text:
How to prevent hydrogen embrittlement in SmCo magnet?
We have a measuring apparatus that incorporates a samarium-cobalt (SmCo) magnet. Normally this is exposed to various gases under test. However, a problem has arisen with hydrogen, namely the magnet shattered.
I am assuming it is hydrogen embrittlement in magnets according to Shin-Etsu Chemical Co., Ltd. — Characteristics of Hydrogen Absorption Characteristics of Hydrogen Absorption on RE Magnets on RE Magnets (PDF).
How to prevent it? The only thing I can think of is some kind of metal plating or encapsulation. Of the two, the former is preferable.
Has anyone come across this problem and knows the solution?
We will be doing tests in Hydrogen of both plated and unplated magnets. The one that shattered was not plated.

Related: nature.com Novel Methods for Prevention of Hydrogen Embrittlement in Iron and wikipedia.org Hydrogen_embrittlement#Prevention

Related: engineering.stackexchange.com how-can-we-prevent-detect-hydrogen-embrittlement

I think embrittlement of RE magnets is something different from H embrittlement of steels. For example , austenitic stainless can absorb significant amounts of H and do not embrittle.

@black you'll still get the corrosion even if there is no carbon to react with. Very expensive corrosion.

Carbon does not react for embrittlement . Carbon reaction is High Temperature Hydrogen Attack.

Low-tech >> Hi tech
How about cling film? Hydrogen absolutely loves to adsorb and react with noble-ish metals and other d-orbital protruding heavies. Deny them the surface.
Wrap it in plastic. It won't solve anything long term, and it won't gain you any credits with the high brow community, but I bet it'll turn your problem into a nuisance.

If you want to be more scientific, look up the hydrogen diffusivity in various polymers (before still using cling film).

@Karl Oh, I know. My chemical plant produces some 2000 m3/h by electrolysis. OPs biggest problem is that his compound looks very tempting for hydrogen to dissociate on. Limiting access to that should help by magnitude or two. But hydrogen penetrates everything in the end.

OK, it is not hydrogen gas per se, but an embedded hydrogen atom. The latter chemically behaves as an mono-atomic hydrogen radical (where .H <-> H+ + e-). It (namely, .H) will, and per a source does, convert PbS, for example, to Pb and H2S. So, think of .H as operationally a (H+, e-) pair. Your suggested method implies less contact than, for example, a high pressure N2O treatment or soaking in a solvent with a large amount of dissolved N2O. As such, I would expect your process to be much less effective.

@AJKOER It might be less effective. Also, it is done by the time you call the first guys over to your desk to brainstorm the idea for how to do a gas coating in situ. And it could be effective enough.

Yes, possibly effective on H2 gas!  The latter could be formed from .H + .H = H2 (g). However, that also implies that the .H has already been self-removed (aka, dissipated with time) with an implied marginal improvement, per your suggested process, of precisely zilch. My comment also suggests importantly, if the degree of metal degradation does not improve with time, there is but little hope for a solution.

There appears to be a lack of understanding on what is occurring. I quote a source "The cathodic reaction produces Hchemisorbed by picking up an electron that released in the anodic reaction (H+ + e = Hchemisorbed ) in Al corrosion in HCl. In such acidic solutions, the Hchemisorbed on the metal surface reacts by combining with other adsorbed Hchemisorbed to form H2 gas molecule, which bubbles from the metal surface. A very small amount of the uncombined Hchemisorbed will remain; however, this amount does not affect ...“ Reference link: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5458939/ .

Try exposing the magnet to N2O either as a gas or dissolved, given its highly soluble in select solvents, especially alcohol.
Logic:

$ { H˙ + N2O -> OH˙+ N2 }$

Source: See "Reaction of the hydrogen atom with nitrous oxide in aqueous solution – pulse radiolysis and theoretical study. To quote:

...reaction of the H˙ atom with nitrous oxide (N2O) producing N2 and ˙OH in aqueous solution. The H˙ atom attacks the oxygen atom in the N2O molecule resulting in the formation of the [H–ONN]‡ transition state and its decomposition into ˙OH and N2.

The reaction requires energy so UV light may assist. See also this dated work and as a cautionary note, this work "The Explosion of Nitrous Oxide--Hydrogen Mixtures" noting, in general, a potential explosion issue, albeit here, the amount of hydrogen is expected to be low.
Also, considering adding a hydroxyl radical scavenger as per this article "Sugars as hydroxyl radical scavengers: proof-of-concept by studying the fate of sucralose in Arabidopsis" to address possible metal corrosion.
I expect it to be, at least, an interesting experiment.
[EDIT] More insights from this 2008 doctoral thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena" by Saikat Adhikari, as an example:

Hydrogen embrittlement, a process in which brittlement and subsequent cracking of metals is caused by ingress of atomic hydrogen into the metal, is also sometimes thought of as a type of stress corrosion cracking. Due to the detrimental effects of such corrosion processes, it is important to have a fundamental understanding of associated alkaline or acidic dissolution mechanisms.

And also:

Apart from activation of the surface, dissolution of aluminum in alkaline solutions at open-circuit also leads extremely high rates of H-absorption into the metal, [9-14] and also formation of nanoscale voids or hydrogen bubbles [15, 16] near the aluminum surface.

This depiction of hydrogen embrittlement leaves me a little more hopeful on a possible solution, but the alluded to formation also of nano-scale voids (and unspecified possible subsequent pervasiveness when H is removed) could suggest a more unrecoverable condition.

The question is how easily N2O gets to atomic hydrogen ( or alternative reactive hydrogen forms ) within the alloy.

The question is there any treatment possible to address the atomic hydrogen? My suggestion is to try safe non-toxic laughing gas based on the suggestive chemistry. Experimentation with various means of possible application may prove somewhat successful (or not).  To be honest, I have not rule out, in my expectations, even if the extraction is successful, whether the brittleness is eliminated or even may worst. Trying may suggest other paths or provide a definitive answer (including none).

See 'my Hchemisorbed' provided comment above for a better understanding on what is occurring,

My note was about the thought the atomic H (or similarly reactive bound H) would eventually occur in the close to surface bulk alloy volume, where N2O would not reach because of its size and limited metal diffusibility.

My understanding is that the process is largely a surface phenomenon. Interestingly, one can crack a glass surface starting with a long scratch. Do you agree that cling film has no impact on surface imbues H atoms?


text:
Regioselectivity of hydroboration-oxidation of enol ethers
In the book Hydroboration and Organic Synthesis, on p. 99, it is written that:

hydroboration [of enol ethers] proceeds regioselectively, placing essentially all of the boron atoms at the β position

On p. 100 of the book, a specific example reaction is provided to illustrate the point:

The book does not explain the chemistry behind this specific regioselectivity for enol ethers and in this post, I would like to ask if anyone has a plausible explanation to offer. My hypothesis is that mesomeric electron pair donation from the oxygen atom adjacent to the $\ce {C=C}$ bond enhances the electron density on the β carbon, thus the electron-deficient boron atom preferentially attaches to it.
Reference
Dhillon, R. S. Hydroboration and Organic Synthesis. Springer-Verlag Berlin Heidelberg. Germany, 2007.

There is this https://chemistry.stackexchange.com/q/50311/16683 which covers pretty much the same thing, but I'll warn that I think my answer there needs some updating. The idea is correct but the devil is in the details...

Just follow the polarity of the reacting bonds: Me2C=CHOEt and H2B-H. For the former the double bond is polarized with the negative charge on the carbon bearing the methyl groups and the positive charge on the other carbon stabilized by resonance fro the oxygen. Boron is more electropositive than hydrogen. Do a cycloaddition such that boron is delivered to the tertiary site. Your intuition is correct. Follow Markovnikov rules with the understanding that hydrogen is not always positive.

@orthocresol Thanks, I think your answer is what I'm looking for

I think you can self answer it. This will make the post complete.

source: Stack Exchange

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

sort: votes


text:
Small mercury spill
few weeks ago a thermometer for home use (temperature measuring) broke under our bed on our wood floor 11/18. We cleaned the spill using paper and safely removed it from the house. We ventilated properly for 24 hours.
On 11/21 when I was cleaning I noticed something silver under the bed. Did not come to mind it could be mercury left over. I cleaned properly, vacuumed, and steam mopped.
Today 11/29 on the same place I saw silver beds and line, and like a fool I cleaned with wet wipes and threw it out.
I am almost 9 months pregnant and super worried. How can I clean up properly, is it dangerous and why is it showing up. It looks like a large quantity, from such a small thermometer.

I encourage you to seek information from reputable, real-life sources. What you are describing is potentially quite dangerous but the internet is not the proper resource for you here.

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


text:
Ritalin/Concerta/Methylphenidate is an amphetamine?
FIDE (the governing body of international chess competition) says here:

The most relevant banned substances for chess are:

• Amphetamines – e.g. Adderall, Ritalin

(...)

Image:

I think either Ritalin should be in a separate category called methylphenidate or Adderall and Ritalin are categorised under 'stimulants (that you need a prescription for but don't have)' (then you can add vyvanse, desoxyn, concerta, etc) or something because afaik Ritalin/Concerta/Methylphenidate is not an amphetamine.
Or am I wrong: is Ritalin/Concerta/Methylphenidate actually an amphetamine?

Firstly, I'll mention that I'm not an expert on psychoactive drugs and their classifications; this is just based on my understanding of the situation as a chemist.
As far as chemical structures go, Ritalin (methylphenidate) probably should not be considered as an amphetamine:

This comparison (where I've highlighted the amphetamine core in blue) might make it look as if Ritalin is derived from an amphetamine. To a certain extent that is true, but the general class of amphetamines doesn't actually allow arbitrary substitution of the amphetamine core. In particular, the name amphetamine is derived from α-methylphenethylamine, which explicitly specifies the methyl group at the α-position (i.e. next to the $\ce{NH2}$ group). To stick something on the end of that methyl group, as is the case in Ritalin, would invalidate this.*
A more technically correct, and general, term that would encompass both the amphetamines as well as Ritalin would just be something that took away the requirement for the α-methyl group. In other words, we are looking for just phenethylamines.
As for the legal technicalities: I am pretty sure that most phenethylamines are controlled drugs because of their psychoactive properties, and hence are likely to be on the list of banned substances. However, since I did not (and probably won't) read the rules in great detail, I'll avoid asserting that "phenethylamines" could be used as a drop-in replacement in there.

* The table on Wikipedia does actually show one example where the methyl group is modified to an ethyl group. However, I guess that would still be considered a very small change, compared to the piperidine ring in Ritalin.

thanks orthocresol♦. God bless you for this. did you add the arrow yourself? or did you get the image (including the arrow) somewhere?

I appreciate your thanks, but it's getting a bit much now. You don't need to overdo it. I drew the entire image myself using ChemDraw.

The formatting is indeed beautiful.


text:
Gibbs energy of electron transfer
The reaction Gibbs free energy of a cell is
$\Delta_\mathrm{r}G^\circ = -nFE^\circ, \tag{1}$
and the Gibbs energy of photoinduced electron transfer according to IUPAC is
$$\Delta_\mathrm{ET}G^\circ = N_\mathrm{A}\Big\{e\big[E^\circ(\ce{D^{+.}/D}) - E^\circ(\ce{A/A^{-.}})\big] + w(\ce{D^{+.}A^{-.}}) - w(\ce{DA})\Big\} - \Delta E_{0,0}. \tag{2}$$
Are both the same except for the vibrational zero electronic energy of the excited partner $\Delta E_{0,0}/\pu{J mol^-1}$ (energy of the excited state)?
The Gibbs energy of photoinduced electron transfer accounts for the effect of Coulombic attraction in the products $w(\ce{D^{+.}A^{-.}})$ and reactants $w(\ce{DA})$. Why does the reaction Gibbs energy of a cell not have these terms?


text:
How does deprotonation works?
I've been studying organic chemistry and came across this example :
CH3-CH2-CH(+)-CH3 ---> CH3-CH=CH-CH3, the explanation the textbook gave was related to deprotonation. I've tried drawing lewis structures to understand this. But it looks incomplete.
My work :
From my basic understanding of deprotonation, it'll be
CH3-CH(-)-CH(+)-CH3, but how does a double bond form in between these 2 atoms? This is the part I'm getting confused with.
Any suggestions to point in the right direction? I'm pretty sure I'm missing something.
P.S. I came across this issue while dealing with a reaction of butan-1-ol heated with concentrated H2S04.

$\ce{C^--C^+}$ and $\ce{C=C}$ are resonance structures of the same thing. By the way, "deprotonation" doesn't necessarily imply just breaking a C–H bond; it really only says that "a proton is lost". In this case, the proton is lost, and a double bond is formed; it's still a deprotonation.

@orthocresol   Thanks a lot! Everything makes sense now :)

If you got your answer, feel free to self answer it. Self answer are always appreciated and it also help future readers.


text:
How to calculate/estimate the contamination rate of ultra pure water (UPW) in the presence of metals?
Metals are known to release ions into ultra pure water. Do any equations or empirical data sets exist for estimating the rate of contamination with different metals? I don't need a specific solution; I just need better vocabulary in this topic to further research.
For example say we have 1 liter of >18.18 MΩ·cm ultra pure water. We expose this water to 1 cm^2 area of polished pure titanium metal for 100 hours at STP. What is the final resistivity or number of titanium ions in solution? (Feel free to make up a different scenario and assumptions if needed; I just need a starting point)

Hard to see how this could be answered. A piece of flat titanium (or other metal) exposed to still water is not serving as an electrode. But if it is used for, e.g., conductivity measurements, then the situation is different because various processes are happening at the (now) electrode’s surface. Maybe the still water and non-electrode case would set a lower contamination limit?

This is true. If I could just get a non-active baseline that would be of value. Sampling doesn't need to be continuous so the sensor would only need to run for say 10 milliseconds every 10 minutes or something like that. Just not sure where to begin.

Have you searched contactless conductivity methods? I don't know how sensitive it is to pure water but is has been used and shown chromatography sensors.

You could start with this (see Table 1, and the references on which it is based), then use Google Scholar to check all the (more modern) papers that have cited this paper:  https://pubs.acs.org/doi/pdf/10.1021/ac60301a006?casa_token=W2ipNRS3VLUAAAAA:efmus8wqfiDQQ9HT0DUfqU6MWrVKjUrKJJ-9JeVx0n-tCwSeU8pbu_bmsNh3Jz0fDg4OzVlJ7zjU3JI  , namely https://scholar.google.com/scholar?cites=1829773286810311078&as_sdt=2005&sciodt=0,5&hl=en .  You could also check Millipore's website.

@theorist, Thanks for the article link! This is helpful.

While it is claimed that "Metals are known to release ions into ultra pure water", in my understanding, even pure water has some, albeit low amounts, of dissolved oxygen, for example, as noted in this previously cited source.
Now, select metals (especially, for example, Cu, Fe,..) will undergo an oxidation-reduction reaction introducing metal ions and consuming O2 along with H+ (from the water or as further provided by say a CO2 presence from air contact).
Further, more than one metal presence can augment the process in a so-called  Galvano-assisted redox (see, for example, this 2020 work, "Kinetic pathways of iron electrode transformations in Galvano-Fenton process: A mechanistic investigation of in-situ catalyst formation and regeneration").
So, exactly how one separates out the alluded to chemistry and possible multi-metal contribution to the reputed metal ion formation in actually not precisely ultra pure water, and ascribe a direct singular metal direct process, is unclear/difficult, in my opinion.
So, as to your request "Feel free to make up a different scenario and assumptions if needed", removing oxygen is one of my key concerns along with using a pure metal source (not an alloy).

Excellent! That makes sense that there must be some reactions to create the metal ions in solution. Since calculation is not an option here; is it fair to say that, in a qualitative sense, a metal lower on the reactivity series would contaminate ultrapure water less than a metal higher on the reactivity series? With platinum contaminating the least?

A pure metal immersed in pure water establishes a potential with the liquid.  Current understanding is that ions migrate into the water and electrons accumulate on the metal. If[when] the electrons react corrosion ensues. A potential would be developed against a reference electrode say a hydrogen electrode that should give an idea of the metal ion concentration. How this potential is to be measured accurately I have no idea except possibly as a limiting process.  Since different metals have very different potentials at reasonable concentrations it seems reasonable that the ionization process would differ for each. These thoughts do not require a reaction rather this process is what causes reactions.  For ions to accumulate in the solution the electrons must react with either water or the low concentration of H3O+ ions[corrosion].

Interesting, I will have to do some more research to understand. I present you with the same question though. Is it fair to say that, in a qualitative sense, a metal lower on the reactivity series would contaminate ultrapure water less than a metal higher on the reactivity series? With platinum contaminating the least?


text:
The reason for selection of wavelengths in the spectrophotometry of Quinoline Yellow SS
The absorption spectrum for Quinoline Yellow SS has peaks at 224 nm and 414 nm, and a trough at 337 nm.
This experiment (see page 31), selects 224, 337 and 414 nm to be used in the spectrophotometric analysis.
As far as I have read, the wavelengths of maximum absorption (224 nm and 414 nm), give the best accuracy and least percentage error. However, why is 337 nm chosen?

Troughs and such are often used to improve selectivity, if you use absorbance peak-trough difference. Similar techniques are used in satellite images in false colours when some of the basic colours is defined as difference of signals for 2 wavelengths.

If there is a peak-trough pair, it helps to deal with the spectrum baseline. If there is peak-trough-peak, it helps  even with a slope.

Not the main issue usually, but at an absorption maximum the slope in the spectrum is zero, just like at a minimum. Thus, if you are a little inaccurate with the wavelength scale, error propagation is not so bad in those two cases compared to measuring at a wavelength with a large slope in the spectrum.

@Poutnik I don't quite understand what you mean by improving selectivity through the absorbance peak-trough difference and the spectrum baseline. Could you please explain or point me to a resource that explains it?

Higher peak absorbance is not caused by the dye if it is not followed by higher peak-trough absorbance difference. IF e.g a peak has absorbance A_0 + A_p and trough A_0 + A_t, that peak-trough difference is (A_0 + A_p) - (A_0 + A_t) =  A_p - A_t. This ways can be compensated the sample matrix absorbance baseline. 2 peaks with trough could be used to estimate and compensate the slope of such baseline.

The experiment in your link pg 31 is "Path Length Dependence of Absorbance Values". As you stated, quinoline absorption spectrum has a dip near 337 nm. The reason for choosing the wavelength is not something fundamental. You already know that most sensitive analysis is carried out at the peak maximum rather than a wavelength corresponding to a valley. All Agilent wanted to show here is that Beer's law is valid for all wavelengths in the absorption spectrum, as long as A > 0.
If you are carrying out this experiment, check how the slopes vary with different path length and wavelengths. Are all the slopes the same?

I graphed Absorbance vs Path length for the three wavelengths (224, 337 and 414nm). The slopes are similar for 224 and 414nm (the peaks), the slope is close to 0 for the trough, which doesn't seem useful. So just to clarify, the choice of the trough is simply to demonstrate that the linear relationship holds for all wavelengths?

@StrugglingChemistryStudent, Yes! You got it right. In analytical chemistry, sensitivity is defined as the slope. Higher the slope, the more sensitive the analysis is.


text:
Determination of energies in MO diagrams
How do we determine the relative placements of molecular orbitals in their diagrams? I understand that we can use point groups to find symmetry-allowed interactions between orbitals that we write on either side of the diagram.
For example, with the $\ce{BH3}$ in the image below, I think we could say one of the $\ce{2p}$ orbitals on the left has $\ce{a_2''}$ symmetry and thus is non-bonding, and the other two $\ce{2p}$ orbitals match the $\ce{e'}$ symmetry on the right. Finally, $\ce{2s}$ matches with $\ce{a_1'}.$
What I don't understand is how we know that $\ce{a_1'^*}$ has a higher energy than $\ce{e'^*}.$ Perhaps there could be some argument from the geometry, but is this generalizable to any combination of symmetry labels? And furthermore, does this mean that any MO diagram can be derived entirely from theory, without looking up any energy charts or through spectrometry?

The figure in the OP's question seems to come from the article "A brief introduction to molecular orbital theory of simple polyatomic molecules for undergraduate chemistry students" by Baibich and Butler. In it (for the first example, the MO diagram of water), they say:

The students know that linear combinations of the central atom atomic orbitals and the LGOs have to be accomplished with orbitals with the same irreducible representations and that, in the four examples being considered in this paper, the orbitals have the energy requirements to combine. For more advanced students, some papers that deal with more detailed energy calculations are suggested.7,8 (Comment: it should be noted it is an approximation to state that a1 and b2 ligand orbitals have the same energies.)

In one of the cited papers (ref. 7, concerned with teaching MO theory), Lawson and Harrison state:

When two identical atoms interact, the atomic orbitals
give rise to molecular orbitals that frequently, but not always,
have  energies  that  are  above  and  below  the  energies  of  the
constituent  atomic  orbitals.

Without a calculation, you can not even predict the relative order of atomic orbital vs. bonding and antibonding MO.
They go on to state that in its simplest form, MO calculations are better at predicting bond lengths than dissociation energies, using the diatomics from $\ce{H2}$ to $\ce{Cl2}$ as an example:

Note that the calculated dissociation energies are often in poor agreement with experimental values and in particular in the MO theory $\ce{F2}$ and $\ce{Na2}$ are both predicted to be unbound and should not exist, while $\ce{Li2}$ is predicted to be marginally stable. Most textbooks do not discuss the direct predictions of MO theory as reflected in Table 3 but focus  on  the  correlation  between  the  bond  order  and  the experimental  $D_e$.  How  well  the  theory  is  doing  should  be gauged by comparison between the predicted and experimental $D_e$ values, not surrogate quantities like bond orders.

Finally, they caution that the energies of the unoccupied orbitals are difficult to interpret:

Note  also  that  in Table  1  there  is  no  energy  given  for
MOs that are not occupied because in ab initio MO theory (9) the energy of an empty orbital corresponds approximately to the energy of an electron in the negative ion and not to an  electron in that orbital in the neutral ground state. For homonuclear diatomics these empty orbital energies are positive and bear no direct relation to the occupied orbital energies. The practice of plotting the empty orbital energies as a function of the atomic number of the atom in the diatomic (13) and showing them merge with the occupied orbital energies has meaning in Huckel-like theories where the energy of an orbital is independent of its occupation, but not in ab initio MO theory.

How powerful is the MO method?

[OP...] does this mean that any MO diagram can be derived entirely from theory, without looking up any energy charts or through spectrometry?

Yes, you do a calculation and get a diagram. You would probably start with atomic orbitals calculated elsewhere, and you would probably want to compare your results with experimental data, if available. One of the strengths of computational chemistry is that you can study molecules that have not yet been synthesized, or gain evidence that a putative molecule is unlikely to exist because it would be expected to fall apart accordning to your calculation. One of the weaknesses of simple MO theory is that the estimated energies are sometimes off by a lot, and that in some cases, the electron configuration (which comes down to the order of the molecular orbitals) is in conflict with experimental data.
See also: https://chemistry.stackexchange.com/a/128550/72973

To tack onto Karsten Theis's answer: Symmetry is inherently a qualitative tool, because it is binary. Two things (AOs) either have the same symmetry and can overlap, or they have different symmetry and can't overlap. But you can't say something is "more symmetric" and thus leads to "more overlap"; that would just be complete nonsense. So, symmetry tells you which lines to draw, but it doesn't tell you the extent of the overlap, i.e. how far up or far down the lines need to go to.
Formally, this is expressed in terms of integrals involving atomic orbitals $\phi_i$:
$$\langle \phi_1 | \hat{H} | \phi_2\rangle \begin{cases}
\neq 0 & \text{if }\phi_1 \text{ and } \phi_2 \text{ transform as the same irrep} \\
   = 0 & \text{otherwise}
\end{cases}$$
The energies depend on exactly what this number evaluates to, and to find that out, you need to actually evaluate this integral (and many others). Conceptually that's not hard to do, because we know how to write the AOs $\phi_1$ and $\phi_2$, and we (kind of*) know what the Hamiltonian $\hat{H}$ is. And it turns out that it isn't too hard for a computer to perform this integration. There are various software packages that will do this sort of calculation for you. But doing it by hand is not feasible, especially for any non-trivial molecule, and that's why if you limit your approach to a symmetry-based one, you can only ever get qualitative information out of it.
Note that I'm not trying to equate the energy of the MO with that integral above. The expression for the energy will (in general) depend on many of those terms. (Exactly how? Well, the MOs are linear combinations of the AOs, or at least we postulate that they are:
$$|\psi\rangle = c_1|\phi_1\rangle + c_2|\phi_2\rangle + \cdots$$
so the energy associated with the MOs is
$$\begin{align}
E &= \langle \psi | \hat{H} |\psi\rangle \\
  &= \langle c_1\phi_1 + c_2\phi_2 + \cdots | \hat{H} | c_1\phi_1 + c_2\phi_2 + \cdots\rangle \\
  &= c_1^*c_1\langle\phi_1 | \hat{H} | \phi_1 \rangle + c_2^*c_1\langle\phi_2 | \hat{H} | \phi_1 \rangle +
c_1^*c_2\langle\phi_1 | \hat{H} | \phi_2 \rangle +
c_2^*c_2\langle\phi_2 | \hat{H} | \phi_2 \rangle + \cdots
\end{align}$$
Yeah, you don't want to do that by hand. And actually finding those coefficients $c_i$ is an entirely separate problem which involves similar considerations.)

* There is a "slight" problem, in that the electron–electron repulsion term in $\hat{H}$ depends on the electron distribution which in turn depends on the form of the MOs. But that's a technical detail. You can worry about it if you're actually writing computer software to do those integrals.


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How to give different color to each type of heteroatom during substructure matching in RdKit
Substructure matching in RdKit generally highlights the matched component in red color over the original molecule. The code for achieving the same is as follows.
# MWE for a Jupyter notebook:
from rdkit import Chem
from rdkit.Chem.Draw import IPythonConsole
from rdkit.Chem.Draw import rdDepictor

rdDepictor.SetPreferCoordGen(True)
IPythonConsole.molSize = (250,200)
from rdkit.Chem import AllChem

# additional import:
from rdkit.Chem import Draw
IPythonConsole.ipython_useSVG=True

# ----

smi = 'CCC1NC(=O)C(C(O)C(C)CCCC(=O)OC)N(C)C(=O)C(C(C)C)N(C)C(=O)C(CC(C)C)N(C)C(=O)C(CC(C)C)N(C)C(=O)C(C)NC(=O)C(C)NC(=O)C(CC(C)C)N(C)C(=O)C(C(C)C)NC(=O)C(CC(C)C)N(C)C(=O)CN(C)C1=O'
core = Chem.MolFromSmiles('C1C(=O)NCC(=O)NCC(=O)NCC(=O)NCC(=O)NCC(=O)NCC(=O)NCC(=O)NCC(=O)NCC(=O)NCC(=O)N1')

newm = Chem.MolFromSmiles(smi)
rdDepictor.Compute2DCoords(newm)
newm.GetSubstructMatch(core)

newm

Source of the code
Here, if we want to highlight different heteroatoms (like N and O) with different colors, what would be the best way??
Edit:
For clarification purposes. The requirement is to change the color of the highlight and not the color of the font of the atoms.

For future reference: Please enclose snippets of code by three grave characters.  Not only will code be visible as such, without unwanted line breaks applied to normal text, there will be (language specific) syntax highlighting, too.  And it would be better to present a self-sufficient minimal working example (MWE), like e.g. in posts on tex.se.

https://chemistry.stackexchange.com/questions/160056/how-to-give-different-color-to-each-type-of-heteroatom-during-substructure-match

If you scale the visual representation of the molecule, i.e. replace the current definition of the image by
IPythonConsole.molSize = (400,400)

instead of (250,250), you see that the Dreiding-model like colour scheme is applied as anticipated (oxygen red, nitrogen blue, etc.):

It might be a little bit difficult to discern from the rosé transparent highlight (especially for oxygen) since the «core structure» you defined runs all around the cyclic peptide.

In addition, you can also change the color of the highlight: IPythonConsole.drawOptions.setHighlightColour((0.8,0.8,0.8))

Thanks. I am able to see that the atoms font color have changed. But the overlying red color is still the same. Is it possible to give varying color of highlight based on atom???


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Can an atom bond with any other atom?
I am designing a game for students which has a sandbox for atoms to create molecules like building blocks. I have the valance of each atom mapped and they can bond with other atoms based on their valency. My question is, is it possible for an atom to bond with another atom just because they can technically form an ionic or covalent bond? Researching about this matter made me believe energy levels contribute to forming of molecules in a reaction but taking out temperature, pressure and energy from the equation, is it valid to say an atom can bond with another if there are free valence electrons?
If you would like to try out the game (this is not an advertisement, the app is still in testing) check ChemistryCat on Google Play Store.

Well, if you don’t care about whether the resulting compound is stable or not, then yes, everything can bond with everything.

@orthocresol thank you for the reply. I see the question is downvoted a lot. I guess I cannot ask a question without someone downvoting it. Defiles the purpose of this platform I guess.

Well, the real question is do you care whether the molecules you make by this method are realistic or not? Making molecules by randomly connecting atoms together according to their valencies is quite trivial. On the other hand, making molecules that stand a chance of existing in the real world is quite non-trivial. Learning what's realistic and what isn't realistic is why people study chemistry for years. If you don't care (as in if that's not the aim of your game), then you can do whatever you want.

The main motive is to make chemistry interesting. Of course there will not be molecules in existence that could be created by simulation. My real question is in reality, atoms do not combine with other atoms just because there is an electron to bond with. But is it possible for such atoms to bond in different physical conditions such as temperature or pressure? I mean there is no known upper limit to temperature right?

As I mentioned, everything can bond with everything. The real question is not whether atoms can bond or not; the question is whether the resulting compound is stable or not. Stability is most certainly affected by temperature. However, I would not consider an "upper bound" of temperature to be useful, because stability usually decreases with temperature. Going into any more detail as to what is stable and what isn't would be way too long, I'm afraid. That would basically be explaining the entirety of chemistry as a science.

The key word is stability. I understood your gist with that word. Thank you very much. As for the temperature, you are right. I was just making a suggestion like there is a possibility so it might occur in some condition we don't know yet. Like I heard about scientists creating metallic hydrogen so it got me interested in molecules we don't know yet that only exist in such conditions.

In simplified chemistry (high school level or general public knowledge) temperature limits would usually be from 123K to 3000K, exclude the cryogenic and super heated. Above 3000K there is a lot that can happen, but if you don't include pressures you're going to start dealing with plasmas and those lack the valence electrons you are using. Actual max temperatures that we can describe is the Planck temperature at 1.417×10^32 K.

Rare gases can make molecules with a lot of atoms, but not with all the $91$ other atoms of the periodic table.

Valence is not always a good criteria to decide whether a compound is possible. For example, iron and carbon make a compound $\ce{Fe3C}$ which is impossible to reprenset with the possible valencies of these atoms. The same is valid for alloys.

@Maurice you are correct. I would like to focus on feasible and simple molecules like H2O, NaCl, etc. Alloys, polymers, lattice structure would get more complicated but will take time to implement. Most of the complex molecules cannot be expressed in two dimensions.

Good intent and probably a fun game that has nothing to do with the reality. On the screenshot from the Google Play only ammonia and benzene look OKish. Either the geometry is flawed (water, acetic acid), or connectivity is wrong (triangular ozone), or the compounds that are not molecular in nature at all (ionic solids) are depicted as molecules.

Since you are using real elements, this actually makes the game counter-educational as it simply teaches wrong things. Other games such as SpaceChem also do this, and it's evil from a chemistry teacher's perspective. Knocking out of the head the colorful nonsense that kids saw on their screens takes time. I suggest to either go full Imaginationland with simplified made-up chemical elements and rules for a dopamine kick for toddlers, or invest some time (or a lot, in fact) in studying simulation of bonding and molecular structures used in real life.

@andselisk thank you for the feedback. I must say a few things. I have mentioned that atoms do not look like this irl in the app itself. You can assume that this is an "abstract" representation of an atom or something. Secondly, the game area has only 2 dimensional fidelity and so it throws out "realistic" molecular structure out the window (not intended to show real molecules anyway). You can use this app as a visual motivator for students and others wanting to know what atoms are. I think the app at this state provides a way for such idea.

@andselisk the molecules in the screenshot are created by me (you can create whatever structure you want, the app only marks "recognized" molecules. So of course it requires a bit of chemistry knowledge to play with. For toddlers it will be just a ping pong colorful circles game.

Valence in itself is a terribly simplified and often useless concept. You might have gotten that many downvotes for this question because it tries to take out the chemistry out of what you claim is educational. I absolutely agree with @andselisk, this will teach wrong things and if your intention really is to get students interested in chemistry, this is the completely wrong approach. Visual stimulators might be nice, but there is nothing gained when what they stimulate is wrong.

I am thinking if one could simplify the task and make an educational game. I am afraid that one would be required to know chemistry before playing even if the above could be accomplished.


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ortho-bromoanisole + NaNH2 + Liquid NH3 =?
when we add ortho-bromoanisole in NaNH2 + NH3 we form products using the benzyne mechanism.
here, we form the products considering only -I effect of -OCH3, if we consider its +M effect we would form, o-Anisidine, why is +M effect ignored here? is it because of sp2 hybridization of carbanion, if so, how?

Is the anion electron pair in conjugation with the lone pairs on the oxygen atom?

First of all,you need to know the basics of Resonance. It is the transfer of π electrons between p orbitals parallel to each other. If they are not parallel,then there will be two cases:-
1)if atleast one of the p-orbital is free to rotate (generally in sp³ atom),then it will align its p-orbital parallel to it,then it's π electrons participate in Resonance.
2)In the second case, if the atom is not free to rotate (generally in sp² atom) then resonance do not happen.
Your Phenyl carbanion lies in the second case.The negative charge resides in the sp² hybrid orbital which lies in the plane of ring. But the resonance is happening in the p-orbitals  perpendicular to the plane of benzene ring. Therefore there is an angle of 90° between the orbital containing negative charge and other p-orbital participating in resonance. Also that carbon is not free to rotate.
Thus,there is no effect on reaction intermediate of Benzyne mechanism due to (+M)-effect. So Inductive effect is the deciding factor for the stability of phenyl anion.


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enthalpy of formation of hydronium
Can someone explain how the enthalpy of formation of aqueous hydronium was determined. It appears to be exactly the same as the enthalpy of formation of liquid water. I know this must have something to do with the fact that the aqueous hydrogen cation has an enthalpy of formation of 0 by definition, but I can't see how.
I'm looking in the appendices of two different general chem texts, the ones by Atkins, Jones, and Laverman; and the one by Oxtoby et al. In both cases, liquid water and aqueous hydronium are both shown as having enthalpies of formation of -285.83 kJ/mol.
I think I understand the answer to my question based on Maurice's reply.
Here's an example.
Consider the following two reactions:
$\ce{Cl^-(aq) +H_3O^+(aq) -> HCl(aq) +H_2O(l)}$
$\ce{Cl^-(aq) +H^+(aq)-> HCl(aq)}$
(Thank for you for the formatting tips @orthocresol. Very helpful.)
Since aqueous hydronium and aqueous hydrogen ions are the same, we expect these two reactions to have the same reaction enthalpy.  And since aqueous hydrogen ions have an enthalpy of formation of 0 (by definition), the only way the two reactions will have the same reaction enthalpy is if hydronium and water have the same enthalpies of formation.

The aqueous hydrogen cation and the aqueous hydroniun are synonym. They are not different from one another. $\ce{H3O^+}$ may be called aqueous hydronium or aqueous hydrogen cation.

Because nobody is sure that aqueous hydrogen cation is always made of $\ce{H3O^+}$. Some scientists think that Hydrogen cations may also be making more complex ions in water, like $\ce{H5O2^+}$ or even$\ce{H_{2n+1}O_n^+}$ with n>2.

Well, let's say you have the reaction $$\ce{H+ + H2O -> H3O+}.$$ Then we have that $$\Delta_\mathrm{r} H = \Delta_\mathrm{f} H(\ce{H3O+}) - \Delta_\mathrm{f} H(\ce{H2O}) - \Delta_\mathrm{f} H(\ce{H+}).$$ It follows that if—and only if—the reaction enthalpy $\Delta_\mathrm{r}H$ is zero, then $\Delta_\mathrm{f} H(\ce{H3O+}) = \Delta_\mathrm{f} H(\ce{H2O})$. But I don't immediately see why that should be zero (I welcome any pointers!).

It would be useful if you could share the source that you are quoting, i.e. what book (or website or whatever) gives you a value for $\Delta_\mathrm{f}H (\ce{H3O+})$. My usual reference (the appendix of Atkins Physical Chemistry 9th ed) doesn't provide this value.

On the solvation models, Wikipedia has a good set of references discussing a few works that seek to ID what those structures look like - https://en.wikipedia.org/wiki/Hydronium#Solvation

@orthocresol - bottom line is that the reaction enthalpy for the reaction you wrote out is 0 because the aqueous hydrogen ion and the aqueous hydronium ion are the same thing and so the reaction is essentially a "null reaction".

I thought of that, and I can see why, but I just have a bad feeling that that opens a separate can of worms. It should be possible, at least, to construct a series of formation enthalpies that are internally consistent using this logic (such as that shown in your question). So my concern might be unfounded. BTW, you can use $\ce{...}$ to typeset chemical formulae/equations correctly (and more easily): $\ce{H+ + H2O -> H3O+}$ $\ce{H+ + H2O -> H3O+}$ is correct, $H^+ + H_2O \to H_3O^+$ $H^+ + H_2O \to H_3O^+$ is italicised (not right).


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Meaning of chemical potential in chemical reactions
In my thermodynamics course, we introduced the chemical potential as a modification of the first + second law of thermodynamics in the case of a system that can exchange particles with its surroundings (consider only PV work):
\begin{equation}
dU = TdS - pdV + \sum_{j=1}\mu_j dN_j
\end{equation}
where the sum is over all the components of the system. This makes sense to me - even if we keep everything else constant, just by adding more matter to the system the energy should increase. Now, this should be true for a closed system undergoing a chemical reaction, so the 3rd term on the RHS could be non-zero. If we consider a quasistatic process, we can identify the first and second terms on the RHS as the heat and work. So the chemical potential term appears to represent some sort of energy transfer separate from heat and work, but how is this possible? Aren't heat and work the only ways to transfer energy?
And if the chemical potential term represents some sort of work in this case, how come the term which we defined to just explain the increase of the internal energy with the increase of the amount of matter in the system also coincidentally can now be interpreted as a work term?
In other words, my question is how do I physically interpret the chemical potential term?
I asked this question in Physics Stack Exchange (I am a physics student), but I wasn't able to obtain an answer that made complete sense to me (I was told that internal energy doesn't include bond energy, but I couldn't find a source that confirms this).

The chemical potential is just as you state: it's the response of the energy of the system to a change in the number of particles. Its use and interpretation are pretty broad: It can be used to determine the gradient of density due to pressure or acceleration. Also the derivation of freezing and boiling point suppression and elevation use the chemical potential. In chemical equilibrium, the mixing entropy of the reactants and products equals the chemical potential of the system. Probably the best way to get aquatinted with it is to do some exercises which involve the chemical potential.

@Peter - Cross-posting is frowned upon here, we are happy to have the question on chem.SE as long as you remove it from physics.SE.

I am afraid that saying that the only way to transfer energy is heat or work is too much confined in the mechanics realm

@Alchimista But doesn't that contradict the first law of thermodynamics?

@Peter I see what you mean. But I didn't want to say that it's not. Actually I sent the comment before finishing and I am still thinking of it ;)

@Peter I am not much versed on formalism but in the first principle it seems we forget to say "at fixed composition". This is at the eye of mine as a chemist is very much clear, as for different chemical species must have different internal energy whatever sum of energies this latter is (here i must mention that U does include bond energy, opposite to what you have been told at Physics SE. Change in this latter component are what are affected by reactions) "Aren't heat and work the only way to transfer energy?" Yes and not, a substance carries its own internal energy.

Isn't pdV a specific type of work? If energy is capacity to do work, as the chemical environment of atoms changes so does their energy. Atkins ch. 15-17 is a great source on this topic.


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How to determine the visual properties of a specific steel compound?
The material that is described in a paper by Li et. al.[1] is interesting to me, but they don't go into any details on what the end result would look like visually. Out of curiosity, I was wondering if there was any way to determine these properties.
As an extremely brief summary of the relevant part of the paper, the material in question is stainless steel sintered with copper-coated graphene.
Additional details that may be important:
Gr-Cu is first produced in sheets and then ball-milled and added to the stainless steel up to 0.2 percentage by weight.
I am also aware that the color of most types of steel can be somewhat controlled based on heat or chemical treatment, but I'm not sure if those options would be be viable for this kind of material.
References:

Li, Z.; Ni, H.; Chen, Z.; Ni, J.; Chen, R.; Fan, X.; Li, Y.; Yuan, Y. Enhanced tensile properties and corrosion resistance of stainless steel with copper-coated graphene fillers. Journal of Materials Research and Technology 2020, 9 (1), 404–412. DOI: 10.1016/j.jmrt.2019.10.069.

Better asking in Physics SE unless there is an advanced Optics SE site. I think you have no chance to get an answer, btw.

I could try wording it as a request for wavelength reflectivity and absorption spectrum, but I suspect they would just tell me to ask here since it's in relation to the results of mixing metallic compounds. I'll agree that I'm probably not going to get an answer though.


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In a galvanic/voltaic cell diagram, does convention require for the anode to be on the left?
It seems as though in every diagram, illustration, and demonstration I've seen of a voltaic cell, the anode is placed on the left and the cathode is placed on the right. Is this a rule, or a convention? Would it be frowned upon to publish a diagram where the cathode is placed left of the anode?

It's apparently from a 1953 IUPAC recommendation, referenced here and here. I, for one, would frown upon what you propose if I were a referee.

There is no such rule that in a schematic/illustration/cartoon of a galvanic or an electrolytic cell the anode should be on the left or on the right. The electrons do not care nor the electrolytic reactions depend on how humans show pictorial galvanic cell schematics.
However you will notice that when electrochemists describe a galvanic cell in a very specific notation, there is a common agreement among them.
Reduction is always shown on the right hand side and this is how I remember it, which means anode or oxidation must be written on the left as follows.

See Wikipedia for more examples Cell notation. Single bar shows a phase boundary and the two vertical bars indicate a salt bridge.
If you are just drawing beakers for just an galvanic/electrolytic cell, it does not matter whether right electrode is the cathode or not. All we have to do is to label the diagram correctly.

See also the etymology and Faraday's East-West mnemotechnic reasoning. It may be the convention since then.

Quoting Wikipedia link above:  .....In that paper Faraday explained that when an electrolytic cell is oriented so that electric current traverses the "decomposing body" (electrolyte) in a direction "from East to West, or, which will strengthen this help to the memory, that in which the sun appears to move", the anode is where the current enters the electrolyte, on the East side: "ano upwards, odos a way; the way which the sun rises".  The use of 'East' to mean the 'in' direction (actually 'in' → 'East' → 'sunrise' → 'up') may appear contrived. ......

P.S.: So perhaps a long standing history-based habit than mandatory convention.

Similar to catod and anod in relation to material falling or be taken up. So that the terms are somewhat confusing

When plotted on a horizontal $0x$ axis, with the zero point at some arbitrary point, the positive numbers are always on the right-hand side of the zero point. And the negative numbers are on the left-hand side of the Zero point. This is arbitrary, but universally respected, in mathematics and in all other common applications.
The same convention is applied for reduction potentials. When reported on a horizontal scale, with Hydrogen in the middle, negative redox potentials are on the left-hand side of hydrogen, like zinc (at -$0.76$ V). And positive reduction potentials are on the right-hand side of the horizontal scale, like Copper (at $+0.34$ V).
With this convention, electrons are running from left to right, from the most negative electrode (anode) to the most positive electrode (cathode). Electrons are running by alphabetic order, from A(node) to C(cathode)


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Why is neutrality so important in a voltaic cell?
I'm having trouble understanding the purpose of a salt bridge in a voltaic cell. I keep reading the same thing: without a salt bridge/porous cup, the anode will continue to lose mass and become aqueous cations, and the cathode will continue to gain mass and lose aqueous cations, building a negative charge at the cathode and a positive charge at the anode. The salt bridge is used to maintain neutrality.
What I don't understand is why do we need to maintain neutrality? Why would the cell stop running if the half-cells aren't neutral?

Without the salt bridge, only a very tiny and declining current initially flows. This creates a back-potential that opposes the current. So you would essentially have a pathetically small and leaky capacitor.

If you realize lightnings are caused by charge disbalance of typically just dozens of coulombs, you will get the picture why is charge balance important.

The bridge allow current to pass and this maintains neutrality would be more appropriate that maintain neutrality alone. This naturally resolves the question about the cell that otherwise stops.

without a salt bridge/porous cup, the anode will continue to lose mass
and become aqueous cations, and the cathode will continue to gain mass
and lose aqueous cations, building a negative charge at the cathode
and a positive charge at the anode.

This is totally wrong. No current will flow without a salt bridge and electrolysis will not occur.
We sometimes undermine the effect of a build-up of small charges. Let us take an example from electrostatics. If you recall Van de Graaf generator which is popularly used to demonstrate air sparks and even electrification of hair in air etc. How much charge are we talking about? I quote one solved example from a physics website here.

A demonstration Van de Graaff generator has a 25.0 cm diameter metal
sphere that produces a voltage of 100 kV near its surface. (See Figure
1.) What excess charge resides on the sphere?

Watch this video, if you are not familiar with Van de Graaf generator YouTube
Now, 100,000 V is really high voltage. What is the charge on the sphere? It is only 1.39 $\mu$C. This sounds like a small charge but it corresponds to a lot of electrons. The moral of the story is it is impossible to store even very small amounts of charges on objects in air. The objects will try to get rid of excess charges as soon as possible. This is basically the reason that matter does not like charge imbalance, in other words charge neutrality is maintained in everyday life. Everyone hates the electric shock after touching a door knob in winter after walking on a carpet. The reason is same. Matter does not like charge build up. You are electrically charged and the moment we are close to a conductor, like a door knob, electron transfer takes place, which is often accompanied by a spark.
Now the voltage on an ordinary battery is nowhere near 100,000 Volts. Still, Nature does not allow the build up of charges on the electrodes due to a 1.5 V home battery. Coming to your thought experiment... Take two beakers with one electrode each, dipped in an electrolyte. There you connect the two electrodes to a battery with the help of wires. No current will flow in the solutions as well as the external circuit because the two beakers are placed apart, and current cannot pass through air easily (very high resistance). When there is no current in the solution, there will be no electrolysis.
What a salt bridge does is that it provides a flow path for current between two separate beakers. In other words, it allows the movement of charges (ions) from beaker to the other.
Just as a mental exercise, replace the salt bridge with a piece of copper wire whose each end dips in the two beakers? What do you expect will happen?

Just to put it another way, the charged van der Graaf generator in the example will have a charge imbalance measured in the range of parts per trillion (e.g., for every one trillion electrons with a counterbalancing proton, there will be only a handful of "extra" electrons or protons). For objects in more normal conditions, any macroscopic charge imbalance is multiple orders of magnitude smaller still.

Would it be correct to say that without a salt bridge electrolysis happens but due to charge build up it very quickly slows down? Or would it be more correct to say it doesn't happen at all?

@AccidentalTaylorExpansion, No it does not occur at all. The very very small current if it ever flows is polarization current just like a capacitor.

@AccidentalTaylorExpansion If both half cells are physically separated, potential of their electrolytes would follow the potential of electrodes with some constant difference.

Neutrality is necessary in all chemical operations occurring in aqueous phase, not only in voltaic cells. It is an experimental result.

I understand, but why?

@FouadSaffar The opposite charges of value that pass in 1 s at current 1 A, if 10 cm apart, are in water attracted by force equivalent to the weight of about 1 100 000 000 kg. That is why. In air, it would be even about 80 times more.


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Is the rate-determining step equal to or the lower limit of the overall reaction rate?
I’ve heard many conflicting points on this single question alone. Even the usually reliable sources I run to are pretty ambiguous on the matter. So now I’m here, after 1 and a half weeks of confusion.
As I understand it, there is usually a faster elementary step and a slower elementary step. The slower one is called the “rate-determining step.” My intuition tells me that the actual overall reaction rate (the reaction rate for the whole process) should be higher than the RDS, since the faster step, although it may be magnitudes faster, still takes up time. Thus, it is only the lower limit of the overall reaction rate.
However, I’ve heard plenty of times that the RDS is the overall rate, and when I ask why, the answer is always the same analogy of the motorway with two toll booths. I get the analogy, but I’m not exactly sure if it’s a complete analogy to how a reaction works. No reaction is instantaneous, and this should show especially if the fast step isn’t much faster than the slow step. Yet, I still hear this all over the place, and this wouldn’t be the first time I was wrong about chemistry.

I am not sure if this shall be a comment or an answer. Are you thinking about the duration of a reaction at molecular scale? It is certainly true that the duration of the overall event is the sum of two or more (extremely short) time intervals. But this is not dn/dt for a large ensemble. You can think of a Baker making 100 breads daily and a delivery boy able to deliver 200 breads. Of course it takes someone more than 1 day to be delivered with his ordered bread. The difference would be noticeable about the first day only. The same here, except that events take times in the order ~ 10^-13 s

Do you mean overall rate or rate constant?. The overall rate has to be the appearance of product; in $A-B-C$ then $dC/dt$. The overall rate constant is harder to define, some sort of average but perhaps the half-life of the rise-time of the product would do. You could calculate this in limiting cases for the ABC scheme to satisfy yourself as to how this depends on rate constants.

Is not an complete analogy an oxymoron? It would be an equivalent, not an analogy.

The overall rate can be temporarily  slower than RDS rate, if the system is not in the steady state yet, so the rate of the fast reaction after RDS has not reached RDS rate yet. Typical case would be generation of uranium decay chain products from purified uranium.

The rate determining step approximation is only a just well enough approximation if there is one step, and exactly only one step, which is many magnitudes slower than every other possible step. A better approximation for the rate of a reaction is the energy span model.

The step called rate determining because the rate over all steps is limited by this single step.
Say the reaction would be
$$\ce{A -> B -> C}$$
with $\ce{A -> B}$ as rate determining.  Because $\ce{B -> C}$ only can happen when $\ce{A -> B}$ takes place and must await the formation of at least one intermediate molecule $\ce{B}$ (in an analogy, can't process this item until it actually arrives on the conveyor), there is no faster advancement than $\ce{A -> B}$.

OP is probably aware of this. The question is why the reaction isn't slower. Eventually it is, for the first 10^-13 s or something like that. The point is that reaction kinetics deals with the transformation of about 10^20 or more particles, and includes the time needed for them to find each others and effectively react. It has little to do with the duration of single events or a sequence of them in the situation when they actually occur.

@Alchimista So you think OP points to rate of the elementary step (contact by collision $\ce{->}$ detachment [if it is a substitution]) vs. observed rate of a diffusion controlled reaction?

The question seems to be formulated having in mind how long it takes for a reaction at molecular scale to occur, and that is obviously the sum of the various steps duration.. It seems to me the only way to give sense to the question. Let us see if OP will comment.

@Paulemic A reason for my referring to diffusion control are surveys by the Mayr group, Munich like this one.

@alchimista Your interpretation is a bit closer to what I was trying to express. I’m just going off of my intuition at a molecular scale to justify my conception.


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What is the relationship between electrodes and ions in a voltaic cell?
From what I understand, a voltaic cell involves two metals in two solutions. The metal and solution picked requires for the metal and the solution to interact. Atoms from the anode lose electrons and become cations, and cations from the cathode gain electrons and become solid metal atoms (I know I'm generalizing).
This explanation seems as though it requires for the electrodes and their respective solutions to be of the same initial atom, e.g. a zinc anode in a Zn2+ solution, and a copper cathode in a Cu2+ solution. This seems to be the only thing that makes sense.
However, in my homework there are various problems asking for a drawing of a "cobalt-hydrogen" standard cell. This seems ridiculous because there can't be a hydrogen electrode; hydrogen isn't a metal. I looked at the solution, and apparently the correct cathode to use is platinum. This seems completely arbitrary to me, what does platinum have to do with hydrogen? Virtually nothing, it's just some random metal that doesn't even appear on any electrode potential tables.
Also, how would this work? How is it possible for hydrogen ions to gain electrons, become hydrogen atoms, and then join platinum solid? That doesn't make any sense at all. Hydrogen is a gas, and even if it wasn't, there's no way it can become a platinum atom,  of all things.
There are more examples than this: there are problems involving a tin(IV)-zinc standard cell (tin(IV) isn't a metal as it is written), a chromium (III)-hydrogen standard cell (both electrodes don't make sense in this case), and a slightly different problem which already writes the cell notation: Pt(s) | Fe2+(aq) || MnO2(aq), HCl (aq) | C(s), and I have no idea how MnO2 and H+ would react with carbon to form a solid.
Maybe an "X-Y" cell doesn't mean what I think it means, and X and Y don't have to be metals. In any case, I need these clarifications to be cleared up, with emphasis on the question: how is it possible for an electrode to be placed in a solution of ions that appear to have no relation to the electrode?

So does that mean that the cell notation they used, cobalt-hydrogen, was correct? They weren't supposed to use "cobalt-platinum"?

Please stop downvoting, I simply lack knowledge in this field and would like to gain it.

The cell notation describes what happens chemically at the electrode. If you call your experiment  "cobalt platinum", you state that cobalt produces electrons and platinum absorbs them. No ! Platinum has no chemical effect. In the cobalt hydrogen cell, the anode is made of Cobalt metal that loses electrons according to : $\ce{Co -> Co^{2+} + 2 e-}$ and on the platinum electrode, the ions $\ce{H+}$ are reduced according to : $\ce{2H^+ + 2 e^- -> H2}$ So some new $\ce{H2}$ bubbles will appear on the platinum sponge.

Q/A sites are not for gaining knowledge, just for filling gaps in existing knowledge. Get textbooks or online courses or similar materials. Wikipedia is a great source of info and references. See also http://chem.libretexts.org

Hydrogen is not a metal, but it can make an electrode like a metal, which produces or absorbs electrons. A hydrogen electrode is obtained with a platinum plate which is covered by platinum sponge, and dipped in an acidic solution. The platinum metal has no chemical effect. It is used as an electron carrier. It has been chosen because platinum does not interfere with aqueous solutions. Any other metal may transmit electrons to the cathode. But it may react and produce its own ions that interfere with the desired reaction : $\ce{H2 <--> 2 H^+ + 2 e^-}$. Platinum has not this drawback.
Introducing now a glass tube under the platinum sponge, hydrogen can bubble in the solution. Of course, most of $\ce{H2}$ will go up the solution and burst at its surface. But a small proportion will be absorbed by the sponge. This will make a hydrogen electrode : $\ce{H2}$ can lose electrons to be sent into the platinum electrode, and the $\ce{H+}$ ions will go into solution. Here $\ce{H2}$ behaves exactly like a metal : it produces positive ions and electrons. And it can also act reversibly.

However, in my homework there are various problems asking for a
drawing of a "cobalt-hydrogen" standard cell. This seems ridiculous
because there can't be a hydrogen electrode; hydrogen isn't a metal. I
looked at the solution, and apparently the correct cathode to use is
platinum. This seems completely arbitrary to me, what does platinum
have to do with hydrogen? Virtually nothing, it's just some random
metal that doesn't even appear on any electrode potential tables.

In electrochemistry, electrode can be anything as long it can participate in electron transfer reactions. Later in your classes you will find glass electrodes! Yes, glass is a non-conductor but a very very thin glass can still serve as an electrode. The working of the glass electrode confused scientists for 100 years. These are very special cases.
The gas electrode is another strange case but not a ridiculous one. It is an interesting case. As you pointed out gases are not conductors of electricity under ordinary conditions, but it turns out that many gases when bubbled over platinum electrode coated with some platinum black (oxide), they start to participate in electrochemical reaction as if the electrode were made of the gas itself.
If I remember correctly, gases adsorb on the surface and participate in electrochemical reduction or oxidation. In electrochemistry, sometimes there are observations and but no explanation exists, so don't be surprised by the mysteries and wonders of this field.
In future you might encounter semi-conductor electrodes and some fancy experiments have been done with Teflon "electrodes".


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Bond angle about oxygen in alcohol
What should be the C-O-H bond angle in an alcohol be, would it be greater or smaller than 109°28’?
According to me, the angle should be greater since the steric factor of R will overcome the lone pair-lone pair and lone pair-bond pair repulsion.

In what alcohol? Steric effects vary tremendously.

See: https://chemistry.stackexchange.com/questions/91102/bond-angles-in-a-oxonium-cation ... https://chemistry.stackexchange.com/a/60645/17368

The $\ce{COH}$ bond angle is $104$°. This can be explained by stating that the two lone-pairs in Oxygen atom are more "cumbersome" than the two covalent bonds $\ce{C-O}$ and $\ce{O-H}$.

Perhaps better explained by Bent's Rule which suggests that the two oxygen orbitals containing the lone pairs will contain more s-character causing the $\ce{C-O}$ and $\ce{O-H}$ bonds to have more p-character.  More p-character results in the $\ce{C-O-H}$ angle moving towards 90 degrees.

@ron I'd like to encourage you to submit this as an answer. Maybe link to your other posts about Bent's rule.

The $\ce{C-O-H}$ angle in ethanol is reported as 105.4°, smaller than the tetrahedral angle of 109.5°. As I mentioned in my comment, Bent's Rule provides a reasonable explanation for this slight angular compression.
Simply stated, Bent's Rule tells us that orbitals with higher electron density will have more s character than orbitals with lower electron density. This is because s orbitals are lower in energy than p orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more s character mixed in.
The electrons in the $\ce{C-O}$ and $\ce{O-H}$ bonds are shared between oxygen and another atom; the two lone pairs on oxygen are not shared, they are concentrated on the oxygen atom.  According to Bent's Rule, we would expect the lone pair orbitals, with their higher electron density, to have more s character mixed in and the $\ce{C-O}$ and $\ce{O-H}$ bonds to contain less s character (consequently more p character).
Placing more p character in two bonds will reduce the angle between them (consider the change in bond angle as we move from sp to sp2 to sp3 hybridization; the ultimate case is pure p hybridization where the bond angle is reduced to 90°). Therefore, as we mix more p character into the $\ce{C-O}$ and $\ce{O-H}$ bonds in ethanol we would expect the $\ce{C-O-H}$ angle to be smaller than the tetrahedral angle as experimentally observed.


text:
Suppressing flammable foam formation in electrolysis
I often use electrolysis to remove rust and other junk off of steel parts. My general setup is:

Electrolyte: Sodium bicarbonate + distilled water.
Anode: Plain carbon steel
Cathode: Carbon steel, iron oxides, and "junk" (e.g. stubborn bits of paint, small amounts of oils / greases, etc.)
Power: 9-24V, 2-10A

Every once in a while, the "junk" on a part (cathode) leads to the formation of foam on the surface of the solution. Of course having a well-cleaned part reduces or prevents this but ... sometimes these things happen.
The problem is: The foam is mostly hydrogen and oxygen gasses, and it's explosive (or flammable or whatever, but it goes boom). And it's near wiring that could spark. So I'd really like to not have any foam.
For example (this is mild compared to some of the situations I've seen) (video of me igniting it):

The pH of the solution in that particular example is ~9.3, if it's relevant.
So my question is: Is there something I can add to the solution that will remove the foam (or prevent it from forming) but that also won't interfere with the reactions at the cathode? Ideally, something that also won't leave me with a hazmat to dispose of, either (NY, USA).
I'm not really sure what keywords to use in searching, so even some terminology tips would be helpful. (Update: "Antifoaming agents")

Update: Simethicone emulsions are the current best contender; discussion in comments below; will post if I can get consistent good results.

Might try over the counter “anti-gas” stuff sold in drug stores. It is intended for human stomach use, so no disposal issue. If it works to suppress the bubbles, then case closed.

@EdV Good idea. Simethicone, iirc. Will have to be wary of additives for testing but if it works the lab grade stuff is a little pricey, but not prohibitively expensive (especially if it doesn't take much).

Are you sure the foam is made of $\ce{H2}$ ? In my opinion I would guess it is a mixture $\ce{H2 + O2}$ ?

It's almost certainly got both H2 and O2 cells.

@EdV Simethicone acquired. Internet says it's very insoluble in water but it's also slightly less dense so maybe it'll float on top and do its thing. Fingers crossed. Got softgels; the drinkable stuff is mixed with aluminum and magnesium hydroxides and also has less of the stuff in it.

Hmm. I ended up adding about 1.5g of it. It does change the characteristics of the foam a bit (larger cells, but not uniformly), somewhat slows the formation rate, and helps it clear significantly faster when the reaction is stopped, but requires frequent agitation and isn't quite there. It tends to ride up the foam, and also behaves very much like an oil on the surface. I'm certain the solubility is the issue. But not calling it a failure yet. Not right this moment, but next try will be to add more of it (enough to cover surface) then I'll try the drinkable stuff after that. Stay tuned.

Alternate Plan B, if the simethicone fails: a quick short spritz from a can of engine starting diethyl ether (aka engine starter fluid). Reference: early on in Fear and Loathing in Las Vegas by Hunter Stockton Thompson.

https://en.m.wikipedia.org/wiki/Defoamer gold mine of plan Bs btw. I knew there was a word for this.

So I put about a 5mm thick layer of mineral oil on top and it's not perfect either but it's surprisingly effective at limiting foam quantity (foam still forms but disappears more quickly). The effectiveness decreased with temperature, and it becomes completely ineffective around 130F or so (although that's way hotter than I normally run at). ... It's a bit messy though, and also I'm wary of putting the electrodes in through an existing oil layer, but with a larger container maybe I can put a circular ring in to act as a caisson of sorts.

Perhaps just provide a steady stream of air across the surface to an outside vent. This is what is (or was, pre-Li-ion) done where large lead-acid cells are charged, such as aboard ships and in telcos's.

Yeah. Actually that was my first attempt (I cook a lot of pasta, I know what's up lol). But for some reason this foam is stubborn and I couldn't find a balance between knocking it down and blowing it all over the place. But if it was a more fragile foam then thatd 100% be my method of choice. I have been tossing some similar physical strategies around in my head though, e.g. air with some sort of overflow trap, etc. If I think of anything that works I'll post it here.

The easiest partial solution is to have separated anode + cathode compartments ( e.g. using a salt bridge ) to have separated H2 bubbles and O2 bubbles. Pure H2 does not explode, just burns - remember the airship Hindenburg. H2 + O2 mixed  in the (near) ideal electrolytic ratio like in your original arrangement would explode magnificently.

@EdV I ended up doing a ton of research into antifoaming agents. Simethicone is a common choice and the insolubility is the big challenge (validating, heh). Long story short though emulsion-type antifoamers are the go-to for aqueous solutions. Simethicone based ones are usually an emulsion prepared with methods I don't understand that seem to involve stearates, sorbates, and glycerides but ultimately I found some patents that lead to searches that led to .. drum roll .. OTC simethicone infant anti-gas drops. And they are giving very promising results. Details once I can characterize it.

(There's a bunch of commercially available defoamers that are reasonably priced, too, although the ones I really wanted were only available from China in very large quantities e.g. Shin-Etsu KS-540; pH is an issue and there are definitely defoamers specifically designed to be stable at high pH's; it's actually a pretty interesting field. I'm not sure if the baby drops will struggle long term at pH > 9 or not but we'll see; if so I'll try to look for something commercial. The anti-gas drops are very cost-effective and working so far. Sigma-Aldrich has some too but their details are unclear.)

Also, pro-tip: Do not mix mineral oil, simethicone, and water, unless you're prepared to spend the rest of your life cleaning up. Alone, they are weak and helpless, but with their powers combined, they're like the kryptonite of soap.

@Poutnik That's a great point, thank you. That must explain the relative violence of the reaction when I ignited it, and (I didn't mention this above) might also explain why I occasionally have observed a spark at one of the electrodes that doesn't ignite the foam (those sparks must have only caught one of H2 or O2, since they're concentrated around their respective electrodes). I might try a salt bridge setup just for kicks (I've been meaning to mess with that and other membrane setups for a while) but, practically, I'd have to think about how to make it convenient for normal use.

(Also, tbh, the fire hazard is my lesser concern; the main one is the splashing of not-always-nontoxic electrolyte solution if it does ignite, and other resulting physical annoyances; separate containers like you describe can definitely help, and also if I can keep the foam down enough, then splash guards and securely mounted electrodes will mitigate the rest. That little cup in my video isn't my normal setup -- my bigger tank is way less sketchy.)

Comments are not meant for extended discussions, you can use [chat] for that. Please don't edit the question to include the answer. It is not necessary, and not recommended, to include commentary of how the question changed, the edit history takes care of that. The question should be concise and have the first-time reader in mind.

@Martin-マーチン Sorry; I'll clean it all up (including this comment) when I post an answer.

In the beverage industry (think sugar-based fizzy drinks and beer) foaming has to be reduced prior sealing the cans, this is done by ultrasound.  The cans and bottles underpass an airborne ultrasound probe which pointing downward.

(Screen photo from this youtube video, edited.)
Since there is no contact between the probe and the bottle, nor immersion in the liquid, it seems likely you may keep the electrodes attached.  These probes come both in larger (example corporate documentation) as well as mobile and hand held size (example video).

Comments are not for extended discussion; this conversation has been moved to chat.


text:
How do fullerenes withstand external pressure from collapsing the molecule?
As Fullerene's have an empty shell inside their structure how they withstand the pressure from the outside?
As a sidetone, are there any methods for calculating the force that the "vacuum" exerts on the outside? At some point as Fullerene's get too big I imagine that the internal force becomes overwhelmed, so this could give different perspective on this question already on the website:
What is the maximum amount of carbon atoms that can comprise a fullerene?

If you believe this question is better re-directed at a different SE Website, then feel free to move it at your discretion.

Fullerene isn't a balloon, it's like empty ball of steel - why would a ball of steel "collapse", whatever it would really mean?

Due to a pressure difference. A ball of steel has air inside it. For the same reason that we do not have vacuum "airships"; the hull would collapse from the difference in pressure. I am baffled by your comment.

I'm baffled by your question. 1 bar of pressure difference won't crush a steel sphere, much less a fullerene, for which you need hundreds of GPa!

Mithoron, I would suggest for you to see the practical limitations of a system having a vacuum. There is a reason why, as mentioned above, no vacuum airships are functional. The thickness requiered to prevent compression and buckling from the exterior atmosphere would negate the density difference needed for buoyancy. A fullerene is a one-atom layer molecule, not centimetres of steel. Big magnitude difference.

Steel is very resistant to both extensional and compressional pressure. Otoh, while stretching at some points breaks covanlet bonds, compressing them is extremely hard, if you look at a Lennard-Jones potential. Ergo: A "fullerene" needs a really big diameter before its local curvature is so low that a local outside pressure fluctuation can push in the surface so much that bonds get extended and bent, and not just compressed.

If the fullerene was made to be  some tens of microns in radius I could well believe that it could be compressed by pressure, of course assuming that such a size could be made in the first case and that it could actually be made with nothing inside such as solvent vapour or whatever. The force needed to buckle groups such as pentacene, phenanthrene and larger aromatics is probably known from spectroscopy and will be small compared to bond extension or compression.

The shape of the fullerene $\ce{C_{60}}$ is close to the one of a sphere. Forces acting from the outside of this body, initially directed inward of this body, are deflected laterally (like e.g., gothic arches, cupolas; submarine vessels) as pressure and tension. So the question might be closer to civil engineering/architecture (limits of traction, torsion, shear and bending like for I-beams used in construction of bridges, and rebars), or vector algebra/finite element analysis (depending on the level of detail if one want's to compute it ahead of time), than chemistry only.

@Mithoron but it can crush a barrel, or a plastic bottle, etc. The question is sensible, although it can somehow be dismissed saying that fullerenes are relatively stable and that is. However, I will try to write a qualitative and simple answer. But basically it is matter of organising all the above comments

Pressure is a pretty meaningless term at the level of single molecules. The real question here is how does a fullerene not deform when interacting with other molecules, perhaps with significant kinetic energy.

@matt_black I partially disagree. The fact that placing a single molecule under P high enough to have effect is difficult or perhaps not feasible is true, though. Experimental data are indeed for not isolated fullerenes and how they do transform. But one can certainly apply forces and P to a molecule, actually the collision you mention is even a possible way providing the fullerene is charged. However I am more interested on writing an answer about hydrostatic P. I am not a specialist, but at least the above comments can be collected in one piece.

@Alchimista I don't think we disagree as much as you think. My point was that "pressure" as normally defined is a bulk property emerging from the statistical aggregation of many molecular interactions. In a gas, it is all about collisions among molecules, in a solid it is about forces among molecules. So the question reduces to something about how fullerenes deform under realistic forces.

In addition to my take, based on what seems qualitatively reasonable and somewhat easy to explain without much technicalities, I have tried to organise the various comments that followed the question.
This answer includes two aspects, and assume a single molecule (it seems what the question refer to), and for some parts a single molecule immersed in a fluid, so the pressure is hydrostatic.
One general aspect is that chemical bonds can be relatively hard to be broken by elongation but compressing them past a certain point is almost impossible. To figure out why it is so, there is no really need for quantum physics or to invoke the Pauli's principle, as for Coulomb repulsion between the electrons in the outer shells suffices, at least qualitatively.
Even for a force coaxial with the bond, if a breaking would occur during compression it will be a kind of slipping or shearing of the bound atoms. Think of the situation in which you try to push together two magnets by the same pole, it should give a good picture.
At this point, you may ask why compression along bonds is relevant to the discussion given that pressure exerts forces which are orthogonal to them, in the current case. One point is that if the atoms are bound in a two-dimensional array, a departure from their equilibrium position in the plane is never bending alone, but geometry implies that bond lengths are altered as well. Typically for a flat array this will be elongation (if you buckle a rubber sheet, the elongation would be radial around the pushing point, to give an example). But keep reading...
The second aspect is due to the particular shape of fullerene, the latter being what has prompted you to pose the question (indeed similar ones can be posed about the structural integrity of other molecules or covalent solids, as for the latter contain "empty space", too). Objects with a high radius of curvature are effectively redirecting      and distributing the forces perpendicular to the tangent towards the latter. This property is well know from ancient times and found application in the construction of archvolts, cupolae, pressurised tanks or vacuum containers as well as the light and resistant domes by Buckminster Fuller (the name Fullerenes is after him, as the molecules and the domes show remarkable similarity). We have seen that bending bonds of a surface implies stretching. However, for a spherical surface, applying outside hydrostatic pressure equates to the simultaneous compression of all bonds.
Now, it should be clear that the two aspects work together, and a buckyball is more resistant than an hypothetical sheet having the same  bond strengths but lacking curvature. The more closest would be a graphene, or a monoatomic layer of graphite.
So, more you push on a fullerene, it gets harder and harder. While in principle a P leading to its implosion - whatever it means, perhaps it is better to call it rupture - could be reached, its value would be very high, certainly order of magnitude bigger than atmospheric P.
The paper at https://arxiv.org/pdf/cond-mat/0610007  gives an estimated figure of 8000 GPa (!) for the critical pressure at which one molecule of C60 fullerene becomes unstable. Rather than on the value, it is worth noting that at sufficiently high pressure the coordination number of the carbon atoms changes even upon "symmetric", hydrostatic compression. The molecule cannot simply shrink but loose its identity as fullerene.
This points again to the tremendous strength of repulsion forces, so that in short and pictorially, we could say that "there is even not that much space for the molecule to implodes in".
It should be clear that, while the shape of fullerenes makes this type of questions almost natural, the scale involved as well as considering the behaviour of a single molecule makes them devoid of sensible, or at least accessible, consequences,with perhaps the exception of atomic force spectroscopy.
The fact that the interior of fullerene is vacuum it does not mean much. What count is the difference in pressure, and we handle fullerenes at about 1 atmosphere. So, from this point of view, it is perhaps more surprising that my vacuum dessicator doesn't implode.
The question is more interesting if taken, as in the linked paper, or as I take it, for how much external pressure fullerene can sustain.
The case of an inflating fullerene (also treated in the linked paper) is even less sensible. There is nothing we can do by suction, nor a kind of needle to directly inflate it. Perhaps one way to pump the fullerene from inside would be some optically pumped endohedral fullerene. As people on optics do quite some magic, never say never...

Besides theoretical work as that linked above, much work is done on fullerenes as very little but macroscopic solid samples. Under high pressure they are known to undergo oligo- and polymerisation, or transitions leading to graphitic or diamond-like phases, depending also on the nature of the sample and temperature. Data are usually attained in the range 5 - 10 GPa. These pressures, as well the higher achievable with diamond anvil cells (about 500 GPa) are already millions times the atmospheric pressure and still one or two orders of magnitude smaller than the figure mentioned above.
Further note. The original question contained reference to virtual particles, which I have removed. A part of it perhaps remains where OP refers to "force that vacuum exerts to the outside". There is not such a force. What might be interesting and connected to the stability of higher term fullerenes is to ascertain if a smaller curvature is so detrimental to the mechanical strength that the cage is indeed unable to sustain pressure. Given the discussion and values mentioned above I would be astonished of such a big difference. But the idea is intriguing.

+1 and I wonder if this or the linked arXiv paper have any bearing on What is the maximum amount of carbon atoms that can comprise a fullerene?


text:
Number of electrons in a 2D box for degenerate energy levels
As we know that the energy of a 2-D box depends on two quantum numbers say $n$ and $m$ so
$$
{E \propto n^2 +m^2}
$$
Now if we say that $n=1$ and $m=1$ so only one energy level is there and 2 electrons of different spin can be accommodated

if we take $n=1$ , $m=2$ or $n=2$ , $m=1$ , the energy level will be the same for these quantum numbers, but the wavefunctions will be different, as well as the probability distribution functions.
So my question is : Can we only  2 electrons in this energy level, or 4 electrons considering that there are two different wavefunctions for this energy level ?

This suits more to Physics SE, as there are not even atoms or ions to say there is some relation to quantum chemistry. // Aside of that, try to reconsider the question readability and eventually reformulate it.

There are always two energy levels, which is easier to see if the box is not square so you could look at the full formula. The levels are accidentally degenerate for the cases you list only because the box is square .

Ever heard of Pythagoras?


text:
Listing all possible structures for a given Markush structure automatically
I have Markush formulas described as mol blocks, v3000, even if that does not matter that much because I could convert it into another format if needed.
I need the possibility to automatically list all possible structures for a given Markush. More specifically I often have aromatic rings with a substituent that can be in any position, and I'd like some automated way of listing the structures for that.
I realize there is this Knime node called "Markush Enumeration" (https://docs.chemaxon.com/display/docs/markush-enumeration.md), but it needs a license.
Is there a 'free' way to easily do it? I am using Python and Knime, if that matters.

Preface:
Since you mention familiarity with/access to Python, and I assume you want to process batches from .sdf from a script rather than from a GUI (note, ChemAxon offers a dedicated Markush Editor in addition to a plugin for their sketcher, Marvin (a tutorial)). Thus, I suggest using RDKit. In case you have access to a Linux distribution of Debian (or a relative of Debian like Ubuntu), there is a repackaged version by DebiChem (tracker), too.  Then you do not need to install a separate instance of Miniconda/Anaconda (some notes about this on Matter Modeling Stack Exchange).
The following works well within a Jupyter notebook by Linux Debian 12/bookworm (branch testing) offering Python (version 3.9.9) and RDKit (version 2021.09.2).  The credit is due to Greg Landrum, who posted the bits and bolts required for this to work and only rearranged/paraphrased for this answer.
Ad rem. In the first cell, initiate the Jupyter notebook:
from rdkit import Chem
from rdkit.Chem.Draw import rdDepictor
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole
from rdkit.Chem import rdMolEnumerator
import rdkit
print(rdkit.__version__)

Next, load .sdf to process, e.g.
pv2 = Chem.MolFromMolBlock('''
  Mrv2007 06242006032D

  0  0  0     0  0            999 V3000
M  V30 BEGIN CTAB
M  V30 COUNTS 10 8 0 0 0
M  V30 BEGIN ATOM
M  V30 1 C -1.7083 2.415 0 0
M  V30 2 C -3.042 1.645 0 0
M  V30 3 C -3.042 0.105 0 0
M  V30 4 N -1.7083 -0.665 0 0
M  V30 5 C -0.3747 0.105 0 0
M  V30 6 C -0.3747 1.645 0 0
M  V30 7 * -3.042 0.875 0 0
M  V30 8 F -5.0434 0.875 0 0
M  V30 9 * -1.0415 2.03 0 0
M  V30 10 Cl -1.0415 4.34 0 0
M  V30 END ATOM
M  V30 BEGIN BOND
M  V30 1 1 1 2
M  V30 2 2 2 3
M  V30 3 1 3 4
M  V30 4 2 4 5
M  V30 5 1 5 6
M  V30 6 2 1 6
M  V30 7 1 7 8 ENDPTS=(2 2 3) ATTACH=ANY
M  V30 8 1 9 10 ENDPTS=(2 1 6) ATTACH=ANY
M  V30 END BOND
M  V30 END CTAB
M  END
''')
pv2

to yield

where grey clouds mark where variation of the position of the substituents is anticipated.  By subsequent
pv2_bundle = rdMolEnumerator.Enumerate(pv2)
Draw.MolsToGridImage(pv2_bundle)

you launch the variation of the positions, with the results displayed as

If you want, a subsequent list comprehension offers you to access the result e.g., in SMILES strings (export as .sdf (v3000) equally is possible, check RDKit's documentation and cookbook):
listing = []
listing = [rdkit.Chem.MolToSmiles(entry) for entry in pv2_bundle]

for element in listing:
    print(element)

with output of
Fc1cnccc1Cl
Fc1cncc(Cl)c1
Fc1cc(Cl)ccn1
Fc1ccc(Cl)cn1

Again, the above only is rearranging an example by Greg Landrum's blog post detailing out multiple additional equally interesting examples of application.

As for the generation of the .sdf (v3000) with «Markush syntax» with a GUI, beside mentioned Marvin by ChemAxon, BIOVIA Draw by Dassault offers this functionality, too.  The later bears some similarity to MDL's ISIS Draw the more senior readers of chemistry.se still might remember/use.

@Martin Thanks for the edit.  If encountering code blocks on chemistry.se, I'm going to stick with the more explicit declaration about the language you deployed.

You're welcome. And it is the safer bet, because it is included in common mark iirc. If you are using multiple code blocks in an answer, it might be more convenient for you to use <!-- language-all: lang-python --> at the start of your post. This should turn it on for all of them. (Also, I believe I am one of these senior readers.)

Thank you so much for the detailed answer. I will try it out in a few days and accept the answer afterwards.


text:
What material should my evaporation dishes be made of to combat static charge buildup?
I have great difficulties moving very fine powders from a drying dish (glass) into small plastic containers using a metallic spatula due static charging. I have to scratch a lot inside the glass beaker to get all the powder together, I guess that the charge builds up that way. When I lift the powder out, the majority or even everything on the spatula flies off to stick to the glass. Due to lab safety I also wear nitrile gloves if that is important.
I want to order better equipment now so that I do not have to combat static charges. I will order plastic single-use spatulas - advertised as anti-static - and some new evaporation dishes. But there is no indication in the webshop, which material might be suitable for that. I can select from porcellain or the plastics PTFE and PFA. There are some anti-static weighing dishes, but they are made from PS and very lightweight, and can thus not hold acetone or ethanol, which is a requirement.
So I am stuck with the question, which material should I choose for the new evaporation dishes?
Edit to answer the comment: The powder, typically less than 100 mg, is washed into the evaporation dish with either acetone or ethanol. When the powder is completely dried it is transferred into a small plastic container. The Aluminium foil would be an option, but Platinum is a bit too expensive for this purpose.

Gloves are causing this static build up. Can you keep the beaker on a conductor like a aluminum foil? And frequently discharge the spatula by touching the Al sheet. Add more details to clarify your question. How much volumes are you talking about? What solvents are present? If the department can afford, there are platinum evaporating dishes.

Depending on the chemistry (solvent to evaporate, sample per se, mass and volume to carry), metal dishes made out of aluminum may be suitable and less expensive than those out of Pt (example 1. example 2), too. For sand baths under the hood, the stainless steel variant (example) work well, too.

Depending on the hazard you can remove the gloves. I mean just for that transfer operation, especially if it is done in a very clean corner of the lab (scale room, preparation room, etc.)

Thanks for this question as static electric charges are largely ignored, while unlike photo/sunlight, the latter effects are well known. A major exception being, of course, when a static induced charges creates an explosion (for example with N2O which is not especially photo sensitive).

Definitely don't opt for fluorinated polymers (PTFE, PFA) as they are one of the best materials for collecting the charge.

Do you have a rotary evaporator? some have adapters to vials which would mean that you can remove the solvent directly from the vial.


text:
Can Ion-Dipole Force Occur Between Lecithin and Water?
So, ion-dipole force occurs between an ion and a polar molecule. I would like to ask if an ion-dipole force would occur between water molecules and lecithin as seen in the picture, since it has a negatively charged oxygen atom on the phosphate group and a positively charged nitrogen atom on the choline group?
Would the attraction between them be considered an ion-dipole force, or is it only applicable between a lone ion and a polar molecule? I understand it seems like a simple question, however, all examples and explanations I've found just stop there.

It can be reduced to the physical scenario of the + charge, the - charge and the dipole as 2 adjacent opposite charges. Each of them interact with each.


text:
What drives hydrophobic effect/interactions? (multiple choice question)

Water molecules cause hydrophobic interactions because
A- they can disrupt surfaces with no hydrogen bonds.
B- they prefer to
be close to hydrophobic surfaces
C- they are restricted in their rotation when they are next to hydrophobic surfaces
D- All of the above

I'm confused regarding the answer of this question. Sometimes I feel it's "all the above" and sometimes that there is no clear good answer! If there was an answer saying that waters are more tightly bound to each other next to hydrophobic surfaces would it be better? Or is the answer "all the above" the correct one?
As I know, the hydrophobic effect will make water molecules surround the hydrophobic substances as much as they can to minimize the surface of it! So, the rest of the answers are the properties of the water but not related to the question.

Where did you find this ridiculous question? No wonder you're confused, it's based on the premise that water has hydrophobic interactions. Water has NO interactions with hydrophobic material, and that's it! (NO of course means "much less than with itself"). A, B,C are all wrong. ?!?

The question is worded a bit strangely, but the answer is C. I presume the question is not asking about the "hydrophobic interactions" of water itself, but rather why water causes what we call the "hydrophobic effect".
A - they can disrupt surfaces with no hydrogen bonds - I'm not entirely sure what is meant here. It sounds like it means disrupting hydrophobic surfaces. Water does not break these up, it causes them to aggregate.
B - they prefer to be close to hydrophobic surfaces - this is the opposite of how water behaves. Water prefers to interact with polar molecules
C - they are restricted in their rotation when they are next to hydrophobic surfaces - This references the fact that the hydrophobic effect is entropy driven. Water around a non-polar surface forms ordered cages (restricted rotation), and thus this is entropically unfavorable.