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Electrochemical Cell
Now in an electrochemical cell , we make use of the fact that whenever there is a potential drop , electrons will flow from lower to higher potential as for a potential drop , there must be an electric field .
Now suppose there are two single electrodes , what creates the electrostatic potential on each of them in the first place ?
And just to confirm , a redox reaction is used just that we can maintain a steady flow ?And so why is it that we always put in that electrolyte that has one of the ions same as the electrode element ?

I wasn't quite sure what you meant by "a redox reaction is used just that we can maintain a steady flow." Can you rephrase that?

It may help you.This explanation gives more insight on what happens at each electrode before connecting wires.
https://www.chemguide.co.uk/physical/redoxeqia/introduction.html

Creating a Galvanic Cell
To simplify, let's forget galvanic cells for a moment. Instead, just picture an oxidation/reduction reaction in aqueous solution.  If we had a solution of $\ce{Cu^+}$ and $\ce{Co^{3+}}$, the metal ions would exchange an electron, and we would end up with $\ce{Cu^{2+}}$ and $\ce{Co^{2+}}$. This happens because of the relative stability of the two ions; the electron is at a lower energy with the cobalt than with the copper (at these oxidation states). Now, if we separate the two metals into two different solutions, insert an electrode into each, and connect the two with a conductive wire, we can still transfer an electron. The potential between the electrodes comes from the same source as the mixture: the chemical potential between the two ions.
A salt bridge is also necessary. The salt bridge balances charge in the two solution. Otherwise, we would have a net positive charge in the copper solution and a net negative charge in the cobalt solution. This would create a potential difference and the reaction would eventually and stop.
Galvanic Cell using Non-ionic Species
Galvanic cells do not have to use species that are ionic in the oxidized and reduced states (or either, for that matter). Consider the cell pictured below. In this example, zinc metal is being oxidized to $\ce{Zn^{2+}}$ and $\ce{Cu^{2+}}$ is being reduced to copper metal. In other words, the zinc electrode is getting dissolved, and the copper electrode is growing. The anodic salt solution is used to create what is called a poised electrode. If a zinc electrode were inserted into pure water, according to the Nernst equation, the system would have an infinite potential.
$$\mathcal E = \mathcal E^0 - \frac {RT} {nF} \ln 0 \rightarrow -\infty$$
As soon as even a few molecules per liter are in solution (which would happen very quickly) the potential drops to around 50 V, and would continue to drop as more zinc was oxidized (although there are certainly deviations from the Nernst equation in an observed system). The salt solution puts the system into a steady state. In terms of the cathode, the metal does not matter. Anelectrode made from any conductive material would behave the same as a copper electrode in a solution of $\ce{Cu^{2+}}$; copper would plate onto the electrode as it is reduced. The system potential is not altered because the activity of a pure solid is unity. It is normally just convention (and for learning purposes) that the electrode material is the same as the oxidant.


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Gas Electrode Working
How does a SHE electrode work?
Does the $H_2$ gas get absorbed on the platinum electrode? How is it reduced? How are electrons transferred from the adsorbed gas to the solution? How is $H^+$ liberated or how does the $H^+$ in the solution take electrons from the adsorbed $H_2$?
If platinum doesn't adsorb the $H_2$ gas, what is its use? How does it provide a surface for reduction/oxidation of hydrogen?
Also in a concentration cell, why is it, in evaluating Q(reaction quotient)
you take pressure of a gas numerically and put it directly as a concentration?
And in evaluating log(Q) in nernst equation, Q isn't always unitless. But inside log, you can put only a unitless number, isn't that so?
And in concentration cells, if [c1] > [c2], then c1 acts as cathode. In Gases, considering Pressure=k[C], where k is a constant fixed at a particular temperature. then in gas concentration cell opposite happens if pressure 1 > pressure 2 then pressure 1 part acts as anode.
What is P? is it pressure of the gas above the solution?
In net, how does a gas electrode work?

Standard Hydrogen Electrode
A SHE consists of four parts: a platinum lead wire, a platinum foil contact, an acid solution, and a hydrogen gas flow. The word standard refers to the fact that all concentrations and pressures are at standard state, thus the acid is 1M and the gas flow is 1 atm. At these conditions, the electrode has a potential of 0 V (this is how potential is defined). If there are any deviations from these values, the electrode is no longer considered a SHE. The potential can then be calculated using the Nernst equation.
The electrode is one half of an electrochemical circuit. The other half cell determines whether the SHE will behave as the cathode or the anode. If it is the anode, the electron flow comes from the external system to the platinum contact. 2 electrons reduce 2 acidic hydrogens from the solution to form one molecule of hydrogen, as in $2H^+ + 2e^- → H_2$. The molecular hydrogen then bubbles up above the solution. If the electrode is cathodic, hydrogen gas is absorbed on the platinum contact plate and oxidized to two acidic hydrogens which remain in solution. The electrons then flow to the external circuit. Electrons do not flow from hydrogen gas to the acid, or vice versa.
Reaction Quotient
The reaction quotient is indeed unitless. While it is common to use concentrations and pressures to calculate Q, it is more properly calculated by using activity of each species, which are also unitless. The activity is the apparent concentration or pressure of a species (can be viewed as a correction to concentration or pressure). Because in most chemical applications activity is very close numerically to concentration, using concentrations or pressures directly for calculations is a close approximation.

""platinum foil contact"" This is wrong, you need an electrode made from "platinized platinum". Ordinary platinum foil will not work.

@Georg I decided to leave out that detail for simplicity and heuristics.

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How to write down the composition of eluent mixtures for TLC
I obtained some Rf values via TLC by adding two or three drops methanolic triethylamine to an eluent mixture of let's say DCM/MeOH = 9:1. I couldn't find an example of how a mixture like that is written down in short in publications. I think that calculating the absolute content of dissolved ammonia wouldn't be suitable.

Describe what you did. The key is make sure your communication allows your experiment to be reproduced.

... Rf 0.34 (10 mL of 9:1 DCM/MeOH + 3 drops of 1.0 M Et3N in MeOH).

If you do all of your TLC with the additional of triethylamine, then you can include a sentence in the general methods about this procedure.

All TLC plates were developed in a 9:1 mixture of DCM and MeOH that was made alkaline by the addition of three drops of 1.0 M Et3N in MeOH for every 10 mL original solution.

The procedure you describe is common, especially in the elution of basic molecules (since silica is slightly acidic). You might find examples of how to state this sort of procedure in the synthesis of amines and alkaloid natural products. Ultimately, in a publication, you should err on the side of thoroughness for your experimental methods. If it's too long, a reviewer of the editor will tell you.

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What compounds do Nor1 and Nor2 citalopram refer to?
In Goodman and Gilman's The Pharmacological Basis of Therapeutics 12th Edition pg. 406, two drugs, nor1-citalopram and nor2-citalopram are referred to but I have no idea what they are. Could somebody tell me a more commonly used name of these drugs or their IUPAC names if common names do not exist. Also please show me what exactly the nor1 and nor2 refers to. I'm guessing there's methyl groups missing in them at different positions because I know that norpethidine (or normeperidine) refers to the N-demethylated metabolite of pethidine (or meperidine).

Those names are referring to desmethylcitalopram and didesmethylcitalopram, which, as you correctly predicted, are the N-demethylated and N-didemethylated metabolites of citalopram. Per IUPAC conventions, the nor- prefix is used to refer to any structural derivative of some parent compound in which a single carbon and (possibly) its hydrogens have been replaced (typically by a hydrogen, but not exclusively). See the relevant Wikipedia page on the prefix for more detail.

Thanks, I saw these pages on wikipedia and I had my suspicions but I wasn't certain so thanks for confirming that.


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Result of glucose with acetic anhydride and pyridine?
What is the result when glucose is mixed with excess acetic anhydride and excess pyridine?

Related: What are the roles of pyridine and DCM in the acylation of an alcohol?

Most probably it forms glucose pentaacetate. The acetic anhydride esterifies with all the alcohol groups on the glucose ring.


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What analytical methods are used to quantify C, O, N, and H in a sample?
A very common analytical method for detection of more than 70 elements is ICP-OES (Wikipedia page). But apparently, it's not suited for some substances (cited from this paper, section 6.2, page 12):

The elements that are
  not usually determined by ICP/OES fall into three basic
  categories. The ﬁrst category includes those elements
  that occur either as trace contaminants in the argon gas
  used in the ICP/OES ($C$ from $CO_2$), constituents of the
  sample solvent ($C$, $O$, $H$), or as contaminants from the
  environment or atmosphere ($N$ for example). The second
  category encompasses those elements that require high
  excitation energy, such as the halogens. These elements
  could be determined with poor LOD, however. The third
  category is the family of short-lived radioactive elements
  that are commonly determined by $\gamma$-ray spectrometry.

So, apparently, when I'm working with, say carbon-based fuels, and I want to determine $C$ contents, ICP-OES would not be the method of choice, at least not the standard variety. So, what else would be the typical choice? Or are there modifications to ICP-OES, such as using very pure $Ar$, or $He$, as the plasma gas and sealing the plasma chamber from the atmosphere?

For N:
Total N = N from NO2 + N from NO3 + Total Kjeldahl N
The methods from here are:
Liquid Chromatography: NO2, NO3
Specific electrodes: NO3, Kjeldahl
Spectrophotometry: NO2
This for water including wastewater. I don't know the aplicability for other samples.
Also ICP-MS is worth researching.

The analytical method preferred for CHNO (and also S, X, etc) determination is elemental analysis, usually by combustion.
For example, if a molecule has the generic formula $\ce{C_wH_xN_yO_z}$, combustion would produce:
$$\ce{C_wH_xN_yO_z + (w +\frac{x}{4}+\frac{y}{4}+\frac{z}{2})O2 -> wCO2 + \frac{x}{2}H2O +\frac{y}{2}N2}$$
Determining the masses of $\ce{CO2}$, $\ce{H2O}$, and $\ce{N2}$ given off by the combustion is the first step in determining how much carbon, hydrogen, and nitrogen was in the original sample. The remaining mass is oxygen.


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(How) Does ICP-OES determine the quantity of detected elements, or just their occurance in the sample?
ICP-OES is a very common analytical method (Wikipedia page here), and I have heard a number of times (from chemists, too) that it is used for quantitative analysis.
However, from the description of the process, I'm not certain if it really does give quantitative information, how it does so, and whether it can do so on its own, or whether there are additional methods run in combination (like with gas chromatography and mass spectrometry) to achieve that result.

Although fp's answer is absolutely correct and I see that you have something more than high school chemistry, I'd like to give a more elaborated, yet simplistic answer, if you or anyone else reading this has any interest on the basic principles of an ICP-OES - you only need high school chemistry to understand it.

I'm not certain if it really does give quantitative information

It does, and very "accurately". Detection limits can be in ppb, that is, it can detect 0.0000001g (0.1 µg) of something in a 1L sample. That's really, really impressive if you stop to think about it.

how it does so

Wizardry! Well, almost. Read on and you'll see.
First of all the sample must be prepared, you can't simply toss something in it. Depending on what your sample is, it must be mixed with a known amount of this or that chemical (you'll see later why). This answers your next question:

and whether it can do so on its own.

Although the actual analysis and preparation can indeed be done purely by the machine in some cases (here's a video), and there are methods such as laser ablation which make it possible to almost just toss something in the ICP, the know-how of a trained analyst is still require to tell the machine what to do, to interpret the results it outputs, and to tell what chemicals to mix in the preparation process, if there is one at all. Equipments such as gas chromatographs can (and might need to) be coupled to assist in the sample preparation.
Back to how it does so:
When you introduce the prepared sample into the machine (it basically sucks it) some processing is done to it, and at some point it reaches the atomizer - the plasma torch. The atomizer does what it's name suggests: It turns stuff into separate atoms (by heating them to absurd temperatures). Here's a picture I took from an ICP atomizer:

Note: Not 100% of the sample is atomized. In fact, only a small percentage is.
When it's turned on it looks like the picture you saw at Wikipedia. You can see by the size of the parts that only a few milliliters of sample - if not less - is atomized at a time. Now that it got our sample atomized, it heats the sample up just a bit more, so the electrons get exited and go from their current level of energy to a higher one. While the sample passes through the equipment it gets cooler, so the electrons get back to the lower energy level, and by doing so, the atom emits light, hence the name optical EMISSION spectroscopy.

(source: wikimedia.org)
But what elements are emitting light, and at what wavelength? The principle of every spectroscopy analysis is that each element emits and absorbs specific, known, predictable wavelengths ranges, and only when heated to specific temperatures (in the case of emission). Because the temperature was set to a specific one for the analyte and a specific set of chemicals were mixed with the sample, only the atoms of the element we want to quantify will emit light at a certain element-specific wavelength range. We set the equipment to detect only that wavelength range and ignore others and that's how we know how much of that element there is in the sample.
But if you kept track of the proportions you might be thinking: "if it detects ppb amounts in only a few ml of sample where only a few % of it is atomized and only certain elements will emit only a certain wavelength, how the hell can it detect that little light?". That's the photomultiplier's job. It's a fantastic piece of equipment and you should definitively read about it.
After it detects the light emitted by the analyte in the sample, it compares it to a blank, does some calculations and displays the results in the form of a graph. Here's a screenshot of an ICP-OES software:

As I mentioned, this is a simplistic explanation, most of the process isn't that simple in reality. So as you can see, it's basically witchcraft.

The actual quantification is not possible without a set of calibration standards. The instrument does not "know" that the concentration is, for example, 100 ng/ml. The instrument must compare the detected signal with the regression curve from the calibration standards.

Quantitative results are obtained by:

Measuring the emited light of known samples (standards) and making a curve of emission for elements of interest.
Comparing the emission of the unknown samples and calculating the concentration of analytes with the curve.
Using samples fortified with standards and duplicates.
Using statistics to measure accuracy, precision, etc. of the results, and in general evaluate the quality of the results. Most of this (if not all) is done by the software that controls the equipment.

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Can there exist any element with an 8th principal shell?
I am a high school student. While studying the atomic structure a question appeared to me. We know that an atom can have the highest limit of 7 Bohr energy shells or principal energy shells, respectively labelled K,L,M,N,O,P,Q. Can any element that contains more than 7 energy shells exist? Equivalently, does there exist any element that has an 8th energy shell (which would presumably be the R shell)?

http://en.wikipedia.org/wiki/Extended_periodic_table

I would like to add that there will likely be an answer soon as réactions for element 119, or ununennium, are currently being attempted. Although no definite success was seen, the new element will be the first member of the 8th period and will be an alkali metal. Because of this, it will probably be the new largest element as well, although it’s radioactivity will most probably not topple cesium from being the largest atomic radius.

That is certainly possible, and in fact very likely. Nuclear physics predicts an island of stable elements around an atomic number of 120 $^{[1]}$, with the two heavier of the three elements predicted to be most stable having an $\ce{8s^2}$$^{[2]}$ and an $\ce{8s^2 8p^1}$$^{[3]}$ configuration, respectively.

Then that would be another milestone in the history of chemistry. Thank you for the help.

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Confusion about the charge of electrons while orbiting through bohr's energy shells
I came across the following part in my text book. Bohr's atomic theory explained the stability of an atom by the idea of energy shells. He said that while orbiting the nucleus through a particular orbit or energy shell the energy of electrons neither increases nor decreases. The electron does not carry energies while orbiting. The shells are charged and the energy of electrons orbiting through the shell equals to the energy of the shell and as the shells are stable therefore the energy of electrons are also stable . Now, the energy increases from shell to shell like(K,L,M,N,O,P,Q)[ascending order]. So,the energy of an electron in K shell should be less than that of an electron in Q shell. But we know that the energy of an electron is constant and is equal to (4.8*10^-10) esu. Then how can this is possible that an electron with a certain energy does not carry any energy while orbiting and it gains the same energy of its orbit while orbiting?
Note that there may be some mistakes in my informations as there may be many false informations printed in my book but still I am curious about the question. Thank you!

Welcome to Chemistry SE. Your question is not very clear. Why do you think electrons don't have energy while orbiting? Also, the energy of an electron is not always constant (it depends on which shell the electron is present in) and it is certainly not $4.8\times10^{-10}esu$. esu is a unit of electric charge so what you have written is the charge on an electron (which is constant - $1.6\times10^{-19}C$ in SI units).

Actually I am having confusion.The informations may have some faults. The principle shells such as (K,L,M,N) are called energy shells because they carry energies. If electrons carry certain amount of energy then while orbiting they would loose energy and tend to the nucleus and have a collision with it(according to the general electrodynamics). But the atom is stable. Why? I read my book but it says all the things that I mentioned in the question. How does the electron make the atom stable while orbiting through energized shells (principle shells)and having energies of its own(1.6*10^-19 C)?

The main confusion I can see in your question is charge vs. energy. Energy is the capacity to do work. Energy comes in many forms. Kinetic energy is the energy of motion. Potential energy is the energy of the system due to its arrangement. Electrons in Bohr shells have energies dependent on the mass of the electron, which is constant, the charge of the electron (constant), the distance from the nucleus (fixed at certain allowable values), and some fundamental constants. Thus, an electron in a shell has unchanging energy. The only way for the electron to change its energy is to move to a different shell. $\Delta E=0$ for an electron in a Bohr shell; $E\neq0$.
An electron in a Bohr shell certainly has kinetic energy because it has orbital angular motion. An electron in a Bohr shell certainly has potential energy because it is being held apart from the oppositely charged nucleus. The total energy of the electron in a Bohr shell ($E_n$) is a function of the principal quantum number, $n$. $Z$ is the number of protons (1 for hydrogen, the only atom for which the Bohr model works) and $R_E$ is the Rydberg energy (related to the Rydberg constant by $R_H=R_E/hc$).
$$E_n=-\frac{Z^2 R_E}{n^2}$$
The Rydberg energy is the smash up of some other more fundamental constants, like the mass of an electron ($m_e = 9.109\times10^{-34} \text{ kg}$), the fundamental unit of charge ($e=1.602\times 10^{-19} \ \text{C}$), Coulomb's constant ($k_e = 8.987 \times 10^9 \ \ce{N m^2 C^{-2}}$ also $k_e=(4\pi\varepsilon_0)^{-1}$, where $\varepsilon_0=8.854\times 10^{-12} \ \text{F/m}$ is the permittivity of a vacuum), and the Bohr radius ($a_0 = 5.29\times 10^{-11} \ \text{m}$). The Bohr radius is itself equal to a mashup of other constants  $a_0=\hbar/(m_e c \alpha)$, where $c=2.998\times10^8 \ \text{m/s}$ is the speed of light in a vacuum, $\hbar=1.054\times10^{-34} \ \text{J s}$ is the reduced Planck constant, and $\alpha=7.297\times^{-3}$ is the fine-structure constant. Eventually, $R_E$ simplifies down to a relativistic energy:
$$R_E = \frac{k_e e^2}{2a_o}= \frac{1}{2}m_e c^2 \alpha$$
Electric charge is something else. It is a unit of electrostatic interaction with other charged particles. The electric charge of an object influences the energy of that object in the presence of an electric field, see Coulomb's Law. The (magnitude of) charge on an electron ($e$) is part of the Rydberg energy, so it influences the total energy of the electron.
However, charge$\ \neq\ $energy.


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Bonding in a bridged metal carbonyl
Can someone please explain the bonding in a bridged metal carbonyl (like diiron nonacarbonyl)?
I cannot understand what kind of bonds (sigma/pi) exist between the bridging carbonyls and the metals.
Also why aren't the metals charged in such a case?
What is the bond order in such a case?

I don't like to reinvent wheel -- if you go to Wikipedia's  Metal carbonyl and follow just two 1st literature position, you'll find your answer concerning bridging.
For MO calculations and a general scheme applicable to systems featuring “dative” MM bonds and non-compensating, semi-bridging carbonyl ligands, see: R. D. Barr, T. B. Marder, A. G.
Orpen, I. D.Williams, J. Chem. Soc. Chem. Commun. 1984, 112–114


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Did Mendeleev predict the existence of neutrinos?
According to this Wikipedia page, Mendeleev originally believed that the inert gases belonged in Group 0 (to the left of the alkali metals). Thus, helium would be placed in the second period to the left of lithium and argon would be placed in the third period to the left of sodium. Apparently this made sense to him at the time, since atomic theory was in its infancy (and Mendeleev didn't like it) and quantum mechanics was only a twinkle in the eyes of Planck, Bohr, and others.
Based on this arrangement, there is an open spot above helium and to the left of hydrogen for an ultralight inert gas he called "proto-helium". We now know that hydrogen must be the lightest known element, meaning Mendeleev was wrong about the placement of the inert gases. However, host of the other holes in Mendeleev's table were filled with real elements having properties nearly identical to what Mendeleev predicted.
Mendeleev predicted that proto-helium would have a miniscule mass (ca. $ 5.3\times10^{-11}$ amu) and high (near relativistic) velocity. Mendeleev believed that this near massless particle would be able to permeate all matter, rarely interacting chemically. He thought that they might be responsible for radioactive decay.
A neutrino is massless (or nearly so), and travels at the speed of light (if massless) or just under (if it has mass). Otherwise, it ignores the electromagnetic force, and so has no chemical interaction. Neutrinos are often produced during radioactive decay.
The relevant references point back to Mendeleev's original writings (in Russian), which I cannot access, so I cannot verify if he made a prediction about "proto-helium". Did he predict this particle? Has anyone previously made the connection between Mendeleev's proto-helium and neutrinos?

The free neutron is the element with atomic number 0. It's unstable, of course, but it is still reasonably denoted at $^1_0\mathrm{n}$. It also doesn't really engage in chemistry on account of the lack of electrons, but it captures quite handily on—for instance—Hydrogen-1 at room temperature.

Conceptually, I agree with you, but a neutron is far from massless. It's mass (1.008664 u) is almost identical to hydrogen (1.007825 u). Free neutrons are also not stable. They undergo beta decay with a half-life of 881.5 seconds.

While it's an interesting thought, I don't think it's very useful to suggest Mendeleev predicted neutrinos. It's far more sensible for it to be a curious coincidence. His suggestions of ghost-like properties similar to neutrinos (very hard to detect, tiny mass, little interaction with matter, etc) are more easily explained as an attempt to justify (properly) why it was so hard to detect: if it had a much larger mass, someone should've weighed it by then; if it interacted more, someone should've managed to capture it physically or chemically, and so on. It's not uncommon for theoretical predictions to temporarily "take refuge" just beyond the limit of observability (for example, proton decay), with varying degrees of justification. In this case, Mendeleev's prediction was wrong in its original terms, and happened to almost match something else entirely because neutrinos really are hard to detect! Also, the connection with radioactivity could very likely have been made merely due to good timing with its discovery. Since no one had a good explanation, why not forward your own!
And as a sidenote, neutrinos definitely have mass, otherwise they could not suffer the observed flavour oscillation.

The presumption that flavor oscillation implies mass is  not obviously true (although likely true in the context of the standard model).

@Lighthart how do you argue that? I haven't heard of any theoretical framework, especially not one neutrino physicists take seriously, that produces flavor oscillation from massless neutrinos.

I would love to see any direct physical evidence that refutes my claim.  Otherwise, 'not one neutrino physicists take seriously' is just politics, not science.

@Lighthart Massless particles have velocity $c$ in every frame of reference and experience no proper time in their travels. Oscillations are the result of a time evolution operator. It follows that oscillating particles are non-luminal which implies mass. QED. Only special relativity and QM required.

And the experimental evidence?

Neutrinos are chargeless, which is a concept that does not fit nicely in Mendeleev's periodic table; it seems improbable he would predict this particle.  The closest bet would be positronium.

Positronium could not be Mendeleev's "proto-helium", since it would definitely interact chemically. It does meet the apparent mass requirement.

Its not so obvious it would interact.  It has a full 1s shell.

@Lighthart Positronium does not have a "full 1s shell", as far as I know. Unlike the 2 electrons in the helium atom, the positron and electron do not experience Pauli repulsion because they are not identical fermions. If we are considering positronium to be atom-like (in the sense of electrons around a positive nucleus), then positronium can be considered to be like a hydrogen atom with a half-full 1s shell. The wavefunction and energy levels are similar to hydrogen but differ in scale due to the difference in mass of the positron compared to a proton.

I understand the positron and the electron are not identical fermions.  However, it is not obvious the Pauli exclusion does not apply here, that is to say is charge a criteria for applying Pauli's rule?  As I'm aware the antisymmetric property which 'naturally  arises' [per all the QM books that don't really elaborate] does not use a charge component in the formulation.

Good point. Though it is my understanding that the Pauli principle does not apply for electron-positron interactions, I don't know why! Atkins' Molecular Quantum Mechanics suggests that the need for antisymmetry with respect to the interchange of identical fermions can be rationalized using relativistic arguments, so maybe a relativistic formulation of quantum mechanics is required to explain this?

Well I think so...  You also might be right about the Pauli principle being necessary here... It just not so obvious to me.  I don't exactly have a bottle of positronium to experiment with ;)... which means my speculation epsitomologically violates the firest law of quantum mechanics (integral of wavefunction times its conjugate normalized is non-zero)

Note that positronium exists in both singlet and triplet states. That is to say that the positron and the electron are decidedly not identical fermions, which is no surprise because the have different intrinsic quantum numbers. The deciding factor is the symmetry of the combined wavefunction under particle exchange (it must be anti-symmetric).

It is antisymmetric with respect to charge.  Can you offer more about this?


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In aminobenzoic acid, do we have three different pKa's?
In para-aminobenzoic acid, how can I know whether the amino group loses a proton or not when $\mathrm{pH}$ exceeds $8.5$? I have only found two values for $\mathrm{p}K_\mathrm{a}$ but I never found info about the pKa (in this case it would be the third $\mathrm{p}K_\mathrm{a}$ at $\mathrm{pH}>8.5$) of the amino group.
In any case I guess at basic $\mathrm{pH}$, negative charge over $\ce{NH2}$ would be stabilized by aromatic ring and $\ce{COO-}$.

The two acidity constants in 4-aminobenzoic acid are due to the loss of $\ce{H^+}$ from the protonated form of 4-aminobenzoic acid (the 4-carboxyphenylammonium cation), transforming it into 4-aminobenzoic acid, and then a further loss of $\ce{H^+}$ turning it into the 4-aminobenzoate anion.

The first two ionizations have a $\mathrm{p}K_\mathrm{a}$ of around 5, while the next ionization quite likely has a $\mathrm{p}K_\mathrm{a}$ well below ~25 (compare with other substituted anilines), so you won't find any appreciable amount of deprotonated nitrogen atoms in an aqueous solution. Groups with N−H bonds are far less acidic than their O−H counterparts due to the lower electronegativity of the nitrogen atom, which is less capable of stabilizing the extra negative charge.

After a year and a half, I've realized there's a small but important mistake in my answer. The third ionization should have a $pKa$ above 25, not below it. Larger values of $pKa$ indicate a lower tendency for deprotonation.

The two marcoscopic pKa for 4-aminobenzoic acid are 2.4 and 4.9.  When the microscopic pKas are close to each other, it is not a particular group that deprotonates. The neutral form of 4-aminobenzoic acid will included the zwitterion and non-zwitterion forms.

@DavePhD Once again you are quite correct. I shall leave a recent related answer of yours here for others to see.

The third $\mathrm{p}K_\mathrm{a}$ would refer to the deprotonation of the amino group of the $\ce{NH2-Ph-COO-}$ (singly deprotonated) species.
This would be very difficult to deprotonate since the molecule is already negatively charged.  Deprotonation would make a doubly charged species which is normally unfavorable.
Contrary to your assertion, the $\ce{COO-}$ provides no stabilization for the negative charge on amine, and is likely to modestly inhibit delocalization into the benzene ring as well.
Very strong bases $(\ce{BuLi}$ or similar) might be able to deprotonate this compound.

The carboxylate would still stabilize a negative charge on the nitrogen relative to, for instance, a hydrogen or methyl group. Otherwise, carboxylic acid dianion chemistry wouldn't work. (Three years later, I know, but if I found it while searching other may, too.)


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When is it okay to use $Q=mc\Delta T$, is this equation only for calorimetry questions?
When is it okay to use $Q=mc\Delta T$, is the equation only for calorimetry questions? When a question talks about heat how do I know if i'm measuring $\Delta H$ or Q?

The equation $Q = mc \Delta T$ is appropriate for calorimetry at a constant volume. $Q$ here refers to the net heat transfer of the isolated system. Change in enthalpy, $\Delta H$, on the other hand, is really the total change in energy of the system, that is:
$$\Delta H = \Delta U + P\Delta V + V\Delta P$$
Where $P$ is pressure, $V$ is volume, and $U$ is the internal energy of the system, which is the sum of the system's net heat transfer ($Q$) and the work done on the system ($W$): $\Delta U = Q + W$. If we substitute the definition of $\Delta U$ into our equation for enthalpy, we get:
$$\Delta H = Q + W + P\Delta V + V\Delta P$$
Because any mechanical pressure-volume work performed on a closed system is equal to $-P\Delta V$, the equation further reduces to:
$$\Delta H = Q + V\Delta P$$
From this it becomes apparent that $Q$ and $\Delta H$ are only equal if pressure is constant (i.e., $\Delta P = 0$). The upshot of all this is that you should not assume $\Delta H$ and $Q$ are equal unless you're explicitly instructed to assume constant volume and pressure. When a problem refers to heat, it's almost certainly referring to heat transfer, $Q$, and not enthalpy change ($\Delta H$).

There needs to be a $V \Delta P$ term somewhere, otherwise, why is constant pressure important? Work is ($w=-P\Delta V$), so your second equation simplifies to $\Delta H = q$ even if the volume changes. At constant volume $q_v=\Delta U$ ($w=0$) and you don't have to invoke enthalpy at all.

You can use $q=mc\Delta T$ for any process in which heat is transferred, not just calorimetry.
Generally, when a question talks about heat transfer, that means $q$. If the question meant enthalpy change, the question would have used the word "enthalpy".
Under constant pressure conditions, the enthalpy change is equal to the heat transfer. Enthalpy equals internal energy plus $PV$. Thus, change in enthalpy is:
$$H=U+PV$$
$$\Delta H = \Delta U + P\Delta V + V\Delta P$$
Since the change in internal energy is equal to the sum of heat transfer and work (and work is $w=-P\Delta V$):
$$\Delta H = q -P\Delta V + P\Delta V + V\Delta P = q+ V\Delta P$$
At constant pressure, $q=q_p$ and $\Delta P = 0$, so
$$\Delta H = q_p = mc_p \Delta T$$
Constant pressure is common in the lab when the reaction vessel is open to the air. At constant pressure, if you are measuring $q$, you are also measuring $\Delta H$.


text:
Glassware Test Instruments CAD blocks library
I am designing a fairly simple distillation for dichloromethane.
Being an engineer I feel more comfortable working with precise dimensions of both the separate glass vessels (condensers, evaporators, flask, etc) and of the complete apparatus.
I have been looking to find a CAD library that has various glassware blocks.
This way I can visually assemble my set and be able to show a drawing to my supervisor.

What are you trying to separate dichloromethane from? At what scale? If the distillation is simple, then there is no need for you to reinvent the wheel and design your own system. Depending on the scale and the impurities, there is likely an extant system (or several) that you could use.

I am dissolving <1g of tars in about 400ml of DCM.
Where might I find examples of such systems?


text:
Chemically removing rust without leaving any unwanted residues
I have this iron pan that got rusty from not being properly dried. Scrubbing it I was able to get rid of most of the rust, but there's still some I just can't remove. I thought I could chemically remove it, but I myself am a bit rusty on the inorganic chemistry subject, so I couldn't think of any way of doing it without leaving any unwanted residue on the pan.
The rust is the brownish-red type, which according to Wikipedia consists of $\ce {Fe2O3·nH2O}$ and $\ce{FeO(OH)·Fe(OH)3}$. Most rust-removal "recipes" I found involve all sorts of acids, but I can't tell what the reaction(s) would be with $\ce {Fe2O3·nH2O}$ so I can't tell if there would be any residues left behind that I could not remove. Anyone with any experience on the matter?

I contacted the manufacturer and the pan is hand-made, just Iron and natural coating (fat).

After you remove the rust, dry out the pan on a hot burner on the stove after washing. This will prevent future rust. You have to keep a close eye on it so you can remove it from heat as soon as it is dry.

I would emphasize that for an iron pan, you need to re-coat "season" before drying to prevent rust. Otherwise moisture from the air will eventually start to rust the pan again.

The common DIY rust remover is cola of some sort -- most colas contain a significant quantity of phosphoric acid, which reacts with the iron oxide to form iron (III) phosphate which you can then scrub off. Any remaining iron phosphate should come off with time and/or more water, and is harmless in domestic quantities.
(Wikipedia:Phosphoric acid - Rust Removal)

...and a third option would be citric acid (citrus juices).

Oxalic acid or citric acids works just fine and are not too toxic (I tested oxalic acid myself, it works great! much better, then most mineral acids) . It is preferable to not use strong mineral acids as they will eat some iron too (in fact, they have problems dissolving iron (III) oxide in acceptable time), and oxalic and citric acids are not strong, but gives complexes with iron ions and dissolves rust just fine. Acetic acid cannot give complexes, so I wouldn't bet on it.

If you're worried about residues, just use a common kitchen item that is acidic. Vinegar is just acetic acid. That should do the trick.


text:
Removing wine stains from a decanter
I have a decanter which I use fairly often to pour out wines. I typically rinse it with hot water after the last wine has come out, but I cannot prevent it from forming 'residual' wine stains over time.
The last time I wanted to clean it, I looked on the internet for advice on cleaning. There are tons of it (Stained wine decanter), but I noticed that there are quite a lot of things around that don't work (or don't work well) and I figured that the question is basically plain chemistry.
Therefore, my question is, can somebody explain how I can best clean the wine stains from my decanter? Please, also explain the chemistry behind the method you propose so I know why it should work.

It's a bit of a blunt instrument, but I'd tend to just go straight for hydrogen peroxide if it's available where you are. It breaks up all sorts of organic dyes via oxidation, and it slowly decomposes by itself, especially in sunlight, so you can just rinse (thoroughly) and leave it to dry in the sun.
(NB: I have never owned a wine decanter, and take no responsibility if your decanter dissolves, explodes, or spontaneously gains the ability to sing operatically.)

I can see whether I can get some at the local pharmacy. I think a dilute version of $\ce{H2O2}$ is used to desinfect infections of the mouth. Do you think a dilute version of $\ce{H2O2}$ will do the trick as well?

@michielm: Dilute solutions in water are the typical way to get it, and should work perfectly well. You might want something slightly stronger than the 3% solutions used as mouthwash but that might be okay if it's the easiest to get hold of -- 6 to 15% solutions seem fairly typical for bleaching and cleaning. Use gloves and wash your hands immediately after use, and if you use a strong solution and it makes white marks on your skin, don't worry, it's only the very top layer and it'll rub off after an hour or so. (I speak from experience with 30% on that.)

3 % peroxide for a mouth wash is much too much! Some drops of that 3 % solution into a glass of water is the recipe! BTW, any household bleach would remove those stains. No need to buy the peroxide, if you happen to have some eau de Labarraque or eau de Javelle at home. :=)


text:
Why is there a charge on COOH⁻¹?
The dissociation of formic acid ($\ce{HCOOH}$) is:
$$
\ce{HCOOH -> H+ + COOH-}
$$
Why is there a charge on $\ce{COOH^{-1}}$
 though?  Is it achieved through:
$$1\times(\ce{C^{4+}}) + 2\times(\ce{O^{2-}}) + 1\times(\ce{H-})?$$
Does carbon exist as a 4+ ion?
Also, as a somewhat unrelated question, can an acid donate multiple $\ce{H+}$ ions (or can a base receive multiple $\ce{H+}$ ions)?
Thanks in advance!

Yes, dibasic acids exist, for example $\ce{H2SO4}$, and similarly we have diacidic bases like $\ce{Ca(OH)2}$

Not all bonds are ionic bonds. While saying that "Carbon is 4+ valency and is a $\ce{C^{4+}}$ ion" works for carbides like $\ce{Be2C}$, it will not work for a covalent compound like $\ce{HCOOH}$.
Instead of ionic bonds, here we have covalent bonds. Whereas in an ionic bond one atom donates an electron to the other, in a covalent bond, two atoms each contribute some electrons and share them.
For example, with $\ce{O2}$, each atom shares two electrons with the other. Neither "give up" any electrons, they are just shared and contribute to the octets of both atoms:

The formate ($\ce{HCOO-}$) ion is like this:
The central carbon shares one electron with a hydrogen, and two electrons with an oxygen. It also shares one electron with a negatively charged oxygen. This completes the octets (or doublets in the case of hydrogen) of all four atoms.

With electrons drawn on:

Note that while the charges are not of the $1\times(\ce{C^{4+}}) + 2\times(\ce{O^{2-}}) + 1\times(\ce{H-})$ form, the oxidation numbers are.

You have convoluted a simple chemical into a ponderous problem. The formula as written [H-CO-OH] implies that is there is a H attached to carbon and a H attached to oxygen. The conjugate base as written [COOH-] implies that the H attached to carbon was removed without its electron leaving a negative charge on the COOH that resides on the carbon. Whether this was intentional to confuse or simply an inanity is hard to tell, but it does make it hard to assign oxidation numbers, but I can try:  O is -2, the first O is double bonded to the carbon sharing 4 electrons so its ox no. is -4. The second O is single bonded and shares 2 electrons so is -2; total negative is -6 from the O's.  We supposedly know that the total charge on the ion is -1 so must come up with 5 +charges; so C+4 and H+1 fill the bill.  [Another way to look at a carbonyl, C=O, is to hydrate it making C[OH]2.] Alternatively
let the oxygens be -2, then the remaining H is +1 and the Carbon is +2. This is in line with the hydrated structure H-C[OH]3: 3 O's = -6, 4 H's = +4, C = +2. This is in agreement with a +2 oxidation number for the carbon in carbon monoxide, CO, which happens to be the anhydride of formic acid. My advice is to not get involved with oxidation numbers with organic compounds.
Now some chemistry! formic acid HCO2H is a weak acid with a proton attached to carbon and a proton attached to one of the oxygens. The hydrogen attached to oxygen is MUCH more acidic and the reaction is H-COO-H +:B = H-CO-O- + H:B+. Now you see where the electrons are [if you were interested, at least you will be less enticed to make H  -1 when attached to O.]

why the downvote?


text:
Why do we need to incubate this system for 18 hours at 60 degrees Celsius?
I am conducting a bio-chemistry related experiment and I have been unable to understand a step which is commonly performed. The procedure that is confusing me is outlined in the supporting information, beginning on page S2.
My aim in this step is to apply a PEG (Polyethylene glycol) silane layer.
After immersing ITO (Indium Tin Oxide) slides in a PEG-concentration, the slides are incubated for 18 hours at a temperature of 60 degrees Celsius.
Can you tell me why the incubation is performed? And why is it for 18 hours at 60 degrees Celsius?

Could you provide a link to the original procedure you are trying to repeat? There is probably some information at that source that can help us help you.

Thanks a lot for your response! It gives me some insight already :). This is the link to the original procedure: pubs.acs.org/doi/abs/10.1021/nl2045414 (see supporting info document).

Until you post the original source procedure, my answer is going to be generic.

When I am repeating an experiment I found in the literature, why do I need to do [insert step that seems unnecessary, difficult, or just plain long]?

The short answer is, because it works.
You need to incubate the slides at 60 oC for 18 hours because that's what the author of the paper found to be necessary to guarantee that the slides were properly PEGylated. It's possible that the author of your procedure is also using a procedure that he or she found in the literature, and the the rationale for the specific incubation procedure will require some digging.
The longer answer is, the reaction probably needs some incubation to run to completion. The reaction between the PEG and the ITO surface is not instantaneous. If you want the coating to be good, you need to give it time to cure, and mild heating can accelerate the process. While 18 hours seems like a long time, it would likely take interminably longer at room temperature. Could you heat at a higher temperature for less time? Probably, but I would wait until you are certain you can make the existing procedure work before messing with it. That way you have some successfully coated slides to compare to.
The 18 hour period has one of two origins. The second is more likely.

The developer of the procedure maximized the efficiency of the process: He/she carefully experimented with the conditions until he/she discovered the exact time/temperature combination that maximized the efficiency of the reaction - getting the best coating in the shortest time.
The developer of the procedure maximized the convenience of the process: He/she determined 1) that the coated slide needed to cure for several hours (3 or 4) at 60 oC,  2) that higher temperatures were detrimental to the quality and integrity of the coating, and 3) prolonged heating at 60 oC is not detrimental. In this case, the researcher puts the slides into the oven in the afternoon to cure overnight and picks up the cured slides the next morning. 4 PM on Monday afternoon to 10 AM on Tuesday morning is 18 hours. As a synthetic chemist, I often wrote down '18 hours', when I really meant 'overnight'. The reaction may have been done at 6 hours or 8 hours, but 18 hours is more convenient. I set up the procedure one afternoon a few hours before I left for the day so that I could make sure it was stable. I worked it up the next morning.

Thanks a lot for your response! It gives me some insight already :). This is the link to the original procedure: http://pubs.acs.org/doi/abs/10.1021/nl2045414  (see supporting info document).


text:
Books for a very advanced 3rd grader (elementary school)
My daughter is a 3rd grader, but she reads at 8th grade level, and she is 2 grade levels ahead in math.
She reads everything she gets her hands on.
She took some chemistry activities, and she loved them. ("Mad Science", "Ooo, goo and stinky too",...)
She knows that matter is made of atoms, and that atoms are made of protons and electrons.
I'd like to find some chemistry books that would be appropriate for her. A good book would be one that starts from beginning, help her connect together the stuff she already knows, and then take her further.
My reason is that I don't want school to be too easy for her. I want to nurture her curiosity, love of challenge, and motivation to study.
On the other hand, I don't want to push her even further ahead in the subjects that she has to take in emementary or middle school anyway -- because that would make our dealing with the school system even more complicated... So, I think Chemistry and Physics are good subjects.

While recommendation questions are in general off topic network wide, this seems well-defined enough for me to allow it. If anyone disagrees, feel free to voice your opinion :)

Not long enough to be an answer, but since she reads at 8th grade level and is advanced in math, why not look for a middle school level physical science textbook? Check to see which book your district uses and then check it out from a library.

One of the best is The Golden Book of Chemistry Experiments.  By today's standards, this book is considered "dangerous", but keep in mind that it was written for teenagers in the 1960s.  In any case, parental supervision should always be required in chemistry, and if you carefully research the experiments before doing them you should be fine.  It takes you all the way through technique, building apparatus, and safety.  The only problem is obtaining chemicals: the sources cited therein are hopelessly outdated (i.e. your will NOT find most of this stuff in drugstores anymore), and you will need to use educational supply companies to obtain small quantities of chemicals.  Have fun!

Thanks for the recommendation. But books with experiments are actually easy to find. I want a book that covers the theory.

It does cover some theory. Very simple, but some theory. For example: Faraday's candle (nice example of scientific method), measurement, industrial and home processes, brief history, periodic table, mixtures, solutions, acids/bases, element chapters are nicely contextualized, basic equation balancing, and even some very basic organic chemistry. I think this is an awfully good introduction to the whole topic. In contrast, Thompson's excellent Illustrated Guide to Home Chemistry Experiments, while more up-to-date, is mainly about experiments.

@Marjeta Experiment is frequently the best way of explaining the theory.


text:
Why do sodium halides react so differently with sulfuric acid?
Why do sodium halides react so differently with sulfuric acid?
\begin{align}
  \ce{NaF + H2SO4 &-> NaHSO4 + HF}
    \tag{1a}\label{NaF}\\
  \ce{NaCl + H2SO4 &-> NaHSO4 + HCl}
    \tag{1b}\label{NaCl}\\
  \ce{2 NaBr + H2SO4 + 2H+ &-> Br2 + SO2 + 2H2O + 2Na+}
    \tag2\label{NaBr}\\
  \ce{8 NaI + H2SO4 + 8H+ &-> 4 I2 + H2S + 4 H2O + 8Na+}
    \tag3\label{NaI}
\end{align}
Conventional explanation: $\ce{NaI}$ is a strong enough reducing agent to reduce the sulfur, and $\ce{NaBr}$ is a little less strong, so sulfur is not reduced as much. Equivalently, $\ce{H2SO4}$ is not as strong an oxidising agent to reduce $\ce{NaF}$ and $\ce{NaCl}$.
My ensuing question: Why is reaction \eqref{NaI} more favourable than \eqref{NaBr} and $(\ref{NaF},\ref{NaCl})$ for sodium iodide, so that it reacts with sulfuric acid like it does. What factors are involved: what is more stable, for instance, (or there is lower Gibbs free energy or something else) about reducing sulfur as much as possible (as $\ce{NaI}$ 'chooses' \eqref{NaI} over $(\ref{NaF},\ref{NaCl},\ref{NaBr})$?
Ditto for sodium bromide. I understand why \eqref{NaI} is impossible, but why is \eqref{NaBr} chosen over $(\ref{NaF},\ref{NaCl})$?

A factor you didn't touch upon which may have some weight in determining the reactions is the acidities of the hydrogen halides formed by reactions of the first type (which is fundamentally a coarse qualitative analysis of the free energy of reaction). In aqueous solutions, the order of acidity is $\ce{HF} \ll \ce{HCl} < \ce{HBr} < \ce{HI}$. While concentrated sulfuric acid is obviously not the same as water, perhaps it is fair to say the relative acidities of the hydrogen halides are approximately the same, given they are both polar protic solvents with very high dielectric constants.
As you go down the family, the acid formed in reactions of the first type become stronger, and there is less thermodynamic drive to follow such a reaction pathway. $\ce{HI}$ is somewhat close in acidity to $\ce{H2SO4}$ since neither conjugate base is very coordinating. Meanwhile, as you say, the reducing power of the halides increases, and since sulfur is in a high oxidation number in $\ce{H2SO4}$, redox reaction pathways become more attractive. A detailed analysis of free energies of reactions should agree with the change in the mechanism since it is unlikely these reactions are kinetically controlled.
Edit: Another neat consequence of the acidity strength argument is that it explains why the reactions stop at the bisulfate anion instead of going all the way to the halide sulfates; $\ce{HSO4-}$ is a much, much weaker acid than $\ce{H2SO4}$, so there is no free energy to lose out of consuming it to produce a much stronger acid (in the case of $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ at least). $\ce{HF}$ is similar in acidity to $\ce{HSO4-}$ (in water), so it's not as readily explainable in such a qualitative analysis.

'The acid formed in reactions of the first type become stronger, and there is less thermodynamic drive to follow such a reaction pathway'.
Sorry, but I don't quite understand why stronger acids have lower entropy (that's what I assume makes it more thermodynamically favourable).

Free energy is what thermodynamically determines the paths of reaction, and it is a combination of enthalpy changes and entropy changes. Stronger acids have higher free energy content mostly due to enthalpy. You can think of strong acids as entities with a very weak tendency to hold onto $H^{+}$ ions, either because the $H^{+}$ is poorly bound or because the acid would be much stabler without it. Therefore, it becomes strongly exothermic for the $H^{+}$ to migrate to another substance that can better accept it, creating a weaker acid. The inverse reaction is highly endothermic, and disfavoured

"acidity" is something which happens in aqueous solutions only! Think of NaCl + SiO2 makes Na2SiO3 + HCl if temperature is sufficient. This kind of reaction is about volatility, not "acid strength"!

Not at all. The concepts of acidity and basicity can be extended to any ionizable liquid. As long as a solvent molecule can ionize into a "solvonium" cation and a "solvate" anion, one can define acidity, even if it seems strange at first. Check this wiki article. Your example doesn't fit an acidity analysis because neither $NaCl$ nor $SiO_2$ are ionizable compounds ($NaCl$ is dissociable, not the same thing! And $SiO_2$ is an extended solid. I'm actually not sure what its liquid molecular structure is like)


text:
How many of different types of chemical species are in a certain volume of aqueous solution of an acid
The acid in question has $\mathrm{p}K_\mathrm{a}$ about 1.29. Density of 98% w/w solution is $d=\pu{1.323 g/cm^3}$. I would like to find the number of undissociated acid molecules ($\ce{HX}$), water molecules ($\ce{H2O}$), dissociated acid molecules ($\ce{X-}$), and hydronium ions ($\ce{H3O+}$).
The molar mass of the acid ($\ce{HX}$) is $\pu{125.1 g/mol}$. Thus 98% w/w solution corresponds to $\pu{10.57 M (mol/L)}$. If I'm not missing sth, the solution contains 1.5 mole of water and 10.57 mole of acid. What I'm not sure is how these numbers convert into the number of ions/molecules which are, let's say in a volume of $1~\pu{\mu L}$ (one microliter).
Though my attempt would be:

Firstly I have to determine $\mathrm{pH}$ of the 98% (w/w) solution in question; I did some rough calculations and got $\mathrm{pH}$ = 0.509

Then, I need to establish the ratio $\dfrac{[\ce{X-}]}{[\ce{HX}]}$; this can be done using Henderson-Hasselbach equation; I got 0.166

Thus, for let's say 1000 $\ce{HX}$ molecules and 172 $\ce{H2O}$ molecules (this stems from 10.57:1.5 molar ratio), I have at equilibrium: 834 $\ce{HX}$, 166 $\ce{X-}$, 166 $\ce{H3O+}$ and 6 $\ce{H2O}$ molecules.

The solution, seems to be correct. 834 $\ce{HX}$, 166 $\ce{X-}$, 166 $\ce{H3O+}$ and 6 $\ce{H2O}$ molecules should have a total mass of $2.129\times10^{-19} \pu{g}$. Since the density is known, such a group of molecules should occupy a volume of $\pu{160.9 nm^3}$ which is perfectly reasonable.

The following is an approximate calculation, since among other things it makes the approximation that the activities of each substance are equal to their molarities, which is not true in such concentrated solutions. Also I take some shortcuts to make the calculation quicker.
Let's look at a volume of $1\ L$. The mass would be $1323\ g$, 98% of which ($1296.54\ g$) is due to the acid, and 2% ($26.46\ g$) due to water. This initially translates into $10.36\ mol$ of acid, and $1.47\ mol$ of water. However, some of the water and acid are consumed to create hydronium and conjugate base.
To calculate the dissociation of the acid, you may want to apply $K_{a}=\frac{[H_{3}O^{+}] [A^{-}]}{[HA]}$ (the Henderson-Hasselbach equation is derived from this by taking the log on both sides, so they're equivalent), but this is actually a very bad approximation, since you need to take into account the concentration of water in non-dilute solutions. We'll do it anyway and I worked it out to be $[H_{3}O^{+}]=[A^-]=0.704\ mol.L^{-1}$. Since the volume we're working with is $1\ L$, that means $n_{H_{3}O^{+}} = n_{A^-} = 0.704\ mol$. Subtracting those amounts from the initial amounts of water and acid, we reach $$ n_{H_{2}O}=0.77\ mol$$ $$ n_{HA}= 9.66\ mol$$ $$n_{H_{3}O^{+}}=n_{A^-}=0.704\ mol$$
To find the actual number of each entity, just multiply by the Avogadro number $6.022\times 10^{23}$ and you have the amount of each in a litre of solution.

Thanks for your prompt answer. It's embarrassing to note that I made several mistakes in my calculations (concentration conversion and solution of the quadratic eqn with $K_a$). I'm aware that I should use activities and not concentrations ($c_i$) in the eqn for $K_a$ but I was simply unable to find proper activation coeffs ($f_i$) for alkyl phosphates. For such condensed solutions, Pitzer eqn for $f_i$ should be used. But its solution is not trivial so I'm wandering to what extent, the values of $n_{\ce{H2O}}$, $n_{\ce{A-}}$, etc. could change.

I thought this question was asked as lower level chemistry homework, however, it seems you're more advanced than that. In that case, perhaps you should investigate how to solve the problem while taking into account the concentration of water. I've never actually encountered such a problem myself, so I'm not exactly sure how to go about it. The problem is in dilute solutions, $[H_2O]$ is implicitly approximated to $\ce 55.55\ mol.L$ in all equilibria. As you can see, this is very far from true. You may have to use $K'{a}=\frac{[H{3}O^{+}] [A^{-}]}{[H_2O][HA]}$ and the expression for $K_w$.

While searching for a correct answer to my own question I performed a simple simulation using PHREEQC. It is critical to realize that the calculation procedure outlined by Nicolau in his answer to my question works properly only for dilute solutions. The highly concentrated solutions have high ionic strength ($IS$) which makes Debye–Hückel theory inadequate. Thus the activity coefficients need to be find in a different way.
It is claimed that Pitzer equations can handle solutions with $IS > 2$. This morning I got accustomed with basic functionality of PHREEQC program (it has amazing possibilities, though it seems to have extremely steep learning curve..). At the moment it seems that it is not possible to perform calculations for alkyl phosphates, which I had been originally interested in, so I decided to determine population of chemical moieties for phosphoric acid. In the calculations described below, I basically start with 14.616 mole of $\ce{H3PO4}$ dissolved in $1 kg$ of water. Then I trigger evaporation to finally get 85% wt solution.
The input file for PHREEQC is as follows (these are my first calculations in this program so I'd appreciate if someone more experienced could comment on the correctness of the input):
SOLUTION 1
  pH 2.0 charge
  pe 0
  P 14616  # 85% wt solution has 14.62 moles of H3PO4
  water 1.0 # start with 1 kg of water
REACTION 1
  H2O -1.0
  41.520 moles in 10 steps # end with 252g of water
PITZER
  -macinnes   true
  -use_etheta true
  -redox      true
DEBUG
  -iterations 200
END

The results include calculated $pH$ ($0.197$), ionic strength ($1.363$), and molality ($58$). The really interesting, however, are final activities calculated for such a condensed solution of $\ce{H3PO4}$:

$\ce{H3O+}$: $0.635$
  $\ce{H2O}$:  $0.346$
  $\ce{H3PO4}$: $56.64$
  $\ce{H2PO4-}$: $0.635$

Using concentrations instead of activities in solving the equation for acid dissociation constant as shown by Nicolau, lead to $\ce{H3O+}$ of $0.324$ and $\ce{H3PO4}$ of $14.756$.

+1 for the effort. Unfortunately I can't exactly repeat your calculations and compare results. The more you play around with the program, the more likely you'll reach the right values. Also, perhaps it is worth searching for the theory behind acid/base dissociation in concentrated solutions, which I think is not necessarily equivalent to the theory of high ionic strength solutions. Honestly, I'm afraid there is not much I can do to help you.


text:
How is hydration free energy of a solute measured experimentally?
I am reading a paper (Shirts, M. R.; and Pitera, J. W. and Swope, W. C. and Pande, V. S. J. Chem. Phys. 2003, 119, 5740-5761).  Near the beginning of the paper (page 5745), the authors state:

Experimental free energies of hydration for weakly soluble solutes are
  determined from concentration measurements made on two phase systems,
  where one phase consists of a vapor with a partial pressure $P_s$ of
  some solute molecule of type $s$, and the other phase consists of an
  aqueous solution with a number density concentration for solute
  molecules of $\rho_s^{\ell}$.  When such a two phase system is at
  equilibrium with respect to transfer of molecules of type $s$ between
  the phases, the solvation free energy is given by $$\Delta G_{\text{solv}} = kT\ln(P_s / (\rho_s^{\ell} kT)).$$

What does this mean in practice?  For example, suppose that I would like to measure the hydration free energy $\Delta G_{\text{solv}}$ of a small molecule (e.g., the amino acid arginine).  That is to say, I would like to measure the solvation free energy of arginine in water.  (Since arginine has a charged side chain at neutral pH, I would expect $\Delta G_{\text{solv}}$ to be negative, because arginine is likely stabilized by water, which is polar.)
Does the expression for $\Delta G_{\text{solv}}$ above imply that I just need an aqueous solution of arginine?  Then would I need to somehow measure the partial vapor pressure $P_s$ of the arginine vapor which exists above the arginine aqueous solution?  How is such a partial vapor pressure vapor measurement done in practice?  Thanks for your time.

Your interpretation is correct. Theoretically, you will need only an aqueous solution and a fixed volume of gas, and then be able to measure equilibrium concentrations. Although, you may not observe very much vapor pressure without using elevated temperatures (as you noted, the zwitterion is very soluble).
Derivation
When a species $x$ is in equilibrium between two states, $$x_{state1} \rightleftharpoons x_{state2}$$ the equilibrium constant is given by the ratio of concentration of $x$ in each state: $$K = \frac {[x_{state2}]} {[x_{state1}]}.$$ The equilibrium constant relates to Gibbs free energy by the following equation: $$\Delta G = -RT \ln(K)=-RT\ln\left(\frac{x_{state2}}{x_{state1}}\right)=RT\ln\left(\frac{x_{state1}}{x_{state2}}\right).$$ Substituting for an aqueous/vapor phase system yields the cited equation. The literature uses Boltzmann's constant, $k$, in place of the gas constant $R$ because they are using units of molecules instead of moles. (The extra $kT$ term within the logarithm comes from the conversion of pressure to number density concentration using the ideal gas law: $\frac{n}{V}=\frac{P}{RT}$, again using molecular units instead of moles will require $k$ instead of $R$.)
Measurement
Determining the "best" method for gas phase measurement depends on what instrumentation you have available to you and how accurate you need/want to your results. Mass spectrometry would be ideal if it employs a soft ionization technique. If the molecular ion doesn't fraction, you could use the raw counts compared to an internal standard to calculate the concentration.
You don't necessarily need to measure the vapor pressure directly, either. If you have a method for measuring the aqueous concentration, you could calculate the vapor pressure based on decrease in solution concentration: $$\Delta G=RT \ln \left(\frac{[arg_{aq}]_i-[arg_{aq}]_f}{[arg_{aq}]_f}\right)=RT \ln \left(\frac{[arg_{aq}]_i}{[arg_{aq}]_f}-1\right)$$ where $[arg_{aq}]_i$ is the initial aqueous arginine concentration (total arginine) and $[arg_{aq}]_f$ is the final concentration (equilibrium amount).  HPLC (or even UV-Vis if that is all you have available) would be a good method, but you would have to either make a calibration curve or use an internal standard (your buffer may work).
Notes
If you have access to J. Phys. Chem. and/or Biochemistry, this article cites articles (ref. 29, 82) that have experimental data for free energies of hydration for amino acids (all but arginine). They may be helpful to consult for methods.


text:
What is meant by "no d-orbital"?
Sulfur and oxygen belong to the same group. Sulfur has a vacant d-orbital while oxygen has no vacant d-orbital.
What does having "no d-orbital" mean? Orbitals are just the spaces around the atom. How can you say that there is no space around the atom which makes the d-orbital?

There is a bit of a philosophical debate as to whether orbitals exist only when they're populated, or if they're always there. In both oxygen and sulfur, there is no occupied $d$ orbital in the ground state (so both $3d$ orbitals are vacant), but in sulfur the promotion energy of an electron from a $3s$ or $3p$ orbital to a $3d$ orbital is much less than the promotion energy of an electron in the $2s$ or $2p$ orbitals in oxygen to a $3d$ orbital ($2d$ orbitals don't exist, of course). Arguably, this means that sulfur can access its $3d$ orbitals under the right conditions since the promotion energy required is relatively low and could be supplied in chemically relevant situations (so the sulfur $3d$ orbital is accessible).
It is not expected that an oxygen atom could ever populate its $3d$ orbital in a stable substance. It is possible to occupy an oxygen $3d$ orbital for a short while, however, by exciting the atom with a photon of the proper frequency.

So, we say that sulphur has a vacant d-orbital because the electrons can occupy that vacancy, right?

Strictly speaking, vacancy merely implies that the orbital is empty. The 3d orbitals are vacant in both oxygen and sulphur, but they're only accessible in the latter.

Actually 3d orbital could be used, if you write down a Full-CI expansion for the ground state wavefunction, isn't it?

@user26143 I don't have a large understanding of computational chemistry, but I have read that calculations can improve if you also include orbitals that aren't formally populated. There are some subtleties which escape me, however. I know it is a point of confusion with calculations performed for "hypervalent" molecules; even though they are calculated to have negligible population of upper d orbitals, including them in the calculation provides lower energies. Hybridization fans will say it's proof of hypervalency, while computational chemists tend to say it's something else entirely.

Sorry, I mean in isolated atom....

As Nicolau hints in his answer, orbitals are not regions of space around an atomic nucleus. They are mathematical constructs formulated as wavefunctions describing the properties of electrons, including their energy, angular momentum, and probabilistic distribution in space. The pictures of "orbitals" that you are used to seeing are the probability density functions (the absolute value of the wavefunction squared).
Ultimately, the only way we can know what a probability density function looks like or what the energy of orbital would be is to pretend there is an electron in the orbital and then do the math. So, orbitals without electrons in them might as well not exist. As Nicolau and ManishEarth point out for oxygen, the orbitals in the 3d subshell are basically inaccessible to the electrons in oxygen. Thus those orbitals do not exist.

Oxygen has a d-orbital as well. Just that there are $3s$ and $3p$ orbitals in between, so extra electrons will get filled there first. The $3d$ orbital has too high energy for it to be of any use.
Also, hybridization almost always (the exception is with coordination compounds) happens for orbitals with the same principal quantum number (shell number). So the $3s$ and $3p$ orbitals are not of any use for accomodating extra electrons/bonds.

Orbitals are uncorrelated spatial quantum states of electrons. This representation makes it intelectually more confortable and allows one to approximate the Schrödinger equation. Nevertheless, for polyelectronic atoms one have to correlate the electrons in order to get to the right energy. On way to do so is to construct a wave function that implies all the orbitals : the one that are occupied (in the fondamental state) and the virtual ones (unoccupied in the fondamental state). For O and S, 3d orbitals are virtual orbitals. By implying those states in the wave function we are able to correlate the electrons and get the right energy for the hole system. Quantum mechanics says that the electron is in all the states at the sametime but with a different probability and the probability to find an electron in 3d orbital for O is much less than for an S electron.

Sulfur and oxygen belong to the same group. Sulfur has a vacant d-orbital while oxygen has no vacant d-orbital.

The most weakly bound electrons in oxygen are in the 2p orbital using a one-electron approximation. There is no 2d orbital, so the next highest orbital is the 3s orbital. You could have an excited state where the 3s orbital gets populated. However, there are no compounds of oxygen where electrons are in the 3s orbital or "make use" of the 3s orbital (in the sense of hybridization or linear combination of atomic orbitals).
The most weakly bound electrons in sulfur are in the 3p orbital using a one-electron approximation. There is a 3d orbital with higher energy, and you could have an excited state where that gets populated. "Making use" of the 3d orbitals has been invoked in compounds of sulfur (like $\ce{SF6}$) but does not really happen either, see e.g. Hypervalency of elements.

What does having "no d-orbital" mean? Orbitals are just the spaces around the atom. How can you say that there is no space around the atom which makes the d-orbital?

Orbitals are solutions (eigenfunctions) of the Schrödinger equation. There is no solution that corresponds to 2d (that would be n=2 and l=2), but there is one that corresponds to 3d (that is n=3 and l=2)
The principle quantum number n (e.g. 2 for the 2p orbital or 3 for the 3p orbital) gives you the total number of the spherical and planar and conical nodes plus one. The d orbitals have two planar or two conical nodes, so the principle quantum number has to be 3 at least.


text:
Diamagnetism in sulfur
Sulfur is a diamagnetic element, meaning that it has no unpaired electrons. But its electronic configuration for valence shell is just like oxygen because they are in the same group i.e ns2, np4.
My question is that why is sulfur diamagnetic? If there are 4 electrons in p-orbital of its valence shell then shouldn't it have 2 unpaired electrons?

You seem to be confusing the elements with the pure simple substances of the elements. In atomic terms, both the oxygen atom and the sulfur atom are paramagnetic, since they contain two unpaired electrons in the ground state. However, the substances oxygen ($\ce{O2}$) and sulfur (generally intended as $\ce{S8}$) are not the same as the atoms.
In the simple picture of Lewis structures, both substances have all atoms forming two bonds, completing the octet rule and therefore without unpaired electrons. This would imply that both are diamagnetic. However, simple valence bond theory is inadequate to explain the bonding in oxygen. Simple molecular orbital theory suggests that there are in fact two unpaired electrons in the $\ce{O2}$ molecule. This gives rise to paramagnetism in the oxygen molecule, also.

So, oxygen is paramagnetic weather it is in atomic state or molecular state. But in Sulphur is paramagnetic only in atomic state. Right?

Correct. One must be take care to be clear when referencing the element itself or one of its allotropes.


text:
Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?
Both $\ce{SF6}$ and $\ce{SH6}$ and $\ce{SF4}$ and $\ce{SH4}$ have the same central atom and the same hybridization, but my teacher specifically mentioned that $\ce{SH6}$ and $\ce{SH4}$ don't exist. I've looked everywhere but I can't figure out why? I'd appreciate some insight into the problem.

related: http://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit

TL;DR Fluorine is electronegative and can support the extra negative charge that is dispersed on the six X atoms in $\ce{SX6}$, whereas hydrogen cannot.

First, let's debunk a commonly taught myth, which is that the bonding in $\ce{SF6}$ involves promotion of electrons to the 3d orbitals with a resulting $\mathrm{sp^3d^2}$ hybridisation. This is not true. Here's a recent and arguably more understandable reference: J. Chem. Educ. 2020, 97 (10), 3638–3646 which explains this. Quoting:

The natural ionicity, $i_\ce{SF}$, of each $\ce{S-F}$ bond [in $\ce{SF6}$] is 0.86, indicating a rather ionic σ bond. Each fluorine has an average charge of $−0.45$, resulting in a sulfur center of charge $+2.69$. [...] In summary, the electronic structure of this system is best described as a sulfur center with a charge somewhere between $2+$ and $3+$; the corresponding negative charge is distributed among the equivalent fluorine atoms. Shown in Figure 12 is the orbital occupation of the sulfur center, $\ce{3s^1 3p^{2.1} 3d^{0.19} 5p^{0.03} 4f^{0.01}}$. The minimal occupation of d-type orbitals eliminates the possibility of $\mathrm{sp^3d^2}$ hybridization.

If not via d-orbital bonding, how does one then describe the structure of $\ce{SF6}$? I'll present an LCAO-MO answer. Here's a "simple" MO diagram (I won't go through the details of how to construct it). It's actually fairly similar to that of an octahedral transition metal complex, except that here the 3s and 3p orbitals on sulfur are below the 3d orbitals.

Just for the sake of counting electrons, I treated the compound as being "fully ionic", i.e. $\ce{S^6+} + 6\ce{F-}$. So sulfur started off with 0 valence electrons, and each fluorine started off with 2 electrons in its σ orbitals. I've also neglected the π contribution to bonding, so the fluorine lone pairs don't appear in the diagram.
You'll see that, for a total of six $\ce{S-F}$ bonds, we only have four pairs of electrons in bonding MOs. The other two pairs of electrons reside in the $\mathrm{e_g}$ MOs, which are nonbonding and localised on fluorine. If we want to assign a formal charge to sulfur based on this diagram, it would be +2, because there are only actually four bonds. We could perhaps use Lewis diagrams to represent it this way:

The "hypervalent" resonance form contributes rather little and does not rely on invoking d-orbital participation; see Martin's comment on my answer below for greater detail about the resonance contributions. I am guessing that its existence can be mostly attributed to negative hyperconjugation, although I'm not 100% sure on this. The trans and cis resonance forms are not equal, so their contribution is not the same, but the contribution from each individual trans resonance form has to be the same by symmetry. Overall, the six fluorines in $\ce{SF6}$ have to be equivalent by the octahedral symmetry of the molecule. You could run a $\ce{^19F}$ NMR of the compound and it should only give you one peak.
(An alternative way of looking at it is that two of the $\ce{S-F}$ bonds are "true" 2c2e bonds, and that the other four $\ce{S-F}$ "bonds" are in fact just a couple of 3c4e bonds, but I won't go into that. For more information on multi-centre bonds, this article is a nice introduction: J. Chem. Educ. 1998, 75, 910; see also refs. 12 and 13 in that article.)
Right from the outset, we can see why $\ce{SH6}$ is not favoured as much. If we use the same framework to describe the bonding in $\ce{SH6}$, then those "correct" resonance forms that we drew would involve $\ce{H-}$. I'll leave it to the reader to figure out whether $\ce{F-}$ or $\ce{H-}$ is more stable.
Alternatively, if you want to stick to the MO description, the idea is that in $\ce{SH6}$, the relatively high energy of H1s compared to F2p will lead to the nonbonding $\mathrm{e_g}$ orbitals being relatively higher in energy. All things being equal, it's less favourable for a higher-energy orbital to be occupied, and $\ce{SH6}$ would therefore be very prone to losing these electrons, i.e. being oxidised.
In fact, if we do remove those four electrons from the $\mathrm{e_g}$ orbitals, then it's possible that these six-coordinate hydrides could form. But obviously we might not want to have a $\ce{SH6^4+}$ molecule on the loose. It'll probably lose all of its protons in a hurry to get back to being $\ce{H2S + 4H+}$. Is there anything better? Well, there's the species $\ce{CH6^2+}$, which is methane protonated twice. It's valence isoelectronic with $\ce{SH6^4+}$, and if you want to read about it, here's an article: J. Am. Chem. Soc. 1983, 105, 5258. While it's hardly the most stable molecule on the planet, it's certainly more plausible than $\ce{SH6}$.
Now, just to come back to where we started from: d-orbital participation. Yes, there is an $\mathrm{e_g}$ set of d orbitals that can overlap with the apparently "nonbonding" $\mathrm{e_g}$ linear combination of F2p orbitals, thereby stabilising it. It is true that some degree of this does happen. The issue is how much. Considering the fairly large energy gap between the $\mathrm{e_g}$ orbitals, this interaction is bound to be fairly small, and is nowhere near enough to justify a $\mathrm{sp^3d^2}$ description of it; Martin's comments contain more details.

Exactly how much the hypervalent structure contributes is up to the method used. On BP86+D3/def2-SVP it is about 5%, on MP2/def2-SVP it is about 8%, and on  HF/def2-SVP it is about 16% (not very surprisingly). There is probably also a basis set dependency. The bonds are 22% S (37%s, 60%p, 3%d) and 78% F (19%s, 81%p, >0.1%d) on MP2. So you remain correct with stating "Not very much at all [...]" contribution from the d-orbitals. (An answer well written btw.)

@Martin-マーチン  Interesting calculations!  I don't know how easy or difficult it would be, but it would be interesting to see what $^1J(\ce{S,F})$ coupling constants the various calculations predict; particularly how the computed coupling constants approach the experimental value as a function of hypercoordinated structure "content".

@orthocresol Have you any knowledge of a resource where (empty) d-orbital energies are approximated for main-group elements? For instance, NIST has this database which includes information on occupied orbitals. Just as a comparison of how large the energy difference is.

From the abstract of your 1st link, re d orbitals: "The sulfur d orbitals are important because they allow strong back transfer from the negatively charged fluorine ligands to the strongly positively charged (+2.9e) sulfur, in turn allowing significant contraction of the S-F bonds and greatly increased molecular stability."
So while the d orbitals may not directly participate in bonding, they do participate otherwise.

@MichaelK, that is true: I will reword the initial assertion to be slightly more careful.

@orthocresol Btw, it's worth noting (at least in this here comment) that the same back transfer stabilization obviously can't occur with positive H atoms.

@MichaelK It would be negative H atoms in SH6, but the point is kind of moot. Actually, I must admit, that this may not have been the best choice of reference anyway. After all, it was 1986, and the calculations were performed using Hartree-Fock theory which is pretty outdated by 2021 standards.

The comparison of $\ce{SH6^{4+}}$ with $\ce{CH6^{2+}}$ is not completely apt. Though they are formally valence isoelectronic, the carbon species is not octahedtal; it's four-coordinate with two of the ligands being side-on $\ce{H2}$. See this answer.

The reason that they do not exist (or at least are not the most stable form) is because the decomposition reaction is exothermic.
\begin{aligned}\ce{
(1) && SF6 &-> SF4 + F2\\
(2) && SH6 &-> SH2 + 2 H2
}\end{aligned}
Reaction $(2)$ is much more exothermic than $(1)$, and it can be argued in two ways: either there is something very bad about the reactants ($\ce{SX6}$) or something very good about the products.
If you look at the bond strength of $\ce{F-F}$ vs. the bond strengths of $\ce{H-H}$ (well-known quantities) you will see that $\ce{H2}$ has a much stronger bond than $\ce{F-F}$.
There is every reason that $\ce{SH6}$ should exist if you are looking at how many "slots" sulfur has.  If it has six slots for fluorine, then surely it could accommodate six hydrogens.
I don't really want to get into it, but I tend to think that "hypervalency" is borderline quackery, preoccupation with non-explanation.  Either the atoms want to be there because their interaction is favourable electrostatically, or they would rather be in some other configuration.

I completely agree with you about hypervalency being near-quackery. All we have are approximate theories about what goes on in these molecules. I doubt anyone has an actual physical idea what's happeing in there.

@Bolt64 Since I wrote my answer I have read up on QTAIM and related theory and I now indeed believe that hypervalency is more a phenomenological model than anything else.

BTW, what do you mean by hypervalency being borderline quackery? I don't really understand what you mean there?

This is the accurate answer.  The accepted one is not even trying to provide an answer to the question and would probably be wrong if it were.

I do very much agree with your point about hypervalency, I also strongly disagree with some of your other statements. A reaction that is exothermic does not mean that the reactant is unstable. Even if that same reaction would be exergonic usually an activation barrier has to be overcome, and that would make the molecule meta-stable and still existing.

That SH6 is unstable is obvious (writing a formula for that however implies that it can constitude a bound state, which you cannot know), sulfur having six "slots" is however even much more a non-explanation as hypervalency. You are reasoning by analogy, ignoring that hydrogen and fluorine have very different possibilities to form bonds.

Incompatibility of $d$-orbitals of S with $s$-orbital of H.

This is caused by the molecule $\ce{SF6}$ being hypervalent, which means that the main element (in this case sulfur) has more then 8 valence electrons.
The reason why this can happen is extremely complex and, to be honest, I am not even sure whether it is a fully solved issue. I do know that the effect is related to the electronegativity of the ligands, which is very high for $\ce{O}$, $\ce{F}$ and $\ce{Cl}$ atoms and somewhat lower for $\ce{H}$-atoms. This might explain why it doesn't happen for hydrogen, but this is just speculation.
An incredibly extensive explanation of hypervalency (and related phenomena ) can be found in this post

Hypervalency isn't an answer.  The comparison here is H vs. F not S vs. O so hypervalency has nothing to do with it.  Ideas related to hypervalency may or may not be relevant.

@Jeff - where do you read that I compare O to S? I compare highly electronegative ligands (O, F, Cl) to H which is less electronegative.

Anyway, since I wrote the answer I have come to believe that hypervalency is just a phenomenological model that can explain some of the symptoms' of breaking of the octet rule' but doesn't explain the reasons at all

This answer is very vague and there are a lot better ones given already. If you yourself came to the conclusion, that there are better theories, you should expand this post. Some people just look for the accepted answer and then leave, they might not read the better ones.

Sulphur does not have more than eight valence electrons; all sulphur compounds abide by the octet rule.

As michielm said, it is because of electronegativity. The bond $\ce{S-F}$ is strongly polarized toward the fluorine (~more electrons are near fluorine), while the $\ce{S-H}$ bond is polarized toward the sulfur.
In $\ce{SH6}$ molecule there would be very high electron density around sulfur. This would increase both the electrostatic repulsion of electron and also the kinetic energy connected with occupation of higher atomic orbitals (this is hand-waving argument).
More rigorously it can be explained by a linear combination of atomic orbitals (LCAO), Hartree-Fock method (HF) or density functional theory (DFT). The increased electrostatic repulsion corresponds to Hartree term in HF or DFT. But even more important is the kinetic energy due to Pauli exclusion principle and requirement of orthogonalization of orbitals of each electron (resp. spin pairs). The orthogonalization requires a certain number of nodes in the wavefunction. Each such node increases the orbital energy (similar to states of particle in the box).
From the LCAO point of view this can be viewed as the contribution of higher atomic orbitals (d-orbitals, for example) to the bonding molecular orbitals.
In the case of $\ce{SF6}$ the contribution of sulfur atomic orbitals to the bonding states is lower (because most of the electrons are localized on fluorines). Because of that the energy of bonding molecular orbitals is not increased much by either Coulomb repulsion or by kinetic energy of higher atomic orbitals of sulfur.
A very rough argument can be given also using Thomas-Fermi model where the kinetic energy of electrons is proportional to $\rho^{5/3}$. However, Thomas-Fermi model is not appropriate for molecules.

Hypervalency is a bit of a touchy subject since there are strong opposing proponents for both the hybridization camp and the 3-center-4-electron bond camp.
While not directly related to your question's compounds, I shall point you to a fun book by Errol Lewars on the computational analysis of uncommon molecules: Modeling Marvels. While some of it isn't displayed on Google Books, a good amount of the relevant chapter is there. Indeed it even mentions the theoretical higher hydrides of the second-period elements.

http://link.springer.com/book/10.1007/978-1-4020-6973-4 is the link to the publisher. I just read some of it and figured it would be worth a try. The two theories you presented are not mutually exclusive. Both deal with linear combinations of atomic orbitals.

$\ce{F}$ is much more electronegative compared to $\ce{H}$. Thus it causes a contraction in the $d$ orbital of $\ce{S}$. The $d$ orbital of $\ce{S}$ becomes much more energetically stabilised. It can then participate in bonding with the filled $p$ orbital of $\ce{F}$. For $\ce{H}$ this is not possible. The cost of hybridisation is much higher in the latter case. Thus $\ce{SH4}$ does not exist but $\ce{SF4}$ does.

Hybridisation is a mere mathematical concept, hence it will not change any energy levels at all.

There is no (or negligible) d-orbital contribution to the bonding in SF6.

As sulfur belongs to group 16 in periodic table, its electronic configuration is ns²np⁴, it can show +2,+4 , +6 and -2 oxidation state. SF6 exists but SH6 doesn't because fluorine is the most electronegative element in the periodic table, its size is extremely small so it has greater polarising power, but if we see in SH6 then we get to know that the electronegativity of sulfur is much more than hydrogen, hydrogen doesn't have sufficient nuclear charge (effective nuclear charge) and its electronegativity is also not much, so SH6 cannot be formed.

Notice also how the Group 1 metals form peroxides much like hydrogen.
Why does potassium form peroxides but sodium does not?
Alkalai Metals bond with Sulfur/Chalcogens in a 2:1 ratio and Hydrogen is behaving like an Alkalai Metal.
Lithium Sulflide = Li2S
Sodium Sulfide = Na2S
Potassium Sulfide = K2S
Rubidium Sulfide = Ru2S
Cesium Sulfide = Ce2S
Francium Sulfide = Fr2S
.
.
.
Hydrogen Oxide = H2O
Hydrogen Sulfide = H2S
Hydrogen Selenide = H2Se
Hydrogen Telluride = H2Te
If Hydrogen were like a Halogen, it would form H4S, and H6S. However Hydrogen doesn't obey the octet rule like halogens as it only needs to get rid of one electron, and yet it shares 2 electrons in a bond with Sulfur. That is exactly what the alkali metals do: they need to get rid of one electron and they share 2 with sulfur.

Na2S adopts the antifluorite structure, which means that the Na+ centers occupy sites of the fluoride in the CaF2 framework, and the larger S2− occupy the sites for Ca2+. -https://en.wikipedia.org/wiki/Sodium_sulfide

Thus an atoms lacking or exceeding by 2 tends to form 2:1 ratios with atoms lacking or exceeding by 1.
Oxygen is really the problematic chalcogen: we are used to the Halogens behaving like hydrogen: See Dioxygen Diflouride and Hydrogen Peroxide.  And then Hydrogen goes out and acts like an alkali metal.


text:
Which d-orbital configurations form an octahedral complex?
With reference to crystal field splitting in octahedral complexes, my textbook shows the following table of distribution of $d$-orbital electrons in octahedral complexes, based on the energy difference of $t_{2g}$ and $e_g$ orbitals:
Table
Do ALL the $d$ orbital configurations (i.e. pertaining to all examples given) form octahedral complexes? I doubt it because not all of them have two vacant $d$ orbitals, especially in the case of weak field ligands where Hund's rule is never violated.

A metal ion doesn't actually need vacant d orbitals to form a coordination complex according to Crystal field theory. All the ions mentioned in the table form complexes with water (which is a weak ligand) - see this page.
Crystal field theory assumes that all bonding between ligand and metal ion in a complex is electrostatic (ionic). That means there is no exchange of electrons between the metal ion and ligand. Instead, the negatively charged (or polar in case of neutral ligands like $NH_3$) ligand is attracted to the positively charged metal ion by purely electrostatic forces. This is why vacant d orbitals are not required for bonding.

Do they all form relatively stable octahedral complexes, though? For most of the intermediate $d^n$ configurations, octahedral compounds are abundant, but what about the extremes?

So,if I want to check the structure of a coordination compound,both the configuration of electrons in $ t_2g $ and $ e_g $ and the field strength of the ligand are required isn't it?Else is all about ionic bonds,no orbital overlap and hybridisation