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Understanding the Jahn–Teller elongation and compression for octahedral complexes (stabilisation/destabilisation of d orbitals)
I am currently learning about Jahn–Teller effect. For elongation I wondered why the d orbitals with $z$ component are stabilised when the metal–ligand bonds are getting longer? I thought a longer bond would mean a less efficient overlap of the d orbital on the transition metal with the orbital on the ligand.
And if yes, shouldn't it be raised so the difference in energy between the d orbital and the ligand’s orbital be greater? I thought the ligands’ orbitals would be lower in energy each time.
Where am I going wrong with my thinking?

JT is usually described as elongation along $z$ and a stabilisation of orbitals with a $z$ component. Not sure what your source on destabilisation has to say.

Ahh gosh, I was getting so confused, that is what I meant to put, but my reasoning doesn’t match up with it being stabilised

Before you consider J-T distortion, take a look at the d orbitals of an octahedral complex. Note that the orbitals that are on the axes where the ligands are are higher in energy. This is because these "d orbitals" are actually anti-bonding orbitals comprised primarily of the metal d orbitals. Because they are anti-bonding, they get higher in energy as the bonding interaction between the ligand and metal gets stronger. Thus, a longer distance between the ligand and metal (ie weaker bonding interaction) results in less destabilization.
Another way to think about it is to consider the interaction of a ligand lone pair (ie negatively charged electron density) with a filled d orbital on the metal (also negatively charged electron density). The closer they are together, the more unfavorable the interaction is.

This is a fantastic answer. Thank you for giving me the time to understand this, I really appreciate it.

There are two ways to approach this: simple crystal field theory or more complex molecular orbital theory. Both can describe and predict Jahn-Teller distortion well.
In simple crystal field theory, one always starts with a set of five degenerate d orbitals on the metal. The ligands are considered as a point-sized negative charge. (This consideration, while not being perfect, is okay, because the main thing the ligands contribute to a complex is electron density or an overall negative charge. The main problem is that it assumes point-shaped or spherical charges which is inadequate to explain more complex bonding – but not required to understand Jahn-Teller.) As the ligands approach the central metal – along the coordinate axes, in an octahedral complex – the d orbitals experience a negative electric field. As electrons occupying an orbital are also negatively charged, the more of a field an orbital experiences the less stable it becomes because the electron is repulsed by the electric field of the same charge.
When considering the orientation of the orbitals, two groups emerge. The first group is $\mathrm{d}_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$ whose lobes are oriented between the coordinate axes by convention. The second group is $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ which both point in the direction of the coordinate axes ($x$ and $y$ in the first case, all three but primarily $z$ in the ‘canonic’ shape in the second). Thus, these two orbitals experience destabilisation and are raised in energy.
If the electronic configuration of a complex happens to be $\text{hs-d}^4$ or $\mathrm d^9$ (other cases are possible but these two provide the strongest Jahn-Teller effect), the two highest-energy orbitals are populated unevenly despite being degenerate. In general, uneven population of degenerate orbitals is not a problem; but in this case the system can distort to provide an overall stabilisation. Assume that the two ligands in $z$ direction are moved away from the metal. This means, that the point-sized electric charge in $z$ direction is further away, which in turn means the electric field experienced by orbitals pointing in $z$ direction is weaker and thus ultimately they are destabilised less. This destroys the symmetry of the two highest orbitals (making one lower in energy than the other); the higher orbital can now be unoccupied while the lower is occupied. This constitutes stabilisation.

Scheme 1: simplified scheme of Jahn-Teller distortion in the crystal field model.
If you were to bring the ligands closer, the effect would be opposite: rather than reduce the energy of $\mathrm d_{z^2}$ you would be increasing it. This would again break symmetry and provide for a clear orbital to occupy less, but the overall energy of all orbitals would be higher. Thus, moving the ligands away provides a net stabilisation while moving them closer does not.
$z$ axis">
Scheme 2: assuming the $z$ ligands were moved closer, this simplified scheme of a Jahn-Teller distortion would result.
In the two schemes note the dashed lines; one pair of dashed lines is always horizontal. This is not an accurate depiction but it serves to get the point of Jahn-Teller across and it is close enough. Also note that the lower orbitals also lose their degeneracy. This is because strictly speaking they will be raised slightly by the incoming point-sized ligands although I have not included that in the diagram.

The more complex way to understand coordination compounds is molecular orbital theory. It is more extensive than crystal field but it provides an easy way to understand many more aspects which simple crystal field theory cannot explain.
The key to creating a molecular orbital diagram for a coordination compound is to understand that it is essentially the same as for any other covalent molecule. One has to determine which orbitals are available on either side of the bonds, how these orbitals transform according to the compound’s symmetry and point group, which orbitals can thus combine in what ways to give molecular orbitals and how strong those interactions will be.
The simplest molecular orbital scheme will consider the full set of 3d, 4s and 4p orbitals on the central metal (these transform as $\mathrm{a_{1g} + e_g + t_{1u} + t_{2g}}$) and six s-type orbitals on the six approaching ligands (these transform as $\mathrm{a_{1g} + e_g + t_{1u}}$). Immediately, one can notice that $\mathrm{t_{2g}}$ is only present in metal-centred orbitals and thus these will not budge upon bonding. Furthermore, $\mathrm{e_g}$ is present both in the ligands and in the metal, meaning that there will be a bonding-antibonding interaction between these two groups of orbitals.
Simple electronegativity considerations (ligands are more electronegative which usually corresponds to lower-lying ground state orbitals) will tell us that the metal’s orbitals are higher in energy than the ligand. Thus, upon forming a bonding/antibonding pair of linear combinations, the bonding interaction will be more closely associated with the ligands while the antibonding one will be centred on the metal. This means that the metal $\mathrm{e_g}$ orbitals will be raised in energy, as they are now antibonding ($\mathrm{e_g^*}$). We have already derived the same picture as the crystal field model predicts.

Scheme 3: Simplified general molecular orbital scheme of an octahedral complex including only σ-type interactions. Taken from Professor Klüfers’ coordination chemistry course’s web scriptum (LMU Munich) and first presented in this answer.
To explain the effect of increasing or decreasing the distance of ligands along one axis, the simpest way is probably to again derive it from a simple covalent bond. There, increasing the distance between two atoms will reduce the orbital overlap and thus reduce the energy difference between bonding and antibonding orbitals, moving both closer towards the baseline. On the other hand, reducing the bond length will have the effect of increasing bonding/antibonding interactions and raising antibonding orbitals while lowering bonding orbitals.
If you remember that the $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals are antibonding orbitals in coordination complexes, moving the $z$-ligands closer will mean that $z$-containing orbitals are raised (while $x$ and $y$-containing ones remain unchanged). Moving the ligands away will decrease the energy of $z$ orbitals (and not change that of $x$ and $y$ ones). This again is the same result as in the crystal field model albeit using different considerations and arguments.
Strictly speaking, one would have to consider the new complex under the new symmetry which will be $\mathrm{D_{4h}}$ in both cases. This places the orbitals into different irreducible representations ($\mathrm{2a_{1g} + b_{1g} + a_{2u} + e_u}$ for the ligands, $\mathrm{2a_{1g} + b_{1g} + b_{2g} + e_g + a_{2u} + e_u}$ for the metal), meaning a different set of interactions would take place. Ultimately, this consideration leaves $\mathrm d_{xz}$ and $\mathrm d_{yz}$ as well as $\mathrm d_{xy}$ where they are whatever happens (they transform as $\mathrm{e_g}$ and $\mathrm{b_{2g}}$, respectively, and have no ligand counterpart), while everything else is somehow affected. Without doing a full development on paper it is probably impossible to see how a shorter $z$ coordination distance would affect the energies compared to a longer $z$ distance but the final outcome should be equivalent to the above: a longer distance will lead to an overall stabilising effect.

This deserves at least 1 million upvotes. Thank you!

@MarinaCalder This isn’t even my final form yet ;)


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Coordination compound nomenclature for [Cu(acac)(bpy)]BPh4
The title says it all really.
I thought it would be acetylacetonatobipyridinecopper(II) tetraphenylborate. Is this correct? I feel as though it looks a bit clumsy and wasn’t sure if I was supposed to include as (2,2-bipyridine)? Is it ok to use brackets like this for naming a coordination compound the IUPAC way or is my first naming correct?

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Beer Lambert law when complex is tetragonal (Jahn-Teller elongated)
I wondered what modifications there would be to the Beer Lambert law to find the molar extinction coefficients for the solution if the solvent contained the ligands which bind to the metal centre of a metal compound (ie 2 ligands add along the z axis only) and the complex was tetragonal?

As far you can define a concentration, that is as far you can identify a molecular or heavily bond entity nothing change. Stoichiometry  always count of course, if you are following a process.

You must find a wavelength that only the complex absorbs, then its the same as normal

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Is there an 'easy' way to determine if a ligand is a pi acceptor, donor or neutral?
First of all I am confused how something can be pi neutral? something either does or doesn't have filled p orbitals for example.
given something I have never seen acting as a ligand before like trimethylphosphate how can I work out whether this is pi accepting, donating or neutral? I am thinking the donor atom would be the oxygen bonded to the Me group and P but do I only consider the orbital of the donor atom?
For a tetragonal Cu(II) complex how can a ligand even act as pi donating if since the t2g orbitals on the metal are filled?
Sorry about this messy question, I am very confused about this in general, I would be so grateful if someone could help me understand this

If you ask for an "easy" way to determine if a ligand is a pi acceptor/donor or none, look at the spectrochemical series. A quick-and-dirty rule (which means that it is correct most of the time but not always) is that weak ligands (iodide, bromide, hydroxide etc) are pi-donor ligands. The medium ligands (water, ammonia etc) are pi-neutral, and the strong ligands (cyanide, carbonyl, bipyridine, etc) are pi-acceptor.
But this method is some kind of reversing here. Because we normally use a ligand's pi character to guess if it is a strong or weak ligand.
If you can, ligand field theory is your best friend in this problem.
Generally, a pi-donor ligand has an extra electron pair that lies in an orbital that has approximately the same symmetry as the metal center's orbital (usually t2g). This allows the two orbitals to interact with each other and creating two new sets of orbitals: t2g and t2g* (as in the molecular orbital theory). The ligand's pi electron will reside in the  lower t2g orbital, whereas the metal's original d-electrons will reside in the t2g* one. That explains your question of the copper complexes.
In the other hand, a pi-acceptor ligand has an empty molecular orbital of low energy and approximately the same symmetry as the orbital of the metal center. In this case, it will interact with the metal's orbitals and the d-electrons of the metal will be delocalized to the ligand. These ligands usually have a highly conjugated system that greatly decreases the energy of the LUMO (bipyridine, triphenylphosphine etc). In the cases of carbon monoxide (called carbonyl as a ligand) and cyanide, they have low-energy LUMO for other reasons.
A pi-neutral ligand (also called sigma-donor ligand) does not process any of the above characteristics. For example, ammonia has only one electron pair on the nitrogen, and that pair is already used in the formation of the sigma bond, so no more pi electrons to donate. The LUMO of the ammonia is also high in energy, so no pi accepting here.
I think trimethyl phosphate is mostly a pi-neutral with the donor atom is the oxygen double-bonded to phosphorus. The molecule is not conjugated at all, and the oxygen can only donate one pair of electron since the other's orbital does not have the same symmetry.

Is the ethylene ligand in Zeise's Salt also a π-donor?


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Maximum absorption wavelength directly proportional to the donor number of the ligand for a copper(II) coordination complex
I am looking at a set of results showing a directly proportional relationship for an increasing maximum absorption wavelength (UV-Vis spectra) with increasing donor number of solvent ligands. The complex has a $\ce{Cu^2+}$ centre with $\mathrm{d^9}$ electronic configuration. Assuming a tetragonal configuration, how does this tie in with ligand strengths?
I am not sure what I can deduce here as I thought the more available a lone pair (basicity) would be able to interact more strongly, a higher wavelength indicates the energy gap between the $\mathrm d_{x^2-y^2},$ and $\mathrm d_{z^2}$ orbital has decreased. So, has there been more significant Jahn–Teller distortion, or is the splitting parameter somehow smaller?
Where does my reasoning fall down?


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How is indigo carmine used in Lowenthal-Permanganate titration as a redox indicator?
In a Lowenthal-Permanganate titration, cider tannins are quantified by titrating the cider with indigo carmine indicator and sulfuric acid against potassium permanganate. Tannins are antioxidants, so that means they give up electrons to reduce the potassium permanganate. The endpoint of this titration is when the blue analyte turns a yellow colour. According to Wikipedia, indigo carmine turns yellow when reduced.
What I don't quite understand is why the indigo carmine didn't reduce when it was in the presence of the antioxidative tannins before the potassium permanganate was added and why it turned yellow only after potassium permanganate was added.

Could there be an error of switched colours ? Titration blue to yellow makes little sense, analysts always try light to heavy color transitions. Check independent sources.

Thanks for the comment @Poutnik, I actually carried out this titration myself and the colours were correct. I based my procedure on procedures used in academic journals such as this one: https://onlinelibrary.wiley.com/doi/pdf/10.1111/1750-3841.14713.

Hm, I was just assuming that as an option.

Can it be gallates are not strong enough to reduce indicator, which is oxidized farther by permanganate ? Similarly as high pH discolour phenolphthalein ?

It's very possible I'm just not understanding this fully, but if the indigo carmine was oxidized by permanganate, wouldn't it be blue?

I mean $$\ce{IC(red, yellow) <=>[oxidation][reduction]IC(ox, blue)->[Strong oxidation][KMnO4]IC(oxox,yellow)}$$ As permanganate is heavy weight for most organic dyes, even  redox indicators with low redox potential see https://en.wikipedia.org/wiki/Redox_indicator

Hmm, that's a good suggestion. Thanks for the help! I'm in high school, perhaps this is why it's confusing for me :(

After the endpoint of this titration you have an excess of permanganate, so your indicator is going to be oxidized, not reduced. It is confusing because most everything you find about indigo carmine as a redox indicator is about its reversible reduction to its leuco form, which is indeed yellow. However, like most dyes, indigo derivates can also be irreversibly degraded ("bleached") by strong oxidizers like acidic permanganate, and that is what's going to happen here.
The tannins do not reduce the indigo carmine before or during the titration, as that reaction would be either unfavorable or very slow, and are not supposed to. They do react rapidly with the much stronger oxidizer permanganate as it is added. Once all tannins are consumed, the excess permanganate will attack the indigo carmine, oxidizing it to, presumably, isatin sulphonic acid which in dilute solution is also yellowish.

I did some brief looking but given that the carmine is also pH indicator that remains blue at below 11.4 pH it would appear as though your sulfuric acid is preventing the reduction from occurring in your indicator. Under acidic conditions KMnO4 is a strong oxidizer where carmine is a weak oxidizer under these conditions. However I am not sure how the carmine gets reduced to yellow upon equilibrium. Perhaps the pH changes. It appears as though permanganates acid conjugate is strong so its a weak base. Just throwing out ideas there though.
http://chem.lapeer.org/Chem1Docs/IndigoCarmine.html
https://www.sciencedirect.com/topics/earth-and-planetary-sciences/permanganate


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Aromaticity of different compounds
I'm new here and I'm still learning, I understand that there four rules for aromaticity:

cyclic
conjugated double bonding (every atom in the ring should be sp2)
huckle number should be 2, 6, 10, 14, 18, ..18
planar ring

The 2nd question should be 4 electrons and is antiaromatic, but how about the first question?

https://chemistry.stackexchange.com/questions/76142/which-rings-of-this-heterocyclic-compound-are-aromatic

Cytosine is aromatic, there are 6 electrons in the ring. Two pairs of pi bonds(thus 4 electrons) and a lone pair on Nitrogen(the one which has H written below it). The other Nitrogen (which is forming a pi bond) in the ring, does have a lone pair, however it is "localised". That lone pair cannot participate in resonance, and hence it is not counted when considering aromaticity.

I put 6, but I got wrong answer, that is why I'm asking here, is 6 or 10? @Kaiser

My answer would be to refuse to answer this case.
The 4n+2 rule applies only if the electrons are conjugated in a cycle -- nothing more, nothing less.  If you have pendant pi bonds in the conjugated system, like the carbonyl group in cytosine and many similarly structured compounds, you don't have the right conditions to apply the 4n+2 rule at all.
Interior conjugated atoms, as in pyrene, also mean you don't have specifically conjugated cycle.

Cytosine is proven to be aromatic, how do they calculate the electrons inside the ring, it seem they add 2 if there is HN: present as hetero, but not the case in NH2, I want to know exactly how many pi bonds and electrons inside the cytosine ring. @OscarLanzi

The issue, of course is not whether it's aromatic.  No doubt it is.  Rather the issue is whether we can really apply the 4n+2 rule that they rendering in many a textbook.  No doubt we can't;  IUPAC recommends the rule only for single, fully symmetric rings with no additional conjugated atoms.

hmm..that's interesting...thank you for pointing this out!, would the answer be 10 electrons in resonance? @OscarLanzi

Yes there would be ten electrons in resonance.  But without them being in a ring, the figure being 4n+2 is not really a predictor of aromaticity.  Cyclopropenone has a pendant carbonyl group and shows aromatic character too -- with four conjugated electrons.

Thank you so much for these tips, I know now that Huckel rule has its limitation like Lewis acid base concept, or octet rule in carbon, so when we have pendant carbonyl group or methylene such in fulvene, do we consider the conjucated electron to be zero because there is a missing conjugation fulvene https://en.wikipedia.org/wiki/Fulvene @OscarLanzi


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Still struggling with Huckel rule and aromaticity
I'm still learning huckel rule and aromaticity in different compound.
For the 1-methoxy-1,3,5,7-octatetraene, is it correct that it has 8 electrons inside the cycle, but since they will be 8, then it is antiaromatic?
And for the Imidazole, is it 6 electron inside because the N: doesn't contribute to the pi, or is it 8?

Why would someone downvote a question? I'm new here in chemistry.stackexchange and I'm still learning...can someone explain?

https://chemistry.stackexchange.com/questions/27099/when-do-lone-pairs-on-nitrogen-contribute-to-aromaticity https://chemistry.stackexchange.com/questions/33332/why-is-cyclooctatetraene-non-planar-but-the-cyclooctatetraenide-anion-planar

I’m voting to close this question because there are two parts to the question, and both are duplicates (see Mithoron's comment). If this were just one question, we could close it as a duplicate, but because there are two, we can't.

I'm voting otherwise.

Advice
Take a closer look at the lone pairs and double bonds that participate only and only in the delocalization of electrons.
1-methoxy-1,3,5,7-octatetraene
As you can see, 1-methoxy-1,3,5,7-octatetraene has four double bonds in an octagonal cycle. The lone pairs of the adjacent oxygen atoms, even though they directly influence the delocalization by means of the resonance effect, aren't actually part of the cycle and cannot be included amongst the pi electrons of the double bonds. Therefore, your molecule is antiaromatic. Its structure is not planar, in fact it looks something like this (the picture below includes only the cyclic backbone):

For aromatic molecules, planarity is absolutely necessary. But Huckel's rule which I'm sure you're familiar with, provides the reasoning behind the planarity of the molecule.
The imidazole
As for the imidazole, your assumption is absolutely correct. Here we get to know a scenario similar to pyridine:

The lone pair does not participate in the aromatic cycle. This can be easily checked by taking a look at the hybridization of the nitrogen atom. If the lone pair were to be part of the aromatic system, it should have been perpendicular to the other orbitals of the cycle. However, this is not the case. The 2p orbitals of the nitrogen are actually involved in the aromatic cycle, while the aforementioned paired electrons get to be embedded in an sp2 orbital, placed perpendicular to the plane of the cycle. This actually has significant consequences on the basicity of pyridine.
On the other hand, in order not to interrupt the aromatic cycle, the lone pairs of the other nitrogen atos must come into play when counting pi electrons. Thus, the molecule is aromatic.

Thank you @TheRelentlessNucleophile I just want to make sure for 1-methoxy-1,3,5,7-octatetraene it is nonaromatic, not antiaromatic (because antiaromatic should retain planar feature), and it is nonaromatic because it is not flat like the figure shows.

In case of 1-methoxy-1,3,5,7-octatetraene, the answer would be zero for the number of electron, right? because since it not planar, then it conjugation will be accounted from because resonance is not happening anymore, please correct me if I'm wrong? @TheRelentlessNucleophile

@user432797 Yes you're right.

I turned out that @TheRelentlessNucleophile is right in everything even with antiaromatic categorization

Apologies, I didn't see the comments you had posted after my answer. If a molecule has an uninterrupted network of pi electrons and 4n pi electrons, it is usually antiaromatic. Because they lack the stabilization that would otherwise be present in the aromatic molecules thanks to the pi orbital that delocalizes over the whole molecule (see Huckel's rule), the molecule stabilizes itself by adopting non-planar configurations.

Rule of thumb: uninterrupted network of pi electrons, planar molecule, 4n+2 electrons, extended pi orbital that delocalizes over the whole molecule: aromatic. Violate one rule and the molecule is rendered either antiaromatic (4n pi electrons) or non-aromatic. Now, I lack the appropriate knowledge to tell you exactly why do we bother with antiaromatic molecules in the first place, but let's just say that, as far as I know, if you were to draw the Frost circles in a Huckel diagram, their HOMOs would be at the ground energy level, rendering the orbitals non-bonding.

Also, keep in mind that several of these molecules prefer to exist as diradicals, instead of completely stable molecules

thank you @TheRelentlessNucleophile ! Can you refer me to the reference that you used? I still don't get why my question was downvoted!

Your question was downvoted becausw it looked like homework and also because of lack of details involving YOUR work in answering the question ans sharing YOUR knowledge. Please refer to the rules of ChemSE.


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What are the products of alkene oxymercuration with diols?
I was wondering what would be the product of the reaction of alkene with a diol (such as glycol) in alkoxymerucuration/demercuration, i.e. $\ce{(CF3CO2)2Hg/NaBH4}$?
I tried to look online but I couldn't land on a paper or website that goes through such reaction. I also tried Scifinder, I marked both compounds as reactants, but zero results stemmed from the reactants!
Any input?

I'm thinking if these possibilities (I have more possibilities in my mind, but for example I selected three) are structurally accurate and energetically stable to be produced in such reaction and which is favored?

To classify them more ... to single and di addition, here is what I'm thinking:

Oxymercuration is often run in the presence of excess water or alcohol, but only one RO/OH adds. Therefore I presume in the presence of a diol you get mono-addition by one of the 2 available OHs, probably the least hindered. https://en.wikipedia.org/wiki/Oxymercuration_reaction

@Mithoron so Alkoxymercuration/demercuration (CF3CO2)2Hg/NaBH4) won't force this reaction? if it doesn't what can force this reaction? I want to see whether such reaction will add seperate glycol on each carbon of the double bond, or the glycol will form a cyclic with the two carbon of the double bonds?

I think the top left option is most likely. I see no reason why you would get di-addition to the alkene

@Waylander I updated the question with single and di-addition possibilities, I'm not sure if excess of glycol or other factors will move the reaction toward single or di-addition or toward open or cyclic formation in either case.

In your original scheme, the product in the upper left would likely have the hydroxyl attach to the benzylic position. Two of the remaining three products involve C-C bond formation. How is that accomplished with oxymercuration?

@user55119 that's my question: could glycol move the reaction toward single or di-addition or and could it proceed toward open or cyclic formation in either case. For instance if it is a single addition, could the glycol, after forming -OR, form a cyclic orientation?

Alkyl mercurials react via radical processes but not to form the products you show.

There are so many structures here that it is at times difficult to keep things straight. The best way to approach your question is to take advantage of what is known about oxymercuration. If you enter "oxymercuration" at Search on Chemistry at the top of the page you will find a wealth of information at your disposal right here on ChemSE.
To start, the alkene is the (E)-propenylbenzene 1 and the diol is butan-2,3-diol 3, devoid of stereochemistry. Formation of mercurinium ion 4 with mercuric trifluoroacetate (2) is drawn as unsymmetrical because in the transition state for addition the carbon bearing the greater positive charge is the site of SN2, C-O bond formation. Mercurial 5 is a racemate, ignoring any stereochemistry of 3, which is what is formed after the first reaction. The second reaction, reduction with NaBH4, proceeds via radical 6 to form alcohol 7. A minor product may be bis-ether 9 arising from alkene 1 and mercurial 5. For mercurial 5 to form 1,4-dioxane 8 requires an oxidation of the carbon bound to mercury and a reduction of Hg+2 to Hgo as shown by the red arrows in structure 5. How do we know that this process does not occur? Because if it did, NaBH4 would not be required! Consider the oxymercuration of alkene 1 in the presence of water. If what you are suggesting were true, then 1 would form a diol rather than the observed mono-ol prior to the participation of NaBH4.


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Determining the number of triphenylphosphine ligands in nickel(II) complex

A complex is synthesised by combining hydrated solutions of $\ce{NiCl2}$ $(\pu{0.6 g})$ with $\pu{1.35 g}$ triphenylphosphine $(\ce{PPh3}).$ After cooling the product precipitates as a solid with formula $\ce{NiCl2(PPh3)_x}.$ Determine $x.$

I first found the molar ratios
$$n(\ce{NiCl2}) = \frac{\pu{0.6 g}}{\pu{129.59 g mol^-1}} \approx \pu{4.6 mmol}\tag{1}$$
$$n(\ce{PPh3}) = \frac{\pu{1.35 g}}{\pu{262.29 g mol^-1}} \approx \pu{5.1 mmol}\tag{2}$$
The smallest ratio these go into gives $x \approx 1,$ but I don’t think this is right.

Have you taken into consideration the possibility of having nickel chloride as a hydrate? x=2 for the hexahydrate.

Oh gosh what a silly mistake to make,  I didn’t take that in thank you!

Nothing looks silly here. Only one issue: the question contained contradictory information: trimethylphosphine $(\ce{PPh3}).$ I assume it was supposed to be triphenylphosphine judging by the molar mass you are using. $x = 2$ looks the most correct answer to me as it both justifies a typical C.N. = 4 for nickel, but still leaves enough room for two bulky triphenylphosphine ligands.


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How easy is it to make animal horns transparent?

I never knew that HORN in HORNbook refers to animal horns, or animal horns can be made transparent? If Medievalists can make it transparent, the transparency process must've been cheap easy and quick, right?
The Horn Book | Why is it called "The Horn Book"?

Back in the sixteenth century, English monks began to make hornbooks to help their pupils learn to read. Usually a wooden paddle with an alphabet and a verse glued to the surface, hornbooks derived their name from the piece of transparent horn protecting the verse. The picture to the right shows a modern replica of a hornbook.

Hornbook - Wikipedia

A hornbook is a book that serves as primer for study. The hornbook originated in England as long ago as 1450,[1] or earlier.[2] The term has been applied to a few different study materials in different fields. In children's education, in the years before modern educational materials were used, it referred to a leaf or page displaying the alphabet, religious materials, etc., covered with a transparent sheet of horn (or mica) and attached to a frame provided with a handle.[3]

Could they have shaved off a really thin layer? If thin enough, it might be transparent without chemical treatment.

Cheap, easy, and quick do not relate at all to parchment production, for example. Yet it was made and used in medieval times. Don’t let modern ideas obscure history.

I found this description of how the horn is prepared in a post by Tammy L. Austin, at the University of Notre Dame's website (https://www3.nd.edu/~rbarger/www7/hornbook.html):

The horn of oxen and sheep were used to make the laminating structure. The horn was left in cold water for several weeks, which separated the usable part from the bone. It was then heated, first in boiling water then by fire, and pressed by plates and machines to make it smooth and transparent.


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Why are alkenes considered nonpolar?
Alkenes have C–C sigma bonds, C=C pi bonds, and also C–H sigma bonds.
I know that C–C and C=C bonds will not give any polarity, but what about the C–H bond?
I think it should give some polarity because C is slightly more electronegative than H, so the electrons in the C–H bond will be slightly polarised towards C.
But alkenes are typically referred to as non-polar. Why?

It depends on the resultant of the electrical dipole moments, which anyways aren't so important. Of course you might find something if you look for small things.

"Alkene" is not a compound. You are overgeneralising. Look at specific examples, I recommend you start with the two simplest, ethylene and propylene.

"Non-polar" in chemistry means "not particularly polar", rather than "not polar at all".

Two reasons:

On the grand scale of things, the polarity of a $\ce{C-H}$ bond is very low. The electronegativity difference is less than $0.5$ which is generally the cutoff of a mostly unpolar bond.

Typically when looking at an alkene at large, the various (small) polarities of the different $\ce{C-H}$ bonds cancel each other out or come very close to doing so. In ethene, they cancel out completely for symmetry reasons. In propene, they should likewise cancel out to almost zero total dipole moment. The larger your alkene gets, the less exact this cancelling will be but the general trend remains the same.

A compound or a molecule being polar or non-polar is rather a fuzzy term.
Being non-polar does not mean no polarity of chemical bonds.
It means this polarity is low enough for atoms of such a molecule not to  interact with other molecules by relatively strong inter-molecular interactions. These include strong dipole-dipole interactions, hydrogen bonds or donor-acceptor bonds.


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Geometry of the transition metal complex in solid dipotassium bis(oxalato) copper(II) tetrahydrate
I was thinking of the geometry of the transition metal complex in solid $\ce{K2[Cu(ox)2] · 4 H2O}$ solid around the $\ce{Cu(II)}$ ion.
At first I thought square planar, but is it possible or is there any chance that it could be octahedral with two water molecules since it is a hydrate?
Or do brackets always indicate what is directly complexed in this case?

Yes, square brackets are used to denote a coordination sphere, but they may sometimes be used incorrectly. Ideally you want to check the crystal structure, luckily there is one [1]:

The $\ce{Cu(II)}$ atom has square-planar geometry with two bidentate oxalate ligands. Each $\ce{K+}$ cation is coordinated by eight oxygen atoms from one bidentate and two unidentate oxalate groups and four water molecules.

Fig. 1. (a) Crystal structures of $\ce{[K2Cu(ox)2 · 4 H2O]}$ (1) and (b) $\ce{[K2Cu(ox)2 · 2 H2O]}$ (2) with atom numbering scheme. The thermal ellipsoids are drawn at 50% probability.

Reference

Fan, J.; Sun, W.-Y.; Okamura, T.; Yu, K.-B.; Ueyama, N. The X-Ray Crystal Structural Characterization of Dipotassium Bisoxalato Copper(II) Tetrahydrate, $\ce{[K2Cu(Ox)2·4 H2O]}$ $(\ce{Ox}$ = oxalate Dianion). Inorganica Chimica Acta 2001, 319 (1), 240–246. DOI: 10.1016/S0020-1693(01)00466-2.


text:
Which has more internal energy, a bucket of hot water or a bucket of cold water
Online answer I found :
The bucket of hot water has more internal energy.
Internal energy depends on the amount (mass) of substance and the kinetic energy of the molecules of the substance.
My thinking to it :
Since it is written dependant on mass, then swimming pool has more mass than bucket of hot water.Since it has even higher K.E , then how can Bucket of hot water have more Internal energy than swimming pool?

One thing is total amount of energy in J, other thing is specific energy in J/kg or molar energy in J/mol. The latter two depend on temperature, the former one on temperature and amount of substance. So hot water in the bucket has higher specific energy in J/kg, the pool higher energy in J.

Ok thnx.I understood

If you thought and searched about it little longer, you would not propably post your question at all.


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Can Steel be Created from Germanium?
I've tried searching with Google as well as on this website, but I haven't been able to find even a question asking about it. It was something that I was just curious about, but my limited knowledge in Chemistry has prevented me from testing anything or realizing how to phrase my question in a way that an answer can be found.
All in all, can steel be made also from the other elements in the carbon group, such as silicon or germanium?
And would the steel be stronger or weaker if it can be made?

Germanium is not in the same group as Iron. Neither ruthenium nor osmium has the same phase diagram with carbon as iron does.

No ! Steel is made of iron, plus some other elements

Well, hypothetically, if we would break heavier germanium nuclei to nuclei of light elements e.g. by high energy proton bombardment, and then if we  fused these light nuclei in particle accelerators to iron nuclei, we could make steel from germanium by some carbon addition. The problem is, current technology would be able to provide just few iron atoms this way.

Since this question has been revived, a simple answer is that even carbon is slightly too large for the interstitial voids in iron which it occupies in steels. Germanium is larger even than iron and would require much greater distortion of the iron lattice in order to fit between the iron atoms.

No.
Steel is a generic term referring to an alloy containing iron and carbon (with iron as the main component) and various types/qualities of steel contain many other additional elements, e.g. nickel, chrome, manganese, etc. to give or enhance different properties in the alloy.
If it is not iron-based, it is not steel.
As for alloys using metals from the same group as iron, firstly, germanium is not part of that group.
All of the elements that do form part of that group (Fe, Co, Ni, Ru, Rh, Pd, Os, Ir, Pt) can be and are used in alloys, (although carbon addition does not have the same influence on their properties), osmium not so much as it is a pain to store or process.  Some of those alloys are indeed "stronger than steel", for some definitions of stronger, but most of these elements are a lot less abundant and a lot more expensive than iron.
No one is going to use platinum, palladium, rhodium, etc. to make rebar or railway lines.
Edit:
Germanium, as pointed out by Martin below, is however part of the same group as carbon.  It and the other elements in that group can also be used in alloys, like carbon, but their properties are different (carbon is non-metallic, silicon and germanium are mettaloids and the heavier elements are metals) and they do not have the same effect on the properties of iron in an alloy.  Germanium, when added to tin alloys, however, has been shown to significantly increase the hardness of the alloy, and as discussed below, it helps tarnish and firescale resistance in silver. One also needs to remember that while silicon and carbon are abundant, germanium is not, and as it is critical in many high tech applications, what is available is not likely to be squandered on experiments for improving steel.
Now, I am going to go away and think about simple, not too technical ways of explaining why, it works this way or finding references that will do it better and will further edit this answer appropriately later.

And if you thought cable theft is bad, imagine how quickly thieves would make off with your railway lines if they did use platinum in the alloy ...

Platinum-osmium alloys are a thing (very hard), but certainly aren't steel.

@Jon Custer Yup. But seriously. Storing and processing osmium is a b ... bear. Plus you have to practically replace the walls every few months and you don't wanna know what the filters for the respirators cost for the personnel working with it. Ask anyone who had the joy of working in a precious metals refinery.

for sure, the whole OsO4 sublimating at room temperature and being teratogenic is pretty off-putting...

Indeed. Most refineries tend to just store it under lower temperature, because there really isn't that much of a market and safely disposing of it is a whole other can of worms. And the walls still slowly turn green regardles of the refridgeration ...

@Ed V But it isn't steel and isn't stronger than steel (and germanium still isn't part of the same group as iron) so I don't think it's really relevant. The OP wasn't asking if germanium could be used in alloys with other metals, they specifically wanted to know if "it and other elements in that group" could be used in place of iron to make steel.

The question (admittedly unclearly phrased) makes sense if you realise that germanium is in the same group as carbon, not in the same group as iron.

@Martin Kochanski ugh true. Yeesh. I suppose because the concept steel is linked to iron so firmly in our minds no one even considered that might have been what the OP was after. I'd better go edit that answer. Although the why in this case will be a pain to try explaining when the OP claims to not know much chemistry ...

Silicon, the element in Grouo 14 directly below carbon, is widely used in steels, especially in electrical steels and more recently in high-strength automotive steels.
Silicon does not replace carbon; it's a separate alloying element just as chromium or manganese would be. Although silicon is in the same group as carbon, the difference in atomic size means it does not fit interstially into the fcc iron lattice (where carbon has significant solubility when hot) or form the hard phases that carbon does upon cooling for strength. Instead, like most metals used in steel, it substitutes for iron atoms in the lattice. But it can impact the surface texture grain structure that impacts electrical steel applications, as well as the phase structure created in the presence of the carbon for the high-strength applications.
Heavier Group 14 elements are less commonly alloyed into steel. Germanuum is expensive, and tin and lead are often undesirable because the formation of low-melting liquids and impact hot mill processing or welding (hot shortness). However, they may be used in specialty applications such as lwad in free-machining steels.


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Why don't atoms with expanded
My title may be confusing, so let me explain. Let's take the molecule SO2 for example...

Above, you can see that the Sulfur atom has 6 electrons from its original valence shell involved in bonding/in a lone pair. However, if you add up these bonds, the atom has 10 electrons, not 8 (which disobeys the octet rule). I was told this is because sulfur has access to the d orbital of electrons. That makes sense, but why does it not get a 2- charge if it is getting 2 more electrons in it's valence shell?
Thanks!
Will

It is not getting any extra electrons. Why would it? Where from?

If this puzzles you, you can remove one of the two bonds between S and O and put it entirely on the oxygen atom which becomes negatively charged. The sulfur atom, having lost one electron, will then become positively charged, but it is now surrounded by 8 electrons.

This is because covalent bonds are when electrons are SHARED BETWEEN THE ATOMS, not all given to 1 atom (that's an ionic bond). The sulfur may gain a partial charge (positive in this case), because more electronegative atoms hold electrons closer, but never a full charge.


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How many electric vehicles can our current supply of lithium power?
This Reddit comment contends

There was 870 tons of lithium mined in 2019 in the US. That enough for under 20,000 cars, assuming you're using ~50 kg per car. So that would be 40,000 in 2023, assuming all of it is used in cars. In reality only a fraction is; most is used in consumer electronics etc.

The global car production is about 100,000,000 per year, meaning you'd need like 5,000,000 tons of lithium just for cars. Current WORLD production is 77,000 tons, although there's a large reserve.

Proliferation of electric cars is, as far as I can see, completely dependent on a large step forward in battery tech.

What exactly does there's a large reserve entail? How much Li does the "large reserve" contain?

Are these calculations correct?

See https://en.wikipedia.org/wiki/Lithium#Reserves

You assume there is X lithium and all of it cost Y. And then there would be no lithium to mine. Thats not quite right. As you are ready to pay more, harder to get sources become profitable and become part of available reserves. Sea has almost unlimited amount of lithium, but it is about x5 as expensive to mine.
The total lithium content of seawater is very large and is estimated as 230 billion tonnes
(wiki)

+1 and it's rather simple to leach practically all Li out of old batteries, so on the long run, recycling serves an increasingly big part of the demand, with battery efficiency also increasing.

Lithium isn't currently the limiting factor in battery tech. Cobalt is really the biggest Li-ion battery issue from a resource standpoint. That's why a lot of current research is focused on developing nickel-rich formulations (which, if successful, would make nickel an even bigger concern than lithium). Of course, this answer is absolutely correct (+1) and the same rationale applies to the other materials.

@PGnome I think you can go ahead and post that as an additional answer, just add a supporting link which shouldn't be hard.


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Permeability of containers to poop smells
I do a lot of hiking trips that involve driving for an hour or two to the trailhead in the predawn hours. I'm a morning pooper, so I need to poop during the drive. Before COVID, I would usually plan a stop along the way at a Starbucks, get coffee, and use their bathroom. Then if necessary I'd also use a bathroom at the trailhead. Right now, during COVID, these bathrooms are all closed, so my new normal is that I pull over somewhere in the countryside, poop, collect my poop using one of my dog's poop bags, and bring it with me to a town or trailhead where I can dispose of it.
Sometimes this was stinking up my car, even though the bag was tied shut, so I tried putting the poop bag inside a large plastic (PETE) can that some nuts came in, and sealing it with the screw-on lid. Even then, there have been times when the smell seeped out within 30 minutes. Sometimes the smell has been like sour milk, sometimes like sulfur or eggs.
If I want to make this work better, what should I be focusing on? Is ~0.5 mm of PETE likely to be permeable to the smelly volatile compounds involved? Or is the issue more likely to be that there's a poor seal at the screw threads? Would I be better off trying something like the glass jars with clamping lids that people use for their coffee? A tupperware-style container with an o-ring seal? What kind of material should I be looking for in a gasket or o-ring?
(The other option I've tried successfully is to clamp the poop bag under a windshield wiper, so it's outside. But I'm worried this would, uh, catastrophically fail if I then drove at freeway speeds.)

Wow, three downvotes and a close vote, none of them with any explanation whatsoever. Are people just that offended by the subject matter? This is just a question about permeability of plastics to organic chemicals. Please think before having this type of knee-jerk reaction of hostility.

Yeah, Chemistry SE is one of the least welcoming communities I've seen. They downvote/close everything that moves. What doesn't move - they push and then downvote/close.

I am against this down voting phenomenon which is only observed here in chem. The problem is chemistry gets a lot of lazy homework or medical questions. On the other hand, many good genuine and innocent questions  get a habitual down vote by some stubborn offenders. I rarely down vote but this question because it was too vividly described with unsolicited details. It could have been asked in better and formal terms.

The downvotes are one thing, and I imagine that’s related to the subject matter, although I can’t guess for sure. But I don’t understand the close votes, to be honest. How / why is this “opinion-based”?

On the other hand, though, Buttonwood’s answer does suggest that you might find more practical answers elsewhere on the SE network, too. There is not really all that much... chemistry in the answer. Perhaps somebody has studied the relevant permeabilities......

I chose the "opinion based" option on VTC, largely because Professor Crowell has not provided us with details about the composition of the feces in question, which certainly is required information to begin to address this question. Also, it's not focused: there are at least 5 sub-questions contained herein. Finally, the superfluous information concerning the morning stooling and unfortunate closure of the nearby Starbucks detract from the question.

@orthocresol True, there were no headspace GC-MS data provided, as sometimes used by Firmenich and Givaudan to offer new components for perfumes.  Despite some of the ingredients have «some» history (e.g., 3-methyl indole, by the way not liquid at ambient temperature), the revision aimed for something «ready to be used», with less sorrows than the chemical toilets seen on construction sites / festivals.

@ToddMinehardt, opinion-based question would be "which producer of goggles do you like the most" - there're no correct answers to such questions. If a question is missing details - you can ask for them. The question doesn't become invalid simply because you need more details to answer it (moreover some people are asking for extra details which aren't actually needed for answers, I see this all the time. This isn't specific to ChemSE though). As for "not focused" - that's when a post contains multiple unrelated questions. The question above is perfectly focused on one problem.

@StanislavBashkyrtsev Thank you for sharing your opinion.

Out of curiosity, is it against the law to dispose of poop bags in public garbage bins (as in a park) in your area? Does the same apply to human feces?

@BuckThorn This (obviously) depends on where you are.  In Switzerland, for example, you find dedicated garbage bins, robidog by Robidog AG: bins, bags, and even phone apps to localize the next bin. The local municipality decides if dog poop (only in a bag) may enter the normal public bin. And no, thankfully never seen to be used for human feces, there are public toilets (https://pee.place/en).

@Buttonwood I have been in similar situations to that described by the OP and have never had this problem using standard poop bags (double bag and tie a tight knot!). Maybe he's special. I guess my thoughts are that this is both logistical problem regarding access of a disposal site as much as of odor containment.

@BuckThorn Possibly yes, «logistical problem regarding access of a disposal site». I read about his in the newspapers, e.g. comparing the climb by Norgay and Hillary back in 1953 to today's situation in some of the camps today (https://en.wikipedia.org/wiki/Mount_Everest#Waste_management).  Esthetics aside, poop, especially in high altitudes, may affect a local ecosystem e.g., as an unwanted fertilizer.  But this is a different story.

@Buttonwood Reminds me of one of the ultimate poop bag/logistical disposal problems ever: http://www.spacesafetymagazine.com/aerospace-engineering/nuclear-propulsion/will-anyone-recover-apollo-13s-plutonium/

The problem likely is not (so much) the plastic material of your container, either solid, or bag.*)  Assuming there is no contamination on the outer side of the container, most likely the smells are noticed because of closing the container by either knot or lid is not tight enough.
After some search on sibling sites LifehacksSE and OutdoorsSE I learned hikers found multiple solutions to this problem (example).  Assuming you stay on ground and don't climb, a) a poop tube may be suitable for you. In this tube, you carry the waste with you until you find back to a place connected to a waste management / waste water facility.

(credit, source describes both construction and use)
Step #3 of this guide includes an important clue how to reduce the smell.  To quote:

«Before you go out camping, hiking, or climbing, preload the airtight bags with 2-3 small scoops of kitty litter or powder detergent to remove moisture. This will also keep obnoxious odors from leaking out of your poop tube.»

b) The idea to catch liquids and thus smell in a bag filled partially with some of (advanced?) cat litter is commercialized elsewhere, too (example).
Maybe this avenue is working well enough that additional sealing of the lid by an O-rings of suitable diameter and material (viton is a material resistant to many chemicals, but I don't dismantle a vacuum deposition to check this field of application) is not necessary.
*) Although permeabilities of liquids and gases across plastics may be determined experimentally.


text:
Which mechanism would be followed by a certain reaction with good leaving group?
In my class test, I was asked this question:

Which mechanism would be followed by a certain reaction with good
leaving group?

E1 reaction

E2 reaction

No reaction

both E1 and E2

I answered 4 because they both occur simultaneously the main difference comes in the amount of product formed by them. But, I was awarded with negative marks.
Can anyone please tell if I was wrong on my part or my teacher?

One point is: Your comparison of E1, E2 and E1cb should equally consider if there is synchronicity (e.g., the approach of a base and leaving of an other group across a transition state) or one after the other (passing intermediates, e.g. a base deprotonates the substrate, the carbanion then expels an other group) to yield e.g., an alkene.  Which is not the same as letting $\ce{X-}$ leave to generate an intermediate carbocation eventually loosing H to yield an uncharged alkene.

the reaction can follow any pathway just the conitions make one case or the other more probable

Yes there are reactions which possibly could proceed by more than one pathway, already because on occasion it may be easier to reject a suggested mechanism than to collect supporting experimental evidence.  Thus the suggest that you add to your question the reaction in question, to set it in specific context to your question.

Take note also of the temperature at which the reaction is allowed to move— E1 reactions usually perform under high temeperatures (based from my Org Chem class in first year).

IMO this question is nonsensical because you can't generally make a strong statement on ideas in organic chemistry. There are no hard and fast rules.

@Buraian but then the options were also nonsensical. Either provide appropriate(non debatable options)

Well of course, nonsensical questions can only have nonsensical options :-)


text:
Can copper sulfate react with terpenes (such as limonene and terpinenes) at room temperature?
I am just wondering, can I make a metalorganic or organometallic complex of copper and limonene/terpinene/phellandrene in the form of lime oil? If so, what is the formula of the complex formed? I tried the reaction with lime oil in iPrOH, and a copper sulfate solution. The result was a blue crystalline solid. What are the crystals? I'd like to know.

I have a video of the reaction that I performed: https://www.youtube.com/watch?v=PeJhFKj1nuw

By the way, the compound is an opaque blue solid, and I used iPrOH to dissolve and extract the terpenes, as well as to wash the crystals.

The terpenes in the oil, I mean. It's poorly soluble in water, and insoluble in iPrOH.

Most of the evidence you present, together with solubility data from Wikipedia, seems consistent with just precipitating the copper sulfate.  Note that because of spectrochemical effects, we would expect the complex to change color if organic ligands were to displace water.

Yes there is nothing in terpenes to coordinate with copper ion. You got copper sulfate again.

For the record, after M. Farook's agreement I am making my comment an answer.
Most of the evidence you present, together with solubility data from Wikipedia, seems consistent with just precipitating the copper sulfate. Note that because of spectrochemical effects, we would expect the complex to change color if organic ligands were to displace water.


text:
(Probabilistic) entropy associated with many equally probable microstates
This is a pretty naive question. The probabilistic definition of entropy is (for a mole of particles)
$$-R\sum_i p_i\ln p_i $$ in which $p_i$ is the probability of a given microstate and $\sum p_i =1, p_i\geq 0.$
If we have a large system whose microstates are equally probable the terms of $-R\sum p\ln p$ will trace a curve like that below (image on Wolfram Alpha), the area under which is (up to some multiplicative constant, omitting R) $ A=0.25.$

My question is whether there is any physical intuition or significance to the shape of the curve. In particular, why would the entropy associated with identically probable  microstates not be the same?
In case it sharpens the question, I would expect that as we add up entropy associated with equal-probability microstates the entropy would accumulate in a linear way. That is evidently not the case.
Thanks for any insights.

The conversation has been moved to chat.

Time does not appear here so the idea of adding bits of entropy seriatim is artificial. Maybe better to think of it as an instantaneous picture of an existing system. Does the picture make physical sense in that case?


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Role of calcium oxide in soda lime decarboxylation
Why do we use calcium oxide $\ce{CaO}$ in soda lime decarboxylation?
$$\ce{\underset{sodium acetate}{CH3COONa} + \underset{sodium hydroxide}{NaOH} ->[\Delta][CaO] \underset{methane}{CH4} + \underset{sodium carbonate}{Na2CO3}}$$
It seems as if it is dehydrating agent to keep NaOH dry), but why is the dehydrating agent required if the heat would be enough to dehydrate the container as the temperature is very high (expected to be over 100 °C)?

If you dehydrate with P205 you get the anhydride. Sodalime first forms the sodium salt. Sodalime is used in preference to sodium hydroxide for ease of handling https://www.chemguide.co.uk/organicprops/acids/decarbox.html

@Waylander I am working with sodium salt already

@Poutnik it feels no difference the reaction i provided is just one step ahead

@Poutnik sorry i didn't catch that

Calcium oxide is the typical catalyst used for this reaction. Caesium salts also catalyze this reaction. wikipedia - sodium acetate

NaOH does not need an extra note it is sodium hydroxide, CH4 is methane etc.

Calcium oxide is not necessary for carrying out this reaction. It only prevents NaOH from touching the glass of the container where the reaction is done. If a solid mixture of sodium acetate and NaOH is heated to high temperature, some NaOH will attack the glass before reacting with acetate, and make the glass melt down. If NaOH is intimately mixed with CaO, this trouble is avoided. The glassware is saved.


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How to separate copper(II) chloride from 4-iodophthalic anhydride?
I am planning a reaction in which I iodate phthalic anhydride. My reaction is I2, phthalic anhydride, and copper(II) chloride as a catalyst. The solvent is 99.9% isopropanol. How do I separate the product (4-Iodophthalic anhydride) from the copper(II) chloride and remaining iodine?
Edit: I planned to boil away the solvent, and sublimate away the iodine from the remaining solid. How do I separate my product from the resulting mix? At the moment, I only have isopropanol.

I would recommend just running a normal flash column since you will likely have unreacted phthalic anhydride after running your reaction. If the reaction did have a very high yield (near 100%), a column may not be necessary. You could just add a bit of sodium carbonate to convert the iodine to sodium iodide, remove the isopropanol using a rotovap, and then use a separatory with methanol and ether (or hexane) to separate the salts (sodium iodide and copper chloride) from the product.

I'm an amateur chemist, and I don't really have a flash column, so I can't really do that, but thanks for suggesting that.

@Eli Jones: How does sodium carbonate reduce iodine to iodide? Sodium sulfite or sodium thiosulfate should do the trick.

Is there any concern that the  isopropanol  may open the anhydride?

@user55119 I agree that sodium thiosulfate would probably be better for neutralizing a bunch of iodine, but since only a small excess of iodine will be present, I read that sodium carbonate will convert iodine to sodium iodide and sodium iodate. I'm not sure how effective this is, so I can't guarantee this method, but it would probably be better than dealing with the messy sulfur byproducts using the other method.

I read that isopropanol does not open the anhydride.

I just realized I can add sodium hydroxide, and filter off the resulting insoluble copper(II) hydroxide and sodium chloride, then boil down the result. If you can think of a better answer, feel free to suggest it to me.

The sodium hydroxide would probably destroy your phthalic anhydride (hydrolyze the anhydride moiety), and your product would still be contaminated with iodine and leftover starting material.

Okay, then what should I do? I don't have any ether. Also, would sodium acetate work instead of sodium hydroxide?

I just mentioned ether because it was a common nonpolar organic solvent. You could try something like gasoline, but I cannot guarantee how effective this will be. Also, be safe while removing your solvents from the product! Since you probably do not have fancy equipment, you should only use small amounts of solvent while outside and away from any open flames.

I would recommend trying to use either sodium carbonate (you might have to heat the solution a bit for neutralization) or sodium thiosulfate. Sodium carbonate is sold commonly as washing soda, and sodium thiosulfate is also cheap and readily available. Sodium acetate might result in some transesterification of the anhydride. Also, I haven't read of sodium acetate being used to neutralize iodine.


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Electrolysis of copper chloride in solution of Vinegar + H2O2 + NaCl after copper etching
I did copper etching (to make a PCB) using a solution of vinegar + H202 + NaCl and now I need to dispose the solution. There is no waste collection for this kind of stuff where I live, so after some research I saw that the cleanest dispose process seams to be to make electrolysis.
I used a copper electrode at the negative pole (cathode), a nail on the positive (anode) and plugged a 9V power supply. I was expecting some bubbling on the nail (Cl2 gas) and the Cu2+ to accumulate on the copper electrode, but the bubbling happened only around the copper wire and after  15 min the solution started to get really dark brown.
My goal is to get the copper out of the solution, so that I can throw the solution away in the drain and the copper in trash.

Is this what is supposed to happen, or did I do something wrong? How do I know when it's done taking the copper from the solution? when done, can I just filtrate and then pour the solution in the drain?

Why the bubbling happened at the cathode and not the anode and why the solution turned brown?

Update
the result after some 2-3 hours. how to know if it is done?

My goal is to get the copper out of the solution || Make a solution of baking soda, mix it with your remains, leave it for a few days, decant clean water, filter off the precipitate, dry it. The result is basic copper carbonate which is mostly harmless, and the solution would be sodium chloride with minor impurities, reasonably safe to go down the drain || You CAN try to reduce it to get metallic copper out, but it is quite slow and isn't easy to do right (I tried).

thanks for reply. but what is mostly harmless? can I just throw it in the trash or is there a "cleaner" option? and how to know how much baking soda is necessary? only by the color?

You should not eat it. Otherwise, it's safe. In my country I would wash it down the drain. If I wanted to be particularly responsible, I would dry and heat the solids until it becomes completely black before washing it down the drain. As for amount of backing soda, you need need one molecule of soda per one $HCl$ molecule in the the etching solution and then a slight excess. Assuming ~35% $HCl$, this should be around 90-100 g of solid baking soda per 100 ml of $HCl$ solution used for the etching solution.

looking at density table Actually, to be safe, I would go for about 110 g. I didn't check the solution density.

A simple test to know whether a solution does contain copper ions is to filtrate it and then to add some milliliters of ammonia solution. If the solution contains copper ions, it will take a deep blue color, and no precipitate. If not it will make a brown precipitate, the solution contains Fe ions.

Mix with sodium hydroxide drain opener (outside!) The metal ions (copper and iron now) will combine with the OH- ions and drop out of solution as harmless iron and copper hydroxide and can be thrown out. The Na+ ions will combine with the Cl- ions and form aqueous NaCl and can safely be put down the drain.
Even if you put too much sodium hydroxide in your solution, it’s still safe to put down the drain because it’s less caustic than it started as when it was drain opener.
If you’re still uneasy about disposing of the copper and iron hydroxide, burn it and the hydroxides will turn into metal oxides.

interesting. should the copper and iron hydroxides formed be filtered out of the solution? or can I throw them together down the drain?

I wouldn't put the hydroxides down the drain - there's no need to add additional metals to your wastewater for your local water treatment facility to deal with. Also, the inside of most people's plumbing isn't as smooth as people expect it to be. Put solids in the drain and it's going to cause you problems when it will be most inconvenient (i.e. when you cannot get in touch with a plumber!)

The best thing to do is to put some nails in the copper solution, or anything made of iron. You will soon see that the nails ge. covered by metallic copper, which is red-brown, After maybe one hour, the solution is not blue any more. So it does not contain copper ions anymore. You can throw this solution down the drain : it will contain iron instead of copper, and it looks like rust. And you can throw away the nails covered with copper. The equation of the chemical reaction is :  $$\ce{Fe  +  Cu^{2+}->  Fe^{2+} + Cu(s)}$$ Quite often, these $\ce{Fe^{2+}}$ions are then oxidized in the air to produce a brown deposit of $\ce{Fe(OH)3}$ according to $$\ce{4 Fe^{2+} + 3 O2 + 6 H2O ->  4 Fe(OH)3}$$ It is a typical chemistry demonstration that probably all chemistry teachers in the world present in their high school classes. It never fails.

Even better could be a galvanic cell , formed by connected iron and copper plates. This would have advantage iron would not get covered by copper.

@Poutnik. You are right. But my method is quickly done, and does not require any equipment.

Iron +3 is very good in masking copper coloration. BTW, copper-chloride solutions are often green.


text:
Determination of Cell constant
I read that cell constant is measured indirectly using a solution with known conductivity(like KCl).
But then how was the conductivity of that solution measured? (Is it by some theoretical consideration or another accurate experiment?)

The gist is that  using a detailed procedure a chemist should be able to make the standard solution very reproducibly.

That is not is question. His is more like a chicken and egg query. Which came first?

The conductivity or the cell constant?

Conductivity measurements were done in late 1800s by Friedrich Kohlrausch, who painstakingly measured the conductivities of solutions so accurately that they are still valid today. Most of his work is in German was published in relatively unknown journals. In his time, conductivity theory was not advanced enough to predict the conductivity of simple salt solutions in water, so there was no a priori theoretical prediction of conductivity, from the first principles. I don't think it is doable today after 100 years (may well be wrong(.
However, the accurate measurement of resistance was well established in physics. See below.
A paper in the J. Am. Chem. Soc. 1933, 55, 5, 1780 describes what Kohlrausch and his meticulous colleagues they did. Note the level of details which is remarkably missing from papers today.

After Kohlrausch had developed his method of measuring the electrical
conductance of solutions by means of an alternating current bridge, he
undertook with the aid of Holborn and Diesselhorst’ the determination
of the specific conductance of seven standard reference solutions in
absolute units in order that the results might be available to himself
and to other experimenters for the calibration of conductance cells.
For this purpose they used two cells whose cross sectional areas, a,
and length, l, were determined mechanically. The cell constant,
l/a, could then be computed in absolute units (cm$^{-1}$) from the measured dimensions.
The cell constants of a considerable number of other cells with
sealed-in electrodes of the usual type were determined electrically by
comparison with the primary cells by measurements of resistance when
filled with a common solution. These cells, whose cell constants were
thus determined indirectly, were then used for the determination of
the specific conductance of the seven standard reference solutions at
approximately 0, 9, 18, 27 and 36 C. The specific conductance of each
of these solutions for every degree between 0 and 36 C was obtained by
interpolation. No actual measurements were made at either 20 or at
25’. The water-bath used to fix the temperature of the cells was not
provided with an automatic thermostatic control. The majority of the
data on conductance in the literature are based on the values ascribed
to these standard reference solutions by Kohlrausch, especially the 1
N and 0.1 N potassium chloride solutions.
In the original paper by Kohlrausch, Holborn and Diesselhorst it is
stated that the normal potassium chloride solution was prepared by
dissolving 74.555 g. of potassium chloride weighed in air and making
up to 1 liter at 18 C. The 0.1, 0.02, and 0.01 normal solutions were
prepared by dilution by volume. The book “Das Leitvermogen der
Elektrolyte” by Kohlrausch and Holborn, which was published in the
same year as the paper referred to, contains alternative directions
for the preparation of these solutions by weight. This book has
probably been used by other experimenters more than the original paper
because it is more generally available.


text:
Chirality and Optical activity
Do all chiral objects rotate the plane of polarization of EM waves? For example my hand is chiral, will it rotate the plane of polarization of EM waves of suitable wavelength?

"optical activity" is a problematic term if only because it implies EM radiation in the visible range. Unfortunately there isn't a simple general term. Though you could say, "can chiral macroscopic objects rotate the plane of EM waves of a suitable wavelength?"

Of course it will, provided you have a billion of it. (One molecule does not rotate much, after all.) Also, each should come without the rest of the body, or at least without the other hand, otherwise that would make a racemic mixture. (I'll leave the grisly details to the reader's imagination.) Also, each hand should form the same gesture, otherwise the destructive interference will ruin the effect.
Also, light of suitable wavelength would be microwave radiation, and a cursory search seems to indicate that the methods of polarizing it and measuring its polarization are known.
With all these conditions fulfilled, I don't see any difference from the situation with the molecules and ordinary light.

Slightly off-topic, but I wonder if it's possible to formalise some concept of "maximising chirality", such that a single interaction between a chiral photon and a chirally-maximised object would be significantly different depending on the relative chiralities. With molecules this may be hard to investigate smoothly, but perhaps macroscopic objects could be more freely adjusted and studied.

Screw-shaped things like helicenes come to mind.

and some types of achiral liquid crystal forming molecules have chiral helical phases.

Human hand is a macroscopic object, even if there are billions, they will not rotate light. Optical rotation is a molecular phenomenon not a macroscopic phenomenon.

Microwave radiation is not exactly light.

Still a lot smaller than a macroscopic object.

How so? Its wavelength is rather macroscopic, as to me.

@IvanNeretin, My understanding has be smaller than the wavelength. Let us roughly say microwaves have 1 cm wavelength, in order to see microwave rotation, the objects have to less than that. See this linked paper,  they used springs to show optical rotation with microwaves but the springs were smaller than the center wavelength. https://ieeexplore.ieee.org/abstract/document/7775673https://ieeexplore.ieee.org/abstract/document/7775673

Great! You have found the evidence that the effect does exist and can be measured for microwaves, after all! As for the size, why bother; just take a wavelength 10 times longer (those still would be microwaves).

The OP should not call it an "optical" rotation. Rotation of polarized long waves is more than century old phenomenon. Blocking of polarized radiowaves with wire "polarizer" is a nice demo in physics. Still a human living hand would not rotate microwave radiation. It will simply absorb it quite strongly. I have worked with microwave spectrometers, and trace of water is a headache.

Who said "living"? As for the rest, well, yeah, the OP's terminology needs correction in that the word "optical" typically refers to the visible light and not to the other ranges of the EM spectrum. Ditto for the word "light", BTW.

The wavelength has to be comparable with the object size in order to see those effects. Your macroscopic hand will not rotate ordinary light even if you have millions hands around, because the wavelength of light is so small. Yes, it will scatter light as a large object. This idea of optical rotation by long wavelengths and by large objects was shown more than 100 years ago by J.C. Bose, using twisted pieces of jute (a type of rope made from plants). Bose was in India and jute was a common material there to play around.
See this article.
https://www.jstor.org/stable/115973?seq=1.

I think you got the the 'Bose's mixed up. The Bose whose you're referring to is JC Bose, who was the one who did many experiments with EM waves. The other Bose is SN Bose who made contributions to Quantum Mechanics.(Bose-Einstein)

You are right Milan. I corrected it.


text:
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.


text:
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


text:
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.


text:
Aromatic Character in Percentage?
I was just searching whether Uracil is aromatic or not and in this link$^1$ it is said that (from the abstract):

High-level computational methods were tested for the enthalpy of hydrogenation of uracil, but only with G3 was possible to obtain results in agreement with the experimental ones. It was found that uracil possesses 30.0% of aromatic character in the gaseous phase.

What is the meaning of this sentence? I have only basic knowledge on Aromaticity limited to Huckel's rule and the names of a few aromatic compounds. I didn't expect Aromaticity to be expressed in percentages. Can someone explain why?
Reference

Tiago L. P. Galvão, Inês M. Rocha, Maria D. M. C. Ribeiro da Silva*, and Manuel A. V. Ribeiro da Silva. J. Phys. Chem. A 2013, 117, 28, 5826–5836.
https://doi.org/10.1021/jp404938u

Nothing is truly black or white. Consider pure ionic and pure covalent bonds to be unobtainable extremes. Every real bond is somewhere in between.

I found this article but I do not have  access to it.

Technically, a good point is made here. What is the meaning of "100% aromatic?"

@Zhe I'm sure the paper explains the meaning of the quoted percentage. ;) Ratio of prefactors in a linear combination of orbital models?

Assuming it is more or less clear to all of us what aromaticy means, and the ways it can be expressed or assessed, then one can construct many comparative scale. One can think of termochemistry, another of chemical shifts, and relate these to the corresponding in benzene, just to give an example. Or, it can come directly by MO calculations at various levels. The most straightforward would be that the phi of aromatic orbital has coefficient of 0.3; or as a comparison is likely made, a ratio between prefactors as @Karl said. This is to say that you can get the message without overthinking.

I took the liberty of expanding the quoted text for clarity. Feel free to rollback the change if you disagree this improves your post.

@MilanPaul if you'd like to access that paper you linked to in the comments, you could use Sci-Hub. Disclaimer: it's legality is questionable

G3 is generally a term used to describe a particular computational model used in calculating molecular bonding in quantum chemistry.  The G refers to Gaussian and 3 is just a third iteration of including additionally complex terms to the molecular orbital calculations being used.  There's a brief discussion on wikipedia, but you'll find more extensive discussion of these types of calculations in Quantum Chemistry text books or directly from the researcher's works just an example, a G3 Theory site.
Aromaticity can actually be measured quantitatively via Aromatic ring current and calculated with [NICS] nucleus-independent chemical shift.  The wikipedia article included a link to a paper describing the quantitative scale.  It's just a way of saying how aromatic is the molecule compared to another (call it a standard / perfect example molecule.)  The problem with such relative standards is there may always be one that exceeds the standard, so you could potentially end up with results larger than 100%.

Looks like I need to learn a lot....

Can you also explain the need for quantitative measurement of Aromaticity(in specific percentages). When we say that a compound X is y% aromatic w.r.t a particular method of quantitative measurement, what does it actually say about the molecule(other than the experimental/computational result itself)? If a compound X  has a higher 'percentage of aromaticity' than a compound Y, is X more stabilised by aromaticity than Y?


text:
Disposal of Silver Halides dissolved in Ammonium Hydroxide
I was reading the Lab Manual for Chemistry prepared by the NCERT . While going through the the test for Chloride ions (pg 88) I came across the following Hazard Warning :

The solution obtained after
dissolving AgCl/AgBr/AgI
precipitate in ammonium
hydroxide should be
acidified with 2M HNO3
and should be discarded
quickly to avoid serious
explosion.

How does it cause a serious explosion?

See these questions: https://chemistry.stackexchange.com/questions/43441/is-tollens-test-safe .... https://chemistry.stackexchange.com/questions/67395/mirror-reaction-avoiding-silver-nitride ... https://chemistry.stackexchange.com/questions/122666/will-disodium-edta-effectively-destroy-tollens-reagent

In the test for Cl- the chloride ions are converted into AgCl by the following reaction,
Cl- + AgNO3
→
NO3- + AgCl
which is then dissolved in ammonium hydroxide by the formation of diammine silver chloride by the following reaction:
AgCl + 2NH4OH
→
[Ag(NH3)2]Cl + 2H2O
On standing, the diammine silver complex subsequently breaks down to Ag3N.
The standard free energy of Ag3N is about +315 kJ/mol . Hence it is unstable and decomposes explosively to metallic silver and nitrogen gas .
2Ag3N → 6Ag + N2
So the diammine silver complex has to be destroyed after use so that it does not break down into Ag3N.
One way of destroying the complex is by adding nitric acid.
[Ag(NH3)2]Cl + 3HNO3
→
AgNO3 + 2NH4NO3 + H2O
So in all reactions involving the formation of diammine silver complex, the complex must be destroyed using acids after the purpose.
The complex is formed when AgCl, AgBr(sparingly soluble)and AgI(the complex will be formed only in trace amounts) are dissolved in NH4OH . The comlex is also formed in the well-known Tollen's test. Tollen's reagent is [Ag(NH3)2]OH. Ammoniacal Silver Nitrate is also used in Fontana–Masson Stain and also in silver mirroring(such as inside insulated vacuum flask).
The Wikipedia page on Silver Nitride
A reported accident

Actually, the mechanism of $\ce{Ag3N}$ creation (and also the associated products, Silver amide $\ce{AgNH2}$, and the imide $\ce{Ag2NH}$), have not been definitely accounted for, either by accident investigation reports or per a recent review of the literature, on possible paths to explosive Silver nitride.
I have ideas of a wider chemistry spectrum based answer which accounts for the apparent role involving $\ce{Ag2O}$.
To begin, I cite the more common background description, per a government source, to quote:

Silver nitride is an explosive chemical compound with symbol Ag3N. It is a black, metallic-looking solid which is formed when silver oxide or silver nitrate is dissolved in concentrated solutions of ammonia, causing formation of a silver-amide or imide complex which subsequently breaks down to Ag3N.

So, of interest is the cited reaction:
$\ce{Ag2O(s) + 4 NH3 + H2O ⇌ 2 [Ag(NH3)2]OH}$
where I would expect any Silver oxide formed from the reverse reaction of dissolving Ag2O in ammonia would then likely create a high surface area (with enhanced reactivity) $\ce{Ag2O}$. It should also importantly be noted that the reaction, itself, is reversible (see as a source, "Second year college chemistry" by William Henry Chapin, page 255), to quote:

As might be expected, the silver-ammonium complex dissociates slightly into its constituents as indicated by the equation

$\ce{[Ag(NH3)2]+ ⇌ Ag+ + 2 NH3}$

This is a reversible reaction, very much like the ionization of a very weak acid or base.

So, once any Ag2O is formed (by say moving the equilibrium to the left from a possible loss of water on standing in an open vessel or upon addition of dry alcohol), a radiation source could further accelerate the process (see, for example, Ag2O as a New Visible-Light Photocatalyst: Self-Stability and High Photocatalytic Activity). To quote from the abstract:

Ag2O is unstable under visible-light irradiation and decomposes into metallic Ag during the photocatalytic decomposition of organic substances. However, after partial in situ formation of Ag on the surface of Ag2O, the Ag2O-Ag composite can work as a stable and efficient visible-light photocatalyst.

As such, the introduction of appropriate light, or other irradiation paths (like X-rays commonly associated with radical formation) may be instrumental. In fact, here is an interesting pertinent radiation study: The X-ray activated reduction of silver (I) solutions as a method for nanoparticles manufacturing.  To quote from a passage discussing an associated radiation treatment:

The next step was aimed at eliminating glucose-containing ammonia solutions. The reactions taking place in these solutions resulted sometimes in the formation of precipitates with unstable composition. These precipitates contained considerable amounts of silver nitride Ag3N, which formed in some ammonia solutions of the silver salts and exhibited explosive properties in a dry state [15]. The results from analysis of the deposits obtained for three selected types of solutions are presented in Table 1.

As the irradiation of glucose is a path to radicals, my proposed mechanics incorporates an ostensible promotive radical pathway for the allude to creation of $\ce{Ag3N, Ag2NH and AgNH2}$, which also appears to be accelerated in the presence of $\ce{OH- and Ag2O}$ in the presence of ammonia.
The proposed reaction path is:
$\ce{NH3 + H2O ⇌ NH4+ + OH-}$
$\ce{NH4+ ⇌ NH3 + H+}$
$\ce{[Ag(NH3)2]+ ⇌ Ag+ + 2 NH3}$  (per above)
In the further presence of light:
$\ce{Ag+ + hv -> •Ag + h+}$
Source: See Photodecomposition and Luminescence of Silver Halides
$\ce{OH- + h+ -> •OH + hv}$  (same source as above)
$\ce{OH- (aq) + hv -> •OH + e- (aq)}$
As a reference, see: Flash photolysis in the vacuum ultraviolet region of sulfate, carbonate, and hydroxyl ions in aqueous solutions
$\ce{NH3 + •OH -> •NH2 + H2O}$
From a related work: Kinetics and Mechanism of the Reaction of •NH2 with O2 in Aqueous Solutions
$\ce{NH3  + •H <=> •NH2 + H2}$
As noted in this work.
$\ce{•NH2 + e- -> NH2-}$
$\ce{Ag+ +  NH2- <=> AgNH2}$
$\ce{AgNH2 -> Ag2NH + NH3}$
$\ce{AgNH2  + Ag2NH -> Ag3N + NH3}$
Note, my suggested pathway is largely promoted from the recognition of Ag2O as a recent (article was 2011) visible-light photocatalyst with the possible introduction of contributing radical based reactions.
In summary, I am suggesting chemically diverse mechanisms, involving both light and radical based reactions, as a more likely explanation.

In layman's terms, silver would have to be in the form of an oxide to set the creation of silver nitride, when dissolving in ammonium hydroxide. It is a simple tradeoff... the silver oxide takes on the nitrogen from the ammonia and donates it's oxygen to the hydrogen. Then you get water and silver nitride. Without silver being bonded to the oxygen, silver will not form an explosive compound when ammonium hydroxide is added to the reaction.

Welcome to Chemistry Stack Exchange. Please visit here - https://chemistry.stackexchange.com/help/how-to-answer - to see how you can improve this answer.


text:
How to find the equilibrium constant under non-standard conditions?

Given the following thermodynamic data at $\pu{298 K}:$
$$
\begin{array}{lrrr}
\hline
\text{Compound} & \Delta_\mathrm{f}H^\circ/\pu{kJ mol^-1} & S^\circ/\pu{J K^-1 mol^-1} & C_p^\circ/\pu{J K^-1 mol^-1}\\
\hline
\ce{HI(g)} & 26.5 & 206.6 & 29.2 \\
\ce{I2(g)} & 62.4 & 260.7 & 36.9 \\
\ce{H2(g)} & 0.0 & 130.7 & 28.8 \\
\hline
\end{array}
$$
determine the equilibrium constant $K$ for the following reaction at $\pu{475 K}$ and $\pu{1 bar}:$
$$\ce{I2(g) + H2(g) <=> 2 HI(g)}$$
NB: Do not assume that the enthalpy and entropy of reaction are independent of temperature. What assumption(s) do you need to make?

At first I thought I could work out the enthalpy change using Kirchhoff’s law of thermochemistry
$$\Delta H(T_2) = \Delta H(T_1) + \int_{T_1}^{T_2}\Delta C_p\,\mathrm dT$$
and then find the entropy change of reaction similarly. Then apply
$$\Delta G = \Delta H - T\,\Delta S$$
and use the equation relating $\Delta G$ and $K,$ but I know that is only for the Gibbs free energy change under standard conditions. The question tells me to assume the enthalpy and entropy depend on temperature, so I’m not sure what my approach should be for this.

The first statement about $\Delta_rG$ is not correct. It should be correct if $K$ is replaced by $Q$

Wouldn’t it be -RTlnK+RTlnQ?

I don't think so. Look ! If $Q$ is approaching $K$, your formula would get $\Delta G = 0$, and $K = 1$.  This is not correct.

Do you know how to quanify the effect of temperature on the equilibrium constant?

The Van‘t Hoff equation? Not sure how I should use the Cp

That's correct.  But first you need to use the Cp's to express the $\Delta H^0$ as a function of temperature.  Then you have to integrate the Van't Hoff equation taking into account the temperature dependent of the $\Delta H^0$.

That makes sense. So I don’t use the deltaG= deltaH-TdeltaS at all? I don’t need to find the entropy of reaction?

You need that to get K at 298K.

Ok. So is the assumption that Cp is independent on temperature? Or is it that deltaH and deltaS only vary slightly with T?

And is it ok to just substitute the modified deltaH into the already integrated regular form of the Van’t equation?

They expect you to assume that the heat capacity is independent of temperature, so that $$H^0(T)=H^0(298)+C_p(T-298)$$It isn't OK to use the integrated regular form.  You need to integrate the new version.

Ok, thank you so much for this. Has this been done before so I can check the final form of my integrated expression?

Have some confidence and integrate it yourself.

I have certainly given it a go, but I have no idea whether it is right or not

Using the Gibbs- Helmholtz equation with deltaG(standard)=-RlnK and integrating with the equation with the modified deltaH i end up with the same as I would have gotten had I just substituted it after. Is the T in the (T-298 ) not just a constant and not a variable? It would be the T we are wanting to find the new K at. Sorry I am taking up so much of your time. I really appreciate it

Ok with a fresh mind I realised what I said was just silly. I integrated properly (i hope) and got K=137. I would write out my full expression but I’m not sure how to do that on here. Does my final answer look right?

@MarinaCalder You can post an answer to your own question, it's a good practice on SE sites. If you are still having difficulties with formatting and typesetting math, post it the best way you can assure it's readable, and someone will edit it. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

There should be used the similar integrating formula for temperature dependence on reaction entropy, with the temperature in the integral denominator ( Cp . dT/T = dQ/T = dS ).


text:
Use of powder of kernel of tamarind seeds for water purification
In the chapter Environmental Chemistry in our textbook it's given that Powder of Kernel of Tamarind seeds can be used for water treatment as a substitute for alum.
But if I understand correctly, alum ionizes in water and these ions neutralize the electrical double layer surrounding colloidal particles. How does a substance like 'Powder of Kernel of Tamarind seeds' produce the same effect?

The key is to know the composition and behaviour of this powder what is mostly unknown to chemists. My understanding is that alum hydrolyzes, forming colloidal hydroxide, which adsorbs these colloidal particles. The powder may form organic colloids of similar properties, settling later  down as aluminium hydroxide does.

mostly unknown to chemists – better expression may be "unknown to most chemists".

Alum is a flocculating agent which causes  suspended solids, like fine clay particles, to form aggregates and precipitate.

In my understanding,, Al^3+ in near neutral environment hydrolyzes faster than it gets chance to interact with colloids.