Is HF the only method for trying to solve the many-electron Schrödinger equation?

So is this a correct assessment? Do we truly only use one method of trying to solve the many-electron Schrödinger equation? Is there any research being done on alternative methods (other than DFT)?

No, it is not correct. Saying that Hartree-Fock(HF) theory is the only method is falling short. Hartree-Fock is only one of many existing methods. While it is true that some single reference methods have a dependence on the HF solution (because they build upon it) when they include a reduced number of determinants, it is not the case for other. Particular mention deserves Coupled Cluster since it is insensible to the reference determinant found by HF. The proof is given by Thouless theorem (see Joshua Goings' post, or via The Internet Archive).
Thus, to give you an overview of the grand landscape of possible approaches I tried to compile the families of methods better established nowadays. I did not intend to specify individual methods because there are so many ... and people continue to investigate new ones. These are:

Deterministic

Semi-empirical (non ab initio): Hückel1/Tight binding2, Hubbard, ...
Wavefunction methods [Trygve Helgaker, Poul Jørgensen, Jeppe Olsen:
Molecular Electronic‐Structure Theory. John Wiley and Sons, Ltd: 2000. DOI: 10.1002/9781119019572]

Hartree method. Caution: mathematicians use this name for a related method.
Hartree-Fock (HF): can be Restricted (RHF), Restricted Open (ROHF), Unrestricted (UHF) or Generalized (GHF).
Configuration Interaction (CI), up to order $N$, which is equivalent to Full-CI1/exact diagonalization2/direct variational approach3.
Multi-Configuration Self Consistent Field (MCSCF). [Chem. Rev. 2012, 112 (1), 108–181.]

Complete Active Space (CASSCF).
Two Configurations.
Manual selection of configurations or other.

Multi-Reference Configuration Interaction (MRCI)
CASPT2, NEVPT2, partially (PC) and strongly contracted (SC)
variants of the N-electron valence state second-order
perturbation theory
Multi-Reference Coupled Cluster methods (MRCC)
Many Body Perturbation Theory methods (MBPT) [Isaiah Shavitt, Rodney J. Bartlett: Many-Body Methods in Chemistry and Physics, MBPT and Coupled-Cluster Theory. Cambridge University Press: Cambridge, 2009. DOI: 10.1017/CBO9780511596834]

Coupled Cluster methods and its simplifications
Möller-Plesset perturbation theory of order $n$ (MPn)
Random Phase Approximation (RPA)
GW-Approximation

Explicit correlated methods R12/F12 [Chem. Rev. 2012, 112 (1), 75–107.]
Valence Bond Theory [Sason Shaik, Philippe C. Hiberty: A Chemist's Guide to Valence Bond Theory. John Wiley & Sons, Inc.: 2008. DOI: 10.1002/9780470192597]

Density Functional Theory (DFT). [Robert G. Parr, Yang Weitao: Density-Functional Theory of Atoms and Molecules. Oxford University Press: USA, 1994. ISBN: 9780195092769]

Thomas-Fermi theory. Orbital-free methods.
Kohn-Sham. Many density functional approximations developed, families are:

LSDA
GGA, hybrid GGA
MGGA, hybrid MGGA
OEP

Pair density functional theory [Comput. Theor. Chem. 2013, 1003, 91–96.]
Natural Orbital Functional Theory (NOFT) [S. Goedecker, C. J. Umrigar: Natural Orbital Functional Theory. In: J. Cioslowski (eds) Many-Electron Densities and Reduced Density Matrices. Mathematical and Computational Chemistry. Springer: 2000, Boston, MA. DOI: 10.1007/978-1-4615-4211-7_8]
Geminal Functional Theory [J. Chem. Phys. 2000, 112 (23), 10125–10130.]
k-particle density cummulants methods [Int. J. Quantum Chem. 2003, 95 (4-5), 404–423.]
Density Matrices methods

Density Matrix Renormalization Group (DMRG) [Annu. Rev. Phys. Chem. 2011, 62 (1), 465–481.]
Density Matrix Embedding Theory. Can be combined with several methods. [J. Chem. Theory Comput. 2016, 12 (6), 2706–2719.]
One-body Reduced Density Matrix Functional Theory (1-RDMFT) [J. Chem. Phys. 1985, 82 (12), 5604–5607.]
Variational 2-RDM Theory [Phys. Rev. Lett. 2012, 108 (26).]

Dynamical Mean Field Theory (DMFT). [J. Chem. Phys. 2011, 134 (9), 094115., arXiv:1012.3609 [cond-mat.str-el]] Starts with a Hubbard model of a lattice problem, then connects it to an impurity problem that can be solved with other methods.

Non-deterministic (stochastic)

Quantum Monte Carlo [Chem. Rev. 2012, 112 (1), 263–288.]

Variational Monte Carlo (VMC)
Fixed-Node Diffusion Monte Carlo (FNDMC).
LR-DMC, lattice-regularized diffusion Monte Carlo
Self-Healing Diffusion Monte Carlo
Green Function Monte Carlo (GFMC). GF2, self-consistent second-order Green’s function
Auxiliary Field Quantum Monte Carlo (AFQMC)
Reptation Quantum Monte Carlo
Full CI Quantum Monte Carlo (FCIQMC) [J. Chem. Phys. 2009, 131 (5), 054106.]
BDMC, bold diagrammatic Monte Carlo
Time-Dependent Quantum Monte Carlo

There are also mixtures of various methods.
The methods above have been tested with molecules by the quantum chemistry community but there are others that have been proven only for one dimension
or toy models. Operator methods are one example.

Many of them try to approximate the Full CI tensor that is described in a basis of determinant functions. It is impossible to treat such tensor for medium to large systems so the point is to take the minimum number of determinants that contribute significantly. The difference lies in how they choose those determinants. There are also other important distinctions among them, like if they are variational, perturbative or none, size-extensive, size-consistent, ...

In summary, each has its advantages and disadvantages. I recommend anyone interested to read
SOLVING THE SCHRÖDINGER EQUATION. Has Everything Been Tried? 2011, Imperial College Press. Ed.: Paul Popelier.

Furthermore, the analytic properties (read regularity) of the wavefunction can be exploited to devise new methods. As Harry Yseretant puts it in his book:

  Approximating the solutions is thus inordinately challenging, and it is conventionally believed that a reduction to simplified models, such as those of the Hartree-Fock method or density functional theory, is the only tenable approach. This book tries to convince the reader that this conventional wisdom need not be ironclad: the regularity of the
  solutions, which increases with the number of electrons, the decay behavior of their mixed derivatives, and the antisymmetry enforced by the Pauli principle contribute properties that allow these functions to be approximated with an order of complexity which comes arbitrarily close to that for a system of one or two electrons.

Also, I would like to note that there is a collaborative project under the hood of the Simons Foundation whose purpose is precisely finding better methods to solve the electronic structure problem. And, of course, there are many more people working on the same task.

Footnotes

Naming adopted by the chemistry   community
Naming adopted by the physics     community
Naming adopted by the mathematics community

References

Mario Motta, David M. Ceperley, Garnet Kin-Lic Chan, John A. Gomez, Emanuel Gull, Sheng Guo, Carlos A. Jiménez-Hoyos, Tran Nguyen Lan, Jia Li et al. Towards the Solution of the Many-Electron Problem in Real Materials: Equation
of State of the Hydrogen Chain with State-of-the-Art Many-Body Methods. Phys. Rev. X 2017, 7, 031059. DOI: 10.1103/PhysRevX.7.031059


How to determine the hydroxide ion concentration in an acidic aqueous solution?

The problem that you are facing comes from the fact that technically, not all of the $\ce {OH-}$ reacts with the $\ce {H+}$.  It is only because the amount of OH- that doesn't react is so negligible compared to the amount that does, that the $\ce {[H_3O+]}$ can be taken to be $\ce{[HCl]-[$\ce{OH-}$]}$.

So, yes, to find the remaining (negligible) amount of $\ce{OH-}$, you must use the "very small value" given by either subtracting $\ce {pH}$ from $14 =\ce{pOH}$ or doing $\ce K_w/[\ce{H+}]=[\ce{OH-}]$ (which amounts to the same thing).


What drives vibrational cooling in an excited stated?

Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality  'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another.  Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon weighted density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.

In many types  molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum  from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.

The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al,  J. Jortner et al. and K. Freed et al.


Ideal gas equation derivation proportionality

I understand your concern because this is the way it is taught in South Asian schools and colleges and even general chemistry texts taught elsewhere are silent about it. They state all the individual gas laws using a proportionality sign "$\propto$" and magically all gas laws are combined and that becomes $R$- the universal gas constant. This is done in a single paragraph.
This is not the proper way how the ideal gas law was derived. Out of curiosity, I checked the original paper " Mémoire on the Motive Power of Heat" where the ideal gas law appeared in 1834!!. It is in French, but one can see the equations. Not once he used any proportionality sign but there is a lot of calculus there.
https://gallica.bnf.fr/ark:/12148/bpt6k3414331n/f34.image.r=M%C3%A9moire%20sur%20la%20puissance%20motrice%20de%20la%20chaleur
You should scroll the original paper and feel the length of it to appreciate that the ideal gas law was not derived as shown in your school / college textbooks.
You would see a formal derivation of the ideal gas law in higher classes using the kinetic theory of gases.


Is the valency of carbon satisfied in graphite?

In the hexagonal graphite structure the carbon atoms are $\ce{sp^2}$ hybridized, just like in benzene.  In graphite, this p-orbital is used for bonding just as it is in benzene, resulting in an extended pi system in the graphite structure.  Graphite is really just a large number of benzene rings annelated together to form a continuous, ring structure.  Just as carbon satisfies its valence requirement of 4 in benzene (3 sigma bonds and 1 pi bond), carbon's valency requirements are satisfied the same way in graphite.


Changes in entropy of the universe in a reversible reaction

First, a defintion of terms:  The meaning of reversible in "reversible reaction" is different from the thermodynamic meaning of reversible.  A reversible reaction is one that is kinetically reversible.  I.e., one where you can adjust the relative concentration of products and reactants back and forth as you change conditions.   By contrast, an example of an irreversible reaction would be cooking an egg, which causes the proteins to denature.  The proteins can't be readily undenatured, even if you change the conditions to thermodynamically favor denaturation because, kinetically, the proteins can't find their way back to their undenatured state.

Anyways, let's suppose we have a "reversible reaction", $$\ce{A <=> B},$$  in which the equilbrium constant is, under certain conditions (e.g., at a certain temperature and pressure):

$$K_{eq} = \frac{[B]}{[A]} = 10,$$ i.e., at equilibrium, we will have 10x as much B as A.

This represents the state at which the entropy of the universe is a maximum under those conditions.  Thus if, under those conditions, $\frac{[B]}{[A]} <10$, the reaction will proceed in the forward direction until it reaches equilibrium, increasing the entropy of the universe in the process.  Likewise if, under those condtions, $\frac{[B]}{[A]} > 10$, the reaction will proceed in the reverse direction until it reaches equilibrium, also increasing the entropy of the universe in the process.

So it is not the case that if, say, $\frac{[B]}{[A]} = 9$, then the entropy of the universe would increase in both directions. Rather, it would increase in the forward direction (until it reached equilibrium), and (theoretically) decrease in the reverse direction.  If we instead had $\frac{[B]}{[A]} = 11$, the opposite would happen: decreased entropy of the universe in the forward direction, increased entropy of the universe in the reverse direction (again, until it reached equilibrium).


In your comment, you say you are interested in "liquid water".

The values in your self-answer are for gas phase water, and the second source is citing the first source, so it is really just one source.  The value is determined from gas phase rotation spectroscopy.

For liquid water, the value is not as precisely known:

105.5° (calculated) and 106° (experimental) as reported in Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572.


How to calculate bond angle of simple molecules quantitatively?

This has a simple reason: To the best of my knowledge there is no quantitative way to predict bond angles without resorting to computational chemistry.

This is because the entire subject of predicting bond angles is a lot more tricky than it was displayed to you. Take phosphane and hydrogen sulphide, the heavy analogues to ammonia and water. Their bond angles are nowhere near $109^\circ$ but much closer to $90^\circ$.

If you dig deeper into the matter, you find that ammonia and water are more the exception than the rule, that the VSEPR model fails spectacularly on the long run, and that their geometries are actually finely balanced between an ideal $90^\circ$ angle and the hydrogen-hydrogen repulsion.

In fact, not even with computers can you calculate the bond angle directly with a certain set of axiomic assumptions. Rather, computational chemistry would empirically calculate energies for different angles and deem the angle that has the lowest energy to be ‘correct’ and ‘calculated’.

So no simple quantitative rule exists nor will it exist.

Note: There are, of course, semi-quantitative rules like the one mentioned by Ringo in a comment. But I don’t consider them quantitative, because they still contain approximations.


How can I determine the quantum number of a electron lost during ion formation?

The correct answer has been given in the comments section of this post. I am providing the answer here in accordance with a meta post discussion.

The question you were asked was in reference to the quantum numbers that correspond to the electron lost when $\ce{Rb}$ goes to $\ce{Rb+}$. The electronic configuration of $\ce{Rb}$ is $[\ce{Kr}]\ 5\mathrm{s}^{1}$; thus, you need the set of quantum numbers for a $5\mathrm{s}^{1}$ electron. That set of quantum numbers is:

$$n = 5;\ l= 0;\ m_{l} = 0;\ m_{s} = +{1\over2}$$


During the development of aspartame, why was the methyl ester chosen?

This question raises more important issues than just the technical "why methyl ester," so I'll address those too.

The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. This report says it is approximately $10\times$ less sweet (see Table VI on page 2689 and the entry '$\ce{Asp-Phe-OEt}$' with "$++$" being defined on pg. 2684). It's an easy economic choice between the two, assuming the only difference is sweetness/g, but you're also worried about the perceived safety concerns of the methyl derivative.

A toxicology class I once took can mostly be summed up by a cliche phrase, "the dose makes the poison." Even too much of a good thing is by definition a bad thing, for example too much oxygen, water, or even the sun can lead to death. We also say that too little of any compound won't cause measurable harm.

It's correct to say that aspartame will be metabolized into amino acids and methanol. Methanol itself is toxic in high doses, in the same manner ethanol is, by acting as a CNS depressant. This acute poisoning can cause respiratory failure or other medical emergencies such as kidney failure. Ethanol poisoning occurs in microorganisms too: it has been proposed that some yeast produce ethanol as a means of eliminating competition. This would have been potentially deadly for a prehistoric mammal too, if it were not for an enzyme, which metabolize ethanol. This enzyme is called alcohol dehydrogenase (ADH), which leads to biochemical pathways that allow the safe removal of ethanol. However, methanol is a different case. ADH instead make formaldehyde, which is then converted to formic acid by ALDH. Formic acid, in large quantities, is used in the animal kingdom as a weapon, most notably in bee venom and from the bites of a fire ant. In low quantities it is fairly benign. Formaldehyde is just a nasty chemical. Both are produced in the body as a natural response to methanol and it's easy to see how too much of either couldn't possibly be desirable. The body also has ways to eliminate these toxic metabolites assuming you don't overwhelm these systems. Hence why a sufficiently small amount of methanol won't make you blind.

Researchers have measured the amount of these undesirable metabolites in the blood after having people, like me, ingest nearly $\pu{1.8 g}$ of aspartame in one sitting. Then they looked at urine and blood concentrations and they couldn't detect any formic acid in the blood but they could clearly see it in their urine for, up to, $\pu{8 hours}$ after a dose. This evidence suggests that acute methanol poisoning isn't going to happen after a ~$\pu{200 lbs}$ person ingests $\pu{1.8 g}$ of aspartame. Let's stop for a second and put this in terms of something we can imagine.

I don't have any reliable numbers, but let's just say that for a $\pu{12 oz}$ diet pop there are $\pu{200 mg}$ of aspartame. That means for every liter of pop, there are approximately $\pu{564 mg}$ of aspartame. That means $\pu{3 L}$ of pop would get me to $\pu{1.8 g}$ of aspartame. That means I would need to drink more than $\pu{3 L}$ in one sitting to overwhelm the excretion systems. This suggests an acute poisoning isn't very plausible.

There is, however, long-term toxicity to consider as well. This is a much more difficult question to probe. "Sure, there is no acute toxicity, but what happens in the long-term with a low dose?" For the answer to this question, we have to turn to toxicologists who have not reported long-term issues. They are looking too. The general consensus is that there is no long-term harm known for the general population. It is hard to quantitate these things, because to perfectly do the experiments necessary would require controlling a bunch of parameters that human experimentation laws prevent, let alone funding constraints. So, researchers make the best with what they have available and using these tools they cannot find any measurable problems with this particular compound in the long-term for the general population. Their analyses are limited by uncertainties in their measurements and this is just the general nature of science.

Tomorrow the majority of scientists in this area of interest could be proven wrong by a very careful experiment, but over forty years of research hasn't shown this chemical to be dangerous at the doses we're likely to be exposed to for any duration of time. We have to make decisions off of the best evidence we have available. I think it's okay for people to apply some level of skepticism to these things, because in science we're continually correcting our accepted knowledge base. Once in a while the general community is proven wrong about something long thought to be true and everybody benefits when that is the case, but we can't say the general community is wrong without substantial evidence. It is for this reason that we have the FDA to evaluate the best available evidence and via a network of reliable strangers, we can enjoy a very cheap/sweet drink that is low in calories. Frankly, I am more worried about the amount of sugar I take in than I'll ever worry about the amount of aspartame I consume -- we have already good evidence that sugar, with a dose too small to cause acute poisoning, is probably killing hundreds of thousands of people due to long-term exposure.


How to rationalise the protonation state of amino acids

Let's assume you drop the amino acid in water at $\ce{pH} = 7$.
You'll have $$\ce{ H3N^+{}-CH(R)-COO^-}.$$
Then you can go to $\ce{pH}= 12$ and the amino acid loose the proton.

You don't get rid of the second proton because the $\mathrm{p}K_a$ is defined as $$\mathrm{p}k_{a}=-\lg\frac{[\ce{NH2}]\cdot[\ce{H+}]}{[\ce{NH3+}]}.$$
This means $\ce{NH3+ -> NH2}$.

You could define another $\ce{p}K_a$ for $\ce{NH2 -> NH^{-}}$.
The $\mathrm{p}K_a$ for this would be much higher than 12 - at least around 35.
You may look at this site: How to use a $\mathrm{p}K_a$ table
You can compare the amine $\ce{NH3}$ (without the rest of the amino acid) with your $\ce{NH2}$ group of the amino acid.


How is the charge of covalently bonded atoms determined?

As you point out, formal charge is a "bookkeeping" method of assigning electric charges to atoms in a way that lets us easily approximate where the electron density is.

Chemistry is made up of layers of successively less detailed approximations. The concept of "formal charge" is near the top of the stack - very little detail is involved. It is only slightly more detailed than Lewis structures, in that we are at least trying to acknowledge that covalent bonds don't really share electrons equally.

With that in mind, we will look at the next higher and next lower level of detail, to see why formal charges are useful.

Whenever we draw a Lewis structure, we imply that each line is made up of two electrons that share a molecular orbital that exists somewhere in the space between the two atoms that it connects. Without knowing any better, we would assume that each electron can be wherever it "needs" to be to fill the valence shell of an atom. So, in the case of $\ce{CO3^2-}$, the carbon has four bonds and no lone pairs, giving it a full octet. The oxygen atoms have either three lone pairs and a single bond, or two lone pairs and a double bond, again giving each a full octet.

Note that we counted each electron twice for the purpose of checking the valence configuration. It's as if we are saying "each atom can borrow electrons from the other whenever it needs them." We are effectively ignoring any tendency of atoms to pull electrons towards each other - they float between atoms as needed, as if maintaining a full octet were the most important job in the world.

Next we look at the formal charges. The rules for determining formal charge say that we count the number of valence electrons an atom normally has (without bonds), then subtract the electrons we assign to it. We assign to an atom all of the electrons that are in lone pairs, plus one half of the bonding electrons. In other words, instead of saying that each atom gets the electrons whenever it wants them, we are saying "each atom in the bond divides the electrons equally between themselves." When you do this, you find that some atoms have "extra" electrons (more than they would normally have in the valence shell) and some have too few - thus, a formal charge.

This starts to give us a slightly more correct picture of how electrons are distributed inside of a molecule, but it doesn't really tell the whole story. Notice that we didn't care about what type of atom it was when we split up the electrons. We only looked at that to find the "natural" number of electrons it would have.

The next layer of detail involves electronegativity. In any covalent bond between two different elements, one element "likes" electrons more. The reasons for this are complex, and you have to go down another couple of layers to get to the reason. For now, just know that it is related to quantum mechanics.

If we know the relative electronegativities of elements, we can start to account for this tendency as well. We do this when assigning oxidation numbers. With oxidation numbers, you assign to each atom all of its loan pair electrons. Then, for each bond, you give all electrons in the bond to the more electronegative atom. Then we subtract the total from the number of valence electrons the atom normally has.

Notice what we are doing. First we said "all electrons are shared equally all the time." Then we said "Divide the electrons equally." Finally we said "the most electronegative atom gets all of the electrons."

None of these is actually true, but each gives a quick approximation that is useful for a particular type of problem. To get a real picture of what is happening to the electron density, you have to do detailed quantum mechanical calculations. They are really hard to do! We typically use computers to solve them. But, if we know what the limitations are, we can use quick approximations to get an idea of where the electrons are at that is good enough to, for example, make reasonable predictions in organic chemistry.

To summarize:

Valence count - assume all electrons are shared equally
Formal charge - assign one half of bonding electrons to each atom
Oxidation number - assign all bonding electrons to the most electronegative atom


Potassium is in the 4th period and therefore has at least 4 shells occupied with electrons while lithium is in period 2 and thus only has 2 shells occupied with electrons. Perhaps you mean they both have 1 valence electron that must be ionised before they become a cation? Anyhow, ionic bonds are formed due to the strong electrostatic attraction between oppositely charged ions. All elements would like to achieve a noble gas electron arrangement due to its stability. However, to be able to achieve this, elements must either gain or lose electrons depending on how close their electron arrangement is to their closest noble gas, e.g. Fluorine would like to gain 1 electron while Magnesium is more likely to lose 2 electrons to form its closest noble gas.

For instance, NaCl: Na would like to lose an electron and Cl would like to gain an electron. Together this is possible. The nucleus of the Cl atom electrostatically attracts the valence electron from the Na atom– this attraction is strong enough the overcome the the attraction this valence electron feels from its own nucleus. As a result, the Cl gains a negative charge due to it gaining a negatively charged electron and the Na atom now has a positive charge as it has one less negatively charged electron. The Cl- and Na+ are oppositely charged and therefore attract each other. This is known as ionic bonding and results in the formation of an ionic lattice.

Now to your question. All elements in group 1 have approximately the same effective nuclear charge, i.e. the attraction the valence electrons feel from the nucleus of their atom after the shielding of the inner electrons is considered. Hence, the fact that they are in group 1 should not significantly affect the ability to lose their valence electrons. However, as one goes down periods, the number of occupied outer shells increases, which means that the valence shells experience a weaker attraction from their nucleus. This is confirmed by the lower first ionisation energy trends. To answer your question frankly (I perhaps overcomplicated with my verbose writing style), it is the strong affinity for electrons of the non-metals that result in the loss of the valence electrons from the metals. Even if the the metal has a smaller atomic radius due to occupying less shells, the affinity for the electrons from the non-metals is strong enough to take the metal's valence electrons and make it into a cation.

References:

Pearson Higher Level Chemistry Textbook, 2nd Edition. By Catrin Brown and Mike Ford.


Where to put the number in organic nomenclature?

The IUPAC recommendations on this matter have been changed.

According to the 1979 recommendations, the correct name was still 4-hydroxy-2-pentanone.

According to the 1993 recommendations, however, locants were placed immediately before the part of the name to which they relate. This rule is still included in the 2013 version. Therefore, the correct name is now 4-hydroxypentan-2-one.


Working with stainless steel in 3 M HCl solution

The austenitic, chromium-nickel based stainless steel SAE 316 approximately corresponds to the material number 1.4401 or 1.4436. Especially due to the addition of molybdenum in its composition it is more resistant to corrosion overall than the usual type 304 (1.4301). However, it is generally not intended to be used in hydrochloric acid.

Hydrochloric acid lacks the oxidizing properties that are needed to maintain the passive corrosion-resistant surface layer on stainless steel.

The typical corrosion rate in $1\ \%$ hydrochloric acid at room temperature is about $0.1\ \mathrm{mm\ a^{-1}}$, which is approximately $0.1\ \mathrm{g\ m^{-2}\ h^{-1}}$. At higher concentrations or higher temperature, the corrosion rate is strongly increased. Note that your concentration of $c=3\ \mathrm{mol\ l^{-1}}$ $\ce{HCl}$ corresponds to about $10\ \%$ hydrochloric acid.

In addition to this uniform corrosion, stainless steels in media containing chloride ions are particularly susceptible to pitting corrosion, i.e. highly localised corrosion with discrete pits.

Furthermore, stainless steel SAE 316 is prone to chloride-induced stress corrosion cracking at elevated temperatures (over about $50\ \mathrm{^\circ C}$).


Why does the water dissociation constant Kw remains constant?

Your question reflects your knowledge in chemistry so I'll try to keep this explanation simple as possible.
Water conduct electricity because it contains $\ce{H+}$ and $\ce{OH-}$ ions by auto-ionization:
$$\ce{H2O <=> H+ + OH- \tag1}$$
It has been defined that this ionization is a constant at $\pu{25 ^\circ C}$. Thus by definition:
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14} \tag2$$
Accordingly, by definition for pure water, $ [\ce{H+}]=[\ce{OH-}]= 1.0 \times 10^{-7}$. And then, there is a widely accepted principle in chemistry called Le Chatelier’s principle:

Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

Accordingly, increasing the concentration of reactants (species on left hand side of the reaction) will drive the reaction to the right (more products), while increasing the concentration of products (species on right hand side of the reaction) will drive the reaction to the left (more reactants). At constant temperature, even though concentrations of species (e.g., here they are $\ce{H+}$ and $\ce{OH-}$ in equation $(1)$) at new  equilibrium state is changed, the equilibrium constant remains same. In other word, the equilibrium constant is depends only on temperature.
According to Le Chatelier’s principle, in in equation $(1)$, if you add more $\ce{H+}$ or $\ce{OH-}$ (RHS), the equilibrium would be adjusted to reduce that amount by making more water. But since this addition is at constant temperature, $K_\mathrm{w}$ remains constant.


Can gas be made to block radiation better?

Your question seems to be about the ozone layer, but shows some misunderstanding.

First, ozone, $\ce{O3}$, absorbs some "radiation", specifically electromagnetic radiation, e.g. visible light or ultraviolet light (UV), as do many other gases. This absorption is not the same for all wavelength ("colors"), but in the UV region peaks about 250 nm, which is a good thing for us, because plants and animals have not evolved good protection from that light, and the small amount of $\ce{O3}$ in the upper atmosphere is enough to block most of that UV. Sulfur dioxide, $\ce{SO2}$, released in volcanic eruptions, is also able to absorb 250 nm UV. Molecular oxygen, $\ce{O2}$, absorbs "vacuum UV", 10-100 nm, which otherwise would be a problem.

Second, Earth is continually bombarded by another type of "radiation", particles such as cosmic rays and solar energetic particles. Rather than electromagnetic energy, these are made of particulate matter such as protons or nitrogen nuclei. All gases, in fact all matter, help block these particles. Magnetic fields such as that of the Earth help divert them around the planet, too, but there is some question whether a magnetic field is essential to blocking particulate radiation. Here at the bottom of the atmosphere there is a column of air weighing about a kilogram over each square centimeter, which is sufficient to block primary cosmic rays, though a showers of weaker secondary particles do reach the surface. There is no advantage to a conductive Faraday cage to protect the Earth.

Third, all gases conduct electricity when ionized. The majority of air is nitrogen, oxygen and argon, which produce various colors in an aurora. Though magnificent to watch, aurorae are simply evidence of energetic particles trapped in a magnetic field, ionizing the upper atmosphere. This layer of the atmosphere is called the ionosphere, and is responsible for long-distance radio communication. It's height varies from day to night and with solar storms.


The general method to solve problems of this type can be found in the answer here, How to set up equation for buffer reaction?. In your case you need to modify this equation. To follow the text below you need to look at the main result in the linked answer (rather than repeat all this here). The derivation is straightforward although the equation is a bit complicated.

The base constant $K_\mathrm{B}$ is given but as $K_\mathrm{w}=K_\mathrm{A}K_\mathrm{B}$ we can start with  main eqn in linked page and convert to $k_\mathrm{B}$ later on. In this case as ammonia is a base (with concentration $c_b= 0.1$ M) we may suppose that any acid concentration $c_\mathrm{a} \approx 0$ then,

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} + \mathrm{[H^+]} - \mathrm{[OH^-])} }{(- \mathrm{[H^+]} + \mathrm{[OH^-]} )}  $$

Additionally we can reasonably assume that $\mathrm{[OH^-] \gg [H^+]}$ and so

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} - \mathrm{[OH^-])} }{ \mathrm{[OH^-]} }  $$

This can be solved by substituting for $K_\mathrm{A}=K_\mathrm{w}/K_\mathrm{B}$  and for $\ce{[H^+]}$ as $K_\mathrm{w}=\ce{[H^+][OH^-]}$ and solve the quadratic equation.  But as $c_\mathrm{b} \gg \ce{[OH^-]}$ we can simplify by letting $c_\mathrm{b} - \ce{[OH^-]} \to c_\mathrm{b}$ then the equation becomes

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{c_\mathrm{b}  }{ \mathrm{[OH^-]} }  $$

We can now get

$$\frac{K_\mathrm{w}}{K_\mathrm{B}}= \frac{K_\mathrm{w}c_\mathrm{b}} { \mathrm{[OH^-]^2}}   $$

from which

$$\ce{[OH]^- } =\sqrt{K_\mathrm{b}c_\mathrm{b}}$$

and so you can obtain the pH.

When you dilute, the concentration of base $c_\mathrm{b}$ is halved.


Are all concerted electrophilic addition reactions of alkenes syn?

If they are concerted, there is a very, very large probability of them being syn-specific. For explantion, let’s look at the π orbitals of ethene (scheme 1). (Remember that double bonds are planar and have two distinct sides when viewed across an entire molecule.)

Scheme 1: π and π* orbitals of ethene, viewed from the side.

Reactants have one of two possibilities, typically. They are either electrophilic and will thus attack the highest occupied molecular orbital (HOMO). Or they are nucleophilic and will attack the lowest unoccupied molecular orbital (LUMO). It just so happens that the π bond is the weakest bond in a typical alkene and thus the bonding molecular orbital is typically the HOMO and the antibonding one is the LUMO.

To attack this orbital, you somehow need to interact with both lobes. If you do so, you will either attack syn-stereospecificly, i.e. arrive from one side due to simplicity. Or you are a very large attacking molecule and are able to grab around the double bond, interacting with one atom on one side and with the other one on the other side.

Most reactants are not that large. Typically, there will be a single attacking atom (e.g. halogens, meta-chloroperbenzoic acid), two atoms bonded together directly (e.g. borane) or two atoms that are still in very close proximity (e.g. $\ce{OsO4}$ — two bonds — or a diene for a Diels-Alder reaction — three bonds). None of these even have the chance to reach across, especially if you consider that the largest of these, the diene, is in fact a large π system and inherently restricted to being planar (within very small possible deviations).

Of course, that does not make it impossible. If you use a large enough π system in a Diels-Alder like reaction, that may be able to wind itself downwards like a spiral staircase. I haven’t seen that type of reaction yet, though.


What is the biggest known difference between rₑ and r₀?

Let me start by saying that this is a very difficult question to answer because it's not very hard to come up with pathological cases which I will address at the end. Also, as you point out, it is somewhat non-trivial to calculate $r_0$ from theory and also somewhat non-trivial to determine $r_e$ from experiment. That being said, I think I have a good first suggestion that I am guessing will be hard to beat. I will just answer with theory because this data is more readily available.

$\ce{F^-(H2O)}$:

As has been pointed out, this question is very closely related to the anharmonicity of the vibration in question. The $\ce{O-H}$ stretch of $\ce{F^-(H2O)}$ is the most anharmonic known vibration for a water molecule. This has an experimentally observed vibrational frequency shift of about $2000~\mathrm{cm^{-1}}$ which I'm almost sure is the largest known vibrational frequency shift in a molecular complex, where the molecule in question doesn't dissociate.

Now, that being said, this will vary based on the method used, but at the MP2/AVTZ level, $r_e=1.0653$ angstroms while $r_0=1.1248$ angstroms resulting in a difference of $0.0595$ angstroms.

This $r_0$ value is calculated from VPT2, which is probably the easiest way to calculate this quantity, but this can fail for certain pathological cases like those I'll discuss below.

$\ce{H5O2^+Ar}$:

Ok, now I know that 0.06 angstroms doesn't seem like that huge of a difference, but generally both of these numbers will increase together, so it's hard for them to be all that different. Now, another way you could consider finding a very large difference would be to look for a double-well system where the zero-point energy of the vibration in question is larger than the barrier height.

Now, the zundel cation is a proton shared symmetrically between two water molecules. This is a single well system, however, so $r_e$ and $r_0$ will be the same and they have a calculated $\ce{O-H}$ distance of $1.202$ angstroms at the CCSD(T)/aug-cc-pVDZ level of theory.

If, however, we disturb this system ever so slightly, we will break the symmetry and might get the desired behavior. This can be done by tagging the complex with a rare gas atom, which is often done in experiments as it reduces the noise in spectra for reasons I won't get into.

$\ce{H5O2^+Ar}$ has a hydronium like complex with an elongated $\ce{O-H}$ bond length of $1.108$ ansgtroms. It is very likely that the barrier to proton transfer in this case is very small because there is no barrier in the absence of an Argon atom. Therefore, it is likely that the proton will be close to symmetric after vibrational averaging, so if it were perfectly symmetric, we would get an $r_0$ of $1.202$ angstroms again. It is likely less than this, but it is possible this would lead to a difference greater than $0.06$ angstroms which we got from $\ce{F^-(H2O)}$.

I should point out that calculating $r_0$ for this system is pretty difficult because using VPT2 will almost certainly give nonsensical answers because the harmonic reference is a terrible one, so the perturbation theory will break down. The best option is probably to do diffusion monte carlo and extract the average from the wavefunction directly. It's possible this has been done, but a cursory search of the literature didn't turn anything up.

Also, the data for the first part comes from the SI of ref.[1] and the data for the second part comes from ref.[2].

References:

[1]: Boyer, M. A., Marsalek, O., Heindel, J. P., Markland, T. E., McCoy, A. B., & Xantheas, S. S. (2019). Beyond Badger’s Rule: The Origins and Generality of the Structure–Spectra Relationship of Aqueous Hydrogen Bonds. The journal of physical chemistry letters, 10(5), 918-924.

[2]: Park, M., Shin, I., Singh, N. J., & Kim, K. S. (2007). Eigen and Zundel forms of small protonated water clusters: structures and infrared spectra. The Journal of Physical Chemistry A, 111(42), 10692-10702.


IUPAC definition

The IUPAC definition is the following:

  The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').
  An increase in the oxidation number of any atom within any substrate.
  Gain of oxygen and/or loss of hydrogen of an organic substrate.

  All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation
  of an organic substrate that can be rationally dissected into steps or primitive changes
  . The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons
  or hydroxide ions, or by nucleophilic
  substitution by water or its reverse and/or by an intramolecular molecular
  rearrangement
  . This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation
  transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1.

So that is the official definition, and it also gives multiple rules (and admits that the third rule does not always work). In order to use the second definition, you also need to look up their definition of oxidation state.

Definitions in your textbook

  I can't understand why they are changing the definitions again and again.

They are changing the definition to apply the concept to more and more reactions.

In the explanation of the IUPAC definition I posted above, they mention "the original idea of oxidation (combination with oxygen)". That's where your first textbook definition comes from. So the following reaction is an oxidation because there is a reaction with dioxygen:

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

If I want to call the similar reaction with fluorine an oxidation as well, I have to generalize the definition to loss of electrons (your second definition). So this would be an oxidation of iron(II) as well:

$$\ce{2Fe^2+(aq) + F2(g)  -> 2Fe^3+(aq) + 2 F-}$$

Finally, if I want to generalize further to reactions where I want to discuss loss of electrons in atoms that are part of a covalent molecule, I need some accounting scheme to assign valence electrons to individual atoms (this is called oxidation number or oxidation state). Once that is in place, I can discuss reactions such as the following in terms of oxidation (and reduction):

$$\ce{H2C=CH2 + H2 -> H3C-CH3}$$

Here, your first definition does not apply (no oxygen) and your second definition is difficult to apply (how do we know which species lost electrons - they are all neutral, and they all have different composition?). So that is where you need some rule similar to your third rule ("decrease in electron density"). Usually, though, this is stated using oxidation states so that you are back to gaining or losing an integral number of electrons (i.e. oxidation state changed by +1 rather than "the electron density decreased slightly).

At this point, it becomes a bit fuzzy and there will be situations where the oxidation state is ambiguous. You gain clarity in the context of electrochemistry when you separate reactions into half reactions in electrochemical cells, and can actually measure the transfer of electrons that pass through the wire connecting the two cells.


What is the nature of the Fe–O2 binding in oxymyoglobin and oxyhemoglobin?

TL;DR: It is not really "one or the other" - it is more of "somewhere in between", like a resonance hybrid. If I had to choose, I'd personally lean towards Weiss's $\ce{Fe(III)-O2-}$ model. However, there will always be people who disagree, and there is still no general consensus amongst the scientific community.

There is an excellent - and recent - perspective article by Bren et al.,1 describing the research carried out on the structure of hemoglobin ever since Pauling and Coryell's initial report2 on the diamagnetism of oxyhemoglobin.

Model systems

A common technique in bioinorganic chemistry is to synthesise "inorganic versions" of enzyme active sites; for example, much work has been done on dinitrogen complexes in the hope of understanding the enzyme nitrogenase. Myoglobin and hemoglobin are not enzymes in the strict sense of the word,3 but the same technique applies: if stable $\ce{Fe(II)-O2}$ complexes could be synthesised, they might prove to be more amenable to experimental investigation.

The main problem was the fact that dioxygen readily oxidised simple Fe(II) complexes to Fe(III). Initially, Co(II) complexes (which only have one unpaired electron) were used: EPR studies showed that the unpaired electron was localised on oxygen,4,5 and the length of the O–O bond resembled that of superoxide:6

$$\begin{array}{ccccc}
\text{Species} & \ce{O2^2-} & \ce{O2^-} & \ce{O2} & \ce{Co(II)-O2}\text{ complex} \\ \hline
\text{Bond order} & 1 & 1.5 & 2 & ? \\
\text{Bond length / Å} & 1.49 & 1.28 & 1.21 & 1.26
\end{array}$$

Of course, Co(II) complexes do not necessarily have the same electronic structure as their Fe(II) counterparts, so such data can only be of limited use. The key to overcoming the early difficulties (of Fe(II) oxidation) was the fact that such oxidations tend to proceed via dimerisation; therefore, if a sterically hindered Fe(II) complex could be made, the oxidation could be prevented. Based on this, Collman and coworkers synthesised so-called "picket fence" complexes,7 which reversibly bound dioxygen, the concept of which is shown below:

Access of other iron centres, as well as the imidazole ligand, to the top face of the metal ion was blocked by the bulky substituents on the porphyrin ring (in one example, $\ce{-C6H4NHC(O)^tBu}$ was chosen). The reported data from different "picket fences" strongly favoured the iron(III)–superoxide model: in particular, the O–O stretching frequencies ranged from $1163\text{–}\pu{1150 cm−1}$ (cf. $1200\text{–}\pu{1050 cm−1}$ in superoxide complexes, $\pu{1556 cm-1}$ in $\ce{O2}(^3\Sigma_\mathrm{g}^-)$, and $\pu{1484 cm-1}$ in $\ce{O2}(^1\Delta_\mathrm{g})$).8

A controversy over magnetic data

It was realised that a key difference between the two schemes was in their predicted magnetic susceptibilities. For the $\ce{Fe(II)-O2}$ binding scheme, one would expect the compound to be diamagnetic at all accessible temperatures (to obtain a magnetic state would require electronic promotion). On the other hand, the $\ce{Fe(III)-O2-}$ binding scheme would be expected to have a temperature-dependent magnetic susceptibility, as antiferromagnetic coupling could be overcome with higher temperatures. In other words, there is a thermally accessible triplet state.

In 1977, Cerdonio et al. published a paper claiming to have observed a temperature-dependent magnetic susceptibility for $\ce{HbO2}$.9 However, their findings were refuted in a later 1984 paper by Savicki et al., which found that $\ce{HbO2}$ was in fact diamagnetic at all temperatures.10 The incorrect magnetic data was attributed to residual deoxyhemoglobin, which is paramagnetic, and in a private communication with Savicki, Cerdonio agreed that his earlier findings were in error.

Spectroscopic data on the proteins

Using infrared spectroscopy, the O–O stretching frequencies in both oxymyoglobin11 and oxyhemoglobin12 were identified as $\pu{1103 cm−1}$ and $\pu{1107 cm-1}$ respectively, again consistent with superoxide.

Furthermore, $\ce{^57Fe}$ Mössbauer spectra of $\ce{HbO2}$ showed large quadrupole splittings, which is consistent with iron being in the +3 oxidation state.13

Computational studies

Early quantum mechanical calculations by Karplus14 showed that the properties of $\ce{HbO2}$ could be explained by a mixture of two $\ce{Fe(II)-O2}$ resonance forms: one with both bonding partners in the singlet state, and one with both in a triplet state. The latter was actually put forth as a separate - third - model, most notably by Goddard in 1975.15

Of course, these are fairly primitive by today's standards. In 2008, Shaik et al. revisited the topic,16 and provided an overview of the numerous computational studies performed up until then, which had failed to reach a conclusion. Instead of trying to discern the appropriate bonding model from indirect parameters, such as the net charge on oxygen or the form of the occupied MOs, the authors opted to transform a CASSCF wave function into a generalised valence bond wave function. They concluded that the Weiss $\ce{Fe(III)-O2-}$ model was dominant for a gas-phase oxygen–heme complex, and that the interaction with the protein further reinforces the $\ce{Fe(III)-O2-}$ character of the wave function.

More recent experimental work

In an entirely separate endeavour, Gray and coworkers found that polyhalogenated iron porphyrins, which possessed unusually high Fe(III)/Fe(II) reduction potentials, did not bind to oxygen.17 This has been taken as evidence for the necessity of electron transfer from Fe(II) to dioxygen in $\ce{Hb-O2}$ binding, and hence the identification of the bonding as being $\ce{Fe(III)-O2-}$ charge-transfer in nature.

X-ray absorption spectra by Wilson et al. further complicate the matter, though. Firstly, their investigation of picket fence porphyrins led to the conclusion that these generally showed characteristics of both models, but were more in line with an Fe(II) description. The effective nuclear charge on iron was found to be closer to that of Fe(II) than to that of Fe(III), and the spectra lacked a low-energy transition characteristic of Fe(III) complexes (arising from the hole in the $\mathrm{t_{2g}}$ orbitals on iron).18

A further study on the protein itself revealed that crystalline $\ce{HbO2}$ mainly has $\ce{Fe(II)-O2}$ character, whereas solution $\ce{HbO2}$ has $\ce{Fe(III)-O2-}$ character.19 They therefore concluded that the bonding was multiconfigurational in nature, and that "secondary changes in the protein environment, crystal packing, and so forth, can tip the electronic structure in either direction".

Notes and references

1. Bren, K. L.; Eisenberg, R.; Gray, H. B. Discovery of the magnetic behavior of hemoglobin: A beginning of bioinorganic chemistry. Proc. Natl. Acad. Sci. 2015, 112 (43), 13123–13127. DOI: 10.1073/pnas.1515704112.

2. Pauling, L.; Coryell, C. D. The Magnetic Properties and Structure of Hemoglobin, Oxyhemoglobin and Carbonmonoxyhemoglobin. Proc. Natl. Acad. Sci. 1936, 22 (4), 210–216. DOI: 10.1073/pnas.22.4.210.

3. Hemoglobin is famously known as an "honorary enzyme"; for an overview of its history, see e.g. Brunori, M. Hemoglobin is an honorary enzyme. Trends Biochem. Sci. 1999, 24 (4), 158–161. DOI: 10.1016/S0968-0004(99)01380-8.

4. Hoffman, B. M.; Diemente, D. L.; Basolo, F. Electron paramagnetic resonance studies of some cobalt(II) Schiff base compounds and their monomeric oxygen adducts. J. Am. Chem. Soc. 1970, 92 (1), 61–65. DOI: 10.1021/ja00704a010.

5. Getz, D.; Melamud, E.; Silver, B. L.; Dori, Z. Electronic structure of dioxygen in cobalt(II) oxygen carriers, singlet oxygen of dioxygen(1-). J. Am. Chem. Soc. 1975, 97 (13), 3846–3847. DOI: 10.1021/ja00846a064.

6. Gall, R. S.; Rogers, J. F.; Schaefer, W. P.; Christoph, G. G. The structure of a monomeric oxygen carrying cobalt complex: dioxygen-N,N'-(1,1,2,2-tetramethyl)ethylenebis(3-tert-butylsalicylideniminato)(1-benzylimidazole)cobalt(II). J. Am. Chem. Soc. 1976, 98 (17), 5135–5144. DOI: 10.1021/ja00433a015.

7. Collman, J. P.; Gagne, R. R.; Reed, C. A.; Halbert, T. R.; Lang, G.; Robinson, W. T. Picket fence porphyrins. Synthetic models for oxygen binding hemoproteins. J. Am. Chem. Soc. 1975, 97 (6), 1427–1439. DOI: 10.1021/ja00839a026.

8. Collman, J. P.; Brauman, J. I.; Halbert, T. R.; Suslick, K. S. Nature of O2 and CO binding to metalloporphyrins and heme proteins. Proc. Natl. Acad. Sci. 1976, 73 (10), 3333–3337. DOI: 10.1073/pnas.73.10.3333.

9. Cerdonio, M.; Congiu-Castellano, A.; Mogno, F.; Pispisa, B.; Romani, G. L.; Vitale, S. Magnetic properties of oxyhemoglobin. Proc. Natl. Acad. Sci. 1977, 74 (2), 398–400. DOI: 10.1073/pnas.74.2.398.

10. Savicki, J. P.; Lang, G.; Ikeda-Saito, M. Magnetic susceptibility of oxy- and carbonmonoxyhemoglobins. Proc. Natl. Acad. Sci. 1984, 81 (17), 5417–5419. DOI: 10.1073/pnas.81.17.5417.

11. Maxwell, J. C.; Volpe, J. A.; Barlow, C. H.; Caughey, W. S. Infrared evidence for the mode of binding of oxygen to iron of myoglobin from heart muscle. Biochem. Biophys. Res. Commun. 1974, 58 (1), 166–171. DOI: 10.1016/0006-291X(74)90906-1.

12. Barlow, C. H.; Maxwell, J. C.; Wallace, W. J.; Caughey, W. S. Elucidation of the mode of binding of oxygen to iron in oxyhemoglobin by infrared spectroscopy. Biochem. Biophys. Res. Commun. 1973, 55 (1), 91–95. DOI: 10.1016/S0006-291X(73)80063-4.

13. Lang, G.; Marshall, W. Mössbauer effect in some haemoglobin compounds. Proc. Phys. Soc. 1966, 87 (1), 3–34. DOI: 10.1088/0370-1328/87/1/302.

14. Case, D. A.; Huynh, B. H.; Karplus, M. Binding of oxygen and carbon monoxide to hemoglobin. An analysis of the ground and excited states. J. Am. Chem. Soc. 1979, 101 (16), 4433–4453. DOI: 10.1021/ja00510a001.

15. Goddard, W. A., III; Olafson, B. D. Ozone Model for Bonding of an O2 to Heme in Oxyhemoglobin. Proc. Natl. Acad. Sci. 1975, 72 (6), 2335–2339. DOI: http://10.1073/pnas.72.6.2335.

16. Chen, H.; Ikeda-Saito, M.; Shaik, S. Nature of the Fe−O2 Bonding in Oxy-Myoglobin: Effect of the Protein. J. Am. Chem. Soc. 2008, 130 (44), 14778–14790. DOI: 10.1021/ja805434m.

17. Grinstaff, M. W.; Hill, M. G.; Labinger, J. A.; Gray, H. B. Mechanism of catalytic oxygenation of alkanes by halogenated iron porphyrins. Science 1994, 264 (5163), 1311–1313. DOI: 10.1126/science.8191283.

18. Wilson, S. A.; Kroll, T.; Decreau, R. A.; Hocking, R. K.; Lundberg, M.; Hedman, B.; Hodgson, K. O.; Solomon, E. I. Iron L-Edge X-ray Absorption Spectroscopy of Oxy-Picket Fence Porphyrin: Experimental Insight into Fe–O2 Bonding. J. Am. Chem. Soc. 2013, 135 (3), 1124–1136. DOI: 10.1021/ja3103583

19. Wilson, S. A.; Green, E.; Mathews, I. I.; Benfatto, M.; Hodgson, K. O.; Hedman, B.; Sarangi, R. X-ray absorption spectroscopic investigation of the electronic structure differences in solution and crystalline oxyhemoglobin. Proc. Natl. Acad. Sci. 2013, 110 (41), 16333–16338. DOI: 10.1073/pnas.1315734110.


It will be more difficult for sterically hindered bases to remove the proton of the β-carbon having the fewest hydrogen substituents, since this carbon atom has more bulky groups around it compared to the the others. So base 3, being the less hindered base, will provide a higher percentage of the major Zaitsev’s product and base 2 will provide a higher percentage of the minor product compared to the other bases.


Given a theoretical combustion engine, how can I calculate the output of this engine depending on the fuel it uses?

The answer to this heavily depends on whether you have an internal combustion engine or an external combustion engine.

External

Because you're just heating the liquid from the outside, you can approximate the final output of the engine as the amount of energy gained from combusting the fuel multiplied by engine efficiency (as you like, but no greater than 100%). In other words:

$$E=kfV$$

Where E: Energy Output; k: Efficiency absorbing the energy; f: Thermal Energy obtained from burning a unit volume of the fuel; and V: volume of fuel combusted.

Internal

For an internal combustion engine, far more factors need to be taken into account. Wikipedia is always a good start to get a sense of the factors that play a role, and whether or not you want to consider them. De
pending on how dedicated you are, and how much the player can 'mess with' the settings of the engine, you could do something as simple as the External Engine (if all the user can do is select fuel and amount of fuel) or as complex as is required based on the variables the user can change.

Some Other Factors to Consider (Optional):

Compression Ratio (internal only)
Gear Ratio (for high torque applications)
Working Fluid (external only)
Ambient Temperature (mainly external only)
Heat Exchanger Material (external only)


Why is the concentration of protons equal to the concentration of a strong acid?

The statement $[\ce{H^+}] = [\ce{A^-}]$ is an approximation for not too diluted acids. However you can pretty fast tell that this won't hold if you add a strong acid to water so that the acid has a concentration of let's say $\pu{10^-8 mol//l}$. Via the formula $\mathrm{pH} = \log\left([\ce{H^+}]\right)$ this would result in $\mathrm{pH} = 8$ so if that was true, you just made the solution less acidic by adding an acid.

In fact the total concentration $[\ce{H^+}]$ is given by $[\ce{H^+}] = [\ce{A^-}] + \pu{10^-7 mol//l}$ where the last term stems from the autoionization of water. However in most use-cases $[\ce{A^-}]$ is far greater than $10^{-7}$. In those cases the autoionization of water can be neglected and thus the approximation $[\ce{H^+}] = [\ce{A^-}]$ gives sufficient results.
However, if you are working with really diluted solutions of acids this approximation is no longer applicable (as demonstrated above) which forces you to actually respect the autoionization of water.

If you want to be on the safe side you can always respect the autoionization. This will always give you the correct pH (even for not so diluted solutions) but you may neglect it if the acid-concentration is high enough (this of course depends on the accuracy of the results you need).

EDIT:
As @MaxW pointed out in the comment a general expression would be $[\ce{H+}] = [\ce{A-}] + [\ce{OH-}]$ which for a neutral solution at 25°C (thanks to @Martin for pointing this out) becomes the formula proved above.


What are the products of a reaction between copper sulfate and sodium bicarbonate?

Copper sulfate reacts with sodium bicarbonate reacts to form basic copper carbonate due to  the great affinity of the $\ce{Cu^2+}$ ion for the hydroxide anion $\ce{OH−}$. From its Wikipedia article:

  Basic copper carbonate can also be prepared by reacting aqueous
  solutions of copper(II) sulfate and sodium bicarbonate at ambient
  conditions. Basic copper carbonate precipitates from the solution,
  again with release of carbon dioxide:

  $$\ce{2 CuSO4 + 4 NaHCO3 → Cu2(OH)2CO3 + 2 Na2SO4 + 3 CO2 + H2O}$$

Here, the "ambient condition" is very important because if we alter the temperature and/or pressure , the reaction may tend to form copper(II) carbonate or any non-stoichiometric copper compound.

Heating the basic carbonate in an atmosphere of carbon dioxide (produced by the decomposition of silver oxalate at 500 °C) produces copper(II) carbonate but it is itself quite unstable and react with moisture to convert back to basic carbonate.

For advanced reading, you can refer to reference 2 below in the reference which describes Copper/bicarbonate equilibria in solutions of bicarbonate ion in details (which clearly shows formation of basic ion $\ce{CuCO3(OH)^2-}$ in the solution).

Notes and References

http://www.sciencemadness.org/talk/viewthread.php?tid=24056
https://www.sciencedirect.com/science/article/abs/pii/0043135471900492
http://science.wonderhowto.com/how-to/synthesize-copper-ii-carbonate-sodium-bicarbonate-302466/ (Although this site claims to make copper(II) carbonate, it is actually making the basic carbonate. You can clearly see its bluish green color of basic copper carbonate).


Confusion regarding $\delta$ charge compared with electronegativity

Although, OP indicated the confusion regarding $\delta^\pm $ charge compared with electronegativity, the text body indicated whole lot of confutations over interatomic/intermolecular forces. Thus, I start with the electronegativity.
According to IUPAC Gold Book, electronegativity is the concept introduced by Linus Pauling as the power of an atom to attract electrons to itself. There are several definitions of this quantity. However, Wikipedia gives a simple to understand version better suited for OP:

Electronegativity is the measurement of the tendency of an atom to attract a shared pair of electrons (or electron density).

The relative electronegativity of $\ce{H}$ atom has given a value of $\pu{2.1 eV^{-\frac{1}{2}}}$ and the electronegativity for each of other atoms in the periodic table have been calculated relative to this value for $\ce{H}$ atom (IUPAC Gold Book). These calculations are beyond OP’s field of study so should be disregarded. The important point for OP to consider is the higher the associated electronegativity, the more an atom or a substituent group attracts electrons. For instance, the electronegativity of $\ce{F}$ is $\pu{4.0 eV^{-\frac{1}{2}}}$ and hence, shared electron fair in $\ce{H-F}$ bond stays more closed to $\ce{F}$ than to $\ce{H}$. As a result, $\ce{F}$ carries permeant $\delta^- $ charge while $\ce{H}$ carries   permeant $\delta^+ $ charge in the $\ce{HF}$ molecule, creating a permanent dipole. Hence, $\ce{HF}$ is a polar molecule. Therefore, among $\ce{HF}$ molecules, there are always dipole-dipole interactions (the image 2 of OP's text). Since the positive end of these dipole is $\ce{H}$ atom, the relevant dipole-dipole interaction is called $\ce{H}$-bonding, the strongest intermolecular force among dipole-dipole interactions. These dipole-dipole interactions occur only in molecules.
Now, I think I have addressed the OP's main confusion:

My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$, then, the electron should be present more towards the $\ce{H}$ side.

According to the above explanation, PO's understanding of electronegativity is not correct. In reality, if an atom is relatively more electronegative than that of the other atom it attaches to, that means it attracts the shared pair of electrons, increasing its electron density compared to its protons (gaining $\delta^- $ charge). Meanwhile, the other atom feels decreased electron density compared to its protons (gaining $\delta^+ $ charge). This is completely opposite of what OP's indicated understanding.
On the other hand, the London dispersion force is also one of the intermolecular forces, but is the weakest among them and occurs between atoms as well. For example, why do you think gaseous $\ce{He}$ become a liquid when the temperature is lowered sufficiently? You can find the answer in chem.purdue.edu:

The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms/molecules occupy positions that make the atoms/molecules form temporary dipoles. Thus, this force is sometimes called an induced dipole-induced dipole attraction. London forces are the attractive forces that cause nonpolar substances to condense to liquids and to freeze into solids when the temperature is lowered sufficiently.

An atom or a molecule can develop a temporary (instantaneous) dipole when its electrons are distributed unsymmetrically about the nucleus because of the constant motion of the electrons (e.g., the image 1 of OP's text). The explanation can be depicted in an image:

For instance, assume two helium atoms approaching each other. When they get closed to each other, the electrostatic attraction force between the nucleus of first $\ce{He}$ atom with two positive charges and the two electrons in the second atom causes each atom to polarize  (creating an induced dipole in each atom). Note that at the same time, there is  an electrostatic repulsion acing as well between electrons of each other as depicted in the image above.
Dispersion forces are present between all atoms or molecules. In molecules, these forces are present regardless of whether they are polar or nonpolar. London dispersion forces depends on few factors:

Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles (e.g., $\ce{Cl2}$ is a gas at room temperature and $\pu{1 atm}$ pressure, while $\ce{Br2}$ is a liquid and $\ce{I2}$ is a solid under the same conditions).
The strength of London dispersion forces varies according to molecular shapes. For example, $n$-$\ce{C5H12}$ and $neo$-$\ce{C5H12}$ are both pentanes, but former is more cylindrical shape while the latter is more spherical shape. Although, both have the same molar mass, $n$-pentane is a liquid (average boiling point of pentanes: $\pu{36.1 ^\circ C}$) at room temperature and $\pu{1 atm}$ pressure, while $neo$-pentane is a gas (boiling point: $\pu{9.5 ^\circ C}$) under the identical conditions.

Also, London dispersion forces are depend of the molecule's polarizability (The ease with which the electron distribution around an atom or molecule can be distorted is called the polarizability). London dispersion forces tend to be stronger between molecules that are easily polarized, but weaker between molecules that are not easily polarized.


Application of the electronegativity effect on bond angles

(I will simply be converting my comments above into an answer here)

Yes, generally, you can apply this in cases with similar geometry. You can see why from the following rationale:

Imagine that the bond pairs look like a ring-shaped cloud, and one is worn on your index finger and another on your middle finger, both towards the fingertip. For example, to represent the two  $\ce{O-H}$ bonds of $\ce{H2O}$.
More electronegative central atoms will pull the ring towards the knuckles and it will cause the middle finger to move aside, thereby increasing the angular separation between the index and middle finger

Here's an illustrative image for $\ce{H2O}$, with blue rings representing bond pair density :

Hence, you can see why the bond angle of $\ce{H2O}$ would be bigger than, say, $\ce{OF2}$ and $\ce{H2S}$.

In case of the former, the fluorines won't led the ring move towards the depression easily, hence the index and middle fingers can come closer to each other in contrast to $\ce{H2O}$

For the latter, $\ce{S}$ is less electronegative than $\ce{O}$, so again, the rings will show less movement towards the knuckles.

This will also help you to understand that you can't do this in cases of radically different molecular geometries, as this visualization will not be very useful there


What databases or repositories are there that have generalized SMILES for functional groups?

I'm currently working on a content store for a resource like this because I've realized I need this as well (and came across your question ~2 weeks ago).

Content Vairable Store

I am writing a python package called global-chem to actively support variable common storage (especially for SMILES/SMARTS) strings that can be easily distributed via pip.

https://github.com/Sulstice/global-chem

The code is fairly simple and can be a one-liner:

from global_chem import GlobalChem
functional_groups = GlobalChem().functional_groups_smiles

GlobalChem is a class and properties on that class correspond to different sets of SMILES/SMARTS strings.

As of right now, I have

93 functional group for SMILES
85 functional groups for SMARTS
19 Amino Acid SMILES strings

I suspect it will grow more over time and probably as I, and others, contribute to it.

If you want to read the docs: https://globalchem.readthedocs.io/en/latest/?badge=latest

Validation

Since these strings can often be hard to validate I've been using a blend of MolVS and RDKit to make sure the strings will match as expected. It probably needs to be also manually tested each one and verified so use at your own discretion while I get some guard rails up.


Why are activities of solids and liquids taken to be unity?

It is not that case that the activity of solids and liquids is always taken as unity.

Instead, activity is unity for a pure solid or pure liquid at standard pressure.

This is a matter of definition of "activity" and "standard state".

According to the IUPAC definitions:

activity (relative activity), a

  Defined by the equation
  $a=e^\frac{\mu-\mu ^0}{RT}$
  where $R$ is the gas constant, $T$ the thermodynamic temperature, $\mu$ the chemical potential
  and $\mu ^0$ the standard chemical potential the definition of which depends on the choice of the standard state.

and standard state

  State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p °, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p ° ...

So activity is 1 almost by definition when a substance is in its standard state, because $\mu - \mu ^0 = 0$  and $e^0 = 1$


In which of the following chemical compounds are the building blocks molecules?

Answer B is correct because nitric acid is the only compound whose building blocks satisfy the IUPAC Gold Book definition of a molecule, a neutral (uncharged) entity consisting of at least 2 or more atoms. The building blocks of all other given compounds are atomic ($\ce{K+}$, $\ce{Na+}$ and $\ce{Ag+}$) or molecular ions ($\ce{NO3-}$), which by their nature do not fulfill the requirement of being electrically neutral. Per definition, they are molecular entities, i.e. distinct particles which can be identified as separate entities. This term is more general and also includes charged species like ions.


n is the number of molecular units in a chain, also called the degree of polymerization. This varies based on the molecular weight of the polymer. The molecular weight is the degree of polymerization times the molecular weight of a monomer.


Does the solubility of iron (iii) oxide increase when the pH of the water it is in decreases?

Iron (III) oxide is insoluble in water. If you add an acid (e.g. $\ce{HCl}$) it might react with the oxide and form a soluble iron salt.

How much of the oxide gets dissolved depends, amongst others, on what acid is used, its concentration and amount, and the time of reaction.


Canonical MOs vs. Localized MOs: Do both represent reality in the same way?

NOTE: In the below, I'm implicitly discussing a ground-state, closed-shell wavefunction, where all occupied orbitals are doubly occupied.  The discussion would be similar for open-shell wavefunctions, but there are complexities that I won't address here. Also, once one starts noodling at excited states, things get complicated pretty quickly, and (AFAIK) some manipulations that are valid for a ground state wavefunction are invalid for excited states.

In my understanding localized molecular orbitals (LMO) are equivalent to "standard" molecular orbitals, often called canonical orbitals (btw. why are they called canonical?). We can produce LMOs by linear combination of MOs and they represent the same physical state.

The "canonical" MOs (CMOs) are called such because they are in a certain sense the "default" MOs for a wavefunction. The sense in which they are the "default" is that they diagonalize the Fock matrix that defines the energy of the system and is used in the course of the self-consistent field (SCF) calculation of, e.g., the Hartree-Fock method for solving the approximate many-electron problem.
Occupied localized MOs (LMOs), when all considered together as a unified wavefunction, in total contain the exact same information as the wavefunction composed of the occupied CMOs.  This is because the LMOs are obtained from a unitary transformation of the occupied CMOs, and the linear algebra indicates that the overall wavefunction is invariant to such a transformation.

If LMOs are equivalent and have the same energy then why do calculations always end up with canonical molecular orbitals and never with LMOs?

The italicized phrase of this question is false. LMOs individually are not equivalent to individual CMOs, and they do not have the same energy. Strictly speaking, the energies of individual LMOs are actually undefined. Only the total wavefunction composed of LMOs is equivalent to the total wavefunction composed of CMOs.
As noted above, the reason why we always get CMOs out of an SCF energy calculation is that they are the set of orbitals naturally associated with the solution of the SCF problem.

And wouldn't there be an unlimited amount of equivalent solutions?

Yes. Every possible unitary transformation among the occupied canonical orbitals, of which there is indeed an unlimited number, would provide a set of orbitals that gives an equivalent total wavefunction.  Not all of these transformed orbital sets are especially useful, however. Specific examples of useful ones include the various localization transforms that are the topic of the question, the series of variously defined "natural orbitals", and (somewhat related) the Löwdin symmetric orthogonalization for atomic orbitals.

Is this just a mathematical gimmick or does this represent the reality? ... Or is this just a thing we do because it makes it easier to understand?

The transformed MOs do represent reality, and often provide a very powerful and/or convenient means for analysis and understanding of the overall wavefunction. You do have to know what is or is not a valid use of any particular set of transformed orbitals, however.
One specific application of transformed orbitals I know about that seems very powerful is the family of local pair natural orbital (LPNO) methods for accelerating correlated calculations. The v4.0 release of ORCA includes implementations of their advanced domain-based (DLPNO) version of the method for several calculation methods (MP2, CASSCF-NEVPT2, etc.; see various of the 2016/2017 Neese citations listed here).  The DLPNO-CCSD method has already proven to be extraordinarily powerful in enabling correlated calculations of large systems—the below figure is a reproduction of Figure 5.5 from the ORCA v3.0.3 manual showing the enhanced performance of (D)LPNO-CCSD methods for a series of benchmark systems (click to enlarge):

Is it possible for a molecule to be in such a state where every electron is in such a localized orbital?

Yes, this is always true, as long as the transformation that is applied to the occupied orbitals includes no unoccupied orbitals. No matter what set of orbitals you use, whether canonical or transformed, every occupied orbital in the wavefunction is always fully (here, doubly) occupied. Transformation among the occupied orbitals doesn't change the occupancy.
Inclusion of some un-occupied orbitals (often called virtual orbitals) in the transformation, though, will lead to orbital occupations between zero and two. This is what happens in the classic natural bond orbital analysis of Weinhold & co., as can be seen in the sample natural population analysis output (pdf link; "Occupancy" column) provided on the GENNBO website .


What is wrong with the Aufbau principle that results in so many exceptions?

It's basically because you can't just add up electronic energies using a shell-by-shell, electron-by-electron approach.  As soon as you have two or more electrons in close quarters, they interact with each other in ways that cannot be described by simple electrostatic laws because of the quantum mechanical nature of the electrons.  The comments (specifically Mithoron's) refer to other questions whose answers explain this much better than I can.  Please look at those.


The following are ladders of oxidation states for two elements. Simple examination shows there is no hard and fast rule for valence and stoichiometry overall. Valence need not be integral, either, for example, the Creutz-Taube ion.

Fe(-2), $\ce{Na2[Fe(CO)4]}$, Collman's reagent
Fe(0), $\ce{Fe(CO)5}$
Fe(0), $\ce{Fe3(CO)12}$
Fe(+2), $\ce{FeCl2}$
Fe(+3), $\ce{FeCl3}$
Fe(+4), $\ce{K4[FeO4]}$
Fe(+5), $\ce{K3[FeO4]}$
Fe(+6), $\ce{K2[FeO4]}$

V(-2), $\ce{[Na(diglyme)2][V(CO)6]}$
V(-1), $\ce{(Cp)V(CO)4}$
V(0), $\ce{V(CO)6}$

Chemistry seeks to establish rules.  Then the fun starts - breaking the rules.


How to prevent Ag2O from decomposing to Ag on Aluminum

First,  as you mention, $\ce{Ag2O}$ starts to decomposes from ≥200 °C. Measure the temperature of the aluminum substrate with an IR thermometer or attached thermocouple to be sure you stay well below that. You can substantially speed drying of the ink by lowering the air pressure, e.g. using an aspirator pump, perhaps eliminating need to heat the substrate.

You may also need to anodize the aluminum prior to printing  to prevent chemical reaction with the $\ce{Ag2O}$ in the ink. Aluminum is much higher in the electromotive series and would rapidly replace silver in the oxide.


Low-spin / high-spin arguments: why overlooking pairing energy?

Regarding your question about the non-linear change with ligands, I suspect this isn't that easy of a question to answer, but the data is probably out there somewhere.

Regarding the quote and question in the title, when these kinds of arguments are made, they aren't ignoring the pairing energy, but they also aren't really making any predictions.

Essentially, Ligand Field Theory (LFT) lays out a simple way that one can rationalize the geometry of a particular transition metal complex based on the energy of the d orbitals. I can see that you know this. What this means though is that when they observe whatever change happens for the iron complex, LFT says that this must correspond to some change in the energy of the electrons associated with shifting changes in the energy of occupying orbitals in a specific way.

That is, in LFT the only two terms that describe the energy of a particular electronic arrangement are the ones describing the potential energy of the electrons while ignoring their interactions, and the increase in energy from pairing electrons in orbitals.

So, when any geometrical or similar observation is made, there's really no other explanation than to say the a high-spin complex became low-spin because the orbitals split more ("considerbale change in LFSE") which guarantees this decrease in potential is larger than the cost of pairing electrons.

I agree that it's quite naive to always explain things in this way, but that's why crystal field theory and ligand field theory are not meant as real computational or predictive tools these days. They are useful for connecting geometry, spin, and energy as well as giving a first-order explanation of observations.

It's like when you'll sometimes hear inorganic chemistry students describe the behavior of a catalyst as being due to a "combination of electronic and steric effects." That's true... but those are also the only possibilities, so it doesn't tell you much.


Colour intensity of transition metal complexes

At first, it is necessary to ask why some transition metal complexes are colourful at all — remember that alkaline metals, alkaline earth metals and earth metals usually form colourless solutions, e.g. sodium, magnesium, aluminium. The transition metals’ colour stems from the d-electrons they have and the shape of the complexes they create. A typical orbital scheme of a 3d-metal in an octahedral complex will look something like this (image originally taken from Prof. Klüfers web scritptum to his coordination chemistry course at the LMU Munich):

On the left are the metal orbitals, from top to bottom it is 4p, 4s and 3d. On the right are the ligand group orbitals, assuming six ligands each only donating electrons in a $\unicode[Times]{x3C3}$ fashion only. Any remaining metal electrons will be filled into the $\mathrm{t_{2g}}$ orbitals which correspond to $\ce{d}_{xz}$, $\ce{d}_{xy}$ and $\ce{d}_{yz}$ and into the $\mathrm{e_g}$ orbitals corresponding to $\ce{d}_{z^2}$ and $\ce{d}_{x^2-y^2}$

The energy difference marked as $10\,\mathrm{Dq}$ in the image would then be the energy of a photon required to excite an electron from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$; this photon would thereby be absorbed, which is where we get our absorption coefficient from. It is usually in the visible range of light.

There are a few quantum mechanical principles that allow or disallow excitation from one energy level to another. One of these is the spin-allowance rule: Excitations are only allowed if the spin of the entire system remains unchanged. (So excitation cannot involve a spin exchange.) This is why $\ce{Mn^{II}}$ or high-spin $\ce{Fe^{III}}$ complexes are generally colourless: all d-electrons are of identical spin so there is no freedom to be excited.

A second rule is the Laporte-rule whose formal definition I never properly understood, but which states that d to d excitation is forbidden in centrosymmetric species such as octahedra. This rule would mean that all d-element complexes should be colourless. However, complexes are not static but the ligands vibrate back and forth, often in a non-centrosymmetric way. Whenever a photon of the correct energy hits a complex that is not ideally centrosymmetric, the excitation is allowed and the photon absorbed. Because this is a rather rare phenomenon, the colour of such a complex is weaker than it could be.

The $\ce{[Co(H2O)6]^2+}$ case is more complicated again, because in a perfectly centrosymmetric world it would be degenerate: Since it is $\ce{d^7}$, two spin-down electrons share three $\mathrm{t_{2g}}$ orbitals. It would be favourable to somehow make these orbitals dissimilar. This can happen by distorting the octahedral shape in a Jahn-Teller manner: The two water ligands on the $z$ axis are moved slightly further away, creating a slight stabilisation of all orbitals which contain a $z$ component, most notably $\ce{d}_{xz}$ and $\ce{d}_{yz}$. But since the final product is still centrosymmetric, this paragraph only explains the wavelength of absorption slightly better and can be disregarded.

Next question is why the chlorido complex of cobalt(II) is not octahedric but tetrahedric. This is due to a hypothetical $\ce{[CoCl6]^4-}$ being unbalanced in terms of large chloride ions around a small cobalt ion that only has a low charge to combat the $-6$ the chlorides have to offer. The tetrahedral complex $\ce{[CoCl4]-}$ has everything that the octahedral $\ce{[Co(H2O)6]^2+}$ has except the centre of symmetry. Because centrosymmetry is missing, the Laporte rule is no longer valid and every photon of the correct energy can excite electrons. More excitation equals a stronger extinction.

Also, the colour of both complexes is notably different because the orbital scheme is, too. $\mathrm{t_2}$ and $\ce{e}$ are now much closer together and their order is flipped. Combining that with chloride ions that usually generate low values of $10\,\mathrm{Dq}$, the energy difference is even smaller therefore the excitation wavelength longer therefore the colour we see blueshifted (complementary colour).


Actually, the numbering is the same, at least for the fructose portion of the molecule (which is the right-hand monosaccharide in your sucrose image - the one on the left is glucose).

The key point in carbohydrate numbering is to look for the carbon which is an aldehyde or a ketone (the carbonyl carbon). This will be the end of the monosaccharide which is numbered lowest. For aldoses, this carbon would be carbon 1. For ketoses, it's the lowest number that can be reached in the linear, ring-open form (e.g. for fructose, it's 2). The rest of the carbons are numbered accordingly.

When you have a linear Fischer projection (like the "D-fructose (linear)" images on the second image) the carbonyl carbon should be obvious. In the ring form, it's less obvious.

Instead of looking for an aldehyde or a ketone in the cyclic form, you're looking for an acetal or hemiacetal. That is, you're looking for a carbon with multiple bonds to oxygens. In the first image, this is the carbon which is making the oxygen bridge between the two monomers (the one on the left of the furanose) - the oxygen of the bridge is one oxygen, and the oxygen of the furanose ring is the other. In the second picture, it's the carbon on the far right. You have the ring oxygen of the furanose, and then a free hydroxyl oxygen. Both of these carbons are numbered "2", and if you imagine rotating the furanose ring 180 degrees to flip over that acetal position, you'll see the other carbons match up as well.

(One thing that might be tripping you up is the directionality of the 1 carbon with respect to the 2 carbon. Keep in mind that the hemiacetal form can interconvert with the ring-open carbonyl form, and the "anomeric position" of the monosaccharide is free to invert in chirality - that is, it can readily flip between "up" and "down". It's only when it's in the bonded acetal form that it's locked into either the alpha or beta configuration.)


As Oscar Lanzi suggested, both $+M$ and $-I$ applies here, but $\ce{Cl}$ stabilizes carbocation, meaning $+M$ is more effective than $-I$. This fact was confirmed by this peer-reviewed paper (Ref.1):

  The lowering of $\ce{C_\beta–H}$ stretching frequencies in carbocations 1a–d and 2a–c induced by hyperconjugation was tested as a possible probe for estimating the electron donating ability of $\alpha$-substituents. Conclusions are based on the results of high level quantum chemical calculations confirmed with experimental FT-IR spectra. Because the decrease in the $\ce{C_\beta–H}$ stretching frequency is comparable in 1b and in 1c, and in 2b and 2c respectively, it follows that $\alpha$-substitution $\color{red}{\text{by a methyl group}}$ or $\color{red}{\text{by chlorine}}$ stabilizes a carbocation with $\color{red}{\text{almost the same effectiveness}}$.

The effect of the chlorine $n$-electron back donation is evident by a partially double bond character of the $\ce{C+–Cl}$ bond in $\alpha$-chlorocarbocations observed experimentally in IR spectra (Ref.2). In this experiment, FT-IR spectrum of $\ce{Cl3C+}$ cation has displayed the $\ce{C–Cl}$ stretching frequency at $\pu{1040 cm–1}$, which is $\pu{250 cm–1}$ higher than in neutral alkyl chlorides (e.g., characteristic $\ce{C–Cl}$ stretching frequency of $\ce{CCl4}$ is $\pu{785 cm-1}$). This is indicative of partial double bond character of $\ce{C+–Cl}$ bond suggested by the resonance structures:

$$\ce{Cl2C+-Cl <-> CH2C=Cl+}$$

The pioneering work of Olah and coworkers has also predicted this phenomena by $\ce{^{13}C}$-NMR studies of trihalocarbocations (Ref.3):

$$
\begin{array}{c|c|c|c}
\ce{HCX3} & \delta\ce{^{13}C}\text{ of }\ce{CHX3}  & \ce{^+CX3}  & \delta\ce{^{13}C}\text{ of }\ce{^+CX3} & \Delta \delta\ce{^{13}C} \\
\hline
\ce{HCCl3} & 77.7 & \ce{^+CCl3} & 236.3 & 158.6 \\
\ce{HCBr3} & 12.3 & \ce{^+CBr3} & 207 & 194.7 \\
\ce{HCI3} & -139.7 & \ce{^+CI3} & 95 & 234.7\\
\hline
\end{array}
$$

Olah has suggested that the decreasing trend of $\Delta \delta\ce{^{13}C}$ by going $\ce{I}$ to $\ce{Br}$ to $\ce{Cl}$ is in agreement with the positive charge stabilization by back-bonding in order of $\ce{Cl > Br > I}$.

References:

Milan Mesić, Igor Novak, Dionis E. Sunko, Hrvoj Vančik, "$\ce{C–H}$ Hyperconjugation in $\alpha$-chlorocarbocations," J. Chem. Soc., Perkin Trans. 2 1998, (11), 2371-2374 (https://doi.org/10.1039/A805772I).
Hrvoj Vančik, Ksenija Percač, Dionis E. Sunko, "Chloromethyl cations in cryogenic antimony pentafluoride matrixes and the generation of carbocations from hydrocarbons," J. Am. Chem. Soc. 1990, 112(20), 7418–7419 (https://doi.org/10.1021/ja00176a065).
George A. Olah, Ludger Heiliger, G. K. Surya Prakash, "Stable carbocations. Part 276. Trihalomethyl cations," J. Am. Chem. Soc. 1989, 111(20), 8020–8021 (https://doi.org/10.1021/ja00202a056).


Triphenylphosphine (PPh3) vs. 1,2-bis(diphenylphosphino)ethane (dppe) ligand?

As far as I am aware, the selection of the best ligands for transition-metal catalysed reactions is still trial and error. In fact, not only is the selection of ligand trial and error but so is the choice of starting materials to result in that metal–ligand combination.

What the original discoverer of a reaction method typically do is run a number of screens keeping most parameters constant but varying one; this one varying parameter will be ligand, solvent, time, temperature, additives and metal salt. (Usually, they will have decided before running the experiments which metal they want to use for their reaction so the only reason to include another one is to directly compare the new method to an existing one and e.g. show that nickel is better than palladium in this reaction.) Whichever reference you’re looking at, the authors most certainly performed such a screening and received a slightly higher yield for one set of reactants compared to another; they then published that as their final choice.

After having found the optimal conditions, many researchers are tempted to ‘explain’ why a certain ligand is better than another ligand. Our knowledge does include how electron-rich or electron-poor the donor atom of a given ligand is, how sterically demanding a set of substituents is and so forth. If ligands of one general type form a general trend, one could well be tempted to draw from that general trend and exclaim that e.g. an electron-rich phosphane works best for this reaction. In ideal cases, that prediction is then tested experimentally with a previously unused but more electron-rich phosphane; a reaction that may or may not work. The point, however, of this paragraph is that more often than not the ‘explanation’ is little more than a post-factum rationalising ballpark and should be treated as such. Years to decades of research will likely be required until we actually fully understand the intrinsic properties at play and are able to make good predictions.


Does the Bohr model violate the uncertainty principle?

The Heisenberg uncertainty principle states that

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$

When you think about it, the whole of classical mechanics violates this principle. Think of a particle with mass $m$, a defined position $x_0$ at time $0$, a defined velocity $v_0$ at time $0$, and a constant acceleration $a$. The laws of kinematics state that at any time $t$, the position $x_t = x_0 + v_0t + \frac{1}{2}at^2$, and the momentum $p_t = m(v_0 + at)$, with zero uncertainty in both quantities, i.e. $\Delta x \Delta p = 0$. That works in most real-life cases because on a macroscopic level, quantum effects are negligible.

The Bohr model was derived via the classical assumption that the electron orbits the nucleus, much like how the Earth orbits the Sun, except that the origin of the centripetal force was electromagnetic instead of gravitational. It then went on to derive the energy levels of the electron in a hydrogen atom using the laws of uniform circular motion and electrostatics. This means that it is theoretically possible to know the position and momentum of the electron simultaneously with no uncertainty - a violation of the uncertainty principle.


Meaning of m2 in the symbol for an isotope of an element

As per this Wikipedia page on isotopes:

  The letter $m$ is sometimes appended after the mass number to indicate a nuclear isomer, a metastable or energetically-excited nuclear state (as opposed to the lowest-energy ground state), for example $\ce{^{180\mathrm{m}}_73Ta}$ (tantalum-180m).

Note that the lower subscript to the left of the chemical symbol is the atomic number of the element (in the example above, 73 is the atomic number for tantalum) and is omitted in your example.


Acidity is directly proportional to the stability of conjugate base. Aromatic compounds are very stable.

In essence only one compound after deprotonating is aromatic - which is the conjugate base of (A).


Why is the eclipsed conformer more stable than the bisected conformer in 1-butene?

In alkanes staggered conformers (all anti) are local minima, while eclipsed conformers represent transition states. The first is "stable", while the latter is not. The rotational barrier has been measured and calculated often, it is around 12 kJ/mol.[1]

In alkenes the eclipsed conformers are local minima, while the staggered, or more accurately bisected, conformers are transition states. The rotational barrier for propene is only about 8 kJ/mol.[2]

I have performed a full, relaxed rotational scan of 1-butene around the central carbon-carbon bond axis (blue). The minima and maxima of this scan have been reoptimised and thermal corrections at 298.15 K and 1 atm have been added (orange).

The chosen level of theory, DF-BP86+D3(BJ)/def2-SVP, is not the most accurate, but it will give you a general idea. You can see the bisected conformations are maxima along the rotational coordinate and the eclipsed ones are minima.
There is dispersion correction included, which makes the E1 conformer slightly more stable than the E2 conformer.
Including the thermal corrections, the energy difference between these two structures is negligible.
$$\begin{array}{crrr}
\text{Conformer} &
\angle_\mathrm{D} &
\Delta E~/\mathrm{kJ\,mol^{-1}} &
\Delta G~/\mathrm{kJ\,mol^{-1}}\\\hline
\mathrm{E_1}     &   0.0             &     0.0  &   0.0   \\
\mathrm{B_2}     &  60.0             &     9.9  &  10.1   \\
\mathrm{B_1}     & 120.0             &     1.2  &  -0.2   \\
\mathrm{E_1}     & 180.0             &     9.7  &   9.4   \\\hline
\end{array}$$

The eclipsed forms are stabilised by two π-σ/π-σ hyperconjugation interactions. Here is the visualisation for E1 in the natural bond orbital picture. Occupied orbitals are depicted blue/orange, virtual orbitals are depicted red/yellow. The phases are assigned randomly.

In the B2 conformer the overlap between those orbitals is not ideal. The nodal planes in the anti-bonding orbitals are in the way.

The representations for E2 are analogue to E1, and B1 is similar to B2. Because of time reasons I have not been able to create the pictures yet.

Taking the maximum difference in the energy we can approximate the rate of the reaction with transition state theory (assuming $\kappa=1$):
$$k = \kappa\frac{k_\mathrm{B}T}{h}e^{\frac{- \Delta G^{\ddagger }}{RT}}
\approx 1\times10^{11}~\mathrm{\frac{1}{s}}$$
This is ultrafast and at room temperature we consider it a free rotation. With spectroscopic methods we will not see a difference. We would probably have to cool the system down to below 1 K to actually see two (E1, E2) conformations.

Steric effects are often electronic effects in disguise, they offer tempting explanations, but are often rooted in a more complex electronic explanation. In this example the possibility of hyperconjugation explains the slightly higher energy the bisected forms compared to the staggered forms.
Hyperconjugation can come in different forms, the most common explanation is the overlap of an empty anti-bonding orbital with a filled (non-)bonding orbital. In this scenario it does not matter which elements are involved in these interactions.
There is even negative hyperconjugation, but this is a little beyond the scope of this post - if there are more questions about the topic, you are always welcome to ask them here.

Footnotes

For a quite recent molecular orbital study see: Ramiro F. Quijano-Quiñones,  Mariana Quesadas-Rojas,  Gabriel Cuevas and  Gonzalo J. Mena-Rejón, Molecules 2012, 17(4), 4661-4671. (open access)
For an educational approach see: Henry Eyring, D. M. Grant and H. Hecht
J. Chem. Educ. 1962, 39 (9), 466.
I did not have the time to search thoroughly, but the following publication contains the value. Leland Cullen Allen, Eugene Scarzafava, J. Am. Chem. Soc. 1971, 93 (2), 311–314.


You should not be looking at $\ce{-NH-CO\bond{-}}$ as two different functional groups i.e. as $\ce{-NH-}$ and $\ce{-CO-}$.

Since as in $\ce{-(NH-CO)\bond{-}}$: $\ce{N}$'s lone pair are actually in resonance with the carbonyl group.

The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.

The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.

Ortho position is hindered due to steric hindrance of the another phenyl ring.

And therefore

   the attack takes place at para position of the left handed phenyl ring

so answer should be option B.


Ok, let’s take it one step at a time:
Percarbonate is actually not a single molecule, but a double-salt of sodium carbonate and hydrogen peroxide.
The sodium carbonate reacts with sulfuric acid, resulting in carbon dioxide, sodium sulfate and water. The reaction scheme is: Na2CO3 + H2SO4 -> Na2SO4 + H2O + CO2

For safety reasons I will strongly discourage you from adding carbonate to concentrated sulfuric acid. It will most likely explode in your face!

But in theory, after having added sufficiently carbonate, Sodium sulfate will precipitate.
But getting a drying effect will be uphill, since water is produced alongside the sulfate, as you can see in the reaction scheme, on top of this is any water that comes from hydrogen peroxide, not to mention the fact that concentrated sulfuric acid is strongly hygroscopic itself.
As for the cleaning of impurities, I fail to see what neither carbonate nor hydrogen peroxide can do. HP can oxidize some organic materials, but most of these will probably already have been destroyed by the sulfuric acid. It can also oxidize some inorganic ions like sulfite (SO3), but do you suspect these to be a problem?
I don’t have detailed knowledge of the conditions needed to form peroxo(mono)sulfuric acid. Undoubtedly some will be formed when HP is added to sulfuric acid, but I am also convinced that only a fraction of the sulfuric acid/HP will react.