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How do I calculate enthalpy change calculations questions?
Below is a question I'm struggling with.

When $\pu{0.1 mol}$ of aqueous potassium hydroxide was added to $\pu{0.1 mol}$ of nitric acid, $\pu{5200 J}$ were transferred to the surroundings. What is the enthalpy change, in $\pu{kJ mol-1}$ for this reaction?
$$\ce{KOH(aq) + HNO3(aq) -> KNO3(aq) + H2O(l)}$$
A -52
B -26
C +26
D +52

This is how I tried to solve the question.
$$Q = \pu{5200 J} = \pu{5.2 kJ}$$
\begin{align}
\mathrm{Moles} &= 0.1 + 0.1 \\
&= \pu{0.2 mol}
\end{align}
$$\Delta H = \frac{Q}{\mathrm{mol}}$$
$$\Delta H = -\frac{5.2}{0.2} = \pu{-26 kJ mol-1}$$
The answer is actually A. What is the reason for this?

Not all numbers, that can be added, should be added. // Would you say "Kilograms = 0.1 kg + 0.1 kg = 0.2 kg"? Wouldn't you use mass or m?

You divide by 0.1 mol only. Think of it this way. Let's change this question to using 0.1 mol $\ce{KOH}$ and 0.2 mol $\ce{HNO3}$. Technically, this reaction should only produce $\pu{5200 J}$ still because you only used 0.1 mol of $\ce{KOH}$ which is the limiting reagent as the reagents are in a 1:1 ratio. By your logic, however, this could compute to $\pu{\frac{5200 J}{0.3 mol}} = \pu{17.3 kJ}$ which is now different than you answer of $\ce{-26 kJ}$. Therefore, you don't consider the moles as the sum of the moles of reactants. Instead, you have to look at what's limiting in this specific calculation. In this case, they are 1:1 ratio and there is 0.1 mol of each so just divide $\pu{5200 J}$ by 0.1 mol.


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Bonding in hypervalent molecules
I don't know where they go the electrons of the extended-valence in molecules like $\ce{SF6}$ or $\ce{PCl5}$ for instance because my teacher said that the d orbitals can't interact with s and p (or hybridised $\ce{sp^3}$) as they are far away energetically. So if the hybridisations $\ce{sp^3d}$ and $\ce{sp^3d^2}$ don't exist, how does this work?
I've also read about it in "Petrucci Ed.11" and this is presented more or as an unsolved mystery that will continue creating controversy.
Do you know any paper that disapproves the $\ce{sp^3d^2}$ model?
Furthermore, there's a picture that shows the molecular orbitals of $\ce{SF6}$. I can't understand the process to deduce them but I hope this is a more accurate version of the molecule (rather than using hybridisation). It seems to demystify the octet rule and just focus on electron pairing. All this gives me a different vision of chemistry, like for example, what if we had diatomic alkaline metals? (I know that there are some calculations about them, but I'm not pretty sure if they can be made at the lab.)

https://en.wikipedia.org/wiki/Hypervalent_molecule

The easiest way around this is to invoke 3 center - 2 electron bonds. Alternatively, you can also have resonance with two ionic bonds in $\ce{SF6}$ or one in $\ce{PCl5}$, with appropriate formal charges on sulfur and phosphorus.

Related: https://chemistry.stackexchange.com/a/32268/72973

@karsten 3-center 2-electron bobds? I thought it was 3-center 4-electron bonds.

Well, but 3c-2e is used in hypovalent molecules like diborane, so why in hypervalent ones?

Related: $\ce{PCl5}$ bonding, https://chemistry.stackexchange.com/a/18544

Wikipedia says about 3c-4e, yes

@JorgeBonifaz The common $\ce{PCl5}$ Lewis structure shows 10 electrons around the central atom. If two of the chlorine atoms and phosphorus are in a 3 center 4 electron bond, you only "need" 8 electrons from the central atom and stay within the octet. Describing it as $\ce{PCl4+ + Cl-}$ has the same effect.

Ok, but if it's like that, it's ionic, and not a molecule anymore....

With resonance, you get a bond order of 4/5 for PCl5, pretty covalent. Here is a paper (behind a paywall, but the abstract is available) on the topic of d-orbitals not participating: https://pubs.acs.org/doi/pdf/10.1021/ja00160a022

https://chemistry.stackexchange.com/questions/6785/how-can-the-bonding-in-if7-be-explained-using-lcao-method https://chemistry.stackexchange.com/questions/86275/why-is-there-only-one-bond-length-for-sulfur-hexafluoride

@OscarLanzi I read up on the nomenclature. Everyone says 3c-4e bonds, so sorry for the (my, mostly) confusion. I did not count the electrons in the non-bonding MO, but everyone else does, so I stand corrected. As Jorge points out, 3c-2e should be reserved to hypovalent molecules.

The starting point for a discussion should be the observed structures. $\ce{SF6}$ has a regular octahedral geometry with bond lengths of 1.56 Å. Below is a depiction compared with some other compounds showing fluorine sulfur bonds:

Source: https://www.researchgate.net/publication/318478861_Sulfur_-_fluorine_bond_in_PET_radiochemistry/figures?lo=1
The structure of $\ce{PCl5}$ in a trigonal bipyramid:

Source: https://commons.wikimedia.org/wiki/File:Phosphorus-pentachloride-2D-dimensions.png
Trigonal bipyramids have distinct axial and equatorial ligands, but they can swap using the pseudorotation mechanism. While this is sometimes faster than the methods to characterize the molecules (e.g. the NMR spectrum of $\ce{PF5}$ suggest 5 identical ligands), the electrons are even faster "figuring out bonding", so we can ignore pseudorotation (Born-Oppenheimer approximation).
So those are the structures we would like to rationalize, with the simplest model possible (so it still fits in our head and we can be convinced by it - "just" solving the problem ab initio doesn't convince most folks).

[OP] Do you know any paper that disapproves the sp3d2 model?

Chemical bonding in hypervalent molecules. The dominance of ionic bonding and negative hyperconjugation over d-orbital participation, Alan E. Reed and Paul v. R. Schleyer, Journal of the American Chemical Society 1990 112 (4), 1434-1445
DOI: 10.1021/ja00160a022
Hypercoordinate molecules of second-row elements: d functions or d orbitals? Eric Magnusson, Journal of the American Chemical Society 1990 112 (22), 7940-7951
DOI: 10.1021/ja00178a014

[OP] So if the hybridisations sp3d and sp3d2 don't exist, how does this work?

The logic is that if the only atomic orbitals on the central atoms you consider for bonding are s-orbitals and p-orbitals, you can't make more than 4 bonds (this also goes for sulfate and phosphate). If you want to stay close to Lewis structures, with a line representing two electrons, you have to invoke ionic bonds and resonance. For $\ce{SF6}$, you write $\ce{SF4^2+ + 2F-}$. This gives

Source: Wikimedia via Wikipedia article on hypervalent molecule

The bonds are a combination of a covalent bond (bond order 2/3) and an ionic bond (attraction of 1/3 charge on fluoride with +2 charge on sulfur).
If you want to use MO-theory, you match the fluoride p-orbitals (the s-electrons don't participate) with the symmetry of the central atom's s-orbitals and p-orbitals (see illustration in the question). The four lowest combinations are shown below. If you use subtraction instead of addition, you get anti-bonding MOs (with more nodes; just like we learn for atomic orbitals, more nodes mean higher energy of electrons in that state).

These get filled with the 8 electrons available. 8 electrons would make 4 standard localized bonds, but here you connect 6 ligands, so the bond order is 4/6 = 2/3). This matches the bond order of 2/3 for the VB model above (localized bonds plus ionic interaction).
If you compare the S-F single bond lengths in other molecules to the bond length in $\ce{SF6}$, they are all very similar, even though e.g. the bonds in $\ce{SF2}$ are "real" single bonds and we just said the bonds in $\ce{SF6}$ have a bond order of 2/3. This is rationalized by the partially ionic character of the latter.
The $\ce{PCl5}$ molecule can be explained in a similar manner. The bond order is 4/5.

[Oscar Lanzi in comments] PCl5: bond orders may be different for equatorial and axial atoms, I would use 1 and 1/2 respectively.

Ok, thanks for the answer. I still have some questions. You said in a comment that the bond order was 5/6 but then 2/3. How can I calculate it by my own?

Sorry but I can't clearly understand where the electrons go when you use MO theory.

@JorgeBonifaz I fixed my comment above. Bond order for $\ce{PCl5}$ is 4/5 and for $\ce{SF6}$ 4/6. The 4 is always from having available one s and three p orbital on the central atom, and using all of them to make bonding MOs.

I edited the answer, adding some more information. The MO theory will make sense when you study it from the ground up (this is more at the hand-waving level). The idea is that you can take N orbitals and mix and match them (linear combination) to get N combined orbitals, sometimes with different symmetry.

Ok, but are there formed 4 bonding and 4 anti-bonding MO?

And you said that there are 8 e- participating, from "s" and 3 "p" atomic orbitals (from sulphur), but it has only 6 valency e-, so does this mean that the other 2 e- are just picked from any fluorine atom so it's delocalized?

PCl5: bond orders may be different for equatorial and axial atoms, I would use 1 and 1/2 respectively.

If you want to use MO-theory, you match the fluoride p-orbitals (the s-electrons don't participate) with the ... I think that is not quite correct. The same image or bonding description works if the s orbitals of fluorine are involved and even if you start to invoke hybridisation. This is all a bit handwaving and flip-flopping back and forth between localised bonding description and molecular orbital theory.

Hypervalency was discounted for sulfur and row 2 elements at least 20 years ago (see the references cited in an earlier answer and related citing works.) Such sulfur compounds have ionic character with the sulfur atom +2 and the negative balancing charge distributed over the more electronegative atoms with delocalized resonant bonds, or similar.

See "Addressing the Hypervalent Model: A Straightforward Explanation
of Traditionally Hypervalent Molecules" J. Chem. Educ. 2020, 97, 3638−3646,
https://dx.doi.org/10.1021/acs.jchemed.0c00368

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Does the metal complexing of crown ether influence its reactivity?
I am conducting a SN2 reaction experiment with 1-aza-18-crown-6 and bromide compound. I used 1 equivalent of bromide compound and 1.1 equivalent of aza crown, but in the TLC plate the aza crown spot disappeared and the bromoide compound spot still showed. I used Na2CO3 as base, and KI as catalyst. Similar papers use NaI, but our lab doesn't have it, I thought that KI can act in similar way. Recently I found that 1-aza-18-crown-6 can form complex with potassium ion. So with the incomprehensible result in spite of the excess amount of the aza crown, I wonder if the complex of the potassium ion and the aza crown doesn't participate in the SN2 reaction anymore.

More details on the reaction you were attempting and the conditions would be helpful.

That seems a significant possibility. Can you run the reaction without KI, or substitute nBu4NI?

I think the problem is Na2CO3 as it reacts with KI to produce K2CO3. https://www.youtube.com/watch?v=LfzxUPF6Mm4

@Reihani Can't K2CO3 also act as base?

K2CO3 can be a very effective base depending on what solvent you are in.

I use acetonitrile as solvent. I think that the reaction of Na2CO3 and KI is not the problem. First I would try not using KI.

K2CO3 has significant solubility in MeCN and is a very effective base in it. I have used it many times. I think you are likely correct that the K is complexing the crown ether and deactivating it.

@KrangLee are you trying to do SN2 ON the crown ether, like your crown ether is the nucleophile? or is it just to increase the rate of the reaction?

@thelocalsage I'm trying to use aza crown as nucelophile.

@KrangLee so the nitrogen in the nucleophile? if so i would bet money the nitrogen is coordinating the potassium ion and is unavailable for nucleophilic attack. potassium is the PERFECT size for 18-crown-6 ethers and that doesn't really change if it's an aza-crown (at least for 1-aza crowns). If so, I'll draft a formal answer for you to give whole context.

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Is it possible that the chemical composition of Greek fire could have been an alkali metal?
"Greek fire" was an ancient weapon used by the Greeks noted as being inextinguishable by water. The chemical composition of this weapon has been debated in the historical scholarship; due to the fact that alkali metals such as sodium and potassium ignite when exposed to water, is it possible that the chemical composition of Greek fire included alkali metals to achieve this property?

They explode in water, so no it couldn't, in fact it was something like napalm.

In a word, no. The electrolytic technology for making elemental alkali metals was far beyond anything the ancient Greeks possessed.

Regardless of ancient Greece inability to produce alkali metals, these elements cause Coulombic explosion in contact with water, but do not burn under water.

Sodium, potassium, and every other alkali metal react explosively on contact with water. The presence of water, however, was not prerequisite for the ignition of Greek fire. This fact is highlighted by the disproved hypothesis on Greek fire composition suggesting it was some reaction between water and calcium oxide (CaO), colloquially called "quicklime". This hypothesis is disproved by accounts in the historical literature, specifically by Emperor Leo's account in Tactica noting that the substance would be spread directly on the surface of enemy ships and thus (even with trace water on deck) would need to spontaneously ignite without water in order to work.
There are other variables that confound your alkali metal hypothesis. Firstly, the instability of alkali metals to air (through oxide formation and aforementioned blowing up in water) makes the materials difficult to store should they even have been produced at all. The account of alkali metals in the pertinent time period all come from salts and minerals which do not necessarily react the way the unstable metal might. Once bound as a salt, the properties of the metal change drastically.
The alkali metals also have extremely characteristic coloration of their flames; this is due to the quantization of electronic energy levels in atoms and molecules which causes substances to emit very specific wavelengths of light upon combustion. This phenomenon is often used for spectacle—like in the coloration of fireworks—and thus would almost certainly have been fervently noted in any historical account on Greek fire; such accounts are absent.
It is generally agreed upon by scholars that Greek fire was—at least primarily—composed of crude oil or some distillation thereof. This is consistent with resistance to extinguishing by water as well as the access of civilizations to crude oil at the time: the Byzantines had clear access to naturally occurring wells of crude oil near the Black Sea, and accounts suggest that 1) this oil was called "naphtha" by the Persians, 2) what the Persians called "naphtha" was the same as what the Greeks called "Median oil", and 3) the term "Median fire" was used alternatively to "Greek fire" in some cases.
It is probable that other components like sulfur and resin were added to change the properties of this substance—especially in places where there were similar weapons documented—but no evidence of specific alkali metals or their salts appear there. Regardless, the water-resistance of the flames is almost certainly due to its identity as a derivative of crude oil given the weapon's properties and the resources of Mediterranean/Middle Eastern civilizations at the time.

From a more recent chemistry perspective, so-called 'spontaneous combustion' as detailed, for example, in this incident: "Spontaneous Combustion of Oily Rags Cause of Lexington Fire", to quote:

The fire started in a dumpster against the outside the garage that contained oil soaked rags and empty linseed oil cans, intermingled with wood, paper, cardboard, and other trash.

More extensively and recently with respect to ancient times, per Bretherick’s Handbook of Chemical Hazards, one finds that "calcium oxide: H2O" is cited as an explosion hazard likely as a heat accelerant as the hydration reaction is reportedly exothermic.
The same sources also cites fine powders (like Milk Powder) after wetting and air exposure, as capable of producing explosive events. More likely here, however, are the reported spontaneous ignition hazards for cellulose and associated dusts of sulfur, wood flour and more generally natural "fuel dust", which would include fine powdered charcoal. Also, "cotton waste soaked in linseed oil", and perhaps, most pertinent "presence of petroleum products, fire and explosion can be spontaneous".
So, I would surmise that the ancient weapon "Greek fire" was a possible composite of crude fuel oil, quicklime, fine charcoal, flour from finely crushed grain and perhaps, even a touch of sulfur, which in the presence of moisture, would form a highly spontaneously combustible product, but even if this is not exactly the "Greek fire" of old, those testing this formulation are advised to exercise extreme caution.


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Functional groups in graphene oxide
There is already a large number of literature on graphene oxide stating that the number of epoxy groups is quite similar to the number of ethyl groups. However, I was wondering if the number of carboxyl groups attached to the edges of the graphene oxide is either more or less than these groups attached to the basal plane of the graphene oxide. Or does it usually depend on the process of graphene oxide synthesis?
I am only looking into the perspective of these three major functional groups present in graphene oxide. Although, I know that many functional groups can be present in graphene oxide.

I think it all depends.


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Electrons in Bohr's atomic model
Why electrons don't fall into the nucleus in bohr's atomic model?
Is there a reason or explanation or is just a consideration?

Is just a postulate, I think anyone can actually explain why.

Related: https://chemistry.stackexchange.com/questions/54914/stability-of-bohr-orbits

@JorgeBonifaz That was a postulate - quantisation of energy. Now it's quite more than that and learning quantum mechanics gives some explanations.

So you mean it "was" in the past? Or only that was in some way the base of future advances in quantum mechanics?

@JorgeBonifaz I mean Bohr just postulated it, but future advances changed this situation.

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It helps preventing  others to tell you what you already know or what you could easily find yourself, being answered many times in multiple ways.

As far as I know, it's a consideration that the electrons must describe circular orbits around the nucleus of the atom without radiating energy.
So this is because is a "postulate", since according to classical electrodynamics a charge with an accelerated movement must emit energy in the form of radiation.
To maintain the circular orbit, the force experienced by the electron (the Coulombian force due to the presence of the nucleus) must be equal to the centripetal force. So here it is:
Taken from Chemistry Libretexts - 6.2 The Bohr Model

$$k\frac{Ze^2}{r^2} = \frac{m_ev^2}{r}$$

Note that this model is only valid for hydrogen-like systems (only with 1 electron).

How does that relate with the "quantisation of energy"?

since according to classical electrodynamics a charge with an accelerated movement must emit energy in the form of radiation. - This is valid only for nonzero time derivative of acceleration vector. Linear constant acceleration does not lead to radiation, what is used in linear accelerators.


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Why is real volume not smaller than ideal volume in the van der Waals gas equation correction terms?
The following equation is valid for ideal gases
$$pV=nRT$$
Please mind the following notation, any parameter that is related to an ideal gas would be denoted with a subscript $i$ and anything related to a real gas with a subscript $r$
For example, the pressure exerted by a real gas is denoted by $p_r$ and for an ideal gas, $p_i$
For real gases, when using the van der Waals equation, the following correction terms are introduced,
For the pressure correction,
$$p_i = p_r + \frac{an^2}{V^2} \tag{1}$$
For the volume correction:
$$V_r = V_i - nb\tag{2}$$
What I think is, since the relation holds for an ideal gas, we should multiply $p_i$ & $V_i$ with the necessary rearrangements.
But the equation taught to us as the van der Waals equation, if you think about it, is multiplying $p_i$ and $V_r$.
Which is correct and why?

Please see the following possible duplicate: https://chemistry.stackexchange.com/questions/37029/why-is-there-a-subtractive-volume-correction-in-the-van-der-waals-equation-for?rq=1 as well as many other earlier posts on the subject

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It helps preventing  others to tell you what you already know or what you could easily find yourself what was provided many times in multiple ways.

@Poutnik  the equation you have wrote above is wrong in my opinion specifically where you mention V(i) = V(r) - nb , how can the ideal volume be less than the real volume.

@Poutnik From what i understand , the volume considered in the equations is the volume of free space in which gases can move, and not the volume of the gases itself. Now in ideal case it's almost equivalent to all of the space in the container since assumption is made that molecular volume is negligibly small. In the real case however the molecules carry significant volume hence the volume for movement is actually lesser than what we thought to be, therefore the subtractive term. And now the equation becomes V(r) = V(i) - nb , but you have wrote it other way around.

@Poutnik That doesn't answer my question, to be specific i asked why are we multiplying P (i) with V (r)

I have moved my topical comments to an answer.

Now cross posted at physics SE: https://physics.stackexchange.com/q/725929/313612. Voting to close.

Have you not read my answer? We multiply what we observe. In case of ideal gas it evens out to be what the ideal gas shows

This was the way I understood the corrections in the van der Waals equation for real gasses.
Before which, a bit of history. How was it derived?

TL;DR: An empirical guide as to how the ideal gas law was proposed
The derivation of the gas equation
The ideal gas equation in its most simplified derivation is the combination of three of four different laws - Charles's Law ($V \propto T$), Boyle's Law ($p \propto V)$, Avogadro's Law ($n \propto V$) and Gay-Lusaac's Law ($p \propto T$).
Combining them, we get the following equation:
$$pV = nRT \tag{1}$$
This is the gas equation that follows. Notice that I haven't used the term ideal gas equation yet. This is because we now move on to the assumptions of an ideal gas to understand why this is the ideal gas scenario.
What makes a gas ideal?
A gas is considered to be ideal if it follows the following criteria as taken from this book

For a gas to be “ideal” there are four governing assumptions:

The gas particles have negligible volume.
The gas particles are equally sized and do not have intermolecular forces (attraction or repulsion) with other gas particles.
The gas particles move randomly in agreement with Newton’s Laws of Motion.
The gas particles have perfect elastic collisions with no energy loss.

Here, properties 1 and 2 are important for this discussion. An ideal gas occupies no volume and doesn't have any forces between particles
Now that we know what an ideal gas is, let's go back to the discussion at hand.
The ideal(?) gas equation
As mentioned in equation 1, the gas equation uses the formula $pV = nRT$. But what is $p$, $V$, $n$ and $T$ in terms of what we can measure? Well, $p$ is supposed to be the pressure exerted on the container by the gas itself. $V$, the volume of the container where the gas can move around in, $T$ is the temperature of the container and $n$ is the amount of gas within the container. But then why do we say equation 1 is the ideal gas equation?
If you look at it carefully and think about it from a molecular standpoint, the above gas equation only holds when the gas occupies no volume and has no inter molecular forces between particles. This means that the pressure observed is the same as the pressure exerted and the volume observed is the same as the volume where the gas can move around. Hence our gas equation has become the ideal gas equation

Time to make things real (or closer to real)
Our gas equation holds because they were derived from empirical relations. This means that it should work for ideal as well as real gasses. However, real gasses don't have the the same two properties that made the ideal gas law so simple to write down. They interact between molecules as well as take up space.
Going back to the definitions of the parameters of the gas equation as well as the above paragraph, we notice one thing. What we measure is not what is actually what we need. Intermolecular forces pull back on the molecules nearer to the edges of the container reducing the actual pressure observed and the molecules actually take up space reducing the volume that they can move around in
Van der Waals came up with the corrections whose proofs can be found online. They were,
$$p_\mathrm{correction} = \frac{an^2}{V^2} \tag{2}$$
$$V_\mathrm{correction} = nb\tag{3}$$
Applying this corrections using the discussion so far, we get:
$$\left(p + \frac{an^2}{V^2}\right)\left(V - nb\right) = nRT \tag{4}$$
Here $p$ and $V$ stand for the observed values. This is why the term for pressure is additive and the term for volume is subtractive.

$$nRT = p_\mathrm{i}V_\mathrm{i} = (p_\mathrm{r}+a\frac{n^2}{V^2})(V_\mathrm{r} - nb)$$

@Poutnik the equation you have wrote above is wrong in my opinion specifically where you mention V(i) = V(r) - nb , how can the ideal volume be less than the real volume.

The root cause of your confusion is that you consider V as the free volume between gas molecules and not as the total gas volume, as intended.

When real volume approaches the minimal volume of free molecule motion, it is like if an ideal gas volume approaches zero. In both cases, the pressure would shoot upwards.
By other words, if there are negligible cohesive forces, a real gas has bigger volume than an ideal gas would have at the same p and T.
By yet other words, ideal gases have infinite compressibility, real gases do not. At extreme pressures, real gas volume can be orders bigger than ideal gas volume.

@Poutnik From what i understand , the volume considered in the equations is the volume of free space in which gases can move, and not the volume of the gases itself. Now in ideal case it's almost equivalent to all of the space in the container since assumption is made that molecular volume is negligibly small. In the real case however the molecules carry significant volume hence the volume for movement is actually lesser than what we thought to be, therefore the subtractive term. And now the equation becomes V(r) = V(i) - nb , but you have wrote it other way around.

The volume in the van der Waals state equation for real gases and in the ideal gas state equation $pV=nRT$ means in both cases the volume of the container.
$V_\mathrm{r} > V_\mathrm{i} = \frac{nRT}{p}$ if cohesion has lower impact than molecular volume. Both terms in parenthases of van der Waals equation means corrections what would be p and V of an ideal gas, if there was no cohession and zero molecular volume.

@Poutnik That doesn't answer my question, to be specific i asked why are we multiplying P (i) with V (r).

We do not. You do.
If you isothermally compress 1 L of helium at 100 kPa by pressure 100 MPa, will be the final volume more or less than the expected 1 mL for ideal behaviour? Is then real volume more or less then ideal volume?
The own volume of gas molecules causes more frequent collisions of molecules with container walls, compared to an ideal gas at the same conditions. That means

either higher real then ideal gas pressure in the same volume,
either bigger real than ideal gas volume at the same pressure.

Van der Waals equation pretends a real gas is an ideal gas, following principles of Perturbation theory(*). It masks the real gas non-idealities by corrective factors:

Increases the real gas pressure by $+a\frac{n^2}{V^2}$
(as real pressure at walls is decreased by cohesion).
Decreases the real gas volume by $-nb$
(as real volume is increased by lower compressibility due molecular volume).

That leads to modification of $V_\mathrm{r}$,$p_\mathrm{r}$, pretending to be ideal gas values $V_\mathrm{i}$,$p_\mathrm{i}$, plugging the latter to the ideal gas equation:
$$nRT = p_\mathrm{i}V_\mathrm{i} = (p_\mathrm{r}+a\frac{n^2}{V^2})(V_\mathrm{r} - nb)$$
Real volume is decreased to the smaller ideal volume ($V_\mathrm{i}=V_\mathrm{r} - nb$), because an ideal gas is more compressible. The extrapolation of the van der Waals model to extreme conditions $V_\mathrm{r} = nb$ leads as well as for $V_\mathrm{i} = 0$ to infinite pressure for $n \gt 0$, $T \gt 0$.

(*) Perturbation theory offers an approximate solution of a complex scenario by corrections(perturbations) of the known, exact solution of a related, but simpler scenario.


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Molecular orbital mixing in second period elements
Why do the $\pi_u(2p)$ and $\sigma_g(2p)$ diatomic MO mix within the second period elements as it's shown in the image below?

If we consider the OM from both 2s and 2p AO, we have the following diagram, which is only valid when the 2s and 2p are far in energy (O2 and F2). This is because as they have a higher Zeff, the 2s orbitals are more stabilized.
For lighter homo-diatomic molecules the AO are very near so they mix

This is a zero order answer but hope it is helpful.
The model that is commonly used to describe the MO is based on the results of expressing the MO as a linear combination of atomic orbitals (LCAO), thus obtaining the MOs and their energies. The contribution of each AO to a particular MO depends on its shape and energy. In general, the LCAO coefficients are optimised to minimise the energy of the molecule and its occupied MO.
Since the atoms of different elements have different nuclear charges and number of electrons, the magnitudes of the attractive nucleus-electron and repulsive electron-electron forces are different and, in consecuence, the energies of AOs of the same type for different elements differ. Thus, when the LCAO method is applied to calculate the MO of a homonuclear diatomic molécule, differences in the energies of the MO are observed.


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How does pH affect the rate that CaCO3 dissolves in freshwater?
It is my understanding that calcium carbonate dissolves faster in lower pH environments. I am interested to know how higher PH levels affect the rate of dissolution. So I have broken this question down into 2 parts:

Starting from a neutral pH, how will each 1 unit increase in pH affect the rate of dissolution?
I believe that other factors likely play into this equation - for example, KH, GH and magnesium levels. How does this work?

What do you mean by KH and GH?

Aquarium talk: "There are two types of water hardness: KH and GH. KH measures an aquarium's carbonate hardness, or the degree to which it is able to buffer against pH fluctuations. GH stands for general hardness, which measures the number of dissolved calcium and magnesium ions. GH is crucial for the growth of certain fish species."

As solid phase is involved, it is much easier to evaluate thermodynamic equilibrium (involving pH and total dissolved carbonate forms) than kinetic rate.

Calcium carbonate dissolves by falling apart into ions (calcium cations and carbonate anions):
$$\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-(aq)}$$
The rate of dissolution does not depend on pH. However, there is a back reaction, which depends on the concentration of the calcium cation and the carbonate anion. The carbonate anion concentration, in turn, depends on the pH because you have carbonate, bicarbonate and dissolved carbon dioxide in fast equilibrium:
$$\ce{CO3^2-(aq) + 2 H+(aq) <=> HCO3^-(aq) + H+(aq) <=> H2CO3(aq) <=> CO2(aq) + H2O(l) }$$
You can take a look at what happens to carbonate at different pH values here (it is called speciation).
The bottom line is that in the pH range above 10 or 11, pH does not matter for carbonate (although the calcium ion might precipitate as hydroxide). When you go more acidic than pH 10, for every pH step, about 10 times less carbonate exists because it turns into bicarbonate of carbonic acid/carbon dioxide. So the back reaction is slower, potentially speeding up the net forward reaction.
Whether the back reaction has an effect on the net rate of dissolution depends on what is in the water before you add the solid calcium carbonate. It also depends on how much you add.
Given two specific scenarios (the entire composition of the system), a quantitative answer would be possible.

Thank you, Karsten for this detailed answer. So results of a water test, say from a local aquarium shop, would allow for a quantitative answer? I thought of posting with that info but refrained because on my main SE site (Webmasters) questions that only apply to one's specific situation are off topic for us. Seems that's not the case here.


text:
Why can't RX add to alkenes / alkynes like HX?
With an analogous mechanism: C bonded to X has a δ+ charge and attacks the pi electrons to form a carbocation intermediate, with X- as a leaving group. X- then adds to the carbocation to neutralize it.
Why doesn't this occur?
If steric hinderance is the answer, what about MeX adding to a terminal alkyne?
RCH2X + OH- → RCH2OH + X- proceeds (with comparable/less hinderance?) so it seems unlikely that this the reason.
H-X bonds are more polar than C-X, but the electronegativity difference between C and H is small enough that we consider C-H bonds nonpolar. Electron-deficient C can pull some electron density from adjacent carbons if there are any - but again, what about MeX? Is the presence of 3 hydrogens enough to stabilize the C-X bond to the extent that the addition doesn't take place?

Well, something like that happens in Friedel-Crafts alkylation, but you need a catalyst. To protonate alkene (or alkyne) you need a very strong acid. Similarly you need very strong alkylating agent and MeX isn't strong.

@Mithoron What factor necessitates use of a catalyst? Autoionization of water seems less favorable and occurs at RTP. Adding R-X has the potential to generate a more substituted carbocation than H-X, with the same leaving group.
What makes the transition state for this reaction so high-energy?
I am just trying to understand this conceptually. If I saw this on paper, how would I know no reaction occurs?

"Autoionization of water seems less favorable"? Only seems. Oxonium cations are waaay more stable than carbenium cations (typical carbocations like CH3+). For starters they have full octet.

@Mithoron That's what I was missing. Thanks.


text:
Are volatile impurities an issue for sublimation as a purification method?
Recrystallization depends on identical solute molecules colliding so that a crystal can grow. An impurity present in low concentration will generally remain in solution because kinetics will favor crystallization of the high-concentration species, all other factors being equal.
But what about when vapor deposits on a cold finger? I would assume there is no such selectivity, with small crystals of the impurity growing alongside those of the target compound (assuming similar phase diagrams). The issue is never mentioned in any of the sublimation demos I have watched.
Is this correct? Should recrystallization be done first in a case like this?

Yes, volatile impurities are quite an issue for sublimation and should be removed in some other manner.

Perhaps look at zone refining as a way to reduce impurity levels.


text:
Why can we use the arrhenius equation to measure the number of vacancies in a material?
In a materials science video I was watching today the prof. said we could use the arrhenius equation to estimate the number of vacancy point defects in a material. And gave the following equation:
Nv = N exp(-Qv / KBT)
Where Nv = number of vacancies
N= # of atom sites
Qv = energy of vacancy formation
KB = Boltzmann constant
T = Temp
I don't understand how we got to that equation from the Arrhenius equation. How do the two relate? How is the above equation derived from the Arrhenius equation?

I'll try a heuristic undetailed explanation. The formation of vacacies has a cost in enthalpy (because atoms are no longer tightly bonded), although it naturally increases entropy (because there are more possible microstates). It follows that there is a certain Gibbs energy (G = H - TS) that induces an equlibrium, where G is minimized. In chemical equlibrium, G is associated with a given K (equilibrium constant), which is temperature-dependent according the Van't Hoff equation (which can be derived from basic thermodynamics), that is similar to the Arrhenius equation.

This looks like the Boltzmann equation, perhaps Arrhenius was a slip of the tongue.


text:
How to store ammonium perchlorate?
How should ammonium perchlorate (AP) be stored safely? The AP at PEPCON was stored in plastic and aluminium containers. I seem to remember that the very material of the storage containers (i.e. aluminium and plastic) contributed to the size of the final explosion, although I do not believe it had much to do with the actual cause of the explosion. Since many metals and plastics such as aluminium and HDPE function as a fuel this kind of makes sense to me.
I can think of many ways of making storage safer: Bunker, inert environment, reduced Temperature, explosion proof installation etc... But how should AP be stored properly? What material should be used for the storage of AP?

According to the state of New Jersey, "Ammonium Perchlorate is not compatible with STRONG REDUCING AGENTS; STRONG ACIDS (such as HYDROCHLORIC, SULFURIC and NITRIC);
METALS (such as ALUMINUM, COPPER, and POTASSIUM); METAL OXIDES; PHOSPHOROUS; and COMBUSTIBLES."

@Karsten  I would like to see the person who got idea somebody could want to store NH4ClO4 with metallic potassium, to explicitly warn about it. I have wanted to add they could mention phosphorus, but they did. But it goes along the warning for Microwave oven sold in US, not to dry pets inside.

Yet this document does not say what compatible choices of containers are. @Poutnik

I have not pretended they do. But the OP does not mentions  origin nor amount nor packaging mass.

You can check the MSDS of AP.

@NilayGhosh No idea why I did not think of this before. I immediately checked, but to my surprise there is no reference to storage materials specifically. I checked 4 different MSDSs. Thanks for the tip though.

@Poutnik I did not think the amount would make much of a difference, but I was just thinking about lab use. It could well be that you can safely store it in a PP, HDPE or even a PET bottle, but if AP is soo sensitive, maybe there are requirements to put it in an anti-static bottle. Or the AP is prone to react with certain materials that are commonly used for storage containers. There are acids that need to be stored in PTFE. I simply have no idea. not a clue.

FRP containers might be a good choice at scale. Ones with conducting layers might be even better My work with explosives is limited to TATB, PETN, HMX and RDX and in all cases, its either A) LDPE/HDPE Bags in HDPE Drums or wooden boxes or at scale in these big FRP tanks or Plastic lined tanks.


text:
How do clay particles form a colloidal mixture?
I have been learning about water coagulation in my chemistry textbook. Apparently, before being treated, water has clay particles, primarily silicates and aluminosilicates, which apparently have a negative charge. These particles repel each other, forming a colloidal mixture.
My question is: how exactly do the silicates and aluminosilicates have a negative charge? I understand that silicates and aluminosilicates have a negative silicate/aluminosilicate anion, but wouldn't that negative charge be counteracted by their bonding to a positive cation, causing the silicate/aluminosilicate to have an overall neutral charge?
My current conjecture is that, in water, the silicate/aluminosilicates particles dissociate, and the cations are solvated by water however the silicate/aluminosilicate anions are too large to be solvated, thus they remain in the water and repel each other. This is the only way that it currently makes sense to me because other ionic compounds like sodium chloride do not form colloids in water despite obviously having cations and anions; likely because both the cations and anions are solvated by the water particles because they are both small enough.
If anyone can correct my conjecture/explain to me how this works, I would really appreciate it.

There is related argentometry indication by - I guess - eosin dye. The colloidal or finely precipitated AgCl is initially negatively charged by adsoption of excessive Cl-. After exuivalence, it gets positively charged by excessive Ag+, what is followed by adsorption of negatively charged anionic dye. (Never personally tried, I have heard the Ag2CrO4 indication is better).

@Poutnik Maybe it’s because I’m in year 12, but I have no clue what any of this means sorry

@ScratchCat   Colloidal particles may cause adsorption of preferably particular ions, what may lead to slight nonzero net charge.  //  If you do not write in advance what knowledge level to expect, expect info at levels you may not have expected.  :-) The implied level of the question seamed to me to be OK with the level of my comment. :-)

You are actually pretty close to answering your own question.
Key here is clay particles are very fine, meaning they have a very high surface area to mass ratio.
So, even though they are not dissolved, interaction with water plays a prominent role in how they act as a colloidal suspension.
The oxygen bears the negative charge in silicates.  In suspension (as you said) they will repel one another and attract the positive part of water molecules.  So you also have water between the clay particles as well, keeping them from clumping together (flocculating) and falling out of solution.
As particle size increases, there is less tendency to be held in suspension, even for clay because the difference in density becomes more and more a factor in separation over time.


text:
Synthesis and retrosynthetic analysis of dicyclohexylamine
I need to consider the retrosynthetic analysis and synthesis of dicyclohexylamine.

For the retrosynthetic analysis, I have relied on the method of disconnections, which consists in splitting the target molecule into several fragments with one or more functional groups.
So, first of all, I have decided to disconnect (i.e., virtually break a bond) the C-N of the amine.

The reagent equivalents of these synthons I thought could be: aniline and cyclohexylmagnesium bromide. In such a case, the synthesis could be given by:

Aniline could be prepared by nitration of cyclohexane by reaction with nitric acid and sulfuric acid.
And, the organomagnesium reagent, from chlorocyclohexane with Mg in anhydrous ether.
Would this retrosynthetic analysis and synthesis method be appropriate?

Did you learn that in retrosynthesis, you usually make polar disconnections, in that one side is a nucleophile and the other side is an electrophile? You have chosen two nucleophilic starting materials, which will not work at all. Radical disconnections (like the one you have drawn, where each fragment has an unpaired electron) are possible, but more advanced, and certainly not needed in this case.

Aniline is not cyclohexylamine. Aniline is an aromatic derivative of benzene not cyclohexane. And you certainly can't make either by nitrating cyclohexane.

Grignards reacting with primary amines act as bases giving the amine anion.
This can be done in one step by reductive amination of cyclohexanone with cyclohexyl amine. Further information here
Both of these starting materials are widely available so you do not need to consider their preparation.

React formamide with 1-chlorocyclohexane. To make the chlorocyclohexane, you can hydrogenate chlorobenzene. To make formamide, you can add ammonia to ethyl formate or heat ammonium formate to dehydration. Then, just heat with NaOH to deprotect the amide protecting group and you have your compound. In my experience, its best to make N-C bonds using Halogens. I believe chlorocyclohexane would be the solvent itself or you can add something like water and a bit of K2CO3 to drive the reaction forward. And heat at reflux maybe if its going too slow. Monitor pH to see if the pH decreases, that means reaction is completing. Use pH meter in a rubber cork, samples with pH paper or just put in some universal indicator solution

Do you have a reference for the hydrogenation of chlorobenzene to chlorocyclohexane? I can only find references to hydrogenation giving loss of chlorine.

I'm with @Waylander!


text:
Excited states and emission lifetimes
I have some doubts about the nomenclature of the lifetimes. Are singlet and fluorescence lifetimes the same? As well as the triplet and phosphorescence lifetimes?
Also, can the triplet/phosphorescence lifetimes be measured by phosphorescence decay as well as by transient absorption of the triplet state by means of laser flash photolysis?

Yes, singlets fluoresce and triplets phosphoresce. So the singlet lifetime is the fluorescence lifetime and is the inverse of singlet excited state decay rate constant and similarly for phosphorescence.
Both singlet and triplet decay can be measured by flash photolysis more often nowadays this is called pump-probe spectroscopy or pump-probe transient absorption (or some variation of this) but it is the same technique as flash photolysis. It has kinetic (decay measured at one wavelength at a time) and spectroscopic (spectrum at sequence of times after excitation) variants.
If the excited state is, say, a quartet or quintet etc. we use the term luminescence.

Thank you for your clear answer. However, it makes me feel uneasy.  In some molecules the ground state is a triplet -e.g $O_2$ ($X ^3\Sigma_g^- $)- and the emission from its triplet excited states is called fluorescence taking into account that the transitions are allowed and the lifetime is short. Also the OH radical ($X ^2\Pi$) has fluorescence and the radical is frequently detected using Laser Induced Fluorescence (LIF).

I wonder if as a general statement (without specifiyng a context) this definition maybe an oversimplification, and I wonder if it misses the importance of distinguishing between an allowed (high-probability, fluorescence) versus  a forbidden transition (low-probability, fosforescence)?

The length of an excited state emission  has nothing, per se,  to do with whether it is fluorescence or phosphorescence, although phosphorescence is normally longer in the same molecule. Symmetry can also be involved in determining emission probability, thus there are many molecules whose singlet is longer lived than other types of molecules triplet state.

Completly agree,  Did not explain myself very well. I wrote a more detailed comment as an answer below.

Thanks @Achem and @porphyrin for your comments. I do not know if an Answer is the right format for this Comment but the comments do not allow enough character to develop my argument.
I think we are all in the same or close pages. My reservation is that we have to be a little nuanced when using this terminology, since otherwise it may leed to confusion.
I rewied this morning the the IUPAC's 3rd edition of Glossary of Terms Used in Photochemistry and look for the definitions of luminescence, fluorescence and phosphoresnce. I quote from the document:

Luminescence.
Spontaneous emission of radiation from an electronically excited species or from a >vibrationally excited species not in thermal equilibrium with its environment.
Fluorescence
Spontaneous emission of radiation (luminescence) from an excited molecular entity >with retention of spin multiplicity.
Phosphorescence
Phenomenologically, term used to describe long-lived luminescence. In mechanistic >photochemistry, the term designates luminescence involving change in spin >multiplicity, typically from triplet to singlet or vice versa.
Note: For example, the luminescence from a quartet state to a doublet state is also
phosphorescence.

Thus, both fluorescence and phosphorescence are examples of luminescence, independent of the spin of the states coupled in the transition.
Both triplet-triplet transitions (for example in O$_2$) and doublet-doublet transitions (for example in OH and NO) are examples of fluorescence. I am not familiar with any molecular transitions involving quartet or quintet states but that is my ignorance.
I guess that in some fields the molecular ground states are singlets and higher spin ground states and oddity, so I understand that the terms fluorescence, phosphorescence and luminescence can be used more loosely, even if it makes me unconfortable, but in that case I feel it is better to state in what context you are using the definition.

I think the definitions are self-consistent, a triplet to triplet in oxygen is fluorescence because the spin multiplicity is the same. I am sure there must be plenty of examples in atomic fluorescence. Like Na fluorescence is doublet to doublet.

Agreed. I meant molecular systems involving transitions between quartet or quintet states. I will edit my answer.

Regarding porphyrin's response and PAEP's reservation about oxygen emission from an excited triplet to a ground triplet being labelled as fluorescence, the issue can be addressed by defining fluorescence as the excitation and emission having the same spin multiplicity. This removes the condition of fluorescence as singlet to singlet transition and then it is not based on short lifetimes. In a similar manner, phosphorescence can be defined as emission from a different multiplicity from the ground state. IUPAC's gold book seems to silent on this topic but other IUPAC documents agree with above.
To the OP, you will have to look at the context. Singlet state does not necessarily imply fluorescence. All it means that all electrons are paired in that system. Anyway, these are man-made definitions/ labels. Luminesence to analytical chemists is a catch all term for emission of light from molecules or atoms by means of various excitations. It has no connotations associated with the spin multiplicity.


text:
How do I interpret the dissolution rate unit?
I have found a paper (link) that is interesting for my research project.
The dissolution rate unit in this paper is described as $\pu{mol/cm^2/s}$.
For my project I would need the dissolution speed measured from the surface of the solid (for example $\pu{cm/s}$). Can I calculate the latter from the first?

I bet you know how to calculate mass from molar amount and molar mass, volume from mass and density, thickness from volume and area. If not, learn it, it is a basic skill.

:/  mol/cm2/s is actual kinetic unit for scientific experiment. Some cm/s could be used to for bomb fuses or something, and even then only for practical purposes.

The linear dissolution rate can be calculated from molar specific dissolution this way:
$$\frac{\mathrm{d}l}{\mathrm{d}t}=\frac{\mathrm{d}V}{A \mathrm{d}t}=\frac{\mathrm{d}m}{\rho A \mathrm{d}t}=\frac{M\mathrm{d}n}{\rho A \mathrm{d}t} = \left(\frac{\mathrm{d}n}{A \mathrm{d}t}\right) \frac{M}{ \rho}$$

l is thickness
t is time
V is volume
A is surface area
m is mass
$\rho$ is density

Although this is a problem for the OP (and not you) I don't see how this would be easily implemented, because it requires monitoring the change in A. If you are dealing with a flat surface (as you assume) it is ok, but for a heterogeneous mixture of (say spherical) particles?

Well, if the dissolution in mol/cm2/s is given.....  I was not addressing how it is being obtained. :-) It was not even asked for. I do agree it may be an essential problem.

I also agree rough surface would distort calculation, if initial quantity relates to a nominal surface area, while the surface is rough. But it can be considered as the best effort to approximate the value, if no better data are available.

@Poutnik I belive $M$ should be in the numerator: $dm = M dn$.


text:
Dependence of rate on the nature of reactants and other factors
Consider a reaction
$$\ce{aA + bB -> cC + dD}$$ whose reactants are given by A and B and the products are C and D.
The rate of this reaction is given by $$\frac{dx}{dt} = -\frac{d[\ce{A}]}{adt}= -\frac{d[\ce{B}]}{bdt} = \frac{d[\ce{C}]}{cdt} = \frac{d[\ce{D}]}{ddt}$$
The rate of the reaction also depends on the surface area of reactants, nature of reactants, stirring, temperature. Does the rate equation account for all the factors (I could find that by changing these factors the change in concentration for a time interval would vary)? If in a lab we are given different samples of reactants and measure the rate of the reaction (but same concentration), wouldn't each get a different rate value? How is the rate uniquely defined?

The rates above assume homogenous reactions, so surface and stirring is out.

So what if its homogenous? Say the reaction happens in the liquid state, stirring would still help ?

Well, stirring increases temperature..... but generally is used to reach a homogenous state.

Oh, I could do it slowly enough so that I could increase the collisions only and not much of the temperature? Isn't that negligible?

The  rate is given by the differentials you present and these have values depending on concentration, temperature etc. as you mention. What defines a reaction is the rate constant.  In your example $$
rate = \cdots etc. =\frac{d[D]}{ddt}=k[A]^{\alpha}[B]^{\beta}\cdots$$ where $k$ is the rate constant and $\alpha+\beta\cdots$ define the reaction order. The rate constant and exponent powers have to be determined by experiment.

Can those vary by changing the reaction conditions?

The Arrhenius eqn. describes how the rate constant depends on temperature for example.


text:
Physical interpretation of transition density matrix
I am studying natural transition orbitals (NTOs) and I am following the Martin's article [1]. I understand the mathematics, but I am a bit confused about what exactly is the transition density matrix defined for the electronic excitation of the ground state $\Psi_0$ to an excited state $\Psi_{ex}$ as follows:
$$
T_{ia} = \langle \Psi_{ex} | c_i^{\dagger} c_a | \Psi \rangle,
$$
where $c, c^{\dagger}$ are destruction and creation operators. Here, it's assumed that we realized a SCF calculation and obtained a set of occupied and virtual orbitals $\psi_p$. The indices $i$ and $a$ represent occupied and virtual orbitals, respectively.
The article said that this is the relevant quantity associated with the electronic excitation $\Psi_0 \rightarrow \Psi_{ex}$, but I don't know what it means. Is the $\mathbf{T}$ matrix related to the probability of the excitation?
Reference

Martin, R. L. Natural Transition Orbitals. Chem. Phys. 2003, 118 (11), 4775–4777. DOI: 10.1063/1.1558471.

I like to see the transition density matrix as the expansion coefficients of an excited state into singly excited configurations. The expression
$$
c_i^\dagger c_a|\Psi_0\rangle \equiv |\Psi^i_a\rangle
$$
is basically a singly excited configuration. If we assume that $|\Psi_0\rangle $ is given a by single determinant with doubly occupied orbitals, then $|\Psi^i_a\rangle $ represents all single excited determinants when $a$ runs over occupied orbitals and $i$ over unoccupied orbitals. We can then expand an actual excited state $|\Psi_{ex}\rangle$ into the basis of singly excited configurations and the expansion coefficients are basically the elements of the transition density matrix.


text:
Why Scandium (Sc) doesn't show variable oxidation states (especially +2), while other transition elements show variable oxidation states?
I have read answer to this question on various sites, but was not satisfied by the answers, because I am not able to understand, that if we remove initially 1 electron from scandium, then 4s would be ome unstable, but if we remove 1 more electron, there remains only one electron in 3d subshell. But we have already observed in many compounds, that heavier metallic elements and highly electronegative atoms force stability of unstable configurations. Also, I read somewhere that in [Sc (H20)6 ]2+   Scandium possesses +2 oxidation state, but when I tried to research more on it, some sites displayed the same compound but a +3 complex instead of +2. I am confused on all of this. Please throw some light on this, and please explain why Scandium cannot exhibit +2 oxidation state, is it just because of instability of 3d subshell? And what are the reasons for instability of 3d subshell, is it only the presence of a single electron? Please detail about this.
P.S. : It was also mentioned on some sites that after removal of one electron from 4s subshell of Sc, very high amount of energy is required for removal of second electron (i.e. I.E. 2 >> I.E. 1). How is this possible?

https://chemistry.stackexchange.com/questions/102331/oxidation-state-in-transition-metals

Scandium very much does exhibit oxidation states lower than +3. See https://en.wikipedia.org/wiki/Scandium.

The claim that the second ionization energy of scandium far exceeds the first is simply not true, or as least not specifically true of scandium. A quick look at the Wikipedia page reveals that the first ionization energy is 633 kJ/mol and the second is 1235, a ratio of about two to one. This is quite normal for metals with two outer $s$ electrons and the figures are actually lower than those for magnesium. Moreover, the third ionization energy is low enough for that to occur readily (unlike magnesium and all other alkaline earth metals), explaining why +3 is the most common (but not exclusive, as evident from the Wikipedia article) oxidation state for scandium.
Scandium in oxidation states lower than +3 is known primarily in organometallic compounds, where most other transition metals also tend to exhibit lower than "normal" oxidation states. One inorganic compound reported by Wikipedia is $\ce{KScCl3}$, in which the scandium has oxidation state +2 but does not consist of separate $\ce{Sc^{2+}}$ ions. Instead this compound retains scandium-scandium metallic bonding. The $\ce{Sc(H2O)6^{2+}}$ ion suggested in the question is probably a misprint of $\ce{Sc(H2O)6^{3+}}$.

The +3 oxidation state for s2, p or dx seems to be the sweet spot for high lattice energies and or solvation energies in aqueous solution.  As p or d electrons are added +2 becomes more stable. The higher ox. nos. involve increased covalent character and charge feedback giving polyatomic ions. Intermediate oxidation states almost certainly are involved in reaction mechanisms actually involving short lived intermediates. This raises the likelihood of more stable compounds of these. Mn, Cr, N and Cl are good examples of this.


text:
Why are vapours of xenon hexafluoride coloured?
According to Wikipedia,

[...] It is a colorless solid that readily sublimes into intensely yellow vapors.

It is clear from the above statement that $\ce{XeF6}$ is colourless in solid form. But why should it be coloured when it converts into vapour form?

https://chemistry.stackexchange.com/questions/82350/what-is-the-molecular-structure-of-xenon-hexafluoride https://chemistry.stackexchange.com/questions/50187/what-is-hybridisation-of-xef6-in-solid-state

Bottom line is it depolymerises.

There is nothing unusual about vapors of elements/compounds appearing colored. It is a common phenomenon. Potassium is a silvery white metal but its vapors appear green! Sodium also shows colored vapor. People have spent their trying to understand the structure of xenon hexafluoride.

Related for oxygen: https://chemistry.stackexchange.com/questions/15035/does-o2-have-a-color-in-the-gas-phase Oxygen is colorless at gaseous state but is slightly colored in liquid and solid state.

@NilayGhosh, This oxygen issue is perhaps a different phenomenon. It is simply a concentration effect (Beer's law). Just like  heavy water appear blue when viewed in a very long path length tube but in an ordinary beaker it is colorless pure white liquid. Xenon hexafluoride is complicated in the sense that there must be a change in the nature of bonding in the solid phase and gas phase. Just like potassium and sodium vapor. One would never assume that potassium has a beautiful green vapor.

H2O is definitely blue in large quantities by transmitted light. D2O is colorless because the stretching frequency overtones do not reach into the visible.

O2 gas is definitely colorless. How long a path is necessary to see that? There are condensed species in the liquid and various solid phases.


text:
Electronic configuration of Cu⁺ and Cu²⁺
I understand that copper has the electronic configuration $\ce{[Ar] 4s^1 3d^10}$ because a fully filled $\ce{4d}$ orbital provide a more stable configuration.
But I don't understand why the two copper ions have the following electronic configurations: $\ce{4s^0 3d^10}$ and $\ce{4s^0 3d^9}$.
Specifically, why is the electron from the $\ce{4s}$ subshell is removed before the electron from the $\ce{3d}$ subshell?
It does have a higher value of $n$ placing it farther than the latter, but it also has lower energy. And how does the extra stability provided due to the fully filled orbital accounted for?
My knowledge of chemistry is very basic at this point, so I would appreciate if any mistakes in the logic of my question or otherwise were brought to my attention.

While adding electrons, see the energy level (generally, and add some $d^5$ and $d^{10}$ considerations too). While removing electrons, the OUTERMOST electron gets removed first.

@insipidintegrator thank you for your comment. This might be a dumb question but does "outermost" simply mean the highest subshell of the valence shell? Ex: 3p in Cl, and 4s in Na?

Yes.  That’s it.

highest $n+l$ value

Don't waste your time in hair splitting of electron configurations. Most chemists remember electron configurations like a parrot and when you ask the details-nobody knows! I have yet to encounter one who knows the experimental details of electron configuration determination.

The electrons will be removed first from the 4s orbital and not the 3d orbital. It is because the 3d orbital is less in energy than the 4s orbital. The 3d electrons screen the outer most electrons, so electrons can be easily removed.


text:
Le Chatelier's Principle - Increasing concentration of coloured ion
Take the following equilibrium:
$$\ce{[CoCl4]^2- + 6H2O <=> [Co(H2O)6]^2+ + 4 Cl^-}$$
Where $\ce{CoCl4^2-}$ is blue and $\ce{[Co(H2O)6]^2+}$ is pink. My understanding is that Le Chatelier's Principle states that the system would act to partially offset any changes. K value is ~ 588, thus heavily favouring products.
Thus, if I add more $\ce{CoCl4^2-}$ and thus increase its concentration within the system, the equilibrium would initially turn much more blue, before the partial offset of this change which would lead to a new equilibrium being re-established which, whilst less blue than the solution immediately after the $\ce{CoCl4^2-}$ was added, is more blue than the original equilibrium since the system can only partially offset this change.
However, on a few sites on the internet, I have read that the solution will actually turn pink after equilibrium is re-established as the equilibrium is pushed to the right hand side as a result of Le Chatelier's Principle, due to a greater concentration of $\ce{[Co(H2O)6]^2+}$. Whilst I understand that $\ce{[Co(H2O)6]^2+}$ concentration will be greater at the re-established equilibrium, I believed that its increase in pink colour would not be enough to fully counteract the increase in blue colour as a result of the added $\ce{CoCl4^2-}$.
Can someone clarify on whether my existing beliefs are correct or whether the solution does actually turn pink?
Thanks

It depends on the equilibrium constant. Does it favor products or reactants?

@BuckThorn I was not aware it made a difference. Could you please elaborate?

Le Chateliers principle simply says that the system tends in the direction that takes it back to a state of equilibrium. It doesn't make much sense to discuss changes to a system if you don't know where this equilibrium lies in the first place.

@BuckThorn Edited. Pink side heavily favoured according to https://chemdemos.uoregon.edu/demos/Cobalt-Chloride-Equilibrium-Influence-of-Concentration-and-Temperature

You forget that to get the pink, blue is sacrificed. So at that point there is no blue left

@SafdarFaisal It's curious, however, that in all videos I've seen of this experiment being conducted (however none do the trial that i describe in the question as far as I've seen) all of the equilibriums are purple (e.g. https://www.youtube.com/watch?v=R0z8Ya-_kI4&ab_channel=ChemistrywithDrH). Blue is sacrificed, however shouldn't the excess blue be enough to counteract the increase in pink (since LCP only partially offsets changes to the equilibrium)?

Borrow a diode array visible spectrophotometer and do the experiment and see what happens. You have Keq so do some calculations to determine reasonable concentrations. Then let us know if you can see the difference.

You have the constant. You can  calculate, you need not to believe.

@jimchmst I don't think I need to know it in such detail - I just need to know if it becomes 'more' blue or 'more' pink if that makes sense.

If you add blue it will become more blue and since some of the blue becomes pink it will become more pink.   blue + pink = ? be a chemist and do the experiment and also get your eyes checked for color

@jimchmst If I had the materials I certainly would. The reason I'm asking is because of this powerpoint slide: https://ibb.co/DQ3tzc6

If $\ce{CoCl2·6H2O}$ is dissolved in water, the solution is pink, because of the pink ion $\ce{[Co(H2O)6]^2+}$. Matter of fact, the amount of blue $\ce{CoCl4^2-}$ ion produced by the following equilibrium (as proposed in the original post) is negligible  $$\ce{CoCl4^2- + 6 H2O <=> [Co(H2O)6]^2+ + 4 Cl-}$$ because $6$ $\ce{H2O}$ will push the equilibrium to the right-hand side.
If the same substance $\ce{CoCl2·6H2O}$ is dissolved in ethanol, the reaction is $$\ce{2 CoCl2·6H2O -> "2 Co^2+ + 4 Cl^- + 6 H2O" -> CoCl4^2- + [Co(H2O)6]^2+}$$ This reaction produces the same amount of both cobaltic ions. But as the blue color of $\ce{CoCl4^2-}$ ion is much more intense than the pale pink color of $\ce{[Co(H2O)6]^2+}$, the mixture of both ions appears blue. In this system, water is lacking. So the mixture turns deep blue, for entropic reasons : there are more independent species ($1 + 6 = 7$) on the left-hand side than on the right-hand side ($1 + 4 = 5$).
Now, adding a few drops water to this blue ethanol solution makes it pale pink, because the equilibrium is pushed to the right-hand side by the excess of water.
Now it you heat a bit the obtained pink solution in aqueous ethanol, it turns blue again. This is because heating a system where an equilibrium exists favors the endothermic direction of the reaction, which is the reaction right to left. And as soon as the hot solution is cooled down, the solution turns pink again.

I'm not exactly sure I follow on the comments about addition of water - I believed that water did not have any effect on LCP when it was a liquid as it did not affect concentration? By adding more water, you don't get more concentrated water, you're just increasing volume

No ! By adding water to the blue ethanol solution, containing practically no water, you change the position of the equilibrium, as written by you. When adding water, the system reacts by consuming a part of this water, according Le Chatelier's principle applied to the equilibrium equation. So the system consumes some $\ce{H2O}$ and also some $\ce{CoCl4^2-}$. And it produces more $\ce{[Co(H2O)6]^2+}$ plus of course some $\ce{Cl-}$

@bio If you formulate the equation for the reaction constant, you will realize changing concentration of all by water dilution strongly disbalances the reaction quotient from equilibrium, which has to be re-established.


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Would it make sense to filter air through oil?
I have seen water-based air purifiers, which work by having air bubble through water supposedly leaving water-soluble molecules behind.
I understand that this concept also humidifies the air due to the contact with water.
Would there be any benefit in doing the same thing, but using oil as the liquid instead?
Does the air contain oil-soluble molecules & would the air become "oily" (in contrast to humid)
Unfortunately my english lacks the correct terms for this and any searches result in information about how to change oil filters in cars.

The compounds in the oil that might end up in the exhaust are the volatile fraction (by definition). Oil might be used in a pump mechanism as a lubricant or sealant, but then shouldn't contain large amounts of volatile substances. Vacuum pumps that use oil or other substances to capture volatile substances are called entrapment pumps. A diffusion pump might use oil, but not as you have proposed.

Oil definitely traps some molecules in air. When you run a vacuum distillation using an oil pump, you place a cold trap between the still and the pump so you don't have to change the oil so often.

It will A) Smell liek the oil you use and B) Carry some oil with it. I have used straight air from an air compressor to breathe in a confined space and yeah I am fine after 4 hours of it but it smells and tastes/feels like oily air. You won't like that air


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How to determine the concentration of citric acid in a carbonated drink
How to determine the concentration of citric acid in a carbonated drink? Does it make sense to use titration if the soda is heavily colored? Or is it possible to get citric acid salt somehow?

Very important: Must leave drink opened in order to ensure complete decarbonation and minimum dissolved CO2. Read here: https://www.researchgate.net/publication/264159629_Determination_of_Citric_acid_in_Soft_drinks_Juice_drinks_and_Energy_drinks_using_Titration

is it possible to precipitate acetic acid using calcium hydroxide somehow?

I haven’t read the whole paper I linked, but your comment seems off-topic and out-of-the-blue. Could you tell me how it is linked to the post please?

perhaps there is a way to precipitate citrate acid from soda without using titration

Soda does not have acetic acid AFAIK

I meant a carbonated drink like coca cola

if you add calcium hydroxide to carbonated drink, won't you get calcium citrate that can be filtered out

I made a typo I wanted to say citric acid

There can be precipitated calcium carbonate too, from remaining CO2 or HCO3^- . // Are you interested in citric  acid alone, or in total citrates? As for usual pH, acid will not be IMHO the majority form.

but the resulting calcium citrate can react with sulfuric acid, this will restore the pH and release citric acid

High Pressure LC, anyone?

The determination of citric acid in carbonated soda is not an easy task because you have at least three different acids, (aq) carbon dioxide, small amounts of phosphoric acid and then citric acid, tons of sugar, and color. Most of the online teaching experiments which rely on simply titration are erroneous. If we simply titrated a soft drink with a base without any pre-treatment/sample preparation, the result is total acidity not citric acid concentration! You cannot use a calcium salt because calcium phosphate and calcium citrate will partially precipitate out. The proper procedure for soft drink analysis is ion chromatography. Only specialized labs have this facility.
If the experiment is to do a simple titration, just call it total acidity.

Citric acid solutions can be titrated by potassium permanganate, as citric acid is $\ce{CHOH(CH2COOH)2}$. It has a central alcohol group which is oxidized into a ketone group by $\ce{MnO4^-}$ ions at boiling temperature. It produces acetonedicarboxylic acid $\ce{CO(CH2COOH)2}$. This technique was used in old literature, like the French book "Guide de chimie biologique", Dunod, Paris, 1913". Unfortunately not too many practical details are given in this book.

Idea is nice but this will not work in the presence of sugar.


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Henry’s Law Pressure Limitations
I’m trying to decide if I can I expect Henry’s Law to be an effective way to estimate how the solubility of a gas in a liquid solution will change if I decrease the pressure.
I know Henry’s is best applied where temperatures are not too low, pressures are not too high, and the solute is very dilute in the liquid solution.
My application is about 40 C and very dilute in the liquid phase, so I’m not really worried about those factors as much. But the pressure is 430 psig (with the gas of interest being a large portion of that).
I know it’s different for every system, but what do we generally consider low/moderate/high pressures for ideal gas/Henry’s application? Do you think Henry’s would still be an accurate predictor here? Okay but starting to deviate a bit? Or likely quite off base?
I haven’t really found any good rules of the thumb on this. Working on getting this modeled so won’t ultimately be assuming simple proportionality for design, but just trying to do a quick check ahead of those results.

You might want to at least give a hint as to the nature of the system. Deviations from ideality will depend very much on this.

I believe Henry's law would be good for a rough estimate and indeed might be very accurate in your case. At your suggested pressure, though, the gas phase will likely deviate from ideality and require some correction.
For very high pressures, Krichevsky and Kasarnovsky (J. Am. Chem. Soc., 57, 2168–2171 (1935)) derived a pressure-corrected version of Henry's law that they found accurate for "nitrogen and hydrogen in water at a pressure up to 1000 atmospheres and temperatures from 0 to 100°"C. It adds another parameter (the partial molar volume of gas in liquid), values of which may or may not be available for your system of interest. The Krichevsky-Kasarnovsky equation still has the same caveats as Henry's law, except for the pressure restriction.


text:
Solid state chemistry current in tunnel diode
I am interested in this subject: Applied Mathematics in Electrical Engineering. I am interested in finding an expression for the tunnel current of a tunnel diode in the negative differential resistance region.
I know the basics of quantum mechanics and how to find the wave function of a single electron inside a potential well.
The potential function V(x) of an electron inside the P-N junction can be split into 3 regions.1 region(A) is the region of increasing potential due to the fact electrons have to jump from the valence band to the conduction band of the N-type semiconductor.1 region(B) is the region of increasing potential due to the electric field of the P-N junction and 1 region(C) is the region of decreasing potential due to the fact the electron loses its energy and goes in the valence band of the P-type semiconductor.
Questions: for region A can I consider the graph of the potential to be a straight vertical line with height equal to the band gap?
for region C can I consider the graph of the potential to be a straight vertical line with height equal to the band gap?
for region B does someone know the function of the potential V(x) inside the P-N junction?I need to solve the time-independent Schrodinger equation for an electron.Can I approximate the function V(x) with a line?
Can I approximate V(x) to be something like this?

Help appreciated.

I am afraid, it is not a question for chemists.

The RCA Tunnel Diode Manual, linked in the answer here might provide a bit of help. But this question really should be migrated to the electrical engineering stack exchange. Or the physics one.

The potential across the p-n junction depends on the type of bias applied, but your approximation is valid for extremely narrow p-n junctions. The Esaki diode needs a difference in Fermi levels across the junction in order for an appreciable current to flow. Here is a diagram from a professor at Vassar showing the potential across the diode:

You can see that they also approximate it as linear and that the valence band of the p-doped side is higher in energy than the conduction band of the n-doped side.


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Can heterocyclic molecules form sandwich-like complexes?
As unsaturated heterocycles have the same planar structure as cyclopentadiene ion/benzene "sandwich" compounds, I was wondering if a similar coordination between metals can be achieved.
When asking this question, it is related to the potential geometry formed that is similar to that of sandwich compounds, not similar bonding characteristics. That is, a ion above the centre of molecule, not adjacent to the side.
If there is any literature on this topic, it would be greatly appreciated.

Yes, heterocyclic compounds are viable for the $\pi$-donation necessary to coordinate a metal as a sandwich complex—the electron distribution about the $\pi$-system is different, so bonding characteristics aren't guaranteed to fully translate (you may get for a five-membered ring an $\eta^4$ coordination instead of an $\eta^5$ coordination sometimes, for example). If a heteroatom has substantial p-character in its electron density, then in theory it can participate in the binding to the metal.
Here is a link to a related 1993 review on the subject if you can find a way around the paywall.


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Do dessicants work inside of a freezer?
I learned that water sublimates when frozen, which prompts me to ask:
Do dessicants work inside a freezer?
While dessication inside of a freezer is unlikely to remove large quantities of water fast, it intuitively seems to me that maybe it could enhance the depth of dessication (i.e. how little traces of water ultimately are left) by certain dessicants (maybe by improving the water-dessicant affinity and/or stabilize higher hydration states of the dessicant used).
Would it be a good strategy do the bulk of dessication inside a dessicator, and throw in a product in the freezer alongside a well selected dessicant expecting it to slowly dessicate to very low trace of water over time this way?
Are some dessicants better suited for this (e.g. that dessicate by trapping away water by adsorbtion?) than others (e.g. that dessicate by trapping away water by forming higher hydration states?)?

Water does not sublime when frozen, you need a very low pressure as well.

First google (highlighted) result says: "For those of us interested in the water cycle, sublimation is most often used to describe the process of snow and ice changing into water vapor in the air without first melting into water."

You are reading out of context and this is why random searches are dangerous. Frozen water does not sublime under ordinary pressures. You need very low pressure for ice sublimation to occur. Your freezer will not have sunlight. When we talk of sublimation in very cold climates, sunlight is required for this process. This is a separate mechanism. I do not know the mechanism of that. You need about "0.00604 atm" to cause sublimation of frozen water in a laboratory!

I don't understand this well. But surely if it occurs at atmospheric pressure with sunlight it occurs to some degree in a freezer? As you state, wikipedia reads: "The loss of snow from a snowfield during a cold spell is often caused by sunshine acting directly on the upper layers of the snow. Ablation is a process that includes sublimation and erosive wear of glacier ice."

@AChem, water most definitely does sublime at low temperatures! For example, look at an ice-cube tray kept for a few months in a freezer at -30°C, and notice how little is left. See this table, https://biopharma.co.uk/blog/2015/03/30/vapour-pressure-of-ice/, for example that show ices still has a comparatively high vapor pressure well below freezing

@DrMoishePippik, Indeed, tt is a game of vapor pressure of ice-water equilibrium in the ambient. If it drops below 0.006 atm, it will start to sublime. However, the timescale, the OP is interested in, pressure has to be dropped!

Yes, an effective  desiccant would help lower the amount of water vapor in a container and speed freeze drying in a sealed system. That said, freeze-drying usually employs continuous vacuum, because it is more effective than most desiccants at low temperature. There are many choices, such as silica gel and clays, and reactive chemicals such as calcium hydroxide $\ce{Ca(OH)2}$, that might be useful.
Water is a bit unusual in having a comparatively high vapor pressure at temperatures well below its freezing point. For example, in a home freezer at -30°C (~243 K), ice cubes will sublime from an open tray in a few months without help from a desiccant or vacuum pump. At that temperature, the pressure would still be ~0.4 mbar. However, if you could evacuate that freezer, they'd disappear more quickly.

Thank you for your answer. Do you know which dessicants are more prescribed for this than others? And which ones are proscribed, if any? Am I for instance more likely to get very low residual water in the product with Na2SO4 or with Silica Gel? Thank you!

From the links cited, you might try a large amount of finely divided silica gel or CaO, carefully dried and kept dry, away from moist air, before sealing in a moisture-proof container (even polyethylene allows moisture through, albeit slowly). Note that CaO forms Ca(OH)2 in air, which can attack glassware.


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What is the Volume occupied by gas at extreme or limiting conditions of volume?
We know that

1 Mole of an Ideal Gas occupies 22.4 L at STP.

[real gases also are also close to this volume , even tho there might be deviation, for this experiment let's take both to be same]
But it's also said that :-

Gases occupy the volume of the container they are placed in.

So what if we take one Mole of a gas at STP and put in a container of let's say 100 L or maybe a container of volume much much greater than 22.4L
Then how much volume does the gas occupy ?
Does it occupy 22.4L or close to it (real gases) ? Or does it actually occupy the whole volume of container and thus in this condition even real gases would start to behave as ideal gases .
(Also does the effect of gravity matter here ?)

Note that, since STP is defined as 273.15 K and 100 kPa, the molar volume of an ideal gas at STP is actually 22.7 l/mol.

@Loong yea ik that, the IUPAC changed the definition recently

well, "recently" means 40 years ago

We're taught that definition and it hasn't been updated in reputed books and exams, so I didn't know

@ParasGupta, These STP conditions are chosen by human organizations, there is nothing heavenly or canonical about them. There are scores of other ways of expressing standard conditions. STP is IUPAC's.

Gases occupy the entire volume of the container, so the gas particles are everywhere in the container. Imagine you transfer 1 mol of gas to an empty container with a volume of 100 L, and if you sample the gas from any region of the container, you find gas particles everywhere. Therefore, the density of the gas will be the same in any region.
Gases begin to display gravitational effects when the container is large, say the size of the Earth. At the bottom of that container, you will find more gas particles than at the top. Moon-sized containers are small.

whatever you said did occur to me and makes sense, but then what is the significance of saying that the Molar Volume of an Ideal Gas At STP is 22.4L.(22.7L according to the updated definition)

Imagine doing an experiment with container which has a frictionless piston. Initially, it has perfect vacuum and its volume is zero. Now inject 1 mol of a gas at STP into this cylinder. The piston will rise to a certain height and if you calculate the volume (Recall the formula of cylinder volume), the volume will be 22.7 L.


text:
Which 1,2-methylene shift is more favourable in ring expansions where ring has substituents

In the above reaction, first the -OH group is protonated after which it leaves creating a carbocation intermediate as shown
Now, ring expansion is a more favourable rearrangement, however there is a confusion between which methylene shift should take place (the two possibilities are shown in orange and yellow) each of which would give two different products -
3-methyl cyclohexene or 2-methyl cyclohexene.

Please explain which methylene shift should occur and why?


text:
Balancing a redox reaction equation with a mystery reactant
The answer to the following question from NSAA 2021 Chemistry Section 2 is C.

Consider the following chemical equation:
$$\ce{vQ + wP4 + x H2O -> yPH4I + zH3PO4}$$
where Q is a binary compound.
The molecules of Q are hexatomic and contain phosphorus in the +2 oxidation state.
Using the lowest integer values for all the coefficients $v, w, x, y$ and $z$, what is the value of $w$ when the equation is balanced?
A. 1
B. 2
C. 13
D. 16
E. 24
F. 26

I assumed that the unknown compound Q would be composed of P and I, and more specifically to be $\ce{P2I4}$ as it is stated that the compound is hexatomic and the oxidation state of P is +2. However, I can not find a way to balance the equation. Does anyone have a solution for this?

Well, $\ce{P2I4}$ has to be reduced to form $\ce{PH4I}$, and $\ce{P4}$ must undergo disproportionation. Otherwise, you don't have enough phosphorus to balance the amount of iodine in forming $\ce{PH4I}$ where the phosphorus has an oxidation state of -3. So you can write three different equations:
$\ce{P2I4 + 8H+ + 10e- -> 2PH4+ + 4I-}$ (eq.1)
$\ce{P4 + 16H2O -> 4H3PO4 + 20H+ + 20e-}$ (eq.2)
$\ce{P4 + 8I- -> 2P2I4 + 8e-}$ (eq.3)
Let's combine eq.1 and eq.3 to cancel out the $\ce{I-}$ ions. Note: $\ce{2PH4+ + 4I- \equiv 2PH4I + 2I-}$:
$\ce{4P2I4 + 32H+ + 40e- + P4 + 8I- -> 8PH4I + 8I- + 2P2I4 + 8e-}$
$\ce{4P2I4 + 32H+ + 32e- + P4 -> 8PH4I + 2P2I4}$
$\ce{2P2I4 + 32H+ + 32e- + P4 -> 8PH4I}$ (eq.4)
Now we need to combine this eq.4 with the eq.2 to get rid of the $\ce{H+}$ in the equations, which should also cancel out the electron terms. You can simply multiply eq.4 by a factor of 5 and eq.2 by a factor of 8 and then combine. This would result in:
$\ce{10P2I4 + 160H+ + 160e- + 5P4 + 8P4 + 128H2O -> 40PH3I + 32H3PO4 + 160H+ + 160e-}$
Or finally the fully balanced equation:
$\ce{10P2I4 + 13P4 + 128H2O -> 40PH4I + 32H3PO4}$
So this is where you get the 13 from.

I don't have 4 disproportionation reactions of P4. Only eq.1,2,3 are relevant. I just combine them in a systematic way to get the final balanced equation which can be verified.

Once we accept that the mystery reactant is $\ce{P2I4}$, this is the equation we are asked to balance (this might or might not be a fantasy reaction - the question does not teach you much about chemistry):
$$\ce{vP2I4 + w P4 + x H2O -> y PH4I + z H3PO4}$$
This is a problem with 5 unknowns. You can come up with 5 constraints (the number of $\ce{P, I, H, O}$ atoms on either side has to be equal, and the coefficients should be the smallest non-zero integers). So unless any of these conditions are redundant, there should be a single solution.
I will show a simple way of balancing this without using half-reactions (even though this is a redox reaction). Then, I will discuss the complexity of trying to use half-reactions.
Using the atom balance to find the coefficients
Notice that all species are neutral, so we have an automatic charge balance. I will start with an atom present in only two species, oxygen, to eliminate the variable $x$, then proceed to hydrogen, iodine, and finally phosphorus:
$$\ce{vP2I4 + wP4 + 4zH2O -> y PH4I + z H3PO4}\tag{O}$$
$$\ce{vP2I4 + wP4 + 4zH2O -> 5/4z PH4I + z H3PO4}\tag{H}$$
At this stage, I will multiply by 4 to remove the fraction:
$$\ce{4vP2I4 + 4w P4 + 16z H2O -> 5z PH4I + 4z H3PO4}$$
$$\ce{5/4z P2I4 + 4w P4 + 16z H2O -> 5z PH4I + 4z H3PO4}\tag{I}$$
At this stage, I will multiply by 4 to remove the fraction:
$$\ce{5z P2I4 + 16w P4 + 64z H2O -> 20z PH4I + 16z H3PO4}$$
$$\ce{5z P2I4 + 13/2z P4 + 64z H2O -> 20z PH4I + 16z H3PO4}\tag{P}$$
At this stage, I will set $z$ to 2 to get the lowest integers:
$$\ce{10 P2I4 + 13 P4 + 128 H2O -> 40 PH4I + 32 H3PO4}\tag{done}$$
Less formal way of writing
If you solve this by inspection, you would leave out the unknowns and just write the coefficients in the correct ratios:
$$\ce{4 H2O -> H3PO4}\tag{O}$$
$$\ce{16 H2O -> 5 PH4I + 4 H3PO4}\tag{H}$$
$$\ce{5 P2I4 + 64 H2O -> 20 PH4I + 16 H3PO4}\tag{I}$$
$$\ce{10 P2I4 + 13 P4 + 128 H2O -> 40 PH4I + 32 H3PO4}\tag{P}$$
Half-reaction method
The only atoms that have distinct oxidation states are the phosphorus atoms; there are four species, each one with a different oxidation state. The two reactants could either be oxidized to the +5 state (in $\ce{H3PO4}$) or reduced to the -3 state (in $\ce{PH4I}$). Knowing the coefficients of the balanced equation, we can make arbitrary decisions about how to write the half-reactions.
If we want to reduce all of the $\ce{P2I4}$, we get the following scheme:

The half-reactions would be:
$$\ce{10 P2I4 +100 e- + 80 H+ -> 20 PH4I + 20 I- }\tag{red}$$
$$\ce{13 P4 + 20 I- + 128 H2O -> 20 PH4I + 32 H3PO4 + 80 H+ + 100 e-}\tag{ox}$$
If we want to oxidize all of the $\ce{P2I4}$, we get the following scheme:

I will leave it as an exercise to figure the half-reaction for this scenario. Or you can choose anything in between these two schemes, for example:

Now, both half-reactions contain both reactant species.
So you have non-unique half-reactions, but a unique solution to the combined reaction. To get from the half-reactions, you would have to combine them in a way that no ions are "left over". This makes the half-reaction scheme complicated and confusing in this case.
Redundant conditions
This problem is different than being asked to balance a reaction like this:
$$\ce{H2(g) + O2(g) + NaCl(s) -> H2O(l) + NaCl(aq)}$$
Like the original problem, we have four different types of atoms and five coefficients. However, the atom balance for sodium and for chlorine counts as only one constraint because they are redundant. In this case, the equation can be separated into:
$$\ce{H2(g) + O2(g)  -> H2O(l) }$$
$$\ce{NaCl(s)  -> NaCl(aq) }$$
and there is no unique way to balance the sum of two independent reactions. For the reaction given by the OP, balancing the reaction has a unique solution, but writing down the half-reactions doesn't. This is unusual and probably disturbing to some.

What is unusual is using a chemical reaction to generate phosphine to react with the HI, phosphine could have been added independently.

@jimchmst I'm not sure that whoever wrote the problem knows about the chemistry behind it. I certainly don't; I took the chemical equation as correct (no intermediates that accumulate etc) and balanced it.

@jimchmst You still have access to the post, and you are encouraged to edit your original answer, merging the two post to communicate your insight. Your second post wasn't an answer, it was a comment on the first post. At least that is what it looked like, so it was deleted.

Thank you I am learning the protocols, slowly it seems! I rewrote my post Please read it.

This involves two independent reactions that, if added properly, will result in a neutralization of the hydriodic acid formed in the first reaction with phosphine formed in the second by conversion [on paper at least] to phosphonium iodide. I think that this is a much simpler approach and that it makes more chemical sense. Whether it actually happens??
The first equation is the actual redox reaction: this is balanced by choosing one as the oxidant and the other as the reductant [it makes no difference which is which], writing the half-reactions and balancing. [it is easier to have phosphorus as P then convert to P4]
$\ce{16 H2O + P4 + 2 P2I4}$ = $\ce{4 H3PO4 + 4 PH4I +4 HI}$
The second equation is the disproportionation of P4: where P is the oxidant and the reductant; simply write a half-reaction for each and combine.
$\ce{12 H2O + 2 P4}$ = $\ce{3 H3PO4 + 5 PH3}$
To react the phosphine with the HI the two reactions must be kinetically similar or somehow be mechanistically combined. But to neutralize the HI with the phosphine the first reaction must be multiplied by 5 to give 20 HI and the second multiplied by 4 to give 20 PH3; they react to give 20 PH4I. Multiply, add, and collect:
$\ce{128 H2O +13 P4 + 10 P2I4}$ = $\ce{32 H3PO4 + 40 P4I}$
Mechanistic studies are necessary to determine what actually happens. I think that this is an improper use of a chemical equation because the two processes are not [or most likely not] mechanistically linked. The same result can be obtained by independently adding phosphine and if needed phosphoric acid.
As an added thought, these reactions, if actually possible in some form, destroy the possibility that the detection of phosphine in Venus's atmosphere MUST be linked to life.

I must apologize the exact relationship furnishes the proper amount of Phosphine to react with the HI from the redox reaction.

You can always edit your answer. When you edit it, you might want to format the equations using mhchem.


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Improper dihedral angles: how to tell between 0 and 180 degress
How can I tell the difference between a 0 and 180 degree value when looking at an improper dihedral. For example, in the molecule below would the dihedral angle 4-2-3-1 have a value of 0 or 180 degrees?

It is obvious that the two planes defined by 4-2-3 and 2-3-1 are coplanar, but how do I define the angle between them? If I understand correctly, it should be the angle between the norm vectors of the two planes, but I am not sure how to choose the direction of the norms. Is there a standard convention for this?

This can be sorted quite easily, have a look at the site  https://applying-maths-book.com/chapter-6/vectors-F.html. for a worked example.  If you cannot see this I can give details as an answer

Yes. Think of the hinge 2-3 as dividing the plane into two half-planes. If 1 and 4 are in the same half-plane as if the atoms had been folded over, that is a zero dihedral angle. If 2 and 4 are in opposite half-planes like an unfolded sheet, that is a 180° dihedral angle. Your drawing looks like the latter.

Yes, following that through, you can define a dihedral 1 3 4 2 alternatively, being 0°.  That doesn't make much sense in the case of ethyne, the model shown here, but in BH3 the definition would make more sense.

I'm quite sure there is a mathematical way of showing this, but that is just a bit too much for me to formalize. Essentially when you go clockwise, the normal vector sticks out of the plane to the front, counter clockwise to the back. Then it's easy to form an angle.

Exactly @Martin. Given four vertices of a tetrahedron you can define any of six distinct dihedral angles!

@OscarLanzi how about 180 and -180 degree? Are they same?


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Can other metal salt compounds besides copper sulfate/zinc sulfate be used to create chlorophyllin? E.g. the magnesium center is replaced
There has been a lot of research done into sodium copper chlorophyllin. The synthesis is quite simple and can be done at home. It involves extracting the chlorophyll, saponifying the solution, adding the metal salt, and then adding actual sodium.
Most synthesis guides call for copper sulfate, a copper salt. Zinc sulfate can also be used. I was wondering which other copper/zinc 'salts' can be used? I am asking because I was looking at purchasing some OTC zinc/copper supplements, instead of having to spend a lot on a bulk size of FCC copper sulfate, and just using the powder from the capsules/tablets. I see things like copper/zinc bisglycinate, copper gluconate, etc. Copper acetate I could make by just adding vinegar to a copper pot!
It would be most ideal if I could make the substance at home with somewhat easy to find chemicals and a copper bowl. I'm not too wild on burning sulfur to make sulfuric acid at home though :s
Thanks!

Synthesis of many different types of porphyrins with very many different metals are known and I suppose that the chlorophyllins are not so very different. Many metal centred  phthalocyanins are also known, so searching for  these types of syntheses may be a useful starting point.