text:
Protein/polypeptide nomenclature, residue number superscripted or not?
When referring to a specific residue number of a given amino acid, should the residue number be superscripted or simply affixed?
For example, When referring to a Tryptophan residue at the 147th position, which is the best notation?

$Trp^{147}$ vs Trp147 vs Trp-147

I can't seem to find any guidelines that strictly prohibit the superscript notation, and I think aesthetically it's easier to read, but it seems less common and perhaps incorrect.
It seems to me that Trp147 is the preferred notation, while the superscript version is reserved to indicate derivatives, such as replacements and deletions, but not expressly prohibited; Trp-147 is discouraged, I believe. It also seems that superscripts are used in special circumstances, to denote all sorts of things (for example: GCPRsGCN numbering), so perhaps this is why superscripts aren't used for ordinary residues.
So, is there any consensus on this? Is it a hard rule or more of a stylistic preference?
Ref: https://iupac.qmul.ac.uk/AminoAcid/AA22.html
I've seen Trp147 notation quite commonly, but I've also seen the superscript version used for ordinary notation of typical protein residues.
Example of Trp147 notation style:
https://www.nature.com/articles/nature07938
Example of superscript ($Trp^{147}$) notation style: https://www.sciencedirect.com/science/article/pii/S002192581894785X


text:
Why are fluorine-19 NMR spectra less affected by paramagnetic centers than other common nuclei?
I know that 19F NMR spectra are less sensitive to paramagnetic centers as an empirical fact: it’s my experience and that’s what people in my lab have always been saying.
For instance, some of my samples are NMR silent in 1H, 13C, and 11B, but not 19F NMR. Why does 19F NMR appear to be an exception? Any helpful literature will also be appreciated.

$\ce{^{19}F}$ has similar NMR properties to $\ce{^1H}$ except in particular for the much larger chemical shift tensor terms and resulting larger chemical shift anisotropy (CSA). As a result CSA relaxation is a significant contributor to $T_2$ relaxation at all field strengths/tumbling rates. For instance, considering a simple model
$$\frac{1}{T_{2,\text{CSA}}}=\frac{2}{15}(\sigma_| - \sigma_\perp)^2 \omega_F ^2 \tau_c \left( \frac23 + \frac{1}{2(1+\omega_F^2\tau_c^2)}\right),$$
the $2/3$ term in parentheses means that even under slow tumbling (large $\tau _c $) the contribution of CSA to $T_2$ relaxation can be significant in $\ce{^{19}F}$ , since the CSA term $(\sigma_| - \sigma_\perp)$ can be 50 times larger than for protons. CSA for protons is typically a minor source of relaxation at intermediate to low fields and in liquids (fast tumbling).
Since the CSA term already makes line broadening an issue for $\ce{^{19}F}$ at all fields/tumbling rates, it follows that adding another relaxation mechanism (due to paramagnetic interactions, for instance) results in a relatively small change in the total linewidth.
References
(1) Lu, M.; Ishima, R.; Polenova, T.; Gronenborn, A. M. $\ce{^{19}F}$ NMR Relaxation Studies of Fluorosubstituted Tryptophans. J. Biomol. NMR 2019, 73 (8–9), 401–409. https://doi.org/10.1007/s10858-019-00268-y.

I should add: $\ce{^{19}F}$ also has much greater sensitivity (slightly lower than $\ce{^{1}H}$) and is present at 100% natural abundance, which would explain why it is detected while the lower gamma  nuclei are difficult to observe.

Excellent answer, thank you. Here's a ref for your comment regarding abundance from my original answer that I've since deleted after seeing your answer/comment http://chem.ch.huji.ac.il/nmr/techniques/1d/row2/f.html

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Mystery GC peak appearing between 1H2 and various molecular combinations of 1H, 2H and 3H?
The following gas chromatograph, taken from "Review of the different techniques to analyse tritium", labels all but the largest peak.  Given that the analyte is likely to be from some source enriched with heavy isotopes of hydrogen, what is does that largest peak most-likely correspond to?  3He?
The aforelinked secondary source cites only two plausible primary sources, neither of which is available online:

Bachelet, Fabre, Salmon, Brix, Tritium impurities measurements
based on gas chromatography, Tritium 2016 conference, Charleston
USA

Tetsuo Tanabe Editor Tritium: Fuel of Fusion Reactors Springer
2017

A follow on question:  para-H2 is about 25% of ortho-H2 and appear at different times.  Why don't similar splits occur between the other isotopic combinations?

The link you provided is a secondary resource. Please consult the original work cite there. Nobody in the world can tell the identity of a peak by looking at this chromatogram. It is very much unlike NMR, IR or any other spectroscopic technique.

It may be HD, or hydrogen deuteride.
Hydrogen has a proton only nucleus.  Ortho H2 nuclei are said to "spin" in the same direction, para H2 nuclei in opposite.  Deuterium and tritium also have neutrons, which may prevent this type of subatomic behavior.
Apparently the difference between "ortho" and "para" H2 is enough to affect its affinity to the GC column$^1$.
$^1$ temperature has an effect on the ortho to para ratio.  At 0 degrees K, it is all para.  At room temperature, it is 3 ortho to one para.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
According to Le Chatelier's principle, is it possible to fully consume a reactant if we increase the concentration of the other reactant(s)?
Suppose there is a hypothetical chemical equation: $$\ce{ A + B <=> C}$$
If we increase the concentration of B to a large extent with the intention of fully consuming A in the reaction, will A actually be fully consumed in the reaction (no traces of A left), and the reaction completely favors the forward reaction to produce C?
Because the formula for equilibrium constant is K= [C]/[A][B], where in this case, A is fully consumed, therefore K = [C]/[0][B], which is not possible because it is not divisible by 0.

No, backward reaction will still be happening and there will be always some very little amount of A in the container even if you observe no A to be left. For example. If you have a bottle of water, there is always some H2O molecules turning back to H2 and O2, But at very very slow rate to notice. Le Chatelier's principle is not the cause, its just the result of whatever formula you have posted in your question.

In fact, it is just a mathematical task. If a=x/(y.z), can z = 0 for x>0 and  large enough y ?

You can make [A] and [B] very small by removing [C]!
Conversely, you make C at a greater rate by increasing A and B.
Just increasing B will not help, except possibly in the case where you have a 2 phase reaction. Increasing [B] in one phase may make more of it available to A, which is dissolved in the other phase.
[C]/0 cannot exist precisely because the equilibrium constant =

K = [C]/[A][B]

if you have alot of C and no A or B than [A] and [B] will increase until equilibrium is reached.

At equilibrium, all species have to be present at some concentration other than zero. This is because at equilibrium, the forward and reverse rates have to be equal and non-zero. This is the property of a dynamic equilibrium.
Adding lots of B will not remove A completely. If the zero in the denominator bothers you, write the equation like this:
$$[A] * [B] * K = [C]$$
Unless [C] is zero, [A] and [B] have to be non-zero. [A] could get very small, but not zero.
On the other hand, if you continuously remove C until there are no atoms left, there will be no C and no A (assuming B is in excess). However, that situation is not called equilibrium.

Because the formula for equilibrium constant is K= [C]/[A][B], where in this case, A is fully consumed, therefore K = [C]/[0][B], which is not possible because it is not divisible by 0

The division by zero is a technical problem of the description of the equilibrium constant, but not the underlying explanation. You can try to calculate the reaction quotient Q for a situation where one of the reactants is absent, and you run into that problem. This does not mean, however, that you can't have that situation. It just means that your model (concerning the definition of Q) is not able to capture it in a meaningful way.


text:
Calculating the percentage of dissociation degree
The question is as such:

Notice the equilibrium expression. $$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}.$$ Initially, there were $0.28$ moles for each reactant, after it reaches equilibrium, the number mol of product produced was $0.014$ mol. What is the percentage of dissociation degree?

Through an ICE table, I calculated the number of mol of $N_2$, $H_2$, and $NH_3$ at equilibrium. There are $0.273$ mol of $N_2$ ($0.007$ mol is used), $0.259$ mol of $H_2$ ($0.021$ mol is used), and $0.014$ mol of $NH_3$, but I'm not sure how to calculate the percentage of dissociation degree.
My teacher taught that $\alpha=\dfrac{n_{\text{change}}}{n_{\text{initial}}}$ (referring to the ICE table), but I'm not sure which $n$ to use. Help would be appreciated, thank you.

It is not a dissociation degree. It looks like a copy-pasted term from acid-base equilibrium context. // The equation uses amount concentrations, but there are provides amounts, not their concentrations.

That's where I got confused. I thought dissociation degree/degree of ionisation is used in acid-base context, but I'm not sure about this one. My teacher also said that whenever the volume isn't mentioned in the question, we are to use $V = 1L$. I'm sorry that I forgot to mention that.

For a reaction of the form:
$$\ce{aA +bB<=>cC +dD}$$
You can calculate the dissociation fraction $\alpha$ using its relationship with $K_C$:
$$K_C=\frac{\left(\frac{C_{Co}}{C_{Ao}}+\frac{c}{a}\alpha\right)^c\;\left(\frac{C_{Do}}{C_{Ao}}+\frac{d}{a}\alpha\right)^d}{\left(1-\alpha\right)^a\;\left(\frac{C_{Bo}}{C_{Ao}}-\frac{b}{a}\alpha\right)^b}\;\left(C_{Ao}\right)^{\Delta n}$$
In your case, the relevant reaction is:
$$\ce{3H2(g) + N2(g)<=>2NH3(g)}$$
Where:
$\ce{H2}$ is the limiting reagent A
$\ce{N2}$ is B
$\ce{NH3}$ is C
Using the molar balance equation for $\ce{NH3}$, you can calculate the change in moles $x$ that takes place after equilibrium is reached:
$$N_C=N_{Co}+cx\implies x=\frac{N_C-N_{Co}}{c}=\frac{0.014-0}{2}=0.007$$
Substituting $x$ into the molar balances of $A$ and $B$:
$$N_A=N_{Ao}-ax=0.28-3x=0.259$$
$$N_B=N_{Bo}-bx=0.28-x=0.273$$
Since it is mentioned that volume is $\pu{1L}$, all amounts are equal to their respective concentrations, and we can calculate $K_C$:
$$K_C=\frac{C_C^c}{C_A^a\;C_B^b}=\frac{N_C^c}{N_A^a\;N_B^b}=\frac{(0.014)^2}{(0.259)^3(0.273)}=0.04132$$
By simplifying and using the equation that relates $K_C$ and $\alpha$, we can calculate the latter:
$$K_C=\frac{\left(\frac{2}{3}\alpha\right)^2}{\left(1-\alpha\right)^3\;\left(1-\frac{1}{3}\alpha\right)}\;\left(0.28\right)^{-2}=0.04132$$
Solving for $\alpha$:
$$\alpha=0.075$$
Or equivalently:
$$\alpha=7.5\%$$
Alternatively, you can also use your teacher's equation, although I would argue it's incomplete since it does not account for the molar coefficient $a$ of the limiting reagent A. The complete form would be:
$$\alpha=\frac{ax}{C_{Ao}}$$
We can verify we get the same result:
$$\alpha=\frac{(3)(0.007)}{0.28}=0.075=7.5\%$$
However, the wording of the problem is in conflict with the definition of $\alpha$, since $\ce{NH3}$ should be dissociating into its constituent gases from a non-zero initial amount with no initial amounts of $\ce{H2}$ or $\ce{N2}$ present, not vice-versa.
Perhaps your teacher overlooked this when designing the problem.


text:
pH and degree of dissociation of drugs
I have a doubt, i hope not so stupid.
Suppose we consider a buffer solution of acetic acid/acetate at pH = pKa = 4.76 and we add aspirin (pKa = 3.5): given that the pH of the solution is higher than the pKa of the drug, then the aspirin prevails in its ionized form, from Henderson-Hasselbalch equation. That said, does the degree of dissociation of aspirin actually depend on the pH of the solution, or does it depend on the ability of the aspirin to react with the conjugate base of the buffer to some extent?
AspH + CH3COO-  <=>  Asp- + CH3COOH
In fact, given that the pKa of aspirin is higher than that of acetic acid, the reaction is shifted to the right, so that aspirin will be dissociated in a certain percentage which depends precisely on this equilibrium, rather than on the pH of the solution (?).
Similar doubt regarding an unbuffered solution: suppose we want to extract aspirin from an organic solvent, with a saturated bicarbonate solution: the reaction between the two species is, also in this case, shifted to the right, as aspirin is a stronger acid than carbonic acid, therefore aspirin will prevail in the deprotonated form and there is a certain degree of dissociation. Again, does the degree of dissociation of aspirin actually depend on this reaction, or on the pH that is reached AFTER the reaction, again using the Henderson-Hasselbalch equation?
Thanks.

How much acetic acid, sodium acetate and aspirin do you dissolve in your water ?

pH=pKa, so [CH3COOH]=[CH3COO-]=0,5 M for example. About aspirin, every quantity. I think that this Isn't much important (?), I Just wanted to know the base principle about what i wrote.

The proton exchange between aspiring and acetic acid respective forms is just an equilibrium dependent of the 2 dissociation equilibrii. The pH given by the acetic buffer ( in excess wrt aspirin) determines [A-]/[HA] ratios for both.  It is like K1 = x.y1/z1 and K2 = x . y2/z2, where x = [H+].

@Poutnik so Is It the pH of the solution that determines the degree of dissociation, and not directly the reaction between aspirin and acetate?
Consequently, is the same also true in the case of bicarbonate? That is, is it the final pH, after the reaction between aspirin and bicarbonate, that determines the degree of dissociation of the aspirin itself, and not the reaction itself?

It depends what has the "upper hand", if the solute presence affects/determines pH or if pH affects/determines solute behavior. In our case, the PH buffer and therefore pH has the upper hand.

@Poutnik so you mean that, in the case of buffer, the pH of the solution determines the degree of dissociation, so no reaction Is needed to be considered to evaluate the degree of dissociation. Instead, in the case of bicarbonate, It Is the solute (aspirin) that has the upper hand, so It reacts with bicarbonate and this affects the degree of ionization of aspirin. The consequent pH lowering has no effect on the degree of ionization (?).

As Ka1=[H+][Asp-]/[HAsp] and Ka2=[H+]]A-]/[HA] then for your supposed reaction between aspirin and acetate:  K=[Asp-][HA]/([HAsp][A-]=Ka1/[H+]. [H+]/Ka2 = Ka1/Ka2.

Let us continue this discussion in chat.

(Partial answer yet)
If we take the H-H equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {\frac{[\ce{A-}]}{[\ce{HA}]}}$$
we can construct this:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a,Asp} + \log {\frac{[\ce{Asp-}]}{[\ce{AspH}]}} = \mathrm{p}K_\mathrm{a,\ce{CH3COOH}} + \log {\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$$
$$\mathrm{p}K_\mathrm{a,Asp} - \mathrm{p}K_\mathrm{a,\ce{CH3COOH}} =  \log {\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}} - \log {\frac{[\ce{Asp-}]}{[\ce{AspH}]}}$$
$$\mathrm{p}K_\mathrm{a,Asp} - \mathrm{p}K_\mathrm{a,\ce{CH3COOH}} =  \log {\left(\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} \cdot \frac{[\ce{AspH}]}{[\ce{Asp-}]}\right)}$$
$$10^{\left(\mathrm{p}K_\mathrm{a,Asp} - \mathrm{p}K_\mathrm{a,\ce{CH3COOH}}\right)} = \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} \cdot \frac{[\ce{AspH}]}{[\ce{Asp-}]}$$
As  $\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}$  is given by the acetate $\mathrm{pH}$ buffer,  $\frac{[\ce{AspH}]}{[\ce{Asp-}]}$ is given too.

Thanks a lot for the answer. So from this discussion I understand that it is the pH that influences the degree of dissociation and not the reaction between aspirin and acetate which would take place, in the case of a buffer.
But in the case of the saturated bicarbonate solution (unbuffered) what happens? Is it the reaction that takes place between bicarbonate and aspirin that directly influences the degree of dissociation or is it the final pH, which is obtained as a result of the reaction, that influences it?

Well, bicarbonate is a kind of pH buffer too.

In theory, in order to form a buffer, it is necessary that the concentrations of acid and conjugate base are close enough. When you create a saturated solution of bicarbonate, tiny amounts of carbonic acid and carbonate are formed, so it's not really a buffet.
To make it easier, I'm assuming that I use NaOH instead of bicarbonate to extract the aspirin from an organic solvent. Returning to the previous question: on what does the degree of dissociation depend in this case? pH can change now.

pH of acidic salts of biprotic acids is approximately the average of both pKa, with the capacity depending on pKa difference.


text:
According to molecular orbital theory, is the antibonding orbital empy when electrons are found in the bonding orbital?
According to molecular theory when 2 atomic orbitals combine,  2 molecular orbitals will be created. So if the electrons are in the bonding orbital will the antibonding orbital exist and just be empy? But that doesn't make sense, I mean orbitals are just a measure of space and energy of the electrons. If there is no electron in an orbital then that orbital exists just in theory right?
Wouldn't it be more right to say that when two atomic orbitals combine they form either a bonding orbital or an antibonding one and the electrons are found in either one of them respectively?

All orbitals exist just in theory.

Molecular orbitals are solutions of the Schrödinger time independent equation for a molecule. Some of these solutions lower the energy of the molecule (bonding molecular orbitals) and some raise the energy of the molecule (antibonding molecular orbitals).


text:
Can molecular hydrogen form hydrogen bonds?
Basically, I want to know if molecular hydrogen (H$_2$) can form hydrogen bonds with other molecules (CH$_4$, CO$_2$, N$_2$). I'm trying to model the mixture of these molecules with SAFT equations of state, and I want to know if I'll have to add the effects of associating fluids (capable of forming H-bonds).

I could imagine that in a solid or dense fluid with a very particular structure, H-atoms in some hydrogen molecules might be so disposed that one was negatively polarized (closer to a hydride) and the other positively (closer to a proton) setting up conditions for H-bonding. I've never heard of this happening though.

Molecules that do hydrogen bonding are always fairly polar. Dihydrogen is not polar. So, no, whatever interactions are happening will not be hydrogen bonds.

@matt_black nonpolar molecules can form hydrogen bonds. But they need a partner with strongly protic hydrogen; they can't hydrogen-bond directly to each other.

I suspect not under normal conditions but maybe under ultra high pressure where solid hydrogen can be formed a hydrogen bond is very much possible.

@OscarLanzi Examples under non-extreme conditions? And would we normally classify any as "hydrogen bonds"?

You don't need ultra high pressure to solidify hydrogen. You just need to get it really cold. In practice the hydrogen ion clusters alluded in my answer are found in stellar nebulae, not in the solid state.

With methane, clathrate-hydrates are not extreme. They are accessible enough to impact deep sea oil and gas drilling, for example.

Something like hydrogen bonding with $\ce{H2}$ molecules appears to occur in hydrogen ion clusters. Clampitt and Gowland[1] study these clusters in connection with stellar nebulae and report that the $\ce{H_{15}^+}$ ion is especially stable; this likely has a $\ce{H3^+}$ core (which acts as the protic hydrogen) surrounded by six associated $\ce{H2}$ molecules.
Other nonpolar molecules, with lower ionization energies and greater polarizability, may form weak hydrogen bonds in more ordinary chemical environments. Such an interaction between methane and water, which may occur in methane clathrate-hydrates, is reported by Isaev[2].
All such hydrogen-bonding interactions with nonpolar molecules require a strongly protic hydrogen source with which the electrons of the nonpolar molecules can overlap. A simple mixture of low-polarity, nonprotic gases would not of itself generate any hydrogen bonding.

CLAMPITT, R., GOWLAND, L. (1969) "Clustering of Cold Hydrogen Gas on Protons". Nature 223, 815–816. https://doi.org/10.1038/223815a0

Isaev, A.N. (2016) "O–H···C hydrogen bond in the methane–water complex". Russ. J. Phys. Chem. 90, 1978–1985. https://doi.org/10.1134/S0036024416100150

By definition, no, they can't. According to IUPAC's definition of a hydrogen bond:$^1$

The hydrogen bond is an attractive interaction between a hydrogen atom
from a molecule or a molecular fragment X–H in which X is more
electronegative than H, and an atom or a group of atoms in the same or
a different molecule, in which there is evidence of bond formation.

That isn't to say that you could not get dispersion interactions forming due to the instantaneous changes in electron distributions, but they will probably be negligible. There is also the possibility of other types of non covalent bonding (e.g. halogen [F, Cl, Br,...], chalcogen [O, S, Se,...], pnicogen [N, P, As,...], tetrel [C, Si,...], etc. bonding),$^2$ but be careful when you use the term 'hydrogen bond'.

Arunan, E.; Desiraju, G. R.; Klein, R. A.; Sadlej, J.; Scheiner, S.; Alkorta, I.; Clary, D. C.; Crabtree, R. H.; Dannenberg, J. J.; Hobza, P.; et al. Definition of the hydrogen bond (IUPAC Recommendations 2011). Pure Appl. Chem. 2011, 83 (8), 1637-1641. DOI: 10.1351/PAC-REC-10-01-02

Alkorta, I.; Elguero, J.; Frontera, A. Not Only Hydrogen Bonds: Other Noncovalent Interactions. Crystals 2020, 10 (3), 180. DOI: 10.3390/cryst10030180.


text:
Can cooling rate affect the yield of crystals?
Cooling an acetanilide mixture with an ice bath has resulted to a heavier amount of acetanilide crystals than cooling the container slowly (setting it aside in the lab). I have gotten different explanations from this. Some say that increasing the cooling rate can include impurities which increases the mass of the crystals. But according to others, decreasing the cooling rate allows the crystal to grow which increases its overall yield.

The experimental results seem to refute one of hypothesis in this case, or at least confirm which of the phenomena is prevalent.

The two cases need not be a contradiction. For example, the first describes a slurry of amorphous solid and small cristallites with solvent adsorbed (occlusion)/incorporated - upon exposure to ambient air, the solvent eventually may evaporate off while the second seeks to use good conditions to seed, and then to grow good crystals.  For the later, the two sets of parameters can differ (see e.g., https://chemistry.stackexchange.com/questions/156251/why-is-it-so-difficult-to-crystallise-anything-in-a-glass-capillary) regardless if you aim for crystals easily filtered off, or an XRD analysis.

What is the quantitative difference between the yields when comparing both processes ? Is it a couple of percents ? More ? Less?


text:
Why are certain lattices compatible with only certain point groups and not all point groups?
I will limit this discussion to 2 dimensions for ease of intuition. My understanding of the 17 crystallographic plane groups is that these 17 groups represent all the possible symmetry groups of any 2d crystal structure we may think of (crystal structure=lattice + motif). Thus, if we think of an arbitrary periodic 2d crystal structure, its symmetry group will necessarily correspond to one of the 17 crystallographic plane groups. My problem is that I can think of numerous 2d crystal structures whose plane group is not one of the 17 crystallographic plane groups. Take for example the one shown below

In this example, we have combined a square lattice with a motif that has a 3 fold rotational symmetry. The resulting structure does not have any 3-fold rotational symmetry axis's nor does it have any of the characteristic 4-fold rotational symmetries present in a square lattice. But it still is a periodic repeating structure. It still has the same translational symmetry as the square lattice. It even has horizontal mirror lines. So its overall  symmetry group is the combination of a mirror line with a square lattice. Yet if we look at all 17 of the crystallographic plane groups below, we find that a single horizontal mirror plane may only be combined with a rectangular lattice (plane group pm in the image below).

Apparently, if we have a structure with only one mirror plane, the underlying lattice has to be a rectangular lattice. But this is absurd. I have a clear counterexample shown in first image of this post. This counterexample should then be an 18'th crystallographic plane group. But its not for some reason. Why?
I understand that when we create the 17 plane groups by combining point groups with bravais lattices, we only combine lattices with point groups when the lattice has at least as much symmetry as the point group. But why do we limit ourselves to only these combinations? Why do we not combine say a threefold rotation axis with a square lattice as shown above? Does nature somehow discriminate against these combinations? I guess my issues boil down to the following

What is the plane group of the structure shown above in fig 11.29 (if it has one)?
If it doesn't have a plane group, why do we not include its symmetry group in the list of all 2d plane groups?
Does nature ever allow crystals to exist if they do not have one of the 230 space groups or 17 plane groups?

What is the plane group of the structure shown above in fig 11.29 (if it has one)?

It is pm. In this plane group, you are free to choose the lattice parameters $a$ and $b$ (they can be different or the same), and the angle has to be 90 degrees. You can prove that the lattice is rectangular by combined translation and mirror operations.

If it doesn't have a plane group, why do we not include its symmetry group in the list of all 2d plane groups?

It does. Whenever you think you found a new plane group, you can probably shift the origin or take linear combinations of "your" cell axes to get to one that already exists.

Does nature ever allow crystals to exist if they do not have one of the 230 space groups or 17 plane groups?

It has more to do with math and geometry. If you define a crystal in the way it is defined, and assume space is Euclidean, you can enumerate the space and plane groups. You could do that slightly differently, but the way it is done, any crystal can be assigned to one of the plane groups or space groups in the international crystallographic tables.

In this example, we have combined a square lattice with a motif that has a 3 fold rotational symmetry. The resulting structure does not have any 3-fold rotational symmetry axis's nor does it have any of the characteristic 4-fold rotational symmetries present in a square lattice. But it still is a periodic repeating structure. It still has the same translational symmetry as the square lattice.

In the absence of four-fold symmetry, we don't call it a square lattice. It is a coincidence that the length of $a$ and $b$ are equal. If you look closely at your crystal, the packing in the vertical direction is tighter than in the horizontal direction, and it would be possible to push the motifs closer in the horizontal direction. This would then match your expectation that the lengths of $a$ and $b$ are different.

If your crystal had 4-fold symmetry, the packing in the vertical and horizontal direction would be equivalent. If for some reason the 4-fold symmetry breaks down (maybe instead of having four identical proteins coming together, I have pairs of two distinct proteins), the necessity for equal lengths would break down, too. So one comes with the other.

Apparently, if we have a structure with only one mirror plane, the underlying lattice has to be a rectangular lattice.

It does not have to be. A square lattice has $a = b$. A rectangular lattice may have   $a = b$ or not, because a square is a special case of a rectangle.

Thanks for the excellent response. You've pretty much cleared everything up for me. I now understand my textbook when it says "Note that here and throughout this book, in reference to symmetry, $\neq$ means need
not be equivalent while = means are required by symmetry to be equivalent.". In the case of structure belonging to pm, we have $a \neq b$ and so we are free to adjust $a$ and $b$ however we please without losing the mirror symmetry. We might even set them equal as in the case in my post.  Just to check my understanding though... (1/n)

(2/n)  Suppose we take any one of the 5 2d lattices (i.e a hexagonal plane lattice for example) and populate it with a motif consisting of an asymmetric capital "R". In this case, the plane group would be p1 regardless of which of the 5 2d lattices we choose to begin with correct?


text:
Does KMnO4 Oxidize Alpha-Beta Unsaturated Aldehydes?
I suspect that since $\ce{KMnO4}$ is a strong oxidizing agent it will react with the $\alpha,\beta$-unsaturated part of the carbonyl forming an epoxide. Are there any secondary reactions that one might have to worry about? Or how can I tell if my oxidizing agent will oxidize or not in a systematic way? Take as a hypothetical example, the oxidation with $\ce{KMnO4}$ of cinnamaldehyde.

I think you are very likely to get the carboxylic acid from the aldehyde

$\ce{KMnO4}$ will not produce an epoxide when acting on a double bond.

Basic hydrogen peroxide will do this, see https://www.tandfonline.com/doi/abs/10.1080/00304940309355863

As Waylander predicted, in aqueous solutions, the first reaction would have been the oxidation of aldehyde to corresponding carboxylic acid (in here, cinamaldehyde to cinnamic acid). When that happens, the resulting $\alpha,\beta$-carboxylic acid would undergo degradation reaction as explained by following example (Ref.1):

According to the authors:

Permanganate ion reacts rapidly with cinnamic acid in aqueous perchloric acid solutions to form an intermediate which appears yellow and which exhibits an absorption maximum at 415 nm. The intermediate then decomposes at a slower rate to give benzaldehyde and manganese(III) as products.

According to the results, the first reaction is clearly a second order for trans-cinnamic acid with $$k_2 = (1.56 \pm 0.03) \times 10^3 \ \pu{M-1 sec-1}$$. The decomposition of this intermediate is a somewhat slower first-order reaction with:
$$k_1 = (2.60 \pm 0.02) \times 10^{-2} \ \pu{sec-1}.$$
The nature of these reactions was further defined by a study of the kinetic isotope effects (Ref.1). When the solution of reaction products was made basic by the addition of sodium hydroxide, a brown precipitate of $\ce{MnO2}$ was indeed obtained, confirming that the intermediate decomposes to give benzaldehyde
and $\ce{Mn^{III}}$ as the initial products (since $\ce{Mn^{III}}$ is known to disproportionate to $\ce{Mn^{II}}$ and $\ce{MnO2}$ under alkaline conditions; Ref.2).
Note: It is recommended to also read Ref.3 for alkene oxidation by $\ce{KMnO4}$.
References:

Donald G. Lee and James R. Brownridge, "Oxidation of cinnamic acid by permanganate ion. Spectrophotometric detection of an intermediate," J. Am. Chem. Soc. 1973, 95(9), 3033–3034 (ODI: https://doi.org/10.1021/ja00790a060).
R. Stewart, In Oxidation in Organic Chemistry, Part A: A series of monographs,; Kenneth B. Wiberg, Ed., Academic Press: New York, N. Y., 1965, p 1—68 (ISBN: 0127480501).
Kenneth B. Wiberg and Richard D. Geer, "The Kinetics of the Permanganate Oxidation of Alkenes," J. Am. Chem. Soc. 1966, 88(24), 5827–5832 (ODI: https://doi.org/10.1021/ja00976a025).


text:
Oxygen tolerant RAFT polymerization using AIBN?
I want to do some RAFT polymerizations but don't have a nitrogen line at the moment. It seems that people have recently implemented oxygen scavenging strategies to enable open-air RAFT polymerization. In Ref. 1-4 and many similar references, people have used ascorbic acid or trithiocarbonate/blue light as the scavenger and transfer the scavenged electrons to photocatalysts to act as the initiator for  polymerization. My question is, can I add ascorbic acid + copper or ferrous iron to a reaction mixture to serve as oxygen scavengers and separately just use AIBN as a radical initiator instead of the photocatalysts?
(1)  Zhang, T.; Yeow J.; Boyer, C. A cocktail of vitamins for aqueous RAFT polymerization in an open-to-air microtiter plate. Polym. Chem. 2019, 10, 4643-4654. DOI 10.1039/C9PY00898E.
(2)  Yeow, J.; Chapman, R.; Xu, J.; Boyer, C. Oxygen tolerant photopolymerization for ultralow volumes. Polym. Chem. 2017, 8, 5012-5022. DOI 10.1039/C7PY00007C
(3) Shanmugam, S.; Xu, J.; Boyer, C. Aqueous RAFT Photopolymerization with Oxygen Tolerance. Macromolecules 2016, 49, 9345-9357. DOI 10.1021/acs.macromol.6b02060
(4)  Fu, Q.; Xie, K.; McKenzie, T. G.; Qiao, G. G. Trithiocarbonates as intrinsic photoredox catalysts and RAFT agents for oxygen tolerant controlled radical polymerization. Polym. Chem. 2017, 8, 1519-1526. DOI 10.1039/C6PY01994C


text:
Electron Affinity of Carbon and Silicon
Which has the greater affinity for an electron, Si or C and why? When thinking about the exception with F and Cl I would expect Si to have a greater electron affinity than C, but when I look at electron affinity values online I get mixed results. Some charts put C at -122kJ/mol and Si at -134kJ/mol which confirms my assumptions (https://images.app.goo.gl/6dpbc44DjnStpaLk9), but others place C at 153.9 and Si at 133.6 (https://college.cengage.com/chemistry/intro/zumdahl/intro_chemistry/5e/students/protected/periodictables/pt/pt/pt_ea2.html) (also not sure about the units for this chart, I would assume kJ/mol but it doesn't say). Which is correct and why?

153.9 and 133.6 could be the ionization energies of C and Si.

https://en.wikipedia.org/wiki/Electron_affinity_(data_page) is quite clear about that.

@Volpina thank you for explaining that to me! My professor actually teaches electron affinity as EA= -(ΔH), so I'm used to seeing the positive values and associating it with energy released. I was actually quite confused when I first looked up electron affinity values and saw the negative values since it's the opposite :'(.


text:
Acid dissociation of two polar liquids
I was thinking in a case where the solute and solvent are two different polar liquids and according to Does the acidity constant depend on the solvent? the constant of dissociation change with the solvent, but changing the amount of liquid of the solute and solvents change their roles so my question is: Do the two liquids influence each other and their constants of dissociations of both liquids change at the same time while you change the amount of liquids?

No, constants don't change, that's why we call them constants. Say, if you were to mix H2SO4 and HClO4, there will be some dissociation constants of both acids (different from those in water), and they won't depend on the relative amounts.

@IvanNeretin thanks for replying, so even if the constants of dissociation don't change, their activities change.

They do, and so does the degree of dissociation, but not the constants.

True TD constants would depend on temperature only, using activities of both conjugate pairs  $\ce{HA + B <=> A- + BH+}$. These activities may change a lot in different mixtures. But the point is, most dissociation constants are not true TD constants, but they implicitly involve simplifying assumptions with limited scope of applicability.

@IvanNeretin As you increase the fraction of solute beyond 50%, the role of solvent and solute (and with it the definition of the standard state) changes, so you would expect the constants to change as well (just like when you switch to a different physical state).


text:
Electron affinity of isotopes anomaly
According to the Wikipedia article the electron affinity of $\ce{^{2}H}$ is higher than the electron affinity of $\ce{^{1}H}$ however the electron affinity of $\ce{^{18}O}$ is lower than $\ce{^{16}O}$. How is this anomaly explained?

I'll be impressed if someone has a qualitative but accurate explanation for an effect measured at the sixth decimal place.

Yet it is still a anomaly.

Heavier nuclei have less spatial spread than light nuclei (in chemistry, we often ignore that because electrons have many times more). For these elements, that have the extra electron constrained in its orbital to add more repulsion to the outer shell, such as oxygen and fluorine, more spread in the position of their nucleus helps to lessen that repulsion. For these elements, that have less constrained outer shell (hydrogen and sulfur in Wikipedia's table), the  stabilization of extra electron at heavier nucleus comes from the reduced mean distance from the common center of mass of the nucleus and the electron.
That said, it is not obvious from looking at the mentioned table only. To know, which elements have more "pressure" in their electron shells, we need to also know a lot about ionic radii, for example.


text:
What is the origin of the spin selection rule?
Why is the change in total spin ∆S = 0 during an absorption of a photon, in contrast to the ∆L = ±1, where the photon transfers its angular momentum?
I am aware that spin-orbit coupling breaks this, I want to know why its a rule in the first place.

Photons have one unit of angular momentum only ($S= 1$), and on absorption or emission the angular momentum of the atom or molecule+photon is always conserved. For example singlet to triplet transfers occur but only if angular momentum from elsewhere (other than spin) is used, as in your example of spin-orbit coupling.

Thanks @porphyrin Do photons also have orbital angular momentum %l = 1%?  If they do, why does transferring this to the molecule not change their orbital angular momentum to 0?  If they don't, how can absorption of a photon cause a change the orbital angular momentum of the molecule?

yes photons have angular momentum, quantum numbers $S=1, m_s=\pm 1$  and yes it changes the angular momentum of the molecule so that the total in the molecule plus photon is the same as in the  excited molecule after the photon is destroyed by being absorbed. The angular momentum of the atom/molecule is changed by $\pm 1$ unit from whatever it was before. You see this clearly in a vibrational-rotational spectrum such as from HCl, where  only $J$ changes by $\pm 1$ are observed and $J'=0 \to J''=0$ is not observed.

Sorry, I'm a little confused.  You're saying photons have $S = 1$ (spin angular momentum), the change in spin orbital angular momentum is usually 0 during the absorption of a photon.  Do they have orbital $L = 1$ as well as  $S$ in order to get $\Delta L = 0$?   If they can freely re-allocate angular momentum between $L$ and $S$ to get the general change in $J$, then why is $\Delta S$ required to be 0?

I'm not sure I understand your question, but it is quite simple in outline. Photons have 1 unit of  angular momentum, however, the electron spin is not affected by this but in a dipole transition the atom/molecule orbital angular momentum has therefore to change by 1 unit, i.e. orbitals $s \to p,; p \to d$ etc for a dipole transition. Similarly in molecules but now you need to look at the point group to determine which symmetries are involved.

So if angular momentum is being transferred from the photon to the electron, but it does not affect the spin of the electron, does that mean that the photon's angular momentum is not spin angular momentum $S$, but orbital angular momentum $L$?

Well, its easier to think of the angular momentum of an orbital and so not get confused with electron spin's angular momentum. The photon has angular momentum, thats it, (not spin or orbital, just ang. mom. ) and this is transferred to the atom/molecule's orbital angular momentum so that overall ang. mom. is conserved.

This would make sense, however I have been reading recently (and I do not fully understand the explanations) about light carrying both spin and orbital angular momentum.  My limited understanding is that spin is rotation of polarisation vector (like what we use to measure chirality), while orbital is related to helicity of wavefronts.  Wikipedia has formulae, https://en.wikipedia.org/wiki/Spin_angular_momentum_of_light
but I still do not understand the $\vec{A}$ field, so they don't make sense to me.

What I get out of it, is that they are two properties, light can have both, and they're not the same.  I think spin is carried by all light (what if it's linearly polarised?) so I don't understand why spin angular momentum of photons turns into orbital of molecules.  Maybe angular momentum is angular momentum, and it doesn't matter where it comes from.  But if that's the case, then the selection rules wouldn't be that the transfer of angular momentum from photons can Only change molecular $L$, and must change $L$, and not $S$.  Right?

The electron's spin is separate from its orbital angular momentum, also sometimes also called 'spin'.  The true electron spin was discovered with the anomalous Zeeman effect and the Stern-Gerlach experiment. But the electron is a structureless elementary particle of no measurable size so its best to think of it having property which has equations that have the  same form as angular momentum. V. odd I know.

continued...The electron 'orbital angular momentum' is what is used in transitions as is its polarisation in electric dipole transitions. The integral $\int \psi_i,\mu,\psi_fd\tau$ has to be evaluated. $\mu$ is translation operator, $x,y,z$ in the point group, and $\psi$ the wavefunctions. This has to be non-zero if a transition is allowed. A really good text is 'Atomic Physics' by CJ Foot (publ. OUP) p35.

Are you saying that whatever type of angular momentum the photons have, $S$ or $L$, when they interact with an electron, because the interaction involves electron polarisation and the dipole operator this turns into $\Delta L$ for the electron?

effectively yes, the photon does not change the electron spin quantum number


text:
How is SO3H added to a benzene ring with two substituents present
If a benzene ring with an isopropyl group in the first position and bromine in the 3rd position was mixed with SO3 and H2SO4, where would the SO3H attach? Since Br is a deactivating agent that directs ortho/para and isopropyl is a weak activating agent that's also an ortho/para director, would it be in between the two groups in position 2 or in position 4?

In an EAS reaction the strongest activator will control the directing effects. This means that on a benzene ring that contains an isopropyl group and a bromine you can expect the isopropyl group to control the directing effects. In the sulfonation reaction that you are describing you must also take into account steric effects of the isopropyl group. Because the isopropyl group is rather bulky the group being added will add para in regards to the isopropyl group. This is also convenient because if the group is added para to the isopropyl group it will also be ortho the bromine.

According  to a Chem. Absts. summary of a patent, 1-bromo-3-(tert-butyl)benzene sulfonates para to the tert-butyl group. No need to worry about the bulk of the isopropyl group.


text:
Molecular orbital picture in conjugated molecules - double bonds
I am a not a chemistry student but a physics student. Nevertheless, I am quite familiar with molecular orbital theory and similar quantum chemistry concepts. However, I have problems understanding the ‚double bond‘ nomenclature, especially in conjugated molecules.
But let’s start simple: If we have a molecule like $C_2H_4$, there is a double bond between the two C atoms. This double bond is due to the overlap of the p orbitals and the overlap of the $sp_2$ orbitals. Until here, everything is clear for me.

$$$$
Now, let’s look at a conjugated molecule like 1,3 Butadiene:

There are also double bonds between every second carbon atom (structural formular in the top left). In the picture above (b) we see, that all p-orbitals of all carbon atoms overlap, producing four different molecular orbitals. In (a) we also see that there are $sp_2$ bonds between all carbon atoms.
My question is:
Why do we draw double bonds exactly between the two first and the last two C atoms in the structural formula? How does this follow out of the MO-picture (where we have four different $\pi$ type MO)? Why isn’t it possible to draw the double bond between the second and third carbon atom instead of between the first two, or even add an electron and draw a double bond between all carbon atoms (since all p-orbitals overlap)? Does it have something to do with the fact that only for the HOMO-orbital there is a node between the central two carbon atoms (where there is a also single bond drawn).
Pictures are taken from LibreTexts

This isn't MO picture it's VBT, and you don't consider resonance structures.

In general, structural drawings are not intended to capture the full electronic structure of a molecule. They are simply agreed upon representations that allow a knowledgeable person to discern the structure.

Oh these pictures are terrible! They try explaining one thing (the sigma framework) with Valence Bond Theory and another thing with Molecular Orbital Theory (the pi framework).  While these two theories are equivalent in their outcome, you cannot just mix them in the interim stages. For example: there cannot be a HOMO in VBT as there are no MO.

The simple pictures chemists draw are used for convenience not to correctly represent the MO pictures of bonds
Chemistry is complicated. The simple pictures chemists draw to explain the structures of molecules are a trade-off between describing the structure unambiguously and misrepresenting the details of the bonds involved. We could draw pictures giving a better view of the bonding, but they would often be complex, ambiguous or make other things harder. So, often, we draw simple pictures to maximise clarity while assuming that other knowledge will avoid misinterpretations of the real nature of the bonding.
One of the things we want to be easy is unambiguously counting electrons. For example the commonest way to draw benzene is this:

This is clearly a misrepresentation of the actual structure of benzene (where all the bonds are equivalent and the reality involves a bunch of molecular orbitals with delocalised electrons). But this picture does allow a chemist  to count the electrons (each line represents a notional two electron bond) which may be important and may help describe simple reaction mechanisms. Some chemists do the same drawing as a hexagon containing a circle to represent the fact that we know the electrons are delocalised, but this doesn't make it easy to count the number of bonding electrons accurately which can make explaining reaction mechanisms much harder.
A similar issue occurs with butadiene. We could choose a more "reliable" drawing of the molecule (or, as Karsten shows in his answer, several drawings) but this loses a lot in clarity without generating a lot more insight that helps in  most situations. We assume that chemists have enough knowledge about bonding to recognise the situations where a better understanding of MO theory and its implications is actually needed to understand a molecule or its reactions.
For many structures and many reactions the "simple" drawing plus some chemical knowledge is good enough for the task at hand. And we don't need to use extra complexity in what is drawn unless it is vital for adding extra insight to the structure or the reaction mechanism.

As stated in the comments, the Lewis structures are representations of a localized bond picture, and you need so-called mesomeric contributors to show delocalized bonding (some of which do show the double bond in the center):

You might ask why the charges, and not single electrons. Single electrons on carbon 1 and carbon 4 don't fit the properties of the molecule (it is not a diradical).
An empirical way of distinguishing double and single bonds is to look at bond lengths.

For butadiene, the formal C=C bond length is 1.338 Å, and the formal C-C bond length is 1.454 Å.

Source: N. C. Craig et  al.
So the structure in the middle is a better representation if a single contributor has to be picked.

[OP] Why isn’t it possible to draw the double bond between the second and third carbon atom instead of between the first two, or even add an electron and draw a double bond between all carbon atoms (since all p-orbitals overlap)?

You could add two electrons, and would think that you have a double bond between carbon two and three. However, this would mess with the pi-bonding of carbon 1 and 2 as well as carbon 3 and 4 (there are nodes there). In other examples, you do get "extra" bonding  by adding electrons (going from cyclopentadiene to cyclopentadienyl anion, for example).
Craig, N. C.; Groner, P.; McKean D. C. Equilibrium Structures for Butadiene and Ethylene:  Compelling Evidence for Π-Electron Delocalization in Butadiene. J. Phys. Chem. A 2006, 110, 7461-7469. DOI 10.1021/jp060695b

Thank you, this helps me quite a lot. But it is still not clear to me how I could determine if a single or double bond is present out of an ab-initio calculation like with MO theory or valence bond theory.

Well, it is neither. For benzene, the bond order is 1.5 for all carbon-carbon bonds. For butadiene, the pi bond order is 0.45 for the central bond 0.89 for the other two bonds according to one source and another.

@karsten if you do a Huckel-type calculation on the pi orbitals of benzene you discover that the total bond order in the ring is actually $1\frac23$ per linkage. Aromatic rings generally have a resonance stabilization equivalent to about one extra pi bond distributed around the ring.

@OscarLanzi I was wondering why for butadiene, it did not add up. That’s unexpected from a VB view, and also not captured in the Lewis structure.

Bond order is an entirely made up concept with arbitrary numbers. Without the proper reference frame the calculations are worse than guessing. Since sigma bonds are always stronger than pi bonds (choose your reference frame well) there could be an argument made, that the bond order of a sigma bond is actually 1.1 while an accompanying pi bond is 0.9; again be careful with the reference frame. The sigma bond in ethane is not the same as in ethene. And so it goes on and on and one…

@Martin By extension, single bond and double bond is also an entirely made up concept. We have a lot of those in chemistry: oxidation states, partial charges on atoms, electronegativity, empty orbitals, electron pushing. Sometimes they are helpful, though.


text:
Exact calculation of pH of diprotic acid plus strong base
Another post has an excellent "exact" calculation for the [H+] (or pH) of a generic diprotic acid dissociating in water.
Combining acid dissociation constants to determine pH of diprotic acid
I am interested in what modifications would be needed to do a similar "exact" calculation with the addition of a strong base to effectively create a buffer solution.
When I look at the equations, my only thought is to modify equation (5) by adding in some strong base "counter ion" to the charge balance equation.  For example, if NaOH was added to the diprotic acid, some of the acid would be neutralized by OH- to produce some initial conjugate base(s) in the mix.  So there would be Na+ to account for in the charge balance.
However, I don't think equation (4) would change, since the total amount of H2A, HA-, and A2- would not change -- just the ratios would change, which should fall out of the calculation.
I can't think of any other changes to make to the set of equations.  I proceeded to solve the new equations (with major help from the Python Sympy forum).  To test the equation, I assumed that the total "I" from equation 4 was 1.0 molar, and the initial concentration of NaOH and therefore the final concentration of Na+ was .5 molar.  For Ka1 and Ka2 acid dissociation constants I used the values suggested, Ka1 = 5.9e-2, Ka2 = 6.4e-5, and for water Kw = 1.0e-14.
Using the nroots() function in Python, an equilibrium value for [H+] of 0.04872 was obtained.
If the diprotic acid is approximated as a monoprotic acid, I think the equilibrium should represent a "half-equivalence" point buffer, where ideally pH = pKa, and [H+] = Ka1.  Comparing the [H+] of 0.049 with the Ka1 of .059 seems not close enough.  I think typically the contribution of the second dissociation to [H+] is small and would only add to the approximation and increase the difference.
I'm not confident that my changes to the original set of equations is valid for the scenario.  If anyone else is interested in these "exact solution" cases and this modification I'm interested in discussion.  The equations are quite complicated, ending up with a quartic polynomial, but I can try to provide them.  My primary interest is in what kind of modifications to the original formulation are valid for the "buffer" case.
For anyone interested, below is the quartic equation (to be set equal to zero) that includes $[Na^{+}{]}$ which essentially represents adding that much strong base, converting that much acid to the same amount of conjugate base, forming a buffer.  I'm not clear yet on what ranges of initial conditions it is valid for (such as adding more NaOH beyond the final equivalence point).
$\displaystyle - K_{a1} K_{a2} K_{w} + K_{a1} K_{a2} [H^{+}{]} [Na^{+}{]} + K_{a1} \left([H^{+}{]}\right)^{3} + K_{a1} \left([H^{+}{]}\right)^{2} [Na^{+}{]} + \left([H^{+}{]}\right)^{4} + \left([H^{+}{]}\right)^{3} [Na^{+}{]} + \left([H^{+}{]}\right)^{2} \left(- I K_{a1} + K_{a1} K_{a2} - K_{w}\right) + [H^{+}{]} \left(- 2 I K_{a1} K_{a2} - K_{a1} K_{w}\right)$
(as mentioned in a comment below, I believe I erroneously assumed the Henderson-Hasselbach equation would be accurate for this case, but in fact $K_{a1}$ is sufficiently large so the small-x approximation causes too much error, so $[H^{+}{]} = K_{a1}$ at half-equivalence does not hold accurately.)

Even for cases where it is hard to find the solution to a set of coupled equilibria, testing the answer is straightforward. Just plug in the equilibrium concentrations into the equilibrium constant expressions, and see how close their values are to the equilibrium constants. If they don't match, you know there is a problem (if they do, it might be luck, so try a different set of initial conditions).

For exact calculations, you should involve ion activity coefficients.

Thanks, those are great suggestions.  I think in this case my reasoning had a flaw.  The Hendelson-Hasselbach equation (expecting H+ to equal pKa at half-equivalence point) does rely on the "small x" approximation.  In this case, pKa is rather large, .059, which casts doubt on using H-H accurately.  I tried a simple quadratic solution, treating the acid as a monoprotic acid, and actually got a value for H+ in good agreement with the exact quartic solution, approximately .04855 .  I am more inclined to trust the quartic formulation now, though I'm not sure it will be valid after the final equiv

At such a high ionic strength, calculations without activity coefficients  are far from exact. Even the modified Debye-Hueckel equation   for activity coefficients has applicability limit I=0.1 mol/L.

I will look into activity coefficients, especially for understanding how to better match experimental results with these kinds of theoretical calculations.


text:
Is it possible to replace 3-mercaptopropionic acid with 2-mercaptoethanol to solubilize avian eggshell membrane to soluble eggshell membrane protein?
Is it possible to replace 3-mercaptopropionic acid with 2-mercaptoethanol to solubilize avian eggshell membrane (ESM) to soluble eggshell membrane protein (SEP)?
Most literature suggested the use of mercaptopropionic acid but I only have mercaptoethanol in the lab. This is because both of them works as reducing agent to cleave the disulphide bonds to solubilize the protein.
Will it works without denaturing the protein?

If you are breaking the disulfide bonds, you are denaturing the protein.

Mercaptoethanol is the reagent used as denaturing agent for protein thus you can use it for this purpose. The mechanism for denaturing proteins by mercaptoethanol is to disrupt the disulfide bonds found in the protein. Therefore, using it will denature the protein in concern (I think same result you would achieve by using 3-mercaptopropionic acid).
Any mercapto-group $(\ce{R-SH})$ would disrupt disulfide bond $(\ce{R-S-S-R'})$. For example, cysteine and glutathione (GSH) have been used in vitro to denature disulfide bonds (Ref.1). In a typical procedure:

The compounds were incubated at $\pu{10 mM}$ with $0.03$ and $\pu{0.2 mM}$ cysteine or $\pu{4 mM}$ GSH, in $\pu{100 mM}$ Tris buffer, $\mathrm{pH} \ 7.0$, containing 5% methanol at $\pu{37 ^\circ C}$. Aliquots were taken at $0, 1, 4,$ and $\pu{24 hours}$, and the samples were analyzed by liquid chromatography with tandem mass spectrometry (LC-MS/MS).

Also read Ref. 2 & 3 for more details.
References:

Donglu Zhang, Aimee Fourie-O’Donohue, Peter S. Dragovich, Thomas H. Pillow, Jack D. Sadowsky, Katherine R. Kozak, Robert T. Cass, Liling Liu, Yuzhong Deng, Yichin Liu, Cornelis E.C.A. Hop, and S. Cyrus Khojasteh, "Catalytic Cleavage of Disulfide Bonds in Small Molecules and Linkers of Antibody–Drug Conjugates," Drug Metabolism and Disposition 2019, 47(10), 1156-1163 (DOI: https://doi.org/10.1124/dmd.118.086132).
Oliver Smithies, "Disulfide-Bond Cleavage and Formation in Proteins," Science 1965, 150(3703), 1595-1598 (DOI: 10.1126/science.150.3703.159).
Navin Kumar D. Kella and John E. Kinsella, "Disulfide-Bond Cleavage and Formation in Proteins," Journal of Biochemical and Biophysical Methods 1985, 11(4-5), 251-263 (DOI: https://doi.org/10.1016/0165-022X(85)90007-7).


text:
Why is kerosene being used in PUREX (plutonium uranium reduction extraction) as a solvent?
I am searching for a reason, why organic solvent in PUREX consists of 30% tributyl phosphate (TBP) in kerosene. Why kerosene (or another hydrocarbon)? What properties does it have? What exactly does kerosene do in the organic phase?

The actual extraction agent in the Purex process is tributyl phosphate (TBP). It is one of the most versatile extraction agents for many inorganic ions and has been made the basis for a number of various processes.
However, because of its high viscosity and a density very similar to that of water, it is customary to use TBP as a solution in an inert solvent such as kerosene.


text:
Glycation of peptides (Maillard reaction) by reductive amination
I would like to link glucose to the primary amines in a peptide containing lysine. Can I just mix glucose and the peptide and add some cyanoborohydride to perform a reductive amination? As long as my peptides dont have any additional aldehydes it shouldnt polymerize, right?
Its difficult to get a confirmation that this will work from google because of the massive amount of literature on the in-vivo/in-pan reactions that takes place naturally

looks like you can  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2629553/


text:
Carbon monoxide question
When reading lots of articles about carbon monoxide poisoning cases I came across to it being referred as "silent killer" many times. Now, I know that carbon monoxide is a byproduct of an incomplete combustion with insufficient oxygen available in the air. Also that we as people only have a carbon dioxide (a product of the combustion) build-up alert system and not "lack of oxygen in the body" alert (unless you consider oxygen deprivation symptoms as such). So, my question is: how come people don't wake up when being poisoned by carbon monoxide if carbon dioxide is a main product and in an environment when oxygen is insufficient for carbon monoxide to be produced, carbon dioxide would still be in the space and not vented out?

That is a question rather for human physiology. Tolerance for CO2 is much higher than  poisonous level of CO.

CO binds more tightly to hemoglobin than oxygen does, so your internal transport of oxygen is reduced. Your CO2 alert is for buildup in your lungs, not in the incoming air. It takes very little CO to make a big difference by taking hemoglobin out of action compared to external CO2 setting off your alarms.

Our bodies are designed to live in caves and hunt woolly mammoths. CO poisoning was not a thing back then, so no alert for it was needed.

Current levels of carbon dioxide are about 400 ppm outside, and inside it can go up to ~2000 ppm (stuffy air, no health problems), while above ~40,000 ppm might be deadly. Carbon monoxide is deadly at ~200 ppm within hours, even with plenty of oxygen. So if 10% of the exhaust is carbon monoxide, it is a potentially fatal problem.

Part of the resolution of your question rests on the definition of "insufficient oxygen". Part of the question relates to the relative threshold for oxygen deprivation and CO poisoning.
A chemist using the term "insufficient oxygen" is referring to the process of combustion. You appear to assume it applies to the whole room. In real combustion reactions, it is very easy to have insufficient oxygen in part of the flame even when there is a good supply of oxygen to the fuel being burned.  Even well regulated heaters based on burning fuel often produce some CO even when the carbon dioxide dominates. This is why regulations insist on good ventilation of the exhaust gases from heaters.
The scenario you describe is neither common nor relevant to worries about CO. If a flame used up all the oxygen in a room and replaced it with carbon dioxide, you would surely be asphyxiated, but you would also notice  very early that the carbon dioxide level was worrying long before that point.
But what happens with badly ventilated heaters is not that. Leakage of CO into the room is the dominant problem not a lack of oxygen. CO is very poisonous because it forms a relatively stable compound with haemoglobin, blocking the blood's ability to move oxygen around the body. This happens at very low levels of CO, long before any external shortage of oxygen has happened. When town gas supplies consisted of hydrogen/CO mixtures (phased out in favour of natural gas which is mostly methane in the 1960s in the UK) "putting you head in the oven" was a favoured method of suicide. As was piping a car exhaust into the car (which became harder post catalytic converters which remove CO and some other gases).
The point is that a small level of CO in a space where there is plenty of oxygen is deadly. And the deadly levels are far lower that the levels of carbon dioxide that would trigger any physiological warnings in the body.


text:
How do the coefficients in an equilibrium question affect the shifts in position of equilibrium?
Let's say I have this equation:
$$A_{(g)}\leftrightharpoons 2B_{(g)}$$
According to Le Chatelier's principle, if I decrease the pressure, then more of B will be made in order to increase the pressure again, because the forward reaction produces more moles of gas.
However, what if my $K_C$ was very small, so that the concentration of A was much greater than B. A would have way more moles of gas than B, so how can you say that the forward reaction produces more moles of gas? I don't understand what the coefficients of the equation actually represent since you can't say that the ratio of concentration or partial pressure, A:B, is always 1:2.
Could you say that the coefficients just tell you which species is most affected by a change?

@Poutnik Can I say, in general, that the larger the coefficient in front of a species is, the more it will be affected by an imposed change (e.g. change in pressure or dilution), due to the coefficient becoming a power in Kc calculations?

Yes, I only really understood it then. I wanted to check my understanding. Sorry if this is not how Stack Exchange works, I'm not familiar with the community yet.

I don't have enough reputation to comment on the other post you wrote, so I will ask here, if that's okay. So, if I have $3A \leftrightharpoons 2B$ then would the frequency be expressed as $k[A][A][A]=k[A]^3$? I get that k represents that rate is proportional to [A], and the second [A] represents how the number of other 'skaters' also affects rate. So what does the third [A] represent?

some details can be found here that shows what happens when the pressure changes in terms of the amount of A and B https://chemistry.stackexchange.com/questions/98770/what-is-the-relationship-between-the-constant-of-equilibrium-kc-and-the-positi/98785#98785

The reaction stoichiometric coefficients do not say which component is most affected by the change, but rather which component pressure/concentration has the biggest impact on the reaction quotient(*)
With temperature change of the equilibrium constant, the component with higher order of power have equlibrium presure/concentration less  affected, as it is boosted by the exponent.
The reaction stoichiometric coefficients say nothing about the ratio of pressures nor concentrations of $\ce{A}$ and $\ce{B}$. They just say 1 molecule of $\ce{A}$ forms 2 molecules of $\ce{B}$ and vice versa.
But, they determine the exponents in the expression for $K_c$ or $K_p$.
$$K_c = \frac{{[\ce{B}]}^2}{[\ce{A}]} \tag{1}$$
$$K_p = \frac{({p_{\ce{B}})}^2}{p_{\ce{A}}} \tag{2}$$
Increasing of pressure causes the right side becoming bigger than $K_p$ (and $K_c$ for gaseous phase) and some $\ce{B}$ reacts to form $\ce{A}$ to make them equal again. And vice versa if the pressure decreases.
Imagine two times bigger pressure concentration means four times more frequent $\ce{B-B}$ collisions and four times faster $\ce{A}$ production While $\ce{B}$ is produced just two times faster.

Note that the value of $K_c$ or $K_p$ is irrelevant here, it is covered by the proportionality constants for reaction rates. As the fraction of collisions leading to reaction varies greatly.
See the ice-skating analogy and another question Why are the stoichiometric coefficients the powers in the rate law?

REferring to the linked ice-skating analogy and further elaborating it:
The 3rd $[\ce{A}]$ represents another skater, colliding with a heap.
The frequency of the collisions of a 2-skater heap and another skater is $$k_3[\ce{H}][\ce{S}] \tag{3}$$
The "concentration" of heaps is $$[\ce{H}]=k_1[\ce{S}][\ce{S}], \tag{4}$$
It gives together:
$$k_3k_1[\ce{S}][\ce{S}][\ce{S}] = k_4[\ce{S}]^3 \tag{5}$$
So, the frequency of collisions of skaters with a 2-skater heap is proportional to the third power of skater "concentration".
But better not to interpret how to divide 3 skaters between the same 2 parts.

Be aware of the note in the other post that it is valid for elementary reactions. Complex reactions with various intermediates have more complex relations in reaction kinetics. Bringing the terms from the equilibrium constants to rate equations is not generally valid, as they do not generally describe the opposite directions of the same elementary reaction.

(*) - the expression for reaction equilibrium constant, but with current values for the reaction, being  generally not in equilibrium


text:
Why do H and Li bond so differently with a hydroxide?
I am sure I am asking a very basic question but before coming here, I have made multiple attempts to find an answer elsewhere. Or maybe I am simply asking the wrong question, in some sense. If someone could nudge me in the right direction, I would be infinitely grateful to them. Here is what I am trying to understand:
A hydrogen cation added to a hydroxide ion forms water, which is neutral. A lithium cation added to a hydroxide forms LiOH, which is a strong base completely dissociating in a solution. Now, I understand why LiOH should be more alkaline than H2O. What I cannot understand is why it is so much more alkaline? Intuitively, one might expect a more gradual transition going from H to K. After all, H and Li are neighboring elements from the same group, with the same structure of the electron outer shell. Moreover, H in H2O is locked in a covalent bond whereas Li in LiOH is locked in an ionic bond. Ionic bonds are generally considered to be (much) stronger. Why is it not so in this case?

Well, hydrogen was sometimes also put over chlorine in the Table. In this context it could be appropriate.

The chemical elements of the same group have similar properties except H , H doesnt have the properties of alkali metals so a H+ added to a OH- (hydroxide anion) doesnt have any reason to be similar to the addition of a alkali metal to a OH-.

Notice the ionisation energies of H 1312 kJ/mol , Li 520 kJ/mol, Na 496 kJ/mol ( see https://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements )

https://chemistry.stackexchange.com/questions/129008/why-is-lithium-oxide-so-different-from-water

I dont understand the downvote,am I saying something wrong?

You're basically saying it's different because it's different.

https://chemistry.stackexchange.com/questions/46882/do-bare-protons-exist-even-transiently-in-aqueous-solution may be of use because H+ would be a bare proton, while Li cation is not nearly as bad (it still coordinate strongly - not really bare).

Yes but I say why H is different from the alkali metals.

So basically you are asking why H and Li are more different than He and Ne, right? H and He are the only period 1 elements, so there is little you can compare them to otherwise. You can describe hydrogen atoms as having one electron in the outer shell (like other group 1), as having a half-filled outer shell (like carbon and the rest of group 14) or being one electron away from a filled shell (like fluorine and the rest of group 17). Also, hydrogen is considered a non-metal and lithium is a no-doubt-about-it metal. Other groups have non-metals and metals, but with metalloids in between.

I want to say a big thank you to all who responded. I guess that the answers suggested by Mithoron and Poutnik are the closest to what I was looking for. Still, what makes the ionic bond in LiOH so much weaker than the covalent bond in H2O? So far as I understand, the distance between Li and O in LiOH molecule is only about 1.6x greater than the distance between H and O in H2O. So the difference in electrostatic interaction should not be that big, on the face of it. There must be something else that I am missing. Is it the shielding by the inner electrons of 1s shell in Li that does the trick?

@Volpina Aim to provide an answer that explains in more depth. Why is H unique? It helps also to cite authoritative sources if just providing an answer that summarizes a larger body of work.

If I understand your question correctly, it sounds like you are interpreting one of the hydrogens in water as being similar to the alkali metals. In other words, you expect to see a similar change as in LiOH and KOH.
Well the reason you don't is because hydrogen is a non metal. It is significantly more electronegative than the alkali metals.
Therefore, H2O is not really hydrogen hydroxide. It's just water. Also, when water dissociates in water, it forms hydroxide ions but also protons. So, concentration of both is equal and pH is neutral. As a result, water is not alkaline.
Meanwhile, LiOH has a metal ionically bound to a covalently bound hydroxide ion.  When it enters water, the strong hydrogen bonds rip apart the covalent bond between lithium and the hydroxide ion. The positively charged lithium ion is hydrated by water molecules with the oxygen end facing the lithium. And vice versa for hydroxide.
As you probably noticed, it does not dissociate into protons and a hydroxide ion. So it is alkali.
The reason hydrogen doesn't fit into the mould is because it is a non metal. And besides, the logic kinda breaks down since pH is a measure of how many protons are present relative to hydroxide ions.


text:
Which reaction would take place if we sent electric current IN a voltaic cell, electrolysing it?
I know how the voltaic cell normally works, producing current. It is not what I'm asking about. I know that normally, the Zinc electrode slowly dissolves (it is anode, $\ce{Zn -2e^- -> Zn^{2+}}$), passing electrodes to the Copper electrode at which $ H^+ $ ions from the acid electrolyte get reduced to $ H_2 $ (it is cathode, $\ce{2H^+ +2e^- -> H_2}$) Thereby creating electric current.
Now my question is, what if we passed current from an external voltage source to that cell?
Not in the same direction of current, like an in series battery would be connected, but in the opposite direction?
By doing this, I hope to block the flow of electrons from the Zinc electrode (and to get it to stop dissolving?) Then reduce some cations, like $ H^+ $ or even $ Zn^{2+} $ after I run out of $ H^+ $, which is more easily reduced?
And oxidize the anion of the acid of the electrolyte on the copper electrode?
Are any of my ideas correct?
I am sorry if this is a bad question, I am a beginner.
Please don't hate me, it's my first time posting here and I tried googling.

No question is a bad question as long as there is some effort. Ignore negative down votes too. This is habitual. I think you are not searching the right terms. Search charging and discharging a battery for an overall concept. Also search null point potentiometry. It does quite the same thing which you are thinking. If you apply an external potential to a voltaic cell in the opposite direction, the cell reaction will stop and depending on the external applied voltage, that voltaic cell can become an electrolytic cell.

@AChem Thanks. Do I understand correctly though, that the voltaic cell cannot be recharged, as the first cations to be reduced would be H+, so during the "recharging" attempt they would all form hydrogen gas? And that there will be no acid as electrolyte left then.

Can you add the chemical equations of the cell you are thinking about? What are the half cells made of? The voltaic cell can be recharged if it is rechargeable.

@Achem I meant the classic battery Volta made. Zn and Cu electrodes, some acid as electrolyte (say dilute H2SO4, but I saw this battery being made even with vinegar). Reactions: Zn -2e -> Zn2+ and 2H+ + 2e -> H2

Petr L, No that is not a classical cell. A standard cell is one where the metal is in contact with its ions in solution.

@Achem Have you never heard of the original battery Alessandro Volta has made?  The Voltaic pile? You're talking about Daniell cell, which was made because Voltas original battery bubbled hydrogen gas

Petr L, Yes, I have read about Voltaic pile for ages. What I was trying to tell you is that Voltaic pile is not amenable to Nernst equation treatment. Also, please write the half cell reactions you are thinking about in your original question.

@Achem Thanks. When researching the topic of electrolysis, I came across explanations explaining products formed using reduction potentials. (for example NaCl electrolysis with Fe electrodes, Fe in anode being oxidized instead of Cl-). Some chemists even used the Nernst equation to calculate potentials for non standard concentrations. Could I please ask, why is Nernst equation not applicable at all?

L, Nernst equation is applied when a metal is in contact with its own ions (as far as I know). For example, Cu electrode in CuSO4 solution, Zn electrode in ZnSO4 solution

@Achem Thanks for the explanation. What equation would you use instead?

Petr L., I do not know of another equations besides the Nernst one.

$\ce{Zn - Cu}$ is not much used these days.The common Leclanché  $\ce{Zn - C}$ "flashlight" cell is not rechargeable, for the reason you give.
However, alkaline cell chemistry, such as $\ce{Zn - Mn}$, has less outgassing of $\ce{H2}$ during reverse current, and there are rechargeable $\ce{Zn - Mn}$ cells. To further reduce $\ce{H2}$ production, or to recycle it back to water,

"The cathode has a catalyst to recombine any hydrogen that forms..."
"Zinc oxide is added to the cathode mix to,... on charge,... form oxygen."

So, at useful charging voltage, both recharging (i.e., plating zinc metal back) and evolution of hydrogen take place. However, one branch of that reaction can be made to be favored over the other.

Very interesting, thanks! How does zinc oxide being added help? EDIT: I see that it can be dissolved in bases, but why does it take priority over reducing H+?


text:
Hydrate forms causing complications with gravimetric analysis of ferric phosphate
I'm planning to determine phosphate concentrations in a solution via adding Iron(III) and weighing the precipitated ferric phosphate.
However, I read that ferric phosphate has a dihydrous form, which is slightly soluble. Furthermore, temperatures of over 200 degrees C are required to dehydrate this.
This is inconvenient for me, and I'm wondering if there is a way to correct for the presence of the dihydrate which means not all the phosphate will be precipitated, as well as the fact that the precipitate will be a mixture of the anhydrous and dihydrous forms. Another method to dehydrate it would work, or maybe a table of what proportions the anhydrous and dihydrous forms exist in under different conditions, so I can just factor it in.

Gravimetry should be used only with established methods, verified over the classical time of analytical chemistry or truly well investigated later. The fact a compound has low solubility should not be enough. Furthermore, If heating is inconvenient, gravimetry should be rather rejected. Consider colorimetry via molybden blue, that is established method to get phosphate content.

As Poutnik suggested, you shouldn't use unestablished methods for gravimetric analysis for analytic determination. The Ref.1 have showed one of the original published method for phosphate analysis by gravimetric determination. Accordingly to Ref.2, which developed this method for undergraduate experiment:

This method consists of first converting all the phosphorus containing species in the sample to soluble orthophosphate $(\ce{PO4^{3‒}})$ ion by oxidation and hydrolysis in acid solution. Then an acidic quimociac reagent is added to the prepared orthophosphate sample and a bright yellow precipitate forms. The resulting
precipitate is filtered, dried and weighed. Precipitation follows the reaction:
$$\ce{H3PO4 + 12H2MoO4 + 3C9H7N -> (C9H7N).3H3PO4.12MoO3.H2O + 11H2O}$$
When the precipitate is dried, the water of hydration is removed, leaving a stable, anhydrous, yellow product with a molar mass of $\pu{2213 g mol-1}$ (Ref.3).

Since resultant product is a fairly large molecule, the accuracy of the method is well established.
References:

H. N. Wilson, "The accurate determination of “phosphoric anhydride” by means of quinoline phosphomolybdate," Analyst 1951, 76(899), 65-76 (ODI: DOI https://doi.org/10.1039/AN9517600065).
Lee Alan Shaver, "Determination of Phosphates by the Gravimetric Quimociac Technique," J. Chem. Educ. 2008, 85(8), 1097-1098 (ODI: https://doi.org/10.1021/ed085p1097).
Wesley W. Wendlandt and William M. Hoffman, "Thermal Properties of Quinolinium Phosphomolybdate," Anal. Chem. 1960, 32(8), 1011–1012 (DOI: https://doi.org/10.1021/ac60164a035).


text:
Why does potassium bifluoride exist whereas bichloride does not?
Why does the reaction \eqref{rxn:R1Q} take place, but the reaction \eqref{rxn:R2Q} does not?
$$
\begin{align}
\ce{HF(aq) + KF(aq) &-> KHF2(aq)}\label{rxn:R1Q}\tag{R1} \\
\ce{HCl(aq) + KCl(aq) &-> KHCl2}\label{rxn:R2Q}\tag{R2}
\end{align}
$$
I found that $\ce{[HF2]-}$ forms due to hydrogen bonding among $\ce{HF}$ molecules. How exactly is the hydrogen bonding responsible for the formation of bifluoride ions?

Think of how electronegativity might play a role in making fluorine an outlier among the halogen group.

Related: Is potassium bifluoride an ionic or a covalent compound?

https://chemistry.stackexchange.com/a/21800/9961

The bichloride ion $\ce{HCl2^-}$ does exist, but it requires other environment than usual solutions in water. E.g. solution of $\ce{[N(CH3)4]Cl}$ salt and undissociated $\ce{HCl(solv)}$ in nitrobenzene solvent, as mentioned in the ACS link of the Oscar's answer. It is the scenario that makes the partial hydrogen charge in $\ce{H^{(+)}-Cl{(-)}}$ quite favourable for chloride anions.

For water solutions scenario, the key is the presence of undissociated $\ce{HF}$ and the ability of $\ce{F-}$  to make hydrogen bond with $\ce{HF}$. Neither is available for $\ce{HCl}$ and $\ce{Cl-}$. There are just very minor traces of undissociated $\ce{HCl}$ in water and $\ce{Cl-}$ would not have bound to it even if it had been there.
There are no $\ce{HCl}$ or $\ce{KCl}$ in water as molecules. There are hydrated ions $\ce{H+(aq)},$ $\ce{K+(aq)}$ and $\ce{Cl-(aq)}.$ Hydrated molecules $\ce{HF(aq)}$ are there in hydrofluoric acid or in  acidified fluoride solutions, reacting:
$$\ce{F-(aq) + HF(aq) <=> HF2^-(aq)}\tag{R1}$$
Hydrofluoric acid is weak mainly because of forming the stable ionic pair, so ion $\ce{H3O+(aq)}$ is mostly not free:
$$\ce{HF(aq) + H2O(l) <=> H3\overset{+}{O}\bond{...}F-(aq) <=> H3O+(aq) + F-(aq)}\tag{R2}$$
The structure of $\ce{HF2-}$ is $\ce{[F\bond{...}H-F]-}.$ There is also $\ce{[H-F\bond{...}H]+}$ in concentrated hydrofluoric acid or liquid  hydrogen fluoride:
$$\ce{3 HF <=> H2F+ + HF2^-}\tag{R3}$$
Liquid hydrogen fluoride has such a high boiling point due $\ce{HF}$ chain linked by hydrogen bonds:
$$\ce{[\bond{...}H-F\bond{...}H-F\bond{...}H-F\bond{...}H-F\bond{...}]}$$

I'm guessing that, more generally, true $\ce{MHX2}$ compounds with well-defined $\ce{HX_{2}^{-}}$ anions can exist for a number of halogens and metals, but only in non-aqueous conditions at low temperatures and/or high pressures. It's interesting to consider where the decomposition boundaries lie.

@NicolauSakerNeto Yes, I suppose that is true. I am aware that many of statements done in context of chemistry have limited scope of validity. At extreme enough environment, they are often may not valid any more. Typically for low T solid matrices or interstellar space.

@andselisk when + is used with overset (I agree with it). should not be - at F in overset too? Like visually more consistent.

@Poutnik Matter of taste, although the top right of F would be the preferred position, an overset is an acceptable one (https://doi.org/10.1351/pac200880020277, pp. 372–374). Besides, you have superscripts in the same equation right next to the drawn formula which sort of undermines the idea of consistent macro for the charge placement here.

The authors from the 1954 paper were able to isolate the solid salts as well as  the nonaqueous solution species, but again this would be without water or any other good B-L base.

The bichloride ion, including isolated salts, has in fact been known since 1954 [1]; but it appears to require relatively bulky counterions whereas bifluoride has a broader range of stable salts.
To form a complex ion of the type $\ce{HX2^-}$ the extra bond strength derived from a three-center four-electron bond (in the complex ion) must exceed the more favorable electrostatic attraction of the counterion to a more compact monatomic halide ion or solvent dipole (which favors dissociation into $\ce{X^- + HX}$), a situation similar to that for homoatomic trihalide ions $\ce{X3^-}$. But unlike homoatomic trihalide ions, $\ce{HX2^-}$ has the central atom fixed at hydrogen which favors strong covalent bonding with lighter, more compact atoms. Thereby $\ce{HF_2^-}$ is more favored than its heavier halide ion counterparts; whereas (at least in water solution and isolated compounds) triiodide ion is favored over lighter homoatomic halogen counterparts.
Reference

Harry F. Herbrandson, Richard T. Dickerson Jr., and Julius Weinstein (11954). "The Bichloride Ion". J. Am. Chem. Soc. 76(15), 4046.
https://doi.org/10.1021/ja01644a066


text:
Colour properties of nickel complexes
According to Crystal Field theory (CFT), complexes with stronger ligands must absorb light with higher frequency hence would transmit corresponding complementary colour.
The colour of $(\ce{[Ni(H2O)2(en)2]^2+})$  blue - purple and colour of $(\ce{[Ni(H2O)6]^2+})$  is green, whereas the colour of $(\ce{[Ni(en)3]^2+})$  purple. (Where $\ce{en ->}$ ethylene diamine.) (According to NCERT chemistry class 12)

But the above example goes against the CFT statement. Any plausible explanation available?

There is no emission of color. It is just transmitted light minus the absorbed portion of the spectrum. I hope the teacher is not explaining it this way.

The absorption spectrum of each would have much higher experimental value.For now,one can just estimate what wavelengths are absorbed and how much.

Colour in complicated. As are the absorption lines in molecules. There is no simple relationship between the overall colour and the energy or frequency of absorption bands (not least because there are often multiple bands). Same problem with the narrow emission lines on flame tests: many lines contribute to colour meaning overall judgements about energy levels based on perceived colour are impossible.

OP's argument:

According to Crystal Field Theory(CFT), complexes with stronger ligands must absorb light with higher frequency hence would transmit corresponding complementary color.

First of all, the different colors we see in these solutions are not emission colors. These solutions do not emit as AChem pointed out in his comment. As he correctly put,
the colors are due to the transmitted light minus the absorbed portion of the spectrum. On other words, The color you see in s solution of a transition metal compound is due to the wavelengths of the visible light that isn't absorbed by the $\mathrm{3d}$ electronic transitions of the metal. For example, nickel(II) aqueous complexes $(\ce{[Ni(H2O)6]^2+})$ often absorb strongly in the red region of the visible spectrum, so the resulting color observed is green. That means the complexes with weaker ligands such as $\ce{H2O}$ absorb light with lower frequency (larger wavelengths) such as color red (see a representation of a color wheel below):

That means, color of the solution is light blue $(\ce{[Ni(H2O)4(en)]^2+})$ means it absorbs orange color, blue $(\ce{[Ni(H2O)2(en)2]^2+})$ means it absorbs orange-yellow color, and pink (to me it looks pink, not purple) means $(\ce{[Ni(en)3]^2+})$ it absorbs yellow-green color (if it is purple, it still absorbs yellow light). If you look at these absorbed light, you'd see frequencies of yellow-green (or yellow) $\gt$ orange-yellow $\gt$ orange $\gt$ red, a good agreement with your statement.
Note: According to Ref.1, the $\ce{Ni^2+}$ complexes exhibit a broad band in the region $\pu{15000-23000 cm-1}$ with a second band in the range of $\pu{23000-27000 cm-1}$. These bands are assigned to the transitions $\ce{^1A_{1g} -> ^1A_{2g}}$ and $\ce{^1A_{1g} -> ^1B_{1g}}$. Also, color change due to number of ethylenediamine molecules $(\ce{H2N-CH2CH2-NH2})$ is published recently as an undergraduate experiment.

References:

Sangamesh A. Patil and Vasant H. Kulkarni, “Complexes of nickel(II) with ethylenediamine and 1,3-diaminopropane derivatives of 2,2′-dihydroxychalkones,” Inorganica Chimica Acta 1983, 73, 125-129 (ODI: https://doi.org/10.1016/S0020-1693(00)90836-3).
Mauro Ravera, Alessandro Nucera, and Elisabetta Gabano, “Freshening up Old Methods for New Students: A Colorful Laboratory Experiment to Measure the Formation Constants of Ni(II) Complexes Containing Ethane-1,2-Diamine,” J. Chem. Educ. 2022, 99(3), 1473–1478 (ODI: https://doi.org/10.1021/acs.jchemed.1c01186).


text:
The need for accurcy in constrained molecular dynamics simulations
In constrained molecular dynamics a system of differential algebraic equations
$$\begin{align} \dot{q} &= v, \\ M\dot{v} & = f(q) - G(q)^T \lambda, \\ g(q) &= 0 \end{align} $$
is solved. Here $(q,v)$ is the position and velocity of all particles, $M$ is a nonsingular diagonal mass matrix, $f$ represents the real forces acting on the particles, $g$ is the (holonomic) constraint function, $G$ is the Jacobian of the constraint function and $\lambda$ is a vector of Lagrange multipliers. In computational chemistry, the standard algorithm for this problem is the celebrated SHAKE algorithm:
$$\begin{align}
  v_{k+1/2} &=  v_{k-1/2} + h  M^{-1} \left(  f( q_k) -  G( q_k)^T \lambda_k \right) \, , \\
   q_{k+1} &=  q_k + h  v_{k + 1/2} \, , \\
   g(q_{k+1}) &= 0
\end{align}
$$
The final equation is a nonlinear equation for the vector $\lambda_k$ of Lagrange multipliers. It is typically solved using the nonlinear Gauss-Seidel method which converges linearly under certain mild conditions. Over the years, there have been several attempt to replace the Gauss-Seidel with quasi-Newton methods that require fewer, but more expensive iterations.
This question concerns the need for accuracy when solving the nonlinear equation
$$
g(q_{k+1}(\lambda_k)) = 0$$
with respect to the Lagrange multiplier $\lambda_k$.
The SHAKE algorithm has several nice properties that all hinge on the constraint equation being solved exactly. These properties are lost when we solve the constraint equation with a tolerance that is greater than zero, and quality of the simulation tends to degrade as we increase the tolerance. In the absence of detailed knowledge of the constraint function $g$, the only really "safe" value for the tolerance is the unit roundoff used by the computer. In particular, we lose the second order accuracy of the method and we loose the ability of compute reliable error estimates using Richardson extrapolation when the tolerance is not tiny.
The SHAKE algorithm has excellent conservation of energy when we use a tiny tolerance, but the total mechanical energy can increase or decrease when the tolerance is not tiny. Moreover, if the tolerance is too large, then it is possible to venture so far from the constraint surface, that the constraint solver is unable to find a solution and the simulation fails. The statements in this paragraph can easily be demonstrated in the case a simple pendulum, i.e., a single constraint.
Mathematically, errors associated with the Lagrange multiplier are equivalent to an external force field. Hence, if the objective is to simulate an isolated system, then we can hardly accept anything but the smallest possible tolerance that can be delivered by the hardware.
I would like to know a few numbers that can be reliably computed from constrained simulations that use a large tolerance when solving the constraint equations. I am primarily interested in numbers that have real life applications such as the development of medicine and vaccines.
Similarly, I would like to know a few numbers that cannot be trusted unless the underlying simulation are executed using a tiny tolerance. Again, I am primarily interested in numbers that have real life applications.
I have a fair understanding of differential equations, numerical analysis, numerical linear algebra and parallel computing, i.e., the skills needed to produce numbers reliably and quickly. On the other hand, I know very little about chemistry, so I may need more basic information than is customary here.

There isnt such physical process where energy isnt conserved because according to the laws of thermodynamics any change in the internal energy of a system requires that system not to be isolated.

But not all MD is done on an isolated system! Calculations at constant temperature couple the system to a heat bath, and thus while the temperature is conserved the energy of the system being studied is not  - rather the energy of the extended system is what is conserved. Constant temperature MD is very common.

The OP says:"I would like your help in assembling a short list of "relevant" calculations that can be performed on the basis of MD simulations, but only if the total energy of the simulations is conserved."


text:
Number of significant figures to use in reporting the concentration of the titrand in acid-base titration
If you perform an acid-base titration yourself, how do you determine the number of significant figures to use when reporting on the concentration of the titrand?
Assume you perform seven titration trials and the titrand concentration varies between $\pu{0.495 mol l^-1}$ and $\pu{0.505 mol l^-1}.$ You use a $\pu{50 ml}$ burette, class A glassware, and you know the concentration of the titrant with many significant figures.
It is incorrect to say that the concentration of the titrand is $c = \pu{(0.500 ± 0.005) mol l^-1}$ since more trials would increase the interval of values and thus increase the uncertainty in the experiment, which is not true.

You said that you have done seven titrations, i.e., you have the set of concentrations $\{c_1, c_2, \ldots, c_7\}$. I offer you the following standard statistical approach to inform your concentration and know how many significant figures to use:

Calculate the mean value of the concentration given the set of your concentration values:
$$\mu = \sum_{k = 1}^N c_k \tag{1}$$
Calculate the standard deviation $(N = 7$ in your case):
$$s = \sqrt{\frac{\displaystyle\sum_{k = 1}^N(c_k - \mu)^2}{N - 1}} \tag{2}$$
Inform your concentration given by the confidence interval
$$c = \mu \pm \Delta c = \mu \pm t_{N - 1}^\alpha \frac{s}{\sqrt{N}}, \tag{3}$$
where $t_{N - 1}^\alpha$ is the $t$-value of the Student’s probability distribution with a confidence interval of $\alpha$ with $N - 1$ degrees of freedom. In your case, this will correspond to $N - 1 = 6$, so looking up in the table you will have to choose within the corresponding row (data source: Harris’ Quantitative Chemical Analysis, Table 4-4. Values of Student’s $t$ [1, p. 72]):

$$
\begin{array}{rrrrrrrr}
\hline
N - 1 & 50~\% & 90~\% &  95~\% &  98~\% &  99~\% & 99.5~\% & 99.9~\% \\
\hline
 1    & 1.000 & 6.314 & 12.706 & 31.821 & 63.656 & 127.321 & 636.578 \\
 2    & 0.816 & 2.920 &  4.303 &  6.965 &  9.925 &  14.089 &  31.598 \\
 3    & 0.765 & 2.353 &  3.182 &  4.541 &  5.841 &   7.453 &  12.924 \\
 4    & 0.741 & 2.132 &  2.776 &  3.747 &  4.604 &   5.598 &   8.610 \\
 5    & 0.727 & 2.015 &  2.571 &  3.365 &  4.032 &   4.773 &   6.869 \\
\to 6    & 0.718 & 1.943 &  2.447 &  3.143 &  3.707 &   4.317 &   5.959 \\
 7    & 0.711 & 1.895 &  2.365 &  2.998 &  3.500 &   4.029 &   5.408 \\
 8    & 0.706 & 1.860 &  2.306 &  2.896 &  3.355 &   3.832 &   5.041 \\
 9    & 0.703 & 1.833 &  2.262 &  2.821 &  3.250 &   3.690 &   4.781 \\
10    & 0.700 & 1.812 &  2.228 &  2.764 &  3.169 &   3.581 &   4.587 \\
\hline
\end{array}
$$
Now we can answer your question. Say, your data with some chosen $\alpha$ gave you $\mu = \pu{0.50145 M},$ and $\Delta c = \pu{0.00627 M}.$ Then, the position of the first non-zero decimal number in $\Delta c$ informs you how many you must use:
$$c = \pu{(0.501 \pm 0.006) M}\tag{4}$$
Always inform $\Delta c$ with only one number, never put something like
$$c = \pu{(0.50145 \pm 0.00627) M}\tag{5}$$
Some people like rounding up the absolute uncertainty $\Delta c$, and state that $\Delta c = \pu{0.01 M},$ and will inform
$$c = \pu{(0.50 \pm 0.01) M}\tag{6}$$
Rounding up the absolute uncertainty is just a matter of personal like, some do, some do not. Of course, this enlarges the confidence interval, now between $\pu{0.49 M}$ and $\pu{0.51 M}.$
If you want more details, you can check the lovely Quantitative Chemical Analysis by Harris in the sections on error or statistics.
Reference
1.Harris, D. C. Quantitative Chemical Analysis, 9th ed.; W. H. Freeman: New York, 2015. ISBN 978-1-4641-3538-5.

There are two approaches to this.

Estimate the error from the estimated error of measurement (how good are you at reading a buret?).

Estimate the error from the variance of multiple measurements.

You can even do both and see if they match (do a chi-square test). If the variance is higher than what you expect from your error model, either the experimenter is having a bad day or the error model is unrealistic. If the variance is lower than what you expect from your error model, it might be beginner's luck or some kind of bias (titrating to the same volume while looking at the volume rather than looking at the color of the indicator).
Neither analysis tells you about systematic errors (like an incorrect assumption about the purity of the substance you make the titrant from, or using the glassware outside of the prescribed temperature range - there is glassware appropriate for the tropics, for example).

[OP] ... since more trials would increase the interval of values and thus increase the uncertainty in the experiment, which is not true.

We distinguish between the standard deviation, which is roughly independent of sample size, and the standard error, which decreases with sample size. You would have to specify which one you are calculating or estimating.


text:
Strange behavior of equilibrium constant when most of the components are solids / pure liquids
Consider the reaction $\alpha A(g) + \beta B(s) +\gamma C(s) +\cdots \rightleftharpoons \delta D(s)+\epsilon E(s)+\cdots $, so the chemical equilibrium constant should be $\dfrac{1}{[A]^\alpha}$, which is only dependent on the concentration on $A$. However, if I throw in a huge amount of $A(g)$ into the system, then it's possible that at least one of the reactants will be depleted before even reaching the equilibrium. So I have a few questions:

Is my deduction correct and does this situation exist in reality?
And furthermore, in more extreme scenarios when all the reactants and products are solid or pure liquid, then is the reaction still considered reversible since adding anything into the system won't change the reaction quotient and any state can be considered as an equilibrium?

These aren't extreme scenarios. It's just that all this plugging concentration thing you think about is limited. There's a tonne of reactions that don't depend on concentration at all. Well, at least not in the way this stuff^ works.

As a simple example, consider melting a solid. At the melting point, you can add or remove any amount of liquid or solid (assuming that what you add is already at the correct temp) and the system will remain in equilibrium as long as there is non-zero solid and liquid. This is of course also very much a reversible reaction.

An example would be $\ce{B(s) + (1/2)N2(g) = BN(s)}$ which has a equilibrium constant $K_e = p_{N2}^{-1/2}$ (=$173.5$ at $1900$ K) taking the activities of the solids to be $1$. There is nothing unusual about this, the equilibrium is always established even if one species is there in minute amounts as a result. Adding one species results in the others changing also.