text:
Difference between HCD and CID collision induced dissociations?
With the Thermo-Fisher Fusion Orbitrap mass spectrometer I can do both CID (collision-induced dissociation) and HCD (higher energy C trap dissociation) collision-induced dissociation studies.
When I apply HCD on inorganic molecules I observe more fragmentation products than with just CID suggesting multiple fragmentation steps. However I don't get why physically this is possible.
Is HCD design to keep the fragmentation products in the trap so that they collide many times with the inert gas?
EDIT: according to this manual found on the Termo Fisher website:

HCD tends to produce EI-like fragmentation
  and records ions from multiple steps of collision as seen
  in Figure 3.

https://assets.thermofisher.com/TFS-Assets/CMD/Application-Notes/AN-540-LC-MS-Small-Molecule-Characterization-AN63396-EN.pdf
And according to a post found in research-gate:

In beam type CID ("HCD") multiple collisions with a collison gas take place under an elevated pressure regime. Ions can be activated multiply, which results in richer spectra compared to resonance type CID ("CID"). In resonance type activation, only the precursor ion is activated - in beam type CID also the fragments (collision gas).

https://www.researchgate.net/post/Why_sometimes_there_are_a-ions_when_you_perform_HCD_fragmentation_for_peptides_on_Q-exactive_plus
From the answer given here the reason of a richer spectra is that:

The reason HCD gives you more fragment ions is that ions do not have time to equilibrate.

So, do I have more fragments when I apply HCD because:
1) I have multiple collisions(multiple activation)where also the fragments collides?
2) It's a non-equilibrium process, so bonds are broken independently of the strength?
3) both?

The new document you list also shows that HCD gives rise to more and smaller product ions, so is consistent in general with the explanation I gave.  I think the discrepancy is about the number of collisions, which isn't really even an important point to understand the differences between the two techniques.  I'll edit my answer to downplay the "multiple collisions" vs. single collision aspect.

There isn't very much about HCD that is "EI-like".  Radical cations are very rare, and unlike with EI, fragmentation pathways cannot be reasonably reliably computed from first-principles.  The only similarity is that in EI and in HCD (and in CID as practiced on qTOFs and QQQs), energy is imparted very quickly to the parent ion, and their is no chance for thermal equilibration.

Thanks for your insight I really appreciate it. So, do you think the statement from this document is not proven? I wonder how they know that with the HCD technique ions follow multiple steps of collision.

I asked about my answer on Twitter.  https://twitter.com/fischer_cr/status/1157421482202390528  Various high-profile mass spectrometry scientists replied with various comments.  The most insightful comments to me are that "HCD" is a "marketing magic" term, which is roughly the same as "beam CID", and different from "resonant CID".  I still think my answer is mostly right -- the difference between beam and resonant CID is equilibration of the excitation energy across the bonds/modes of the ions, but you should check out the tweets and replies for more info.

After further searching it seems that actually there are multiple collisions involved when I apply HCD. I wonder if both your answer and the multiple collision explanation are both true and contribute equally to the richer spectra.

Brief overview of CID vs. HCD
"Collision-induced dissociation" or CID is a much older and more general term in mass spectrometry than HCD.  HCD is a vendor-specific term invented to describe a new modification on how ions can be dissociated in Orbitrap mass spectrometers.  CID is a "universal" term that applies to many different types of mass spectrometers.  Unfortunately, how exactly CID works depends on which type of mass spectrometer you are talking about.
Wikipedia has a good overview of how CID implementations vary according to instrument type.  Understanding that the meaning of "CID" varies strongly by instrument type is important.  The naive assumption that CID on e.g.a triple-quadrupole or qTOF instrument is like CID on an ion trap instrument is understandable -- it is the same term after all -- but very wrong.
CID on an Orbitrap
On an ion-trap instrument like your Fusion instrument, CID is done in an ion trap.  This means that on an Orbitrap, CID is a resonant excitation technique.  Thus, to reach a certain collision energy, ions must be accelerated relatively slowly from their starting energy to the final energy.  (The final energy is a user-controllable parameter.)  They thus collide many times with the inert gas during the excitation process.  Each of these collisions may "heat" or "cool" the ion.  In an ion trap, resonant excitation will only accelerate the precursor ions, so after dissociation has taken place, fragment collisions with the gas molecules will cool the ions, eliminating or at least reducing further fragmentation into smaller pieces.
A 2014 paper in the Journal of Mass Spectrometry by Ichou et al. explains CID on an ion-trap-equipped Orbitrap fairly well:

In ion trap devices, each collision can activate or cool the stored precursor ion. The evolution of this equilibrium depends greatly on the ion kinetic energy. The second step relies on the unimolecular dissociation pathways of the precursor ion. It has been extensively modeled using the Rice–Rampserger–Kassel–Marcus (RRKM) theory18-21 and the quasi‐equilibrium theory.22 Note that the internal energy distribution of ions greatly relies upon the rate of these two steps. So, the internal energy distribution depends on the regime of collisions under study.23-25 The ‘slow’ or ‘rapid’ heating regime is defined according to the unimolecular dissociation constant rate. If the latter is slower than the constant rate of the activation/deactivation process, the regime is defined as a slow heating technique. The slow heating regime in an ion trap instrument for the conventional CID mode may be described as a random walk process,25, 26 where a large number of ion‐neutral gas collisions are needed to ‘heat’ the ions in order to dissociate them. Some studies25-29 have demonstrated that the ion internal energy distribution for ion trap instruments is close to the temperature due to the multiple collision regime.

Thus, on an ion-trap Orbitrap like your Fusion, CID can be at least close to an equilibrium process.  Whether it is or not depends on how quickly ions are ramped in energy to the final user-decided collision energy.  If ramp times are long compared to the unimolecular dissociation rates, ions have the opportunity to distribute the excitation energy across most of the bonds in the molecule.  They bump into many gas molecules, sometimes being heated by the collision, and sometimes being cooled.  These collisions are what allows the energy to distribute itself among all the bonds of the excited ion.  As a result, in slow resonant CID, the weakest bonds break first.  If ramp times are short, then CID (on an ion-trap instrument) ions do not have time to equilibrate and dissociation spectra can be more like what is observed for HCD.
HCD on an Orbitrap
The same paper explains HCD fairly well too:

'HCD' is an excitation mode that belongs to the category of non‐resonant activation techniques like the CID conducted in tandem‐in‐space instruments.

Thus, 'HCD' in an Orbitrap is like CID on a triple-quadrupole or qTOF instrument.  It is a non-resonant dissociation technique.  Ions do not experience a long series of "equilibrating" collisions with gas molecules.  Effectively, a single bolus of energy is imparted to them very quickly. This type of dissociation is a non-equilibrium process.  Both weak bonds and stronger bonds may have a chance to break, depending on which ones are excited by the smaller number of collisions with gas molecules.  Product ions remain excited, and do not experience enough collisions to "cool" them down, meaning that if they do further collide with gas molecules, they may be further fragmented.
Answers to your questions

Is HCD design to keep the fragmentation products in the trap so that they collide many times with the inert gas?

No, actually the opposite, as hopefully the discussion above makes clear.  On an ion-trap instrument (but not on other instruments), it is CID that collides ions many times with the inert gas.

When I apply HCD on inorganic molecules I observe more fragmentation products than with just CID suggesting multiple fragmentation steps. However I don't get why physically this is possible.

The reason HCD gives you more fragment ions is that ions do not have time to equilibrate, so many bonds can be broken, each with some probability.  Parent ions are energized rapidly, in a near-single-step, to highly excited states.  A variety of vibrational modes can get excited, leading to a variety of bonds breaking in the dissociation step.  In (slow) CID, ions are excited slowly; as energy accumulates during the resonant excitation, ions collide with and have time to redistribute energy over the whole molecule, meaning that only the weakest bonds tend to break, resulting in fewer fragments.


text:
Why is sodium peroxide not of white colour even if it is paramagnetic?
Is colour always associated with paramagnetism?

is it really paramagnetic? 2) no

Colour can be due to a 0.1% impurity

Color is related to how the compound absorbs light. For example, highly unsaturated compounds (e.g. beta-carotene) will absorb green to blue light, and thus reflect red and yellow. This is why it appears orange. Somewhat unsaturated compounds (e.g. containing an aromatic ring), on the other hand, mostly absorb UV light and reflect all visible light; they thus appear white.
Paramagnetism isn't really related.

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text:
Electroplating boat hulls with ceramic like material containing copper oxide to prevent barnacles from sticking?
Calcium bentonite can be electroplated onto steel. After it is electroplated on to steel it has a rough and porous like surface that is very hard and tough.
If you were to electroplate a thin layer of calcium bentonite onto steel then brush on a thin coating of copper oxide. The copper oxide would stick to the surface and fill some voids because the calcium bentonite that was electroplated on to the steel is porous. Is it then possible to electroplate more calcium bentonite over top this copper oxide layer? And repeat the process so that you have layers that start with steel then calcium bentonite then copper oxide then calcium bentonite then copper oxide then calcium bentonite etc. The final layer could be smoothed out by sanding/grinding.
I know that most paint for boats use copper oxide to keep barnacles from forming but thot what if there were a more permanent or tougher/longer lasting solution to that problem that causes ships to waste massive amounts of fuel due to drag.

You are going to have a major corrosion problem with copper oxide in intimate contact with your steel hull.

How durable is bentonite clay on steel in water???

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text:
How were x-ray diffraction patterns deciphered before computers?
I know the general idea behind x-ray crystallography is to take a high quality crystal and place it in the path of an x-ray beam. Areas of high electron density will diffract the beam and lead to spots on a detector screen where photons have constructively interfered with each other (and nothing appears where destructive interference has occurred). The crystal is then rotated and another diffraction pattern image is taken and the cycle repeats until there are enough views to reconstruct the atomic structure of whatever you crystallized. My question is what is the general process for doing this without computer software? How was this done in the 20s? The only equation I know for x-ray diffraction is Bragg's Law but is this the only equation used to interpret the data? Surely there must be others? How do you translate the spots on a detector to electron density plots using Bragg's law?

Though it is a fascinating question, it is also quite a broad. I'd recommend to get André Authier's book Early Days of X-ray Crystallography which gives a nice overview of the problems that had to be tackled as well as historical background; brief history on Wikipedia; also, there is a series of free access articles by IUCr; and, of course, Bragg's X-Ray Crystallography (1968).

To put it simply, the bare-bone math apparatus was pretty much ready at that time, but the calculations had to be done manually, by hand. That's why crystal structure determination of a single compound used to be a topic for a Ph.D. and might took several years.

In x-ray diffraction images it is the intensity of spots that contain the information. In the past Beevers and Lipson Strips were used to help with the Fourier transforms; not easy but a clever way of determining intensities on a photographic plate. Wikipedia gives a little more information.

@andselisk As a comment years later... I got the Authier book and it was fantastic. Thanks for the rec

I agree with @andselisk that this question is quite broad. I will focus on two specific questions asked

The only equation I know for x-ray diffraction is Bragg's Law but is this the only equation used to interpret the data? [...] How do you translate the spots on a detector to electron density plots using Braggs law?

Apart from Bragg's law (which tells you where the diffraction spots are for a known orientation of a crystal with know unit cell), it was also known that real space (electron density) and reciprocal space (diffraction pattern) are related by 3D Fourier transform. For structures containing one or two atoms, just knowing the unit cell parameters and the symmetry is enough to get the entire structure. For anything slightly more complicated, the Fourier transform of amplitudes or later, of intensities (Patterson methods, http://reference.iucr.org/dictionary/Patterson_methods), had to be used. The first crystal structure analyses were of crystals with centrosymmetry, where the phase problem is easier to solve.

My question is what is the general process for doing this without computer software?

Diffraction data was measured on film, with gray-scales to assess intensity of signals. To calculate a Fourier transform, pre-computed tables were used, such as the Beevers-Lipson strips. As Andselisk commented, Fourier transform was used late in the 20s, and initially for problems that were one- or two-dimensional. There is a nice account by P.P. Ewald available here: https://www.iucr.org/publ/50yearsofxraydiffraction/full-text/structure-analysis
This was written at a time when the myoglobin structure had just been solved, after "50 years of X-ray diffraction".

I believe Fourier synthesis wasn't used extensively until late 1930s, even though it was proposed in 1925 and Zachariasen first applied 2D Fourier projections in 1929, and 3D Fourier transformation is definitely not about the 20s:)

I also remember an empirical plots useful to identify the crystallographic type or something. Forgot the name of the compiler. Basically one moved his/her stripe to find a correspondence. Do you remember that?

@Alchimista We had to find out the crystalligraphic type and cell dimensions from diffractions patterns in our advanced inorganic lab course. With a bit of practice (I TA'd the course later), you could guess quite many of the films' types at first glance. Typical patterns. (Of course the films were chosen to be rather easy, and I seems to remember that some types were missing, because they're impossible to tell apart by any student?)

Not just in the '20s but up to the '90s at least, the d-spacings were estimated by hand measurements of diffractometer peaks or film lines and applying the Bragg formula. d-spacings were then ranked from the most intense down to the least intense. Starting with the 2 most intense d-spacing values - and allowing +/- 0.02 angstrom error margin to each - a process of searching the Hanawalt Search Manuals was done till a chemically plausible phase with these two values was found. If no plausible phase was found for these two most intense d-spacings, the second value was replaced with each of the third/fourth/fifth/etc most intense d-spacing value and another search made till a chemically plausible phase was found. Once located, the remaining less intense d-spacings listed for this phase were compared with the remaining calculated d-spacings and if found this was supporting evidence for this phase. Remaining d-spacings were relisted according to intensity and the process repeated. It always seemed daunting at first but, with a little initial guidance from a more experienced person (usually another postgrad student) and with a shortlist of all plausible phases plus their d-spacings beside you, it got easy enough in metallurgy and ceramics. It must have been much harder in biochemistry but I guess those working in that métier had their own shortcuts.
For more detailed work, e.g. for residual stress calcs, grain texture effects, part-crystalline/part-amorphous phases, exploration of short-range ordering in amorphous phases, etc one referred to texts by Cullity and Azaroff. Beyond that you had to look at papers by other researchers, talk to x-ray lab technicians and above all try to look at the problem in the way that was most natural to oneself or most amenable to the nature of the problem.


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Labware to handle hot aqueous NaOH
This question regarding appropriate handling of hot concentrated aqueous $\ce{NaOH}$ - with particular concern for avoiding dissolution of borosilicate glassware and contamination of the $\ce{NaOH}$ solution and other reagents - led me to think of two possible solutions:

Coat the glassware with a silane, providing say a polyfluorinated or other inert protective coat
Avoid using glass, choose instead metal or plastic, if possible

Is either of these choices reasonable or common, or is there a better alternative? If so, which is best (considering chemical inertness, safety, complexity (including potentially cost), in that order of priority)?
In the case of choice 2, what commercial sources are there for such labware (for handling hot caustic solution for extended period). A cursory search - eg google and Merck website - did not turn up such products.
Note this question is about relatively small-scale labware (benchtop), not for reactors and the like. Volume max $\pu{5 L}$, $\pu{1 M}$ $\ce{NaOH}$, $\pu{>100^oC}$, exposure time >1 h.

Even a weak alkali solution should not be stored in glass or quartz vessels for an extended time as it will inevitably react with any glass ("hard" borosilicate glass reacts slower, but still). For a hot alkaline solution exposure should be limited to minutes, maybe an hour if the glass is thick enough. Old lab textbooks suggested to coat glassware used for storing alkali solutions from the inside with paraffin; these days teflon seems to be better alternative, as you suggested.

@andselisk Thanks. Thermo-fisher offers plastic-coated glassware ("kimcote") but that's not ptfe and not sure it would hold up. I find it strange ptfe coated products are not more prominent in product catalogues. A websearch shows me hits that cater mostly to manufacturing, not labs.

TBH, I think we never bought teflon labware as it was always ridiculously overpriced. It was always cheaper and faster to craft own reactors with a lathe from a raw piece of teflon which is quite cheap (speaking of central Russia).

Instead of glassware, you might have to think about metallic containers. A 1 M NaOH solution should not be that harsh for stainless steel for a short period of time. Imagine, in the chlor-alkali industry, nothing is made of glass. The electrodes are made of titanium as far as I remember. If we were rich, a platinum lined crucible would have worked! Titanium might work too, very resistant to bases.
Here is a useful link: Selecting Stainless Steel for Caustic Soda
Regarding silanes, the Achilles heel is not the perfluoro functional group on the silane but the siloxane bond itself- i.e., the Si-O-Si ---- (functional group). Base will chew this bond apart into pieces even at pH 8-9. This is the biggest drawback of HPLC silica columns. We have not been able to synthesize any base stable silica bonding.

Just remember not to melt alkali in a Pt crucible.

I am not sure if you have seen Hildebrand's Applied Inorganic Analysis. Amazing book on crucibles choices.

No, I haven't yet, but eventually I found 1940 ed. of Hildebrand's Reference book of inorganic chemistry which also says the same, quoting from p. 415: "Platinum … is almost indispensable in the manufacture of chemical utensils for high temperature ignitions. Such ware, however, must be handled with some care since it is attacked by a number of reagents, e.g. aqua regia, chlorine solution, ferric chloride, and fused alkalies. It alloys with many metals, especially lead, tin, bismuth, and mercury; and unites with carbon, phosphorus, sulfur, and silicon, becoming brittle."

Found a soft copy from the great Internet Archive. My original copy of Applied Inorganic Analysis is sitting in another country. Cannot find it anywhere now. Pretty rare book of the 1940s.

@andselisk  not all alkalis are unsafe in a Pt crucible. Carbonates are just fine. I melt sodium carbonate in Pt crucibles all the time. Halides are often ok as well. But yea- alkali hydroxides are a completely different thing and will violently destroy the Pt.

@Gimelist Carbonates are fine because they are not alkali.

If we wish to be pedantic, alkali is not necessarily a hydroxide or a caustic soda. It is a very broad term. The Arabic word from alkali is derived means soda or potash (OED). Geologists use it differently I guess. The original meaning of alkali as given by the OED is " saline substance, typically consisting largely of sodium carbonate, extracted from the calcined ashes of saltwort (genus Salsola), glasswort (genus Salicornia), or other plants; soda ash. Cf. soda n.1 1a. Now hist. except as passing into sense"

Thanks for your answer. I chose the term "aqueous NaOH" to avoid ambiguity, although the original post that inspired my question is ambiguous about concentrations and temperature. Seems my question ended up being ambiguous too.

I think your question was clear enough about 1 M NaOH. Fusing NaOH was another side topic.

@andselisk they are in geochemistry. Interesting how the usage of terms differs between various fields.

@Gimelist Yes, indeed, you are correct. There are some quirks when it comes to the fine details in terminology between fundamental chemistry and geo-sciences. If I may ask, could you please suggest a good reference book/style guide or set of standards that are being used among geochemists these days, if there are any?

I don't think there is any authoritative source. You guys have IUPAC, we don't. Sometimes nomenclature differs between subfields in geochemistry, or even labs doing the same thing! There are geological dictionaries, but they don't often contain all terms used in geochemistry. That said, I think any undergraduate-level modern geochemistry book would cover most of it.


text:
Extended Hückel method: what are the atomic orbital energies usually used?
I was checking out an extended Hückel (non-iterative) code, and realised that for s-block elements it would be impossible to simply use atomic ionisation energies as an approximation to the orbital potential energies (due to the p orbitals being empty). In this case, what values are used?

Please see the following reference:

A. Herman, Modelling Simul. Mater. Sci. Eng. 2004, 12, 21.

It has $H_{vv}$ terms for different orbitals of all elements of the table.


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Preparation of Effexor
I take venlafaxine (Effexor) as medication, and having wrapped up 2 semesters of Organic Chemistry I thought it might be fun to design a synthesis for it. A picture of venlafaxine:

I came up with the following preparation (beginning with phenol) and was hoping someone could check it for me:

React phenol with $\ce{NaOH}$ to form phenolate, followed by methyl iodide to form methoxybenzene.
This forms a moderately activated ring with ortho/para direction.
Friedel-Crafts alkylation: allyl bromide (3-bromo-1-propene) and $\ce{AlBr3}$ adds propene to the para position.
Ozonolysis with DMS to give p-(acetaldehyde)methoxybenzene.
Reductive amination with dimethylamine and triacetoxyborohydride.
Radical bromination should be selective for the benzylic position.
Finally, Grignard reaction with cyclohexanone.

Don't really need your first step as anisole is readily available

I suspect the product from 5 is going to ring close to the aziridinium.

@Zhe Thanks! I take it that means the nitrogen would attack and displace the bromide? I'll try and change the order to make amination follow the Grignard reaction if I can. Would the oxygen in the carbonyl compound do the same if I brominated the benzylic position after its formation in ozonolysis?

Unlikely. Yours is the basic problem of trying to figure out functional groups (and protecting groups) that have orthogonal reactivity.

@Zhe Thinking of that now though I would then have the problem of it interfering with the Grignard. What if I changed the order to 1,2,5,6,3,4?

4-anisaldehyde is pretty cheap and available if that helps reduce the use of conversions with scale up issues.

@ Beerhunter: I think i is a great idea. p-Anisaldehyde is available from Aldrich for  117.00 USD/1 kg.

@Zhe: Ring closer to the aziridinium would not happen, because at step 5, OP has (dimethylamino)methyl group instead 2-(dimethylamino)ethyl group. So, OP need more work before radical bromination.

@MathewMahindaratne I don't think so. OP ozonlyzed a allyl group, so they added two carbon atoms to the backbone.

You may not necessarily need to start your procedure beginning with phenol. You may start it with readily available (for cheap price compared to the trouble you are going to have by handling methyl iodide, allyl bromide, and $\ce{AlBr3}$ or $\ce{AlCl3}$) 4-anisaldehyde (4-methoxybenzaldehyde). That is the exact product you are getting after your step 3. Although, steps 4-6 are some very thoughtful organic synthesis steps, I doubt it will give you the expected product. Therefore, I suggest following steps to follow:

Buy 4-anisaldehyde (4-methoxybenzaldehyde) from Sigma-Aldrich (available in 98% purity.
Prepare Grignard reagent using chlorocyclohexane/$\ce{Mg}$ in diethyl ether.
React Grignard reagent (cyclohexyl magnesium chloride) with 4-anisaldehyde.
Perform dehydration of resultant $2^\circ$-alcohol to give double bond.
Perform epoxydation of double bond with 3-chloroperbenzoic acid.
React epoxide with nitromethane.
Reduce nitro group to $1^\circ$-amino group.
Finally, convert $1^\circ$-amino group to dimethyl amino group by reductive amination with formaldehyde.

Cyanide may be a better nucleophile to use than nitromethane. Certainly the reduction will be easier, reductions of alkyl nitro are notoriously fickle

I agreed. That is an one excellent suggestion. I suggested nitromethane because I have an experience with that during synthesis for inhibitors for DBM (dopamine-$\beta$-monooxygenese).

@Waylander my thoughts exactly which is where the anisaldehyde suggestion originated


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How do I solve such questions on paramagnetism and ferromagnetism?

Match the type of magnetism given in Group I with the material given in Group II:
$$
\begin{array}{ll|ll}
\hline
 & \textbf{Group I} & & \textbf{Group II} \\
\hline
\text{P} & \text{Ferromagnetic} & 1 & \text{Nickel oxide} \\
\text{Q} & \text{Ferrimagnetic} & 2 & \text{Sodium} \\
\text{R} & \text{Antiferromagnetic} & 3 & \text{Magnetite} \\
\text{S} & \text{Paramagnetic} & 4 & \text{Cobalt} \\
\hline
\end{array}
$$
A) P-4, Q-3, R-1, S-2
B) P-4, Q-1, R-3, S-2
C) P-1, Q-2, R-4, S-3
D) P-3, Q-2, R-1, S-4

It was taken from a mock test sheet for GATE in India.
With regard to whether a material is ferromagnetic or paramagnetic etc. All that I know is to check the magnetic domains and decide. However, with regard to such compounds I have been facing extreme difficulties. Is this something that is totally experimental and that I need to know by information? Or is there a certain intrinsic property of the elements which can help me to decide? If yes, please explain how to solve such questions.

Ferromagnetism is not something you can deduce from first principles.

Basically, you have to memorize which is which.

I believe GATE is a university entrance exam in India, so they will not expect you to solve an extremely complicated equation to predict ferromagnetism or to memorize an infinite list of substances. Sadly, there is some element of rote memorization still lingering like a pest in the educational testing system.
Good news for you is that relatively few compounds and metals show ferromagnetism or ferrimagnetism. Find out the list of ferromagnetic metals from the web, you might end up with very few elements and oxides, such as iron compounds, iron oxide containing materials. Later you can attempt the multiple choice questions by the process of elimination. For example, in the given choices, choose the answers which show that magnetite is ferrimagnetic.
Lo and behold, there is only ONE choice which matches with Q-3!

Historically, the term ferromagnetism was used for any material that could exhibit spontaneous magnetization: a net magnetic moment in the absence of an external magnetic field (Wikipedia). However, in 1948, Louis Néel showed that there are two levels of magnetic alignment that result in this behavior (Ref.1):

One is ferromagnetism in the strict sense, where all the magnetic moments are aligned A few examples of this type are $\ce{Fe}$, $\ce{Co}$, and $\ce{Ni}$.
The other is ferrimagnetism, where some magnetic moments point in the opposite direction but have a smaller contribution, so there is still a spontaneous magnetization. The oldest known magnetic substance, magnetite ($\ce{Fe3O4}$), which contains both iron(II) and ion(III) oxides is a well-known ferrimagnet ($\ce{Fe3O4}$ was originally classified as a ferromagnet before Néel's discovery). The other cubic ferrites composed of iron oxides with other elements such as $\ce{Mg}$ (e.g., $\ce{MgOFe2O3}$), $\ce{Cu}$ (e.g., $\ce{CuOFe2O3}$), and $\ce{Ni}$ (e.g., $\ce{NiOFe2O3}$) are also examples of ferrimagnets.
Néel had discovered a third level of magnetism called antiferromagnetism. In a special case where the opposing magnetic moments balance completely, the alignment is known as antiferromagnetism; but antiferromagnets do not have a spontaneous magnetization. A well-known antiferromagnets are common Iron oxide, hematite ($\ce{Fe2O3}$), the trasition metal oxides such as nickel oxide ($\ce{NiO}$), and alloys such as iron manganese ($\ce{FeMn}$).

All of these three levels of magnetisms have a kryptonite temperature, above which all of them become paramagnetic. For ferromagnetism and ferrimagnetism, it is called Curie temperature while for antiferromagnetism, it is Néel temperature.
Thus, M. Farooq gave an excellent way to answer your question for novices. Mostly, known ferromagnets are metals and metal oxides are usually ferrimagnets, except for $\ce{NiO}$, which is a known antiferromagnet. You have two metals and two oxides in RHS column. Now, it is easy to figured out.
References:

M. Louis Néel, "Propriétés magnétiques des ferrites ; ferrimagnétisme et antiferromagnétisme (Magnetic properties of ferrites: ferrimagnetism and antiferromagnetism)," Annales de Physique (Paris) 1948, 12(3), 137-198 (https://doi.org/10.1051/anphys/194812030137).


text:
How to prove it mathematically that either rhombic or the monoclinic sulfur is denser?

For the transition, rhombic sulfur → monoclinic sulfur, the value of $ΔS$ is positive. The transition temperature increases with increase in pressure. Which is denser, the rhombic or the monoclinic form? Prove your answer mathematically.

I tried by using Clapeyron equation but still failed. How should I approach and solve it? This is so tricky.

$$\frac{\mathrm dP}{\mathrm dT} = \frac{\Delta_\mathrm{trs}S}{\Delta_\mathrm{trs}V}$$
$\Delta_\mathrm{trs}S$ is positive. $\displaystyle\frac{\mathrm dP}{\mathrm dT}$ is said to be positive as well. Thus, $\Delta_\mathrm{trs}V$ has to be positive, i.e. monoclinic sulfur has a greater molar volume, and thus rhombic sulfur is denser.

Note that IUPAC recommends to denote "transition" with upright "trs" (subscript "tr" refers to either  "transmitted" or "triple point") as subscript and there is no analogy to superscripted "transition state" notation (e.g. "≠" or "‡").


text:
Question on ring expansion

In the orange box, I think that the blue-carbon should be a carbocation instead. So, option A should be correct.
But, the answer is B. Why is it so?

migratory aptitude in cationic reactions is secondary > primary > methyl because the migrating carbon has to bear some of the positive charge in the transition state, see e.g. https://chemistry.stackexchange.com/a/16470/16683

Thanks @orthocresol, actually I was starting the migrating-$\ce{C}$ from that quaternary carbon and so I was confused :P

As orthocresol commented (vide supra), migratory aptitude in cationic reactions has been given as: $\ce{H}=\mathrm{Ph} \gt \ce{(CH3)3C \gt (CH3)2CH \gt CH3CH2 \gt CH3}$ (Wikipedia and Migratory aptitude in pinacol-pinacolone rearrangement). In your question, the secondary carbocation formed when protonated has three groups, which can migrate to give more stable tertiary carbocation. They, with their migratory aptitude, are:  $\ce{-R^1CH2(CH3)CH\!- \gt R^2CH2CH2\!- \gt H3C\!-}$ (where $\ce{R^1}$ and $\ce{R^2}$ are continuation of rest of cyclopentane ring).
Now, if  $\ce{H3C\!-}$ group migrated, it'd give option (D); if  $\ce{R^2CH2CH2\!-}$ group migrated, it'd give option (A); and if  $\ce{R^1CH2(CH3)CH\!-}$ group migrated, it'd give option (B). If nothing migrated you have option (C). Although all 4 products are possible in the reaction mixture ((C) would be very trace amount), asking is the major product, and hence is is the product given but migration of group with highest migratory aptitude. Therefore, it should be the product given in option (B).


text:
Strength of bases and acids
Is it possible that strong acids like $\ce{H2SO4}$ behave as weaker ones like $\ce{CH3COOH}$ when the concentration is really high and vice versa for weak acid? And the same question for bases.
By this I am generally referring to pH values.
Why do concentrated acids react more than dilute ones? The degree of ionization of concentrated acids are lower than diluted acids, so they are supposed to react less. Where am I wrong?

Actually, strong acids can only partly dissociate (e.g. act as weak acids) when the concentration is really high (see the Debye-Hückel equation)

Sorry, I'll edit it. However, during titration, the pH of acid solution increases slightly before reaching end point if we add base, doesn't it mean that by lowering the concentration, the strength decreased. Aren't degree of ionization and pH the similar things?

@Raphaël Degree of dissociation and pH can describe properly only diluted solutions. Actually dissociation of strong acids only weakens them as undissociated acid is stronger the H3O+.

Actually, yes, concentrated sulfuric acid passivates iron, chromium and aluminium and thus may appear as  weaker acid.

Also never forget the concept of solvo-acids and solvo-bases. In a system, say water, the solvo-acid, so the protonated form H3O+ is the strongest acid. All other acids added to water will only form H3O+. Which means that if you system was ammonia, something like acetic acid could be considered a super-acid. If we reduce the amount of H2O in there to basically zero our solvo-acid becomes the protonated form of our acid like H3SO4+. But if no stronger acid is present that can protonate H2SO4 it's up to the autodissociation of H2SO4, hence a low H3SO4+ concentration.

If you find effects like passivation in cocentrated acids it's also likely that the compound formed on the surface is protecting the metal while water tends to dissolve it, so without water it remains almost untouched. If you compare the famous acid + carbonate volcano, when using 99% acetic acid it will be much slower than with diluted acetic acid or vinegar as there are less protonated species available.

In the case of weak acids dissociating like strong ones when they are highly dilute, there is a limit to how far this can go.  If you were to try to measure acetic acid dissociation at $10^{-n}$ molar for $n=4,5,6,7,8,...$ at ambient temperature you would at first rise nicely towards $100$% as you might expect, but eventually level off at only about $99.4$%.
What stops you is the autoionization of water.  As you approach infinite dilution the hydrogen ion concentration approaches not zero but the autoionized value of $10^{-7}$ molar.  So the ratio of dissociated acid to undissociated acid levels off at $K_a/10^{-7}$.  Accordingly the limiting degree of dissociation is
$1/[1+(10^{-7}/K_a)]=K_a/(K_a+10^{-7})$
For acetic acid with $K_a\approx1.8×10^{-5}$ this limit is about $99.4$%, but for a much weaker acid such as a typical phenol or hydro-cyanic acid the dissociation is much lower even at extremely low concentrations.  When $K_a<10^{-7}$ the limiting dissociation is less than $50$%; you need a stronger base than water to coax dissociation as a major reaction in aqueous solution.

A strong acid will always have strong characteristics. What makes an acid strong or weak is the minimum pH required to make it's acid hydrogen (the hydrogen that gets ionized in the acidification of the solution) ionized. The stronger the acid the lower the minimum pH, thus making it possible to make a more acid solution (hence the name strong). Same things for bases, but swap the acid hydrogen by the basic OH.
It is not right that diluted acids ionize less. It ionizes by the same amount (acid's kpa) but the solvent makes the average ionization smaller.
For this reason weak acids make tamponing solutions. When they get near their minimum pH they start acting like a base, deionizing the solution as it gets more hydrogen ions, holding the pH.

The characteristic property of all acids and bases is the pH. Their characteristic general reactions are the acid-base reactions. Let's concentrate on this kind of reactions. We don't want to consider here reactions of particular acids and bases (like e.g. oxidation, passivation, complexation).
We have to distinguish between dissociation degree of an acid and acidity (acid strength) of this acid. The degree of dissociation depends on the initial concentration, the acidity not. The acid constant of an acid is a measure for its acidity. An acid is characterized therefore by its acid constant. Acids are grouped into strong acids, medium strong acids and weak acids according to their acid constants.
You are right, higly concentrated strong acids can have lower dissociation degree like weak acids.
But for the behavior of an acid in an acid-base reaction, the reaction velocity is crucial. It depends on the hydrogen ion concentration in the solution. The measure for this concentration is the pH of the solution.
Highly diluted strong acids can have the same pH like less diluted weak acids and therefore the same reaction velocity in acid-base reactions.
$\ $
The analog holds for bases.


text:
What are the factors affecting osmosis (bio)?
Asking from the blood barrier angle. The wikipedia states that glucose, urea, and $\ce{Na/K}$ count. Can there be any other factors, like $\mathrm{pH}$, temperature, hydraulic pressure on the cell walls? Why do minerals not play a role? Because there are only traces of them?

I'm a little confused about your question. Do Na/K not count as "minerals"? And no, "minerals" are not generally present as traces in plasma (some may be, depending on what you mean by trace and by "mineral").

Do you mean "do any other factors, like pH, temperature, and hydraulic pressure on the cell walls, affect the osmotic pressure?"?

Beware dangling participles... https://www.dictionary.com/browse/dangling-participle

https://en.wikipedia.org/wiki/Blood_plasma#Composition

Osmotic flow is the movement of a substance (typically water), across a semi-permeable membrane, from the side on which the chemical potential of the substance (water) is higher to the side on which it is lower.  I.e., just as temperature provides a gradient for the flow of thermal energy, chemical potential provides a gradient for the flow of matter.
Thus anything that affects the chemical potential of water will affect osmotic flow.  This includes dissolved solutes, pH (which of course comes from a particular dissolved solute—H$^+$), temperature, and pressure. For instance, increasing the solute concentration decreases water's chemical potential, while increasing the pressure increases it.
Indeed, that's the basis of osmotic pressure—water will flow toward the side with the greater solute concentration (activity) until the pressure on that side increases enough such that the reduction in water's chemical potential due to the presence of the dissolved solutes is exactly negated by the increase in water's chemical potential due to the increased pressure.


text:
Why can't carboxylic acid + NaH + tert-butyl bromide react to create an ester?

So, the picture I showed above asks for which reagents can form the ester product. I was wondering why D can't form it? I was thinking that first, the $\ce{NaH}$ would react with the carboxylic acid to form the basic form of the carboxylic acid (where the $\ce{-OH}$ becomes $\ce{-O^-}$). Then, the $\ce{-O^-}$ of the carboxylic acid would attack the tert-butyl bromide in a $\mathrm{S_N1}$ reaction where the bromide would leave first. Why is this not possible?
Also, does F not work because Fischer esterification does not occur in basic/neutral conditions?

For conditions F, all you are doing is deprotonation of the benzoic acid with a strong base. Carboxylate is a poor nucleophile because of the delocalisation of the charge. Conditions B is a much better source of tBu+ than tBu-Br.

For condition D, the carboxylate will effect E2 elimination of the tertiary bromide..


text:
Why doesn't the expermintal molar solubility match the one calculated by the Ksp value?
For $\ce{HgCl2}$ ($M = \pu{271.52 g mol-1}$), the reported $K_\mathrm{sp}$ value is $10^{-15.69}$.
This means that the calculated molar solubility of $\ce{HgCl2}$ in water is $(10^{-15.69}/4)^{1/3}\cdot\pu{271.52 g mol-1} \approx \pu{0.0010 g L-1}$, or equivalently, $\pu{0.010 g cm-3}$.
However, in Pubchem, the reported molar solubility is 6.9 g/100 cc. What is the source of the large deviation between the experimental and the theoretically calculated molar solubility of $\ce{HgCl2}$?

It does match, you're confusing HgCl2 - soluble and Hg2Cl2 - almost insoluble.

What do you mean. Both of my data values are for $\ce{HgCl2}$.

Just use, at very least, Wikipedia instead of some obscure and dubious sources.

@Mithoron bruh Wikipedia doesn't have Ksp data for it smh

The $\mathrm{p}K_\mathrm{sp}$ value of $15.69$ for $\ce{HgCl2}$ given by your reference must have an error. First of all, $\ce{HgCl2}$ is fairly soluble in water at $\pu{20 ^\circ C}$. Wikipedia lists it as $\pu{65.7 g L-1}$, yet lists solubility of $\ce{Hg2Cl2}$ as $\pu{3.25e-4 g L-1}$. Meantime, I found an University of Arkansas, Little Rock website, which has listed $K_\mathrm{sp}$ value of $\ce{Hg2Cl2}$ as $\pu{1.43e-18}$ ($\mathrm{p}K_\mathrm{sp} = 17.84$).
The molar mass of $\ce{Hg2Cl2}$ is $\pu{472.09 g mol-1}$. Thus, theoretically, the solubility ($s$) of $\ce{Hg2Cl2}$ can be calculated similar way you did:
$$s = \left({\frac{\pu{1.43e-18 mol3 L-3}}{4}}\right)^{\frac13}\times \pu{472.09 g mol-1} = \pu{3.35e-4 g L-1}$$
I think this is close enough. :-)


text:
Why can Neutral Red (indicator) not test values below its pKa?
It is the only indicator I have found that does not test below its pka, and can't think of any reason for this.
Neutral Red has a pKa value of 6.8.
(
Reference 1,
Reference 2
),
but looking at its pH testing range, it only shows from pH 6.8-8.0 (Reference 3)
What would be the reason for this?
Thank you

When comparing different tables of data you likely will find inconsistencies. Such inconsistencies are impossible to track down since many references  don't give sources of the data.

@MaxW Yes but every pKa value for neutral red I have found is 6.8 (within a few decimals accuracy), and every pH range for it I have found is 6.8-8

I didn't try to solve your mystery, but rather read your statement that you had two sources that said pKa was 6.8 but a third that said pH range was 6.8-8.0.

What do you mean by "cannot test below"? The pH range you cite is the transition range. Neutral red is yellow above, red below, and orange-y inbetween.

The colour of neutral red in aqueous solution changes from violently red (protonated cation) to yellow/mostly colourless (the uncharged amine).
If the $pKa$ is 6.8, then the dye is still very much red at pH 6.8, and you won't be able to see the colour difference to, say, pH 5.
You have to keep adding base up to a pH of eight until the concentration of the colourful cation is so low you cannot see it any more.
Some other indicators change from one strong colour to another, there the $pKa$ will be more in the middle of the transition range.  Or it is in fact a mixture of two indicators.


text:
Mechanism of Bis-decarboxylation of α,β-dicarboxylic acid
According to an internal test series in FIITJEE, it was given that the following compound decarboxylates on heating,

I know that β-keto acids undergo decarboxylation on heating, but I was not sure about the above compound. So, on searching the internet, I found this post, Why are beta-ketoacids better at decarboxylation than gamma or delta? which seems helpful.
As, it says that for γ-keto acids or δ-keto acids, decarboxylation will most likely not happen with a beautiful explanation. So, I drew a possible mechanism and here's what I came up with:

With the help of the reasoning given by @TOLA3HPPA, I can say that decarboxylation will not happen as opposite charges are much separated, but I'm not sure about the mechanism.
Further, while searching on NIST Chemistry Webbook, I've seen an article named,

The Oxidative Bisdecarboxylation of α,β-Dicarboxylic Acids, J. Am. Chem. Soc., 1952, 74, 4370..

which possibly indicates that the above compound undergoes decarboxylation. So, if α,β-dicarboxylic acid undergoes decarboxylation, then what can be the possible mechanism?

Possible but improbable mechanism. If your final structure is electroneutral (: = negative charge), how do you bond 2 hydrogens to the same carbon? In my experience decarboxylation doesn't occur although anhydride formation may.

I agree that this is improbable as explained in related post. And, the 2 hydrogens were added automatically by ChemDoodle, the software I used : j

So much for ChemDoodle!

Don't forget to manually add any charges to atoms so the implicit hydrogen count can be calculated correctly, such as in this case. Would be cool if ChemDoodle could automatically determine implicit hydrogens based on electron pairs if charges aren't provided though.

IMO this would form an anhydride on heating, and woudn't decarboxylate.


text:
Mechanism for Synthesis of Tetranitromethane from Acetic Anhydride and Fuming Nitric Acid
In my undergraduate Mechanisms-class the formation of tetranitromethane from acetic Anhydride with fuming $\ce{HNO3}$ was listed as optional knowledge in a chapter about the reactions of enols, enamines, etc. I can't find it in my notes or in the recommended literature, neither can I find information on molbase, and the two papers I can find, the original 1910 synthesis and This, are not mentioning any reaction mechanism.
I guess it somehow decomposes at the stage when the acetic anhydride is tri-substituted on one side, and somewhere there is a trinitromethane-anion involved which forms the product with $\ce{NO2+}$?

Yeah, probably something like that.

The Organic Synthesis preparation you cited for the preparation of tetranitromethane has some clues that are useful in providing a viable mechanism. Upon slowly adding acetic anhydride (Ac2O) to anhydrous HNO3 with cooling an exotherm is produced. The reaction mixture is allowed to sit at room temperature for 7 days. Assuming that they investigators did not go on vacation for a week, I presume that the formation of tetranitromethane (TNM) is slow and that the exotherm is due to the formation of the mixed anhydride 2. [The nitronium cation is also a possible electrophile as well as the mixed anhydride 2.] The enol of acetic anhydride can undergo substitution to afford species 3 and then repetitively on to the trinitro mixed anhydride 4. Anhydride exchange with acetic acid (or nitric acid) affords trinitroacetic acid 5 which decarboxylates to enol 6 (aci-form). Tautomerization of 6 can form trinitromethane. Footnote 3 in the Organic Synthesis link mentions that trinitromethane 7 can be converted to tetranitromethane 8 with HNO3. The transformation of 6 to 8 does not preclude nitration via mixed anhydride 2.


text:
Why does half-reaction with largest positive potential will run as written while other is reversed?
While balancing the half redox reaction, the textbook indicated that the flip of equation is dependent on the size potential. But I do not understand the reason why it is like that. Please explain for me.

Please update your question with an example. What do you mean by "flip of  equation"? What potential?

Following @Karl , it would be helpful for you to show a fully-worked example from your textbook, complete with half reactions, standard cell potentials, net reaction, full cell potential and free energy change, with correct algebraic signs throughout.


text:
Effect of Synergic Bonding on Bond Lengths in Metal Carbonyls
Consider the metal carbonyl, pentacarbonyl iron(0). There exists synergic bonding between the central metal atom, iron and the neutral carbonyl ligands. This strengthens the metal-ligand bond but because electron density from the filled orbitals of the Iron atom is pushed into the anti bonding orbital of the $\ce{C-O}$ bond, it is weakened thereby increasing bond length. My question was to what extent does the $\ce{C-O}$ bond length increase. Also, is this increase uniform? The reason for my doubt was, the compound has a trigonal bipyramidal type geometry. So the axial metal-ligand bonds are weaker than the equatorial ones. I felt that synergic bonding would not take place to the same extents in the axial and equatorial case - the synergic bonding would be stronger in the equatorial case and weaker in the axial case. Is that so?


text:
What are the constraints on the two electron integral matrix?
In semi-empirical methods, the two electron integral matrix is simplified and the entries are replaced by actual values instead of integrals to evaluate. However, while doing that, the variational principle does not seem to directly hold; in other words, you can put values in the two electron integral matrix that make the energy go down infinitely. To make the variational principle hold, what are the contraints that the constants you put in the two electron integral matrix obey?

General Answer:
The basic premise of semi-empirical methods is to replace the computationally expensive integrals with values fit to experimental data or, in some cases, to high level ab initio calculations of the relevant quantity [1]. You are completely correct to point out that making this substitution will break the theoretical validity of the variational principle.
The short answer to your main question then is that the constraints are given by the experimental values to which one fits. That is to say, if you fit the values of your integrals to experimental data and all of a sudden you get out energies which are unbounded, discontinuous, or have some other undesirable and un-physical feature, then the problem is either in the fitting or in the corollary between the experiment and integral you are fitting. So, as in many physical contexts, the answer is circular. Sure, we could have any value for the integral, but the answers which experiments give live in a very small region of all possible answers to the question.
Approximations:
First of all, there are tons of semi-empirical methods which differ slightly in how they do the fitting, what approximations are made, etc. One important thing to note is that no semi-empirical method is really claiming to get the physics of these interactions correct for the right reason. They all rely on cancellation of errors, but this is built into the method on purpose, so this cancellation tends to be rather robust. One can achieve this because the behavior of the integrals arising in electronic structure is quite systematic, so if you are careful to know what you neglected in one part, you can try to neglect something of the opposite sign in another part. Reading ref. [1] is a good place to start to see how this is done in practice.
DFTB:
I should also point out that density function tight binding (DFTB) is often described as a semi-empirical method, but DFTB does obey a variational principle[2, 3]. That is, one can expand the total energy from regular density functional theory with respect to variations of the charge density (the central object in DFT). To zeroth order, one gets a non-iterative theory which depends only on properties of the individual atoms in the molecules and then corrects for interatomic repulsions. This is trivially variational.
If one includes terms up to second-order in the density variations, then you get self-consistent-charge DFTB (SCC-DFTB). This still defines an approximate Hamiltonian, so one can variationally solve for the energy. So, you might ask why this is considered semi-empirical. I'll just quote from ref. [2] as they give a nice summary.

Comparing SCC-DFTB with MNDO-type methods, the
  underlying philosophy and the derivation of the applied approximations seem quite different at first sight. If one focuses
  on the actual working equations and their implementation, there
  are, however, many similarities: All these methods are valenceelectron SCF-MO treatments with a minimal valence AO basis
  set, only one-center and two-center terms are included, the twocenter two-electron integrals are represented by damped Coulomb interactions with the correct limit at large and small
  distances, and there are repulsive atom-pair terms that correct
  for deficiencies in the formalism. The methods are semiempirical
  in the sense that they employ empirical parameters which are
  adjusted to reproduce selected reference data, i.e., DFT energies
  and potential curves in the case of SCC-DFTB, and experimental
  heats of formation and other experimental data in the case of
  the MNDO-type methods.

Hopefully I've addressed your main concerns in the original question and left you with some resources so you better approach further questions.
References:
[1]: Thiel, W. (2014). Semiempirical quantum–chemical methods. Wiley Interdisciplinary Reviews: Computational Molecular Science, 4(2), 145-157.
[2]: Otte, N., Scholten, M., & Thiel, W. (2007). Looking at self-consistent-charge density functional tight binding from a semiempirical perspective. The Journal of Physical Chemistry A, 111(26), 5751-5755.
[3]: Seifert, G. (2007). Tight-binding density functional theory: an approximate Kohn− Sham DFT scheme. The Journal of Physical Chemistry A, 111(26), 5609-5613.


text:
Can the Galvanic cell use a different ionic solution?
It makes sense that
water-soluble salt when dissolved in water, they conduct electricity. Then when using the old method of the Volta Battery(using copper and zinc) with a different water-soluble salt dissolved in water, can it make electricity too?
Plus, If I use a ammeter to check its current, then does all of the water-soluble salts do not go to zero?

Technically Yes it can, given the solution itself dosen't react with electrode and form a layer around it, which usually happens with almost all the solutions, So practically No ! I guess..

Why does a layer form around it?

Because ions in solution usually react with molecules of electrode and form some other neutral compound which inturn get deposited on electrode, thus forming a layer around it.

See also: https://en.wikipedia.org/wiki/Lemon_battery

Can the Volta Battery use a different ionic solution?

Yes you can use any soluble salt as long as it does not react with electrodes, and certainly such a battery would be able to produce current. My chemistry high school teacher had a clock which ran on salt water. See for example, https://science.howstuffworks.com/environmental/green-tech/sustainable/water-powered-clock2.htm. This technology is at least 400 years old. volta used common salt as well for zinc and copper.
Now the problem is that you can generate voltage by using copper/ zinc and salt water however one cannot predict the voltage by using the Nernst equation. It requires that the electrode dip in its own ions. Hence such batteries are theoretically difficult to deal with.


text:
Why methoxy group acts similar to trifluoromethyl in nucleophilic aromatic substitution
In an online test, the following question was asked:

The answer to this question was:

I totally agree with the answer, as $\ce{OCH_3}$ is an EDG, hence it will tend to form meta product in this reaction.

While reading NAS on Master Organic Chemistry, I came up to this article:

This article says that $\ce{OCH_3}$ behaves similar to $\ce{CF_3}$, but I can't figure out the possible reason. Any help is appreciated!

Related (unanswered) question: Nucleophilic substitution in aromatic compounds

although the use here is technically correct, usually "nucleophilic aromatic substitution" refers to SNAr mechanism. 2) for a benzyne mechanism, resonance effects do not matter, only inductive. This is covered in typical organic chemistry textbooks.

Normally during a nucleophillic aromatic substitution reaction one of oxygen's lone pairs would delocalize across the ring, the existence of a triple bond here prevents this from happening so oxygen actually ends up acting like an EWG.


text:
Why do we need to use oxidation number in order to balance a complex chemical reaction easily in an acidic or alkaline medium?
I have been trying to balance this chemical reaction in an "acidic" medium without using oxidation number or ion electron method: [$\ce{Fe^2+ + Cr2O7^2-  -> Cr^3+ + Fe^3+}$].
Though I can balance number of atoms on both the sides $[\ce{Fe^2+ + Cr2O7^2- + 14H+ -> 2 Cr^3+ + Fe^3+ + 7 H2O}]$, but charge is still unbalanced. I am not sure why? I don't have any intuition about what's going on here.

I think you mean $\ce{Cr2O7^2-}$.

This gives a pretty good step-by-step guide on how to go about the task: https://www.ck12.org/book/CK-12-Chemistry-Concepts-Intermediate/section/22.10/

$\ce{6Fe^2+ + Cr2O7^{2-} + 14H+ -> 2 Cr^3+ + 6Fe^3+ + 7 H2O}$

You need not to use oxidation number, they are just convenient guidance. Good chemists can balance equation by their intuition, as they can easily see the proper way to do it. Other need to follow 2 fundamental rules of nature:  1/ Law of the total charge conservation 2/ Law of the element mass conservation.

A balanced equation must have a mass balance i.e., the masses should be equal on both sides and charges must be balanced as well. Oxidation number is just a way of book-keeping. Nothing fundamental there.
$$\ce{Fe^2+ + Cr2O7 + 14H+ -> 2 Cr^3+ + Fe^3+ 7 H2O}$$
What are you forgetting? It is mass balanced. Does $\ce{Cr2O7}$ exist? Hint: It is potassium dichromate.


text:
Fick's second law atomic diffusion question
I have a question about diffusion and i'm not sure how to tackle it.

A new processing method is being tried for a lithium battery cathode material -LiCoO2. the process leaves the ceramic oxygen deficient, which is causing problems. to counter this problem the LiCoO2 is annealed at low temp (400 K) in oxygen rich atmosphere to encourage oxygen take up. this generates a localised atmosphere of oxygen at the surface of $1.16\times 10^5 \pu{mol/m^3}$.
i. The initial concentration of oxygen in the cathode material is $9.64\times 10^4  \pu{mol/m3}$. The diffusion constant is $4.92\times 10^{-7} \pu{cm^2 s^-1}$ for the oxygen in the cathode material. calculate how long the material must be annealed for it to reach the desired concentration of $1.04\times 10^5 \pu{mol/m^-3}$ across the whole sample (which is 5 mm in depth).
ii. The oxygen is though to primarily diffuse into the material via a grain boundary diffusion mechanism, draw a diagram showing how the oxygen concentration would vary across the material and around the grain boundaries."

For part i. I have tried using $$c(x,t)=\frac{A_0}{\sqrt{\pi Dt}}\exp(-x^2/4Dt)$$ but ended up with negative time which doesn't make any sense, I have also tried solving it using error function table $$c(x,t)=0.5A_0[1-erf(\frac{x}{2 \sqrt{Dt}})]$$ but ended up with negative time too (different value)
Could someone advise me which equation to use

Your concentrations seem to be wrong, should they not be $10^{-5}$ instead of $10^{+5}$? The second equation apples to diffusion across an interface and is the one to use assuming that outside the cathode the oxygen concentration falls. If, instead, it should be held constant then you need to solve Fick's 2nd Law equation with these boundary conditions.  The solution will be a constant plus a sum of sines in $x$  multiplied by an exponential in time.


text:
How can I calculate the equilibrium constant for this reaction?

Silver(II) oxide partially decomposes into solid silver and oxygen gas according to:
$$\ce{2 AgO(s) <=> 4 Ag(s) + O2(g)}$$
with enthalpy $H^\circ = \pu{62.0 kJ}$ and entropy $S^\circ = \pu{133.6 J K-1}.$
a) Calculate the change in standard Gibbs free energy. (Answer: $\pu{22.2 kJ}$)
b) Calculate the equilibrium constant for this reaction.

This is one of the exam questions I got, but I don't understand how I'm supposed to answer question b without the concentration. I also didn't get a mass or volume so I can't calculate it using the molar mass.
I tried using the formula $G^\circ=-RT \ln K,$ but I don't get the right answer with this formula. Could someone help me find the right way and formula to answer this question?
$$\frac{ΔG^\circ}{-RT} = \frac{-RT}{-RT}\ln K$$
$$\ln K = \frac{ΔG^\circ}{-RT}$$
$$
\begin{align}
K &= e^{-ΔG^\circ/(RT)} \\
  &= e^{-22.2/(8.3145\cdot 298)}\\
  &= \pu{8.18545e-9}
\end{align}
$$
The answer I'm supposed to get is $\pu{1.30e-4}.$ The answer I get according to my calculations is $\pu{8.1854e-9}.$

Please show us your calculation.

You need to give more information about what you did. Have you consider using dG= dH - TdS? and then replacing the value for dG in dG = -RTlnKp? Is any other data provided to solve the problem? Also notice that in this eq system you have to consider only O2, since the other reagents are solid. Also you need to define which constant are you going to calculate. Kc, Kp...

What is the temperature? What is "the right answer" for b?

@Rose For calculating $\Delta G^0$ use the relation $\Delta G^0 = \Delta H^0 - T\Delta S^0$, and put $T = 298$ K, as it is in the standard state (Note it should be change in enthalpy that is given in the question not the enthalpy. Change in enthalpy is the ( $\sum_{products} H_f^0 - \sum_{reactants} H_f^0$) which is actually meant in the question). Similarly for the next part use $\mathrm{\Delta G^0 = -RT lnK}$ and put $T = 298 $K, and use the result of $\Delta G^0$ obtained in the previous result.

Thank you very much! I got the correct answer! I appreciate your help!

For the record, this is why we propagate units through the calculation. You could have spotted the error yourself if you had done that in your calculations here.

You might want to check your units.
The energy should be in J/mol, according to the values you put in, and not in kJ/mol.
$e^{-22200/(8.314*298)} = 0.000128397 \approx 1.3*10^{-4}$


text:
Ionization energy of S²⁻ vs S
Which of $\ce{S^2-}$ and $\ce{S}$ has smaller ionization energy?
On one hand, ionization energy of $\ce{S^2-}$ can be predicted to be smaller than that of $\ce{S}$ because it is "easier" to remove and electron from an electron-rich ion, which already experiences a lot of electron-electron repulsion and reduced $Z_\mathrm{eff}$.
On the other hand, $\ce{S^2-}$ has a stable noble gas electron configuration, and removing an electron would get rid of the noble gas configuration.
I could not find relevant data for the second electron affinity of sulfur, so I am unable to check the reasoning.

You have a serious misunderstanding here. There's no such thing as bare $\ce{S^{2-}}$ or any monoatomic ion with -2 charge, as  mentioned under the answer. While you can imagine, say, $\ce{K2S}$ as having such ions, it does not.

Start by considering that electron affinities and ionization energies are related. Thus $\mathrm{EA}$ of $\ce{S-}$ is equal to $-\mathrm{IE}$ of $\ce{S^2-}$. (Note the convention used here is to report the value of the signed energy change associated with the process (not the absolute value), see the Wikipedia for an explanation if necessary).
A good place to continue is reading chem.libretexts.org:

the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons.[...] Similarly, the formation of all common dianions (such as $\ce{S^2-}$) or trianions (such as $\ce{P^3−}$) is energetically unfavorable in the gas phase.

This means that the electron affinity of $\ce{S-}$ is positive. Since $\mathrm{IE}(\ce{S^2-}) = -\mathrm{EA}(\ce{S-})$, this means that  $\mathrm{IE}(\ce{S^2-}) < 0$.
Note in comparison that the first ionization energy of sulfur is $\pu{999.6 kJ/mol}$.
Therefore $\mathrm{IE}(\ce{S^2-}) < \mathrm{IE}(\ce{S})$, with $\mathrm{IE}(\ce{S^2-})$ exothermic.

Indeed. I do not know of any stable -2 charged ions - they all auto-ionize to a -1 charge state. Some atoms (particularly the noble gases) do not have stable -1 charge states.


text:
Is Phosphate (PO4 3-) solube in water?
Josh~
Superphosphate is used instead of just phosphate because superphosphate is a compound whereas phosphate is an ion. This means that phosphate must attach itself with a cation in order to give a plant its phosphorus supplement, it could do this by attaching itself to 3 calcium to make tribasic calcium phosphate but that is insoluble in water and Plants can only absorb materials when they are dissolved in water. In chemistry ‘like dissolves like’ so nonpolar solutes can only dissolve in nonpolar solvents, water is polar and tribasic calcium phosphate is nonpolar so phosphate cannot dissolve in water. This is why Superphosphate is so great because it has phosphorus and it can provide it because it is soluble.

You seem to be in some fundamental disagreement with our reality. You've asked this before, and the answer is not going to change, no matter how much you dislike it or how many times you repeat the question. Being soluble or insoluble is simply not a property of ions. Then again, in my book phosphate is definitely an ion; maybe you are using this word to denote something entirely different?

Ions are always dissolved in their solution. Anything that has charge is an ion. If phosphate ions are present in a solution it is because they came from a soluble parent molecule such as phosphoric acid.

Thanks for you answer, Can you give plants phosphate (PO4 3-) instead of super phosphate for their phosphorus supplement?

When you give a plant super phosphate you are giving it monnoammonium phosphate salt. You can't get "just Phosphate ions" by themselves, it's likely that some other salts of phosphate would work and that others would kill your plant.

Thank you so much, you mean monocalcium phosphate right?

I was looking at a bag that used monnoammonium phosphate, you must be looking at one that uses monocalcium phosphate. What the plant needs is phosphate and various cations bound to the phosphate may work.

Thanks so much for you help, you've helped me in more ways than you can imagine. If you dont mind can you tell me if this paragraph makes sense in respect to Chemistry. I put it in the description under your name. I really appreciate your help.


text:
What are the limitations of the Hendersson-Hasselbalch equation?
Our instructor told us that prior to the equivalence point, the $\mathrm{pH}$ of a solution is dependent on the HH equation. However, when I tried practicing for polyprotic acid titration, I came upon this problem that does not make any sense at all. When I'm using the HH equation, the $\mathrm{pH}$ I get is lower than the initial $\mathrm{pH},$ which is illogical.
The problem goes likes this:

What is the $\mathrm{pH}$ of the solution of $\pu{25 mL}$ $\pu{0.100 M}$ of $\ce{H2SO3}$ $(K_\mathrm{a1} = \pu{1.23e-2},$ $K_\mathrm{a2} = \pu{6.6e-8})$ when $\pu{5 mL}$ of standardized $\pu{0.1000 M}$ of $\ce{NaOH}$ is added?

The initial $\mathrm{pH}$ that I've calculated is $1.53,$ but every time I try to use HH for the computation of the first $\pu{5 mL}$ increment of titration, the $\mathrm{pH}$ is lower than the $\mathrm{pH}$ of the initial conditions.

The derivation in general and the conditions where the HH equation is applicable is given in this answer . https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566

The key approximation made in deriving the Henderson-Hasselbalch equation is that the equilibrium constant can be written as $$K=\frac{c_{H^+}c_{A^-}}{c_{HA}}$$
that is, we assume activity coefficients are unity. If you take the base-10 logarithm of this equation and rearrange terms you arrive at the HH equation:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}\right)$$ Problems usually arise with approximations made beyond the HH equation, in the context of a polyprotic compound particularly  by ignoring one of the equilibria, including perhaps water autodissociation, when this shouldn't be done.
In the case of the problem you outline, you might

ignore the second dissociation equilibrium and
assume that added base neutralizes an equimolar amount of remaining acid $\ce{H2SO4}$ to form additional $\ce{HSO4-}$.

Then
$$\begin{align}c_{HA} = c_{HA,\,initial}-c_{added\,OH^-}&=\pu{0.0705 M} - \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0538 M}\end{align}$$
$$\begin{align}c_{A^-} = c_{A^-,\,initial}+c_{added\,OH^-}&=\pu{0.0295 M} + \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0462 M}\end{align}$$
Applying the HH equation we then have, as the approximate pH
$$\begin{align}pH &=1.91 + \log_{10}\left(\frac{0.0462}{0.0538}\right)\\&=1.84\end{align}$$ which is above the starting pH of 1.53.
A more accurate value (although still approximate) can be obtained starting from the condition of charge balance for the system, and by assuming that the $\ce{OH^-}$ concentration is negligible, from which one obtains pH=1.75.

In connection to your solution, the pH midway from the first equivalence point (12.5 mL of titrant) is defined as -log(ka1). However, utilizing your equation, when 10 mL of titrant is added on the solution, the pH exceeds the pH for the midway point. How can I resolve that?

I'm afraid this answer is incorrect. The key approximation of HH isn't that the activity coefficients are unity. It's that the equilibrium concentrations of HA and A- are nearly identical to their initial concentrations. Thus one can use an "x is small" approximation (say, when using an ICE table) to calculate the pH.  This is a much stronger approximation than merely assuming the activity coefficients are unity, and is what gives the HH equation its functional form.

@theorist it all depends on how you use the equation. Obviously during a titration (not a buffer solution) the concentrations will change from their initial values. To the extent that the equilibrium expression defines that which it is stated to express, it is correct. Please write an alternate answer or provide a link if you disagree.

I downvoted this answer because, while it is strictly correct, it is also misleading, particularly in the context of the OP's problem. As theorist's comment has described, the key approximation here in the OP's problem, as well as in the common use of the H-H equation, is using initial values rather than equilibrium values, not neglecting activity coefficients.

@a-cyclohexane-molecule You are welcome to downvote. I understand that my interpretation of the question was not in line with your understanding of how the HH equation is often used. I did not associate the question with the use of the HH equation as an approximate way to compute pH values during buffer preparation. I interpreted the OPs question as "what approximations are intrinsic to the HH equation", not to its application in approximating pHs during a buffer prep. The OPs posted interest is in following the effect of adding a certain amount of strong base to an acid.

@a-cyclohexane-molecule For reference to others who may wonder about this: https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation

@a-cyclohexane-molecule After re-reading the other answer I notice that it isn't correct, since it states something about what is assumed during the derivation of the HH equation which isn't correct. The HH equation is simply a re-expression of the concentration equilibrium constant in log10 terms. How you use it might involve some approximations which additionally impact the accuracy of the predicted pH (particularly in the case of buffer calcs).

As requested, here's a reference: Tro, Chemistry: A Molecular Approach, 4th Ed., pp 783-84.  Tro writes the following about the derivation of HH (Tro uses italics to emphasize the importance of this, and I reproduce the italics here):  "If we make the same x is small approximation that we make for weak acid or weak base equilibrium problems, we can consider the equilibrium concentrations of HA and A- to be essentially identical to the initial concentrations of HA and A-."  I.e., the whole point of HH is that it allows the calculation of pH from initial rather than equil. concentrations.

@theorist https://goldbook.iupac.org/terms/view/H02781

@theorist My point is you should distinguish between the derivation of the equation and the realm of approximate application. What you state can basically be determined by propagation of errors. What I suggest makes fundamental assumptions about thermodynamic behaviour.

@theorist Thanks however for the effort to dig up the ref. I grant that the realm of application is important.

For the derivation of the Hendersson-Hasselbalch equation, it is assumed that both hydronium and hydroxide ions are minor species, i.e. there concentration is low compared to that of the buffer species.
In your exercise, the hydronium ion concentration (according to your calculation) is about 0.03 M or higher. With a initial concentration of $\ce{H2SO3}$ of 0.1 M, this is not negligible. One easy way to see that is to use the calculated pH to determine the concentration of all the other species, determine the reaction quotient Q and compare it to the equilibrium constant.


text:
How to find out the value of 'n' in the equation (ΔG = −nFE)?
I can figure out the value of 'n' in simple equations like
Fe³⁺ + e⁻ → Fe²⁺ (which would be n=1 here.)
But in equations like 2Cu⁺→ Cu²⁺ + Cu, I can't figure it out. I assumed the value of n would be 2 here as two electrons are involved in the disproportionation reaction, but my book says n=1 here. Can anyone help me understand what is the value of 'n' in this equation? Thank you.

One $\ce{Cu+}$ transfers an electron to the other, thus n=1.

Why two electrons? You strip an electron from one Cu and give the same one electron to another Cu.

Oh, I understand where I was going wrong. I looked at the electrons on both sides of the equation. Thank you!


text:
What Is the Difference Between Radiation and Radioactivity?
Does all radiation glow green and mutates people on contact?

To put it mildly, cartoons are not a reliable source of information about the real world, much less real science.

Radioactivity is a very specific type of radiation
Radiation is a very broad term. Usually it means electromagnetic radiation which includes everything from long wavelength radiowaves to hard gamma rays. If we are going with the even more general usage of the word we would also include the particlulate "radiation" often associated with radioactivity.
Electromagnetic radiation includes the (usually harmless) radio waves that power broadcast television, radio, your WiFi, mobile phones; visible light; infra red light (which we perceive as heat); ultraviolet light (a small part of which is visible only to those who have had the lens of their eye replaced); and X-rays used to detect broken bones in hospitals. Some of this radiation has the potential to cause harm. Overexposure to X-rays will cause mutations (but the doses used in X-ray machines are very low and only worrisome if you have a lot of X-rays). Near-UV light (with wavelengths just shorter than blue visible light) is harmful if you stay in the sun too long (that's why we have sunblock) but is also good for you in small quantities as it promotes the production of vitamin D in the skin. The infra-red radiation emitted by hot objects isn't going to mutate anything but it will burn you if the intensity is too high.
As for radiation associated with radioactivity, that is a mix of high-energy electromagnetic radiation like gamma rays but also particulate radiation like alpha and beta particles (helium nuclei and electrons). All of this has enough energy to harm you and will cause mutations. But only gamma radiation is good at penetrating clothing so the big worry with it is if you ingest something that emits it.
So, in short, most radiation won't make things glow green and won't mutate you. The radiation associated with radioactivity is a very small subset of all the sorts of radiation that exist and is usually harmful. But don't confuse the two. Much of the modern world is enabled by radiation that causes no known harm and you would not be able to see anything if there were no visible light (which is the radiation we encounter most frequently).


text:
Are Slater’s rule and lanthanide contraction contradictory?
According to most textbook and the Wikipedia page of lanthanide contraction, the phenomenon is due to poor shielding of nuclear charge by 4f electrons. However, the σ value of 4f electrons to 6s is 1.00 in Slater’s rule, meaning that if we go through the lanthanide elements, there should be no net change in the effective nuclear charge of 6s electrons. So are they contradictory? Is it that the σ value of 4f electrons should be smaller than 1.00?

Slater’s rules are an approximation. You’re reading too much into them.

Take the example of group-12 chlorides $\ce{ZnCl2}$, $\ce{CdCl2}$ and $\ce{HgCl2}$.
The ionic radii (VI coordination) increases from $\ce{Zn++}(\pu{0.74~\overset \circ A)}$ to $\ce{Cd++}(\pu{0.95~\overset \circ A)}$ to $\ce{Hg++}(\pu{1.02~\overset \circ A)}$. Due to lanthanide contraction or poor shielding of $\ce{4f^14 e-}$ in $\ce{Hg}$, the difference of atomic and ionic radii between $\ce{Cd}$ and $\ce{Hg}$ is very less.
$$\ce{Zn: 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^10, 4s^2}$$
$$\ce{Cd: 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^10, 4s^2 4p^6 4d^10, 5s^2}$$
$$\ce{Hg: 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^10, 4s^2 4p^6 4d^10 4f^14, 5s^2 5p^6 5d^10, 6s^2}$$
Applying Slater's rule to the valence shell $n\mathrm{s^2}$ $\ce{e-}$, the calculation of screening constant $s$:
$s[\ce{Zn(4s)]}$ $\mathrm{=0.35\times1 + 0.85\times18 + 1\times10=25.65}$
$s[\ce{Cd(5s)]}$ $\mathrm{=0.35\times1 + 0.85\times18 + 1\times28=43.65}$
$s[\ce{Hg(6s)]}$ $\mathrm{=0.35\times1 + 0.85\times18 + 1\times60=75.65}$
Calculation of effective nuclear charge
$Z_{eff}=Z-s$
$Z_{eff}(\ce{Zn})=30-25.65=4.35$
$Z_{eff}(\ce{Cd})=48-43.65=4.35$
$Z_{eff}(\ce{Hg})=80-75.65=4.35$
We can see that effective nuclear charge is same in all three cases. But the covalent character in these compounds increases gradually as:
$\ce{ZnCl2 < CdCl2 < HgCl2}$
This might be due to gradual increase in effective nuclear charge in cation.
Therefore, Slater's rules are an approximation to give us the idea about effective nuclear charge and it might be contradictory to lanthanide contraction in some cases.


text:
Probability density "AT" the nucleus?
Just had a conceptual question from radial wave functions.
I wanted to know if probability density AT the nucleus is maximum, or zero.
From the attached images, it is clear that the function is at zero for p and d orbitals, but TENDS to infinity as approaching r = 0 for the s-orbitals. Is the value of the function 0 at r = 0?
Intutively, as the distance from the nucleus is the least, there should be maximum probability density.
Additionally, can there be some other interpretation for the phrase "at the nucleus"? It is quite confusing.
Thanks a lot.
EDIT: Apologies, don't know what was going through my head while posting - I initially posted something about PROBABILITY of finding an electron at the nucleus (I know the answer to that, oops).
I have changed it to what I meant to ask.

Probability density of s orbitals at the nucleus is not 0, nor is it infinite.

@IvanNeretin Wouldn't you say that $\int_0^0 \phi^2(\vec{r}) \mathrm{d}\vec{r}=0$?

@Martin-マーチン Nucleus isn't really a point.

@Martin That's true. But then again, that would be equally true at any point.

@Martin-マーチン The integral of the square modulus of the wavefunction over any range is a probability. The square modulus itself is a probability density. They are not the same thing - they have different units for starters.

Hydrogenic Wavefunctions:
A hydrogenic wavefunction, $\psi_{nlm}(\boldsymbol{r})$, can be written as the product
$$
\psi_{nlm}(r,\theta,\phi) = R_{nl}(r) Y_{lm}(\theta,\phi)
$$
where $R_{nl}$ is a radial wavefunction and $ Y_{lm}$ is a spherical harmonic.

$R_{nl}$ is a radial wavefunction
$ Y_{lm}$ is a spherical harmonic
$n$ is the principal quantum number, $n = 1,2,3\dots$
$l$ is the angular quantum number, $l = 0=s, 1=p, 2=d,\dots,n-1$
$m$ is the magnetic quantum number $m = -l, -l+1,\dots,l-1,l$

Probability/Electron Density
The probability density, $\rho$, is equal to the square modulus of the wavefunction:
$$
\begin{align}
\rho_{nlm}(\boldsymbol{r}) &= |\psi_{nlm}(\boldsymbol{r})|^2 \\
\rho_{nlm}(r,\theta,\phi)&= |R_{nl}(r)|^2 \cdot |Y_{lm}(\theta,\phi)|^2
\end{align}
$$
The spherical harmonics $Y_{lm}$ have angular nodes through the nuclei for all values of $l$ and $m$ except $l =m =0$, so we can say that for all $l>0$ orbitals, i.e. all orbitals except s-orbitals, the nuclei will have $0$ probability density.
S-orbitals
Consider s-orbitals, in atomic units and substituting in the solution of the radial wavefunctions in generalized Laguerre polynomials $L^\alpha_k$:
$$
\begin{align}
\rho_{n00}(\boldsymbol{r})&= |R_{n0}(r)|^2 \cdot |Y_{00}(\theta,\phi)|^2\\
&= |R_{n0}(r)|^2 \frac{1}{4\pi} \\
\rho_{n00}(r)&= \frac{Z^3}{\pi n^5} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\
&= A(n) \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right)
\end{align}
$$
For all values of $n$, $L^1_{n-1}\left(0\right) = n$, so we know that the nucleus never has zero probability density for s-orbitals, and has probability density
$$
\rho_{n00}(r=0) =  \frac{Z^3}{\pi n^4}
$$
Stationary Points
But we can differentiate with respect to $r$ to find stationary points in the radial probability density.
$$
\begin{align}
\frac{d}{dr}\rho_{n00}(r)&= A(n) \frac{d}{dr} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\
&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) +  \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\
\text{if}\;n=1: \frac{d}{dr}\rho_{000}(r)&= -\frac{Z^4}{\pi n^6} \exp\left(-Zr/n\right) \\
&> 0 \;\forall\;r
\end{align}
$$
Hence the probability density is constantly decreasing away from the nucleus.  So as you previously assumed the maximum for probability density for the 1s orbital is at the nucleus.
$n>1$:
$$
\begin{align}
\text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) +  \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\
&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) - L^2_{n-2}\left(2Zr/n\right) \right]
\end{align}
$$
Setting the expression above equal to zero and solving for $r$ will give the positions for the radial nodes and maxima inbetween for a given value of $n$.  I am not aware of a general method/expression in n.
As you asked about nucleus, we can also consider the point $r=0$:
$$
\begin{align}
\text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= -  \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} L^1_{n-1}\left(0\right) + L^2_{n-2}\left(0\right) \right] \\
&= -  \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} n +\frac{n^2-n}{2}\right] \\
&= - \frac{Z^4}{\pi n^5} - \frac{Z^3 \left(n-1\right)}{2\pi n^4} < 0
\end{align}
$$
Conclusion/TL;DR
Hence the nuclei is always a local maximum in the probability density for an s-orbital.  You would expect the exponential decay in the radial wavefunction to dominate over the Laguerre polynomial term, so would expect local maxima further out to be smaller that the maxima at the nuclei and hence that the point of highest probability density is at the nucleus.
This makes physical sense, as classically the negatively charged electron wants to be as close as possible to the positively charged nucleus to minimise its potential energy, but the quantum mechanical nature of the electron prevents it from tumbling directly into the nucleus.

All the math detail is nice, but the simple answer is that the functions simply don't consider that the nucleus has a finite volume. You'd need a more complicated model to account for that fact.

In an otherwise excellent answer there is a small problem with "The spherical harmonics Ylm have angular nodes through the nuclei for all values of l and m except l=m=0, so we can say that for all l>0 orbitals, i.e. all orbitals except s-orbitals, the nuclei will have 0 probability density." The spherical harmonics have no radial dependence so we can't say anything from them about the behaviour at any value of r and r=0 in particular. The point is rather that Rnl is zero for r=0 for all l>0 (i.e. not an s type orbital) from the Laguerre polynomials


text:
Why do the d orbitals contract when pairing of electrons start, because the size should increase due to increase in energy

The energy of an orbital is proportional to its mean radial distance, and since the 3d orbital is much larger it is much higher in energy than the 3s and 3p orbitals

All references from JD Lee Concise inorganic chemistry

However, d orbitals are in general too large and too high in energy to mix completely with s and p orbitals.

In this caption, JD Lee says that the d orbitals find it hard to undergo hybridization because of its too large size and higher energy, but to decrease its energy and size , it employs various solutions, such as:

A second factor affecting the size of d orbitals is the number of d orbitals occupied by electrons. If only one 3d orbital is occupied on an S atom, the average radial distance is 2.46 A., but when two 3d orbitals are occupied the distance drops to 1.60 A.
A further small contraction of d orbitals may arise by coupling of the spins of electrons occupying different orbitals. It seems probable that d orbitals do participate in bonding in cases where d orbital contraction occurs.

One of them was that when electrons start filling in d orbitals, contraction of orbitals occurs. When pairing of electrons commences, d orbitals contract further.
But I find this line confusing, because the pairing/filling of electrons in the d orbitals must increase the energy, and thus the size, of d orbitals. But JD LEE states that the opposite happens. Could someone please explain why?

If electrons are excited from 3p to 3d, then that would mean less shielding from the inner orbitals which would serve to contract outer orbitals, I guess?

There is no size of d orbitals, they extend to infinity. Therefore d orbitals cannot get larger and smaller, and they most certainly do not 'employ solutions'. So wherever these numbers come from, one would need more information to judge the statement by JD Lee. I have a very bad feeling, that the argument leads towards hybridisation involving d orbitals; this would be wrong and you can safely assume that the rest is also phishy.


text:
Why are sulfur trioxide and nitrate not isoelectronic, even though both have the same number of outer electrons?

Though $\ce{SO3}$ has the same number of outer electrons as $\ce{NO3-}$, the two are not isoelectronic.

This statement is from JD Lee, but I could not understand why is he calling these two molecules as not isoelectronic because ultimately being isoelectronic means having the same number of electrons in outermost shell, which they clearly do! Please explain.

I thinks its just a difference of usage. Though not as common, it wouldn't be unreasonable to read "isoelectronic" as meaning "the same total number of electrons". Of course, it leads to confusion when it isn't properly explain.

You need to compare the electron configuration, not just the number of electrons!

Isoelectronic species are those which have similar electronic configuration. for this total no. of electrons should be same not only valence electrons.


text:
Are primitive GTOs normalised before contraction, or is the resultant basis function normalised afterwards?
I'm quite curious about this:
In a basis set (I'll just use minimal-basis STO-nG basis sets for convenience), the basis functions are written as a linear combination of primitive GTOs. Are the GTOs normalised first, like this:
$$\Psi^\mathrm{STO} = \sum_i N_i c_i \phi_i^\mathrm{GTO}$$
where $\phi_i^\mathrm{GTO}$ are un-normalised GTO functions, and $N_i^{2} \left<\phi_i\left|\phi_i\right.\right> = 1$
or are they normalised like this (GTOs are not normalised):
$$\Psi^\mathrm{STO} = N_\mathrm{STO} \sum_i c_i \phi_i^\mathrm{GTO}$$
where $N_\mathrm{STO}^{2} \left<\Psi^\mathrm{STO}\left|\Psi^\mathrm{STO}\right.\right> = 1$?
I thought it was the former and not the latter, but after poking around the internet I'm not so sure anymore.

They don't necessarily have to be normalised at all, as long as the trial wave function you are using them in is later normalised. Since GTO are supposed to mimic (approximate) STO, the first way you have proposed makes not much sense to me.


text:
Will charcoal briquettes (not activated) absorb water out of the air in the house in a subtropical climate?
I live in Florida, an incredibly wet environment where it's very difficult to get moisture out of house without using a dehumidifier 24/7, which is too expensive.  Would charcoal briquettes be of any use?

No, and even if the briquettes were able to absorb water, they'd be saturated in a day, and then you'd have to dry them or replace them, far more expensive than A/C.

I know this wasn't your question, but it might be helpful: Dehumidifiers differ in their energy efficiencies.  You might want to make sure yours is on the high side (you could check the Energy  Star rating; you can also check Consumer Reports if you have a subscription).   And you probably already know this, but there are several things you can do to optimize your dehumidifier's efficiency (keeping it away from walls to ensure maximum airflow, keeping the filter clean, etc. -- you can search for more detailed info. online if you're interested).

The body of your question shows that your primary concern isn't charcoal's hygroscopic character per se but it's suitability as a practical means to lower the relative humidity inside your house.
Let's start with it's hygroscopic character: It seems that charcoal indeed displays hygroscopic properties (the propensity to absorb water to itself) to some degree (at least to the extent that it affects it's burning properties).
The subject was treated by this paper, wherefrom the following citation was selected as part of this answer to your question:

The charcoal samples placed in the environment with high relative humidity showed an
  increase in humidity by the heightened final carbonisation temperature, with the highest average detected for charcoal produced
  at 750 °C. This was the reason why the charcoal produced in
  this temperature showed higher moisture related to higher porosity
  presenting a larger area available for water condensation inside the
  pores.
The difference between the moisture adsorbed by the charcoal packed in
  two environments can be attributed to more water molecules present in
  the humid atmosphere (RH = 98%) and a greater ease of inducing the
  porous region of the charcoal. This means that the more humid the
  environment, the greater the adsorption of moisture by charcoal.
These two different moisture conditions represent different possible
  situations of charcoal storage where high relative humidity is not
  ideal.

Now if one of the reasons you would like to use charcoal over other means of lowering the humidity inside your house is that you were planning to re-use the charcoal later for it's primary purpose (as a fuel), this means that it wouldn't suit that purpose very well, at least if you don't plan to dry your charcoal as an intermediary step in the process (dehumidify > dry > burn). Water requires a lot of energy to evaporate which would waste fuel if you wanted to use it directly as such after using it for your house.
Now for absorbing humidity, there are other (much better) substances you could use out there: You could for instance use the very safe and well known SiO2 form frequently referred to as "silica-gel", which works very well at higher RH levels, and can be conveniently "recycled" by ~ 1h pass in the oven at ~100-120C. The main weakness of silica gel in your case is that it quickly loses efficiency at higher temperatures that are not uncommon in Florida. If you'd like to be a slightly less conventional, one could recommend Calcium Chloride (CaCl2), a fairly cheap substance which will hold up to ~4 molecules of water for each of it's own up to a temperature of 45°C, and at the very minimum 2 molecules of water for each of it's own at any temperature naturally encountered in Florida.
I hope this helps!

Desiccants are not practical for whole house dehumidification.

Desiccants need to be cycled, which I agree can be impractical, but I firmly believe that CaCl would be more practical than coal. Without using desiccants, condensating the water would be the only practical option.

Here's an application that will dehumidify your home and provide running water: https://www.nature.com/articles/s42004-018-0028-9 :-)

A couple of comments: it's $\ce{CaCl2}$, and the mole ratio looks much less attractive when translated into a weight ratio.

@BuckThorn That is correct, not sure why I wrote it wrong (twice!)

Yes, CaCl2 is in fact employed in commercial moisture removal products (example, DampRid). Reheating the hydrate to re-use may be somewhat successful. However, my experience is that once you expose fresh CaCl2 to air/CO2, with time it appears to create, to a limited extent, an insoluble basic hydroxide or mixed carbonate salt therefrom, which limits its usability.


text:
NaOH vs Polypropylene
This source lists polypropylene as suitable for use with NaOH.
However, this paper seems to demonstrate that NaOH slowly degrades polypropylene (even at fairly moderate temperatures).
Does anyone know more about this? Would anyone know for instance what it could degrade into (potential for trace contamination of the NaOH solution)?
There is a small possibility that some of that information is described in the body of the paper, but  unfortunately I only have access to the abstract.

Sadly, the paper does not discuss about the degradation products. It is very observational.