Is sodium borohydride really a better base than an alcohol?

General considerations

It’s not the basicity that is the important factor here, it is the nucleophilicity. The hydride ion, whether present as tetrahydridoborate or as a lone hydride is a very lousy nucleophile while an alcohol is an okay-ish nucleophile. The general mechanism of the addition is given in scheme 1.

Scheme 1: General mechanism of a nucleophilic attack on a carbonyl.

Now in the case of a hydride — a very weak nucleophile — this reaction is generally slow. It is faster, the more reactive — electrophilic — the carbonyl species is. If $\ce{X}$ is a carbon or hydrogen atom, then all we have is a localised double bond without any significant reduction in reactivity.[1] Reactions with those should thus be fast. If $\ce{X}$ is an element that has a lone pair, such as oxygen or chlorine, then there is a notable mesomeric stabilisation as shown in scheme 2.

Scheme 2: Mesomeric stabilisation by lone pairs next to a $\ce{C=O}$ double bond.

This stabilisation is not strong if a chlorine atom is $\ce{X}$ since chlorine is much larger and its lone pairs are more disperse. The stabilisation is stronger in the case of acid anhydrides; however, since a single oxygen has to stabilise two carbonyls it is still not too strong. In esters, carboxylic acids (and their anions) and most so in amides this stabilisation is very strong ad nucleophilic attacks difficult. This explains the general electrophilicity trend:

$$\text{aldehydes} > \text{ketones} > \text{acid chlorides} > \text {acid anhydrides} > \text{esters} > \text{amides}$$

Now let’s look at the reactions in detail. Throughout this post I will assume neutral conditions only; no acid or base catalysis.

The hydride attack

As I already noted, hydride is a bad nucleophile. It makes sense that it can attack the first members of the series — aldehydes, ketones and acid chlorides — but fails to do so for more stabilised members. For completeness, the mechanism of a hydride attack onto an acid chloride is given in scheme 3.

Scheme 3: Mechanism of a hydride attack onto an acid chloride.

We see another key point here: Once the carbonyl has been attacked, the oxide anion has a tendency to ‘rebound’ a lone pair and displace one of the bonded atoms. Generally, the displaced anion must be reasonably favourable for the rebound to occur. Chloride ions are good leaving groups and thus a displacement of chloride is likely to happen. Hydride ions are absolutely absymal leaving groups, so the hydride attack is not reversible.

After the inital attack and displacement of the chloride, we are left with an aldehyde which of course will react quicker and lead to full reduction to the alcohol.

Acid anhydrides are not attacked this way because the initial step does hot happen. If you accidentally got a hydride to attack, the $\ce{^{-}OOC-R}$ leaving group is great and thus a reaction would occur.

The alcohol attack

Alcohol hydroxy groups are still not the best nucleophile but they are better nucleophiles than hydride is. Thus, they even manage to attack acid anhydrides and — to a limited extent — esters and acids. The general mechanism of attack is shown in scheme 4.

Scheme 4: Reaction mechanism of an alcohol attacking an acid chloride including a potential second attack.

The key difference here is that the alcohol is a neutral nucleophile, not an anion like hydride is. Thus, when the rebound occurs, it is susceptible to redisplacement. This is shown in the final equilibrium of the mechanism above. This is also the reason why — under neutral conditions! — an alcohol may attack an aldehyde or a ketone but the attack is not productive. The alcohol will simply be rebounded back out since the hydride is a terrible leaving group.

There is a productive reaction mechanism for alcohols with aldehydes or ketones — the formation of acetals and ketals — however, that generally requires acid catalysis.

Finally, there is also the possibility of base catalysis when alcohols react. It would deprotonate them to give the alcoholate which is a much better nucleophile and even able to attack esters. However, base catalysis is outside of the scope of this question.


Why is acetone and diethyl ether miscible, but not water and diethyl ether?

Acetone's dipole moment is 2.91D while that of ether is only 1.15D.  Water is a very polar substance, so acetone will dissolve in it while ether will not.

Two substances are miscible when their intermolecular forces (IMFs) are similar enough such that the forces of attraction between molecules of different substances are similar in strength to the forces of attraction between molecules of the same substance.  If we look at the miscibility of A and B as an example, if A attracts B about as strongly as A attracts A and B attracts B the substances should be miscible.  If A and A have much stronger attraction than A and B the substances are less likely to be miscible.

In your example it's not that some molecules of ether will not dissolves in water and vice versa, it's just that if the two liquids are in contact the water will stay with water and the ether with ether for the most part.  This is because water's strongest IMF is hydrogen bonding while ether's is dispersion with some dipole-dipole forces.


Why don't alkaline earth metals lose only one electron when they are ionized?

Magnesium can definitely exist as $\ce{Mg^{+}}$. Hence the availability of first ionization energy data for magnesium.

The same goes for other metals; there are only uncommon valences, such as $\ce{Na^{3+}}$.

If you had enough energy, you could make $\ce{Na^{4+}}$ if you wanted to. It would just cost, according to the table below, 9543 + 6912 + 4562.4 + 495.8 kJ/mol.

I think your question is why isn't $\ce{Mg^{+}}$ more commonly encountered. That has to do with stability. Losing two electrons and becoming isoelectronic with a noble gas is more stable than having an unpaired electron. An unpaired electron is available for bond formation, and bond formation tends to be exothermic.

If you want to read up on the +1 valence of magnesium, such as in $\ce{MgH}$, you can read more here:

http://bernath.uwaterloo.ca/media/24.pdf


The strength of Lewis acids relative to your Lewis base has been commented on. A fuller answer to your question needs to include the Lewis acid you are considering.
The original Bronsted-Lowry theory of comparing acidity and basicity, or acids and bases, involved $\ce{H^+}$ and $\ce{OH^-}$ and the transfer of a proton.
The broader Lewis theory allowed consideration of other cations besides the proton as acids, and other nucleophiles than hydroxide ion to be bases. So water can be considered a Lewis acid relative to chloride ion because $\ce{H2O}$ hydrates chloride ions by hydrogen bonding. And water is a Lewis base relative to metal cations because water hydrates the metal ions through electron donation from oxygen.
So the answer to the question is yes and no. Yes, $\ce{HCl}$ can be considered a Lewis base - relative to a very strong acid, as mentioned in the comments - but in the ordinary world, I would not first think of $\ce{HCl}$ as an alkali. Stretching the word "base" by adding "Lewis" in front of it needs more, like adding the reference acid, to make the comment clear rather than confusing.


Electromagnetic energy to Chemical energy

You can certainly turn electromagnetic energy into chemical energy, but the key is 'efficiently.'

A photo-voltaic cell can be used to split water into Hydrogen and Oxygen. The Hydrogen can be burned or used in a fuel cell.

But the key is how efficient is 'efficient.'

Each step will lose energy, but if solar is the input, there is a huge supply. If a solar powered system were continually operating, and we had many of these systems, a huge amount of Hydrogen gas can be generated. Even considering the costs of delivering to where it is needed (which can be paid with the product itself) there is a huge and realistically limitless supply.

That being said, there is much 'waste' which hampers a calculated efficiency, but once established a system that relies on water for a raw material is relatively efficient because there is very little required for raw material acquisition.


Can a tertiary enamine tautomerize to an imine?

Tautomerization is a chemical reaction whereby constitutional isomers interconvert. The interconversion process involves the shift of a hydrogen atom accompanied by the shift of a double bond.  Here are a few examples.

In molecule (a) pictured above, there is no hydrogen attached to a nitrogen (it has been replaced by a methyl group), so there is no hydrogen atom available to shift.

Response to OP's edit:

The picture you posted in your edit is incorrectly labeled.  The charged molecule on the left is an iminium ion, not an imine.  So this is not an example of an imine-eneamine equilibrium.  You are correct that in this equilibrium (the one pictured in your edit) there is no hydrogen transferred to nitrogen, but again, this is not an example of an imine-eneamine tautomerization.  I think this error is the source of your confusion.

Here is a picture of an imine-eneamine tautomerization.

As you can see, a hydrogen is transferred between nitrogen and carbon.  This hydrogen transfer can occur with or without solvent molecules playing a role.  That is to say, the hydrogen can be transferred directly between the carbon and nitrogen atoms, or it can be transferred from one of the atoms to a solvent molecule which can then transfer it back to the other atom involved in the tautomerization. The image source also provides a nice explanation of tautomerization.


Hyperconjugation - concept clarification

Cyclic systems aren't planar. The system orients itself in such a way that the vacant p orbital is in plane in order to gain the ability to engage in hyperconjugation which stabilizes it very well.


How can the trityl tetrafluoroborate function as a Lewis-acid catalyst

One point is that you're assuming that the "p" orbital in the center is the Lewis acidic site. This may not be the case, you can draw a resonance structure that breaks aromaticity but puts the cation at the para position of the phenyl ring. For example, the ortho and para positions are known hydride acceptors (link).

Another issues to consider is that the phenyls are bulky enough that they can't even remain coplanar.

So we can return to your issue of reversibility by looking at trityl via these two properties:

If the Lewis acidic site is via one of the phenyl rings, then the resultant adduct is destabilized by breaking aromaticity. It may be kinetically accessible and allow the trityl ion to serve as a Lewis acid, but the adduct is not stable.
If you do create an adduct with the central carbon, this adduct is greatly destabilized by steric repulsion. Enough so that fragmentation becomes likely.

In both cases, there is a kinetically accessible Lewis acid/Lewis base adduct, but this adduct has some kind of feature that makes it thermodynamically less stable. Therefore, the trityl ion can serve a catalytic role as a Lewis acid.

The exact mechanism of action (1) or (2) is likely dependent on the exact reaction you're looking at.


Is it known for sure that bases feel slippery because of the production of soap/surfactant?

Potential candidates might include:

Soap/surfactant,
Glycerol produced from hydrolysis of "finger lipids",
Broken down protein products, or
the pre-existing [inter-cellular lipid layers][5] that would be exposed by the removal of successive layers of cells within the stratum corneum.

Very good guesses but none of the above actually (i think, what does "finger lipids" mean?). But the reason bases feel slippery to bare skin no matter how much you scrub first, is bases attack the phospholipid bilayer of your cells quite readily not only saponifying them, but breaking the inter-molecular bonds of molecules that already resemble a soap. Take for example phosphatidylcholine:

The phosphochloline already has all of the elements to make a soap and when base is added, the molecule does not need to interact with other molecules to achieve local charge balance, but now have the ions from the base to balance, which breaks intermolecular bonding creating a small layer of soap, making the base feel slippery.


Different chemoselectivity of transition metals in catalytic hydrogenation

Unfortunately, this is extremely complicated and the topic of active research in both computational and inorganic chemistry and chemical engineering. There is no simple answer to your question. We can discuss various forces that go to determining chemoselectivity though.

1) Crystal structure: You said that you are primarily focused on bare metal catalysis, so I will restrict myself to those systems. Not all surfaces of even the same material behave the same. These are commonly described by Miller indices (1), which describe the surface by how a plane cuts through the material's unit cell. The reactivity and selectivity are heavily determined by the exact surface structure (think dangling bond sites, binding sites, multicenter reactivity, all of which are influenced by the surface atomic structure both local and in its immediate environs). Different materials will have different crystal structures based on both electronic effects (like how many valence electrons there are) and size effects (how big atoms are, although there is relatively less variation on this in the 2nd and 3rd rows).

2) Chemistry of binding sites: Some materials are more likely to bind molecules than others. The Pt-H bond is extremely strong, much more than a Pd-H bond (these trends down a period are often explained in terms of the diffuseness of the 5d orbitals vs 4d which lead to better overlap with the main group orbitals of hydrogen and other light elements). This leads to greater coverage of a Pt surface with hydride equivalents than a Pd surface (generally, caveats apply). On the other hand, Pd tends to bind oxygen atoms more strongly than Pt does (one can think about hard-soft acid base theory here to explain that, but the situation is probably more subtle and complicated than that). One can easily imagine a benzylic C-O oxygen bind to the Pd surface which holds the site for hydrogenation close the the surface until a hydride equivalent can diffuse over to react (hydrides on surfaces are quite mobile and diffusion is often quite rapid, of course, hydride diffusion rates are also strongly determined by surface structure and material properties (because almost nothing can be simple)). On the other hand, $\pi$-system can interact strongly with metal surface (especially of soft (again in the HSAB sense) metal). So, for Pt, the aromatic system is brought close to the surface, allowing hydrides to migrate over to it.

So we can hand-wavily discuss why Pd is more likely to hydrogenate benzylic C-O bonds and Pt is more likely to hydrogenate aromatic systems as we did above, but the nitty-gritty detail is quite complicated. This is a massive area of research. Inorganic chemistry is its own field not because everything is so similar, but because metals have hugely varying behavior. The fact that these are "all being transition metals" should NOT imply that their reactivity is similar in the slightest. Being in the same period as a transition metal tells you as much about being in the same period in the main group. C and O and N have totally different chemistry, as do Rh and Ru and Pd.

P.S. f orbitals do not figure into this.


How does di-isobutylaluminium hydride ( DiBAl-H ) reduces carboxylic acid

Reduction of esters of carboxylic acids directly to aldehydes is of great synthetic interest. Zakharkin and Khorlina in 1962 (Ref.1) has shown that, diisobutylaluminum hydride (DIBAL-H) could be very useful in reducing esters to aldehyde in high yields at low temperatures (about $\pu{-70 ^oC}$). They also found that solvents affect reduction significantly. For example, the reduction reaction in toluene or hexane proceeds to give corresponding aldehydes in yields l0-15% more than those in ether as the solvent. Previously, in 1959, Miller et al. (Ref.2) found that DIBAL-H is a more selective reagent than lithium aluminum hydride in the reduction of nitriles to aldehydes, and it can also be used to reduce benzoic acid to give 72% yields of benzyl alcohol (they also showed esters can be reduced to corresponding alcohols). The conditions and yields of these references are summarized in this page. In Ref.2, during acid reduction, Miller et al. have used 3 equivalents of DIBAL-H and observed that a gas (hydrogen) evaluation upon the addition of the hydride to benzoic acid until one mole of the reducing agent had reacted. Thereafter, the addition of more DIBAL-H gave no gas. Thus, the first reaction is, presumably an acid-base reaction. However, DIBAL-H here should be acting as a Lewis-acid instead of a base, as illustrated in scheme 1:

The ist mono-adduct, then further reacted with second DIBAL-H to give di-adduct, which upon acid hydrolysis would give expected aldehyde (Scheme 2):

References:

L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).
A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).


As Ed V points out, the conditions of formation of struvite have many variables and the struvite crystals produced are variable. On the one hand, this could be a pleasant exercise of playing with atomic models, but on the other, it may be unproductive unless you were to use computer graphing to model various configurations.
Perhaps you could start with a simpler, but similar compound and develop a crystallization mechanism from that. Ammonium phosphate seems simpler, but isn't, because the third proton of H$_3$PO$_4$ is so weak that pH is still a significant variable, and you are also dealing with diammonium phosphate and double salts (Ref 1).
Another possibility is plain magnesium phosphate, but it appears in several forms and with several degrees of hydration, so it is not a simple crystal formation. (Ref 2) The anhydrous form is too unrelated to struvite. (Ref 3)
Stretching your imagination further, replacing the ammonium with potassium gives an interesting material that has applications in cementing. One article gives a simple picture of the "components" of KMgPO$_4$.6H$_2$O:

The octahedron of Mg(H$_2$O)$_6$$^+$$^2$, the tetrahedron of PO$_4$$^3$$^-$, and the sphere of K$^+$ are objects that could be played with to make a model crystal. The sphere could also suggest how the NH$_4$$^+$ ion (just assume it is almost spherical) fits into struvite. (Ref 4)
Struvite was the basis for a rapid repair patch for roadways some years ago (Ref 5), and has current applications in immobilizing radioactive waste. (Ref 6)
Ref 1. https://en.wikipedia.org/wiki/Ammonium_phosphate
Ref 2. https://en.wikipedia.org/wiki/Magnesium_phosphate
Ref 3. https://byjus.com/chemistry/magnesium-phosphate/
Ref 4. https://pubs.acs.org/doi/10.1021/acs.jpcc.7b01396#
Ref 5. US Patent 3,960,580 J.M. Gaidis et al.
Ref 6. https://www.sciencedirect.com/topics/engineering/magnesium-potassium-phosphate


Search by CAS Registry name to obtain trackable accession: ie InChI?

Chemical Identifier Resolver can convert several identifiers and has an API so you can convert with a script.

A Python example for CAS to InChIKey:

from urllib.request import urlopen

def CIRconvert(cas):
    try:
        url = 'http://cactus.nci.nih.gov/chemical/structure/' + cas + '/stdinchikey'
        ans = urlopen(url).read().decode('utf8')
        return ans
    except:
        return cas

casno = ['50-78-2', '69-72-7', '95-1E-6', '108-24-7', '31710-30-2']

for casid in casno:
    print(CIRconvert(casid))

Output:

InChIKey=BSYNRYMUTXBXSQ-UHFFFAOYSA-N
InChIKey=YGSDEFSMJLZEOE-UHFFFAOYSA-N
95-1E-6
InChIKey=WFDIJRYMOXRFFG-UHFFFAOYSA-N
InChIKey=CIPFDHFTBYJKQB-UHFFFAOYSA-N


Equilibrium constant for the thermal dissociation of phosphorus pentachloride

Note the equation as follows: $$\ce{PCl5 <=> Cl2 + PCl3}$$

You may begin with an ICE table showing the initial and final pressures of the gases.
$$\begin{array}{|c|c|c|} \hline
 & \ce{PCl5}  & \ce{PCl3} &\ce{Cl2} \\ \hline
\text{Initial} & 2 & 0&0 \\ \hline
\text{Change} & -1 & +1 & +1 \\\hline
\text{Equilibrium} & 1 & 1 &1 \\ \hline
\end{array}$$
Then it is a simple matter to find $K_p$ with the equation $$K_p = \frac{[\ce{P_{PCl_3}}][\ce{P_{Cl_2}}]}{[\ce{P_{PCl_5}}]}$$

If you would like $K_c$ you would need the ideal gas law ($PV=nRT$) but I doubt that is what your question requires.


Can we produce tritium from helium 3? If yes how?

From my knowledge, helium-3 is actually a by-product of tritium decaying. Tritium is usually manufactured industrially by bombarding lithium-6 nuclei with neutrons in a nuclear reactor, and is currently the cheapest way to manufacture tritium.


Help with determining the rate constant of a reaction

This Khan Academy Video explains it very nicely.

So since this is a first order reaction
$$\ce{ -\Delta N2O5/\Delta t = k[N2O5]}$$

This gives you the rate of reaction at the very start. A second later however,  $\ce{[N2 O5]}$ decreases which will in turn change the rate of reaction. So lets apply calculus. We can write this as

$$\ce{ \frac{-d [N2O5]}{d t} = k[N2O5]}$$

Then we can integrate to find out the total change over a period of time.

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= \int_0^t-k dt$$

since $-k$ is constant, we can take it outside of the integral

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= -k\int_0^tdt$$

So then we get

$$ \ln([N2O5]_t) - \ln([N2O5]_0) =-kt$$

So putting in your values

$$\ln(0.20\cdot 10^{-2}) - \ln(1.24\cdot 10^{-2}) = -k\cdot 60\cdot 60$$

So I'm pretty sure the answer is $$5.068\cdot10^{-4}~\pu{s^{-1}}$$


Volume change in positive deviation from Raoult's Law

I'm a little late to the party here, but wanted to chime in.

The link you posted isn't working anymore, but you appear to be making too many assumptions about the system being measured.  Nothing about the measurement of the volume change implies there is necessarily an open space above the liquid for vapor to form.  Imagine an open beaker of ethanol with a small disk floating on top so that no molecules can escape upwards, but it can move up and down with the change in liquid level.  This would also keep the pressure of the liquid constant (although it doesn't really matter since the liquid is essentially incompressible) .

If you were able to inject a certain amount of acetone into this beaker without opening it to the atmosphere for very long, the resulting volume of liquid in the beaker would be greater than the sum of its parts. Hopefully it makes sense that if there is no space for liquid to form, the positive volume change of mixing can show itself.

As far as the idea that more vapor would form to decrease the volume of liquid, realize that evaporation is a time-dependent process, so likely the mixed liquid would reach an effective equilibrium state where it's volume change could be observed before the extra liquid has time to evaporate.  It is true that it would evaporate faster if the effective vapor pressure (technically "fugacity" or "chemical potential") were higher, as the driving force for mass transfer would be greater, but this is also very dependent on the way the system is specified.  Technically if you left the container open in a large room forever, it would all evaporate in either case.


Is there geometrical isomerism in case of 2-methylcyclopentanone?

If you look at the structure of 2-methylcyclopentanone, you'll notice that it is asymmetric about the C-2 atom

This gives rise to two isomers; the R isomer on the left and the S isomer on the right

These are definitely optical isomers, more specifically enantiomers of each other. Whether these are 'geometric isomers' or not is a matter of how you define 'geometric' isomerism; the IUPAC Gold Book strongly discourages the usage of this term, instead preferring to use 'cis-trans isomerism'. If it is defined by the latter, then this compound does not display cis-trans isomerism, as there are no groups for the methyl substituent to be trans to.


How to calculate bond enthalpies from first principles?

Firstly the following basic facts can be helpful :

Ionic bond is stronger than covalent bond in most cases .
A metal non-metal bond is more ionic.
Non-metal non-metal bond is covalent in nature.

By considering Valence Bond Theory , we can draw the fact that overlapping of larger orbitals ( usually of large atoms ) with smaller orbitals is less efficient and will result in a weaker bond .

For example ,

HF>HCl>HBr>HI is the order of their bond strengths .

This is because of the fact that as we go down the group , atomic size increases and so does its p-orbital which takes part in the overlapping with small s-orbital of H , making it a weaker bond .

Electronic configurations of various elements ( which is derived from the atomic number provided in the periodic table ) can be used to make the Lewis structures of most molecules and then the number of covalent bonds can tell comparitive enthalpies. ( more the bonds , more is the enthalpy )

Also , some Periodic tables try to be more informative and provide Pauling EN alongside the atomic number and mass . In that case , bond polarity can be measured by calculating the EN difference . Usually , more the polarity of the covalent bonds, more they act like an ionic bond and are more stable .

Though these tips can be of help in most comparisons , relying on actual values cover anomalous behaviors and exceptions which bump along our ways .


Order of reactivity of substrates towards Wurtz reaction

The Wurtz reaction is an organic reaction, in which two molecules of an alkyl halide (e.g., $\ce{R-Cl}$) are coupled in the presence of $\ce{Na, Mg}$, or other reactive metals  to form an alkane (original classical Wurtz reaction used $\ce{Na}$ metal). The mechanism begins with a single electron transfer (SET) from sodium metal to the alkyl halide, which dissociates to form an alkyl radical ($\ce{R^.}$) in a similar fashion to the formation of a Grignard reagent and sodium halide salt (e.g., $\ce{NaCl}$):

$$\ce{R-X + Na^\circ -> R^. + NaX} \; \text{ where } \ce{X = Cl, Br,} \text{ or } \ce{I}$$

A second molecule of sodium undergoes another SET to the alkyl radical to form a nucleophilic carbanion (e.g., $\ce{R-}$):

$$\ce{R^. + Na^\circ -> R^- + Na+} $$

This alkylmetal intermediate can be isolated (e.g., alkyllithium) and has been isolated in a several cases. The carbanion then attacks another molecule of alkyl halide in a nucleophilic substitution reaction ($\mathrm{S_N2}$) to form the final coupled product (e.g., $\ce{R-R}$) and another molecule of sodium halide salt:

$$\ce{R^- + R-X -> R-R + X-} $$

Thus, the mechanism is believed to be undergoing in both free radical and ionic pathways in particular steps. The depicted diagram of the mechanism is illustrated in below:

In a review of the Wurtz reaction (Ref.1) revealed that the reaction is severely limited in scope as demonstrated by the following observations:

In the synthesis of noncyclic systems (dimerization or cross-coupling) reasonable yields require the use of primary alkyl halides, with iodides giving the best results. Secondary halides give very poor results;
Cross-coupling reactions (using a mixture of two different alkyl halides) give approximately statistical ratios of cross-coupled : dimeric products, due to the high reactivity of the organosodium intermediates involved. Thus, in the synthesis of 3-methyluntriacontane (4 ($\ce{R-R'}$); see above diagram), coupling of the iodide (1; $\ce{R-I}$) with 1-iodooctadecane (2; $\ce{R'-I}$) under standard Wurtz conditions gave a mixture of 3 ($\ce{R-R}$) from dimerization of 1, 4 ($\ce{R-R'}$) from the desired cross-coupling, and 5 ($\ce{R'-R'}$) from dimerization of 2, in the ratio $21:50:29$.

However, This classical reaction is of value in the preparation of cyclic systems, notably [2.2]phanes and bicyclobutanes (e.g., bicyclo[1.1.0]butane), as well as in the preparation of symmetric dimers. The Wurtz reaction provides one of the most successful approaches to these class of compounds, by ring-closure reactions. The reaction may proceed either intra- or inter-molecularly, with intramolecular couplings giving better yields (Ref.1):

The modest yields of [2.2]phane products are due to competing oligomerization/cyclization reactions, which is demonstrated by the product distribution obtained on treatment of the dibromide (6) with sodium/tetraphenylethane (TPE) in THF at $\pu{-80 ^\circ C}$ (See above diagram). A homologous series of products has been obtained under these conditions, the approximate yields of the metacyclophanes being obtained were given in the scheme.

In conclusion, the Wurtz reaction is seldom used because of side reactions and it is intolerant of a range of functional groups. It has limited use to the synthesis of symmetric alkanes and cycloalkanes. If two dissimilar alkyl halides are taken as reactants in cross coupling reactions, the resulting product is a mixture of alkanes that is often difficult to separate. In the case of (1,3), (1,4), (1,5), (1,6) dihalides (when used as a reactant), the reaction leads to formation of cyclic products instead of intended dimerization. The Wurtz reaction fails in case of tertiary halides. In addition, since the reaction involves free radical species in one part of the mechanism, a side reaction occurs to produce an alkene (specially with 1,2-dihalides). This side reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon atom.

There is an organometalic compound in the mechanism at one stage, which undergoes a $\mathrm{S_N2}$ replacement of halide ion of available second $\ce{R-X}$ molecule. Whether these types of organometalic compounds are ionic (cabanionic nature of $\ce{R-}$) has been discussed and reviewed using quantum chemical calculations recently (Ref.2).

Thus, it is safe to say that the reaction is better for primary ($1^\circ$) halides, regardless of whether the mechanism is free radical oriented or carboanion oriented. Even more than 150 years after Charles Adolphe Wurtz introduced the reaction in 1855, the mechanism of the reaction is still up in air, regardless of involvement of both in the mechanism.

References:

David C. Billington, "Chapter 2.1: Coupling Reactions Between $\mathrm{sp^3}$ Carbon Centers (https://doi.org/10.1016/B978-0-08-052349-1.00068-8)," In Comprehensive Organic Synthesis, Volume 3: Carbon–Carbon $\sigma$-Bond Formation; Barry M. Trost, Ian Fleming, Editors-in-Chief; Elsevier Science Ltd.: New York, NY, 1991, pp. 413-434 (ISBN: 978-0-08-052349-1).
Christoph Lambert, Paul von Ragué Schleyer, “Are Polar Organometallic Compounds “Carbanions”? The Gegenion Effect on Structure and Energies of Alkali‐Metal Compounds,” Angew. Chem., Intl. Ed. 1994, 33(11), 1129-1140  (https://doi.org/10.1002/anie.199411291).


Why is a reaction more favorable if one of the products is not soluble/sparingly soluble?

If the reaction product is removed from the phase (liquid solution for example) in which the equilibrium reaction is occurring, significant amount of product can be obtained even if the equilibrium constant for the forward reaction is relatively small ($K<1$).

Aside from precipitation, product could also be removed by vaporization, with subsequent condensation (distillation) to collect the product.  Anything that reduces the product concentration in the phase where the equillibrium occurs favors the forward reaction.


What makes a glowing splinter glow and a burning splinter burn

A splinter can undergo multiple types of thermal reactions depending on both the maximum temperature and amount of oxygen present.  The answer to the question seems to be in the difference between total and partial inceneration.

In total inceneration, which is very likely in the total oxygen atmosphere at high temperatures, all of the organic carbon is converted to carbon dioxide. Carbon dioxide is colorless so the emission of electrons, i.e. the glowing, is observable.

In partial inceneration, which occurs when oxygen is present but deficient, the organic carbon is converted into both carbon dioxide and a variety of volatile compounds observable as smoke. Smoke forms at lower temperatures than thermal emissions. Smoke can also obscure thermal emissions.


Why does ice water get colder when salt is added?

When you add salt to an ice cube, you end up with an ice cube whose temperature is above its melting point.

This ice cube will do what any ice cube above its melting point will do: it will melt. As it melts, it cools down, since energy is being used to break bonds in the solid state.

(Note that the above point can be confusing if you're new to thinking about phase transitions. An ice cube melting will take up energy, while an ice cube freezing will give off energy. I like to think of it in terms of Le Chatelier's principle: if you need to lower the temperature to freeze an ice cube, this means that the water gives off heat as it freezes.)

The cooling you get, therefore, comes from the fact that some of the bonds in the ice are broken to form water, taking energy with them. The loss of energy from the ice cube is what causes it to cool.


Looking at the equation of a reaction, how do we know it is endotermic or exotermic

You are correct. If you have enthalpies of formation of each reagent and each product then you sum up enthalpies of each product multiplied by stoichiometric coefficient and subtract enthalpies of each reagent multiplied by stoichiometric coefficient. If the result is negative then the reaction is exothermic and vice versa.


Rotational frequencies are not close to zero. They are only smaller than vibration frequencies. You may know that frequencies and energies are proportional at the atomic scale. It does not require much energy to make a molecule move or rotate. More energy is needed if you want to deform it and try to stretch its bonds.


Changes of Internal Energy un open system

Suppose you assign an reference state for internal energy such that the internal energy per mole u of the gas is zero at $T=T_r$.  This reference state is going to be held fixed for the material both entering or leaving the open system, as well as the material within the system at any time.  Then, in the initial state of a constant-temperature system, $U_{init}=n_{init}C_v(T-T_r)$ and, in the final state, $U_{final}=n_{final}C_v(T-T_r)$.  So, $$\Delta U=\Delta nC_v(T-T_r)$$Typically, in the overall energy balance on the system, including entering and exiting streams, $T_r$ will cancel out (just as it would for a closed system).

Please note, however, that if chemical reactions are occurring, the equation for   the internal energy and the specification of the reference state has to be extended to take this into account (using the concept of "internal energy of formation").


Oxidation number of oxygen atoms in oxygen difluoride

Oxygen difluoride $\ce{OF2}$ has a single oxygen atom. You must mean dioxygen difluoride $\ce{O2F2}$.
The rules are the same as for hydrogen peroxide.
Both oxygen atoms in $\ce{O2F2}$  are equivalent, so their oxidation number is +1.


Do molecules with axial chirality have "stereogenic units"?

Yes, axially chiral compounds have a stereogenic unit. Biphenyls (such as BINOL) and allenes are common types of compounds that exhibit axial chirality. As you describe, for BINOL the stereogenic unit is the set of atoms including the carbons joining the aryl groups and one carbon further from the biaryl junction. IUPAC's description is simply meant to define a torsional angle, or thought of another way, define two planes around an axis.

  If it does qualify, suppose temperature is increased to the point where spontaneous racemization happens (and suppose that any side reactions or decomposition can be ignored). Is that chain no longer a stereogenic unit?

This is a great question that forces us to think about what we mean by 'stable' and under what conditions is something stable.

Reference to 'stable conformation' means that the two conformations cannot interconvert at ambient temperature. In BINOL, there is restricted rotation around the biaryl bond, which means that the two stereoisomers cannot interconvert under ambient conditions.

Let's think though about a much simpler compound with four carbons in a row that can be used to define a torsional angle: butane. The gauche conformation actually meets the definition of chiral: it's mirror image is not superimposable on itself. However we don't consider butane to have a stereogenic unit (or consider butane to be a chiral compound) because those two conformations can interconvert readily at room temperature.

So those are two ends of the spectrum if we consider room temperature: BINOL has a stereogenic element but butane does not.

Let's say we heat up (S)-BINOL to a temperature that allows for rotation around the biaryl bond. As you say, the sample will racemize. At that temperature, the compound no longer has a stereogenic unit because rotation is no longer restricted. Allow the sample to cool, and the stereogenic unit returns, although the sample will be racemic.


Why is the molecular peak not observed in the mass spectrum of 2,2-dimethylbutane?

The answer probably has to do with how the mass spectrum was obtained.

Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately $70~\mathrm{eV}$ of energy), and then bombard the sample with those high-energy electrons in order to ionise the sample.

The ionisation of these organic molecules doesn't take that much energy - $15~\mathrm{eV}$ is a good approximation. So you have an extra $55~\mathrm{eV}$ of energy in each molecule. How can the molecule get rid of this excess energy? The simplest way is to break chemical bonds; a typical bond strength lies between $3$ and $10~\mathrm{eV}$.

Branched compounds in general do tend to fragment more easily than straight-chain compounds. The reason is simple. If you think about n-hexane, no matter which bond you cleave, you're going to end up with a primary cation (or radical). However, if you cleave at one of the branching points of 2,2-dimethylbutane, you'll get a tertiary cation (or radical). A more stable product means that fragmentation occurs more readily.

However, the fact that the molecular ion is not observed is not really unique to 2,2-dimethylbutane. Lots of EI mass spectra don't have the $\ce{M+}$ peak. If you want to observe the $\ce{M+}$ peak, it's a better idea to use a gentler ionisation technique, such as chemical ionisation or electrospray ionisation.


Nitrogen dioxide fluorescence quenching and lifetime

You are looking for the rate constant $1/\tau^0=k_2$. The rate equation for the $\ce{NO2^*}$ is (using $N^*$ for $\ce{[NO2^*]}$ and N for ground state NO2), $\displaystyle \frac{dN^*}{dt}=-k_2N^*$ and when quenched $\displaystyle \frac{dN^*}{dt}=-k_2N^* -N^*((k_3+k_5)N+k_4[Xe])$ which can both be integrated to give exponentials, i.e $N^*=N^*_0e^{-k_2t}$ and similarly for the other equation. Writing the rate constants as $1/\tau^0=k_2$ and $1/\tau_Q=k_2+(k_3+k_5)N+k_4[Xe]$ you can find the values after looking at the pattern of the quencher concentrations in the question.


Natural Diamonds are formed under high pressure and temperature at depths of 140 to 190 kilometers (87 to 118 mi) in the Earth's mantle. Carbon-containing minerals provide the carbon source, and the growth occurs over periods from 1 billion to 3.3 billion years (25% to 75% of the age of the Earth). Diamonds are brought close to the Earth's surface through deep volcanic eruptions by a magma, which cools into igneous rocks known as kimberlites and lamproites.

According to Wikipedia,

  The formation of natural diamond requires very specific conditions—exposure of carbon-bearing materials to high pressure, ranging approximately between 45 and 60 kilobars (4.5 and 6 GPa), but at a comparatively low temperature range between approximately 900 and 1,300 °C (1,650 and 2,370 °F). These conditions are met in two places on Earth; in the lithospheric mantle below relatively stable continental plates, and at the site of a meteorite strike.

So yes, Diamonds can be produced synthetically if you can simulate almost similar conditions as in Earth's Mantle. (See high-pressure high-temperature process; Synthetic Diamond) An alternative, and completely different growth technique is chemical vapor deposition (CVD).


How to get a precise reading from pipette?

The bottom of the meniscus (in the middle) is your measurement point and should lie on the line that you are trying to read.
This is completely dependent on the type of pipette that you are using. Some pipettes are designed so that you just leave the last drop and some you are supposed to just touch the tip of the pipette to the surface of what you've dispensed already to get the last bit. TD pipettes stand for "to deliver" (leave the last bit). TC pipettes stand for "to contain" (take the last bit out).
Again, this is dependent on the pipette. If you have a classic bulb pipette then you shouldn't hold it by the bulb as it is the weakest part and that's how you end up with glass in your hand.


Changes in pressure for an equilibrium reaction containing a solid/liquid on one side only

There are many examples of such reactions, for instance:

$$
\begin{align}
\ce{S (s) + O2 (g) &<=> SO2 (g)}\\
\ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\\
\ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
\end{align}
$$

However, your assumption

  all that is considered is which side the solid/liquid is on, not the moles of gas present on either side

is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.


What is meant by a "very strong" Be−O bond? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?

You got that absolutely right.
$\ce{Be\bond{-}O}$ is a strong bond because of the small size of Be. Smaller cation size means a stronger pull on the $\ce{O}$ electrons, thus reducing the bond length and hence increasing the bond strength.

[...]the electrons in the O−H bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O\bond{−}H}$ bonds more polar.

You struck at the right point. Let's look at the original question.
I'm sure you must know that a stronger acid has a higher tendency to release $\ce{H+}$ ions. This means that the electron pair should be drawn away from $\ce{H}$ atom in the $\ce{O\bond{−}H}$ bond. Thus, a polarized bond makes it easier for the H+ ion to 'escape' from the molecule.

Why does a polar bond have to be a weaker one?

It is not appropriate to answer this question in definitive terms. The strength of a bond is defined by its bond length, not its polarity. To address the confusion you're having, look back at the context of the question. $$\ce{[Be(H2O)4]^{2+} + H2O -> [Be(H2O)3OH]+ + H3O+}$$
We're talking about a compound with polar O-H bonds dissolved in water. Water has a high dielectric constant, and can easily break ionic/polar bonds since it liberates a high amount of solvation energy. This is because of its high dielectric constant.
So, are polar bonds weaker than non-polar bonds in water? Sure, I can see your line of thinking. But, always remember that ionic bonds can easily be broken by water, but that doesn't mean that ionic bonds are weaker than non-polar bonds.


Decide major product for first order elimination reaction

The product that you have drawn is very unlikely to be the major product of a mixed aldol condensation. The actual major product in said reaction would be (assuming I get the substrates right) would be as follows:

However, your question is a valid one if we take this out of consideration. Now, If we consider just the elimination reaction and your reaction conditions, your mechanism is correct. However, if this reaction were to take place via E2, the rate determining step would be that of the removal of the hydrogen and this would mean that the product formed would be (c). In case of E1, (b) would be the major product since the rate determining step is that of the formation of the carbocation.


Determining the oxidation state of a bridging platinum complex

For organometallic complexes it's often easier to throw the "more electronegative element" idea out entirely and focus on the characteristics of the ligands. This is what @andselisk's comment is referring to. Generally ligands are partitioned into two types, $\ce{L}$ and $\ce{X}$:

$\ce{L}$ refers to neutral ligands: for example $\ce{CO}$, phosphines, $\ce{H2O}$, alkenes (via the pi bond), ... and $\ce{SMe2}$ falls comfortably in this category.
$\ce{X}$ refers to anionic ligands: for example halides, hydroxide, azide, alkyl anions such as $\ce{Me}$, ...

Other ligands can be constructed as sums of these: so for example η4-butadiene would be $\ce{L2}$ (two pi bonds coordinating to the metal), an allyl group would be $\ce{LX}$ (one pi bond plus one anionic centre - ignore resonance), and so on.

The oxidation state of the metal, then, is simply the charge on the metal plus the number of X ligands. In this case each Pt centre is neutral, and has two L and two X ligands. That makes for a total oxidation state of +2.

As another example, the platinum in Zeise's salt has one L ligand (ethylene), three X ligands (chloride), and a negative charge: so its oxidation state is $-1 + 3 = +2$.

Of course, this method will fall apart very quickly when there are non-innocent ligands: $\ce{NO}$ and $\ce{O2}$ are the examples that come to mind, although there are many more. But for relatively simple complexes they suffice.


Do you assume undrawn hydrogens are pointing away from you when determining R/S?

When no bond from the chiral center has up or down graphical indication, it can mean any or unknown stereochemical configuration. Which might be confusing, because it can also mean, that the stereochemical configuration on this atom was mistakenly omitted.

To explicitly indicate that any or unknown configuration, wavy line bond can be used:

(Hydrogen atom with the dashed bond on the (other) chiral center can be can be omitted at all, the stereochemical meaning is unambiguous, much more if the bond wedge was filled with black.)


How to find the percentage contribution of a compound via gravimetric analysis?


Why is a titration necessary? Why can't we just compare the moles of the acid and base?

Man! you must have got impure sample of Mg(OH)2
Moreover, Mg(OH)2 being sparingly soluble may not be finding a way for direct titration, so the method suggested in the problem is a process we call as BACK-TITRATION. This you have to follow i all such situations, where direct titration might not, for any of the reasons, be possible.

Because all 1 g Mg(OH)2 is not pure, you can expect incorrect answer if you are directly comparing the moles of H+ & OH- ions.

Would this be fine?


Isothermal vs adiabatic compression of an ideal gas

There are a number of ways to look at this. One is to look at the equations. Another is to plot the pressure p as a function of volume V. The area under the pressure curve on the plot is the total work done on the system.

It is very easy to answer your second question by inspection of this figure, but perhaps not the first question without actually evaluating integrals (areas).

Yet another way to answer this, but perhaps hand-wavy, is to think of what happens as you compress the ideal gas either adiabatically or isothermally. In the adiabatic compression no energy is lost to the surroundings as heat. All of the energy transferred to the gas as compression work is retained by the gas. But the only way to increase the internal energy of an ideal gas is to raise its temperature. Therefore the effect of the adiabatic work is to increase the temperature of the gas. The increase in temperature is accompanied by an increase in pressure above that due to Boyle's law. This follows from
$$p=\frac{nRT}{V}$$
In the adiabatic process p increases with the higher temperature above the level it would have if the heat had been lost to keep the temperature constant, as it would in an isothermal process. In that way the adiabatic compression is a more efficient way of increasing the pressure. This means less work is required to arrive at the same final p. By the same logic, if you want to achieve a determined final volume, you will have to work harder in an adiabatic setting because the pressure you are working against increases more rapidly as the gas gets hotter.


Where do the electrons that are released in the sun go?

Some of the electrons will react with positrons formed during the fusion processes; they annihilate each other and give off high-energy radiation (also called gamma rays):

Source

But this process  happens at the core, and there is no way that those electrons get out to the surface. So my best guess is that the electrons which don't get annihilated just stay there in the core plasma. I do not think that they combine with the $\ce{^{4}_2 He^{2+}}$ particles formed by this chain, because the temperature is still too high. So you start out with 6 protons and 6 electrons (for electrical neutrality reasons), two electrons get annihilated in the process and you end up with a doubly positively charged helium particle and two protons and four electrons. So the electrical neutrality is not violated.

There is not only the core of the sun but also the outer layers and the corona, from where we observe the solar wind, which is a stream of particles, mainly electrons and protons. So there you have some electrons that escape the sun. But these are not the electrons from the core, they come from the surface of the sun.


What is the word for how polyprotic an acid is?

The term that I learned was (ironically) "basicity."

I'm not sure if that's the standard term, or even if there is a standard term, but this refers to how many bases you need to neutralize the acid completely.

EDIT: Thanks for the correction from LDC3: the "basicity" of an acid refers to how many bases you've already added to it. Thus, $\ce{NaH2PO4}$ would be "sodium phosphate, monobasic," $\ce{Na2HPO4}$ would be "sodium phosphate, dibasic," and so on.

Again, not sure if this is standard terminology.


There are mainly two factors governing this phenomenon of ring expansion and rearrangement of carbocation, mainly Stability factor and Thermodynamic factor. When you have a carbocation which can rearrange within itself, it will definitely prefer a rearranged structure which has the highest stability. Not only that, the carbocation also needs some energy for the rearrangements to occur. If it needs multiple steps of rearrangements to attain a structure which has decent stability, but the loss in the potential energy can't compensate the requirement of energy for rearrangement, it is not also possible for the molecule to undergo that many number of steps thermodynamically.

Based on these two factors, we can at least logically conclude which will be the final rearranged product. In both of your predicted compounds, first there will be a rearrangement by Hydride shift. Now it is highly difficult for a molecule to undergo two rearrangements at once unless it gets  extreme high stability by the second case. In this case, it does by forming a cyclopropyl methyl cation which is unusually highly stable due to non-classical resonance and thus ring-expansion occurs in this case. So, by these two steps you have got your first possible carbocation.

Now , if we look for the third possible rearrangement of ring-expansion, here energy requirement is extremely high as it becomes more and more difficult for a molecule to undergo rearrangement as its number increases as a single reaction intermediates. Also, the newly formed carbocation is tertiary which is obviously less stable than cyclopropyl methyl carbocation. Thus, the last ring-expansion is energetically less favourable.

So, I think the correct one should be the image you have drawn at the end in your question, as further more rearrangements on that system is both thermodynamically and stability wise less feasible.


There are a bunch of of sub parts to these questions; let me see if I can tease some of them out.
Why does a potential difference occur at an interface?
The easiest way to answer this question is through thermodynamics.  If we consider the situation of a reactive metal rod (e.g. magnesium) being exposed to water, the reaction $$\ce{Mg -> Mg^{2+} + 2 e-}$$ is favorable.  In other words, the $\Delta G$ for this reaction will be less than zero.  I've done nothing more than summarize the website you reference in your question, which does a nice job at describing what happens when a piece of metal comes in contact with water, so I don't feel the need to repeat that work.  We then get to your next question:
Why is a second phase required to generate a potential difference?
If we have a universe that consists just of magnesium, then there would be no place for the above reaction to occur.  In this trivialized case, there is no driving force to make magnesium give up its electrons.  That doesn't mean it would not happen, only that any dynamic equilibrium that is established would highly favor the pure metal.  The meat of question one, then, is (may be?):
Why doesn't a potential difference between a metal and gas occur?
Actually it does.  There's an interesting paper (which is behind a paywall if you are not at a university that subscribes to this journal) that describes the measurement of potentials across metal/gas interfaces.  The challenge in studying metal/gas potentials is the ability of controlling this potential externally.  With the metal/liquid scenario, an electrochemist (or budding electrochemist) can add another electrode to the solution and apply a potential between the electrodes.  It's much harder to do the same thing with a gas.  (Well, technically it's *not harder to connect two electrodes to one another and hold them in air, but the redox work functions, resistance and capacitance issues that arise in gas-phase electrochemistry make this a really challenging feat.)
The paper I reference above uses chemisorbed species that have active IR or Raman vibration modes to measure the surface potential using spectroscopy and something called Stark tuning (not the best reference, but it's also not behind a paywall) which states that the vibrational frequency of an harmonic oscillator is proportional to an applied electric field.  Not many metal/gas interfaces have been explored using this method, but those that have display surface potentials fairly close to 0 vs. SCE or 5 eV vs. vacuum.
What about the rest of the stuff in solution?
Back to the metal/liquid scenario, the last part of your first question asks about other ions in solution.  They most definitely play a role in defining the potential of a metal in solution.  In order to accurately predict electrode potentials, one must know concentrations and possible reactions (including states of matter) for all species present.
What about question 2?
I'm afraid question 2 might be based on the false premise that we can measure the potential of an electrode in the absence of anything else (liquid, gas, ions or molecules).  When electrochemists refer to a 'half cell', they have in mind both the reduced (in this case, metal electrode) and oxidized (metal ions) forms of a redox couple contained within the same system.  (Think about a beaker with a copper wire placed in it and the beaker is filled with a solution containing 0.1 M $CuCl_2$.  That half-cell or system has a given potential based upon (primarily) the standard reduction potential of $Cu^{2+}$ to $Cu$ and the concentration of $CuCl_2$.  If I have misread your second question, please edit and I will update my answer as needed.


What makes some chemicals form steep density gradients in solution?

A determination was made that coordination number of cesium ions is 8 only in diluted solutions; in concentrated solutions it is near 4. The form of cation hydrate complexes depends largely on concentration and effect of counterions. https://inis.iaea.org/search/search.aspx?orig_q=RN:27004247

If the cesium hydrates are fairly stable, their volumes, and hence densities would vary with the amount of water bonded. It would seem that after centrifugation and separation, each portion with a uniform density would be a separate phase, with a different composition. Not a liquid crystal, or a crystalline liquid. Maybe a liquid glass? It should be possible to extract some of each layer and determine the total solids, or water content.

Sucrose presents a more complicated situation, but studies indicate degrees of hydration with relaxation times of a few picoseconds to 30 or so (https://www.sciencedirect.com/science/article/abs/pii/S0008621596910198), and so this could indicate hydrates of different stability that might be forced into a certain and constant composition - another liquid glass.

Another separation from a mixture that I have seen occasionally is when a mixture of different polymers in the same solvent evaporates. If the polymers are of sufficiently low molecular weight, or high mobility, or slow enough evaporation rate, they might separate into two layers. I suppose if the mobility is not high enough, the mixed polymer will evaporate with obvious inhomogeneity, or haze.


Why has nature evolved to put a porphyrin (i.e. chlorophyll) as the chromophore in a leaf?

Chlorophyll (chl) is involved in two ways in photosynthetic organisms firstly as a light gathering pigment and second in electron transfer once the light is captured. It is unique in this role.

In the first case, that of gathering light there are pigment protein complexes containing many chl (100's) and work in such a way that energy absorbed in one molecule is transferred among the others until it reaches the reaction centre where electron transfer starts. The feature that makes chl good for the energy transfer bit is that it has a large overlap in energy (equivalently wavelength) between its fluorescence spectrum and the absorption spectrum. This allows very efficient energy transfer between pigments, each step taking only picoseconds. Chlorophyll is among the best if not the best molecules known to do this.

(In some cases carotenoids may assist this energy capture process and in some antenna called phycobilisomes linear tetrapyrroles augment chl (they extend the wavelength range when the organism normally grows in low light conditions) but in these cases there are also chl pigment-protein complexes.)

The second role is that after energy is absorbed it is funnelled to a Reaction Centre. This is done by the energy transfer steps described above but slight changes in the chl environment change its energy levels slightly so that spatially the energy moves towards the reaction centre where electron transfer begins. The electron produced ultimately reduces carbon dioxide to carbohydrate and is replaced by dragging an electron of a water molecule, this is done in the protein in a sequence of complex steps called the dark reactions.

Returning to the Reaction Centre, this consists of a pair of slightly offset face-to-face chl molecules (the Special Pair)  that act as an energy trap and from here an electron is transferred in a couple of picoseconds to a nearby chl then to a pheophytin (a chl less the Mg atom) then to quinone then to an Fe atom all within a nanosecond. The crystal structures are all known and the photophysics of species involved also measured. The structure of these Reaction Centres can be found on the web.

The important property here is that the special pair of chl molecules has the right redox potential to donate an electron to a nearby chl and this the correct potential to transfer the electron to the pheophytin at the same time preserving as much energy as possible and preventing the electron from returning, i.e. making sure it goes down hill in energy but only slightly and thus preserving as much of the initial photon energy as possible. The special pair does not fluoresce to any appreciable extent and so does not loose energy this way.

Chlorophyll has a unique set of properties, excellent ability to transfer energy among other chl with out being quenched until it reaches the reaction centre where the special pair uses this energy to transfer an electron to nearby molecules and start the chemistry leading to reducing carbon dioxide using water as a fuel. All the variant of photosynthesis, algae, plants, bacteria work in essentially the same way, the bacteria have one photosystem, the plants two (PS1, PS2) that work together.

(You should be able to find lost pictures of xray structures of pigment protein complexes on the web but if not I can post some later)


How can we say that one mole of a substance contains avogadro number of particles?

There's a nice Scientific American article that provides an interesting historical perspective on Avogadro's Number.  They go on to point out that one of the early accurate determinations was based on the first accurate measurement of the charge on an electron.  The charge on a mole of electrons (1 Faraday) had been known for some time.  Dividing the latter by the former yielded Avogadro's Number.

As to the volume of 1 mole of gas at STP, imagine the following experiment:

1) weigh out 1 mole of oxygen gas (32 gm)

2) fill a balloon with this weight of gas

3) attach a pressure gauge to the balloon and expand or contract the balloon's volume until the gauge reads 1 atmosphere

4) Immerse the balloon in a water tank and measure the volume of water displaced - 22.4 liters


Plutonium having more oxidation states than samarium?

This is yet another interesting effect of the anomalous compactness of orbitals in the first appearance of each type of subshell ($1s$, $2p$, $3d$, $4f$, $5g$, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with no radial nodes. This means the electrons in these subshells are closer to the nucleus than you might expect, and therefore they are more strongly bound.

The striking chemical similarity of the lanthanides comes from the fact that the $4f$ subshell is both anomalously compact and subjected to a high effective nuclear charge. The electrons end up being held so closely to the nucleus that they effective behave as core electrons, and participate very little in chemical interactions; lanthanides do not form coordination compounds using their $4f$ orbitals, and it is very difficult to go past an oxidation state of +3 in most because few conditions can compensate the large energy required to ionize a fourth electron coupled with the relatively weakly bound compounds they would form.

In the actinides, the $5f$ orbitals do not suffer from the same lack of radial node, and therefore are more chemically available. The first half of the actinides show many compounds with high oxidation numbers, and have significant coordination chemistry using $f$ orbitals. Curiously, as you go form curium to berkelium, there is a sudden increase in reluctance to display high oxidation numbers, and the actinides beyond curium tend to behave similarly (though of course exploration of their chemical properties is significantly hindered by their rarity and instability). This is likely because even though the $5f$ orbitals have a radial node, in the second half of the actinides they are subjected to such a high effective nuclear charge that become too strongly bound to display significant participation in chemistry.

This same effect explains why the metals in the first row of transition metals do not display as many stable compounds with high oxidation states compared to the second and third rows, as the $3d$ orbitals are much smaller than $4d$ or $5d$ orbitals.


Why can Neutral Red (indicator) not test values below its pKa?

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.

Other indicators change from one strong colour to another, there the $pKa$ will be more in the middle of the transition range.


Firstly, I would replace the word "feasible" with "favorable". I'm also going to replace $E_{cell}$ with $E_{OCV}$, where the open circuit voltage (OCV) is the potential of the cell without any applied electric potential. So you're right in that if the $E_{OCV}$ is positive, the net reaction is thermodynamically favorable: it will occur spontaneously if allowed to and will produce energy that can be used by the load connected to the battery (i.e. the cell can do work).

You're correct in saying that the reverse reaction is thermodynamically unfavorable, but that doesn't mean the reaction won't happen, just that you need to do work on the battery in order for it to happen. In most cases, this means applying a bias potential ($E_{cell}$) that is more positive than the $E_{OCV}$. When that happens, the current flowing across this bias potential takes energy into the cell and recharges it.

However, not every reaction will proceed in this way. In many cases, if a high potential is applied, another reaction will occur. If you have water in your battery and apply a high potential, you may electrolyze water, splitting into elemental hydrogen and oxygen. This will then reduce the usable capacity and/or cause rupture of the battery.

Lastly, all of this ignores the role of kinetics. Just because a reaction is thermodynamically favorable doesn't mean that it happens at a rate that useful. If you have a cell with $E_{OCV} = 3.60V$, applying $3.601V$ will begin to charge the cell but at such a slow rate than you'll never notice. You generally need to apply at least a few tens of millivolts of over-potential to a lithium ion battery in order to get any appreciable current. So in the case on non-rechargeable (chemically non-reversible) batteries, the over potential needed to reverse the reaction will be big enough that you will instead hit the potential for some other, undesirable reaction.


Aluminum and gallium are on different periods. When an additional shell is (fictitiously) added to aluminum, the size is seen to decrease. This is, as correctly mentioned, due to the poor shielding offered by the $(n-1)d$ orbitals to the incoming electrons in the ultimate $p$ subshell.

However, between copper and zinc, the matter seems to be different–their electronic configurations are different.

While copper has a $4s^13d^{10}$ arrangement, zinc has a $4s^23d^{10}$. I would presume the fully filled spherical $4s$ orbital to considerably repel (also read as: shield) the d-electrons,  and hence the effect appears as a minute increase in size.

There can be deeper discourses, but I think this should suffice for now.


Does adding water to a reaction mixture shift equilibrium?

Yes, the equilibrium does indeed readjust after a change in concentration. This can be seen by careful examination of the formula of the equilibrium constant $K$. Consider an arbitrary reaction that produces two products from a single reactant and is in equilibrium:

$$\begin{align}\ce{A &<=> B + C}\tag{1}\\[0.5em]
K &= \frac{[\ce{B}][\ce{C}]}{[\ce{A}]}\tag{2}\end{align}$$

Let’s assume initial concentrations of $\pu{2M}$ for each species; that gives us $K = 2$ in dimensionless units (you can also use M as a unit here). Remember that $K$ is a constant so we can examine whether equilibrium has been reached by examining $K$.

Now let’s double the volume of solvent. The new concentration of each reactant is $\pu{1M}$. Therefore:

$$K' = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]} = \frac{1\times 1}{1} = 1\tag{2'}$$

Since $1\ne2$, we are no longer at equilibrium. Thus the concentrations must readjust. How? Obviously, the value of the fraction is too low. To increase its value, we can increase the numerator or decrease the denominator — luckily for us, both of these mean that we must perform the reaction in a forward direction to get think right again. Since $[\ce{B}]=[\ce{C}]$, we can actually calculate the ratio of concentrations and from $2[\ce{A}]+[\ce{B}]+[\ce{C}]=\text{const.}=4$, we can calculate exact values.

$$\begin{align}&&K &= \frac{[\ce{B}]^2}{[\ce{A}]}\\[0.5em]
\Longrightarrow&&[\ce{A}] &= \frac{[\ce{B}]^2}2\\[1.3em]
&&2\times\frac{[\ce{B}]^2}{2} + [\ce{B}] + [\ce{B}]&=4\\
&&[\ce{B}]^2 + 2[\ce{B}]-4 &= 0\\
\Longrightarrow&&[\ce{B}]_{1/2} &= \frac{-2\pm\sqrt{2^2+4\times4}}{2}\\[0.5em]
&&[\ce{B}]_{1/2} &= -1\pm\sqrt{5}\\[0.5em]
&&[\ce{B}]_{1/2} &\approx 1.24\end{align}$$

Thus, the product side has increased.

A more real-world example is maybe the dimerisation of hydrogenchromate $\ce{HCrO4-}$ according to $(3)$:

$$\ce{2 HCrO4- <=> Cr2O7^2- + H2O}\tag{3}$$

This is also pH-dependent, since hydrogenchromate is a weak acid and can deprotonate to give chromate $\ce{CrO4^2-}$. However, it is fairly easy to try this experiment yourself: take an orange coloured dichromate solution and add water. A colour change to yellow should be observed indicating that the reaction has equilibrated to the reactants’ side.


How to disperse salicylic acid in sodium lauryl sulfate solution?

I don't have a lot of experience with salicylic acid as I'm just starting to work with it myself at the moment. However, I find that it disperses better if it is added to the solution at a high temperature (before surfactants) and then homogenised (once thickening agents, if any, have been added).

The other thing which may help is by dissolving in denatured alcohol, although I haven't personally tried this myself. Although it's not a shampoo, I have recently formulated a facial cleanser and as described above I firstly added the thickening agents to the water and then salicylic acid, heated to approximately $70~\mathrm{^\circ C}$ and then homogenised.


Is there a reason for the mathematical form of the equilibrium constant?

Equilibrium is also the state in which the forward and reverse reactions are proceeding at equal rate. Consider a hypothetical reaction

$$\ce{aA + bB} \overset{k_1}{\underset{k_{-1}}{\ce{<=>}}}\ce{ cC + dD}$$

If we assume that the forward and reverse processes are elementary reactions (a big assumption), then:

$$rate_1=k_1 [\ce{A}]^a [\ce{B}]^b$$
$$rate_{-1}=k_{-1} [\ce{C}]^c [\ce{D}]^d$$

At equilibrium:
$$rate_1=rate_{-1}$$
$$k_1 [\ce{A}]^a [\ce{B}]^b=k_{-1} [\ce{C}]^c [\ce{D}]^d$$
$$\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

The ratio of the rate constants is the equilibrium constant:

$$K=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$


If we compare the HOF and HOCl
1) In HOF , the oxygen atom cannot do backbonding since fluorine has no vacant orbital while in HOCl it can do hydrogen bonding it can do backbonding which reduces electron density on oxygen by a slight extent( Remember 2pπ-3pπ backbonding is not very much effective due to size difference of 2p and 3p orbitals) . But yeah this factor makes HOCl bond more polar .
But there are two more important factors-:

2[)INDUCTIVE EFFECT]-  among oxygen and fluorine , fluorine is more electronegative , so it partially pulls the sigma bonded electrons towards itself giving a  delta+ charge to oxygen .
While in HOCl , O is more electronegative than Chlorine so the opposite mechanism happens and O gets a slight negative charge.
So according to this HOF has increased bond polarity than HOCl.
3) This factor is the most important and sufficient to answer almost any acid comparison . Compare the stability of the ion formed after releasing H+ ( formally called as the conjugate base of the acid).
So the question left is simply the comparison between OF- and OCl- ,
 The first two factors again work here but another important factor is the 2p-2p electron repulsions( due to smalls ize of sigma bond the electrons in the orbitals of O and F repel each other. This factor  in itself is quiet dominating) present in OF- makes OF- less stable than OCl-

So according to me HOCl must be more acidic than HOF.

Another important thing i wanted to draw attention to is that HOF is not the common acid that can be kept in laboratory. It is quite reactive and reacts vigorously even at 0°C , measuring its acidity is quite diffcult . I also searched online but couldn't find any data concerning Ka values for ionisation of HOF.
I hope this helps.


Why does tetrachlorocuprate(II) form in aqueous solution even though water should be the stronger ligand?

[Water ligands] being stronger than $\ce{Cl-}$ ligands will form a $\ce{[Cu(H2O)6]^2+}$ complex.

This is incorrect. Water ligands may be stronger field ligands in the spectrochemical series but that does not in any way affect the strength of the ligand–metal bond or the kinetics and thermodynamics of certain complexes, their formation and their breakdown.

Which ligand forms bonds of which strength to which metal is only very generally determinable; and in general you would always denote binding affinities in a $1:1$ manner: ligand $\ce{X}$ forms strong bonds to metal $\ce{Y}$. So sulphide ions will form strong dative bonds to mercury(II) ions but not-so-strong bonds to titanium(IV) ions. Likewise, palladium(II) if presented with both phosphane and amino ligands will preferentially bind to phosphanes while copper(I) will preferentially bind to amino ligands.

The thermodynamic energy differences between different complexes are often relatively small and typically the kinetic barrier for ligand exchange is low, too. Thus, in many cases which complex is formed depends strongly on the concentrations of individual ions. The mere fact that complexes without aqua ligands can and are formed shows that water is not that good a ligand after all. In fact, due to the positive charge of the central metal having an anionic ligand bind is often preferred due to simple electrostatic interactions.

To answer the actual question you posed: $\ce{[CuCl4]^2-}$ will form because it is the predominant species under a given chloride ion concentration. This can be backrationalised by looking at formation constants $K_1$ to $K_4$. However, aside from quantum chemical calculations there is no way to properly rationalise it a priori.


Identifying the number of bonding and lone pairs without a dot and cross diagram

There is a simple, four-step calculation that you can perform:

count the valence electrons the atoms you are bonding have.
count the number of valence electrons the atoms would like to have for a noble gas’ valence shell. (i.e. eight for everything main group, two for hydrogen.)
substract $2.-1.$, i.e. the first (existing) from the second (desired). This is your number of bonding electrons. Divide by two for bonding electron pairs.
substract $1.-3.$, i.e. the third (bonding) from the first (existing). This is your number of free electrons. Divide by two for lone pairs.

Using a simple example such as sulphur dioxide:

$6\ (\ce{S}) + 2 \cdot 6\ (\ce{2 O}) = 18$
$8\ (\ce{S}) + 2 \cdot 8\ (\ce{2 O}) = 24$
$24 - 18 = 6$, i.e. three bonding electron pairs.
$18 - 6 = 12$ i.e. six lone pairs.

Unfortunately, that by itself does not allow you to write the structure; you need to know which element is in the centre and how they are connected. Here, we have a central sulphur bonded to two oxygens. The final result is:

$$\ce{O=S+-O-}$$

With two lone pairs on the left-hand oxygen, one on the sulphur and three on the right hand oxygen. The formal charges are important.


Why doesn't EDTA complex with alkali metal ions?

If you're going to be picky that a "complex" doesn't form without covalent coordination bonds, then that's your answer. Alkali metals will tend to form cationic salts with EDTA.

Personally, I agree with the other comments. EDTA does form complexes with alkali metals.

I'd classify the disodium salt of EDTA as a complex. It has a metal and an organic ligand, and there's no direct metal-carbon bond (which would make it organometallic). As discussed in another question, it's a pretty arbitrary distinction between an ionic bond and a covalent one:

  What happens if the electronegativity difference is exactly 2.1?

This distinction is particularly cloudy in the case of metal complexes, since most metals readily gain positive charge.


Does 50% hydrogen peroxide dissolve gold?

I have no expertise, but Pourbaix does.  Let's take a look at some of his diagrams which offer a hint to what is going on, with both silver and gold.  Or more accurately, why gold is indeed less vulnerable to attack by hydrogen peroxide than silver.
Begin with just the diagram for hydrogen and oxygen in aqueous solution.  Ross[1] gives a diagram that includes hydrogen peroxide as well as the stable species hydrogen, oxygen and water (figure from this reference):

Pourbaix diagrams typically include a pair of parallel lines, here labeled (a) and (b) representing water reduction to hydrogen and oxidation to elemental oxygen respectively.  We see that hydrogen peroxide requires a higher oxidizing potential to form than elemental oxygen, by roughly half a volt depending on concentration.  This is no surprise, of course, since elemental oxygen plus water would be more stable.  Hydrogen peroxide is, of course, only metastable.
Now let's see the diagram for silver.  There is some variation from different sources, but Wei et al. [2] show the following, which is typical.

The boundary between elemental silver and various oxidized species (essentially, dissolved silver ions in acid and silver oxides in basic solution) actually lies below the line for water oxidation to oxygen, so in principle even dissolved air could attack silver given enough time, let alone the more powerful oxidant hydrogen peroxide.  In fact, at the potentials where hydrogen peroxide appears in the diagram from Ref. 1, one may even see the oxide of silver(II)!  We are lucky, actually, that the kinetics is slow enough so the silver electrodes have some lifetime.]
Now it is time to go for the gold.  Huayhuas-Chipana et al. [3] adopted the diagram below from Ref. [4], showing that gold is not oxidized at all when water is oxidized to elemental oxygen (the higher of the two red lines) and that even hydrogen peroxide with its potential at half a volt higher is only on the borderline of beginning to oxidize gold.  More over, the soluble cation $\ceAu^{3+}}$ appears only in fairly string acidic solution, further hindering any dissolution of gold.  Thus gold would indeed be expected to hold up more strongly in your hydrogen peroxide solution.

References
1.
P. N. Ross Jr. (2010), "Oxygen reduction reaction on smooth single crystal electrodes", Handbook of Fuel Cells
Vol. 2.
2.
Weifeng Wei, Xuhui Mao, Luis A. Ortiz and  Donald R. Sadoway, "Oriented silver oxidenanostructures synthesized through a template-free electrochemical route", J. Mater. Chem., 2011, 21, 432-438.
3.
Bryan C. Huayhuas-Chipana, Juan Carlos Morales Gomero, Maria Del Pilar, Taboada Sotomayor, "Nanostructured Screen-Printed Electrodes Modified with Self-Assembled Monolayers for Determination of Metronidazole in Different Matrices", Journal of the Brazilian Chemical Society 25(9) (Sept. 2014).
4.
J. Van Muylder, M. Pourbaix, Atlas of Electrochemical
Equilibria in Aqueous Solutions, 2nd English ed.; Marcel
Pourbaix, National Association of Corrosion Engineers: New
York, 1974.


Computational Chemistry Modelling Redox Potentials

Here is a non-electrolysis scenario that could be of interest. Add deoxygenated water to FeCl2 in a closed vessel filled with oxygen or air. Record the oxygen consumption.

Chemistry:

$\ce{4 Fe(2+)  + 2H2O + O2 (g) -> 4 Fe(3+) + 4 OH- }$

The wet ferrous chloride should be presented in a fashion to maximize oxygen exposure from air.

[EDIT] I would also place the system (till the reaction is completed) under pressure to limit adjustments owing to the speed of oxygen assimilation.

To be clear, as apparently required by a comment, the purpose of this experiment, per my understanding of the Question, is to serve as a frame for real world validation of a sophisticated theoretical/computational based model.


How is a counterpoise corrected geometry optimization done?

Background

For a system consisting of two molecules (monomers or fragments are also used) X and Y, the binding energy is

$$
\Delta E_{\text{bind}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{X}}(\ce{X}) + E^{\ce{Y}}(\ce{Y})]
\label{eq:sherrill-1} \tag{Sherrill 1}
$$

where the letters in the parentheses refer to the atoms present in the calculation and the letters in the superscript refer to the (atomic orbital, AO) basis present in the calculation. The first term is the energy calculated for the combined X + Y complex (the dimer) with basis functions, and the next two terms are energy calculations for each isolated monomer with only their respective basis functions. The remainder of this discussion will make more sense if the complex geometry is used for each monomer, rather than the isolated fragment geometry.

The counterpoise-corrected (CP-corrected) binding energy [1] to correct for basis set superposition error (BSSE) [2] is defined as

$$
\Delta E_{\text{bind}}^{\text{CP}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{XY}}(\ce{X}) + E^{\ce{XY}}(\ce{Y})]
\label{eq:sherrill-3} \tag{Sherrill 3}
$$

where the monomer calculations are now performed in the dimer/complex basis. Let's explicitly state how this works for the $E_{\ce{XY}}(\ce{X})$ term. The first molecule X contributes nuclei with charges, basis functions (AOs) centered on those nuclei, and electrons that will count to the final occupied molecular orbital (MO) index into the MO coefficient array. There is no reason why additional AOs that are not centered on atoms can't be added to a calculation. Depending on their spatial location, if they're close enough to have significant overlap, they may combine with atom-centered MOs, increasing the variational flexibility of the calculation and lowering the overall energy. Put another way, place the AOs that would correspond to molecule Y at their correct positions, but don't put the nuclei there, and don't consider the number of electrons they would contribute to the total number of occupied orbitals. This means that for the full electronic Hamiltonian

$$
\hat{H}_{\text{elec}} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee}
$$

calculating the electron-nuclear attraction $\hat{V}_{eN}$ term is now different. Considered explicitly in matrix form in the AO basis,

$$
\begin{align*}
V_{\mu\nu} &= \int \mathop{d\mathbf{r}_{i}} \chi_{\mu}(\mathbf{r}_{i}) \left( \sum_{A}^{N_{\text{atoms}}} \frac{Z_{A}}{|\mathbf{r}_{i} - \mathbf{R}_{A}|} \right) \chi_{\nu}(\mathbf{r}_{i}) \\
&=\sum_{A}^{N_{\text{atoms}}} Z_{A} \left< \chi_{\mu} \middle| \frac{1}{r_{A}} \middle| \chi_{\nu} \right>
\end{align*}
$$

there are now fewer terms in the summation, since the nuclear charges from molecule Y are zero (the atoms just aren't there), but the number of $\mu\nu$ are the same as for the XY complex. This and the $\hat{T}_{e}, \hat{V}_{ee}$ terms aren't really mathematically or functionally different then; this is more to show where the additional basis functions enter, or to show where nuclei appear in the equations [3].

These atoms that don't have nuclei or electrons, only basis functions, are called ghost atoms. Sometimes you also see the term ghost functions, ghost basis, or ghost {something} calculation. Adding the basis of monomer Y to make the full "dimer basis" means taking the monomer X and including basis functions at the nuclear positions for Y.

Geometry optimization

Now to calculate the molecular gradient, that is, the derivative of the energy with respect to the $3N$ nuclear coordinates. This is the central quantity in any geometry optimization. For the sake of simplicity, consider a steepest descent-type update of the nuclear coordinates
$$
R_{A,x}^{(n+1)} = R_{A,x}^{(n)} - \alpha \frac{\partial E_{\text{total}}^{(n)}}{\partial R_{A,x}}
\label{eq:steepest-descent} \tag{Steepest Descent}
$$
where $n$ is the optimization iteration number, $\alpha$ is some small step size with units [length2][energy], and the last term is the derivative of the total (not just electronic) energy with respect to a change in atom $A$'s $x$-coordinate. Even Newton-Raphson-type updates with approximate Hessians (2nd derivative of the energy with respect to nuclear coordinates, rather than the 1st) need the gradient, so we must formulate it.

Formulation of the energy

We're in a bit of trouble, because we want to replace $E_{\text{total}}$ in the gradient with $E_{\text{total}}^{\text{CP}}$, but all we have is $\Delta E_{\text{bind}}^{\text{CP}}$. The concept of CP correction can still be applied to a total energy, but the BSSE must be removed from each monomer. The BSSE correction itself for each monomer is
$$
\begin{split}
E_{\text{BSSE}}(\ce{X}) &= E^{\ce{XY}}(\ce{X}) - E^{\ce{X}}(\ce{X}), \\
E_{\text{BSSE}}(\ce{Y}) &= E^{\ce{XY}}(\ce{Y}) - E^{\ce{Y}}(\ce{Y}),
\end{split}
\label{eq:2}
$$
which, when subtracted from $\eqref{eq:sherrill-1}$, gives $\eqref{eq:sherrill-3}$. More correctly, considering that the geometry for each step is at the final cluster geometry and not the isolated geometry, the above is [4]
$$
\begin{split}
E_{\text{BSSE}}(\ce{X}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{X}) - E_{\ce{XY}}^{\ce{X}}(\ce{X}), \\
E_{\text{BSSE}}(\ce{Y}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{Y}) - E_{\ce{XY}}^{\ce{Y}}(\ce{Y}).
\end{split}
\label{eq:sherrill-10} \tag{Sherrill 10}
$$

The CP-corrected total energy is the full dimer energy with BSSE removed from each monomer is then
$$
\begin{split}
E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} &= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - E_{\text{BSSE}}(\ce{X}) - E_{\text{BSSE}}(\ce{Y}), \\
&= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right].
\end{split}
\label{eq:sherrill-15} \tag{Sherrill 15}
$$
Note that I have modified which geometry is used for each monomer in $\eqref{eq:sherrill-15}$. All monomers are calculated at the supermolecule geometry. This is convenient for two reasons: 1. We are only interested in removing the BSSE, not the effect of monomer deformation, and 2. a isolated monomer geometry without deformation doesn't make sense in the context of a geometry optimization. I also added the tilde to signify that the supermolecular/dimer geometry used may not be the final or minimum-energy geometry, as would be the case during a geometry optimization. We simply extract all structures consistently from a given geometry iteration. Perhaps $\ce{XY}(n)$ would be better notation.

Formulation of the gradient

As Pedro correctly states, the differentiation operator is a linear operator. Because there are no products in $\eqref{eq:sherrill-15}$, the total gradient needed for $\eqref{eq:steepest-descent}$ will be a sum of gradients [5]:
$$
\frac{\partial E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}}}{\partial R_{A,x}} = \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY})}{\partial R_{A,x}} - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X})}{\partial R_{A,x}} \right] - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y})}{\partial R_{A,x}} \right],
$$
so each step of a CP-corrected geometry optimization will require 5 gradient calculations rather than 1. Note that the nuclear gradient should be included for each term as well, which is a trivial calculation.

Extension to other molecular properties

Although not commonly done, counterpoise correction can be applied to any molecular property, not just energies or gradients. Simply replace $E$ or $\partial E/\partial R$ with the property of interest. For example, the CP-corrected polarizability $\alpha$ of two fragments is
$$
\alpha_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} = \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right]
$$
where I believe it now makes even less sense to have each individual fragment calculation not be at the cluster geometry. In papers that calculate CP-corrected properties, no mention is usually made of which geometry the individual calculations are performed at for this reason.

References

Boys, S. Francis; Bernardi, F. The calculation of small molecular interactions by the differences of separate total energies. Some procedures with reduced errors. Mol. Phys. 1970, 19, 553-566.
Sherrill, C. David. Counterpoise Correction and Basis Set Superposition Error. 2010, 1-6.
One implementation note: Most common quantum chemistry packages should allow for the usage of ghost atoms in energy and gradient calculations. However, as Sherrill states, they do not properly allow for composing the full gradient expression to perform CP-corrected geometry optimizations. Gaussian can, and Psi4 may. For programs that can calculate gradients with ghost atoms, Cuby can be used to drive the calculation of CP-corrected geometries and frequencies.
There is a typo in the Sherrill paper; the subscripts for all 4 energy terms should be $AB$, which here are $\ce{XY}$.
Simon, S.; Bertran, J.; Sodupe, M. Effect of Counterpoise Correction on the Geometries and Vibrational Frequencies of Hydrogen Bonded Systems. J. Chem. Phys. A 2001, 105,, 4359-4364.


Why are sodium and potassium displaced from soap by calcium and magnesium ions in hard water?

I have no external justification to point to on this, but I've always thought one aspect contributing to this behavior is the fact that the divalent $\ce{Ca^2+}$ and $\ce{Mg^2+}$ bind two soap fatty acid molecules, whereas $\ce{Na+}$ and $\ce{K+}$ as monovalent ions can only strongly bind one molecule.

As a result, the alkaline earth cations have two fatty acid "tails" that can 'glom' around them, shielding them better from the water molecules and making the resulting complex more stable, as well as less soluble.

On the other hand, since the alkali cations only have one fatty acid "tail" they're strongly associated with, that one tail can't wrap around nearly as completely, and so there is significantly greater accessibility of the solvent to the cation.


Disproportionation of silver(I) in aqueous solution

It is convenient to solve problems like that with a Latimer diagram, which is a great tool for predicting conditions for the reactions of disproportionation and synproportionation.
A generic Latimer diagram
$$\ce{A ->[$E_1$] B ->[$E_2$] C}$$
posesses the following properties:

If $E_2 > E_1,$ then $\ce{B}$ is thermodynamically unstable and disproportionates to $\ce{A}$ and $\ce{C}.$
If $E_2 < E_1,$ then the mixture of $\ce{A}$ and $\ce{C}$ is thermodynamically unstable and synproportionates to $\ce{B}.$

Now we can visualize your problem using a Latimer diagram
$$\ce{Ag^2+ ->[\pu{+2.00 V}] Ag+ ->[\pu{+0.80 V}] Ag}$$
and the condition of disproportionation resulting from the application of Nernst equation:

if the potential to the right of the species is higher than the potential on the left, it will disproportionate.

Since $E^\circ(\ce{Ag+(aq)/Ag}) < E^\circ(\ce{Ag^2+(aq)/Ag+(aq)}),$ disproportionation is thermodynamically unfavorable, and silver(I) can be considered stable in solution.
You can get the same result from a linear combination of both equations written for the disproportionation:
$$
\begin{align}
\ce{Ag+(aq) + e- &-> Ag} &\quad E^\circ_1 = \pu{+0.80 V} & \tag{1}\\
\ce{Ag^2+(aq) + e- &-> Ag+(aq)} &\quad E^\circ_2 = \pu{+2.00 V} &\quad|\cdot (-1)\tag{2}\\
\hline
\ce{2 Ag+(aq) &-> Ag + Ag^2+(aq)} &\quad E^\circ = \pu{-1.20 V}
\end{align}
$$
Since resulting $E^\circ = \pu{-1.20 V} < 0,$ free Gibbs energy $Δ_\mathrm{r}G^\circ = -nFE^\circ > 0,$ and the disproportionation of silver(I) in solution can be considered a thermodynamically unfavorable process.


Disproportionation reaction of dimeric sulphur monochloride

Compiling my comments into an answer, sulfur monohalide disproportionate to sulfur dihalide and/or sulfur tetrahalide and elemental sulfur.
Wikipedia has the reaction:
$$\ce{2S2F2 ->[180 ^\circ C] SF4 + 3S}$$
Also, there is a reaction scheme from a 1974 entry1 which indicates formation of octasulfur and sulfur difluoride as sideproduct.
$$\ce{12S2F2 -> 3S8 + 8SF4}$$
$$\ce{8S2F2 -> S8 + 8SF2}$$
It also further noted that this disproportionation reaction is useful in the synthesis sulfur tetrafluoride and the decomposition of two moles of $\ce{S2F2}$ into one mole of $\ce{SF4}$ and sulfur yields $\pu{\Delta H^\circ = -24.6 kcal/mole}$.
Furthermore, sulfur difluoride may disproportionate to sulfur tetrafluoride and sulfur or some other oligomer/polymer of sulfur or back to sulfur monofluoride or isomerize to various sulfur compounds like $\ce{FSSF}$,  $\ce{SSF2}$ or $\ce{SF3SF}$. These compounds are extremely unstable and prone to hydrolysis:
$$\ce{16SF2 -> 8SF4 + 8S8}$$
$$\ce{3SF2 -> SF4 + S2F2 isomers}$$
The stability of $\ce{SF2}$ can be considered thermodynamically. The decomposition of two moles of $\ce{SF2}$  to yield $\ce{SF4}$ and $\ce{S8}$(gas) is $\pu{-75.7 kcal/mole}$. Another reason for this decomposition is polymerization which forms new sulfur-fluorine bonds. Also, it is noted that $\ce{SF4}$ is thermodynamically unstable and disproportionate to sulfur hexafluoride and sulfur:
$$\ce{24SF4 -> 16SF6 + S8~~~~\Delta H^\circ = -62.9 kcal/mole}$$
So, sulfur compounds in these intermediate oxidation state are found to be thermodynamically unstable and it is also influenced by its kinetic stability.
Sulfur monochloride disproportionation occurs in a similar way as noted in this patent:

Elevated temperature as used herein refers to temperatures between
about 75 and about 150C; and preferably between about 85 and about
138C. Higher temperatures can be used but the sulfur monochloride will
tend to disproportionate to form sulfur and sulfur dichloride. lf
higher amounts of sulfur in the sulfur monochloride product are not
objectionable, higher temperatures can be utilized

Notes and References

Advances in Inorganic Chemistry and Radiochemistry, Academic Press, 1974
There is a different pathway for disproportionation of selenium monochloride/bromide where it is seen to yield $\ce{SeX2}$ and $\ce{Se_{n}X2 (n =3,4)}$ unlike the general pathway you mentioned which yields $\ce{SeX4}$ and elemental selenium. See this paper for more details.


I am trying to picture how electrons move around in atomic orbitals

I think you miss an important notion to get a better understanding of the behavior of the electrons: wave-particle duality. I believe that a better visualization takes into account that you have to deal even with a wave. From what you say it seems that your picture is shifted only to the particle-side,  keep meditating for a half an hour trying to visualize the electron as a wave. For me is a very interesting mind experiment. Then move to the orbital concept.

At Lund University researches took a picture of a moving electron. This image maybe could be helpful for your meditation:

Lund University. "Electron Gets Film Debut In First-ever Video Of Its Kind." ScienceDaily. ScienceDaily, 25 February 2008. www.sciencedaily.com/releases/2008/02/080222095358.htm.


Scale to be considered for comparing electronegativities of nitrogen and chlorine

Please take into account that there are a lot of different ways for calculating electronegativity.

1) Pauling obtains it by comparing the energies of dissociation of the three diatomic molecules AA, BB, and AB, stating that H electronegativity  is 2.1 !  Unit : √energy

2) For Mulliken, it is the average value between the electronic affinity and the first ionization energy of the atom. Unit : energy

3) For Allred - Rochow, it is the force acting between a valence electron and the charge of the atom when the other valence electrons are removed. Unit : force

4) For Sanderson, it is the average electronic density of an atom divided by the average electronic density of the corresponding atom of inert gas.No unit.