounds
I had given an examination recently in which the question asked was:

Write the decreasing order for the intensity of color in $\ce{[CoCl4]^{2-}}$, $\ce{[Co(CN)4]^2-}$, $\ce{[Co(H2O)6]^2+}$

What my approach was to first find out the geometry of compounds after which I found the geometry as follows
$\ce{[CoCl4]^{2-}}$ : $\mathrm{sp^3}$,
$\ce{[Co(CN)4]^2-}$ : $\mathrm{dsp^2}$,
$\ce{[Co(H2O)6]^2+}$ : $\mathrm{sp^3d^2}$
Now, I approached the question with the thought that more the symmetry less is the intensity of the color of the compound, which gave me a conclusion about what I had learned earlier that intensity of color in $\mathrm{sp^3 > sp^3d^2}$. And as the symmetry of $\mathrm{dsp^2}$ is comparable to that of $\mathrm{sp^3d^2}$, in that the intensity of color would be more for strong field ligand.
This made me conclude to the answer $\ce{[CoCl_{4}]^{2-}$ > $[Co(CN)_{4}]^{2-}$ > $[Co(H2O)_{6}]^{2+}}$.
But I am really confused and not sure about this answer as maybe the intensity of the color is completely determined only by the nature of the ligand, or geometry is involved or not. I also have confusion about how should I compare the intensity of color in $\mathrm{dsp^2}$ and $\mathrm{sp^3d^2}$.
P.S.: If possible please answer the question with basics as I am a High School student and don't know much about Higher Chemistry.

I don't think hybridization helps to predict intensity. Can you add what the textbook says?

Unfortunately: Why is it wrong to use the concept of hybridization for transition metal complexes? There won't be a simple answer; there might be a retrofit of an explanation using simpler models, but that will fit the situation only.

@M.Farooq I have been taught about this by a teacher. He taught me that the intensity of colour in sp3 compound would be higher than sp3d2 as sp3 is more asymmetric.

Bhanu, I never heard of that. Ask your teacher to give a reference.

Qualitatively, asymmetric complex may allow breaking of the selection rules and its intensity may be higher. I do not recall seeing your teacher's explanation.

I'm giving JEE this year too, and so far I've always used this: More the crystal field splitting energy, more is the intensity of colour. Your reasoning of symmetry I'm hearing for the first time. (P.S - If your teacher gives a reference, please share it here. Also, when NTA releases the answer key, please update the final order. Thank you.)

@TRC If it was only on the basis of Crystal field splitting energy then I am definitely losing marks over here as then the answer would clearly be CN>H2O>Cl.

That is why I am scared of this question, losing 5 marks in this would literally bring me down like 700-800 rank.

Forget about hybridization in coordination chemistry. This is nothing to do with colors! This question is really poorly worded by your examination board and please tell them to update this century old chemistry syllabus. By color intensity, are they implying that hue of the color of the complex or the intensity of eletronic transition?

Anyway, let us take it literally that they want you to predict intensity of transitions in complexes. "Intensity" in absorption spectroscopy has a different meaning altogether. It means how strong is the absorption of light (molar absorptivity). It is not a trivial task to predict just by this given information. It is a cruelty to ask this question from college students. The second problem with this question is that it is asking you to compare three tetrahedral or perhaps square planar complexes with an octahedral one.

Try to look at this link. Orgel diagrams
With that tetrachloro cobalt complex should absorb strongly than octaherdral cobalt complex.

@M.Farooq I could probably go on ranting about this exam for an hour. But that's not going to help anyone. If the OP, and I, want to get into a good engineering college, we have to bear with them. The worst part is that they don't even have a fixed syllabus - they just give the broad topics and we are expected to memorize as much as we can (and afterwards? Over 90% of students going into engineering will never need chemistry knowledge (except those in metallurgy branch) - obsolete methods, useless theories that have been long proven wrong.)

@BhanuKiran I'm not completely sure, but yes, that's the option I'd have marked, had I been in your shift. From the feedback you got here you must have realized that it's an utterly nonsensical question they asked. When CBSE conducted JEE Mains till 2-3 years ago the questions were better - they made you think, not score easy marks based on rote memorization. I don't think they'll ask such questions in Advanced either. It's just the gift of our dear N-T-A.

Yep, that's what I was thinking during the examination, how are they even comparing the intensity of colour in completely different hybridisations (considering hybridisation is valid in coordination compounds). I hope the questions are better framed in advance.

@BhanuKiran What is the answer given by NTA?

@TRC I uploaded the question wrong over here, It was for Ni and not Co. The answer for Ni was [NiCl4]2->[Ni(H2O)6]2+>[Ni(CN)4]2-.

I thought the same way, sp3 hybridization would have had the highest intensity but I completely don't understand how they have compared between H2O and CN.

@TRC I mean if they had considered strong field ligand then the answer would have been reversed, most probably something with hybridization?

Seeing the answer now, the best I can do is to hazard a guess - for weak field ligand the energy gap for electronic transitions is less. Less energy gap means it's easier for electrons to jump back and forth. More electrons jumping implies more photons being emitted/absorbed - so more intense colour.

@TRC I think that's right and also splitting energy of octahedral complex is less than square planar hence octahedral will have more intensity, but there's also a problem that Cl is weaker ligand than H2O which becomes contradictory. Feels like this question was data-based and we had to memorise it.

Exactly - it cannot be explained theoretically, as others have already pointed out.


text:
What's the equilibrium pressure of CO2 gas over sodium hydroxide?
Many (too many) decades ago in physical chemistry lessons, I seem to recall that given the enthalpy changes of reactions, and maybe phase diagrams, one could work out what equilibrium a reaction would reach.
I'm contemplating doing an experiment in much reduced pressure, by no means a hard vacuum, around 1 kPa, that might be achievable by flushing the volume with CO2, and then connecting a container of NaOH as a sorption pump, (both of which I have) as an alternative to buying a refrigerant servicing pump. What's the lowest system pressure that's thermodynamically possible this way, at room temperature?

It is perhaps not as trivial. NaOH will almost instantaneously absorb all the carbon dioxide in the beginning, then you will have sodium carbonate. Sodium carbonate can further absorb carbonate dioxide to form sodium hydrogen carbonate. So you have multiple reactions.

Two carbon dioxide scrubbers come to mind: lithium hydroxide and soda-lime.
(https://en.wikipedia.org/wiki/Lithium_hydroxide)"Lithium hydroxide is used in breathing gas purification systems for spacecraft, submarines, and rebreathers to remove carbon dioxide from exhaled gas by producing lithium carbonate and water:
2 LiOH•H2O + CO2 → Li2CO3 + 3 H2O
or
2 LiOH + CO2 → Li2CO3 + H2O
The latter, anhydrous hydroxide, is preferred for its lower mass and lesser water production for respirator systems in spacecraft. One gram of anhydrous lithium hydroxide can remove 450 cm3 of carbon dioxide gas. The monohydrate loses its water at 100–110 °C."
The anhydrous form is preferred because of lower molecular weight, but I doubt that it is entirely anhydrous - not after the first CO2 reacts. I wonder if you would get all the CO2 out, but replace it with water - which could have a reduced vapor pressure because it is being absorbed by the lithium hydroxide.
(https://en.wikipedia.org/wiki/Soda_lime) "The main components of soda lime are
Calcium oxide, CaO (about 75%)
Water, H2O (about 20%)
Sodium hydroxide, NaOH (about 3%)
Potassium hydroxide, KOH (about 1%).
The reaction mechanism of carbon dioxide with soda lime can be decomposed in three elementary steps:

CO2(g) → CO2(aq) (CO2 dissolves in water – slow and rate-determining)
CO2(aq) + NaOH → NaHCO3 (bicarbonate formation at high pH)
NaHCO3 + CaO→ CaCO3 + NaOH (NaOH recycled to step 2 – hence a catalyst)"

The phrase the bothers me is that "CO2 dissolves in water - slow and rate-determining." I wonder if the vacuum would pull of the water before the CO2 had a chance to react with the NaOH. I think the reason soda-lime is used rather than solid NaOH is that the NaOH has very little surface area for interaction with the CO2; soda-lime is more powdery.
As M. Farooq points out, the calculations are not trivial. I don't want to sound negative; give it a go. My point was that NaOH is not generally used to absorb CO2, but rather, soda-lime. But it has that water content that might interfere with your vacuum. Lithium hydroxide - at least from the equations - might devolve into LiOH.H2O plus Li2CO3. There may be other options.


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What chemistry topics should I learn if I have to do research in microfabrication (ex. photolithography) and ionic liquids (QSPR, synthesis)?
I'm starting a PhD and my research topic involves microfabrication (photolithography, etc.) and also ionic liquids (including QSPR and possibly synthesis). The thing is, I know close to nothing about chemistry, my background is in aerospace engineering. I am ready to make a big investment in time to learn whatever I need to reach to the point where I can do research in these topics, but I also can't take all the courses of a Chemistry undergrad at this point. What topics should I learn, and what can I skip? If you could recommend me some books/online classes or courses that would be a bonus! Thank you very much :)

Have you gotten any input from the graduate advisors or faculty in your program?

The present question is probably a bit too general to receive a fitting answer. There is a (limited) survey of recommended resources which may become more useful after discussion and narrowing the addressed target(s) with your supervisor. If focused  in time and topic, a search with keywords on publishers' / societies' websites (e.g., mdpi, ACS) may be an inspiration.

Thank you Todd and Buttonwood. I'll ask faculty too, good idea. Buttonwood, I know the question is a bit broad. I just try to read papers or books related to those two topics and I realize I lack of a lot of background knowledge. I'm wondering what learning path should I take in order to understand these topics better. I'll probably review the basics with Oxtoby's book. But then should I look into physical chemsitry, inorganic chemistry? And thank you for the resources!


text:
Double boiler apparatus with no loss of water?
Is there an apparatus that operates like a double boiler, but loses no steam?
In other words, in a normal double boiler the steam is allowed to escape, so energy is lost and the boiling pot must be periodically renewed with fresh water.
If, however, we make the boiler sealed, then pressure will build up to potentially dangerous levels.
Is there a way to keep the steam captive and reflux it safely back to the boiler without loss of heat except to the vessel being heated?
Note that using a condenser does not solve the problem because energy will be lost to the condenser and even more energy will be required because a pump will be needed for the condenser.
Offhand, the only way I can think to do this would be to have a sealed boiler with a pressure sensor and then reduce the heat when the pressure reached a certain level. That way there would only be a fixed amount of steam so the pressure vessel would (hopefully) never reach dangerous levels of pressure and the heating would be controlled. So, it would be something like this:

I have done searches on "pressurized double boilers" and the only thing that comes up are Espresso machines!

Why are you worried about loss of water? The point of a normal bath is to control temperature of the stuff being heated (so it doesn't exceed the temperature of boiling water) and water is cheap. Pressurised steam doesn't solve the heating problem as the temperature will rise with pressure. And alternative (non-evaporating) solvents will be more useful (and temperature control can use a cheap thermostat).

@matt_black It's not a bath. It's a steam dryer. Losing steam is bad because energy is lost AND not only that but requires that the water be refilled. This process needs to be running autonomously.

If you need to control pressure, steam is going to escape. If you need a vessel that can cope with any pressure, the expense will be huge and you won't control the temperature. If you want continuous operation, you could feed steam into the vessel and let it condense in another vessel, but that would require condensation many industrial processes do this, efficiently). I suspect you have incompatible requirements.

if the contact between steam and the material isnt necessary, if you're going to use temperature control anyway you should use oil instead of water to maintain temperature of 100°C without significant evaporation = without heat loss

How about an autoclave?

Just use a water or oil bath with an accurate temperature control and stirrer such as these ThermoFisher units or from Thomas Scientific. After all, temperature control is the purpose of a double boiler.
Since your question also concerns efficiency, this is a better solution, since the container can be insulated. If energy were at a premium, e.g., at a remote location, then it could even be sealed, though a safety valve would be needed; in normal operation, it would not vent.

Steam is required because the vessel being heated must be above 212F

@ShakaBoom, Incorrect: read the answer - oil bath.

It seems to me that you want to prevent water vapor from escaping the container. There are different ways to stop this, but having an escape duct for the steam is useful in order to prevent an accident. For example, let's consider that your heating apparatus has an issue, and instead of periodically heating the container which heats the water, it keeps on applying energy. Air pressure in your container might accumulate to dangerous levels until one of two things occurs:

The container melts/cracks due to high temperature, and the vapor escapes.
The air pressure ruptures your container (could be violent).

Both scenarios are not ideal. So an air pressure escape duct is useful. In other words, don't make a container that is prone to accidents.
Now, you suggest that a pressure sensor would be a good way to prevent pressure from accumulating. Your apparatus would measure the pressure, and stop applying heat when the level is too high. My suggestion would be to use a heat sensor instead. There is no need to invest in a pressure sensor that will probably be more expensive. Remember that the ideal gas law (assuming you are not doing precise calculations) relates pressure to temperature; so the temperature probe will do just fine.
Now, there is a fundamental issue with what you are trying to propose. You want to achieve something impossible: "without loss of heat". If you stop applying heat, the already present heat will disperse into the environment. It doesn't matter what kind of container you have, heat will simply be conducted and lost to the environment. As DrMoishe Pippik suggests in his answer, insulting your container will reduce heat conduction, but it does not stop it. You will find yourself spending large amounts of money on an apparatus that simply won't work. Additionally, "keep the steam captive and reflux it safely back to the boiler" is not a smart nor cost-effective idea. Firstly, your gas will cool down, and trying to re-heat the water with the "captive" vapor is just overcomplicating things. However, saunas do exactly that, but they don't re-use their water, and they always heat the vapor again.
Finally, if you don't want the vapor to escape, just use a large container that can maintain all the water in its vapor state without safety issues. Then, your probe measures the temperature and stops heating when it is too high. Since $\ce{H2O(l)<->H2O(g)}$ is reversible, you won't lose your water. And make sure to have an escape duct!

Losing heat is fine, but we want the heat to be transfered from the steam to the material being heated. We don't want the heat to escape in the form of lost steam. Obviously there would be a pressure release valve in the vessel. The question is: does this kind of equipment already exist? If so, what is it called?

@ShakaBoom So you want to be able to take out the material without losing the vapor?

I have added a diagram showing an idea of how it could be done, but note that there is not much design for reflux at all, its just a controlled pressurized double boiler. As far as I can tell, this type of device does not exist as standard lab equipment, but maybe I am wrong.

@ShakaBoom This sort of equipment does exist and is common in industry-scale work in chemical plants or pilot plants. They use site wide steam to heat vessels (on 100kg+ scales). But the plants recycle the steam so steam jackets the vessel under constant pressure and condensed water is recovered and reheated in a closed loop. but this makes almost no sense on smaller scales where simpler heating methods are easier and more convenient. What scale are you working on?


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Can a complex reaction have more than one rate determining steps?
I was solving some questions related chemical kinetics and encountered one question with multiple correct answer , the question is :

For a complex (multistep) reaction,
(a) the overall rate of reaction is the rate of slowest step.
(b) the overall molecularity has no significance.
(c) molecularity and order may or may not be same.
(d) the number of rate determining steps may be more than one.

The question have all the four options as correct answer. There's no doubt about first three options, but about the fourth one I want to ask ,
How a complex reaction have more than one rate determining steps? And if it has then how the overall order and molecularity of the reaction may be given as for complex reation we look at the slowest step and with the help of slowest step we decide the order and molecularity .
Then with multiple RDS how it is possible ?

I am not really sure, but maybe its possible that two steps are equally or similarly slow? You can identify a step as rate determining only when its rate is appreciably slower than the other steps.

@S R Maiti So we can take any of the slowest step to determine the order and molecularity ?

No, I think if there are multiple steps which are slow, then you would have a very complicated rate equation. Sometimes you can derive these by steady state approximation. Other times it might not even be possible to write a rate law. For example the decomposition of HBr has a very weird looking rate law with 5/2 exponents and a denominator etc. The order would be undefined in that case. And molecularity is not even defined except for elementary reactions, so there is no question of molecularity for multi-step processes.

There is always one step that will be the slowest even if the difference is miniscule. By definition there can only be one slowest step. I think someone above mentioned that there could be a case where two steps have identical velocities but it's unlikely.

For a sufficiently complicated overall reaction which hides a large interconnected reaction network, the concept of a "rate determining step" may effectively not exist (at least not in any way that we might be familiar with from more simple cases), such that sentence (d) kind of loses meaning. I briefly talk about it in this past answer and the second half of this one.


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Do lithium iron phosphate cathodes change color during charging and discharging while losing lithium ions?
The following reaction takes place  in a lithium iron phosphate (LFP) battery:
$$\ce{LiFePO4 -> FePO4 + Li+ + e-}.$$
According to Wikipedia, $\ce{LiFePO4}$ is a grey, reddish grey, brown, or black solid, and $\ce{FePO4}$ is a yellow-brown solid. If you had a window into the LFP battery, could you see this color change occurring during use?

I am not sure, if there are (colored) salts of transition metals, that have the same color at different metal oxidation states.

Is LiFePO4 darker than FePO4 as described by Wikipedia solo because of the addition of carbon for conductivity in batteries?

Since there are carbon coatings on the $\ce{Li_xFePO4}$ particles and additional graphite/conductive carbon in the electrode, the electrode always appears to be black or dark grey, and color changes within the phosphate are hidden. Chen et al. write [1]:

Cathode materials, such as lithium iron phosphate with lower conductivity, are difficult to be observed because of the high light absorption of the carbon additives that are used to improve the conductivity of electrodes.

$\ce{LiFePO4}$ crystals have a blackish green color [2], cf. the color of the mineral triphylite which is described as bluish- to greenish-gray. I am not sure about the color of the $\ce{FePO4}$ in a battery since the usual $\ce{FePO4}$ described as yellow-brown solid might be influenced by its water content.
References

Chen, B.; Zhang, H.; Xuan, J.; Offer, G. J.; Wang, H. Seeing Is Believing: In Situ/Operando Optical Microscopy for Probing Electrochemical Energy Systems. Advanced Materials Technologies 2020, 5 (10), 2000555. DOI: 10.1002/admt.202000555.
Chen, D. P.; Maljuk, A.; Lin, C. T. Floating Zone Growth of Lithium Iron(II) Phosphate Single Crystals. Journal of Crystal Growth 2005, 284 (1), 86–90. DOI: 10.1016/j.jcrysgro.2005.06.024.


text:
Should I do solvent extraction for Buchwald–Hartwig amination?
I am conducting a synthesis involving Buchwald–Hartwig amination. Catalyst is $\ce{Pd2(dba)3},$ ligand is $\ce{P(o-tol)3},$ base is $\ce{Et3N},$ and solvent is toluene. Image below is the proposed mechanism.

I used $\ce{Et3N}$ as base, so I assume that an ionic compound $\ce{(Et3NH)+Br-}$ is formed. Should I do solvent extraction with water to remove triethylamine hydrobromide?
I couldn't find any procedures in the literature, although some papers mention Celite filtration.

The triethylamine hydrobromide is not toluene soluble so it should be removed by the Celite filtration

Ah, I just forgot that they would not be soluble in toluene so be precipitated. Thank you!

It seems like you figured out that filtration of the reaction mixute through a Celite$^®$ pad would work in this case since you have used nonpolar solvent such as toluene. However, if you have used tert-butoxide (e.g., $\ce{KOC(CH3)3}$) insteard as in the given mechanism, the byproduct is tert-butanol, which is soluble in toluene. In this case, you can do solvent-sovent extraction to remove most of byproducts.
The workup procedures in Buchwald–Hartwig amination reaction is depends on the type of base and the solvent used. For example, see following two examples:

A mixture of 2,6-dichloropyrazine (25 g, 168 mmol), 4-((3-methyloxetan-3- yl)methoxy)-2-nitrobenzenamine (40 g, 168 mmol), palladium (II) acetate (1.5 g, 6.72 mmol), (±)-2,2-bis(diphenylphosphino)-1,1'-binaphinyl [(±)-BINAP] (4.18 g, 6.72 mmol), and cesium carbonate (76.7 g, 235.2 mmol) in toluene (800 mL) was stirred at $\pu{90 ^\circ C}$ under $\ce{N2}$ for 23 h. The mixture was filtered through a pad of Celite and washed with EtOAc. The filtrate was evaporated under reduced pressure and the crude product was purified by flash chromatography on silica (EtOAc : petroleum ether = 1:5) which gave the title compound as a red solid (22.4 g, 37% yield) (Pfizer Patent: WO2010016005).

A mixture of (6-bromo-4H-benzo[d][l,3]oxazin-2-yl)-(R)-indan-l-yl-amine (172 mg, 0.5 mmol), commercially available 2-amino-6-trifiuoromethyl-pyridine (162 mg, 1.0 mmol), tert-Bu-XPhos (34 mg, 0.08 mmol), $\ce{Pd2dba3}$ (18 mg, 0.02 mmol), sodium tert-butylate (53 mg, 0.55 mmol), tert-butanol (0.5 ml) and dioxane (3 ml) was heated in a sealed tube at $\pu{120 ^\circ C}$ for 16 h. The reaction mixture was poured into water (15 mL) and extracted with ethyl acetate (2 x 30 mL). The combined organic layers were washed with water (20 mL) and brine (20 mL), dried (magnesium sulfate) and evaporated. Further purification of the crude product by flash chromatography on silica gel (ethyl acetate/ heptane) yielded the title compound (127 mg, 60%) as light brown foam (Roche Patent: WO2010026110).

I used toluene as solvent. Ionic compound Et3NHBr (although it's unclear that it is really formed) is insoluble in organic solvent such as toluene, so it can be filtrated through Celite pad.


text:
Effect of trifluoromethyl substituent on electrophilic substitution
Problem -

Find which carbon is most susceptible to electrophilic attack (among carbons numbered $1$ to $6$):

Answer given (may not be the correct one) -

 $2$

My question -
$1.$ I know that $\ce{-CCl3}$ group exerts a negative hyperconjugation effect, hence acting as a deactivating and meta-directing group. If $\ce{-CF3}$ has a similar effect (which dominates over $\mathrm{-I}$ effect), then in that case phenyl ring D will be more electron-rich as compared to ring B - since the ring B, being attached at para of the main ring A, will participate in resonance due to the negative hyperconjugation of $\ce{-CF3}$ group. Since resonance does not result in a positive charge at the meta carbon of ring A, so I feel ring C should be more susceptible to electrophile. Within ring C, I expect only inductive effect to operate, so the carbon farthest from ring A will be richest in electrons. Accordingly my answer in this case will be $5$.
$2.$ If $\ce{-CF3}$ does not exert negative hyperconjugation - then in that case, only inductive effect will decide the electron density and consequently the carbon farthest from $\ce{-CF3}$ group will be most electron-rich. My answer in this case will be $6$.
In either case, I don't see how $2$ can be the answer. Is the answer key wrong? If yes, then which of my two reasonings is correct? If no, then what am I missing?

@Ashish Ahuja - it does, it's the same question from the same test paper!

If you don't mind, can you tell me whether, the paper mentioned is for mains or adv?

@Rishi Advanced - and their answer key to this one was wrong btw, the correct answer is 4.


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Solubility of Sodium Chloride
I have come to know that, the lattice energy of NaCl is 788 KJ/mol and hydration energy of NaCl is 784 KJ/mol.We know,if lattice energy> hydration energy,the compound won't dissolve.But how is NaCl dissolved in water?

Dissolving $\ce{NaCl}$ in water is an endothermic process, because of the reasons you mentioned. It simply absorbs some energy from the surroundings; it could be someone physically stirring the solution for example.

You might want to look at https://chemistry.stackexchange.com/questions/317/can-dissolution-reactions-be-endothermic and https://chemistry.stackexchange.com/questions/94261/how-can-endothermic-reactions-be-spontaneous


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Does HIO4 cleave hemiacetals?
I had written in my notes that it does. But I can't find any credible source for it.
A simple Google search only brings up this old and unanswered question.
The questioner failed to specify the source of their extract though. There're not any answers to refer to either. It would be helpful if someone could point me in the right direction.

Since neither questions provide any reliable sources and ask essentially the same using affirmative/negative wording, I edited the origin a bit and closed this question as a duplicate. Anyway, thank you for using search before posting.

I never knew $\ce{HIO4}$ could cleave either acetals or hemiacetals. I thought it only worked on vicinal diols or diketones.


text:
Correct Definition of Absolute Electrode Potential
I thought absolute electrode potential is Galvani potential difference at the interface. However, it is given by following equation in John Bockris - Modern Electrochemistry (Ref.1): $$ E^M \mathrm{(abs)} = \Delta^M_S\phi -  \frac{\mu_e^M}{F} $$
First term is Galvani potential difference on the metal/solution interface and the other is chemical potential of electrons in the metal divided by Farady's constant.
Question 1: Why is chemical potential of electrons in metal included in the definition?
On the other hand, on Wikipedia, absolute electrode potential is defined as: $$E^M \mathrm{(abs)} = \Delta^M_S\psi  + \phi^M  $$
Where second term is work function of the metal. This definiton is a little bit different than in Ref.1 since work function includes surface potential difference on the interface in addition to chemical potential of electrons in the metal.
Question 2: Why are these two definitions different?

Reference

John O’M. Bockris and Amulya K. N. Reddy, In Modern Electrochemistry: An Introduction to an Interdisciplinary Area, Volume 2; Plenum Press: New York, NY, 1970 (ISBN: 978-1-4615-8600-5).

Please look at the equations I edited are the one you meant to say. Also, see the reference 1 whether it is the correct edition you got the equation.

Oh, yes. I didn't add references on reference list. Thank you. Your edits of equations are correct, I just used notation which came to me since I think it is not too important.


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Synthesis of active chemical manganese dioxide (CMD) for primary alcohol oxidation
This paper1 uses chemical manganese dioxide (CMD) instead of activated manganese dioxide for mild alcohol oxidation. I'm looking for a route on how to synthesize active CMD. Could this work?
$$\ce{Mn2O7 + 3MnO → 5MnO2}$$
$$\ce{2KMnO4 + 3MnSO4 + 2H2O→ 5MnO2 + K2SO4 + 2H2SO4}$$
How to make chemically pure $\ce{MnO2}$ active for oxidation reactions?
Reference

Lulinski, Piotr et al. “Chemical Manganese Dioxide (CMD): its application to the oxidative iodination of benzene, halobenzenes and some deactivated arenes.” Molecules (Basel, Switzerland) vol. 9,7 595-601. 30 Jun. 2004, doi:10.3390/90700595

This preparation of active MnO2 from activated carbon and KMnO4 by Carpino (https://doi.org/10.1021/jo00836a091) may be of use.

Pyrolysis of Manganese carbonate at around 410C gives a very active form of MnO2 for use in oxidations

@Waylander that's convenient, I have access to a lot of manganese sulfate, but building something that could hold 410C for 6 hours would be a challenge.

My research centre used to have a muffle oven set up at around that temperature for drying mol. sieves and I used that.

Do not involve Mn2O7. It is a highly reactive and a dangerous oxidizer. It can violently explode and you won't have any control of the reaction. With that being said, check out this paper from 1968: https://pubs.acs.org/doi/abs/10.1021/jo01258a106

@NilayGhosh I really like this one, but I'd rather try it using dean stark and form an azeotrope with toluene/ethyl acetate or ethanol+methanol instead, why do they specifically use benzene?

Your question suggests you did not access Leo Paquette's Encyclopedia of Organic Reagents (14 volumes, some schools have access to the electronic version eEROS), or its one-volume siblings by topic, e.g., Oxidizing and Reducing Agents (this one by Burke and Dannheiser).  These resources cover physical / chemical properties of frequently used reagents and briefly showcases a selection of typical applications.
Citing the entry about $\ce{MnO2}$ of the former source, one of the of early hints preceding the protocols to prepare $\ce{MnO2}$  states for example

«It is worthy of note that the percentage water content strongly influences both the oxidizing power and the selectivity (oxidation
of multifunctional molecules) of active $\ce{MnO2}$. Thus it is well
known that the wet material (40–60% $\ce{H2O}$) obtained after filtration must be activated by drying (heating to $\pu{100–130 ^\circ{}C}$ for
$\pu{12–24 h}$ or, better, at $\pu{125 ^\circ{}C}$ for $\pu{52 h}$). Indeed, an excess of water decreases the oxidation power since, according to
the triphasic mechanism generally postulated, it would prevent
the adsorption of the substrate to the oxidatively active polar
site on the surface of $\ce{MnO2}$.»

The first of the six protocols presented is the basic one by Attenburrow (beside protocols under acidic conditions, by pyrolysis, coating silica gel with active / activated $\ce{MnO2}$, etc.) and reads as

«A solution of $\ce{MnSO4 ⋅ 4 H2O}$ ($\pu{110 g}$) in $\ce{H2O}$ ($\pu{1.5 L}$) and a solution of $\ce{NaOH}$ (40%; $\pu{1.17 L}$) were added simultaneously during $\pu{1 h}$ to a hot stirred solution of $\ce{KMnO4}$ ($\pu{960 g}$) in $\ce{H2O}$ ($\pu{6 L}$). $\ce{MnO2}$ precipitated soon after as a fine brown solid. Stirring was continued for an additional hour
and the solid was then collected with a centrifuge and washed with water until the washings were colorless. The solid was dried in an oven at $\pu{100–120 ^\circ{}C}$ and ground to a fine powder ($\pu{960 g}$) before use.»

Probe which of the protocols works well enough for the application in question.  Since eEROS points to the primary literature which you may combine in a subsequent search in Reaxys, Science of Synthesis, SciFinder etc., you equally may see which of them are applied on substrates similar to yours.
References
Cahiez G.; Alami, M.; Taylor R. J. K.; Reid, M.; Foot, J. S.; Fader L.; Sikervar V.; Pabba, J. Manganese Dioxide in Encyclopedia of Reagents for Organic Synthesis; doi 10.1002/047084289X.rm021.pub4.
Attenburrow, J.; Cameron, A. F. B.; Chapman, J. H.; Evans, R. M.;
Hems, B. A.; Jansen, A. B. A.; Walker, T.. 194. A synthesis of vitamin A from cyclohexanone. J. Chem. Soc. 1952, 1094-1111; doi 10.1039/JR9520001094.

I assume you are looking for activated manganese dioxide (AMD). Apart from the Attenburrow process mentioned in @Buttonwood's answer ($\ce{MnSO4 + KMnO4 + NaOH}$), I found three other process that might suit your need (I noticed OP has loads of $\ce{MnSO4}$. So, the aforementioned process will work. The processes mentioned in this answer doesn't involve $\ce{MnSO4}$ so it can work as alternatives):

Rheological phase reaction method from manganese benzoate and $\ce{KMnO4}$ characterized by chemical titration (Yuan et.al., 2003)
Potassium permanganate and manganese acetate, via a simple alternate drop-feeding method (Han et.al., 2010)
Using decolorizing/activated charcoal and aqueous potassium permanganate  to precipitate a brown-black powder presumably manganese dioxide mixed with carbon which can be separated using Buchner funnel (Carpino, 1970)

The Attenburrow protocol only uses $\ce{MnSO4}$, water, $\ce{NaOH}$, and $\ce{KMnO4}$. Reading the entry by Cahiez et al. just again, none of the other five protocols adds sulfuric acid.  Even the second method Preparation of Active $\ce{MnO2}$ from $\ce{KMnO4}$ under Acidic Conditions. only reads as «Active $\ce{MnO2}$ was made by mixing hot solutions of $\ce{MnSO4}$ and $\ce{KMnO4}$, maintaining a slight excess of the latter for several hours, washing the product thoroughly with water and drying at $\pu{110–120 ^\circ{}C}$.»  Thus, your $+ \ce{H2SO4}$ might be a misunderstanding.

@Buttonwood Oops, I wrote H2SO4 instead of KMnO4. Fixed it.


text:
Cleaning with Hydrogen Peroxide is not a chemical process?
Reading here:

if the peroxide does seem to help is that the bubbles forming and foaming is helping dislodge tiny particles in the cracks of the mineral.

Is this the case in general for cleaning with hydrogen peroxide, i.e. there is no actual chemcial reaction, its just dislodging particles?

Whether there is a chemical reaction depends on the sort of dirt on the rock. The article cites some reactions where it is a chemical process even while claiming that the bubbling also does some good. Organic material,a common contaminant of rock samples, is chemically destroyed by peroxide and some mineral stains will be chemically altered by it. The bubbling is likely a bonus physical effect, loosening soil and dirt, and not the primary effect.

Well, it is a play of words. Chemical process or a physical process do not mean a lot. Dilute hydrogen peroxide decomposes catalytically by many things, including dust/ grime, plant matter, minerals, iron oxides, etc. When it comes in contact with dirty rocks, it begins to decompose and begins to bubble oxygen. This bubbling action may loosen dirt.
Since geologists test many minerals rocks by acids (dilute hydrochloric acid), bubbling indicates carbonate, which is essentially like decomposing the rock/mineral being tested. The author is trying to point out that when dilute hydrogen peroxide is "acting" on the rocks, it is not chemically decomposing them; rather, it decomposes itself into oxygen and water.
It can certainly cause chemical changes, for example if there is sulfide in the "ore" material, it will be slowly converted into sulfate.
Nothing is inert in this world, including water.

I am not a chemist. In my mind, a physical process is as described here, a particle is dislodged. The dirt and the O are not bound whereas a chemical process would be like soap, where one end of the soap molecule binds to the dirt, the other end to water, and the dirt gets carried away as part of the soap molecule. Is this accurate?

Yes, this is correct. Physical cleaning would be akin to blowing air from a dirty surface. It is just mechanical force that "cleans" the surface.