What is the alternative to chromyl chloride test?

Why chromyl chloride test is used as confirmatory test for $\ce{Cl-}$:

According to this Wikipedia page:

  The chromyl chloride test entails heating a sample suspected of
  containing chloride with potassium dichromate and concentrated
  sulfuric acid. If chloride is present, chromyl chloride is formed and
  red fumes of $\ce{CrO2Cl2}$ are evident. If there is no chloride
  present, no red fumes are produced. No analogous compounds are formed
  with fluorides, bromides, iodides and cyanides, so this test is
  therefore specific for chlorides. [1]

Most chemistry lab manuals mention that only red vapour is not the confirmation for presence of chloride ion. Rather, if the red vapour dissolves in a solution of $\ce{NaOH}$ and turns the solution yellow (due to $\ce{Na2CrO4}$), and that solution gives yellow precipitate with lead acetate solution, then only, presence of chloride is confirmed.

However, this paper mentions that chromyl fluoride is produced on heating calcium fluoride with chromate and sulfuric acid. The compound chromyl fluoride is a red-vapour at normal temperatures, and is also produced by reaction of dichromates and $\ce{HF}$.[2].

Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis[3] mentions that for chromyl chloride test:

  Fluorides give rise to volatile chromyl fluoride, $\ce{CrO2F2}$, which
  is decomposed by water, and hence should be absent or removed.

These suggest that there is a possibility of production of chromyl fluoride by reaction of potassium dichromate, concentrated sulphuric acid and a fluoride sample. As chromyl fluoride is red vapour, this would result in a false positive. Moreover, chromyl fluoride is also hydrolysed in water, [4]  $$\ce{CrO2F2 + H2O->CrO3 + 2HF}$$ and thus it should also turn $\ce{NaOH}$ solution yellow, and produce yellow precipitate of $\ce{PbCrO4}$ with lead acetate. So, there is no way they can be distinguished without performing some other test for chloride or fluoride.

It is not possible to know whether chromyl fluoride is actually produced in large quantities like chromyl chloride without performing the actual experiment. But, I think there is a chance of error with fluoride sample.

Also, bromide salts produce red vapours of bromine when heated with $\ce{K2Cr2O7}$ and conc. $\ce{H2SO4}$(this is from my own experience). I don't know why the Wikipedia page specifically mentions that no red fumes are produced in absence of chlorine, and no analogous compound is formed with fluoride. It appears to me that the information is questionable.

  Alternative to chromyl chloride test:

A feasible (and in my opinion, easier) alternative is this procedure—

Prepare the soda extract or sodium carbonate extract of the salt in question. This would produce $\ce{NaCl}$, even from covalent chlorides. For insoluble chlorides, it has to be done in melt.
Make a water solution of the extract, and filter it.
Add $\ce{AgNO3}$ soln. to this solution. A white precipitate of $\ce{AgCl}$ would form.
To confirm that it is indeed $\ce{AgCl}$ and not $\ce{AgBr}$ or $\ce{AgI}$, add sodium arsenite($\ce{Na3AsO3}$) solution, and shake. Yellow precipitate of $\ce{Ag3AsO3}$ is formed[3]. This does not occur in case of $\ce{AgBr}$ or $\ce{AgI}$. [I am not sure about $\ce{AgCN}$ though.]

Another alternative is—

Add $\ce{AgNO3}$ solution to the sodium carbonate extract.
$\ce{AgCl}$ is white and soluble in $\ce{NH4OH}$. $\ce{AgBr}$ is pale yellow and partially soluble in $\ce{NH4OH}$. $\ce{AgI}$ is yellow and insoluble in $\ce{NH4OH}$. To improve colour detection, use a grooved tile made from black granite, and perform spot test[5].
For detecting cyanide, add iron(II) sulphate to the soda extract and then add a mineral acid. If a precipitate of prussian blue does not appear, then there is no cyanide.

References:

[1] https://en.wikipedia.org/wiki/Chromyl_chloride

[2] Pure Chromyl Fluoride. Alfred Engelbrecht, Aristid V. Grosse. J. Am. Chem. Soc., 1952, 74 (21), pp 5262–5264 |http://pubs.acs.org/doi/abs/10.1021/ja01141a007?journalCode=jacsat

[3] Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis, 5th edition, Longman Group Limited, 1979.

[4] https://en.wikipedia.org/wiki/Chromyl_fluoride

[5] Greener Qualitative Analysis for Anions in a Mixture.
Indu Tucker Sidhwani, Sushmita Chowdhury, Veena Tucker, Shweta Bansal, Elipsha Sharma and Parul Johar. DU Journal of Undergraduate Research and Innovation. Volume 2, Issue 2 pp 70-79, 2016| http://journals.du.ac.in/ugresearch/DU%20e-Journal%20-%20UG%20Research_files/DUJ4%20Vol%202%20Issue1/J%201.8.pdf


Create Sodium Hydroxide from Sodium Silicate

The reactions you proposed are pure fantasy. They will never happen. Na2SiO3 does not react with either H2O2 or Hydrogen gas or CH2 (which does not exist, by the way).

The only possibility would be to have silica precipitate by bubbling CO2 in the solution. This will produce a precipitate of Si(OH)4. But this precipitate is so gelatinous that it cannot be easily filtrated. If you are able to filtrate (under pressure), the remaining clear solution contains Na2CO3.
$$\ce{Na2SiO3 +CO2 + 2 H2O  ->  Na2CO3 + Si(OH)4}$$

which could be transformed into NaOH by adding calcium hydroxide (or chalk) Ca(OH)2. This addition creates a new precipitate, CaCO3, which can be filtrated, and the residual solution contains NaOH.
$$\ce{Na2CO3 + Ca(OH)2 -> CaCO3 + 2NaOH}$$


Can very dilute acids exposed to air become more concentrated as water evaporates?

Sulfuric acid (and other acids) can be concentrated through evaporation of water.  It would be a good idea to wash the surface down with dilute, basic solution.  That said, if I'm reading your post correctly, your droplets are already 0.1M.  That is still a strong acid and care should be taken in its clean up.  I was once working with a 0.01M solution of sulfuric acid and when cleaning the flask a few drops of the dilute solution splattered onto my foot, I didn't even notice it at the time.  An hour or so later my foot started to burn.  When I looked, my sock was gone in the affected area.  If you're dealing with anything more than drop or two use a spill kit and let someone know what you are doing.


Can a spectrometer be used to figure out spice ratios in a sauce?

Unfortunately, this will not even qualitatively work because taste which is a sense of perception and UV-Vis are barely related.. The USB spectrometer would be great for color matching, but you are assuming the taste and aroma comes from UV-Vis absorbing compounds only. This may not be the case. Let us assume, the taste comes from UV-Vis absorbing compounds, as you know from Beer's law, an $n$-component mixture will have  absorbance sum from $n$-components. If some of these molecules are in trace quantities, they might have a little effect on the overall spectrum but they might have a huge influence on the taste and aroma of your sauce. Food chemists prefer more of gas chromatography and mass spectrometry & infrared spectrometer which may cost more than 100K.


How is the Rate law of a reaction measured?

Preamble
For a reaction with no intermediate steps, you could derive the rate law from the reaction equation. For a generic equation
$$\ce{$a$A + $b$B -> C}$$
the rate is given by
$$r = k\cdot c^{a}(\ce{A})\cdot c^{b}(\ce{B}).$$
Rigorously, this is only possible for elementary reactions. A prominent example for a strictly unimolecular reaction is radioactive decay. Also quite commonly known is the photolytic or thermal cleavage of hydrogen or bromine:
\begin{align}
\ce{H2 &->[\Delta T] 2H. }\\
\ce{Br2 &->[h\nu] 2Br. }
\end{align}
An example for a bimolecular reaction would be
$$\ce{2NO2 -> NO3 + NO}$$
Reaction mechanism of the combustion of hydrogen bromide
This seems to be a reaction with a fairly complicated mechanism, involving multiple species, most of them radical. I would assume, that it is somewhat similar to the combustion of hydrogen, that I have described here.
A quick search on some forums will turn out a reaction mechanism, which I think is wrong. But I will show it here anyways, since it gives you a first clue, about how complicated this reaction might be. (Note: These are supposed to be gas phase reactions.)
\begin{align}\ce{
HBr + O2 &-> HOOBr\\
HOOBr + HBr &-> 2 HOBr\\
HOBr + HBr &-> H2O + Br\\
}\end{align}
The following mechanism is taken from Lilian G. S. Shum and Sidney W. Benson Int. J. Chem. Kin., 2004, 15 (4), 341-380. I think this is a much more suitable and credible approach to the problem. (Note: Most of the reactions are in equilibrium.)
\begin{align}\ce{
HBr + O2 &~<=> HO2 + Br\\
HO2 + HBr &~<=> H2O2 + Br\\
H2O2 + 2HBr &~->T[wall] 2H2O + Br2\\
Br + Br + M &~<=> Br2 + M\\\hline
4HBr + O2 &~<=> 2H2O + 2Br2\\
}\end{align}
Unfortunately I was unable to determine what $\ce{M}$ refers to in this case. It might be any inert gas used as a solvent in the reaction or the wall.
Rate law of the combustion of hydrogen bromide
As a consequence of the mechanism you cannot find the rate law from theory, but only from experiments. It is important to realize, that most of the reactions are in equilibrium and the whole system is dependent on the concentrations of each of the involved species. Therefore the kinetics of this reaction may be expressed as a complicated system of coupled differential equations.
However, Shum and Benson conclude, that the first step will be the rate-determining and the rate expression is given through

$$\frac{\mathrm{d}[\ce{H2O}]}{\mathrm{d}t} = 2k_g[\ce{HBr}][\ce{O2}].$$

Therefore it can formally be regarded as a bimolecular reaction.


NMR: How the relaxation times T1 and T2 depend on the correlation time / amount of molecular tumbling.

Relaxation is often very counterintuitive and isn't a particularly easy subject to understand, so it's not always useful to argue based on what seems to be logical. I'm not sure where you're studying this from, but it seems to be slightly lacking in the detail needed to properly understand what's going on. Here is a brief overview of what's going on, but I strongly recommend consulting a textbook which will explain it in more detail. Chapter 9 of James Keeler's Understanding NMR Spectroscopy (2nd ed.) is a very good reference. You can find a pdf of the corresponding chapter in the previous edition here.

The link between rate of tumbling (correlation time, $\tau_c$) and the "amount of fluctuations at the right frequency" is given by the spectral density

$$j(\omega) = \frac{2\tau_c}{1 + \omega^2\tau_c^2}$$

This function tells you how much the random fields are fluctuating at each frequency $\omega$. For relaxation, the important frequency to look at is the Larmor frequency $\omega_0$, i.e. the central quantity is

$$j(\omega_0) = \frac{2\tau_c}{1 + \omega_0^2\tau_c^2}$$

By differentiating this with respect to $\tau_c$ you can find that this is at a maximum when $\tau_c = 1/\omega_0$. The rate of spin-lattice (longitudinal) relaxation is proportional to $j(\omega_0)$, so it is fastest when $\tau_c = 1/\omega_0$. That means that $T_1$ will be smallest when $\tau_c = 1/\omega_0$. When $\tau_c$ decreases below this, or increases above this, spin-lattice relaxation will be slowed down and $T_1$ will increase.

On the other hand, spin-spin (transverse) relaxation depends on the sum $j(\omega_0) + j(0)$. $j(0)$ is simply equal to $2\tau_c$, and you can show (again with calculus) as $\tau_c$ increases, the total sum $j(\omega_0) + j(0)$ increases continually. So $T_2$ decreases when $\tau_c$ increases.


If CO2 is nonpolar how come much more dissolves in water than O2?

Taken from my answer to your original question

There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.
$$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$
This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.


Why are group 1 elements called alkali metals and group 2 elements are called alkaline earth metals?

The description of the reasoning of the naming is … let’s call it simplified.

First, concerning your first point. Alkali metals are not bases by themselves, they are metals. Metals typically cannot be classified in the original Arrhenius or extended Brønsted-Lowry acid/base classification. The cannot dissociate to liberate either $\ce{H+}$ or $\ce{OH-}$ because they only consist of metal atoms. And they cannot act as a donor or an acceptor for protons — one because of the lack of protons and two because of the lack of favourable lone pairs.

It is only when we oxidise these metals in some way or another that they form a species that can interact with water in the acid-base formalities. One possibility for oxidation is simple burning, i.e. reacting with atmospheric oxygen. Unfortunately, only lithium is nice enough to ‘do what you might expect’, hence why I’m using that as an example in the following equations:

$$\begin{align}\ce{4Li + O2 &-> 2 Li2O}\tag{1}\\[0.4em]
\ce{Li2O + H2O &-> 2 LiOH}\tag{2}\end{align}$$

As we can see, the oxide we generated, lithium oxide, can react with water to form a hydroxide which dissociates into $\ce{OH-}$ ions — an Arrhenius base. We can also say that the oxide anion is capable of accepting protons donated by water — a Brønsted-Lowry base. Therefore, lithium oxide (and also lithium hydroxide) are the bases while lithium metal is not.

The second question is easily answered. Alkali is a noun and alkaline is an adjective. Thus, alkali metals is a compound word formed by two nouns while alkaline earth metals is also a compound word formed by two nouns where the first noun is further specified by an adjective.

The third question is difficult to answer. There is not really an easy or understandable explanation; it all boils down to churning numbers and finding out that the dissolution is favourable, i.e. the corresponding equation represents a decrease in Gibbs free energy: $\Delta G < 0$.

We can break dissolution of solids into two general processes:

breakdown of the ionic lattice, i.e. liberation of individual ions from a solid

$$\ce{NaOH(s) -> Na+(s) + OH- (s)}\tag{3}$$
dissolution of the individual ions

$$\begin{align}\ce{Na+(s) &->[H2O] Na+(aq)}\\[0.3em]
\ce{OH-(s) &->[H2O] OH-(aq)}\end{align}\tag{4}$$

The first of these steps requires energy because you are breaking bonds (this is always unfavourable). It is typically represented by $\Delta_\text{latt} H^0$ — the standard lattice enthalpy.

The second process is exothermic because the individual ions are now again somewhat balanced by the polar water molecules. It is typically represented by $\Delta_\text{solv} H^0$.

Both processes are favourable from an entropic point of view.

Whether the entire reaction $(5)$ happens, depends on the difference between lattice and solvation enthalpies. If the lattice enthalpy is too high to be offset by solvation enthalpies and the entropy effects, an ionic compound will not dissolve. For alkali metal oxides and alkaline earth metal oxides, the lattice enthalpy is typically low and thus they dissolve.

$$\ce{NaOH (s) -> Na+ (aq) + OH- (aq)}\tag{$3+4=5$}$$

The actual reason for their names is in another castle beyond the scope of this answer. At first glance, I would forward you to the corresponding Wikipedia articles.


Photochemistry and steady state approximation

The processes are absorb light into A, and then A* has only two pathways, i.e. fluoresces and reacts with B to form B*.

The B* has only two pathways, it  fluoresces and converts to C. All B* molecules must therefore follow one of these paths, so the sum of yields of B* (paths fluorescence + forming C) must be the same as the yield to form B* in the first place, thus you don't need to do a calculation.

In general if you have to find a yield of any process x this is, $\displaystyle \phi_x = \frac{\text{rate of process x}}{\text{rate of absorption}} \equiv \frac{\text{rate const x} }{\text{sum of all other rate constants from x}}$.

In your example $\displaystyle \phi_{ET}= \frac{k_{ET}[B]}{k_f^A+k_{ET}[B]}$ where superscript $A$ refers to fluorescence from $A^*$.

To work all this out write down $d[A^*]/dt = +k_{abs}[A]- k_f^A[A^*]-k_{ET}[A^*][B]$ and use steady state.


Can green vehicle antifreeze (Ethylene Glycol) be diluted in a glass container?

As Long as I rinse the previous contents (milk) out, can I safely use this glass container for diluting antifreeze and putting it into my vehicle?

Not sure if you are talking a bout antifreeze for the radiator system or windshield deicers, but both systems are rugged enough that the do not require very pure water to be used. A washed milk container would suffice. The antifreeze wont disrupt the bottle and the bottle wont adulter the antifreeze

  Once I'm done can this go back into the recycle bin, or will trace amounts of antifreeze spoil it for recycling?

This somewhat depends on how it will be recycled. If you are putting it in general glass recycling then the bottle will become cullet that will heated to temperatures high enough to burn off any residue and it wont matter.

If you are returning the milk bottle to the producer for reuse then they have to clean them well enough to remove milk reside which is harder to remove than either your antifreeze or the bitterants in antifreeze. Any reputable bottler would expect some amount of non-milk substances in the bottles and clean them sufficiently well to make sure every bottle is clean and safe. Though I would recommended that you do rinse it before furnishing it to them out of courtesy.


How to identify if it is a buffer solution

Well, it all boils down to the preparation of buffer solutions in the first place. One of the methods of preparation includes the mixing of a salt solution of an weak acid and a strong base with a solution of a strong acid, with the solution of the strong acid as a limiting reagent, which leads us to conclude positively for the third choice.

Alternatively, you can think which of the mixtures will resist the change in pH on mixing a strong acid or a strong base; for instance in the third choice, nitric acid being a strong acid will help to resist the change in pH when a strong base is added, and $\ce{CH3COONa}$ being a weak acid + strong base type of salt resists the change in pH that could've been caused due to the addition of a strong acid.


TL;DR: There is no inner- or outer coordination spheres in this complex. Both cobalt(II) and mercury(II) have nearly ideal tetrahedral coordination environment with $4$ $\ce{N}$ and $4$ $\ce{S}$ atoms, respectively, hence the proper name would be cobalt(II) mercury(II) tetrathiocyanate. Mercury(I) readily undergoes disproportionation, so assuming mercury(II) is the safest option.

Actually, denoting an inner coordination sphere is not correct for this compound. First crystal structure [1] has been assigned to cobalt(II) tetrathiocyanatomercurate(II) $\ce{Co[Hg(SCN)4]}$:

  It appears reasonably certain that the $\ce{S}$ atom is attached to the $\ce{Hg}$ in tetrahedral co-ordination, with $\ce{S-Hg-S}$ angles of 120° and 104°.

But two decades later the structure has been refined again, by the group including the author of the original publication. This time the geometry has been established more precisely (ICSD#36062), and it turned out that there is an infinite network of $\ce{Co^2+}$ and $\ce{Hg^2+}$ cations cross-linked via thiocyanate ligands, and the new suggested name was cobalt mercury thiocyanate $\ce{Co(SCN)4Hg}$ [2]:

  The combination of two tetrahedrally coordinated atoms, $\ce{Hg}$ and
  $\ce{Co}$, has produced a most unusual arrangement in which the $\ce{Hg}$ and $\ce{Co}$ atoms are held apart by four spirals, each containing $4$ $\ce{SCN}$ bridges, which are interlinked so that any one $\ce{SCN}$ bridge takes part
  in eight spirals.

  [...] Each such spiral is a spring holding the $\ce{Hg}$ and $\ce{Co}$ atoms apart and straining the bonds in the process. It is almost certainly this strain which flattens the tetrahedral coordination round $\ce{Hg}$ and $\ce{Co}$ in the $c$ direction. If the arrangement could be reproduced mechanically it would probably provide the ideal spring mattress!

References

Jeffery, J. W. Nature 1947, 159 (4044), 610. DOI: 10.1038/159610a0.
Jeffery, J. W.; Rose, K. M. Acta Cryst B 1968, 24 (5), 653–662. DOI: 10.1107/S0567740868002980.


I figured that I was using the wrong format, the detailed format of .SKF files is given here

The correct extracted parameterization profile is given below:


Difference between polarizability and dipole moment?

First, the quick definitions. A dipole moment ($\mu$) implies a permanent separation of charge across a distance. For example, HF has a positive charge near the hydrogen atom and a negative charge near the fluorine:

Instead, polarizability $\alpha$ indicates the degree to which electrons (and thus charge) re-arrange in an applied electric field. Indeed, when a polarizable molecule experiences any type of electrostatics, there is an induced dipole moment:

$ \mu_{induced} = \alpha \vec{E} $

Now this induced dipole moment (from the polarizability) is completely distinct from the permanent dipole moment. For example, benzene is non-polar (no net dipole moment) but highly polarizable. In the HF case, I could find some large applied field $\vec{E}$ that pushes electrons from the F to H.

Here are my mental analogies:

A dipole moment is like a hill or tilted plane - water (charge) moves "downhill" and it's always there.
Polarizability is like hitting the water with a cannonball. I can drive water from one side of a bathtub to the other!

The two properties can be related, but it's possible to find highly polarizable molecules with no dipole moment - particularly if they are symmetric. Here's a plot with ~80,000 molecules from PQR - note that all combinations occur.


I agree with Ivan Neretin in that the sodium oxidation state is equal in all compounds, and it will not act as oxidizing or reducing agent.

My thoughts:

Na$_2$O will not act as a strong oxidizing or reducing agent, as both ions are in their preferred state.

NaO$_3$, or sodium ozonide, will likely act as an oxidative agent. However, I believe it disproportionates in water: 4 O$_3$ $^-$ + 2 H$_2$O $\rightarrow$ 5 O$_2$ + 4 OH$^-$, so that's kind of a candidate: it goes from oxidation state -1/3 to 0 and -II.

NaNO also seems strange to me - if it's Na$_2$N$_2$O$_2$, I see here that it acts as a reducing agent.

NaNO$_2$ can certainly act as both reducing and oxidizing agent, as the nitrogen atom is in oxidation state +III.

In conclusion, (4) seems to be the safest bet, considering only one answer is expected.


How common are automated pipetting systems

Speed isn't everything. Reproducibility might be much more important!

Imagine that you have to transfer different volumes from five different stock solutions to a 1536 well plate and adjust the total volume by adding water or the common solvent of your choice. A programmable pipetting automat might be a blessing here.


The effect of hyperconjugation on the stability of alkenes with MO theory

Since you are familiar with how carbocations are stabilized via hyperconjugation I will keep this brief: With alkenes, it really is the same principle at work. Instead of the empty p orbital, you have to consider the empty $\pi^{*}$ orbitals of the $\ce{C=C}$ double bond. They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\sigma}$ bonds in $\pi$-bond-like manner

and so from those two fragment orbitals you will get two molecular orbitals: one from the in-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be lower in energy then the fragment orbitals, and one from the out-of-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be higher in energy then the fragment orbitals. Since in this interaction there are only two electrons involved, i.e. the ones in $\sigma (\ce{C-X})$, you get a stabilization from the interaction since those two electrons will now occupy the in-phase MO and have thus lowered their energy. The out-of-phase MO is empty and doesn't contribute to the energy of the molecule.


What is the chemical name of this derivative of aniline?

The preferred name is N-phenylmethanimine, but often it is named N-methyleneaniline (or N-methylenaniline, more correct one would be N-methylideneaniline though).

UPDATE a typo (excessive "e" in bogus "phenylmethaneimine") fixed. (thanks, Loong)


Temperature dependence of molarity and molality

Recall that molarity $c$ is

$$c=\frac{n_\text{solute}}{V_\text{solution}}$$

whereas for molality $b$ is

$$b=\frac{n_\text{solute}}{m_\text{solvent}}$$

While in the extreme case you mentioned, both quantities are not defined because all the solvent is either boiled away (or for heat-sensitive species, the solvent can even decompose under heating), molarity depends on volume, and the volume of the solution depends on temperature due to expansion/contraction.


when do we indicate the "+ or -" sign in uncertainty

You either use both "+" and "-", $\pm$, or neither in which case it is implied that you mean $\pm$.

Usually uncertainty is expressed in a way that is symmetric about the mean, such as $123\pm4$, meaning to a certain degree of confidence the true value is between 119 and 127.

However, you could write $123^{+5}_{-3}$ to indicate the best estimate of the true value is 123 and to a certain degree of confidence the true value is between 120 and 128.

For example, if you were highly certain that an object is a cube, and measure its edge as 100 $\pm 20$, then it might be better to express the volume of the cube as

$1.0^{+0.7}_{-0.5} \times 10^6$


Z-matrix of a series of 5 colinear atoms

You cannot have bond angles of 0 or 180 degrees in a Z-matrix. This is because the dihedral becomes degenerate. To solve this, you can add 'dummy atoms' which provide an auxiliary point of reference to remove the 180 degree angle. These atoms are usually denoted $\ce{X}$ or $\ce{Xx}$ in computational chemistry suites. Usually, dummy atoms are placed at 90 degrees to a pair of atoms and the third collinear atom is then defined with a bond angle of 90 degrees and a dihedral of 0 or 180 degrees with respect to the previous 3 atoms.

Example

To illustrate what I'm talking about, here's a schematic of your pair of molecules, with three dummy atoms:

This could have a Z-matrix like:

O   1
C   1    a
Xx  2  1.0  1  90
O   2    b  3  90     1  180
Xx  4  1.0  2  90     3  0
H   4    c  5  90     2  180
Xx  6  1.0  4  90     5  0
O   6    d  7  90     4  180
H   8    e  6  104.5  7  180

Notes

There are no hard and fast rules about how many you should use (obviously we could simply define subsequent dihedrals in terms of the leftmost dummy atom) however it's best to strive for locality so that small changes in some variable don't correspond to large displacements elsewhere, in which case rounding errors or large gradients can ruin your day. You could also have defined all of the dihedrals in your molecule with respect to that convenient non-collinear hydrogen, however this is the Z-matrix equivalent of spaghetti code.

You will note that I use different letters for each variable. This is because forcing the $\ce{H-O}$ and $\ce{C=O}$ bonds to be of equal length is probably an unrealistic constraint for a geometry optimisation of this system.

Finally, note that dummy atoms do not take part in the electronic structure of a molecule, so feel free to put them in anywhere and set the dummy atom 'bond' length to whatever you want. 1.0 Ångström is traditional as far as I know.

P.S.

Molden has a complete, if idiosyncratic graphical Z-matrix editor.


Determination of products in reactions involving carbocation rearrangement?

All else equal, the smaller group migrates preferientially. The energy required to more the group is related to the mass of the group:

$$W=\int_0^\vec{x}\vec{F}\cdot d\vec{x}$$
$$\frac{d\vec{x}}{dt}=\vec{v}   \ \ \ \ \ \therefore \ \ \ \ \ d\vec{x}=\vec{v}dt$$
$$W=\int_0^t\left(\vec{F}\cdot \vec{v}\right)dt
$$$$\vec{F}=m\vec{a}$$
$$W=m\int_o^t \left(\vec{a}\cdot\vec{v}\right)dt=m\int\left(\vec{v}\cdot\vec{a}dt\right)$$
$$\vec{a}=\frac{d\vec{v}}{dt}$$
$$\vec{a}dt=d\vec{v}$$
$$W=m\int_0^\vec{v}\vec{v}\cdot d\vec{v}=\frac{1}{2}m\vec{v}^2\vert_0^\vec{v}=\frac{1}{2}m\vec{v}^2$$


This is going to be a rehash of Oscar Lanzi's answer, but this point needs to be driven home, so I make no apologies.

"Special" electron configurations - fully-filled or half-filled subshells - are only a relatively minor factor in determining the stability of oxidation states. I've written about this before in a slightly different context, but IMO it would be well worth reading this, as the exact same principles operate in this case: Cr(II) and Mn(III) - their oxidizing and reducing properties?

What determines the "most stable oxidation state"? We can consider what happens when we increase the oxidation state of a metal: firstly, we need to pay ionisation energies to remove electrons from the metal. On the other hand, though, metal ions in higher oxidation states can form stronger bonds (e.g. lattice energy in ionic compounds, solvation energy in solution, or more covalent bonds in molecular compounds). The balance between these two factors therefore leads to a "best" oxidation state.

For example, sodium could hypothetically go up to Na(II) and form $\ce{NaCl2}$. Theoretically, this compound would have a significantly larger lattice energy than $\ce{NaCl}$. However, sodium's second ionisation energy is prohibitively large, which means that it stays put in the +1 oxidation state. On the other hand, magnesium's second ionisation energy is not prohibitively large, and therefore Mg preferentially forms $\ce{MgCl2}$ over $\ce{MgCl}$.

For all lanthanides, the first three ionisation energies are all fairly comparable. There is a graph here which shows the variation in the ionisation energies of the lanthanides, which I reproduce below.

(source: Inorganic Chemistry 6ed, Weller et al., p 630)

For all lanthanides, these three ionisation energies are easily compensated for by the extra lattice energy / solvation energy derived from a more highly charged ion. The fact that Eu(II) has a $\mathrm{f^7}$ configuration only serves to make its third IE marginally larger than that of Gd. This difference is sufficient to make Eu(II) an accessible oxidation state (cf. different electronic behaviour of EuO and GdO), but not sufficient to make it the most stable oxidation state.

Note that "most stable oxidation state" depends on the conditions, too, but I assume we are talking about aqueous solution.


Cooking with vinegar and lemon juice (Carpaccio)

Apparently this question has been covered on the Cooking SE site - in this post, it is reported that:

The Oxford English Dictionary defines to cook as:
To prepare or make ready (food); to make fit for eating by due application of heat, as by boiling, baking, roasting, broiling, etc.
However, from a chemical point of view, what happens when you are cooking meat is that you are using heat, amongst other things, to denature the proteins of the meat. Denaturation can also be achieved by other means, such as marinating the meat in an acid, such as citric acid (contain in lemon juice) or acetic acid (contained in vinegar).
You can also use other agents to induce denaturation, but they are not always fit for eating.
However, it is important to understand that from a purely cooking perspective the two things are not the same: "cooking" with lemon (I guess it is fine to use the verb to cook as long as you put it in quotes), gives a completely different result than cooking with heat. This is because other chemical reactions happen during application of heat and because, obviously, you do not end up with an hot product!

which indicates that the application of acidic media to meat accomplishes the act of denaturing proteins therein in a way mechanistically similar to what cooking with applied heat does.
From Kitchen Daily:

Acids like lemon juice, lime juice and vinegar break down raw meat, allowing the marinade's oil and spices to deeply penetrate and infuse the meat with flavor and moisture. They dig out little pathways in the meat and allow the marinade to seep in. The scientific term for this breakdown is "denaturation," which is the process in which protein loses its structure by application of a strong acid, base, salt, alcohol or heat. Basically, the scientific process that cooks meat with heat in the oven is the same one that cooks it with acid. The results may taste a little different -- fish cooked in the oven is flaky, whereas fish cooked by lemon juice (a.k.a. ceviche) has a more raw-tasting consistency -- but the process is the same.


How long would it take for sucrose to undergo hydrolysis in boiling water?

This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions.  In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, "Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase".
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 50% decomposed into monomers after 440 years -- at 25 °C.  That value comes from thermodynamic extrapolation of experiments done at higher temperatures.  (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after.  Figure 1 from the paper has the data.

The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^{-6}$ per second.  That means that 63% degradation would take $10^{6}$ seconds, or about 11.5 days.  Getting 99% degradation would take about 53 days.
edits
The above has been edited to fix sloppy math errors, as noted by @gremin and @John Bentin in the comments.  Thank you both.


Does the structure of water change when a drop of color is added?

No it does not change the structure of water.
Yes the H-bonds between water molecules are broken to accomodate the dye molecules. However, liquid water does not have a set shape or fix molecular arrangement. So what happens is,
The dye molecules simply diffuse out all throught and among the water molecules. The dye molecules become hydrated to some extent, that is, the charged groups and areas on the dye molecule interact with water to form H-bonds or polar bonds which allow the molecule to remain suspended among the water.

Source, General chemistry course work.


Yours is a primary alcohol and it has $\beta$ hydrogen atoms also. Thus, it will undergo $\pu E2$ mechanism. In this mechanism removal of protonated hydroxyl group and $\beta ~\pu H$ happen at once (in a single step). Therefore, there will be no scope of hydride shifting.
You may compare this with the dehydration of neo-pentyl alcohol.


The Jahn-Teller effect is not relevant here. The triplet state is observed in many molecules to be of lower energy than the singlet. The Pauli exclusion principle, makes the states symmetric or anti-symmetric to the exchanging electrons.
A triplet state has unpaired spins while in a singlet they are paired. The triplet is lower in energy than the corresponding singlet. One reason for this is that the electrons try to ‘avoid’ one another in the triplet; (see Salem 'Electrons in Chemical Reactions', publ Wiley 1982 for a detailed discussion). This means that the spatial part of the triplet wavefunction is asymmetric to exchange of the electrons;
$$\varphi_T = \psi_1(1)\psi_2(2) − \psi_1(2)\psi_2(1) $$
and the singlet spatial part is symmetric which means that the electrons tend to crowd together raising the energy
$$\varphi_S = \psi_1(1)\psi_2(2) + \psi_1(2)\psi_2(1) $$
The numbers in brackets refer to the electrons’ coordinate. As the total wavefunction is anti-symmetric to exchange of electrons the corresponding spin parts have the opposite symmetry to the spatial parts.
If the spins are on two separate atoms then the singlet and triplet energies can be very close to one another and the order can be changes by external magnetic fields. There is more detail on this effect in this answer How does spin flipping of triplet carbenes occur?


Do covalent salts produce sodium carbonate extract?

Soda extract is done to remove multivalent cations from your sample solution, by precipitating them as (mixed) carbonates and hydroxides.

Antimony or tin chlorides are well soluble even in water, and even if the solids have a strong covalent character, they are still ionic in solution.

The sulfates of barium, strontium and calcium are however very badly soluble, so it doesn't work for those. You can however do it in the melt, instead of solution. After cooling and powdering that melt, you add water. What dissolves are sodium sulfate and soda. Add $\ce{BaCl2}$ to detect the sulfate. The residue are the alkaline earth metal carbonates, which dissolve in $\ce{HCl}$.

(Please check your textbook/lab manual for the actual procedure. I have Jander, Blasius, Einführung in das anorganisch-chemische Praktikum. Perhaps someone else can recommend a good English book.)


Reaction between copper (II)chloride and aluminium foil

This is an easy experiment to do at home, so I did it about an hour ago. The first figure shows the starting chemicals and "equipment":

I did not have any copper chloride, so I used copper sulfate pentahydrate, $\ce{ CuSO4.5H2O}$, and $\ce{NaCl}$. (The sodium and sulfate ions are just spectator ions here.)

I added the two chemicals to the $\pu{25 mL}$ of water in the little beaker, stirred until they were fully dissolved, then poured the solution onto the aluminum foil. The second figure shows what happens about every 2 minutes:

The third figure shows what happens around the 10 minute mark: the solution has eaten through the aluminum and the brown copper is evident:

Finally, Fig. 4 shows the result after the solution has leaked away:

So, the redox reaction works as expected and as shown in various videos on the web. Note that copper sulfate pentahydrate crystals are blue, as are aqueous solution of it. The copper chloro complex ion, $\ce{CuCl4^2-}$, is green.

EDIT: Per the comment from @AJKOER, I ran the experiment again, with powdered iron (Fisher, 100 mesh, electrolytic) and copper sulfate solution. The next figure shows the starting materials. I put a strong magnet under the little beaker to prevent iron, which was deliberately in excess, from going over in the decantation that would follow.

Pouring the copper sulfate solution into the beaker resulted in a vigorous reaction and quite a bit of heat. I stirred the reaction mixture and let it go to completion. The magnet still stuck strongly to the bottom of the beaker, indicating that there was still substantial powdered iron remaining.

The mixture was decanted to another beaker, with some mostly useless filtering (no good filter paper at hand), and allowed to settle for about an hour. The copper particles produced in the reaction are very small and settle out very slowly. Some of the supernatant solution was transferred to a sample cell, illuminated from the left via an LED flashlight and photographed. This is shown in the next figure:

Despite the light scattering, the green color of the ferrous sulfate solution is evident. The photo will be updated after more particulate has settled out.


Why is platinum used as a catalyst in fuel cell electrodes (and elsewhere)?

Imagine that a fuel cell is like electrolysis - the other way around.

In the electrolysis of water, you'd apply a certain potential (a voltage) to the electrodes and observe hydrogen and oxygen formed at the electrodes.

In theory, you need a particular potential to run the reaction. In reality - and this depends on phenomena on the surface of the electrode material - you often need more.

With platinum, however, this overpotential for hydrogen is zero.

This means that you have no losses, neither in electrolysis, nor in the fuel cell.

This might be oversimplified but I hope it's not completely wrong ;)


Ok, I will follow the assumption proposed by @Zhe above (possible incorrect as he states, but please do not be confused by that).
In order to solve this problem we need two acidity constants:  $\mathrm{p}K_{\mathrm{a}(\ce{NH4+})} = 9.25$ and
$\mathrm{p}K_{\mathrm{a}(\ce{CH3COOH})} = 4.76$.
First  we state the proton balance (the amount of protons taken up have to be equal to the amount of protons given off in the system): Initially we have $\ce{H2O}$ and $\ce{CH3COONH4}$.
Proton balance:
$[\ce{H3O+}] + [\ce{CH3COOH}] = [\ce{OH-}] + [\ce{NH3}]$
At $\mathrm{pH} = 7, [\ce{H3O+}] = [\ce{OH-}] = \pu{10^{-7} M}$:
At this $\mathrm{pH}$ the proton balance can be simplified as $[\ce{CH3COOH}] = [\ce{NH3}]$.
The simplified proton balance will be true only at a $\mathrm{pH}$ that is exactly in the middle of the two $\mathrm{p}K_{\mathrm{a}}$-values. We get $\mathrm{pH} = (4.76 + 9.25)/2 = 7.005 \approx 7$ (only one significant figure is given, since you have stated the concentration as $\pu{1 M}$).


Is there any book or text where ionic radius of elements are given?

There is "classic" work by Shannon and colleagues [1, 2], also available online as Database of Ionic Radii.

Extended ionic radii for lanthanides with higher coordination numbers were published by Jia [3].

The values from the references above are also compiled in recent editions of CRC handbook of chemistry and physics, e.g. [4, p. 12-12].

More recently, an updated series of calculated ionic radii has been published by Rahm et al. [4].

References

Shannon, R. D.; Prewitt, C. T. Effective Ionic Radii in Oxides and Fluorides. Acta Cryst B 1969, 25 (5), 925–946. https://doi.org/10.1107/S0567740869003220.
Shannon, R. D. Revised Effective Ionic Radii and Systematic Studies of Interatomic Distances in Halides and Chalcogenides. Acta Cryst A 1976, 32 (5), 751–767. https://doi.org/10.1107/S0567739476001551.
Jia, Y. Q. Crystal Radii and Effective Ionic Radii of the Rare Earth Ions. Journal of Solid State Chemistry 1991, 95 (1), 184–187. https://doi.org/10.1016/0022-4596(91)90388-X.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97.
Rahm, M.; Hoffmann, R.; Ashcroft, N. W. Atomic and Ionic Radii of Elements 1–96. Chem. Eur. J. 2016, 22 (41), 14625–14632. https://doi.org/10.1002/chem.201602949.


why do atoms need to be ionised in a mass spectrometer?

why do atoms need to be ionized in a mass spectrometer? why can't you determine the mass of the atoms when they are not ionised?

There are various types of mass spectrometer
https://en.wikipedia.org/wiki/Mass_spectrometry#Mass_selection
but all of them require neutral particles to be charged in order to be accelerated and then detected. The spectrometers use electromagnetic fields to accelerate the ions, this effect only occurs if the species are charged as the field will not accelerate a neutral particle.

For example, in the magnetic sector machine, the ions are formed in an ionization chamber, accelerated in a uniform electric field and shot into the magnetic sector where they are sorted by their path through the field. The higher the charge on the particle, the more it is defected, the higher the mass, the less it is deflected:

http://orgchemguide.blogspot.com/2011/04/magnetic-deflection-or-sector-mass.html

  Also what is the mass-to-charge ratio

So the ions are sorted not by their charge or mass alone, but by the ratio of the two, the mass to charge ratio. The machine cannot tell the difference between an ion with mass 4 amu and charge 1+ and another with mass 8 amu and charge 2+, both have the same mass to charge ratio and will behave identically in the field.

Most detectors also work only if they are struck by a charged particle.

Here is a great guide to the simple principles of how a MS works:
https://www.chemguide.co.uk/analysis/masspec/howitworks.html


Can assumptions made during calculation of ka of buffer solutions be justified?

Answering your first question, $[\ce {H2O}]$ is constant because the square brackets indicate a concentration. Concentration is the number of moles of the solute on the volume of the solvent.

Except in this case, water is the solvent. So the concentration will always be constant, since the volume is proportional to the number of moles.

Besides, while dealing with the law of mass action, for liquids we don't take concentration, we take activity. Which is approximately one for water.

As for your second question, we basically have the following product:

$$\ce{([CH3COO- ]_{CH3COOH} + [CH3COO- ]_{CH3COONa})[H+ ]}$$

which can be rewritten as $(a + b)b$, where $b << a$. Let's say $a$ was $1$ and $b$ was $0.0001$.

The actual value of the expression is $1.0001\times 10^{-4}$. Neglecting only the first $b$ will give $1\times 10^{-4}$, which is not too much different.

Neglecting only the second $b$ will give $1.0001$, and neglecting both will give $1$. Both of these answers are very wrong.

When a small number is added to a large one, the small number can be ignored. But not when it is multiplied.


According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), alphabetical order is used.

Alphabetical order is used in formulae as follows.
(a) Within the group of cations and within the group of anions, respectively, in formulae of salts and double salts. Deviations from this rule may be acceptable if it is desired to convey specific structural information (…)

and

Alphabetical order is used in names as follows.
(d) In compositional names, the names of the formally electropositive components and the names of the formally electronegative components are each arranged alphabetically with the former group of components preceding the latter. (…)

Note that the order of components in the name may deviate from the order of the corresponding components in the formula. For example:
$\ce{KMgF3}$ (since K < M)
magnesium potassium fluoride (since m < p)
$\ce{ZnI(OH)}$ (since I < O)
zinc hydroxide iodide (since h < i)


How do I write many cyclic voltammetry curves in Nova?

Try to use command "repeat n-times", that will make n-data files and each can have 30 000 points.


I suppose that with reactivity you mean a metal's tendency to become oxidized.
There are several reasons and it depends on the metals you are analyzing.

If you are referring to alkaline or alkaline earth metals, reactivity increases as you go down the group. They all have a large oxidation potential since it's relatively easy to remove an electron (or two) of its orbital. As you move down the group, atoms are larger and the valence electrons will be less atracted by the nucleus. Therefore, they are easier to remove and the metal will be more reactive.

For d-block transition metals (most of them are considered "transition" metals) it's a whole different story. There are many factors that contribute to its reactivity and it's harder to specify general trends. Many of them (especially the ones on the second and third transition rows) can be oxidized to different oxidation states depending on the conditions. You need to look at their electron configuration and reduction potentials to start rationalizing its properties. Depending upon the electron configuration, they will be able to lose (or even gain) less or more electrons and with different ease.

I don't know your background with chemistry, so I won't write a more detailed explanation but if you clarify that you have a more concrete interest in any explanation, just ask.


How do I get Mass spectrometry graph, history, and Infra-red spectroscopy graph for Secnidazole?

https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi (or search sdbs database, it should be the first link) is a useful database that contains a variety of spectral data for a large number of compounds. If you search for Secnidazole, they appear to have MS and IR data for this compound.

If you have access to journal articles, you can also do a literature search for this compound. It is likely that spectral data for this compound will be present in any paper where this molecule was synthesized (either in the main article or its supporting information).

Edit: I just did a brief literature search through Reaxys - most of the papers seemed centered around clinical trials rather than organic synthesis, although I will admit I didn't look over any of them in any great detail.  I'll leave that to you.

Chemical companies such as Sigma Aldrich or Fisher Scientific also sometimes supply spectroscopic data for the compounds they sell.

If you have access to ChemDraw, you can simulate $^1$H and $^{13}$C NMR data for any compound you draw.  I'm sure other spectral simulation programs exist, but don't have much experience in that area.


How to convert wedge–dash structure into Fischer projection?

Method 1

Method 2

Quoting from

http://oscar.iitb.ac.in/OSCARPP/Chemistry/UploadedStoryboards/SCH023_3P0431Wedge-dash%20to%20Fischer.ppt

The following is conversion of molecules with two chiral centres from wedge to fischer projection yhrough a series of operations as shown below.

Reference :

http://oscar.iitb.ac.in/OSCARPP/Chemistry/UploadedStoryboards/SCH023_3P0431Wedge-dash%20to%20Fischer.ppt


Guilin Hu, Kim Dam-Johansen, Stig Wedel, Jens Peter Hansen. Decomposition and oxidation of pyrite, Progress in Energy and Combustion Science, Volume 32, Issue 3, 2006, Pages 295-314, ISSN 0360-1285, http://dx.doi.org/10.1016/j.pecs.2005.11.004.

    "In inert atmospheres, the transformation of pyrite proceeds through a multi-step process in the sequence of pyrite→pyrrhotite→troilite→iron. The whole process is determined by the temperature and the total sulfur gas pressure in the system."

Since there isn't any oxygen in an inert atmosphere, the sulfur in the pyrite must be converted to elemental sulfur to maintain charge neutrality and mass balance as you proposed.


Addition of inert gas at "constant volume"?

You should not :)

"does not affect the equilibrium state" means that the thermodynamical constant is still the same. If you add an inert gas $\ce{G}$ in your system it will not react with other chemical coumponds.

For example at the beginning you have $\ce{A} \rightleftharpoons \ce{B}$ and after you add you inert gas you have $\ce{A + G} \rightleftharpoons \ce{B + G}$. So it not afftects the value of $\ce{K_{eq}}$ because $\ce{K_{eq}}$ depends only of $\ce{T}$ the temperature.

If you use a solid container the volume will not change but the pressure will.

Also we have $$\ce{\mu_i(T,P_i)=\mu_i^°(T)}+\ce{RT\ln\frac{P_i}{P^°}}$$

Then $$\left(\frac{\partial\Delta_r\ce{G}}{\partial n_{inert}}\right)_{\ce{T,V}}=0$$

If you prefer calculus.


You are right in saying that the electroplating could equally well be carried out with any electrode. Just offhand though, I could think of a couple of reasons why you would want the anode to be made of copper.

Choosing copper as the anode makes the overall cell potential equal to zero. The reaction at the anode is $\ce{Cu -> Cu^2+ + 2e-}$ and the reaction at the cathode is $\ce{Cu^2+ + 2e- -> Cu}$ - they are identical but one is flipped, so $E^\circ_{\mathrm{cell}} = 0~\mathrm{V}$. Compare this to if you used graphite - the reaction at the anode would then be $\ce{2H2O -> O2 + 4H+ + 4e-}$ and you would have $E^\circ_{\mathrm{cell}} = -0.892~\mathrm{V}$, meaning you would need more energy to run the reaction.
More importantly, it also ensures that the concentration of $\ce{Cu^2+}$ ions in solution remains constant. If you use a graphite electrode, the solution can initially provide the $\ce{Cu^2+}$ ions required for electroplating, but as the concentration of $\ce{Cu^2+}$ decreases, $E_{\ce{Cu^2+}/\ce{Cu}}$ becomes smaller (by the Nernst equation) and it becomes harder and harder to reduce $\ce{Cu^2+}$ at the cathode.


Looking for a mechanism for the demercuration of organomercurial alcohol

The direct displacement of the mercury acetate by borohydride, as prooposed by @xavier_fakerat, looks attractive because it (1) makes some chemical sense and (2) gives the correct products. However, that mechanism can be disproven by looking at the stereochemical outcome of the reduction step.

Pasto and Gontarz did just that (Pasto JACS 1969). When cyclopentene was subjected to oxymercuration, followed by reduction with sodium borodeuteride, the product observed was the trans-configuration of the hydroxyl and deutero groups. The replacement of mercury for deuterium went with retention. This rules out direct displacement (SN2 like mechanism) which would lead to inversion of configuration.

When cis- and trans-butene were subjected to the same process in independent experiments, in both cases the result was a 50:50 mixture of syn- and anti-stereochemistries. This indicates that there is cleavage of the carbon-mercury bond and sufficient lifetime for the resulting intermediate to undergo bond rotation before carbon-hydrogen bond formation. For other substrates, C-C bond rearrangements were observed.

The authors propose that hydride displaces acetate at mercury. This is followed by homolytic C-Hg bond cleavage, leaving a carbon centered radical. Finally, hydrogen atom abstraction from the mercury-hydride gives the reduced product and elemental mercury.


I got the answer, equilibrium favors the right side because (Cd) is a soft metal and (S) is soft too, so it binds better to (Cd) than (O), in accordance with the HSAB theory.


The extent of enol content can be explained on the basis of acidic hydrogen, intramolecular hydrogen bonding.

The stability of the carbanion formed is dependent on its stability after it is formed. In compound $\ce{(c)}$, the carbanion is stabilized by extended conjugation due to presence of phenyl group. Such stabilization does not occur in $\ce{(d)}$. In $\ce{(d)}$, methyl substituent at carbanion increases negative charge via inductive polarization. The resulting carbanion is unstable.

After enolization , enol is stabilized via intramolecular hydrogen bonding, which is a common feature in all examples provided.
The enol in compound $\ce{(c)}$ is stabilized via extended conjugation.

Taking the above factors into account, $\ce{(c)}$ has highest enol content.


How many atoms of helium are in the chamber?

You already mentioned the equation for the rate of collisions of gas molecules with a section of wall:

$$Z_\mathrm w=\frac14\frac NV\sqrt{\frac{8RT}{\pi M}}A$$

We know the values of
rate $Z_\mathrm w=2.25\times10^{15}\ \mathrm{s^{-1}}$,
volume $V=22.7\ \mathrm l=0.0227\ \mathrm{m^3}$,
molar gas constant $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$,
temperature $T=298\ \mathrm K$,
molar mass of helium $M=4.003\ \mathrm{g\ mol^{-1}}=0.004003\ \mathrm{kg\ mol^{-1}}$, and
area $A=1.63\times10^{-5}\ \mathrm{mm^2}=1.63\times10^{-11}\ \mathrm{m^2}$.

Thus, the number $N$ of helium atoms is

$$\begin{align}
N&=\frac{4Z_\mathrm wV}{\sqrt{\dfrac{8RT}{\pi M}}A}\\[6pt]
&=\frac{4\times2.25\times10^{15}\ \mathrm{s^{-1}}\times 0.0227\ \mathrm{m^3}}{\sqrt{\dfrac{8\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm K}{\pi \times 0.004003\ \mathrm{kg\ mol^{-1}}}}\times1.63\times10^{-11}\ \mathrm{m^2}}\\[6pt]
&=9.98\times10^{21}
\end{align}$$


Put your asphaltene source on a silica gel column and elute with hexane and then hexane/toluene mixture to remove paraffins (saturated compounds), aromatics, and resins (small polar compounds). At last, remove the asphaltenes (high molecular weight polar compounds) with pure toluene, evaporate to dryness (a brittle black solid), then with heat you can disolve/disperse the asphaltenes in a simple high molecular weight neutral material like petroleum jelly and let it cool. This will give a high and/or adjustable concentration of what you want to look at. https://en.wikipedia.org/wiki/Asphaltene


Alkaline hydrolysis of ester:

  $\ce{RCOOR' + OH- -> RCOO- + R'OH}$

Almost all organic acids are weak acids:

  $\ce{RCOO- + H2O <<=> RCOOH + OH-}$

Therefore, a lot of $\ce{OH-}$ is consumed during the esterification, with the product $\ce{RCOO-}$ generating little $\ce{OH-}$.

There is a net consumption of $\ce{OH-}$, making the solution more acidic, lowering the pH.


Calcium Carbonate cannot burn. Its already in its highest oxidation state (+2 for Calcium and +4 for Carbon), and cannot be oxidized (combusted) to a higher oxidation state. However, calcium carbonate does decompose on heating to quicklime and carbon dioxide.

$$\ce{CaCO3 -> CaO + CO2}$$

This doesnt give out any flames in the first place.

Now let's talk about how to salvage your project. Why don't you try to heat the calcium carbonate, and show that carbon dioxide is evolved by using a burning match stick, or by using lime water? $\ce{CO2}$ extinguishes the burning flames and forms a milky precipitate with $\ce{Ca(OH)2}$ (limewater). This can be used to show that perhaps egg-shells and marble have carbonate content in them.


Does the third law of thermodynamics apply to all compounds and elements?

I strongly suggest you read this. From there:

  The laws of thermodynamics dictate that absolute zero cannot be reached using only thermodynamic means, as the temperature of the substance being cooled approaches the temperature of the cooling agent asymptotically. A system at absolute zero still possesses quantum mechanical zero-point energy, the energy of its ground state. The kinetic energy of the ground state cannot be removed.

And:

  An even stronger assertion is that It is impossible by any procedure to reduce the temperature of a system to zero in a finite number of operations. (≈ Guggenheim, p. 157)

And:

  Perfect crystals never occur in practice; imperfections, and even entire amorphous materials, simply get "frozen in" at low temperatures, so transitions to more stable states do not occur.


Why does a buffer work best at the pH closest to its pKa?

A buffer works best when there is the same amount of weak acid/base and its conjugate. If you look at the Henderson Hasselbalch equation, and set the concentration of the weak acid/base equal to each other, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.


Why is an alkoxide released instead of hydroxide in the attack of trialkyloxyborane by hydroxide?

Both hydroxide and alcoholate can be displaced from the tetracoordinated intermediate. The difference is that displacing the hydroxide is just the back reaction of the inital attack while displacing the alcoholate will move the reaction forward.

You cannot argue that $\ce{X}$ is a stronger base than $\ce{Y}$ and therefore $\ce{X}$ will always get displaced. At best, you can argue that $\ce{X}$ is ten times stronger so the probability of $\ce{X}$ begin displaced is higher than that of $\ce{Y}$. For all intents and purposes of the argument, the basicities of (primary) alcoholates and hydroxide are too similar to make a real difference.


Are there any significant industrial processes that use liquid ammonia as a solvent?

I might be terribly wrong, but I'm  not aware of any other industrial process involving ammonia as a reactant (fertilizers, caprolactam, amino acids) where liquid ammonia is used as a solvent.

As a side note, liquefication at least plays a role over here (Germany) for the transport in rail tank cars.


I would suggest the Lehninger Principles of Biochemistry, which is one of the best in my opinion since it covers most of the key aspects in biochemistry.
It is quite expensive to buy as new, but since it is quite common you can maybe borrow it from a library. In this way you can get an impression about the book and eventually decide if it is worth buying it or not. If you are interested in having it I would buy a secondhand one and I would say it does not matter if it is not the last version, the previous one should be good as well.


Meaning of polarization pulse in microwave (rotational) spectroscopy

In NMR the sum of the magnetic dipole vectors is called the magnetisation, and this is induced in the sample by the large fixed external magnetic field acting on the nuclear spins. As you mention it is influenced, i.e. moved about,  by the magnetic field of the rf pulses applied to the sample and so ultimately cause the nmr signal. In optics the similar thing  can occur where an intense optical field can align the electric dipoles and cause a polarisation, i.e. macroscopic dipole in the sample, this is the optical equivalent of magnetisation. The experiment is far more difficult to do in optics, however, as the dephasing time (T2 time in nmr) is several orders of magnitude faster than in nmr as is the lifetime of any level optically excited.

The field is generally called optical resonance spectroscopy and/or coherent transient spectroscopy with effects such as transient nutation, photon echos and self-induced transparency.


Why is it possible to image LUMO if these orbitals are, by definition, unoccupied?

Why not? It's not like we poke them with our finger, anyway. All we can really see are the electron transfers. Now, the transfers from the molecule are as good as the transfers to it. If we adjust our setup so as to look at the latter, then we may see the shape of the orbital which is now empty but ready to accommodate our electron.

In one setup, the electrons tunnel from the HOMO to the tip, thus visualizing the HOMO. In another setup, they tunnel in reverse direction (hence the change in the sign of voltage), from the tip to the LUMO, thus visualizing the LUMO.


Why do the physical properties of an egg shell change when the egg shell is exposed to vinegar for a week?

It doesn't make calcium carbonate rubbery, it removes the calcium carbonate.  Egg shells are not purely calcium carbonate they are more like a composite with a continuous matrix of calcium carbonate and a smaller continuous matrix of protien.   When you put the egg in vinegar, you etch away this calcium carbonate matrix leaving the formerly less noticeable protein matrix as a gel filled with vinegar and creates a gel.


Why is it that energy of electrons in higher orbits is greater than the ones in lower orbits?

What I'm having trouble understanding is why the energy level is
  higher and not lower. The electrons are further and further away from
  the nucleus, at infinity their energy would be zero.

You are correct. The innermost electrons are the most tightly bound to the nucleus and the outermost ('valence') electrons have the least energy, explaining why they can be removed (in chemical reactions e.g.)

But these energies are always negative and use the principle that infinitely far away from the nucleus the electron's energy would be $0$.

This way defined, the energy of the hydrogen electron in the ground state ($1s$) e.g. is $-13.6\ \mathrm{eV}$. The energy of the electron in excited states ($2s$, $2p$, $3s$ etc) is less negative, until for infinite distance (a free electron) it is zero ($0$).

Let's take a numerical example to demonstrate this. The ground state ($n=1$) of hydrogen's electron is $E_1=-13.6\ \mathrm{eV}$ and the first excited state ($n=2$) is $E_2=-3.4\ \mathrm{eV}$. What are the energies needed to remove the electron from these states?

Since as the energy at infinite distance is $0$, in the case $n=1$, then:

$$\Delta E= 0-(-13.6)=13.6\ \mathrm{eV}$$

And in the case $n=2$, then:

$$\Delta E= 0-(-3.4)=3.4\ \mathrm{eV}$$

So it takes less energy to remove an electron in a higher orbital than a lower one. That fits our observations.

Equally, the energy needed to excite the electron from $n=1$ to $n=2$ is:

$$\Delta E_{1\to 2}=-3.4-(-13.6)=10.2\ \mathrm{eV}$$


Does the bond dissociation enthalpy vary with temperature?

The bond dissociation enthalpy, like any enthalpy of reaction, will have some variation with temperature. This change in enthalpy can be determined using Kirchoff's Law: $$H_{T_f}=H_{T_i}+\int_{T_{f}}^{T_{i}}\Delta c_p(T')dT'$$
Where $$\Delta c_p(T')=\sum_i\nu_ic_{p,i}(T')$$ so that $c_{p,i}$ is the constant pressure heat capacity of the $\mathrm{i^{th}}$ compound in a reaction and $\nu_i$ is the the reaction coefficient of the $\mathrm{i^{th}}$ compound ($\nu_i$ is positive for products and negative for reactants).

If none of the $c_{p,i}$ have a temperature dependence, we can rewrite the $1^{st}$ equation above as $$H_{T_f}=H_{T_i}+\Delta c_p\cdot (T_f-T_i)$$ If they do have a temperature dependence, you will have to know the functional form of $c_{p,i}$ with respect to temperature to integrate them.

So, in regards to your specific question, the bond enthalpy for an $\ce{H-H}$ bond will be different at different temperatures. However, it is worth noting that for highly energetic reactions like this, the contribution due to $c_{p,i}$ is usually fairly small in comparison.


Does lower activation energy definitely mean higher rate?

The lower the energy of the intermediate, the faster the transition from the reactants to the products, at a given temperature.


How would one determine an element simply by looking at its binding energy?

There are a few clues on how to solve this question:

you are given binding energies. The energies in question are just enough to liberate electrons from their bound state.
The equation you've written $$E_{nl} = -\frac{Z_\mathrm{eff}^2R_\ce{H}}{n^2}$$ describes the binding energy. The important part is that it can be assumed to be only a function of n and l, that is, of the subshell to which an electron belongs ($Z_\mathrm{eff}$ is a function of n and l quantum numbers. Unfortunately this is probably a somewhat complicated function but that is not very important just now)

Finally consider the Aufbau principle, outlined in this libretext:

  In essence, you can think of electrons filling in "n+l " order. If you have an equal "n + l"  value for two orbitals, then electrons will occupy the orbital with the lower "n" value.

which is followed by a nice illustration (shown below). The number of distinct binding energies equals the number of subshells. The energy for emission from each subshell decreases in the same order in which the subshells are filled. All electrons in a subshell share a common binding energy. Following the Aufbau pattern you can identify the highest-energy occupied (valence) subshell, which corresponds to the smallest binding energy. I count 8 subshells and assign the minimum $E_{nl}$ to the 4p subshell (the electron configuration is $\ce{[Ar] 3d^{10} 4s^2  4p^?}$). This means there are six possible atoms that match the number of binding energies, namely Ga through Kr.

If we cheat and check a table of ionization energies we see that the lowest binding energy (-14 eV) matches that of Kr (other values in the table don't match those in the problem since the table lists the ionization energy for multiple electron emissions whereas the problem lists energies for all possible single electron emissions $\ce{A->A+ + e-}$, starting from the neutral atom, an emission from one subshell corresponding to one energy). You should note the pattern of decreasing binding energies (magnitudes) starting from the 1s electrons (−14326 eV) up to the valence shell (-14 eV). The following diagram adapted from the libretext quoted above illustrates how this works:


pKa of Tris corrected for ionic strength

First decide the Ic you want to work with. Second, you should correct the pka with the Debye Huckel equation and finally you use the pka corrected with the Henderson Hasselbach equation.


Why are a calico strainer and chloroform used in a British Pharmacopoeia assay of milk of magnesia?

Ok I understand where you got chloroform from.

This is not the assay of milk of magnesia, but the extemporaneous preparation.

In extemporaneous preparations chloroform is used as:

a general anaesthetic
a carminative
flavouraing agent
preservative

So in the case of mik of magnesia it has two main roles that is a carminative and a preservative.

Hope this helps


Base-promoted rearrangement of β-ketoester

This problem has probably been on everyone's mechanistic chemistry exam at some time or another. Does the alkyl (allyl) just "jump" from one side of the carbonyl group to the other? In most examples, the allyl group is a saturated alkyl group. Under such circumstances I am in full agreement with @orthocresol as to the mechanism of the process (1 --> 9). The reaction is not catalytic in methoxide and requires sufficiently concentrated alkali to assure sufficient ester enolate, lest the reaction stops at the diester 4. Notice that the red hydrogens in β-ketoester 1 retain their position in the product 9.

While @Carlson Miller suggested the formation of an enolate, his proposal brings up an interesting possibility in this particular example. The ensuing mechanisms are improbable in protic media but imagine running the reaction with a kinetic base ((NaH, (TMS)2NNa, LDA) in an aprotic solvent is intriguing. There are two possible mechanisms for the reaction. First, enolate 10 can fragment to a ketene-ester enolate that must lose methoxide and generate diketene 11. At this point the reaction is stoichiometric in diketene and methoxide. These two species lead to enolate 12. Bimolecular hydrogen exchange affords the more stable enolate 8. Although I am not particularly enamored with this option, it is nonetheless possible in spite of the limited amount of base relative to the protic reaction. Incidently, any product of intramolecular ketene dimerization would be susceptible to opening by methoxide. This mechanistic pathway, as with the protic reaction, retains the red hydrogens in their original location.

However, the more likely pathway for enolate 10 is via an oxy-Cope-like rearrangement. Although enolate formation is the driving force for the oxy-Cope-like rearrangement (13 --> 14), the rearrangement of enolate 10 gives the more stable enolate of the β-ketoester 8. In this case, the red vinylic hydrogens of ketoester 1 are now located on the methylene group of the allyl residue. Indeed, the allyl group does "jump" in the oxy-Cope.
Enolate 16 is generally formed by alkylation of the dianion of an acetoacetic ester with an allyl halide. Is anyone in SE-land aware of the transformation 15 --> 16?


Why do atoms need 8 electrons to stabilize?

The valence orbitals of atoms are composed of suborbitals (s and p)  there is 1 s suborbital which is spherical and can hold 2 electrons (one with up spin and one with down spin). There are 3 p suborbitals which are dumbbell shaped (look like two balloons tied together at the ends) and align along the x,y&z axies and hold a total of 6 electrons (2 per axis, 3 with up spin and 3 with down spin).  Since an atom is energetically satisfied when all of the electrons are paired and 2+6 = 8, an atom must have 8 electrons in its valence shell to pair all of the electrons.

The octet rule does not apply to all atoms though, for transition metals and later elements there is a d suborbital which can hold an addition 10 electrons, meaning that now an atom may have 18 electrons in its valence shell to be satisfied.  This allows for materials such as metal complexes and the sulfate poly-atomic-ion to exist despite having more than eight electrons in the valence shell.


Why does lithium move on the surface of water during the reaction?

What happens when you put lithium (or sodium or potassium) in water? What is the chemical reaction that is occurring?

The alkali metals react with water to produce alkali hydroxides and hydrogen gas:

$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$

The two products of the reaction of water and lithium are lithium hydroxide (which dissolves) and hydrogen. Hydrogen is not very soluble in water, and it is much less dense than air. The motion of the lithium on the surface of the water is created by the escape of the hydrogen gas, which provides upward force.


How should I go about picking a functional for DFT calculations?

For Structures and Energies: Answer Hazy, Try Again Later
People can talk on this subject for literally days.
It mostly comes down to computational expense vs accuracy - if you have a larger system, you can't afford the fancier functionals. The absolute highest accuracy state of the art functional-wise would probably be those using the Random-Phase Approximation (e.g.), and also the use of more complex range separation techniques (such as the Coulomb-Attenuated Method mentioned below).
Having said that, a lot of people just use B3LYP for everything. B3LYP is parametrised for low-Z main-group chemistry, and so does reasonably well there. It's a hybrid functional (i.e. it includes the exact Hartree-Fock form exchange operator), so it's not the cheapest. I tend to use the series of PBE, PBE0, TPSS - their only parameters are physical constants, so they should never give surprisingly bad results.
A couple of basic guidelines, though (which should be taken as "as far as I know"s):

If you have a delocalised system, you want exact exchange (so, a hybrid functional like PBE0, B3LYP, M06-2X, etc). Without this, most functionals will over-delocalise the orbitals.
If you have a system in which dispersion forces are important, worry. Then use something like $\omega$B97X-D which includes empirical dispersion corrections. Some packages can also apply Grimme's empirical dispersion corrections to any functional.
If you have a periodic system, you probably want to use screened exchange rather than full exact exchange - something like HSE (which tails off the exact exchange past a certain range) rather than PBE0 should be faster while not significantly less accurate.

For TDDFT: Probably CAM-B3LYP
Fortunately, this is a bit easier.
Firstly, you want something that incorporates exact exchange (a hybrid functional), which quickly rules out many functionals. Pure GGAs and LDAs give qualitatively wrong answers for states involving charge transfer. Secondly, functionals that include long-range exchange corrections, such as CAM-B3LYP, also give better results.
Also, look out for triplet instabilities, which can be a problem with the introduction of HF exchange.
(David Tozer is my go-to author for TDDFT testing, though I would warn that despite the above answer, I've never used TDDFT in "production".)
(Now, bring on the functional flamewar.)


What is the accuracy of scf=qc when convergence criteria is lowered?

Something that I have discovered is that when Gaussian (or other good ab initio software) fails to converge, there is something sketchy about what one is trying to calculate. Sometimes DFT calculations just aren't appropriate, especially in cases that require multiconfigurational SCF for even a zero-th order description.

In this case, given the sheet-like character, I suspect the problem is that the molecular orbitals are so close together, and so similar, that there are serious linear dependency problems that lead to all sorts of havoc in the linear algebra routines and convergence accelerators.

If you are trying to simulate an infinite sheet, then I suggest that you try plane wave codes like those found in solid state chemistry (Gaussian's PBC code, NWChem, abinit, etc.)  These sorts of calculations have gotten easier in recent years, but sometimes they require serious horsepower.

There really does seem to be a problem with SCF convergence, and this could be because you are approaching some kind of band structure limit. I have always used QC as an absolute last resort when I must do something wacky.

(One other thing, you have "capped" your peripheral benzenoid rings with hydrogens?)


Norbert Haider from the University of Vienna has been working on the identification of functional groups for quite a while.
You might want to have a look at his checkmol/matchmol tool, which natively reads from MDL, Alchemy, and Sybyl mol files.
Other input formats may be processed after converting them using Open Babel.


You going along the right track. The stacking attraction occurs when two molecules with $\pi$ orbitals come face to face with one another, typically their separation is 0.34 nm. However, for the interaction to be attractive the two molecules have to be displaced slightly, this prevents direct electron repulsion between $\pi$ electrons on one molecule with those on the other. The attraction is in general between the $\pi$ electrons on one molecule and the $\sigma$ framework on the other$^1$ and vice - versa.

The model assumes that a $\pi$ orbital can be split into three charged parts, separated by a distance $\delta$. The lobes of the orbital have charge -1/2 each and at the position of the atom's nuclei is placed a charge of +1. Thus there are also three interactions $\sigma - \sigma , \pi - \sigma $ and $ \pi - \pi$ and three positions of charge, making 9 interactions in all for each atom. The interaction is taken to be electrostatic, i.e. Coulomb and each term has the form $E_{ij} = q_iq_j/r_{ij} $ for atoms i and j and is summed over all atoms.

In addition a van-der-Waals term is added to the calculation of the total interaction, this has the form $E_{vdw} = \Sigma _{ij}(A_{ij}exp(-\alpha_{ij}R_{ij})-C_{ij}/r_{ij}^r))$ where A, $\alpha$ and C have standard values$^2$. In solution the value of this energy is much reduced vs. its value in the crystal making the $\pi-\pi$ interaction the dominant one.

The charges are shown schematically below for carbon atoms. For other atoms, e.g. nitrogen the charge is given as 1.5 with 1.5 electrons associated with them. (All images are from ref 1 )

charge distribution

optimal geometry for Zn porphyrin $\pi-\pi$ interaction. The molecules are displaced (in x and y) relative to one another at a separation of 0.34 nm.

The contour plot for $\pi-\pi$ interaction and van-der-waals (vdw) energy as a function of their centre to centre offset as shown above (energies in $\pu{ kJ mol^{-1}}$). The black squares are geometries from $\pi-\pi $ stacking in crystals. The shape of the contour plot with or without the vdw interaction is the same as with just the $\pi-\pi$ interaction alone, i.e this interaction determines geometry but  the vdw generally lowers the energy and more clearly show regions of low energy as negative values.

$^1$ The model was developed by Hunter & Sanders J.Am. Chem. Soc. 1990, v112, p 5525-5534

$^2$ Van der waals parameters Caillet & Claverie, Acta Crystallogr. sect A1975, v31, p448


See this website. Essentially it's a misprint in the IB Chem testbank.

https://www.thinkib.net/chemistry/blog/8944/multiple-choice-answers

There's a correct rewording for D that makes C the only correct answer, according to the link I just posted.


From what I have learned about oxoacids, the bond strength determines the strength of the acid.

Yes. But that is not the whole picture. You also have to consider what else is floating around in the solution.

The reaction of the species (I assume aqueous solution) is given below:
$$ \ce{H2SO4 <=> HSO4- + H+ <=> SO4^{2-} + 2H+} $$

As you can clearly see, the concentration of the protons is raised when going from left to right. There is a natural equilibrium that pushes backwards.

However, as you correctly point out, the bond stability (or "strength") directly correlates to the stability of the acid species. It might well be (and I'd even bet on it) that the bond strength of $\ce{H-SO4-}$ is higher than $\ce{H-HSO4}$ (note that the bonds indicated here are not the real bonds as they occur but merely a representation) but this requires an in-depth (possibly computational) analysis, as a quick search of the internet did not yield any useful results.

Edit: I did some small, very basic geometry optimization calculations with turbomole (on default settings, so def-SV(P) basis set for all atoms) to find out the bond length of the hydrogen to its oxygen. Turns out my hunch was right, the bond is shorter in $\ce{HSO4-}$ (95.07 pm) than in $\ce{H2SO4}$ (95.72 pm), indicating stronger bonding once the first proton went flying away.


Are the Phosphate and Sulfate ion diagrams with double bonds an invalid picture?

Recent experimental evidence shows that d orbitals are involved in the extra double bond of sulfate or phosphate ions. Therefore a stable sulfate or phosphate ion has 2 double bonds instead of only one to make the formal charge on sulfur or phosphorus zero. Any Lewis structure with expanded octet (periods 3 and higher) should have zero formal charge on the central atom.


Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene?

Your teacher is correct, some of the resonance structures in your figure are not aromatic.  Still though, as resonance structures they contribute to the overall description of the molecule, just less than the structures that are aromatic.  The resonance structures at the right and left end of your figure are aromatic. The fourth resonance structure (the one in the middle) is referred to as being a "quinone" rather than an aromatic.

o- and p-benzoquinone are examples of quinones; they are not aromatic compounds (ref, note the sentence "Quinones are conjugated but not aromatic").

Likewise, their carbon analogues, o- and p-xylylene are also non-aromatic quinoid-like compounds.

For a compound to be aromatic it must be

cyclic
contain 4n+2 pi electrons
have a continuous loop of p-orbitals
and you must be able to draw double bonds - internal to the ring - from each ring carbon atom.

It is this last point that serves to differentiate quinones and quinone-like molecules from aromatic compounds.

Finally, aromatic compounds need not be planar, deviations in excess of 15° from planarity can be tolerated and the molecule can remain aromatic (see, for example, [6]-paracyclophane)


First, you can use Google Ngram viewer to check out how often and in which context proposed words and their combination is used/has been used in the literature.

Second, I feel that "mono-oxidized" is overly redundant and just "oxidized" will deliver. Even in Russian you would rarely say so ("моноокисленный" or "единожды окисленный" is usually seen as an unnecessary complication in most cases). For "dioxidized" there are plenty examples when it is used in a similar context both in inorganic and bio chemistry, so it should a safe usage from what I see.

There is also an another option to say "two-stage oxidation" ("окисленный по двум ступеням") instead of "dioxidized" if you want to underline that the process has been carried out step by step.