text:
Double dehydration of a geminal diol
I've recently came across this reaction and I am wondering about the mechanism. Could someone please help me? If you could give me the name of the reaction (if it has one) it should also be enough for me

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

Technically, only a single dehydration is happening here; count the atoms in the reactant and product.

For future reference: consider learning how to use a molecular sketcher when reporting an organic reaction.  You find a comparison (e.g., http://www.gunda.hu/dprogs/) and listings (e.g., https://en.wikipedia.org/wiki/Molecule_editor) and even find some not requiring a local installation to export, e.g., .png (https://pubchem.ncbi.nlm.nih.gov/edit3/index.html).

It's a single dehydration with a carbocation intermediate that allows the five-membered ring to expand to a more stable six-membered ring.

While your answer is correct, we discourage here answering questioms violating homework policy.


text:
Using charcoal to clean water
I have read about charcoal being used to clean water before it is then boiled or mixed with a UV light stick. I have some questions about this:

1) Does it have to be activated charcoal?
2) Can I use normal charcoal from any old shop (in the UK) as I assume
  these days everything is synthetic or have added harmful chemicals?
3) Can it purify irradiated water? Yes I am a huge Fallout 3 fan!

If there is anything else that I might not have thought of asking, or might be interesting, feel free to let me know :-).

Activated charcoal cleans water mainly by adsorbing the impurities.
You can also couple normal charcoal with sand and form alternate layers of charcoal and sand and pass water through it. It's a very effective Indian style of filtering water. Nearly all suspended particles gets blocked by the above mentioned filtering style.
You can also convert normal charcoal into activated charcoal to serve your purpose.
Now irradiated water cannot be treated by these methods alone. Normal carbon filtering is used along with several other methods such as ion exchange methods and other such high level water processing methods to treat irradiated water. High levels of radiation can't be treated fully though.
You can refer to this link to learn about effective radioactive water purification techniques.

source: Stack Exchange

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sort: votes


text:
What is the principle of RP-HPLC?
Is the following statement the correct way to explain principle of RP-HPLC?

It works on the principle of hydrophobic interactions hence the more nonpolar the material is, the longer it will be retained.

"It works on the principle of hydrophobic interactions hence the more nonpolar the material is, the longer it will be retained." ?

It is a very crude, first order, approximation. A philosophical then arises, what is "non-polar" material and how would we measure it's nonpolarity? One should immediately start to see problems with the words like polarity...an umbrella term which is good for organic chemists only. A good rule of thumb for any explanation is that can I measure this property? If you can't, that term is just a useless filler word.
As a matter of fact let us say you have a C18 column X, C18 column Y, C18 column Z from three different sources, and you separate a mixture of A, B, C, D, E, F, F etc, which are hydrophobic compounds in general. The order of elution on all three may not be the same. The reason is that hydrophobicity is not the only reason for retention in RPLC.
See http://www.hplccolumns.org/howitworks/index.php
Despite 100 years of chromatography, it has been very difficult to predict to elution order and elution time from the first principles. People try to over rationalize chromatographic behavior. LC is still an experimental science when it comes to real mixtures.

It works on the principle of hydrophobic interactions [...]

That part of the statement is fairly accurate. While "normal phase" chromatography has a stationary phase that is capable of hydrogen bonding, the reverse phase stationary phase is not.
Rather on focusing on the complexity of the process, I will compare reverse phase with ion exchange chromatography, using separation of oligodeoxynucleotides (short pieces of DNA) as an example.
Reverse phase
You would apply the sample in aqueous solution, and it would bind to the stationary phase. Then, you elute with a gradient of increasing acetonitrile concentration. The acetonitrile competes with the solid phase for interactions with the aromatic groups in the DNA.
Ion exchange
You would apply the sample in low ionic strength aqueous solution, and it would bind to the stationary phase (DNA is charged negatively, so you would work with an anion exchange material with positive surface charge). Then, you elute with a gradient of increasing sodium chloride concentration. The sodium ion compete with the anion exchange material for ionic interactions with the phosphate di-ester groups in DNA.

[...] hence the more nonpolar the material is, the longer it will be retained.

If you are trying to rationalize which of two molecules will elute first, you also have to consider the differential interactions with the solvent.

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text:
Simple synthesis of thiocyanates
Can sodium thiocyanate be synthesised by combining equimolar amounts of ammonia, sodium hydrosulfide and carbon dioxide in the presence of phosphorous pentoxide by the following reaction: NH3 + NaSH + CO2 -> NaSCN + 2H2O ?

It has been observed spectroscopically but has not been isolated as a pure substance.

Thiocyanic_acid
Regarding the sodium salt,  creating it in concentrated sulphuric acid cannot be considered seriously.
$\ce{P4O10}$ would not be much better, forming various forms of condensated phosphoric acid.
Furthermore, carbon dioxide manifests  weak oxidative properties at very high temperatures only.

I see. Edited question.


text:
Will NaF + CaCO3 precipitate much CaF2?
If I mix sodium fluoride, calcium carbonate powder, and water, under what conditions (if any) would there be an equilibrium?
I.e. assume we start out with fully dissociated $\ce{Na+(aq) + F-(aq)}$, because it's fairly soluble. But when I add the $\ce{CaCO3}$, it's only slightly soluble, so assume an excess of solid $\ce{CaCO3}$ and a small amount dissolved.
Will that be stable with $\ce{NaF}$ concentration much greater than the saturated concentrations of $\ce{CaCO3}$, or will $\ce{CaF2}$ start to precipitate, and will it continue until exhaustion of a reactant, or reach equilibrium?
I'm wondering if it's the relative solubilities of $\ce{CaCO3}$ and $\ce{CaF2}$ that will drive the equilibrium (if that's what it is) one way or the other. I can't work out how to apply the theory of an equilibrium constant because of the excess solid reactant and corresponding possible precipitate messes up how concentrations balance out in a normal equilibrium… And I'm wondering if the absolute concentration of $\ce{NaF}$ effects things as well.
Any pointers appreciated. Or even some jargon to say what this kind of reaction is called.

I'm voting to close this question as off-topic because this is a personal/medical question since you're try to formulate a toothpaste. To give any kind of advice on a personal care product is just letting the camels nose under the tent.

Removed post script. Now it's a general question

I may be wrong but its not possible to make CaF2 because CaF2 is soluble and CaCO3 is not .

No, they're both slightly soluble

Updated question to be as general as possible so that it is on-topic.

Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate.
The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's well below the solubility of NaF.
However, the presence of additional carbonate could be used to reduce the calcium concentration. Similarly, any addition of acid will convert carbonate to bicarbonate and increase the maximum calcium concentration.
UPDATE:
In the above answer, I assumed a basic familiarity with solubility products. For those not familiar with those, here's more detail.
The key quantitative measure of solubility of ionic compounds is the solubility product, usually indicated as $K_{sp}$, which is the product of the concentrations of the separate ions.
For $\ce{CaCO3}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{CO3^2-}]$. Reported values vary somewhat, but are typically around $2\times 10^{-8}$.
Likewise, for $\ce{CaF2}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{F-}][\ce{F-}]$, and the reported values are around $4\times 10^{-11}$. Thus, the condition for keeping fluoride in solution is
$[\ce{Ca^2+}][\ce{F-}][\ce{F-}]<4\times 10^{-11}$.
If $\ce{CaCO3}$ is dissolved in distilled water to maximum solubility, the calcium ion concentration is $\sqrt{K_{sp}}\approx 0.15$ mM, which is the basis of the calculation above.

Attempt to expand your explanation: CaF2 will start to precipitate as long as the concentration of fluoride ions is > ~0.5mM and calcium ions > 0.25mM. Since there is an excess of CaCO3, the precipitation will continue until either the fluoride ions are reduced to 0.5mM or enough carbonate ions build up to prevent more than 0.25mM of calcium dissolving...? But the 'common ion effect' not only works with the carbonate. Fluoride ions will lower the solubility of CaF2, so that has to be accounted for as well? I'll work on a concrete example and cite your answer assuming my expansion is correct.

One has to consider calcium concentration not originated from CaCO3 dissolution, in the case the solvent contains any.

@Poutnik - I interpreted the question to mean that CaCO3 is the only other source of calcium, and I specified "distilled water" in the answer to make clear that I am assuming no other source of calcium.

I understand, but both cases are worthy to consider

I think the question was pretty clear that CaCO3 is the only source of calcium

Consider three compositions:
A. 2NaF + CaCO3              B. CaF2 + Na2CO3, and
C. NaF + 0.5 CaCO3 + 0.5 CaF2 + 0.5 Na2CO3.
Using data from the CRC Handbook (62nd ed), the heats of formation of A and B are respectively 560.47 and 560.6 kcal, so there is little driving force to make a reaction go to completion. Note that A should be near neutral pH, but B and C would be about 12 pH.
The difficulty with trying to make A become B is that CaCO3 is so insoluble that there aren't enough Ca++ ions to make a new crystal of CaF2 rapidly. And the F- ions from the NaF will just drive the Ca++ concentration down even lower - and it doesn't get better as the reaction proceeds, because the CO3-- ion produced is just as effective at keeping the Ca++ level down.
But if you make up system C, which is just 50% reacted and followed its pH over time, you might be able to tell if the reaction goes forward or backward, or more likely, just remains the same. The reaction just doesn't have enough dissolved Ca++ to react in a reasonable time.
But an acid would increase the concentration of Ca++ ions. CO2 does the least change to the system; then the system becomes D:
D. 2NaF + CO2 + H2O + CaCO3 --> CaF2 + 2NaHCO3.
The left-hand side of the equation now has a total heat of formation of 727.47 kcal and the right-hand side has 743.3 kcal, a change of 2.1%. Not much, but maybe that would force the system at a more reasonable rate to the desired reactants. NaHCO3 gives a pH of about 8.5. Fascinating that Ca(HCO3)2 is more soluble than NaHCO3! The CO2 that becomes entrapped in the NaHCO3 is a reactant, not a catalyst. The CO2 could be applied in excess (pressure) so that it is partially a catalyst, dissolving the CaCO3 and CaF2 until you dry the mix, at which time the CO2 volatilizes.

When you say "And the F- ions from the NaF will just drive the Ca++ concentration down even lower". Isn't that actually a driver for the CaF2 to form? Because them being in solution reduces the solubility of CaF2, but not CaCO3, so the F- does push down Ca++, but more should dissolve from CaCO3 until CO3-- builds up...?

Yes, F- from NaF drives the formation of CaF2. But with CaCO3 alone in H20, or with CaF2 alone in water, the Ca++ concentration is low; adding NaF (case A) or Na2CO3 (case B) drives the Ca++ concentration VERY low because the concentration of F- from NaF is way greater than CO3-- from CaCO3 and the concentration of CO3-- from Na2CO3 is way greater than F- from CaF2. What you say could be the best explanation if the reaction goes rapidly, but I think the rate would be very slow because of the low concentration of Ca++. That's why I suggested adding CO2 - to raise Ca++.

Another reason the reaction might slow down, in addition to low concentrations of Ca++, is that F- might coat the CaCO3 rather than precipitate as a separate crystal of its own. That would stifle dissolution of CaCO3. The same could happen with CO3-- on CaF2.


text:
On the meaning of distinguishability, and wavefunctions for 3 electron atoms
In a 2-electron atom at lowest energy, the $(1s)^2$ is occupied and the electronic wave-function must satisfy anti-symmetry requirements in the particle coordinates, as the spatial wave function is symmetric. How is the situation in a 3 electron system?
In a 3-electron atom (or in a nucleon with one excited quark) of lowest energy, say the $(1s)^2(1p)$ states are occupied, is the fermion in the $(1p)$ state distinguishable? I.e. must my wave-function still treat all 3 electrons indistinguishably, or are mixed symmetry wave-functions now physically viable?

All electrons are indistinguishable, which leads to the concept of a Slater determinant.

Note that there is no $1p$ state. Above $1s$ is $2s$, though the point of your question is still clear.

The only case in which electrons can be considered distinguishable is when they are so well separated that their wave functions do not overlap. Any electrons on the same atom or within the same molecule are considered indistinguishable.
In chemistry we like to talk about electrons being in specific orbitals, but the indisinguishable nature of them means that they are all in all orbitals simultaneously. This is why a mathematically rigorous description of a molecule using hybrid orbitals is just as valid as one using canonical orbitals. Neither is true in the sense of describing where each electron is, and both are true in the sense of describing the collective electron density.


text:
Is aluminium foil pure aluminium?
Is aluminium foil pure aluminium? Or does it have other elements in the e.g: tin (Sn) or lead (Pb)? Because when I try to oxidize it through the process of submerging it in water, it doesn't look like it wants to oxidize to aluminium oxide (AlO+). I did this exact same thing with pure aluminium and it oxidized within minutes.

Aluminium foil is pure Al covered with some polymer. Sn or Pb are definitely not a part of the picture.

You can try cleansing with eg acetone or ethanol or detergent and see if the coating is removed

Aluminium won't oxidise by submerging it in water any more than it does in air. Pure aluminium is covered in a tough but very thin layer of aluminium oxide which will protect the bulk metal from further reaction with air or water unless the protective layer is disrupted. Your "pure" aluminium was probably just treated with something that disrupted the normal protective layer.

@matt_black It was taken out of a emergency water light which had other chemicals like CU and some black coating, it did produce electricity(obviously); the electricity helped it remove the 'protective layer'?

From the Institute of Minerals, Materials and Mining,

Most aluminium foil is made from pure aluminium, but increasingly alloys are used to improve properties and reduce the thickness required.

If you want to make sure that the Aluminium foil you're using is pure, the only way would probably to ask the manufacturer. Other than that, it could be the case that the foil is either pure or alloyed, depending on how it's made.

Reynolds Kitchens says Relnolds Wrap® is $98.5\%$ $\ce{Al}$ with the rest mostly $\ce{Fe}$ and $\ce{Si}$ to improve strength. You know this because when you dissolve it in $\ce{NaOH}$ solution to produce $\ce{H_2}$ a black residue remains undissolved; pure $\ce{Al}$ metal would leave no residue. Perhaps $\ce{Al}$ wiring would be more nearly pure because small impurities decrease electrical conductivity. $\ce{Cu}$ for bus tubing is cast in a $\ce{N_2}$ atmosphere for this reason: $\ce{O_2}$ from the air would otherwise contaminate the metal.

NaOH = Sodium Peroxide? Are there any other chemicals that I could dissolve it in? Like NaHCl?

$\ce{NaOH}$ is sodium hydroxide, commonly called lye. Aluminum is amphoteric so it can dissolve in strong acid or base. $\ce{NaHCl}$ isn't a chemical; maybe you mean $\ce{HCl}$ hydrochloric acid, also called muriatic acid?

dang, got the name wrong I was tryna say NaOCl.


text:
How is bromothymol blue synthetised?
In a classical reaction, phenolphthalein is synthetised from phthalic anhydride and phenol with acid catalyst. When one uses substituted phenols, one obtains similar compounds (e. g. cresolphthalein, thymohlphthalein).
However, some acid-base indicators contain a sulfo- group instead of a carboxy- group of phthaleins. Does that mean that they are synthetised from phenols and the anhydride of o-sulfobenzoic acid? Phthalic anhydride is fairly cheap, but how is o-sulfobenzoic anhydride prepared?

Phthalic anhydride is prepared industrially by catalytic oxidation of naphthalene or o-xylene over vanadium pentoxide https://thechemco.com/chemical/phthalic-anhydride/

@Waylander I meant o-sulfobenzoic anhydride, but it wasn't really clear. I edited the question.

Now a days, sulfonephthaleins can be prepared by reaction of readily available saccharin and the desired plenol compound (Ref.1 & Ref.2). In this method, active reagent, sulfobenzoic anhydride will be prepared in situ as depicted in following diagram:

In this method, it's described the preparation of phenolsulfonephthalein (Phenol Red). If you use thymol instead of phenol, you will get thymolsulfonephthalein, which can be subsequently brominated by bromine/acetic acid to obtain dibromothymolsulfonephthalein (Bromothymol Blue). This procedure is described in Ref.2.
References:

V. H. Tillu, D. K. Dumbre, H. B. Borate, R. D. Wakharkar, V. R. Choudhary, "Solvent-Free One-Pot Synthesis of Sulfonephthaleins from Saccharin and Phenols," Synthetic Communications 2012, 42(8), 1101-1107 (https://doi.org/10.1080/00397911.2010.535946).
B. S. Rao, J. B. Puschett, B. M. Karandikar, K. Matyjaszewski, "Synthesis of Functional Bromothymol Blue Dyes for Surface Attachment to Optical Fibers," Dyes and Pigments 1991, 16(1), 27-34 (https://doi.org/10.1016/0143-7208(91)87018-I).

o-Sulfobenzoic Anhydride is prepared by cyclisation of ammonium o-sulfobenzoate (itself prepared by cHCl hydrolysis of saccharin) with thionyl chloride in benzene. The Organic Syntheses procedure is here: Org Syn

But is bromothymol blue actually synthetised from sulfobenzoic anhydride?

This reference may tell you https://www.sciencedirect.com/science/article/abs/pii/014372089187018I  From the abstract it looks like it is


text:
On the angle subtended between the angular momentum vector and the applied field for a single spin

According to Dr James Keeler of Cambridge University in one of his lectures at Queensland University, the angle theta can be any angle, but I struggle to see how this is true. There are only a certain number of I values that nuclei take and as illustrated in the figure above, each magnetic quantum number corresponds to a fixed theta. Is my logic flawed?

It depends whether you are talking about a single spin or a collection.

a single spin. I'd be interested to know about the collection case too.

As an aside, the coordinate system you drew is left-handed. There's nothing intrinsically wrong with that, but conventionally people work with right-handed systems.

You are correct that a given value of $m_I$ and $I$ does correspond to one single value of the angle $\theta$. For the most commonly used nuclei in NMR spectroscopy, the "allowed" values are $I = 1/2$ and $m_I = \pm 1/2$. The two states are commonly labelled $\alpha$ and $\beta$; these are the eigenstates of the angular momentum projection operator $\hat{I}_z$, such that
$$\begin{align}
\hat{I}_z|\alpha\rangle &= \frac{1}{2}|\alpha\rangle \\
\hat{I}_z|\beta\rangle &= -\frac{1}{2}|\beta\rangle
\end{align}$$
(treating $\hbar \equiv 1$). The problem is that you are assuming that the wavefunction (of a spin) can only be in an eigenstate. This is not true. It is perfectly valid, and indeed necessary (see below), to have states that are superpositions of eigenstates. In general, a spin can have any state
$$|\psi\rangle = c_\alpha |\alpha\rangle + c_\beta |\beta\rangle$$
where $|c_\alpha|^2 + |c_\beta|^2 = 1$ (normalisation), and in general the coefficients are time-dependent (see below). This is no longer an eigenstate of $\hat{I}_z$, and consequently is no longer well-described by one value of $m_I$, or for that matter, $\theta$. But that is not intrinsically a problem. The only circumstance in which a wavefunction must be described by an eigenstate is immediately following an observation, where the wavefunction $|\psi\rangle$ collapses to the eigenstate $|i\rangle$ with probability $|c_i|^2$. Otherwise, superposition states are entirely consistent with quantum mechanics.

A more generalised example is to think of the time-independent Schrödinger equation:
$$\hat{H}|\psi\rangle = E|\psi\rangle$$
which is an eigenvalue equation for the Hamiltonian, $\hat{H}$. It is true that only eigenstates of the Hamiltonian are admissible solutions to this equation. But that does not mean that the only physically permissible states for the system are eigenstates of the Hamiltonian. The time-independent SE only governs systems that are not evolving in time, or stationary states. We need something else to describe systems that do evolve with time, which is the time-dependent Schrödinger equation:
$$\hat{H}|\Psi\rangle = i\hbar\frac{\partial |\Psi\rangle}{\partial t}$$
The most general solution to the TDSE is, indeed, a superposition of stationary states (the Internet has lots of more thorough explanations, or read Griffiths' Introduction to Quantum Mechanics):
$$|\Psi\rangle = \sum_i c_i \exp\left(-\frac{iE_it}{\hbar}\right) |\psi_i\rangle $$
where the $c_i$'s are time-independent. Indeed, it is necessary for us to have superposition states, because this allows the coefficients of each eigenstate to evolve in a time-dependent fashion (because of the exponential term, which evolves at different frequencies for different eigenstates). If a system is prepared in an eigenstate of the Hamiltonian, then it can never evolve into another eigenstate (the exponential evolves, but if there is only one eigenstate, it is no more than a simple phase factor).
This might not seem immediately relevant for NMR, but recall that the (free) Hamiltonian in NMR is simply $-\gamma B_0 \hat{I}_z$. So, the eigenstates of $\hat{I}_z$, $|\alpha\rangle$ and $|\beta\rangle$, are also the eigenstates of the Hamiltonian. If we never had superposition eigenstates, then the spins would simply sit there forever and do nothing.*

* As long as the Hamiltonian doesn't change. But if you change the Hamiltonian, say from $\hat{H}_0$ to $\hat{H}_1$, then the eigenstates change, and so your system (which was originally an eigenstate of $\hat{H}_0$ is no longer an eigenstate of $\hat{H}_1$). Indeed, it would be in a superposition of eigenstates of $\hat{H}_1$, and would evolve in time according to the TDSE, so it doesn't invalidate any of the preceding discussion.

I think your answer (and mine) might benefit from inclusion of a description of how the observed angle $\theta$ depends on the quantum state (or coherence).

@BuckThorn, I considered going into it but felt too lazy...! Imo it would require describing density matrices, Pauli matrices and the Bloch sphere (I may be overcomplicating it, of course). NMR theory uses these tools regularly to describe bulk magnetisation, less so for a single spin. But quantum computing uses Bloch spheres for single spins. Maybe if I have time...

@orthocresol I appreciate the technical correctness of your response, but I must admit that I am not particularly au fait with QM. From what I understand of your answer, the angle theta is constant for an angular momentum vector of a spin with a certain magnetic quantum number only when it's observed. When not being observed it can take any value of theta. This is essentially the Copenhagen interpretation. Is this correct? The bulk magnetization vector can take many values of theta when observed as it's an average of all the thetas of each individual spin. Is this also correct?

@ETS, that’s a pretty accurate summary.

The magnetization is composed of the sum of dipole moments associated with individual spins:
$$\vec{M}=\sum_i \vec{\mu}_i$$
and the component along the magnetizing field (assumed to lie along the z axis) is
$$\begin{align}M_z&= \left( \sum_i \vec{\mu}_i\right)_z \\ &=  \sum_i \mu_{iz}\end{align}$$
Since the individual dipole moment components along z take on discrete values (are quantized, ie for the homonuclear case $\vec{\mu}_{iz}=\gamma\hbar m_z $) we can write
$$M_z =  \gamma\hbar\sum_i m_{i} \tag{1}$$
where the symbols have their usual meanings. Since the sum extends over a large number of spins $i$, the resulting magnetization can for practical purposes be regarded as taking on a continuous range of values, even if the z-component of individual spins is discretely valued.

You are correct, except that individual spins need not always take quantised values of $m_z$. The bulk magnetisation is a valuable point to bring up though, especially in the context of NMR.

@orthocresol in the presence of an external field in an NMR experiment the z component of each spin is assumed quantized, as far as I am aware.

I've gone into more detail about that in my answer; it's not true that individual spins must be either up or down. This is something that is always taught at the introductory level (probably to make things easier), but is inconsistent with QM, where superposition is a key concept.

@orthocresol that's a good point but it brings to mind the confusing distinction between superpositions and coherences. In general measurement is associated with quantization along a measurement axis. Note that since I limit discussion to $M_z$ afaik equation 1 is general. Also, I am trying to explain how an observable macroscopic property could have a continuum of values while made up of discrete entities, but thanks for bringing this to my attention, I will have to reread my copy of Cavanagh.

"how an observable macroscopic property could have a continuum of values" I understand, and that point is entirely valid (hence why I said it's valuable!). It's just that OP is likely confused over the single-spin case. Modern NMR textbooks, like that of Keeler (and also Levitt's Spin Dynamics), point out that superpositions are allowed and that individual spins only being in up / down states is mostly a myth. I'm not sure if Cavanagh mentions it (I've only perused small sections of the book), but I wouldn't be surprised if he does, too.

Also re. "measurement is associated with quantization along a measurement axis": you are correct about that too. But I don't think just placing the spins in a magnetic field constitutes a measurement; a measurement would be something more like a Stern–Gerlach apparatus, where you probe each individual spin to see whether it's up or down (or to be more precise, whether it collapses into up or down).

@orthocresol In section 2.1.4 of the 1st edition they go into the time dependence of the expectation value of the component along a particular axis of the dipole moment  of a single spin, approaching the problem as you do, and an angle $\theta$ can be defined on the basis of this (but is not defined in that section). Immediately after that they jump into the density matrix treatment. As I mentioned, I need to reread, but discussion in terms of the evolution of single spins (while important) is not sufficient to understand magnetization.

Re measurement: I think NMR qualifies! :-)

In NMR we only detect bulk transverse magnetisation; this doesn't collapse the wavefunctions of the individual spins, it only collapses the total wavefunction to one which agrees with the measurement. Sorry, but even though I fully agree with you that bulk magnetisation is the physically important quantity, I cannot upvote now because I don't agree with the treatment of quantisation. I do very much appreciate your good-natured replies, though.

@orthocresol That's ok. The form collapse takes always depends on what you measure. I don't see why NMR should be any different.


text:
How to perform Mulliken population analysis by hand with GAMESS output?
I did a simple calculation on hydrogen fluoride with GAMESS and I wanted to reproduce the Mulliken population analysis for atomic orbitals.
A molecular orbital is a linear combination of atomic orbitals and if I understand correctly the Mulliken analysis uses coefficients to calculate population (overlap or atomic).
In the molecular orbital part of my GAMESS output, if I want to calculate the population of the 1s orbital of the hydrogen, I'm summing every coefficients squared of the contribution of this AO in all the MO, but I'm not getting the right answer.
Would someone be able to clarify this?
INPUT:
 $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE COORD=CART ICHARG=0 MULT=1 $END
 $SYSTEM TIMLIM=90  MEMORY=1000000 $END
 $STATPT OPTTOL=1.0E-4  NSTEP=100 $END
 $BASIS  GBASIS=STO NGAUSS=3 $END
 $SCF DIRSCF=.TRUE. $END
 $GUESS  GUESS=HUCKEL  $END
 $DATA
 HF
C1
H     1.0     0.00000     0.00000     0.00000
F     9.0     0.00000     0.00000     1.0000
 $END

OUTPUT:
      ------------------
      MOLECULAR ORBITALS
      ------------------

                  1          2          3          4          5
              -25.9035    -1.4599    -0.5737    -0.4631    -0.4631
                 A          A          A          A          A
1  H  1  S    0.005341   0.150431  -0.533704   0.000000   0.000000
2  F  2  S   -0.994746  -0.250678  -0.078268   0.000000   0.000000
3  F  2  S   -0.022260   0.946703   0.410905   0.000000   0.000000
4  F  2  X    0.000000   0.000000   0.000000  -0.004070   0.999992
5  F  2  Y    0.000000   0.000000   0.000000  -0.999992  -0.004070
6  F  2  Z    0.002673  -0.078256   0.698060   0.000000   0.000000

                  6
                0.5898
                 A
1  H  1  S    1.054350
2  F  2  S    0.080571
3  F  2  S   -0.515861
4  F  2  X    0.000000
5  F  2  Y    0.000000
6  F  2  Z    0.816436

           ----- POPULATIONS IN EACH AO -----
                         MULLIKEN      LOWDIN
          1  H  1  S      0.80774     0.85856
          2  F  2  S      1.99913     1.99848
          3  F  2  S      1.94822     1.88452
          4  F  2  X      2.00000     2.00000
          5  F  2  Y      2.00000     2.00000
          6  F  2  Z      1.24491     1.25844

Please add the input of your example calculation, as well as the relevant portions of your output, and your attempt in calculating the Mulliken population.

@Martin-マーチン Thank you, I edited my answer and changed the molecule to HF, as it is more interresting with a polar molecule

You need the density matrix and the overlap matrix to do a Mulliken population analysis.

Using ORCA and reverse engineering the calculation provided in the question (HF/STO-3G at a $\ce{H}$-$\ce{F}$ distance of 0.955464 Angstroem), I obtained the following output, to which I added the matrix product of the density and overlap matrices, dubbed the Mulliken Population Matrix. By comparing to the output of other programs, the reader can follow the analysis.
------------------
MOLECULAR ORBITALS
------------------
                  0         1         2         3         4         5
             -25.90350  -1.45985  -0.57365  -0.46312  -0.46312   0.58982
               2.00000   2.00000   2.00000   2.00000   2.00000   0.00000
              --------  --------  --------  --------  --------  --------
0H   1s        -0.005340 -0.150428  0.533708 -0.000000  0.000000 -1.054338
1F   1s         0.994746  0.250679  0.078267 -0.000000  0.000000 -0.080569
1F   2s         0.022260 -0.946706 -0.410898  0.000000 -0.000000  0.515847
1F   1pz       -0.002673  0.078253 -0.698059  0.000000 -0.000000 -0.816434
1F   1px        0.000000 -0.000000 -0.000000 -0.998068 -0.062124 -0.000000
1F   1py       -0.000000  0.000000 -0.000000 -0.062124  0.998068  0.000000

-------
DENSITY
-------
              0          1          2          3          4          5
  0       0.615003  -0.002499  -0.154014  -0.768633  -0.000000   0.000000
  1      -0.002499   2.116969  -0.494672  -0.075355   0.000000  -0.000000
  2      -0.154014  -0.494672   2.131170   0.425379   0.000000   0.000000
  3      -0.768633  -0.075355   0.425379   0.986834  -0.000000   0.000000
  4      -0.000000   0.000000   0.000000  -0.000000   2.000000   0.000000
  5       0.000000  -0.000000   0.000000   0.000000   0.000000   2.000000

--------------
OVERLAP MATRIX
--------------
              0          1          2          3          4          5
  0       1.000000   0.045145   0.423474  -0.335755   0.000000   0.000000
  1       0.045145   1.000000   0.237990  -0.000000   0.000000   0.000000
  2       0.423474   0.237990   1.000000   0.000000   0.000000   0.000000
  3      -0.335755  -0.000000   0.000000   1.000000   0.000000   0.000000
  4       0.000000   0.000000   0.000000   0.000000   1.000000   0.000000
  5       0.000000   0.000000   0.000000   0.000000   0.000000   1.000000

--------------------------
MULLIKEN POPULATION MATRIX
--------------------------

 0,807741631  -0,0113884814   0,1058290434   -0,9751233323  0  0
-0,091108347   1,9991291934   0,0080871908   -0,0745159483  0  0
 0,583325991   0,0055721863   1,9482220861    0,4770899706  0  0
-0,923232404  -0,0088189886   0,0819491725    1,2449063729  0  0
 0             0              0               0             2  0
 0             0              0               0             0  2

-----------------------
MULLIKEN ATOMIC CHARGES
-----------------------
0 H :    0.192259
1 F :   -0.192259
Sum of atomic charges:    0.0000000

Thank you very much, your answer helped me a lot !

I would be very cautious with the term "reverse engineering".


text:
Why are drugs mixed with their sodium salts?
I have come across quite a few drugs that are of the form $\ce{HA + NaA}$, where $\ce{A}$ is your target organic compound. A few examples of this are:

Carmicide which is a mixture of Sodium Citrate and Citric Acid
Fucidin which is a mixture of Sodium Fusidate and Fusidic Acid

What advantages are there in conjugating a drug with its sodium salt?
A similar question says that this increases solubility and also makes the drug easier to crystallize and purify. I'm assuming that the same reasoning does not work for organic acids, considering they will dissolve/ionize quite easily. Please clarify and provide further insight into this topic.

I'm not sure if this is related, but the vinegar flavouring in foods is often added not as acetic acid or sodium acetate, but rather sodium diacetate, essentially a 1:1 molar ratio mixture of the two.

Forming a oH buffering pair, stabilizing conditions at which the drug behaves by predicted way, in context if it's effect and stability. If digested, pure acid or base may not be welcome.

A large percentage of pharmaceuticals are sold as salts, and sometimes even as co-crystals, i.e. a drug cation and a drug anion. You are right in the sense that the basic purpose is to make them soluble but this is not the only one. If it is a intravenous drug, salt form may have less pain. A low molecular weight medicine might be in a liquid form yet its salt is a solid. Sometimes you want to make a hydrophobic salt so that its stability is higher in high humidity. Sometimes you wish to improve the taste. There are plenty of factors.
Most likely the reason for a mixture of medicine and its sodium salt is its controlled release and some buffering action. In case of your carmicide which is a mixture of sodium citrate and citric acid, is a buffering system and secondly citric acid alone is very corrosive and very bad tasting (tart). Fucidin is only allowed as a Na-salt.
This review will tell you the principles behind choosing a salt.
Go to Google Scholar and read:
Gupta et al. Salts of Therapeutic Agents: Chemical, Physicochemical, and Biological Considerations, Molecules 2018, 23, 1719; doi:10.3390/molecules23071719

The choice of a particular salt formulation is based on numerous
  factors such as API chemistry, intended dosage form, pharmacokinetics,
  and pharmacodynamics. The appropriate salt can improve the overall
  therapeutic and pharmaceutical effects of an API. However, the
  incorrect salt form can have the opposite effect, and can be quite
  detrimental for overall drug development.

API= active pharmaceutical ingredient

Good answer, just had one question: Do buffering actions for salts work in vivo given that the body has it's own way of controlling pH? eg. blood has a pH of 7.3 to 7.45: intravenous drugs conjugated with their salts might form a buffer at a different pH.

I don't know what goes on in vivo. Even it is buffered, it should not change the local pH at the site of injection. That is why their pH must be kept close to the blood pH. Many tablets taken orally are not released in the stomach but rather in the intestine. I don't know much about the pharmacokinetics but just for reference a new drug can take up to 5-10 years before it can go from lab to humans (with millions spent in research). They study each an every aspect of it. A trained pharmacist can tell you more details. Chemists don't study these topics in general.

In addition to M. Farooq's answer, there is also the question of the chemical's stability and our body's metabolic route for it. For example, most tryptamines you will find in the fumarate salt form, and this is because most other salts and the free base forms of these chemicals deteriorate rapidly.
On the metabolic side of this answer, take Tianeptine as an example. Simply changing the salt form from Tianeptine sulfate to Tianeptine sodium  is effectively changing the drug from an instant release to an extended release form. This is because our body metabolizes the two salts of the chemical differently, and so one requires a lower dose, has a faster onset and shorter duration, and the other requires a higher dose but takes longer to onset and lasts much longer. Hope this provides some insight to your question.


text:
What repulsion keeps non-hydrogen atoms at a distance: between inner shells or between nuclei?
Bonded or not, when two atoms or ions come closer than the bond distance or the van der Waals distance, the energy increases drastically. What is the underlying cause of that repulsion?

The repulsion of the negatively-charged inner-shell electrons
The repulsion of the positively-charged nuclei (or the effective positive charge of nuclei plus inner-shell electrons)
Any classical explanation solely based on electrostatics falls short, and the increase in energy is related to the Pauli exclusion principle.
There is no single underlying cause, it depends on the specific case. Sometimes repulsion of positive charges is the problem, sometimes repulsion of negative charges.

Option 3.

... enter at least 15 characters? @Mithoron

Bypassing of this become like a way to show off quite some time ago :D

I found a nice figure and the relevant statement in a paper by Frenking and Krapp (Unicorns in the world of chemical bonding models, 2006, https://doi.org/10.1002/jcc.20543):

The crucial term which is responsible for repulsive interactions in chemical bonds except in two‐electron systems such as H2 is the Pauli repulsion. The three terms (a) quasiclassical electrostatic interaction, (b) resonance, and (c) Pauli repulsion give a complete picture of the interatomic interactions which yield a chemical bond.

The figure shows how the distance-dependent repulsion for one scenario including the Pauli repulsion and two scenarios ignoring the Pauli repulsion; the latter do not show the drastic increase in repulsion that is observed.

Source: https://onlinelibrary.wiley.com/cms/attachment/2b2dc015-ca3c-4fd3-b51c-f768a6a924b5/mfig003.jpg
The entire essay is great, connecting the picture of bonding that bench chemists have with the developments of theory and computation in the 90 years since quantum mechanics was first applied to chemistry.

The Hamiltonian for the collection of charged particles (electrons and nuclei) comprising a pair of molecules or ions has terms describing the kinetic energy of individual particles and pair-potential terms describing Coulombic interactions, that is, all interactions between particles are Coulombic. Where a  Pauli repulsion term can crop up is when evaluating the expectation value of the Hamiltonian and factoring the total energy into piece-wise contributions. A  Pauli repulsion term appears because the wavefunction is antisymmetric, encoding how electrons avoid each other due to their fermionic behavior, but this does not invalidate the fundamentally Coulombic nature of the interaction, and all repulsion can be attributed to Coulombic repulsion. Nucleus-nucleus Coulombic interactions are always repulsive, as are electron-electron ones.
As emphasized in the OPs answer, presumably the question is not just about the origin of the repulsion, but also about the functional form of the potential. I would argue that this is a very broad question, since it depends on the specific species that are interacting, and on the value of r. One can agree however that at energies typically relevant to chemists the repulsion at short r is dominated by electron repulsion and specifically by a distribution determined by the fermionic nature of the electrons (Pauli exclusion) which forbids electrons from having the same wavefunction. This is sometimes roughly described by the orthogonality condition captured by the overlap term
$$S=\int\phi_a\phi_bd\tau$$
This term is often written as an exponentially decaying function.

My point is that the reference for comparison is two uncharged fermions. What does their interaction potential look like? eg https://en.wikipedia.org/wiki/Electron_neutrino


text:
Validity of the leveling effect of water in nitration reactions
Wikipedia gives this as the levelling effect of water

Any acid that is stronger than $\ce{H3O+}$ reacts with $\ce{H2O}$ to form $\ce{H3O+}$. Therefore, no acid stronger than $\ce{H3O+}$ exists in $\ce{H2O}$. For example, aqueous perchloric acid ($\ce{HClO4}$), aqueous hydrochloric acid ($\ce{HCl}$) and aqueous nitric acid ($\ce{HNO3}$) are all completely ionized and are all equally strong acids.

So from what I can understand, if I take an aqueous medium, then $\ce{H2SO4}$ and $\ce{HNO3}$ both should be equally strong.
But then comes the nitration reaction, where the nitronium cation acts as the real attacking electrophile, which means that in this reaction, $\ce{HNO3}$ acted as a base and $\ce{H2SO4}$ acted as an acid.

I am also aware concentrated acids are taken in the nitration reactions, but even the most concentrated acids have some amount of water.

How are these two concepts contradicting, where am I going wrong?
What are the limitations of the levelling effect and where all is it applicable? Is it not applicable to concentrated solutions, and if so, how concentrated?

Hi, AFAIK for comparing acidic strengths of strong acids we use other solvents (rather that water, for the same reason as you mentioned) like acetic acid. Therefore, since nitric acid is weaker than sulfuric acid it should get protonated in nitrating mixture. There is a full chance, that I didn't get the question...

Yes I wholly agree that Nitric acid is weaker, but dude in FC we do have water content @Zenix

As you already mentioned that nitrating mixture consists of concentrated sulfuric acid, just a thought, won't it absorb water, if it's still left? Since it's a powerful desiccant...

Nah Nah doesnt work that way, it forms a constant boiling azeotrope and it cant be purified further (anyone correct me here if i am wrong)

Oh! I knew that HNO3 forms a constant boiling azeotrope, do H2SO4 also form constant boiling azeotrope?

Yes is what i learnt in high school

I wouldn't get too hung up on the "levelling effect" - once you start getting familiar with the different types of acidity and basicity you really do not consider "levelling" effect anymore. It is just that when studying the chemical reactions themselves, the acidity and basicity is defined from what reactions happen, not the other way around. And solvent choice is highly relevant for which way you want the reactions to go.

The stable "concentrated sulfuric acid" is about 98% H2SO4, so the water activity is essentially 0, and there are many more molecules of acid than there are of H2O. Thus, the behavior of aqueous solutions is irrelevant. NItric acid has higher water content, but typically only a small volume of it is added to the sulfuric to make the mixed acid.

@StianYttervik could you explain why?

@Andrew I'd raise the same question to you as I raised to Mathew's answer

What you have gone wrong in this question is you completely disregard the fact that mentioned stronger acids than water are all in aqueous medium, in other words, dissolved in water. Thus, fast acid-base reaction ($\ce{H2O}$ acts as a base here) happens to give $\ce{H3O+}$ as the only acid in these mixtures (leveling effect of water).
On the other hand, concentration of sulfuric acid is $\approx 98\% (w/w)$. That means, approximately only $\pu{2 g}$ of $\ce{H2O}$ in $\pu{100 g}$ of concentrated solution. Therefore, mole ratio of $\ce{H2O : H2SO4}$ would be about $0.1 : 1$. Thus, ($\ce{H2SO4}$ is not leveling with water. That extra remaining $\ce{H2SO4}$ can be acted on $\ce{HNO3}$ in nitrating mixture as shown in your scheme (vide supra).
Note: As a rule of thumb, you may simply consider $1:1$ mole ratio of water and any stronger acid than water would be the limiting level for the leveling effect.

Okay, but what we have learnt in ionic equilibrium as that the strongest acid always reacts with the strongest base. So, $\ce{H2O}$ is definitely much much stronger a base than nitric acid.

@HahaHahaha - of course the $\ce{H2O}$ reacts with the sulfuric acid, but there is so little of it that even if it were all protonated to $\ce{H3O+}$, there would still be plenty of protons from $\ce{H2SO4}$ available to react with the nitric acid. You need to think of $\ce{H2SO4}$ as the solvent. Every molecule of $\ce{H2O}$ or of $\ce{HNO3}$ is surrounded by many molecules of $\ce{H2SO4}$.

Okay great that makes sense thanks both of y'all, I am accepting this answer

I didn't see OP's question until now (was in my sleep). Thanks Andrew for answering. :-)


text:
Specific reduction of certain functional groups

Consider the compound $\ce{Alkyl-(4 oxocyclopent-2-ene)carboxylate (A)}$. For example consider-

  Now consider the operations:
$\ce{1. (A) + (i)BH3/THF (ii) H+}$
$\ce{2. (A) + H2/Pt}$
$\ce{3. (A) + H2/Pd-C}$
Products
In $1.$ ester group reduced to an alcohol.
In $2.$ double bond reduced to a single bond, and
In $3.$ double bond reduced to a single bond, ester group reduced to an alcohol.

However, on the next page addition of $\ce{(i)BH3/THF (ii) H+}$ selectively reduced carboxylic acid when both ester and carboxylic acid were present.
My questions:
$(i)$ I know that $\ce{(i)BH3/THF (ii) H+}$ can reduce carboxylic acid, but I am unaware if it can reduce ester.
$(ii)$ I only knew that, the reduction of ester in presence of carbonyl group could be done by first protecting the carbonyl group (converting it to cyclic ketal by reacting with ethylene glycol) and then using $\ce{LiAlH4}$. Is reaction $3.$ correct?
$(iii)$ Why aren't product $2.$ and $3.$ same? I've always seen the two work same.
Note:
My chemistry teacher gave me few notes for Chemistry Olympiad preparation, so I can not cite source. I don't know how to make compounds, so will completely reside on IUPAC nomenclature.

Man reduction is highly specific and its trends are difficult. However, I am sure there should be a reason to this. All I can add is that Pd-C is stronger than Pt as an adsorbent. Nice question!

I'm not sure that differences in Pt vs Pd hydrogenation can be easily explained. I'm also not entirely convinced that H2 with Pd/C reduces esters. Maybe in a few isolated cases, or maybe with higher pressures, but I doubt it's anywhere near as facile as hydrogenation of C=C bonds. As far as I know, this doesn't occur sufficiently regularly to use it in a predictive sense. Also, your IUPAC name is slightly ambiguous: is this a CO2R group sticking off carbon-1 of your cyclopentene?

Yes, it's CO2R group. What's the correct IUPAC name?

@Zenix: from which book is this question? I feel that these questions are interesting!

For (i), it is correct that borane/THF will not reduce an ester. One explanation is here: https://www.ch.imperial.ac.uk/rzepa/blog/

Full link: https://www.ch.imperial.ac.uk/rzepa/blog/?p=5114

1.The selective reduction of the carboxylic group is explained in this article. Link:https://www.ch.imperial.ac.uk/rzepa/blog/?p=5114. It could reduce esters but is not favourable at all and difficult.So if u only have an ester grp and its treated with bh3, then you can reduce it. However if both carboxylic and ester grps are present, carboxylic acid gets reduced .

Regarding reduction with h2/pt and h2/pd-c not much can be said. The thing is at higher temperatures and pressures(extreme conditions)both could reduce esters. Generally esters are NOT reduced via cataytic hydrogenation (it is just not favourable).However reduction has been performed in mild conditions via other complexes like ruthenium.(https://pubs.acs.org/doi/pdfplus/10.1021/op4003278


text:
The second Ionisation energy of the Zn, Cd, and Hg follows the order?
It is observed that the order of second ionisation energy of these three elements is $\ce{Zn > Cd < Hg}$.
Why is there an anomaly in the observation?

Do you know which of the three is anomalous?

@orthocresol yes, the ionization energy should decrease down the group: Zn>Cd>Hg. The behavior of Hg is anomalous.

Excellent. The answer has to do with relativistic stabilisation of the 6s orbital. Probably you will find more material on Stack Exchange, or on the wider Internet, or in a textbook.

@orthocresol It's not in my high school textbook, I'm not familiar with the relativistic stabilisation. I looked it up and found that 6s electrons experience an inward force due to the increment in mass of $e^-$ for moving at a higher speed. Is it correct?

Some authors also claim that the electrons in the f-orbitals do not shield the nuclear charge as well as electrons in other orbitals.

In the lower part of the periodic table, the nucleus charge becomes so high that the electron must rotate at a relativist speed to stay in orbit in the Bohr's model. And even if the Bohr's model is outdated, it can be used to explain the strange chemistry happening in the heaviest atoms.
Relativity says that when an object moves at a speed not far from c, its dimension decreases. That is the case for electrons in the heaviest atoms. And this effect is proportional to the sum of the quantum numbers n and l. Mercury's configuration for example is $(Xe) 4f^{14} 5d^{10} 6s^2$. The corresponding quantum numbers n and l give successively n+l = 4+4, 5+3, 6+1. So for these electrons, n+l is 8 or 7. This is huge. Their orbital dimensions are so heavily contracted that they disappeared inside the inner orbitals, where n+l is not so high. As a consequence, the mercury atom looks like the xenon atom when observed from outside. And indeed it is "nearly a noble gas", as its boiling temperature is so much lower than its neighbors in the table, like gold (Au) or hallium (Tl). This is also why the ionization potential is so high: mercury behaves nearly like a noble gas.
All this theory is discussed in great detail in Pyykkö Pekka's articles, like
- P. Pekka, Relativist Effects in Structural Chemistry, Chemical Review 88, 3, 563 - 594 (1988)
- P. Pekka, Relativity and the Periodic System of Elements, Accounts of Chemical Research 12. 8, 271 - 276 (1979).
 A short summary can be found in Wikipedia, under the title "Relativist quantum chemistry".
This relativistic effect explains also why mercury's neighbors behave like xenon neighbors. On the right-hand side of $\ce{Hg}$, thallium ($\ce{Tl}$) is a near alkali-metal atom, as it may have +1 as an oxidation number. On its left-hand side. Gold is nearly a halogen, as it may form an anion $\ce{Au-}$ in the cubic crystal of "caesium auride" $\ce{CsAu}$. Further on the right-hand side, lead shows some similarities to barium, assuming a +2 oxidation state and having its sulfate $\ce{PbSO4}$ relatively insoluble in water like $\ce{BaSO4}$.


text:
How and why do concentrated solutions react faster?
It seems intuitive that concentrated solutions would react faster, but, when I stopped and thought about it, I got confused. Wouldn't concentrated acids and bases be in contact with less water than their more diluted counterparts, therefore being less dissociated and having less ions (H+ or OH-) available for reactions? How would completely concetrated acids (since bases would be solid) even react as acids?

Acids and bases do not have to dissociate to react with each other. The proton transfer can occur directly between them, though this is rare in practice, because before you run out of water, you typically begin to lose acid to the gas phase.

What if the acid is not gaseous on its own, but liquid like H2SO4?

Strong acid dissociate, because they are stronger acids then $\ce{H3O+}$, and react with its conjugated base $\ce{H2O}$:
$$\ce{HA + H2O -> A- + H3O+}$$
Additionally, the $\ce{H3O+}$ activity coefficient steeply grows in concentrated acids, and so does the acid reactivity.
In concentrated $\ce{H2SO4}$ and oleum, the role of $\ce{H3O+}$ is overtaken by $\ce{H3SO4+}$:
$$\ce{2 H2SO4 <=> H3SO4+ + HSO4-}$$


text:
Why does radioactive matter decay in series of half-lives?
I was studying types of nuclear decay, and I came across the concept of "half-life". And I started wondering why does nuclear matter decay in half-lives? In other words, why does the rate of nuclear decay decrease as a function of time?
Is it because as more atoms decay, the surface area of the radioactive matter decreases, and the number of atoms "exposed" decrease? Does the outer layer of atoms somehow "shield" the inner atoms and stabilize them, preventing them from decay?
Or is there another mechanism at play?

But the rate stays the same over time (the foundation of radiocarbon dating). After $t_{1/2}$, half of what you started with is gone. At $2t_{1/2}$, you are at half of what you had after $t_{1/2}$ (so a quarter compared to $t = 0$). The rate is the same: 50% of what you have right now is gone after another $t_{1/2}$. Radioactive decay is spontaneous and statistical, there is no shielding etc.

The decay is an intrinsic property of an atom so it does not matter if they are in rock or a vacuum they still decay. The rate at time $t$ is proportional to the amount unreacted $dx/dt$ where $x$ is the number, and changes in time as $dx/dt=-kx$ where $k$ is a number specific to each type of atom called the rate constant or given as $\tau = 1/k$ the decay lifetime.The minus sign means the species decay. The half life is a measure of decay similar to the lifetime, the lifetime measures time  to $1/e$ of the initial amount, the half life to 0.5.

There are fewer decays because there are fewer atoms to decay
The simple reason why the number of decays (strictly, the number of decays per unit time) decreases in simple radioactive decay is because there are fewer atoms left to decay.
Nuclear decay is probabilistic. The probability of any given unstable atom decaying is constant (independent of time or the environment). So the reaction rate (the total number of decays per second) per mole is constant. If this were a chemical reaction we would call it a first-order reaction (since it isn't a chemical phenomenon we can call it a first order process but the mathematics are the same).
In any first order process there is a characteristic time when half the original material will have gone (we call this the half life of the process). So, if we measure time in half-lives at t=1 we will have, say, 1 mole of the material; by t=1 we will have 0.5 moles; at t=2 we will have 0.25 moles of material. If we count the number of emissions per second they will follow the amount of material left. So the radioactivity will decay at the same rate as the material.
We don't need to invoke any interactions between the atoms of the material to explain this.
PS it gets a little more complicated if one radioactive compound decays into another. Then the total radioactivity will depend on two competing decays with different half-lives (but the radioactive emissions may well be different as each has a characteristic energy and expensive instruments can tell them apart giving two separate emission counts). But the mathematics remains the same.

It is a general principle, not limited to nuclear chemistry, but is common for many areas, e.g. for the reaction kinetic of the 1st order.
All processes, where the value time rate is proportional to the value, have value time evolution in the form of the exponential function.
$$\frac {\mathrm{d}x}{\mathrm{d}t}= -k \cdot x$$
leads to
$$x= x_0 \cdot \exp { (-k \cdot t )}$$
Radioactive atomic kernels have constant probability of the decay over the given time interval. The actual decay rate is then for large kernel numbers proportional to this probability and the number of these kernels, described by the above equations in the differential and integral form.
When there is twice as  many radioactive atoms of the same kind there is twice as high decay rate.
The shape of the exponential function shows its value decreases to a half by constant time, no matter what the value is. This constant time is the decay half life.
where
$$ t_{1/2}=\frac {\ln{2}}{k}$$

I upvoted this nice answer! Any physical system, with one energy storage element and one decay pathway, must follow the ordinary differential equation, with initial condition, that you give. So it is ubiquitous in the real world: Newton's law of cooling, the RC low pass filter, radioactive decay, etc.

Decaying are like coin tossers
It helps me to compare two different mental pictures.
The first picture, which is not at all like radioactive decay, is of lighting a fuse. It starts burning at the near end and burns toward the far end at some number of millimeters per second. Each section's chance of burning depends completely on the section before it: the second millimeter of fuse can't burn before the first one, and the third can't burn before the second, etc.
The second picture is a lottery based on coin flips. A billion people each have a coin to toss. Everyone who gets tails sits down, and the rest proceed to the second round of play. The last person standing wins the prize. Each person's chance of getting tails is completely independent of anyone else's toss, and every toss has a 50% chance of tails. You can expect that roughly half the people will sit down in each round. The actual number will be closest to 50% when the number of people is still large.
Atoms of radioactive isotopes are like the coin tossers. Each atom will decay at some random time, completely unaffected by what the others are doing. Of course, many rounds of "the radioactive decay lottery" are played per second, so instead of a coin toss you might imagine the atoms rolling a die with very many sides and sitting down if they get a "1". The more stable atoms have die with more sides, but each atom's moment of decay is still unpredictable.
Another way of stating a half life might be this: "given how unstable this isotope is, how long would you have to watch a given atom to have a 50% chance of seeing it decay?"
Just a repeated percentage change
I think a half-life pattern is just a convenient way of stating a recurring percentage decrease.
For example, what happens if you start with 1 million atoms and each year you lose 0.1% of whatever you have left?
I wrote a little computer program to simulate that. I had it run 10,000 rounds and print a message each time it reached a halving - that is, when the remaining amount reached 500,000 or less, then when it reached 250,000 or less, etc. Here's what I saw:
reached 499900.2346477281  at round 693  - 693 rounds elapsed
reached 249900.24460085342 at round 1386 - 693 rounds elapsed
reached 124925.1909144911  at round 2079 - 693 rounds elapsed
reached 62450.132251566276 at round 2772 - 693 rounds elapsed
reached 31218.83576633967  at round 3465 - 693 rounds elapsed
reached 15606.30332502206  at round 4158 - 693 rounds elapsed
reached 7801.594694162155  at round 4851 - 693 rounds elapsed
reached 3900.019018238128  at round 5544 - 693 rounds elapsed
reached 1949.6204223478458 at round 6237 - 693 rounds elapsed
reached 974.6157066056886  at round 6930 - 693 rounds elapsed
reached 487.2106204235443  at round 7623 - 693 rounds elapsed
reached 243.55670347259502 at round 8316 - 693 rounds elapsed
reached 121.75405321597741 at round 9009 - 693 rounds elapsed
reached 60.864879771978956 at round 9702 - 693 rounds elapsed
...

So every 693 rounds, the amount halves. This remains consistent after 10,000 rounds and after 1,000,000 rounds. (Eventually it breaks down, probably because the program has limited mathematical precision.)
If I have it lose 2% each round instead of 0.1%, the half-life is 35 rounds.
Based on this, I think you could state any half life in terms of percentage loss over a period of time. But "percentage loss per year" would hardly make sense for isotopes whose half life is a fraction of a second, and "percentage loss per second" would be incredibly small for isotopes with a half life in the billions of years. Speaking in terms of half lives gives us a single number to compare between those two isotopes.


text:
Comparison of I-F, Br-F, and Cl-F bond strengths
From Chemistry of the Elements by Greenwood and Earnshaw [1, p. 833]:

Figure 17.9 Mean bond energies of halogen fluorides.

This shows the mean bond energy for $\ce{X-F}$, where $\ce{X}$ is a halogen. I have several questions:

Why is the $\ce{X-F}$ bond in $\ce{XF}$ stronger than that in $\ce{XF3}$ or $\ce{XF5}$? I think it is because the $\ce{XF}$ bond has bond order 1, but $\ce{XF3}$ and $\ce{XF4}$ both have 3c-4e bonds. However, I am not too sure about this explanation. It also doesn't really explain why $\ce{XF5}$ is weaker than $\ce{XF3}$.
Why is the order of bond enthalpy generally $\ce{IF} > \ce{BrF} > \ce{ClF}$? I thought it would be the reverse, because smaller atoms would be able to have better atomic orbital overlap.

References

Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.

When appending an illustration from a third-party source, make sure it's readable and the source is properly cited (see my edit; also feel free to use it as an example for your future posts).

For me this graph is very logical and provides insight into actual properties of these compounds. For example it's generally more difficult to get higher and higher oxidation states of halogens.


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Kinetic Theory of Gases and molecular velocities of gases
The assumption that molecules collide with elasticity was postulated in the kinetic theory of gases.
So my question is molecules should have higher tendency to collide obliquely and would therefore result in change of velocity (as the vector components would cancel or add up the velocities of molecules, resulting in change of velocities of molecules) and energy states of electron, are the assumptions not very correct or am I wrong?

Causing an excitation (according to Bohr's theory at least) requires that the colliding atom have a minimum energy of ~10.2 eV. Most colliding atoms at and around room temperature have less energy.

So, the assumption is not very strong?

But, are the collisions perfectly elastic?

Try to ask the above question as it is written eyes to eyes and watch, if the other person understands the question. As I bet the person will forget the beginning until you finish. As a guidance for scientific texts, it is adviced sencentes are 20-25 words long in average, with strong recommendation not to write ones longer than 35 words. Your single sentence has 64 words.

Could you clarify why you think your scenario shows that the process is not elastic? Elastic does not mean individual velocity can't change, it means there is no net loss in kinetic energy across all atoms.

Because in my opinion oblique collisions should exist which are not the same as elastic collisions.

Inelastic collisions are such collisions, that dissipate mechanical energy to other energy forms. It may be e.g. acoustic energy, light, thermal energy ( in macro context only ) or internal energy of an object.
The change of the internal energy in molecular context can be creation/breaking of chemical bonds, or changes of electron energy states. In usual conditions of stable enough gases, these events are very negligible, as average kinetic energy is much smaller than the energy needed to such a change.
Be aware that even fully elastic collision generally leads to a change of the mechanical energy of an object, in whole, or its translational, vibrational and rotational parts.
Imagine 2 identical balls moving at the same speed on a colliding course. After the collision, they have still identical speed if and only if their collision was symmetric. Otherwise 1 ball will slow down and the other ball will speed up.
The justification of the kinetic gas theory is the best shown on very good agreement of prediction and observed behavior of permanent gasses at normal conditions.

The postulate that a particular number of molecules posses a particular magnitude of velocities but oblique collisions should not result in such, if you could explain this keeping the above question in mind that will be great!

It is not very clear what you complain against, or disagree with. Even if by a magic wand all molecules got the same speed, in time shorter than a microsecond speed would get redistributed according Maxwell-Boltzmann distribution.

Speeds are constant, collisions are elastic. How can collisions and speeds be elastic and constant respectively? There is a higher probability of oblique collisions which should not result in the constant speed law.

Speeds are NOT constant. The speed of every gas molecule typically change 10 billion times per second, at each collision. What is constant is the speed distribution over the time at equilibrium.

I meant that only.

I do not get it.

How can you say that speed distribution is constant and collisions are elastic?

I believe maximum collisions should be oblique which should not result in a constant speed distribution.

Easily. Where do you see contradiction ? It is implied the realtime distribution fluctuates around the predicted  distribution. But with the molecule count, the values of the fluctuations are fully negligible.

Be aware of possible mixing (Constant speed) distribution and Constant (speed distribution). I speak about the latter.

Let us continue this discussion in chat.


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Is there a name, or any specific jargon for a precipitation equilibrium involving a solid reactant?
Let's say you have some ions in solution, for example Na+(aq) and OH-(aq), and some partially dissolved solid, MgCO3(s), Mg+2(aq) and CO3-2(aq), which precipitates Mg(OH)2(s) until equilibrium.
So you have two solubility equilibriums interacting with each other:
$$\ce{ MgCO_3(s) <=> Mg^{+2}(aq) + CO_3^{-2}(aq) }$$
and
$$\ce{ Mg^{+2}(aq) + 2OH^-(aq) <=> Mg(OH)_2(s) }$$
overall (with all spectator ions removed):
$$\ce{ MgCO_3(s) + 2OH^-(aq) <=> Mg(OH)_2(s) + CO_3^{-2}(aq) }$$
Is there a name for this type of reaction, and what method would I use to calculate the equilibrium point, with what physical properties?
I would call it something like a "binary solubility equilibrium". And I guess if I know the starting OH concentration and the solubility constants of the two solids, I could work out the equilibrium?
Any clues and key words appreciated. Thanks.

AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $\ce{MgCO3}$ is being converted into $\ce{Mg(OH)2}$.
To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $k_\text{sp}$ of both $\ce{MgCO3}$ and $\ce{Mg(OH)2}$, as well as the $\ce{pH}$ of the solution (which gives us $[\ce{OH^-}]$).
The solubility products of $\ce{MgCO3}$ and $\ce{Mg(OH)2}$ are $10^{-7.8}$, which is approximately $1.58 \cdot 10^{-8}$, and $5.61\cdot 10^{-12}$, respectively.
To calculate how much magnesium hydroxide precipitates, we would first need to find the concentrations of $[\ce{Mg^{2+}}]$ and $[\ce{CO3^{2-}}]$ using the $k_\text{sp}$ of magnesium carbonate. This gives us the solubility of $\ce{MgCO3}$, which is $1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$. That means that we have $[\ce{Mg^{2+}}] = 1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$ in the beaker.
Now, to find out how much $\ce{Mg(OH)2}$ will precipitate, we need to find the reaction quotient $Q_c$ of the reaction.
$$Q_c = [\ce{Mg^{2+}}][\ce{OH^-}]^2$$
(I cannot calculate $Q_c$ here because it depends on $[\ce{OH^-}]$, which will have to be given.)
If $Q_c > k_\text{sp}$, then the reaction
$$\ce{Mg(OH)_2(s) <=> Mg^{+2}(aq) + 2OH^-(aq)}$$
will proceed in the backward direction and magnesium hydroxide will precipitate out. This will continue until the concentration of magnesium ions decreases to such a value that $Q_c = k_\text{sp}$, at which point the reaction will be at equilibrium. Here, I've assumed that the concentration is too large to be appreciably changed by it precipitating out of the solution. The exact value of $[\ce{Mg^{2+}}]$ as well as how many moles of $\ce{Mg(OH)2}$ will precipitate can be calculated with this assumption. (However, if we want, we can do it without this assumption too. The calculations would become more messy and involve solving a quadratic.)


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What is the chemical equation, along with the change in heat of the crystallization process of Sodium Acetate
This is the reaction found in chemical hand warmers, and I can't figure out the equation and the change in heat. Any help would be greatly appreciated

https://en.wikipedia.org/wiki/Sodium_acetate#Heating_pad

There is no chemical reaction.
The heat is released as the latent heat of crystallisation of a supercooled solution of sodium acetate trihydrate; you can obtain similar effects with sodium sulfate decahydrate or sodium thiosulfate pentahydrate.
When heated up, the hydrate dissolves in its own crystal water, which absorbs thermal energy similar to melting; when cooled down, the solution becomes supercooled.
Their state is called the metastable state: this means that it is not thermodynamically stable, but conditionally stable for kinetic reasons, as the crystallization requires a trigger.
The triggering can be done by, for example: insertion of a small crystal of the same salt, bending a steel chess (used in pocket heaters) or vigorous shaking.

Thank you for the clarification! I now understand what I need to do. Thank you!

If satisfied, you may want to mark the answer as accepted, what also marks the question as answered.


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Understanding sodium hydroxide drain cleaner's exothermic dissolution of aluminum; does the aluminum need to be oxidized before it dissolves?
I'm not a chemist but I remember that the standard HF/nitric silicon etch used in micro-fabrication supposedly worked in two steps; the HF would dissolve the thin SiO2 oxide layer on the silicon surface, exposing silicon atoms, and the nitric acid would then re-oxidize the bare silicon, allowing the HF to then dissolve it.
Or at least that's how it was explained to me.
Despite being challenged here, Wikipedia explains that the aluminum bits in sodium hydroxide-based drain cleaner are included to produce heat via exothermic liberation of hydrogen gas so that the heat can loosen some of the clogging materials. The article provides the following two steps.

Breakdown of Aluminum Oxide: $\ce{ \ \ \ Al2O3 + 2NaOH + 3H2O → 2Na [Al(OH)4]}$

Oxidation of Aluminum metal: $\ce{ \ \ \ 2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2}$

Is what is happening similar to my understanding of the HF/nitric etch of silicon; it is the oxidized aluminum that is removed by the solution, and only because it happens to also oxidize the freshly exposed aluminum metal that the oxide is removed?

Snapshot of some drain cleaner with aluminum bits

That last sentence is a bit confusing, maybe you can explain the mechanism you are talking about in a little more detail.

@BuckThorn the first four words of the first sentence are "I'm not a chemist..." so i this case maybe I can not. However the miracle of Stack Exchange allows those who understand better to post answers, and that is exactly what has happened.

The question is about logic and usage of language. Perhaps read this again: "only because it happens to also oxidize the freshly exposed aluminum metal that the oxide is removed?"

@BuckThorn if there is no oxidation, then logically there will be no more oxide to be removed, and therefore aluminum dissolution will cease. I'm not getting what it is that you are asking me to clarify.

Your use of english is confusing but I get it now: "the solution removes solid aluminum through formation of a soluble oxide intermediate".

@BuckThorn except that I've asked it in a form of a question, yes I think so.

If you want to understand the reaction of $\ce{NaOH}$ on Aluminium, you may describe it this way. First, the aluminium reacts spontaneously with water, like sodium or calcium, producing gaseous Hydrogen $\ce{H2}$ and an metallic oxide or hydroxide. With sodium or calcium, the hydroxide produced by this reaction quits the metal surface, and get dissolved, leaving a fresh surface of metal ready to react with more water. The reaction with sodium is:
$$\ce{2 Na + 2 H2O ->   2 NaOH  +  H2}$$
With aluminium, it is not the case. The reaction is:
$$\ce{4Al + 6 H2O -> 2 Al2O3 + 6 H2}$$
But the aluminium oxide is not soluble into water and it makes a waterproof layer on the surface of the metal, which prevents further contact between the metal and water. So the reaction stops, due to this protecting effect of $\ce{Al2O3}$.
There are two ways of allowing the reaction to continue:

Dissolve $\ce{Al2O3}$ as soon as it is produced by adding $\ce{NaOH}$, as @Uhoh tells it in its first equation.
Prevent the $\ce{Al2O3}$ oxide layer from adhering on the metal surface by adding some mercury salt in the water. In this case the mercury ions $\ce{Hg^{2+}}$ may go through the oxide layer, reacts with the metal and produces a mercury deposit which forms an alloy with Aluminium. The reaction is:

$$\ce{2Al + 3 Hg^{2+} -> 2 Al^{3+} + 3Hg}$$
The $\ce{Al2O3}$ layer cannot adhere on the surface of this alloy Al-Hg. So the
 reaction with water continues as long as there is some mercury atoms left on the
 metallic surface.
So, to go back to @Uhoh's question, it is not $\ce{NaOH}$ that acts on Aluminium. It is water. The action of $\ce{NaOH}$ is to remove the oxide layer.

okay so somewhat like the HF/nitric combination example dissolves silicon in two steps, this NaOH/water combination dissolves aluminum also in two steps; the NaOH removes the oxide and the water is sufficient to quickly re-oxidize the freshly exposed aluminum, and because aluminum is so incredibly reactive it happens fast enough that a more powerful oxidizer (like the HF in the silicon case) is not needed. Thanks!

You are right !


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How is the bonding in the [Au6C(PPh3)6]2+ cluster explained?
How can carbon form six bonds in such a compound as $\ce{[Au6C(PPh3)6]^2+}?$ I understand how carbon with five bonds can be formed $(\ce{CH5+},$ for instance), but it shouldn't have enough electrons to form six bonds.

Carbon can well be hexacoordinate, or even hepta. This case isn't all that different from methanium you mention.

For example https://chemistry.stackexchange.com/questions/51209/coordination-number-of-carbon-in-methyllithium-tetramer https://chemistry.stackexchange.com/questions/10415/hyperlithiated-carbon-species

Mildly related: I explain a (relatively simplistic) model of bonding for $\ce{SF6}$ in Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?. The discussion following the MO diagram briefly touches on species such as $\ce{CH6^2+}$. Now, if you treat $\ce{Au(PPh3)+}$ as a replacement for a single proton (the proper term is isolobal), then that becomes $\ce{C[Au(PPh3)]6^2+}$. Note the qualitative similarity in the MO diagram there to the one andselisk has drawn.

This is by no means a trivial case.
You are dealing with a so-called template or interstitial atom placement within a cluster shell.
There is a thorough review available [1, pp. 18–20] (reference numbers updated):

An alternative way of satisfying the closed shell electronic requirements and retaining a positive charge on the cluster is possible if an interstitial atom is introduced.
  For an octahedral gold cluster an additional gold atom is too large to satisfy the steric requirements of the interstitial cavity, but a small main group atom, e.g. C, N or O, is able to satisfy these steric requirements.
  Figure 11 gives the relevant interaction diagram for octahedral $\ce{[Au6C(PPh3)6]^2+}$ and an interstitial carbon atom, which contributes just the right number of electrons to lead to a $\mathrm{[S^σ]^2[P^σ]^6}$ pseudo-spherical ground state conﬁguration.
  It is noteworthy that this simple theoretical analysis was ﬁrst published in 1976 [2], and the existence of $\ce{[Au6C(PPh3)6]^2+}$ was conﬁrmed by Schmidbaur and his coworkers in 1989 [3–5].
  This compound had been isolated and structurally characterised previously, but the interstitial carbon had not been identiﬁed.
  The introduction of the interstitial atom strengthens the radial interactions signiﬁcantly as a result of effective overlaps between the carbon $\mathrm{2s}$ and $\mathrm{2p}$ orbitals and the matching $\mathrm{S^σ}$ and $\mathrm{P^σ}$ cluster molecular orbitals.
  In broad brush terms the stabilisation of the valence orbitals of the central atom are stabilised by $\mathrm{nβ^σ}$/degeneracy of the molecular orbitals.
  If $\mathrm{β^σ(s)} = \mathrm{β^σ(p)} = \mathrm{β^σ(d)},$ the relative stabilisations are $\mathrm{6β^σ(s)},$ $\mathrm{2β^σ(p)},$ and $\mathrm{3β^σ(d)}.$
  Therefore, the greatest stabilisation involves the $\mathrm{s}$ orbitals of the central
  atom and increases as the number of metal atoms, $n,$ increases.
  For filled shells the stabilisation energies are independent of geometry as long as the ligand polyhedron approximates to a sphere.
  It follows that gold clusters with main group interstitial atoms are characterised by a pec of $12n + 8$ (sec 8) valence electrons, since each $\ce{AuPPh3}$ fragment is associated with a filled $\mathrm{d}$ shell containing $10$ electrons and a bonding $\ce{Au–P}$ bonding molecular orbital.

Fig. 11 Molecular orbital interaction diagram for $\ce{[Au6C(PPh3)6]^2+}.$
  Similar analyses may be constructed for trigonal bipyramidal $\ce{[Au5N(PPh3)5]^2+}$ and tetrahedral $\ce{[Au4O(PPh3)4]^2+}$

References

Gold Clusters, Colloids and Nanoparticles II; Mingos, D. M. P., Ed.; Structure and Bonding; Springer International Publishing: Cham, 2014; Vol. 162. DOI: 10.1007/978-3-319-07845-8.
Mingos, D. M. P. Molecular-Orbital Calculations on Cluster Compounds of Gold. J. Chem. Soc., Dalton Trans. 1976, No. 13, 1163–1169. DOI: 10.1039/DT9760001163.
Schmidbaur, H., Grohmann, A., Olmos, M. E.; Organogold chemistry. In: Schmidbaur, H. (ed) Gold progress in chemistry, biochemistry and technology. Wiley, Chichester, 1999. pp 648–731.
Steigelmann, O.; Bissinger, P.; Schmidbaur, H. Assembly of the $\ce{[CAu6]^{2⊕}}$ Cluster with a Tailor-Made Diphosphane Spanning the Octahedral Edges. Angewandte Chemie International Edition in English 1990, 29 (12), 1399–1400. DOI: 10.1002/anie.199013991.
Schmidbaur, H.; Brachthäuser, B.; Steigelmann, O. Direct Observation of the Central Atom in $\ce{[C\{Au[(C6H5)2(PC6H4NMe2)]\}6](BF4)2}$ by $\ce{^{13}C}$ NMR Spectroscopy. Angewandte Chemie International Edition in English 1991, 30 (11), 1488–1490. DOI: 10.1002/anie.199114881.

I wonder, if the LaTeX advanced editors use any handy reference generator, or if all is written down.. I bet for the former. When I was writing my diploma work, top tech were DOS based editors a latex was rubber.

@Poutnik I use Zotero (desktop)/Zbib (online) for references as there is no Bib(La)TeX support on any of the SE sites.

@Poutnik, the references in andselisk's answer are not in LaTeX (apart from the chemical formula in ref 5, which I assume is manually typed in). They are just plain Markdown. See https://chemistry.meta.stackexchange.com/a/4299/16683 for a tool which generates citations from a DOI.

@orthocresol I see. I had rather in mind perhaps just generated by LaTex aware word processing software, as I am not familiar with these systems. I will check the link, thanks.

You beat me to it, +1.  Compare with SF6 where the e_g orbitals are occupied.  As explained here this orbital pair becomes antibonding if the central spacer atom is too small.  But if that level is empty, as in the gold cluster, you want the small central atom to promote ligand-ligand bonding in the still occupied lower orbitals.


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Violation of Fajans' rule
According to Fajans' rules, greater charge in an ionic compound results in more covalent character. But why is it that sodium oxide is more covalent than magnesium oxide? Google says boiling point of magnesium oxide is more than sodium oxide.

https://chemistry.stackexchange.com/questions/86331/when-comparing-boiling-points-of-two-ionic-compounds-what-bonding-factors-shoul........maybe this would help

Kindly look for similar questions before asking..cause these questions are very likely to get reported as duplicate.

Does this answer your question? When comparing boiling points of two ionic compounds, what bonding factors should you assess?

Does this answer your question? https://chemistry.stackexchange.com/a/64974

Sodium oxide has less electrostatic binding than magnesium oxide, if both are 100% ionic.  Higher cation charge and more favorable stoichionetry with magnesium.


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How much electron hybridization and bonding occurs in liquid helium?
Helium is the only element which does not freeze at atmospheric pressure, and only forms a liquid at around 4 Kelvin due to weak van der Waals forces.
I'm interested in knowing roughly how much hybridization/bonding occurs in liquid helium (e.g., energies in cm$^{-1}$ or electronvolts) at ambient pressure and cryogenic temperatures. I would guess that since the enthalpy of formation for liquid helium is about $10^{-3}$ eV per atom, the bonding energies are comparable, but I have not found any numbers on this topic in the literature.
To ask the same question without jargon: how much are the energy levels of Helium smeared when it is in the liquid phase?
For example, in solid Copper, a plot of the density of states shows that the atomic energy levels hybridize to form solid state energy bands. For the $d$-orbitals the overlaps are relatively small, giving rise to roughly 4eV wide bands. On the other hand the $s$- and $p$-orbitals have large overlaps, resulting in these atomic orbitals being broadened to be >12eV wide.

"hybridization energy" sorry, but there is no such thing. Can you explain why you use this term?

I just mean the energy scale of the hybridization, sorry for not using the correct terms. For a single atom the levels are sharp lines, but in a liquid I assume those lines hybridize to form a band (like solids). What is the energy width of that band in liquid helium?

I have never heard the term "hybridisation" together with electronic band structures. And the electronic band structure has little to do with the dispersive attraction between atoms. Metals are kept together by their common electron gas. With a liquid isolator, I think this concept is a bit of a misapplication.

Ok, let me ask the same question without jargon: how much are the energy levels of Helium smeared when it is in the liquid phase?

To those who request closing the question due to lack of details, can you please elaborate? I am using terminology commonly used in the context of physics, but am learning that have slightly different connotations in chemistry. At the very least the bolded question should be well-defined.

@Karl I think your statement is not correct. You can get electron band structures even in insulating solids held together by "van der Waals forces". Hence you find that noble gases (besides Helium) can become solids with electronic band structures. I am using the term "hybridization" in the sense that atomic orbitals hybridize to form delocalized states in both metals and insulators, giving rise to electronic bands. See for example, https://journals.aps.org/pr/pdf/10.1103/PhysRev.124.652

Are you interested in specific liquid phase conditions wrt temp and pressure? Here's a reference for high pressure and temp liquification: https://www.pnas.org/content/105/32/11071

@Andrew thanks for the reference. I have in mind liquid helium at cryogenic temperatures and atmospheric (or vacuum) pressure, like one would find in a laboratory. Let me add that to the question. I think the reference is focused on extremely large pressures and temperature (>10,000 Kelvin, >10Mbar)

Hybridization is a simplification, it doesn't do complicated.

I was guessing that you weren't interested in high P and T, but figured it was worth clarifying. My impression is that at standard P even at cryogenic temps, liquid helium has such low density that the atoms are not close enough together to have significant orbital overlap, but I don't know references off-hand to confirm that.


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electrochemistry experiment
something strange.
i made an electrochemistry experiment in which i have measured the potential of cell:
Ag, AgCl | (1M)KCl ║ [Fe(CN)6] 3-, [Fe(CN)6] 4-  | Pt
with different concentrations of Fe+2/Fe+3 .
from the results i have found the faraday constant:
96181 Coulon /mol
very close to the published value (99.7% accuracy).
the intercept of the graph of E (v) against ln(Fe+2/Fe+3)   is:
intercept:    0.198985714  V

(the intercept is also the standard potential for the cell)
yet, when i tried to find the standard potential of
Fe3+ +e- -> Fe2+

i got the result of +0.423 V which is almost half of the published value (+0.77 V)
E_cathode =E^° + E_anode

E_cathode=0.198985+0.224= +0.423 V

how can it be that the faraday constant is almost accurate and in the same time the result of the standard potential is far from the real one.? i'm sure i don't have any calculation mistakes, if it's necessary i can describe the method of work here.
thanks!

What you are measuring is not the potential $Fe^{2+}/Fe^{3+}$. It is the potentiel of the couple $[Fe(CN)_6]^{4-}/[Fe(CN)_6]^{3-}$ which is equal to +0.36 V in the table.

which table are you using?

This is more suited as a comment Maurice

My table is "Chemistry Data Book" , by J G Stark and H G Wallace, p. 70, edited by John Murray, 50 Albermale St, London W1X 4BD, 1994, ISBN 0-7195-3951-X. It has been reprinted every two years since.

thanks! but still the faraday constant is more accurate by 14% than the standard potential, even though i used the standard potential for obtaining it.


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Solvent for reaction mixture
I'm looking for some suitable solvent for ether cleavage of Guaiacol (o-Methoxyphenol) using aluminium triiodide ($\ce{AlI3}$) with $\ce{DMSO}$. The original paper calls for using hot acetonitrile as a solvent, but I wonder what might be the alternatives. The ideal solvent would be some solvent with only mild toxicity.
What hydrocarbon solvents are a suitable replacement for this preparation?

Would you provide the original reference for reader's benefit?

Can't seem to find the original article, but this seems to have the same point. (https://www.x-mol.com/paper/833347) Similarly here the article discusses one-pot preparation using cyclohexane as a solvent. (https://onlinelibrary.wiley.com/doi/full/10.1002/slct.201803469) Do you think toluene might be suitable?

Aluminium iodide ($\ce{AlI3}$) is first introduced in 1984 as an easily accessible, a highly regioselective versatile ether-cleaving reagent with novel cleavage pattern (Ref.1). Application of $\ce{AlI3}$ in organic synthesis has been reviewed (Ref.2) and it seems like acetonitrile is the best solvent to be used in catechol ether demethylation (e.g., 2-methoxyphenol or 1,2-dimethoxybenzene).
For example, recent article discussed an one‐Pot cleavage of aryl alkyl ethers by aluminum and iodine in Acetonitrile (Ref.3). Their proposed mechanism for the cleavage is depicted in following diagram:

Accordingly, since it needs acid work-up for final product, I'd suggest you may use acetonitrile as the solvent of reaction, and use your solvent of choice to extract the final product from aqueous work-out.
If you are not happy with that, you may use either carbon disulfide ($\ce{CS2}$), which is also toxic, or toluene (their are references have been in literature using benzene as a solvent with mixed results) (Ref.2). Use of $\ce{CS2}$ is applied for specific reactions such as substrates having allyl functionality. Yet, it should work with simple compound as yours.
I'm little curious of your intend to use non-toxic solvent in the reaction since DMSO is also toxic in certain concentrations. In that case, you may use solvent-free conditions (you need to use enough DMSO for convenience, which is not critical according to the mechanism)(Ref.4).
Note: Cyclohexane is not a suitable reagent for this reaction, although it was involved in some publications (Ref.3). For example, when demethylation of eugenol reaction has bee done in acetonitrile at $\pu{80 ^\circ C}$ for $\pu{18 h}$, hydroxychavicol was achieved in 78%. However, when the reaction was run in cyclohexane under identical condition, no hydroxychavicol was achieved.

References:

M. Vivekananda Bhatt, J. Ramesh Babu, "New reagents 3: aluminium iodide - a highly regioselective ether-cleaving reagent with novel cleavage pattern," Tetrahedron Letters 1984, 25(32), 3497-3500 (https://doi.org/10.1016/S0040-4039(01)91058-5).
Juan Tian, Dayong Sang, "Application of aluminum triiodide in organic synthesis," ARKIVOC 2015, (vi), 446-493 (http://dx.doi.org/10.3998/ark.5550190.p009.309).
Juan Tian, Huaxin Yue, Pengtao Yang, Dayong Sang, "One‐Pot Cleavage of Aryl Alkyl Ethers by Aluminum and Iodine in Acetonitrile," Chemistry SELECT 2019, 4(1), 38-41 (https://doi.org/10.1002/slct.201803469).
Mehran Ghiaci, Jila Asghari, "Dealkylation of Alkyl and Aryl Ethers with $\ce{AlCl3 - NaI}$ in the Absence of Solvent," Synthetic Communications 1999, 29(6), 973-979 (https://doi.org/10.1080/00397919908086060).

Thank you for your answer, I have a few questions.
Why cyclohexane is not a suitable solvent in this case?
Do you think toluene or benzene might work with allyl catechol ethers? That could be interesting.
As for the non-toxic solvents, it's out of curiosity and for the sake of convenience.
I wouldn't regard DMSO as a toxic or hazardous solvent since it's LD50 is twice that of ethanol and almost three times that of ethyl acetate.

There was a publication, which states that reaction in cyclohexane has given least conversion among solvents used; 2) toluene  or benzene can be used with allyl catechol ethers, but should be careful with temperatures, since substrate is susceptible to undergo Cleisen rearrangement at high temp; I'm ok with DMSO reaction but still think it is toxic to humans. :-)

I edited the last paragraph to give the literature reference for cyclohexane use.

I agree it's not suitable for one-pot preparation when everything is mixed together. It might be because iodine forms charge transfer complex with ether and it prevents the formation of AlI3 thus failing to produce the desired product. So My thought was to wait for AlIO to form (cyclohexane+Al+I2+DMSO) and then add dropwise allyl catechol ether of choice.
One study achieved very good yield in 85%+ range with RT, so I'm going to try this way. Thanks for heads-up ;)


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Galvanic Cell Temperature Dependence
I'm currently studying the temperature dependence of the following cell: $\ce{Cu_{(s)} + 2Ag^{+}_{(aq)} <=> 2Ag_{(s)} + Cu^{2+}_{(aq)}}$ ($E^⦵ = \pu{+0.46 V}$)
Experimentally, I recorded the initial temperature of the cell at $\pu{273 K}$, before increasing the temperature of the cell and recording its potential each time. This nicely followed the equation:
$$E^⦵ = T\frac{\Delta S^⦵}{nF} - \frac{\Delta H^⦵}{nF}$$
Which apparently can be used to describe cell temperature dependence. It's worth noting that the activities of all species was 1.

The issue is that in retrospect, the entire experiment seemed kind of pointless, as it doesn't appear to actually tell me anything at all. Consider the following logic:
In a 'normal' equilibrium reaction, e.g. the Haber process, the reaction will proceed until it gradually reaches equilibrium, giving an equilibrium constant; let's call it $K_{1}$. Changing system temperature will shift the equilibrium until a new equilibrium is formed, with a new equilibrium constant $K_{2}$... And so on for each temperature change.
In a cell, this is not what happens. If we consider that $E^⦵$ is a 'guide' to equilibrium position, and assumed that the above was true, then changing temperature would've caused a new value of $E^⦵$ to equilibrate each time. However, we know it doesn't do this, as by definition equilibrium in a cell is when $E^⦵ = \pu{0 V}$. Therefore, this seems to me as though a cell will naturally reach equilibrium  with the same equilibrium constant at any temperature.

Thus, apart from giving me rather accurate, experimentally derived values for $\Delta H^⦵$ and $\Delta S^⦵$ (assuming that these parameters aren't temperature dependent, I know), what does the experiment actually tell me? I mean, the cell is predisposed to run to equilibrium and thus have an $E^⦵$ value of $\pu{0 V}$ regardless of temperature, so where does the temperature dependence come into it?
Furthermore, this makes it confusing to study how the reaction quotient (and equilibrium constant) varied with temperature:

It's logical to extrapolate the graph and find the temperature at which equilibrium is apparently achieved; but as discussed, it doesn't work like that.
It's also logical to calculate both parameters at each temperature (which appears to fit nicely as my calculations show that $\log K$ decreases with temperature, vice versa for $\log Q$) and finding the point where $\log K = \log Q$ would surely give me the temperature at equilibrium; but again, as discussed it doesn't work like that, mainly because 'equilibrium' results in $E^⦵ > 0$

So is there any simple way to determine how the reaction quotient changed with temperature?


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Chemistry books for the general reader
I’m quite interested in chemistry, but reading the textbooks are a bit challenging. Are there any good chemistry books for the general reader that go a bit beyond the basics?

Check out Resources for learning Chemistry. Probably you'll find something not only for general Reader, but also for major Curiosity:)

What do you mean by "challenging" ? Too simple ? Too difficult? What ? What is "beyond the the basics" for you ?

Does this answer your question? Resources for learning Chemistry

Do you know Sir Michael Caine's word as father Robert Spritzel in the 2005 movie «The Weather Man» to his son?  To quote: «Do you know, that the harder thing to do, and the right thing to, were usually the same thing?  Easy does not enter in the grown-up life.  To get any thing of value, you have to sacrifice.»  It is around 1:28 min to 1:45 min of this preview here: https://www.youtube.com/watch?v=vZIIKsnM6MM.

I suppose you mean "popular science"? It much depends on what branch of chemistry you are interested in, traditionally: analytical, organic, inorganic, physical. (others include polymer, materials, surface ...)


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Which of the following reaction will have greater enthalpy and why?

Which of the following reaction will have greater enthalpy and why?
$\ce{C (g) + 4 H (g) -> CH4(g)}$ (reaction enthalpy = $x\ \mathrm{kJ/mol}$)
$\ce{C (graphite) + 2H2 (g) -> CH4 (g)}$ (reaction enthalpy = $y\ \mathrm{kJ/mol}$)

I thought that since $\ce{H2}$ is in its aggregation state it would have enthalpy of $0\ \mathrm{kJ/mol}$ and hence $y\gt x$.

The first reaction is the most exothermic rection

Could you complete your argument? If you compare two things, you need to make statements about both to be able to compare them. As it stands, it is not clear why you conclude that y > x.

Maurice has stated why equation 1 is much more exothermic than equation 2. A visualization of the processes may help you understand the details. Equation 1 involves the reaction of a carbon atom with four atoms of hydrogen to form methane in the gas phase. The bond dissociation energy (BDE) of the hydrogen molecule is +104 kcal/mol. Accordingly, the heat of formation (ΔHfo) of the hydrogen atom is half this value or +52 kcal/mol. The ΔHfo of the carbon atom from graphite is +171 kcal/mol. Both hydrogen and graphite are taken as 0 kcal/mol at the Standard State. The right hand side of the chart shows that the sum total of the necessary atoms is +379 kcal/mol above the Standard State. This is your starting point for equation 1. To form methane in the gas phase the reaction will be exothermic by -379 kcal/mol plus the ΔHfo of methane (-18 kcal/mol) for a total of -397 kcal/mol for equation 1. On the other hand, equation 2 is nothing more than the ΔHfo of methane as -18 kcal/mol below the Standard State as shown in the left hand diagram. This diagram also shows the stepwise BDE's of C-H bonds starting from methane. The same value for equation 1 is obtained, -397 kcal/mol.

The heat of formation of $CH_4$ is $-75$ kJ/mol. This is not much : y is small.  As the dissociation energies of $C-H$ bonds are higher than $400 kJ/mol$, the heat of reaction of the first equation is more than $1600 kJ/mol$. This first reaction is about $20$ times more exothermic than the second reaction. x >> y !!