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Squiggly line in chemical bond notation
In general is there a convention as to how one interprets squiggly lines in chemical structure notation?
e.g. see the below sketch. I suppose it has something to do with cis / trans isomerism. But e.g. where do you draw the squiggly i.e. which side of the double bond? Is it used in other contexts as well? And does it mean we don't know which isomer is formed or does it mean we are talking about a cis + trans mixutre?

Larger context here (so the explanation about variable chain length may not make sense?)

I'd guess that the squiggly line is supposed to represent various aliphatic chains of different lengths. I don't know that it is any standard convention.

@MaxW Thanks! I added more context about the sketch I took this snippet from. If you look at the reaction scheme the variable length explanation would not make sense, right? OTOH I recall seeing used squiggly lines in that context as well. Hence the question!

It seems that the squiggly line is supposed to represent the single bond between conjugated double bonds. It isn't a typical aliphatic bond in that it is shorter and has a barrier to rotation.


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What is the rationale for calling something as an 'inorganic [organic compound]'?
While studying the main groups of p-block, I have seen many compounds being labeled as an analog to some popular organic compound. For example, borazine is termed as 'inorganic benzene', while boron nitride has been called 'inorganic graphite'.
Now, my question is, whether these monikers (if that's what they may be called?) reflect just the structural likeness, the chemical similarities or, perhaps, both?
Moreover, are there some other examples which illustrate this?

It is because of the structural likeness, but I wouldn't use the analogies.

@MaxW won't the structural similarities, in some slight manner, cause similarity in certain chemical characteristics? Say, for example, why couldn't borazine show EAS ? I am not saying that it does, and in general I have no clue about its chemical properties, but can't I imagine the substitution of a proton by an equivalently monovalent ion, say, a halogen?

I just think making such analogies is ripe for creating misunderstandings. For example in benzene all the hydrogen are equivalent. That obviously isn't true in borazine.

Yes, I agree. But, then again, many sources cite them as such, so these analogies are quite the trend, rather than some one-off author trying infuse fun into a fact.

I would not call graphite organic, but there is no clear-cut way of defining organic and inorganic compounds.
To your question, the pairs of compounds are isoelectronic. That means that if we assume that molecular orbitals (MOs) arise from atomic orbitals, the corresponding MOs in the two compounds have the same occupancy. In these cases it arises because the number of electrons in the dimer BN is the same as in the dimer CC. Another example of isoelectronic compounds would be N$_2$ and CO. Being isoelectronic, they do share some properties, e.g. both N$_2$ and CO have a triple bond. If only the valence electrons are identically occupied, we call them isovalent, which would include e.g. H$_2$O and H$_2$S.
The bonding of benzene and borazine differs in that not all sites are equivalent, but the form of the MOs is broadly similar and the same applies to boron nitride. Similarity in MOs is then reflected in similarity in reactivity. (Even though they are not that similar in reactivity.)

If one takes the 1 graphite plane structure formally as extrapolation of e.g. coronene toward the infinite 2D size......

My take is that the terms "inorganic benzene" and "inorganic graphite" are meant to relay thoughts of both structural and chemical likeness to organic counterparts; however, graphite is not an organic compound since it is not a compound. It is an allotrope of carbon, just like diamond, buckminsterfullerene and graphene.
The IUPAC standards establish a common language for chemical researchers and hence what is an organic compound and what is an inorganic compound.
Probably the best examples of comparisons of organic and inorganic compounds, their similarities and differences, would be the work to evaluate whether sulfur-, boron-, and silicon-based life is possible. (The most plausible is probably silicon.) See Petkowski et. al. for a recent paper that includes an overview of the basic physico-chemical properties of carbon and silicon, including their respective bond and common functional group reactivity.

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Why would T1 relaxation time change ~20 fold between solvents for the same substance?
I have a simple, completely conjugated pyridinium salt with MW of ~300 g/mol. The molecule consists of regular non-labeled atoms (only C, H, N).

If the 1H T1 relaxation time for this molecule is measured in DMSO-d6 the T1 values of all of the protons are within the expected range (9 sets of protons with T1 values of 1.4-2.1 s).
However if the 1H T1 relaxation time for exacty the same molecule is measured in CDCl3 the relaxation times differ up to 20 times (9 sets of protons with T1 values of 0.04-1.9 s). The 1H relaxation is especially fast at the pyridinium ring (2 sets of pyridinium ring protons with T1 values of 0.04 and 0.07 s and further removed protons from the pyridinium have T1 values of 1.5-1.8s).

From what I have read in the literature the relaxation times usually differ up to 2-fold.
Any ideas why this is the case?

Can't help but wonder if aggregation is to blame. Unless you have a huge non-coordinating anion to go with your pyridinium compound, you can expect some pretty tightly bound ion pairs in chloroform, whereas DMSO would easily pull them apart.

We performed dynamic light scattering experiments on the samples, and aggregate formation was not detected. I would imagine that a dimeric (or similar) structure could form in the weakly solvating solvent such as CHCl3, due to the formation of highly energetic pi+-pi+ or pi+-pi interactions. However even if an associates between the molecules form the relaxation time should not change in such a drastic way.

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Regioselective enolate formation, choosing major and minor products
This question came in my organic chem assignment. The options highlighted are the answers.
The end carbon(adjacent to the carbonyl group) should have more acidic alpha-hydrogen and the alkene formed is more substituted (as compared to the one in F). Still it is given as the minor product. Could someone explain why?

The regioselectivity depends on the conditions that you use. See e.g. https://chemistry.stackexchange.com/q/132890/16683 or https://chemistry.stackexchange.com/q/17103/16683. The major product here is the so-called "thermodynamic enolate".

The original question here is somewhat misleading in that it assumes that these are the only two products. There is E-/Z- isomerism in what's labelled here the "major" product that must also be accounted for.

@orthocresol, thanks for the links.
I understand that the removal of the gamma hydrogen results in resonance (leading to a stable product). However, shouldn't the fact that E has more substituted alkene also come into play?

But, in E one double bond is trisubstituted and the other disubstituted, and in F the same is true. Based on that, I don't see why the alkene substitution should affect the overall stability all that much. You would need a significant effect to overcome the preexisting bias, which favours extended conjugation (F) over cross-conjugation (E). Agreed about E/Z isomers; that's a separate issue, though.

A major factor contributing to the stability of the enolate ion formed as the major product (F) is the stability associated with extended conjugation or extended delocalisation via resonance. Due to this, terming (F) as the major product seems justified.
If the product were to be a kinetically-controlled one (KCP), then the presence of three acidic alpha hydrogen atoms could've caused faster formation of product (E) ; it still would not have been the more stable product, but since the major product in KCP-type reactions is controlled more by the kinetics , rather than the thermodynamics, we would possibly expect to see (E) in a larger yield in that case.
Note: Strictly speaking, the problem should mention conditions that can help us to unambiguously determine which pathway (KCP vs. TCP) should be chosen. But a practical advice, garnered from the experience of working problems which seldom perfectly mention the reaction conditions, is that unless mentioned otherwise, it is always wise to first consider the thermodynamically stable product.


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Frontier orbital analysis of zinc-catalysed carbonyl-ene reaction
Q1. Is the $\sigma_\ce{C-H}$ bond a donor in this reaction? If not, then can we take the alternate combination (HOMO of alkene, LUMO of the $\ce{C-H}$ bond and, HOMO of the carbonyl)?
Q2. We can also draw the LUMO of the carbonyl group in the other possible manner(simply flip the shaded lobe on each atom, hence keeping the node). The reaction would then have to go through an antarafacial manner.
Why do we draw the LUMO in this manner?

Think about the role of ZnBr2: you know it is a Lewis acid. That should give you a hint as to whether the HOMO or the LUMO of the carbonyl component is more important.

I'm not sure I fully buy that the "ene" portion here is part of the LUMO... Put it this way, if this were just a π system adding into a carbonyl leaving behind a carbocation, I think we would argue that the π system was part of the HOMO.

@Zhe, could you elaborate what you're saying? I can't understand it.

If an alkene is reacting with a carbonyl, the most reasonable thing to suggest a priori is that the alkene is the nucleophile (HOMO) and the carbonyl is the electrophile (LUMO).

First of all, a molecule only has one HOMO and LUMO. Don't confuse MOs with localized orbitals from VB theory.
You can basically approximate this mechanism with four MOs:

LUMO+1: Mainly $\pi^*_{C=C}$ with contributions from $\sigma^*_{C-H}$
LUMO: Mainly $\pi^*_{C=O}$
HOMO: Mainly $\pi_{C=C}$ with contributions from $\sigma_{C-H}$
HOMO-1: Mainly $p_O$ with contributions from $\sigma$-orbitals on the carbonyl carbon. We can regard this MO as one of the lone pairs of the oxygen.

The corresponding interactions will roughly look like this:

Hence, the HOMO - LUMO interaction explains the C-C bond formation, while the HOMO-1 - LUMO+1 interaction explains C-H bond breaking and O-H bond formation. Both HOMO and LUMO+1 also explain the shift of the double bond.
Now to your question:

Q1: Yes, $\sigma_{C-H}$ is part of HOMO, but has a lower contribution compared to $\pi_{C=C}$. So I would mainly call the C=C bond as donor. As you can see $\pi_{C=O}$ is not part of the FMOs, so your proposal makes little sense. It mainly contributes to more stable occupied MOs.
Q2: If I understand correctly, you propose an antarafacial [2+2], which is in principle symmetry allowed, but would lead to a very strained transition state with a high barrier. You are free to swap the sings on MOs, but as soon as interact with another MO, you take the one that leads to stabilization of course.

Edit: As reference, I got the FMOs from a HF/6-31G calculation on a slightly simplified system:

a molecule only has one HOMO and LUMO That is strictly speaking incorrect. Please do not confuse localised orbitals with orbitals from VBT. Then you carry on arguing with somewhat localised orbitals; delimitations that do not exist in MOT, especially there are no lone pairs in MOT. If you want to put in this interpretation, localise your orbitals. If HOMO and LUMO of the molecule would interact, there would be different HOMO and LUMO. This answer doesn't make much sense.

How would you obtain more HOMOs/LUMOs for this molecule?

Need to have another look on VBT. I'll get back on that.

Aren't the MOs in MOT always constructed from more localized orbitals? I don't see any issue in trying to weight their contributions to get some intuition.

Indeed, once you generate the combination of HOMO and LUMO, you get new orbitals of course, but don't you plot the original orbitals in FMO analysis?

While you will get notified of any comments on your answer, you will need to ping other users for them to be notified. E.g. @Tyberius


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Free Expansion - Ideal Gas vs Real Gas
By free expansion, I am referring to gas kept in a piston-cylinder arrangement freely allowed to expand against vacuum.
It is clear to me that free expansion is an irreversible process because if it were not then it could get compressed as well at equilibrium, but we know it doesn't as it would be violating the 2nd law of thermodynamics.
I know that all the parameters work 'W', Change in Internal Energy '∆U', and Heat transfer '∆Q' would all be zero in the case of an ideal gas.
But, what I want to know is how will all the parameters vary if gas is real?
Also, I wanted to ask whether the answer will change if the surface between the piston and cylinder is not frictionless?

This is a duplicate of the same question that has been answered in Physicsstackexchange.

I asked it here as well because there are only few people who are active both on Chem and Physics SE.

Also, I wanted to see both viewpoints from chemistry and physics as thermodynamics is one topic that is common to both to a large extent.

I read the wikipedia article. But I think my problem pertains more to Joule Expansion.

Just want to ask this. If T= inversion temperature then will real gas behave like ideal gas.

I mean will temperature remain constant

Joule expansion  and J.-T, effect are about the same phenomena. The system with a piston with friction will be like an irreversible adiabatic expansion Delta U = -W, So all gases would cool down, unless friction was really tiny.. BTW, put the elaboration info rather to the question.

Please, don't post the exact same question on both Physics and Chemistry. You can always post on Physics, then advertise it in Chemistry chat, or vice versa. Or, if you want to post the same question, my suggestion would be to tailor it a bit to both sides, don't just copy paste the same text.

Crossposted to https://physics.stackexchange.com/q/608001/2451

For a real gas, the internal energy varies with both temperature and molar volume.  The variation is described by the equation:
$$\mathrm dU=C_V\,\mathrm dT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]\mathrm dV,$$ where $U$ is the internal energy per mole and $V$ is the molar volume.  This means that, unless the equation of state is such that pressure is directly proportional to temperature at constant molar volume (like an ideal gas), the internal energy will depend on molar volume.

But isn't free expansion an irreversible process? Then how are we using differentials? Also, how can we have an equation of state for it?

I read that we can only define state variables when the system is at equilibrium.

Also will the change in enthalpy be zero in case of real gas?


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What is the 'electronic energy', in units of 'negative' hartrees, of a molecule?
On a NIST (National Institute for Standards and Technology) page on the molecular details of water and it's various isotopologues, I came across this:
Electronic energy:  -76.4089547246 hartree
There is, really, no other context on the page....
I found the page after typing in 'ionization energies of water'....
I put this here on Chemistry SE rather than Physics SE, I hope that is OK....
I DID look around to try to find an answer elsewhere, before coming here and posting...

That's not what you need. Roughly speaking, it is the energy required to remove all electrons.

https://en.wikipedia.org/wiki/Hartree_atomic_units Explains the Hartree unit.

The energy is the potential energy of all electrons in the molecule. Sometimes the term also includes the potential energy of the nuclei but in most cases it doesn't.
As reference or zero energy you can imagine the case where the electrons are non interacting or infinitely far apart from each other. In this sense it tells you the energy required to "remove" all electrons, i.e. pulling them away from each other so that they no longer interact with each other.
It can be interpreted as ionization energy where you remove all electrons in one go form the system but that is an theoretical process. The nuclei would push each other apart before you could ionize water to that degree. Normally the ionization energy refers to the removal of one electron. And the second ionization energy normally refers to the removal of a second electron after the system relaxed from the first ionization. And so on for higher ionization. This consecutive ionization process is different from removing all electrons in one go and would yield a different energy than the electronic energy given here.
This question/answer might also be helpuful :
https://chemistry.stackexchange.com/questions/38589/definition-of-electronic-energy#:~:text=Electronic%20energy%20is%20indeed%20an,nothing%20to%20do%20with%20electrons).


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Advice for a good solvent for fragrance
Hi any suggestions for a solvent for aroma oils, that has low evaporation and high diffusivity (low molecular weight and hence spreads well in the air). The solvent should also be soluble in water and should have a low evaporation rate

2-butanol? Bpt 99C

Thank you so much for responding, but 2-butanol is flammable , sorry i forgot to mention this...I am trying to work on home made reed/ wood fragrance diffuser...need something non toxic too...is Dowanol TPM good ...i am not sure if it will diffuse well in the air

I am not a fragrance chemist but Dowanol is flammable.

Diffusion of the solvent has nothing to do with carrying the fragrances around. It should even be a con as for high diffusion should (on feeling) go like boiling point, at least from molecular mass only. Either you do not want naked fragrances at the beginning, or if so, diffusion of the solvent doesn't count. Rather you need a special size of droplets. It seems to me that you should reproduce what big names or artisans do, rather than starting your own recipes. Many of your requirements won't have easy conciliatory solutions.


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What is the condensed formula for cyclohexane?
I can't find it written anywhere and my guess is that it doesn't exist as it would be impossible to write it out in a straight line.
Could $\ce{CH2CH2CH2CH2CH2CH2}$ be used ?

I don't understand. How does one draw a circle in one dimension as a straight line?

@Zhe What don't you understand ? Kantura just wants to know whether his formula describes correctly the cyclohexane molecule. The answer is yes or no. For me, the answer is no, and the formula should be written $\ce{c-C_6H_{12}}$.

@Maurice I'm trying to state the obvious. You're not going to capture a cyclic structure via an linear formula (without cheating and drawing a bond between the terminal methylenes in the OP's formula)...

@Zhe OK. I see we both agree


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Explanation of the orbital energies resulting from Jahn-Teller distortion of octahedral complex using MO theory
I am confused about the energies of the orbitals after Jahn-Teller distortion of an octahedral complex. Take elongation for example.
To my understanding, since the ligands on the $z$ axis move away, there is less orbital overlap with the $d_{z^2}$ metal orbital, leading to weaker orbital interactions, leading to the antibonding orbital being lowered in energy.
There are two things I don't get: 1. The lowering of $d_{xz}$ and $d_{yz}$ and 2. the raising in energy of the $d_{x^2-y^2}$ and $d_{xy}$ orbitals.
1. The $d_{xz}$ and $d_{yz}$ don't seem to have the correct symmetry to sigma bond with any $p$ orbitals of the ligands. They are nonbonding in a perfect octahedron, so how can the be lowered in energy after distortion? Maybe they only appear lower relative to the $d_{xy}$?
2. Are they not actually raised in energy, only appearing higher relative to the $d_{z^2}$ and $d_{xz}$ and $d_{yz}$ orbitals? Or are the equatorial bonds shortened so there is a stronger interaction, raising the $d_{x^2-y^2}$? If so, then why is the $d_{xy}$ orbital raised?
Is there an explanation for the orbital energies using MO theory? If you could clear up my confusion that would be great. Thanks!

The diagram assumes a bonding pi interaction with xz and yz in addition to the sigma interaction with the z2. That's why xz and yz go down relative to xy. The "raising" of xy and z2 is an artifact of centering both diagrams on the average energy. The actual molecular anti-bonding orbitals that are mostly comprised of z2 and xy don't change energy unless the equatorial ligands get closer to the central atom (which can happen).

Does this answer your question? I would like to point to my answer there which goes into some detail concerning the MO descriptions. If so, I can drope a duplicate close vote ;)

@Jan It doesn't fully answer my question.

I disagree with your statement that

"The dxz and dyz don't seem to have the correct symmetry to sigma bond with any p orbitals of the ligands. They are nonbonding in a perfect octahedron"

Consider the case of a halide ligand like Cl-, which has filled p orbitals perpendicular to the axis of the M-Cl sigma bond. If these are aligned with the coordinate axes, they can each interact with proper symmetry with one of the off-axis t2g orbitals. In the axial position, these orbitals match up in symmetry with dxz and dyz.
In the perfect octahedron, all three t2g orbitals are affected equally (each interacts with the four ligands in its plane), so whether or not the ligand has p orbitals (and therefore pi bonding) does not qualtitatively change the splitting diagram. It just changes the magnitude of the splitting.
When the octahedron is distorted, however, the interactions with each of the t2g are no longer the same in magnitude. Just as with the sigma bond, as the axial ligand moves away from the metal center, the pi bonding interaction also weakens, so the d orbitals involved in pi bonding (which as you noted are anti-bonding orbitals) lower in energy.
The reason that the dxy appears to be raised in energy is likely just an artifact of centering the orbitals on the average energy in both cases. Alternatively, the diagram could be based on the assumption that as the axial ligands move away from the metal, the equatorial ligands move closer, increasing the strength of the sigma and pi bonding of those ligands. In the pi case, this would raise the energy of all three t2g orbitals because all of them can engage in pi bonding with equatorial ligands, but the dxy would go up the most since it can pi bond with all four equatorial ligands, whereas the other two can only pi bond with two of the four equatorial ligands.
UPDATE: Here's an image from some old lecture notes showing the difference in convention of alignment between the "MO" view and "d-orbital view" in which the "d-orbital" diagram recenters the energies around the average, even though only the dxz and dyz MO (or more properly the antibonding orbitals primarily composed of them) energies actually changes:

We are not trying to maximize overlap in this diagram, quite the opposite. You can imagine it rather as pushing negative charges, the negative ligands ions, unto the electrons on the central atom while looking at the orbital energies during this process. That is a repulsive interaction between two negative charges, that raises the energy of orbitals that closer to the negative charges.
The $d_{xy}$ orbital and the $d_{x^2-y^2}$ orbitals are in the xy-plane with 4 ligands close to them. The $d_{yz}$ and $d_{xz}$ orbitals are in the yz/xz -plane where 2 ligands are further away. This means they go down in energy, as the repulsive interaction with the approaching ligands is lower. The $d_{x^2-y^2}$ is actually the "worst" energetically, since it points directly towards the 4 closest ligands. This is why it is the highest in energy. The $d_{z^2}$ orbital goes down since we just increased the distance with the ligands.
"MO" explanation:
The explanation for the $d_{xz},d_{yz}$ and $d_{xy}$  remains the same. You are correct when you say that their symmetry doesn't fit any bonding ligand orbital symmetries. The lowering and raising of the energies is purely due to the new geometry. When we go from octahedral $O_h$ symmetry to $D_{4h}$ symmetry, we lose the energetic degeneracy due to symmetry. I.e. the degeneracy in the octahedral geoemetry is due to the fact that these orbitals form a basis for the  the irreducible  $T_{2g}$ representation. This degeneracy of the set $(d_{xz},d_{yz}, d_{xy})$ is lifted in $D_{4h}$. The orbitals belong in $D_{4h}$ to $E_g$ $(d_{xz}, d_{yz})$ and $B_{2g}$$(d_{xy})$. There is no symmetry reason for these two different sets to be degenerate in $D_{4h}$, i.e. the geometry with the elongated z-bonds. And the argument for there relative energy ordering is just the same as given before. These nonbonding orbitals are more stable if they are further apart from negative charge, i.e. the ligands. When we draw these plots we typically assume total orbital energy conservation, and to achieve this we must push one up if we lower two others. I.e. the ones in the z-plane go down and the other goes up such that the "center of energy" remains constant.
The $d_{z^2}$ orbital shown in your figure corresponds to an antibonding molecular orbital. By increasing the distance of the ligands by elongating the z-distance, the overlap becomes smaller and the energy splitting between bonding and antibonding becomes smaller, i.e. the bonding orbital(which is not shown in your figure as it lies way below the d-orbitals) goes up a bit and the antibonding goes down a bit. The $d_{x^2-y^2}$ was in the octahedral geoemetry degenerate with the $d_{z^2}$ orbital as both together formed a basis for the $E_g$ representation. This degeneracy is lifted in $D_{4h}$. In this case we conserve the center of energy of the $E_g$ set. Since we know that the $d_z^2$ orbital goes down, we must have the remaining $d_{x^2-y^2}$ orbital go up.
You should also rather look at actual molecular orbital diagrams instead of looking at crystal field diagrams. Its easy to get confused by mixing up ideas from crystal field theory, ligand field theory and molecular orbital theory.
I used the following character tables

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=604&option=4
http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=904&option=4

Answer to question in comment:
My wording "more stable" was perhaps a bit vague. I mean that the non bonding orbitals are lowered in energy with respect to their energy in the octahedral geometry. Why is this ? Well, the energy does not change due to overlap with some orbitals. It changes due to the lower magnitude of the coulomb interaction of the ligand orbitals with the non bonding orbitals when the ligand orbitals are further away, which is the case in the elongated geometry.
Even if orbitals can't mix, they still affect each others energy simply due to the fact that they represent negative charge distributions. Bringing non mixing orbitals together raises their energy even if their shape doesn't change at all. You could calculate the potential energy due this interaction by
$$
E_{pot} \propto \int dr_1 dr_2 \frac{|\phi_{nonbonding}(r_1)|^2 \sum_{\text{Remaining orbitals}}|\phi_b(r_2)|^2}{|r_1-r_2|}
$$
This positive energy contribution to the total orbital energy decreases with distance between the orbitals and thus the nonbonding orbital energy decreases as the ligands move away. A term like this is also part of MO theory.

Sorry, but I'm looking for an MO theory explanation.

@user103515 I added a small MO discussion.

Thanks for the MO part. "These nonbonding orbitals are more stable if they are further apart from negative charge, i.e. the ligands." What is the MO explanation for this? How can they be lowered in energy when they are nonbonding?
"When we draw these plots we typically assume total orbital energy conservation." Why is this? Do you have a source I can read up on?

@user103515 I have added an explanation why non bonding orbital energies change,  at the end of my answer. I am afraid that i have no source, my answer is based on lectures that i heard quite some time ago.


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Bonding in 1,4-dihydro-1,4-diborinine
In most aromatic compounds (e.g. benzene), their $\pi$-bonding molecular orbitals are completely occupied and their antibonding molecular orbitals are unoccupied. I'm wondering if there are any compounds where this is the case, but where the highest energy bonding MOs are only half-filled, so they contain unpaired electrons. My best guess at a molecule with that property is 1,4-dihydro-1,4-diborinine, as it has the same geometry as benzene (I think), including the adjacent p-orbitals, but it has two fewer electrons. Does this molecule still exhibit ring currents like benzene? Does it show the same type of delocalization?

A molecule with half-filled HOMO will be antiaromatic.

@ivan You may be right, but it's not as simple as it seems.  Boron without the ligand does an entirely different thing.

The given dihydroborinine is indeed anti-aromatic if is it fully conjugated and planar, but counting pi electrons in a conjugated ring is not as simple as it seems when you have boron in the ring.  If the boron is not joined to a ligand, as in (mono)boronine or unhydrogenated diborinine, it could pass its electron deficiency into the sigma bonding framework to preserve pi electron aromaticity.
The basic (or Lewis acidic) case of borinine
Borinine or borabenzene, $\ce{C5H5B}$, is two electrons short of a well-known isostructural aromatic compound, pyridine.  We might suppose that as in pyridine, the sigma bonding and non-bonding orbitals including the outward-directed orbital from the boron must be filled before pi electrons can be introduced, which would leave borinine with only four pi electrons.  And we might be wrong.
Raabe et al. [1] perform an extensive quantum mechanical study of borinine molecules.  They find that in fact there are six pi electrons, the ring is aromatic, and it is the sigma electron framework that is electron-deficient because the outward-pointing boron-based orbital is empty.
To understand this, we look at the structure of borinine given in the above reference, from which the figure below is taken:

The above figure is one of several calculation results, but the all show similar results.  Not only is the ring highly distorted, it is distorted oppositely from what one might expect from the boron atom being inherently larger than the carbon atoms.  The boron is displaced inwards from where it "should be".
This inward displacement means the lobes from the adjacent carbon atoms are overlapping the "empty" outward boron-based orbital.  This overlap creates delocalized bonding around the boron and makes the empty orbital sigma antibonding, whereupon it becomes the LUMO.  The low-energy sigma orbitals are filled after all, but they do not include that outward-pointing boron orbital.  Thus borinine retains six pi electrons.
We should not conclude that borinine is therefore stable.  The pi electron component is stabilized by aromaticity, but with the overall electron deficiency in borinine this comes with an electron-deficient and highly distorted sigma-electron framework.  The antibonding orbital referenced above is too high in energy to be occupied in borinine as such, but it is low enough to combine with an electron pair from a Lewis base, such an an amine.  The resulting adduct (called a boratabenzene when the base is an anion) is the typically isolated product.  With the outward-pointing boron lobe now occupied, boratabenzenes and neutral Lewis-base adducts resume a more typical benzene-ring structure.  They still have six pi electrons, and thus remain aromatic, because the base contributes both electrons to its bond with the boron.
Going for two
I do not have a similar reference for diborinines or their hydrogenared derivatives.  However, Wahab et al. [2] provide nucleus-independent chemical shift (NICS) values for several monocyclic rings which they use in the simplified computational scheme they discuss; these are shown in the figure below, taken from this reference.

As many readers know, NICS models the inherent magnetic shielding (negative value) or deshielding (positive value) within a conjugated ring, these opposing signs then being interpreted as showing aromaticity or anti-aromaticity respectively.  Thus the unhydrogenated 1,4-diborinine is aromatic and even beats several more familiar aromatic heterocycles.  The delocalized bonding around the boron noted above apparently applies at each boron atom here, enabling the aromatic ring structure to be retained.
However, the dihydro compound shown at the right of the figure has become antiaromatic.  As was also noted earlier, when the boron is bound to a ligand the delocalized bonding around the boron is lost and we can go back to counting electrons as we would in an ordinary benzene ring.  However, whereas in Lewis base adducts both electrons needed to bond the boron to its ligand come from the ligand, in dihydrodiborinenes the hydrogen ligands provide only one electron apiece and thus the remaining two electrons to form these bonds have to come from "robbing" the pi system.  Thereby aromaticity in the pi system of dihydrodiborinines is lost, as seen in the NICS becoming positive.
Nobody else like you
Ultimately, we have to be careful rendering molecular-orbital structures in boron compounds because boron, uniquely among nonmetallic elements, lacks sufficient valence electrons to fill its orbitals through pairwise electron sharing alone. The unique bonding characteristics boron often adopts to compensate for this deficiency must be taken into account in order to achieve a proper rendering.
References
1.
Gerhard Raabe, Wolfgang Schleker, Eberhard Heyne, and Jörg Fleischhauer , "Borinine (Borabenzene): Its Structure and Vibrational Spectrum. A Quantum chemical Study", Z. Naturforsch. 42a, 352-360 (1987).  Link.
2.
Alexandra Wahab, Felix Fleckenstein, Stefan Feusi and Renana Gershoni-Poranne, "Predi-XY: a python program for automated generation of NICS-XY-scans based on an additivity scheme" Electronic Structure 2, 47002 (2020).  Link


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Electron Configuration confusion
I'm getting so much conflicting information!! When I practice electron configuration and google to check my answers I often see configurations that list the 3d orbital before the 4s orbital (apparantly the anomaly that the 4s is actually lower energy than 3d)
and 4d written before 5s which is apparantly another anomaly. I'm assuming the order is important.
1s
2s   2p
3s   3p   3d
4s   4p   4d   4f
5s   5p   5d   5f
According to this diagram 5s comes before 4d and 4s comes before 3d however when I google electron configurations this rarely is shown for example Kr is shown as {Ar} 3d10 4s2 4p6. As Ar finishes at 3p6, the configuration should restart at 4s, but online it's always 3d which restarts.
Another example is Iodine 53. According to my diagram it should be {Kr} 5s2 4d10 5p5 but when I check online its always 4d10 5s2 5p5
Does this matter?

I would suggest that you do not worry too much about it is just an exercise. Practice the simple ones. I would rather ask the teacher, how do you experimentally or theoretically determine the electron configuration even for simple elements. You will be surprised that there will be no concrete answer (I am searching for the answer for years). So why worry about Pt which is a very complex atom.

The list of electron configurations is given here by NIST. https://math.nist.gov/DFTdata/atomdata/configuration.html

While writing out the electronic configurations, we usually write them in the order of 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f and so on. This method of writing the electronic configuration only represents the traditional increasing order of shells ( 1 2 3 4 or K L M N ), and nested within each shell are the subshells ( s p d f).
This is only the usual method of writing the electronic configurations; it does not reflect the order of increasing energy of orbitals. The order of energy is given by the Aufbau's Principle or the building-up priciple.
In essence, according to the rule, for comparing energies of orbitals of all elements, except hydrogen, we compare the values of the sum of (n+l) ; where n is the priciple quantum number ( corresponding to K=1 L=2 N=3 etc. ) and l is the azimuthal quantum number ( corresponding to s=0 p=1 d=2 etc. ). Greater the value of this sum, more is the (relative) energy associated with that orbit. Moreover, if this sum is same for two orbitals, then we compare the value of n for those orbitals.
So, for example, you mentioned the anomaly in the energy order of 4s and 3d. Using this rule, for 4s we have n=4 & l=0, giving a sum of 4, while for 3d we have n=3 & l=2, giving a sum of 5. This gives us a greater energy for 3d, which shows us that it confirms to the rule rather than being anomalous.
Further, consider the case of 4s and 3p orbitals; the n+l sum is same for both (4 in each case), but since 4s has a greater value of n than 3p , it is more energetic.
Now, electronic configurations are important in the order 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f when we are considering cases of ionization ; so, if you want to make a +2 ion, of say, Fe, then the two electron will be removed from the outermost orbitals, i.e. 4s. However, in cases where we are concerned with the actual energies, we must subscribe to the Aufbau's Principle.

I would like to understand your statement. For example, the orbitals $2p$ and $3s$ do have the same value of $n+l$. For $2p$, the sum $n+l$ is $2+1 = 3$. For $3s$, the sum $n+l$ is $3+0 = 3$. So, with your criteria, the $3s$ orbital should be more stable than the $2p$, and should be filled before the $2p$. It is not the case. What is wrong in my reasoning ?

@Maurice the 3s orbital would be more energetic; how does this imply that it would be more stable?

Why is the $2p$ first filled, before the $3s$ ? Both have same $n+l$ value. In that case the principal quantum number decides. So $3s$ should be filled before $2p$. What is wrong in my development ?

@Maurice Exactly, the principal quantum number decides; as a result, 3s would be more energetic than 2p, and hence should be filled after 2p. Clearly, we would first like to fill up electrons in a manner so that first the less energetic orbital (2p, in this case) is filled, and only then the electrons should be filled in the more energetic orbital (3s, here). I don't understand as to which particular argument of my answer leads to the inverted conclusion. Could you please clarify?

I misunderstood the meaning of your word "energetic". For you it means "higher in energy". For me it was "more stable" But now I understand what youweanted to state. Thank you.


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What are the criteria for a molecule to be chiral?
What are the criteria for a molecule to be chiral?
(We frequently get questions of this type, so this is an attempt to construct a suitable general answer for duping.)

Related: What is the perfect definition for chirality?

Mirror images
The very first definition of a chiral molecule is one where it is not superimposable on its mirror image. Therefore, one of the most straightforward ways to determine chirality is to construct the molecule and its mirror image (perhaps via a model), and then to see whether they are superimposable.

Symmetry elements
A fully equivalent definition, rooted in molecular symmetry, is that a species is chiral if it does not have any improper axes of rotation.
An improper rotation is a rotation of the species (say 90° clockwise in the plane of the page), followed by a reflection perpendicular to the axis of rotation (in this case, the page). The improper rotation is generally denoted by $S_{n}$ where $n$ is the number of times you need to rotate to get back to the start, e.g., $n=4$ for a quarter rotation (90°).
So, molecules where you can rotate and reflect them to make them look like the original are not chiral. In other words, it's easier to say when a molecule is achiral. Let's look at some examples of improper rotation that make molecules achiral.
The simplest case to look at is $S_{1}$, but no one calls it that. This is a trivial rotation (a rotation that does nothing) followed by a reflection, or just a reflection, usually denoted by σ. Molecules with planes of symmetry are achiral. This molecule, (2R,4S)-2,4-dichloropentane, is achiral because there is a plane of symmetry that reflects the left side onto the right side.

Another example is $S_{2}$, rotation by 180° then reflection. Generally, we also don't call it that. This is the same as an inversion through the center, which is frequently denoted $i$. Molecules with centers of symmetry are achiral. The following molecule has a center of symmetry. You can also see that rotating it 180° and then reflecting through the page gives the same thing back. For clarity, the mirror image is shown on the right. You can rotate it to see that it's the same.

The above two are the most common symmetry elements that make molecules achiral. Occasionally, you'll run into a higher-order improper rotation axis, like $S_{4}$ in the following example. Rotation by 90° and then reflection through the page gives back the original molecule. The mirror image is shown on the right, and you can rotate it to see that it's the same, thus making this molecule achiral. There are more examples here.

So, looking for the above features will let you know if something is achiral.

Stereogenic elements
To figure out whether or not you even need to consider chirality, you will to find a stereogenic element. Keep in mind that having a stereogenic element doesn't mean that molecule is chiral, because it could have one of the symmetries discussed above and thus, be achiral.
Molecules with stereogenic elements should be considered for chirality, but are not guaranteed to be chiral. In other words, having a stereogenic element is necessary but not sufficient to make a compound chiral.
The main source of chirality that introductory students should worry about is the stereogenic center, usually a carbon with four different substituents:

More generally speaking, a stereogenic center such as this falls into the category of point chirality. Other examples of point chirality include inorganic complexes such as the octahedral $\ce{[Co(en)3]^3+}$ (link, also illustrated below), as well as substituted adamantanes.

Apart from point chirality, here are also two other ways to generate chirality, which are typically seen in more advanced courses.
The first is axial chirality (Wikipedia, IUPAC). This is a twist that can be described as either left-handed or right-handed. These include unsymmetrically substituted allenes, unsymmetrically substituted biphenyls, rings with exocyclic double bonds, and helicenes (image adapted from Wikimedia):

The second is planar chirality (Wikipedia, IUPAC). This is when a plane, which has high symmetry, has its planes of symmetry successively broken by substitution. An example is this borabenzene metal complex. A borabenzene has two planes of symmetry, but these can broken by putting a methyl group on the ring and then binding a metal to the face of the ring.

@orthocresol the 'unsymmetrically substituted biphenyls' noted here is known as an atropisomer.

@z1273 I'm not sure what point you are trying to make here. It's still axially chiral.

My intent was to add a point of precision (emphasis in fact) with the hope of enriching the post, not as a critique. As I worked with an atropisomer molecule, I can assure you it is ill-appreciated out there. I hope you recognize my positive intent here.


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Aromaticity of Caffeine

In the Caffeine molecule, is aromaticity confined only to the 5 membered ring? The entire molecule is planar and the 6 membered ring also seems to have 6π electrons.
The entire molecule has 10π electrons.(inside)

related: https://chemistry.stackexchange.com/q/54268/102629

This question does not mention anything about the 6 membered ring except that it is planar

This link has an answer: Caffeine.

You should condense the content of that link into an answer here.

No. The 6 membered ring is also a part of its aromaticity.
This can be better understood by considering the Zwitterion of Caffeine.

The entire molecule is planar, conjugated and has 10π electrons(4n+2)
Reference:Slides by Nevina Keskineva


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Nature of a1g Molecular Orbital Interactions in Ferrocene
While learning about ferrocene I've come across this molecular orbital diagram.

What particularly confused me is the nature of the a'1g orbital. My professor taught that a'1g and e2g are considered HOMO bonding orbitals, but this MO diagram clearly shows the a'1g as an MO with antibonding character.
To make sure it wasn't just a bad MO diagram, I looked for other MO diagrams, but they consistently display a'1g as a higher energy orbital compared to Iron's $3dz^2$ orbital.[a][b] So, slight anti-bonding character, right?
The problem is, looking at bond-length comparison, the a'1g orbital is slightly bonding. The Fe-Cp distance for ferrocene is 1.66 Å; for ferrocenium, it's 1.68 Å.[c] Given that the removal of an electron from the HOMO a'1g orbital resulted in the lengthening of Fe-Cp distance, it should have bonding character.
So what is going on here? Why is there this discrepancy? How are these orbitals actually interacting with each other?
[a]http://alpha.chem.umb.edu/chemistry/ch612/documents/ConstructingpiMOdiagramssandwichcomplexes.pdf
[b]https://doi.org/10.1021/ja00744a011
[c]https://doi.org/10.1107/S0108270189005883

HOMO stands for highest occupied molecular orbital. Of the nine occupied MOs in that diagram, the one that is highest in energy is a'1g. Your assumption that the HOMO cannot be antibonding is not correct.

@Zhe I understand that. My question is about the seeming discrepancy between its position on the MO diagram (showing antibonding character) and physical analyses (Fe-Cp distance comparison showing bonding character).

But ferrocenium is charged. What's more important in determining structure, changes in electronic structure or charge/charge repulsion?

@Zhe If anything, I would think ferrocenium should have greater coulombic attraction with the negatively-charged cyclopentadienyl ligands and result in shorter bond lengths. Is that wrong?

My intuition says it's more complicated than that. You're assuming that even in the complex, the Cp's are strongly charged negative and the positive is more localized to the Fe center, and I'm not convinced that that's the case.

@Zhe That's a fair point. The paper that I cited for the bond length comparison attributes it to the molecular orbital's bonding nature though, so I think I'll need to see some sort of quantitative evidence for that explanation before I'm convinced. It really does seem like people consider it to be a bonding orbital.

Possibly related: https://chemistry.stackexchange.com/questions/35641/why-is-the-bond-order-of-co-3-5

Just because an orbital comes out higher in energy than another orbital in a different environment does not mean it is anti bonding.

This is a reasonable guess and it can be wrong.
First, the energy of the molecule does not correlate to the bond length. In hyperconjugation, the electron in C-H sigma bond is donated to the empty p orbital of the carbocation. And the consequence is that the carbocation is stabilized and the bond length of the C-H bond is increased.
Does an increase in bond length cause the molecule has higher energy? Definitely not.
A decrease in bond length does not mean the molecule has lower energy. And this also applies to ferrocene and ferrocenium.
$a_{1g}$ is only the symmetry of the orbital which the $d_{z^2}$ orbital interact with the $Cp^-$ ligands. As they are in the same symmetry, they will interact. Electrons in the orbital will bring the ligand closer to the metal which causes a shorter distance.

Your image clearly shows that the orbital is not antibonding. Unfortunately your last sentence isn't really true. Although it often appears this way.

I've recently done an NBO calculation on the staggered conformation of the ferrocene molecule, using the 3-21G basis set, using the embedded Gaussian engine on the free version of the WebMO application.
This calculation is easily reproducible, so I am not going to post the exact details here- the important conculsion from the calculation is that there are three nearly fully occupied "lone pair"s of bonding character (denoted as LP) and six partially occupied "lone pair"s of antibonding character (denoted as LP*) on the central iron atom. Three filled lone pairs correspond to six electrons, exactly what iron(II) should have.
Therefore, the a1g lone pair is likely not of antibonding character.

Is there a particular reason why you chose to make this post community wiki? Usually these are for large-scale community efforts such as compilations. Normal posts can already be edited by all, just like a community wiki, but will retain attribution to the original author i.e. you. Some figures would also be useful.

I posted this on community wiki because, as a chemistry beginner, felt that intuitive, VB-based models of coordination complexes should be more known to chemistry beginners (for the sake of simplicity and interlectual digestibility)) before they venture into the more accurate MO-based models.

Community wiki isn't to do with the level of the post or the target audience; actually the content of the post has no bearing on whether it should be CW or not, they're orthogonal to each other. CW is a marker to indicate that an answer is meant to be a collaborative effort between multiple people.

I'm going to convert all your answers back to regular answers as opposed to community wiki answers. There's no apparent reason why any of this needs to be CW; if you genuinely think it should be a collaborative effort, then make it clearer as to what aspects of it you want people to contribute to.

Can you elaborate your answer please? You say the calculation is easily reporducible, but not everyone is a computational chemist (not to mention that not even all computational chemists deal with QM calculations) so it would be good if you could explain the energy levels and your data briefly here.

I literally pointed out the app used, the engine used, the basis set used, and the conformation of ferrocene used; the bond lengths and angles in the lattermost were omitted in this answer since they are already known and conveniently available from, say. Google Scholar or references linked on Wikipedia.

It doesn't specify the method used and the version of the program. And if it's the built-in version of NBO, then it's outdated by a few decades and known to fail for organometallic compounds.


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Why does water splash if you put it into a hot pan?
I had kept my tea pan on the stove (At high)for about 30seconds . I was making tea and getting the water to put in the tea pan. There was nothing in it before that .After that , I quickly poured some water in it and boom.
The pan kind of like jumped out of the stove. The water from the pan instantly became steam and some droplets of water touched my kitchen ceiling.
The scene looked like a bomb exploding. As a single droplet of water just touched the pan , the pan just went of control.
Also , you hear a very noisy sound if you do it.
What could be reason for this little amount doing such a reaction ?
Yes, I do not advise kids to do without parents or some elders support.

Google: Leidenfrost effect

When you pour water on a very hot plate, much hotter than $100°$C, the water cannot touch the plate. It boils immediately, even before touching the plate. This makes a sudden and huge production of vapor, which takes much more room than a liquid : this throws the rest of the water anywhere in the vicinity with a great speed. It looks like an explosion. Some water may even be thrown to the ceiling.
Another famous experiment can be done directly on the electric plate, without a pan. If you pour maybe one milliliter water on a red hot plate, which has a small cavity in its center, the water floats above the plate in the cavity without touching it. For a couple of seconds, the vapor produced is not strong enough to eject the drop, which stays calmly in levitation. This effect is called Leidenfrost effect, or calefaction.

Ok. But how is it that vapor is able to move the pan?

I don't know the form and the geometry of your pan. Is it like a kettle ? Probably the hot water was ejected with such a high speed that it touched the upper part of the kettle, and pushed it away.

I have added an picture in edit for the pan.

@Srijansingh7 it must be some asymmetry in the way water was ejected, ie pushing the wall in one preferable direction


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Ascorbic acid and aldehydes: Reaction and Influencing Metals
I've researched online and I've read that ascorbic acid can both promote oxidation and reduction depending on the conditions (I've read about trace metals but I've never found which metals affect its chemistry and which ones don't).
So if I mix ascorbic acid with a bunch of different aldehydes, will it react with all if them? Some of them? Will it reduce them (into alcohols?) or oxidate them (into carboxylic acids)? Do metals have any influence on what happens? And if so, which metals do?

So, I can generally just consider it will reduce stuff? Will it reduce all aldehydes? Do you maybe know something about the metals that will reduce it?

Try longer thinking about and  searching for answers to all your questions, before asking. Focus on one question at time. Providing answers is much more resource demanding than raising questions .

You would need very strongly reducing compound to be able to reduce ascorbic acid. Aldehydes are not the case.
Ascorbic acid would not reduce aldehydes like ethanal either, as can be seen by comparison of their respective redox potentials below:

Redox system
$E^{\circ} [\pu{V}]$

Ascorbic/dehydroascorbic  acid
+0.35

Ethanal/ethanol
-0.197

Transition d-orbital  metals with multiple possible oxidation numbers very frequently act as redox catalysts. Both neighbour oxidation numbers may be common ( $\ce{Fe^{+III}/Fe^{+II}}$, $\ce{Mn^{+III}/Mn^{+II}}$,
$\ce{Mn^{+IV}/Mn^{+III}}$ ), or one may be uncommon ($\ce{Cu^{+III}/Cu^{+II }, Ag^{+II}/Ag^{+I}}$)
Catalytic effect is often related to the opposite charges of reagents and catalysts. The typical example is reaction of oxalate and permanganate anions, catalyzed by manganese cations.
A good metal catalyst has standard redox potential comparable with respective redox potentials of the oxidant a/o reductant.
E.g  $\ce{Ag+}$ may act as a catalyst during oxidation by peroxodisulphate, as $\ce{Ag^2+/Ag+}$ has the similar redox potential. Similarly, $\ce{Mn^2+}$ ions act as the catalyst during  oxidation of oxalic acid  by permanganate in acidic solutions, as $\ce{Mn^2+/Mn^3+}$ has similar redox potential as $\ce{MnO4-/Mn^2+}$.

Thank you Poutnik. You're awesome! :)


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Solubility of Aluminium Fluoride in other other fluorine-containing solvents
I encountered a problem which required me to compare the solubility of $\ce{AlF3}$ in various solvents, namely $\ce{HF}$, $\ce{KF}$ & $\ce{KHF2}$ .
Further, this problem neither supplied, nor expected me to utilize  any solubility data. Infact, the problem only asked for the expected order, discounting any exceptional factors that may affect the actual numerical data.
My thoughts regarding this are as follows;

As $\ce{AlF3}$ is a Lewis acid, I expect it to have higher solubility in solvents which have lone pair containing donor ; possibly the fluoride ion from $\ce{KF}$ may be such a species
Further, in aqueous acidic solution, I expect a lower solubilty owing to common ion effect lowering the solubility
I am unsure and unclear regarding the action of $\ce{KHF2}$
Perceptions regarding H-bonding and complex formations also seem to apply, but I am unable to formulate these coherently

Given these points, can someone help me to organize my thoughts? I wish to know the lines along which I should guide my thoughts.

The Handbook of Chemistry and Physics mentions that $\ce{AlF3}$ is soluble in water, and that $\ce{AlF3·3H2O}$ is not soluble in water. So it is difficult to say if any of these substances will be soluble in $\ce{ KF, KHF2}$ or $\ce{HF}$.

See: https://chemistry.stackexchange.com/questions/65861/why-is-alf3-soluble-in-anhydrous-hf-only-in-the-presence-of-kf?r=SearchResults

@NilayGhosh thanks for the link to that question ; the answers over there answer my query.


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Which commonly available UV light source to use with fluorescein
I want to start experimenting with fluorescein.  I have a 353 nm UV light source, but am thinking that it might be too short since it looks like peak excitation occurs at 494 nm.  So am I correct in assuming that I should buy light sources that emit light at around 494 nm (480 nm for example)?

494 nm is not ultra-violet.It is in the blue region. So you don't need an UV source to start experimenting with fluorescence. A purple or blue light would do the job. A source at 480 nm is OK.

You have to first look at the excitation spectrum of fluorescein. Where is the peak maximum? You would rather avoid UV due to safety reasons.

Some blue or white LED source could be good enough.

Your UV source should be fine. If not there are plenty of near-visble UV, low output sources (widely used for revealing the UV security marks in banknotes) and there are even cheap, low power near-visible UV handheld laser pointers or LED sources.

BTW, if you use a blue or indigo light source, look through a yellow filter to greatly increase contrast - it blocks blue, but lets yellow-green fluorescence trhough.

This page shows the spectra https://omlc.org/spectra/PhotochemCAD/html/010.html so you can see that 353 nm is a poor wavelength to use. 450 nm ish would be better, selected with an interference filter and an arc or tungsten lamp would be simplest otherwise find some Wratten filters or of course a 450 nm diode laser.


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Are diamonds as chemically inert as glass?
Every beaker that I have seen in a lab is made of glass. Glass, as I understand, is made of SiO2, or silicon dioxide, which has a crystalline structure that is similar to diamond. This property causes both diamonds and glass to be chemically inert. My question is, (ignoring the costs and technicalities of production), could beakers be feasibly made out of diamonds? Following up on this question, are there solutions that can dissolve glass effectively? What about solutions that can dissolve diamonds?

No feasibility as there is no significant chemical advantage over the glass and the cost involved.

Don't omit that burning diamond is difficult but it can be done.

Diamonds are chemically inert. They burn at very high temperatures, but apart from this, they don't react. They are insoluble in all sorts of solvents, with the exceptions of iron above its melting point (~1500°C}. Glass is not really chemically inert. It is slowly attacked by concentrated solution of NaOH, or by pure NaOH above its melting point. It is also attacked by hydrofluoric acid HF.

Does diamond really dissolve in iron, but as what? By dissolution, I mean one can recover them after dissolving iron in acid. Never

Iron with dissolved carbon is steel. Can you get carbon from steel by dissolving iron in an acid? I don't think so.

When diamond is dissolved into iron, it cannot be recovered later on after dissolving in an acid. The only substance to be recovered is amorphous carbon, or iron carbide.

Diamonds are less chemically inert that lab glassware. For example they will burn at temperatures which would merely melt lab glassware (which is done to manipulate it into different shapes.).

@matt_black It's not a competition. ;)

"functionalized diamond" gives five million hits on google. ;)

The comment above makes me to want point out that dissolving something it just means getting it dissolved. Finer details and mechanisms are not included in asserting that X dissolves Y.


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What's the chemical reaction between NaOCl and Brilliant Blue dye?
Based on my research NaOCl dissociates into Na+ and OCl-. Looking at the photo, how does the OCl- ion and the brilliant blue dye structure end up as the product? How does the OH attach itself to the carbon group and what happens to the Cl- afterwards? I'm pretty sure since the C group has a double bond it needs to be broken by an addition reaction but I'm not sure how.

I don't trust this product structure, to begin with. What's with that trivalent carbon to the right?

I doubt the reaction occurs as written. The transformation from left to right is not due to the $\ce{OCl^-}$ ion. No. It is due to the addition of $\ce{H2O}$ and this addition is improbable because the quinonic ring would loose its aromaticity in the reaction. I would rather think the hypochlorite ion acts on one of  the Nitrogen atoms.

I think you guys are right. It seems strange that the Cl- isn't at all involved in the reaction. This is the source I found the mechanism from if it helps: https://drive.google.com/file/d/1dI9CoBMX14Rk9j4m0atFnw-axXZnV_hm/view?usp=sharing

If you read the reference, they say that the product of the reaction is "not perfectly known". And indeed the formula of the product has no influence on the exercise. The only thing to know is the fact that bleach destroys the dye. And measuring its absorbance decay allows you to know the rate law of this reaction. This is the only important.information.


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Why doesn't ionization energy decrease from O to F or F to Ne?
I know that in general, the first ionization energy increases across a period due to increasing nuclear charge, reasonably constant shielding & decreasing atomic radius.
From N to O, however, the I.E anomalously decreases. The reason usually given is that the spin pair repulsion between the 2px2 electrons in oxygen outweigh the effect of the increased nuclear charge.
However, the additional electrons in F & Ne are also spin-paired. Why doesn't the extra spin pair repulsion outweigh the extra nuclear charge in this case? http://www.chemguide.co.uk/atoms/properties/ies.html#top says

Between nitrogen and oxygen, the pairing up is a new factor, and the
repulsion outweighs the effect of the extra proton. But between oxygen
and fluorine the pairing up isn't a new factor, and the only
difference in this case is the extra proton. So relative to oxygen,
the ionisation energy of fluorine is greater. And, similarly, the
ionisation energy of neon is greater still.

But I don't understand how being a new factor or not makes any difference on its effect. It's not like the atom adapts to the new factor in O, reducing its effect its effect in F & Ne

https://chemistry.stackexchange.com/questions/32363/anomalous-trends-in-ionization-energy https://chemistry.stackexchange.com/questions/91179/is-the-first-ionization-energy-in-oxygen-slightly-more-than-nitrogen

My goodness, that second question and its answers are a real mess, for a topic that is literally one of the first things taught in chemistry. And it has 20k views...

It's a good question, and I agree that it can be quite confusing, but Chemguide is (at least partly*) correct about this.
I'd phrase it this way, which I hope is clearer. First, think about what would happen if the "paired electron" factor was not a factor at all. We would then expect that, going from B to Ne, there would be a roughly constant increase in ionisation energy. It would look something like this. Please don't take the exact shape or numbers too seriously; they don't mean anything, only the trend does.

This is obviously not the reality, because in O, F, and Ne there is the additional factor to consider, which is that the electron being removed is already paired in an orbital, making it easier to remove. This means that the ionisation energies of O, F, and Ne are each decreased by a roughly constant amount (red arrow).

You can see that we recover the observed trend in ionisation energies, and we have indeed accounted for the pairing energy in all of O, F, and Ne.
There are two alternative ways of looking at it which may "click" better for you. I'll describe these in text.

Going from N to O, the nuclear charge increases by 1 unit, which on its own should cause the ionisation energy to increase. The pairing energy consideration, however, overrides this. So far so good.
Going from N to F, the nuclear charge increases by 2 units, which on its own should cause the ionisation energy to increase roughly twice as much. Therefore, to make IE(F) smaller than IE(N), you would need two counts of the pairing energy. But we can't do that: you are only ionising one electron each time, so you can only subtract one count of the pairing energy.
The argument can be easily extended to N vs Ne.

Going from O to F, the pairing energy does decrease the ionisation energy of F. However, the pairing energy also decreases the ionisation energy of oxygen, so this factor on its own cannot be used to say that IE(F) should be lower than IE(O), as it affects both. The same is true of F and Ne.

* Exchange energy, which Chemguide does not cover, is also a consideration. However, that's a smaller effect, and it's not necessary to invoke that to answer this question.

this question has been asked on this site several times and this is the only question giving the correct answer in a simple and obvious way. I think I didn't realize that the spin-pair repulsion doesn't actually stack up and is used just to overcome that 1 proton gained from N to O each time.


text:
Do most elements generally exist in a planet during early formation, or is it a small subset?
I'm interested in whether there are a small subset of elements that are by far the most common in a planet's early stages of formation and a wider range are formed slowly over time through chemical reaction, or if a very wide range of the periodic table can be initially found from the start.
Obviously every planet it unique, but I'm asking about whether there's a common scenario / elemental ratio present in the hot cataclysmic early stages of a planet's formation, or rather it's pretty much diverse from the start. Do most elements generally exist in a planet during early formation, or is it a small subset?

If you mean "stars", this earlier answer may have some useful info and links.

Assuming the planet wasn't created in the Universe's early infancy (say a few hundred million years after the Big Bang), it's more likely that there are less elements on a planet as time goes by. See extinct radionuclides. That is, unless your planet goes on to develop an advanced technological civilisation...

The only process that can create elements on a planet is radioactive decay, but that removes more elements than it creates (and mostly add to the stock of elements already present). But, the mix of elements on the surface of a planet is heavily influenced by geophysical processes which concentrate some in the surface and others in the core.

Elements are not formed through chemical reactions or in planets (except through nuclear decay) . They are formed in stars through nuclear reactions.
Your heavier elements can't even be formed in your typical star. It takes a supernova.
So pretty much everything the planet is made of is there from the start.

Exception is helium, which is from alfa decay ( as helium of big bang origin escaped ) and beryllium + boron from cosmic ray spalation of
nuclei.

Nice answer, also add lead (Pb) which is constantly forming among other radioactive decay products like radon gas.


text:
Change in Enthalpy in Free Expansion of Ideal and Real gas against Vacuum
By free expansion, I am referring to gas allowed to expand freely against vacuum in a Joule Expansion.
If gas is ideal then change in Internal Energy '∆U' and change in Enthalpy '∆H' is zero.
(By ∆H = ∆U + nR∆T).
But I was wondering whether '∆H' would still be zero for real gas?
(I know that ∆U will be zero for real gas. I just want to ask about ∆H)

I know about ∆U. I want to know about ∆H.

I didn't understand your point. ∆H = ∆U + nR∆T is just applicable to ideal gases, right?

So how do I know about ∆H? (for real gas)

See Joule-Thomson effect  for real gas vacuum expansion and van der Waals equation for pV=f(T,p) dependence. As $\Delta H=\Delta U + f(T,p)$. Also look at https://en.wikipedia.org/wiki/Joule_expansion

@Poutnik Joule Thomson effect is a bit different from this as it has constant H but not constant U. Joule expansion is exactly what I am talking about, but its wiki page has nothing on H only U. I had asked this question after reading that only

That's why I asked only about H and not U.

@InfiniteCool23  It doesn't sound like you are asking about a free expansion, since it's not generally the case that $\Delta \text{U} =0$ for free expansions of real gases.  Instead, it sounds like you are asking about an adiabatic free expansion, i.e., a Joule expansion.  If that's the case, you should edit your question appropriately.

@InfiniteCool23 "Technically, it would be called adiabatic only if q=constant throughout the process." Nope.  q is a path variable.  You're incorrectly treating it like a state variable. Initial and final states don't "have heat" the way they have, e.g., internal energy.  q only appears during a process.  And adiabatic doesn't mean q is constant.  It means q is zero (no heat flow) for the process.

@InfiniteCool23  "I think that ΔU=0 is generally the case for free expansion"  Nope.   If you have a free expansion, w=0 but, if it's not adiabatic, q will generally not be zero for real gases.

I am just talking about Joule expansion, then what will be change in enthalpy for real gas?

In a Joule expansion, which is an irreversible adiabatic expansion against a vacuum, $q = 0 \text{ and } p_{ext} = 0$.  Thus, since the only type of work in a Joule expansion is $pV$-work:
$$\Delta \text{U} = q + w = w = -p_{ext} \Delta V = 0 \text{,  always.}$$
And since $\Delta \text{H} = \Delta \text{U}+ \Delta pV$:
$$\Delta \text{H} = \Delta pV \text {, always.}$$
For ideal gases:
$pV$ = constant (at a given $T$), so $\Delta \text{H} = \Delta pV = 0.$
For real gases, we have two cases:
I. At the inversion temperature:
$pV = \text{constant (at a given } T \text{), so } \Delta \text{H} = \Delta pV = 0$.
II. Not at the inversion temperature:
$pV \ne \text{constant, so } \Delta \text{H} = \Delta pV \ne 0$.

In the Joule experiment, $\Delta H$ is not zero for a real gas.  $\Delta H=\Delta U+\Delta (PV)$, and even if $\Delta U$ is zero, there is no physical reason to expect that $\Delta (PV)$ would be zero in this process.


text:
The negative absorbance values in the baseline measurement
In the spectrophotometric analysis(uv-vis-NIR) of ITO Nanoparticles in tetrachloroethane using a double beam instrument baseline was measured with 2 identical cuvette with solvent only. This gives me only negative values for absorbance. Does this mean there is significant difference in transmission between the blank and the reference even if they are the same compound?

With zero measurement noise(s) and an ideal double beam spectrophotometer, the baseline would be zero and perfectly flat. Real instruments are not ideal, measurement noise is not zero and  baselines have some noise, which can give small negative values. Remember that what the instrument does is measure two transmitted light intensities, one for each optical path, then compute a ratio and negative logarithm.

That is why you run a baseline correction just prior of a serie of analogues measurements. And periodically check that all is OK.

Negative absorbance in a double beam instrument has no physical meaning because absorbance, by definition, cannot be negative. It ranges from $0$ to $\infty$.
Instrumentally, all that means is that somehow the sample cuvet is transmitting more light than the reference cuvet. In other words, the measured absorbance of the sample is less than than measured absorbance of the reference cell.  Usually this can happen when the cuvets are not optically matched (i.e., they may look identical physically but their optical transmission characteristics are very slightly different).
Try swapping the reference cuvet with the sample cuvet and see if the readings remain negative again. If you are getting large negative absorbance values with the same solvent, try another cuvet set and if the problem persists, the spectrophotometer must be checked. However, if the negative values are quite small such -0.001 or -0.0002, this is tolerable and you can subtract the baseline from the measured signal in order to correct this systematic error.

As EdV has pointed out the noise issue, check the magnitudes of the negative absorbance. If they are random, small, and you also have positive values, it is not a problem. Make sure there is no drift in the baseline in the scan.


text:
Deciding which group twists out of the plane when ortho effect takes place
I'm going to take a few stills from 3D conformers of the following compounds available on PubChem in which ortho effect is applicable to show what I'm referring to.
2-Nitrobenzoic acid:

N,N-Dimethyl-2-nitroaniline:

2'-Aminoacetophenone:

In the compounds above due to steric repulsions between the groups present, one group needs to be twisted out of the plane of the benzene ring. My question is, how do we theoretically explain which group shifts out of the plane, and which doesn't?
To my understanding, two things would effect this; one would be the extra stabilisation/destabilisation of a group due to not being in resonance any longer once it moves out of the plane of the benzene ring, and the second would be the "bulkiness" of the group itself, which would energetically be measured as the group's A-Value.
For the above three compounds in the same order, from the perspective of A-Values (the text in brackets mentions what actually happens):

($\ce{COOH}$ twists out of the plane) The A-Values are $\ce{COOH} = 1.2, \ce{NO2} = 1.0$, so it makes sense for $\ce{COOH}$ to stay in the plane of the ring.

($\ce{NMe2}$ twists out of the plane) A-Values $\ce{NMe2} = 2.1, \ce{NO2} = 1.0$ so it makes sense for $\ce{NMe2}$ to stay in the plane.

($\ce{COMe}$ slightly twists out of the plane) A-Values $\ce{NH2} = 1.6, \ce{COMe} = 1.17$ so it makes sense for $\ce{NH2}$ to stay in the plane.

(data from here and here)
I understand that A-Values are not the only thing which we need to consider, since resonance plays a part (maybe even other stuff?), but I haven't been able to figure out convincing reasoning as to why in the above examples the group which shifts out of the plane does so. Does the volume occupied by the groups play a crucial part as well, since so far I am yet to come across a compound in which the smaller group (by volume) shifts out of the plane? Are A-Values even relevant to such a scenario?
(I'm looking for a list of factors which need to be considered, whether this can even be explained/predicted theoretically, and/or explanations for the any of the above compounds. I understand that this could be slightly lengthy, so I'd be more than happy for answers which address any one of these, but I feel talking about the three compounds was necessary to explain the question properly, and hence the somewhat broad question.)

I am not sure the 3D conformation stills as reported on PubChem are all that accurate. Although crystal structures are sometimes referenced for specific molecules, some of the displays are in fact models, some of which do not make sense to me.
Here are some observations regarding the three molecules you noted.
2-Nitrobenzoic acid
The PubChem model shows both oxygen atoms of the nitro- group to be precisely coplanar with the phenyl ring. In the crystal structure, these oxygen atoms have a dihedral angle of $125.5^\circ$ and $129.1^\circ$ with respect to the aryl, i.e. away from coplanarity.
The PubChem model shows both oxygen atoms of the carboxylic acid to be precisely normal to the phenyl ring. In the crystal structure, these oxygen atoms have a dihedral angle of $152.9^\circ$ and $162.7^\circ$ with respect to the aryl, i.e. closer to coplanarity.

Pubchem
Crystal structure entry CCSJWHN

N,N-Dimethyl-2-nitroaniline
I could not find the crystal structure of this molecule. However, I’m not sure it makes sense that the N,N-dimethyl amine adopt an $\ce{sp^2}$ character at the nitrogen center, at least as far as the Pubchem model representation is concerned. In such a pose, there can’t be extended resonance from the phenyl to amine as obtained with unsubstituted aniline. Either the methyls are near coplanar with the aryl to allow the pi orbitals to enter resonance or the amine is in $\ce{sp^3}$.
2’-Aminoacetophenone
I could not find the crystal structure of this molecule. What I found was the 4,5-dimethyl variant as 1-(2-amino-4,5-dimethylphenyl)ethan-1-one.
The 2’-Aminoacetophenone PubChem model shows the acetone oxygen to be precisely $30.0^\circ$ out of the aniline plane. In contrast, the crystal structure of 1-(2-amino-4,5-dimethylphenyl)ethan-1-one indicates that oxygen to be precisely in plane with the aniline. This coplanarity may not be surprising in this case as there is an intramolecular hydrogen bond between the two substituents.

Pubchem
Crystal structure entry 1825318 (BEXTUW)

2'-Aminoacetophenone
1-(2-amino-4,5-dimethylphenyl)ethan-1-one

Original Post:

To my understanding, two things would effect this; one would be the extra stabilisation/destabilisation of a group due to not being in resonance any longer once it moves out of the plane of the benzene ring, and the second would be the "bulkiness" of the group itself, which would energetically be measured as the group's A-Value.

I think steric hindrance, you noted as “bulkiness” and estimated with an A-Values is the primary factor. There is little room for compromise on this one as two atoms can’t occupy the same space. The second factor is resonance as you noted. Another influencing factor is the presence of intramolecular interactions, such as a hydrogen bond. Solvation and pH are other influential factors to the ortho effect.


text:
Tanabe-Sugano diagram x-axis question
While reading up on Tanabe-Sugano diagrams I noticed that many diagrams spread over the internet that all seem to come from this Wikipedia article: https://en.wikipedia.org/wiki/Tanabe–Sugano_diagram. Those graphs are all derived from this data: http://wwwchem.uwimona.edu.jm/courses/Tanabe-Sugano/TSspread.html
When I myself create the same graph from the same raw data the x-axis is not the same in regards to scale. My question is, since the creator of the Wikipedia graph left no further documentation, what step am I missing?

related: https://chemistry.stackexchange.com/a/64486/102629

The step that was missing was basically just human error in the creating of the graph (wrong column for x) on top of a 10Dq/B to Dq/B conversion. The Wikipedia graph was correct and the raw data was correct, the creator of the graph on the right (me) made a mistake. My bad.

I'm still a bit skeptical. To me, your plots make more sense; at least they are coherent with what one typically finds in literature (e.g. Atkins' Inorganic Chemistry).

The raw data used 10Dq/B while the wikipedia one used Dq/B. 10Dq/B is more in line with the literature you linked to. That bit is down to a factor 10 difference, and is more personal preference from what I understand. Compare that with the faulty graph that is off by a factor 17 instead and then my error is easier to see.


text:
charge density as a measure of lattice enthalpy & polarizing power?
My official CIE A level textbook, and some mark schemes mention the following:

Ions with the same charge have a lower charge density if their radius is large. This is because the same charge is spread over a larger volume. A lower charge density weakens the electrostatic attraction and thus decreases the lattice enthalpy of an ionic lattice.

From a physics point of view, this can't possibly be right, can it? Modelling the ions as spheres, gauss' law/ shell theorem clearly state that for all points outside a shell/sphere (of uniform charge density), the spheres/shells are equivalent to point particles. Why should the volume of the ion have any affect on the attractive force? And what would, then, be the correct explanation for decreasing lattice enthalpy down the group?
moreover, charge density of an ion is  also said to affect its polarising power. The polarisibing power should depend on the electric field created by the ion which,again, doesn't depend on the size at all. If this too is wrong, why does Be have a greater covalent character than Mg?

"My official CIE A level textbook, …" If it's official, and you quote from it, please add a citation.

https://chemistry.stackexchange.com/questions/145151/charge-density-as-a-measure-of-lattice-enthalpy-polarizing-power

For both of these questions, we can still understand the results by thinking about point charges/applying Gauss' Law.
Charge Density and Lattice Enthalpy
Comparing two ions with the same charge, but different radii - e.g. $\ce{Li+}$ which is smaller than $\ce{Rb+}$, or $\ce{F-}$ which is smaller than $\ce{I-}$ - the size of the radii with determine the effective distance between charges (modeled as points or otherwise).  The larger the radii, the further apart the charges will be, and hence less strongly they'll be electrostatically attracted to each other by Coulomb's law, for energy $E$: $E = - \frac{k q_1 q_2}{r}$
Hence, the smaller ions of $\ce{LiF}$ are more tightly bound with a lattice energy of $\pu{1049 kJ/mol}$ vs the larger ions of $\ce{RbI}$ which has the same crystal structure but a lattice energy of $\pu{632 kJ/mol}$.[1]
Polarising Power of Cations
As pointed out by user Buck Thorn, the induced dipole - a measure of polarisation - is proportional to the applied electric field:
$$
\vec{\mu} = \alpha \vec{ E }
$$
The electric field generated by the polarising cation will be determined by its charge and distance from the anion being polarised.  The electric field, $\vec{E}$, generated by a point charge at position $\vec r$ is given by:
$$
\vec{E} = \frac{1}{4\pi\epsilon} \frac{q \vec{r}}{\left|\vec r\right|^2}.
$$
The larger the cation, the further away from the anion it will be, the weaker the electric field applied to the anion will be, so a smaller dipole will be induced in the anion.
So overall it will have a lower "polarising power" or polarising effect.
[1] Handbook of Chemistry and Physics 101st Edition

[1] Handbook of Chemistry and Physics 101st Edition is paywalled. So many tears shed.

You left a comment below my answer, which was meant for the OP!

Atoms are not static distributions of charge, and the models we use are sometimes heuristic (approximations). It can be useful - if approximate - to describe an atom in terms of a distribution of static charges, and to compare two different atoms by saying they differ by evenly "stretching" the charge distribution in the radial dimension.
Yes, Gauss's law applies (assuming a static charge distribution). Assume we are comparing two ions that differ only in the radial charge distribution (by stretching the distribution). Gauss's law simplifies the comparison, since we need only consider the change in the distance from the ion center to the charged test particle (placed at the surface of the ionic sphere) and not the magnitude of the charge within the sphere (which remains constant; for a uniform charge distribution it can also be assumed to remain centered in the sphere). The Coulombic interaction energy falls as 1/r (the inverse distance between charges) and the electric field as $1/r^2$. The effect of radially scaling the distribution ($r\rightarrow a\cdot r$) is to multiply the energy and the field by a constant ($a^{-1}$ or $a^{-2}$). If you make an ion larger, the distance of a charged particle (outside of the ionic sphere) from the center of the ion increases and both the interaction energy and electric field decrease in magnitude.
Electric polarizability $\alpha$ (here assumed isotropic and written as a scalar - it is generally a tensor) is a measure of the extent to which charges can be separated by an applied electric field, resulting in an induced dipole moment:
$$\vec{\mu} = \alpha \vec{ E }$$
To compare atoms consider the magnitude of the electric field required to induce an equivalent dipole moment. This is determined by the polarizability (or rather its inverse). More polarizable means a weaker field suffices to generate the same dipole moment.
Finally, as explained in the Wikipedia:

Generally, polarizability increases as the volume occupied by electrons increases.[7] In atoms, this occurs because larger atoms have more loosely held electrons in contrast to smaller atoms with tightly bound electrons.[7][8]

I think you misunderstood my second question. I was talking about the polarizing power of a cation, as opposed to the polarizibility of an anion. I get that polarizibility of an anion falls down the group due to lower effective nuclear charge on outer electrons, but why does the polarizing power of a cation also fall?

Could you edit the original question so it's clear your asking about cations, rather than the more commonly discussed polarizability of anions if that is indeed the case?

@OVERWOOTCH I added an explanation on how both the electric field and the potential scale with the distance between ions. That should explain the change in "polarizing power". The other answer dwells into this too.


text:
What factors other than atomic weight affect density of an element?
Iron, for example, has an atomic weight of 55.845

Whereas hydrogen's atomic weight is 1.008, 1.8% of Irons.

And the density of Hydrogen (at STP) is 0.0899 g/L, 1.14% of of Iron's density of 7.874 g/cc, not an exact ratio relationship, but similar.
What other factors affect density of an element?

The other factor is the distance between atoms and/or molecules. These distances are huge in a gas like $\ce{H2}$, and very small in solids like metallic iron. And the density of $\ce{H2}$ is not $1.14$% of iron's density, as you state. It is still $1000$ times smaller !

I forgot to covert between L and cc

@J. Todd. It is exactly what I told you. This does not change my explanation.

right, I understand. Is there a formula where you can input the molecular structure / atomic data and get the density?

No ! Because nobody can guess that the Iron atoms are piled up on a cubic base up to a certain temperature, then it changed to a hexagonal structure at a higher temperature, to go back again to a cubic arrangement at a still higher temperature.

I see. Thanks for your time even though my question wasnt very well thought out


text:
Is this bond Ionic or Covalent, and why? AlBr
We know a compound could form between NaCl because they are +1 and -1 ions which make them both into a complete valence set.
Could a compound form between Al and Br, for example, and what type of bond would this be? Would such a compound never form because it is not 'stable' (both atoms having a complete octet, like NaCl)?
It is said that a bond between a metal and non-metal such as this must be Ionic. If this is the case then it violates the rule that the Electronegativity difference must be > 1.7. The EN difference is only ~1.3 between Al and Br, which would indicate a Covalent bond instead. Which is correct?

There are laws which are precise with absolute mathematical certainty, and there are laws which are about as precise as our measurements, and then there are laws which are very, very approximate and imprecise. Those which you mentioned are of the latter sort. $\ce{AlBr3}$ is mostly covalent, until dissolved in water, at which point it starts looking more like ionic, but not quite.

And @IvanNeretin forgot to mention that there are rules which are more like guidelines than actual rules. Welcome to the Black Pearl!

Neither the 1.7 electtonegativity difference rule nor the metal-nonmetal rule for ionic bonding is really correct, as Ivan points out in the comments.  Following are some well-documented exceptions to both rules:

The chart given here, illustrated below, shows potassium iodide has about 70% ionic character even though its electronegativity difference is only 1.6.  The ionic character matches that of caesium chloride, which has an electronegativity difference of 2.3.  Also note the electronegativity dufference of 1.8 but less than 50% ionic character in hydrogen fluoride.

The low solubility of most transition metal sulfides in water may be correlated with covalent character in their bonding, which goes along with sulfur having a relatively low electronegativity for a nonmetal.  Alkali and alkaline earth sulfides which have more ionic character dissolve readily in water, and also react with it through hydrolysis of the strongly basic sulfide ion.

You need a better rule.  One such rule considers the structure of the compound.  When you are dealing with elements in Groups 12-17, they have only four valence orbitals; thus a coordination number of four or less serves as an indicator of covalent bonding while higher coordination numbers require at least some ionic contribution.  Thus the dimeric molecular structure of aluminum bromide heralds covalent bonding in the neat aluminum bromide.

From Wikimedia Commons


text:
Does an electron's spin change when being excited?
My preconception is that when you fill an empty orbital by exciting an electron (with a photon), the empty orbital should be filled by an electron with the same spin. However, I've seen examples of that happening and not happening. For example, on this University of Calgary
web page there is an image of the electron spin being flipped:

The lowest energy transition is that between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) in the ground state.  The absorption of the EM radiation excites an electron to the LUMO and creates an excited state. The more highly conjugated the system, the smaller the HOMO-LUMO gap, DE, and therefore the lower the frequency and longer the wavelength.

Does it just not matter that much energetically?

@orthocresol Does that mean the spin angular momentum of a photon at fotoexcitation generally leads to change of an electron orbital angular momentum  but not the spin one, correct ?

@Poutnik I think so, that’s a typical selection rule for spectroscopy.

TL;DR Excitation of an electron typically conserves spin; that is to say, the spin must be left unchanged by the process. If the initial state has no net spin (one spin-up and one spin-down electron), then the final state should also have no net spin (one spin-up and one spin-down electron). This rule is reliable for small-ish atoms, but often breaks down with heavier atoms.
Note that this doesn't prescribe which electron gets excited (spin-up or spin-down); in general we can't say anything about that.

The reason is to do with Fermi's golden rule. The transition probability is governed by a quantum mechanical integral (the "transition dipole moment") which looks like
$$\langle \psi_\text{final} | \hat{H} | \psi_\text{initial} \rangle.$$
Here, $\psi_\text{initial}$ and $\psi_\text{final}$ are the initial and final states respectively, and $\hat{H}$ is the perturbation Hamiltonian. For an electromagnetic wave process this is the dipole moment operator $\hat{\mu}$, which can be expressed in terms of the position operators:
$$\hat{\mu} = \mu_x\hat{x} + \mu_y\hat{y} + \mu_z\hat{z}$$
The key things to note are that:

The electron wavefunctions $\psi$ can typically be factorised into a spatial part and a spin part, which I denote as $\phi$ and $\chi$ respectively:

$$\begin{align}
|\psi_\text{final}\rangle &= |\phi_\text{final}\rangle \cdot |\chi_\text{final}\rangle, \\
|\psi_\text{initial}\rangle &= |\phi_\text{initial}\rangle \cdot |\chi_\text{initial}\rangle. \\
\end{align}$$

The perturbation Hamiltonian $\hat{H} = \hat{\mu}$ only depends on spatial degrees of freedom, and not spin.

Therefore, the transition dipole moment can be factorised into a spatial part and a spin part:
$$\begin{align}
\langle \psi_\text{final} | \hat{H} | \psi_\text{initial} \rangle &= \langle \phi_\text{final}\chi_\text{final} | \hat{\mu} | \phi_\text{initial}\chi_\text{initial} \rangle \\
&= \langle \phi_\text{final} | \hat{\mu} | \phi_\text{initial} \rangle \langle \chi_\text{final} | \chi_\text{initial} \rangle
\end{align}$$
In particular, the spin part $\langle \chi_\text{final} | \chi_\text{initial} \rangle$ is zero, and hence the transition probability is zero, unless the initial and final states have the same spin. Processes which involve spin flips are often called "spin-forbidden".
Under certain conditions, the assumption (1) breaks down: that is to say, the wavefunction $\psi$ cannot be cleanly factorised into spatial and spin parts $(\phi, \chi)$. This happens particularly when spin–orbit coupling is present:  heavier elements (i.e. elements with larger atomic number) typically exhibit greater spin–orbit coupling.
Thus, for light atoms like hydrogen, spin-forbidden processes are genuinely "forbidden", and therefore almost never happen. However, so-called "spin-forbidden" processes can actually happen for heavier atoms. Some examples include:

the pale pink colour of $\ce{Mn^2+}$ salts results from a spin-forbidden excitation process. The fact that it is spin-forbidden just means that the colour is a lot weaker than that of a typical transition metal. There is a lot more discussion of this particular case at Why is manganese(II) coloured although the transition should be spin-forbidden?.

the purple colour of iodine $\ce{I2}$ results from a singlet to triplet excitation. The fact that this colour is pretty intense (as opposed to the pale pink of $\ce{Mn^2+}$ simply reflects the fact that iodine has a large atomic number.

The picture shows a triplet excited state returning to the ground state by emitting a photon, i.e. phosphorescence. This can only happen if there is also an interaction that couples angular momentum change with the transition, such as spin orbit coupling.
The electron has two quantum numbers; the spin $S=1/2$ but is not the spin of the electron that is changing but the other quantum number 'z' , 'azimuthal' or 'magnetic' spin quantum number $m_z=\pm 1/2$.

I know it looks as if the photon is being emitted (that was my first thought as well), but especially placed in context of the webpage, it’s depicting an excitation process. I think maybe I shouldn’t have removed the contextual text, sorry.

No apology needed. Well the arrow is the wrong way round if it is excitation,  and if it were absorption it would predict a singlet to triplet process which would still need the coupling I mention to break the $\Delta S =0$ rule.


text:
What happens to dissolved ions when an aqueous solution evaporates?
I'm working on a chemical design project where zeolite crystals containing $\ce{Na+}$ cations is ion-exchanged via an aqueous solution of $\ce{AgNO3}$, $\ce{Cu(NO3)2}$, and $\ce{Zn(NO3)2}$, also containing $\ce{HNO3}$. The solid product now containing $\ce{Ag+}$, $\ce{Cu^2+}$, and $\ce{Zn^2+}$ cations is removed from the solution and dried. I expect the remaining $\ce{Ag+}$, $\ce{Cu^2+}$, $\ce{Zn^2+}$, and $\ce{Na+}$ will precipitate with available $\ce{NO^-3}$ as the water evaporates, but what will happen with any remaining $\ce{NO^-3}$ anions? Can they react with water or air to form any solid/gaseous products?

Have you studied the ion-exchange process on a simple polymer? The polymer is sulfonated (so negatively charged). The negative group is fixed, but the hydrogen ion or any other cation is replaceable. There is always a charge neutrality. Look at this figure: https://en.wikipedia.org/wiki/Ion_exchange#/media/File:CationExchCartoon.png

Once you have understood the cation ion-exchange process, look at the structure of your zeolite and see where Na-ion is attached.