text:
Unknown notation relating to stoichiometry

I have never seen the highlighted symbol before. What is the name of the symbol itself, and what does it mean in this context?

The sign $≏$ is read "is chemically equivalent to". The term chemically equivalent refers to a specific chemical reaction. For the reaction $$\ce{N2(g) + 3 H2(g) -> 2NH3(g)}$$ $$\pu{1 mol } \ce{N2} \, ≏ \,  \pu{3 mol } \ce{H2} \qquad \pu{1 mol } \ce{N2} \, ≏ \pu{2 mol } \ce{NH3}$$
Reference: Atkins, P. W.; Jones, L. L.; Laverman, L. E. Chemical Principles: The Quest for Insight, 6th ed.; W. H. Freeman: New York, 2012. Page no. 96, Section L: Reaction Stoichiometry.
Edit:
It seems that the OP is using Chang's Chemistry. In the same book, on pg. no. 98, the meaning of the symbol is explained as "stoichiometrically equivalent to" or simply "equivalent to".

Well, I must admit that this is the best answer. Great find!

@andselisk Can you make that sign $≏$ large? I tried it using $\Large{≏}$ but it is making the other text large too.

I used $\Large ≏$ ($\Large ≏$) in my post as a dirty hack:). \Large is basically a macro that acts as a switch, and does not take any parameters, so \Large{...} is not going to work.

Yes, I'm using Chang's textbook, and you have a very sharp eye to spot that symbol in the textbook. Thank you.

It is probably supposed to be the 'corresponds to' sign[ref, p 105]; in this case simply representing that for every two moles of $\ce{NaOH}$ one mole of $\ce{H2SO4}$ is converted by reaction stoichiometry.
This sign is difficult to typeset in LaTeX which is why I presume the authors used something else instead. Creating the sign is even more annoying in MathJax. The closest I [got at first] were
$$\overset{\lower.5em\wedge}{=}\ \text{or}\ \hat{=}$$
which are $\overset{\lower.5em\wedge}{=}$ and $\hat{=}$, respectively.
Edit. After some testing, this option $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ looks best in my opinion.
$$\ce{2 mol NaOH $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ 1 mol H2SO4}$$
full code: $$\ce{2 mol NaOH $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ 1 mol H2SO4}$$
Just the bare $\widehat{=}$ leads to
$$\ce{2 mol NaOH \widehat{=} 1 mol H2SO4}.$$

ref: E. R. Cohen, T. Cvitas, J. G. Frey, B. Holmström, K. Kuchitsu,
R. Marquardt, I. Mills, F. Pavese, M. Quack, J. Stohner, H. L. Strauss, M. Takami, A. J. Thor. (2007). Quantities, Units, and Symbols in Physical Chemistry. ('IUPAC Green Book'). Third Edition. International Union of Pure and Applied Chemistry. (2008 IUPAC & RSC reprint). ISBN: 978-0-85404-433-7. [free copy]
Here is a picture of how it looks in the 'Green Book':

Usually one can type ≙ easier with \wedgeq $\rm \LaTeX$ command, which unfortuanety appears not to work with MathJax:)

@andselisk What package is that from?

\wedgeq is from amssymb.

I suppose it's a $\Large ≏$ "difference between" math symbol, which here denotes inequality in amounts between acid and base. Can be typed using $\rm \LaTeX$ command \bumpeq, or Unicode symbol 0x224F.
Also note that there are two more similarly looking unrelated symbols:

$\Large $ "alchemical symbol for sublimation" (Unicode: 0x1F75E);
$\Large ♎$ "libra" ($\rm \LaTeX$: \libra; Unicode: 0x264E).

Also, check out relational symbol for “the same strings”. It seems that some authors use math notations quite liberally, and it might as well be that they assigned this character for something different. In order to prove or disapprove that, it would be ideal to get the full-text.

This alchemical symbol is so obscure, that my browser doesn't view it :D

@Mithoron I added few links to Wikipedia where you can enjoy the glory (or obscurity) of all the symbols in PDFs (e.g. alchemistry charts) :)

Should not it be equivalence of amounts rather than inequality in amounts?

@mamun $\pu{2 mol}$ is not equivalent to $\pu{1 mol}$, I think.

Could you explain how your interpretation of the symbol fits with the context of the sentence? In context I am inclined to go with 'corresponds to' or 'is stoichiometrically equivalent to', as has been suggested.

That's what I was trying to say. They are stoichiometrically equivalent. Albeit, 2 mol is not equal to 1 mol, but 2 mol NaOH is capable of neutralizing 1 mol H2SO4. As a result, they are said to be equivalent (not equal), hence they have this obscure notation to differentiate these two ideas.

@a-cyclohexane-molecule (cc mamun) Trying to dig up what authors might wanted to say with this is a rabbit hole to hell. I edited the answer a bit, but to get a 100% correct interpretation the editor or the author should appear in this topic and explain the intended meaning.


text:
Why is the boiling point of hydrogen sulfide higher than that of hydrogen chloride?
The boiling point of $\ce{H_2S}$ is higher than $\ce{HCl}$. Is this due to the greater number of hydrogen bond that $\ce{H_2S}$  can form compared to $\ce{HCl}$? I found this explanation for the comparison between $\ce{H_2O}$ and $\ce{HF}$ so I tried to use it here also, but I'm not even sure that the intermolecular forces among   $\ce{H_2S}$ and $\ce{HCl}$ molecules can be defined as "hydrogen bond" (which should occur only for  $\ce{F,O,N}$).
So how can the greater boiling point of  $\ce{H_2S}$ compare to $\ce{HCl}$ be explained?

Please don't use mathjax in titles if possible, see: https://chemistry.meta.stackexchange.com/questions/261/is-mathjax-in-titles-a-problem-and-why# Usually I write the full chemical names, or if they're impractically long, just writing naked H2S and HCl works OK.

Definition of the hydrogen bond
In 2011, a modern and more general IUPAC definition of what constitutes a hydrogen bond was recommended (Arunan et al., 2011). The definition is stated as such:

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

It appears that the definition of then hydrogen bond which includes interactions only with $\ce {F, O or N}$ seems too narrow and restrictive, which I personally feel is indeed the case.
Therefore, it is perfectly fine to say that $\ce {H_2S}$ makes hydrogen bonds with other $\ce {H_2S}$ molecules and that $\ce {HCl}$ makes hydrogen bonds with other $\ce {HCl}$ molecules.
Explaining the boiling points
Wikipedia gives the boiling points of $\ce {H_2S}$ and $\ce {HCl}$ as $\ce {-60 ^{\circ} C}$ and $\ce {-85.05 ^{\circ}C}$ respectively. When rationalising boiling point differences, the first consideration is always the strength of the intermolecular forces between the molecules in the liquid.
The Pauling electronegativities of $\ce {S}$ and $\ce {Cl}$ are $\ce {2.58}$ and $\ce {3.16}$ respectively. Evidently, the electron deficiency in the hydrogen atom bonded to $\ce {Cl}$ is more severe compared to that bonded to $\ce {S}$. This suggests that $\ce {HCl}$ molecules are capable of forming stronger hydrogen bonds. However, it is also important to consider the number of hydrogen bonds formed between molecules, as you have already noted in your question. Each $\ce {H_2S}$ can form 4 hydrogen bonds with other $\ce {H_2S}$ molecules, whereas each $\ce {HCl}$ molecule can only form 2 hydrogen bonds with other $\ce {HCl}$ molecules. Therefore, despite the fact that $\ce {HCl}$ molecules can form stronger hydrogen bonds between themselves, the intermolecular attractions between $\ce {H_2S}$ molecules still prevail as they are twice as numerous. Thus, $\ce {H_2S}$ has a higher boiling point than $\ce {HCl}$.
References
Arunan, E., Desiraju, G. R., Klein, R. A., Sadlej, J., Scheiner, S., Alkorta, I., . . . Nesbitt, D. J. (2011). IUPAC definition of the hydrogen bond. Terminology and nomenclature. Pure Applied Chemistry, 83(8), 1619-1636. doi:10.1351/PAC-REP-10-01-01
Electronegativity. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Electronegativity
Hydrogen chloride. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Hydrogen_chloride
Hydrogen sulfide. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Hydrogen_sulfide

Nice :) Still it would be better to cite sources cited in Wikipedia articles.

@Mithoron Will take note of that in the future

source: Stack Exchange

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sort: votes


text:
Comparison of the rate of SN1 and SN2 reaction
It is known that $\mathrm{S_N1}$ (Substitution Nucleophilic Unimolecular) follows the order of rate of reaction:

Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$)

For $\mathrm{S_N2}$ (Substitution Nucleophilic Bimolecular) the order of rate:

Tertiary ($3^\circ$) < Secondary ($2^\circ$) < Primary ($1^\circ$)

But what is the effective rate of reaction, if we consider that both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ may occur?
Let us assume that we are using $\ce{NaOH(aq)}$ as the reagent, and using n-butyl chloride ($1^\circ$), sec-butyl chloride ($2^\circ$), and tert-butyl chloride ($3^\circ$). As water is used, substitution will be favoured, and for this discussion, let us neglect any elimination reaction occurring. In this case, what would be the effective rates of reaction for these cases?

https://chemistry.stackexchange.com/questions/35013/are-bulkier-molecules-more-likely-to-reactThere's also

For SN1, carbocation stability is the main factor, while for SN2, steric hindrance is the main factor. When we think about carbocation stability, EWGs destabilise the carbocation while ERGs stabilise the carbocation.

So say I had tert-chlorobutane, and tert-methoxybutane, the rate of reaction of methoxybutane would be much higher as -OCH3 group is electron releasing in nature

Here is a paper that studies concurrent SN1 and SN2 reactions in substituted Ar2CHBr compounds, showing how the concentration of the nucleophile, the lifetime of the carbocation and the temperature all have a role in the preference for either SN1 or SN2.

@Karsten I am not able to access it. If I could I would have read it and updated my answer accordingly. I am able to see only the supporting info.

An interesting interplay: $\mathrm{S_N}$ mechanisms

The effective rate of reaction in a state of affairs wherein both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions can arise is decided via the interaction of the rates of those two mechanisms.
It's crucial to note that the relative rates of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions are influenced with the aid of factors consisting of the substrate, nucleophile, and solvent.
In general, for the given substrates (n-butyl chloride, sec-butyl chloride, and tert-butyl chloride), we can make a few observations:

$1^\circ$ Substrate (n-butyl chloride):

$\mathrm{S_N2}$ is favored because of the much less hindered nature of the primary carbon.
The effective rate would primarily be undertaken by $\mathrm{S_N2}$.

$2^\circ$ Substrate (sec-butyl chloride):

$\mathrm{S_N2}$ is still preferred, however $\mathrm{S_N1}$ may have a greater substantial contribution as compared to the $1^\circ$ substrate.
The powerful rate might be a mixture of both mechanisms, with $\mathrm{S_N2}$ being extra distinguished.

$3^\circ$ Substrate (tert-butyl chloride):

$\mathrm{S_N1}$ is desired due to the stability of the carbocation intermediate.
The effective rate could in general be governed with the aid of $\mathrm{S_N1}$.

In summation, the effective rates for the given substrates might observe this fashion: [$1^\circ > 2^\circ > 3^\circ$]
This implies that for n-butyl chloride ($1^\circ$), the $\mathrm{S_N2}$ mechanism dominates; for sec-butyl chloride ($2^\circ$), it's a combination of both mechanisms however with $\mathrm{S_N2}$ being extra considerable; and for tert-butyl chloride ($3^\circ$), $\mathrm{S_N1}$ is the number one mechanism.

An important point to be noted:
This answer explains how steric hindrance negatively affects the rate of reaction.
$\mathrm{S_N1}$ mechanism has 2 steps, while $\mathrm{S_N2}$ has only 1.
This is not all!
The example will vary from substrate to substrate. It all depends on various other factors such as resonance, hyperconjugation, inductive effects etc.
Thus we will have to carefully consider all the electron displacement effects and then go ahead with making decisions with respect to what comes first.

Sources:

https://typeset.io/questions/what-are-the-different-factors-that-affect-the-rate-of-sn1-2c406c7yp2

https://www.masterorganicchemistry.com/2012/08/08/comparing-the-sn1-and-sn2-reactions/

https://courses.lumenlearning.com/suny-potsdam-organicchemistry/chapter/8-3-factors-affecting-rate-of-nucleophilic-substitution-reactions/

https://www.chemguide.co.uk/mechanisms/nucsub/whatis.html#top

In $\mathrm{S_N1}$ the r.d.s. is the carbocation formation and rate is independent of nucleophile concentration, while in $\mathrm{S_N2}$ rate is dependent on nucleophile concentration. $\mathrm{S_N2}$ has only one r.d.s., and the transition state formed in between is transient in nature

Going by what @jimchmst said, it implies that the asker of the question itself has incorrectly put the numbering. I'll edit that as well as update my answer.

Also my inference from @jimchmst's comment is that this exercise is more of an experimental nature rather than that of a theoretical nature.

Thanks for pointing that @jimchmst, actually isobutyl chloride will be a bit of a gray zone because it would rather go rearrange itself into a tertiary carbocation. Thus it's a bit complicated; actual secondary and tertiary halides will give a better insight.

Also in which solvent we carry out sub. of isobutyl chloride also will play a major role in that case.

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


text:
What's the longest C=C bond?
There's been ample work and media coverage on how to make a long C-C single bond. To my knowledge, the longest stable C-C bond is currently 1.704 Å, prepared by Schreiner Nature 477, 308–311 in 2011.
All around us, there's considerable plastics, made from polymerizing alkenes like polyethylene. Typical bond lengths for alkenes are ~1.33 Å, with bond dissociation energies around 611 kJ/mol, vs ~347 kJ/mol for C-C.[ref] Clearly, longer C=C bonds could decrease the energy required to polymerize the alkane.
Using either chemical databases, like the Crystallographic Open Database, or computational methods, what's the longest C=C bond you can find?
Please include in your answer:

The molecular structure, with the C=C bond either obvious or highlighted (e.g., if it's a conjugated bond)
The method used to determine the bond length (e.g., DFT geometry optimization using program X, found in database Y, etc.)
Proof that the bond is really a double bond (e.g., clear evidence of substituents, planarity, etc.)
Any appropriate literature citations

http://www.ch.imperial.ac.uk/rzepa/blog/?p=17122

Feel free to use Henry's blog as a starting point. (Consider that I might ping him to add an answer too. :-)

tbh; this is a lot more challenging than I initially thought. C=C does not like to be stretched.

I also must admit that Henry Rzepa did a great job fetching those structures from CCDC. I followed his approach, and oh boy, I haven't seen so many garbage crystal structures with C=C bonds from 1.7 up to 2 Å in years.

Not directly related to the topic, but since you mentioned the longest C-C single bond, the record has now been drastically increased to 1.806(2) Å - "Longest C–C Single Bond among Neutral Hydrocarbons with a Bond Length beyond 1.8 Å".

While the two other (so far) presented molecules look realistic (Martin) or are real (Marko), I couldn't find the constraint, that the molecules actually need to be realistic, so here we go.
This is Gaussian 16 Rev. A.03 using (again, why do we use this niveau?) B97D3/def2TZVPP with Opt=(VeryTight) and SCF=(VeryTight) on to-be-named-by-Loong I (TBNBL I) with some nice double (or so) bond length of $\pu{143.5 pm}$. (At least formally, it needs to be a double bond.)

E(RB97D3) =  -692.564719674
C                  0.00000000    0.71742600    0.00000100
C                  0.00000000   -0.71742800    0.00000100
C                 -1.26096700    1.35299700    0.00000100
C                  1.26096700    1.35299700    0.00000100
C                 -1.26096700   -1.35299900    0.00000100
C                  1.26096700   -1.35299900    0.00000100
C                 -2.47714700    1.44346000    0.00000000
C                 -2.47714700   -1.44345900    0.00000100
C                  2.47714700   -1.44345900    0.00000000
C                  2.47714600    1.44345900    0.00000100
C                 -3.88898600    1.10111200   -0.00000100
C                  3.88898600    1.10111300    0.00000000
C                  4.18780000    0.00000100   -1.08412100
C                  3.88898700   -1.10111100   -0.00000100
C                  4.18780200    0.00000000    1.08411900
C                 -3.88898700   -1.10111100    0.00000000
C                 -4.18780000    0.00000000   -1.08412100
C                 -4.18780200    0.00000100    1.08411900
H                 -4.55239300   -1.97024400   -0.00000100
H                 -3.55876000    0.00000100    1.97407800
H                 -5.24617200    0.00000100    1.36283000
H                 -5.24617000    0.00000000   -1.36283300
H                 -3.55875700    0.00000000   -1.97407800
H                 -4.55239300    1.97024500   -0.00000100
H                  4.55239300   -1.97024300   -0.00000100
H                  5.24617000    0.00000100   -1.36283300
H                  3.55875700    0.00000100   -1.97407800
H                  5.24617200    0.00000000    1.36283000
H                  3.55876000    0.00000000    1.97407800
H                  4.55239300    1.97024500   -0.00000100

It has no negative frequencies and thus is a local minimum on its potential energy hypersurface. This holds true for every following molecule.
BDE according to [1] is 553 kJ/mol.

This led me to change the cyclobutyl rings into double bonds (TBNBL II), but that was not productive, as the bond length is reduced to $\pu{140.4 pm}$. BDE is $\pu{606 kJ/mol}$.

E(RB97D3) =  -535.325912063
C                  0.00000000    0.70224700    0.00000000
C                  0.00000000   -0.70224700    0.00000000
C                  1.29152400    1.32362500    0.00000000
C                 -1.29152400    1.32362500    0.00000000
C                  1.29152400   -1.32362500    0.00000000
C                 -1.29152400   -1.32362500    0.00000000
C                  2.51038500    1.30761800    0.00000000
C                  2.51038500   -1.30761700    0.00000000
C                 -2.51038500   -1.30761700    0.00000000
C                 -2.51038500    1.30761700    0.00000000
C                  3.80696800    0.68032900    0.00000000
C                 -3.80696800    0.68032900    0.00000000
C                 -3.80696800   -0.68032900    0.00000000
C                  3.80696800   -0.68032900    0.00000000
H                  4.73157700   -1.24777100    0.00000000
H                  4.73157700    1.24777100    0.00000000
H                 -4.73157700   -1.24777100    0.00000000
H                 -4.73157700    1.24777100    0.00000000

Ok, if getting smaller does not help, maybe getting bigger again? What about changing the cyclobutyl* into cyclohexadiene*? We get larger again, but not larger as my first try. TBNBL III: $\pu{141.5 pm}$ with BDE of $\pu{587 kJ/mol}$.

E(RB97D3) =  -844.915342833
C          0.00000        0.70775        0.00000
C         -0.00000       -0.70775        0.00000
C          1.25608        1.36005        0.00000
C         -1.25608        1.36005       -0.00000
C          1.25608       -1.36005       -0.00000
C         -1.25608       -1.36005        0.00000
C          2.45242        1.57986        0.00000
C          2.45242       -1.57986       -0.00000
C         -2.45242       -1.57986        0.00000
C         -2.45242        1.57986       -0.00000
C         -3.91584        1.44760       -0.00000
C         -4.32828        0.66483        1.24739
C         -4.32828        0.66483       -1.24739
C         -4.32828       -0.66483        1.24739
H         -4.57791        1.23705        2.13578
C         -4.32828       -0.66483       -1.24739
H         -4.57791        1.23705       -2.13578
C         -3.91584       -1.44760       -0.00000
H         -4.57791       -1.23705        2.13578
H         -4.57791       -1.23705       -2.13578
H         -4.39998       -2.43128       -0.00000
C          3.91584        1.44760        0.00000
C          4.32828        0.66483       -1.24739
C          4.32828        0.66483        1.24739
C          4.32828       -0.66483       -1.24739
H          4.57791        1.23705       -2.13578
C          4.32828       -0.66483        1.24739
H          4.57791        1.23705        2.13578
C          3.91584       -1.44760       -0.00000
H          4.57791       -1.23705       -2.13578
H          4.57791       -1.23705        2.13578
H          4.39998       -2.43128        0.00000
H          4.39998        2.43128       -0.00000
H         -4.39998        2.43128        0.00000

There is also the cyclohexyl analogue (TBNBL IV) which has a little longer bond with $\pu{142.4 pm}$ and a BDE of $\pu{571 kJ/mol}$.

E(RB97D3) =  -849.855064121
C          0.00000       -0.71210        0.00000
C         -0.00000        0.71210        0.00000
C         -1.25688       -1.35508       -0.01308
C          1.25688       -1.35508        0.01309
C         -1.25688        1.35508        0.01309
C          1.25688        1.35508       -0.01308
C         -2.46545       -1.49799       -0.06310
C         -2.46545        1.49799        0.06310
C          2.46545        1.49799       -0.06310
C          2.46545       -1.49799        0.06310
C          3.90819       -1.43687        0.24464
C          4.25416       -0.48923        1.44910
C          4.64897       -0.93627       -1.02206
C          4.64897        0.93627        1.02206
H          5.08644       -0.93074        2.00793
H          3.39579       -0.45561        2.12315
C          4.25416        0.48923       -1.44910
H          5.72105       -0.97274       -0.79444
H          4.47838       -1.63040       -1.84994
C          3.90818        1.43687       -0.24464
H          5.72105        0.97275        0.79444
H          4.47838        1.63040        1.84994
H          3.39579        0.45561       -2.12315
H          5.08644        0.93074       -2.00793
H          4.28399        2.44120       -0.47664
C         -3.90819       -1.43687       -0.24464
C         -4.25416       -0.48923       -1.44910
C         -4.64898       -0.93627        1.02206
C         -4.64897        0.93627       -1.02206
H         -5.08644       -0.93074       -2.00793
H         -3.39579       -0.45561       -2.12314
C         -4.25416        0.48923        1.44910
H         -5.72105       -0.97274        0.79444
H         -4.47838       -1.63040        1.84994
C         -3.90819        1.43687        0.24464
H         -5.72105        0.97275       -0.79444
H         -4.47838        1.63040       -1.84994
H         -3.39579        0.45561        2.12315
H         -5.08644        0.93074        2.00793
H         -4.28399        2.44120        0.47664
H         -4.28399       -2.44120       -0.47665
H          4.28399       -2.44120        0.47664

So, this also did not help. Then I thought about going back to TBNBL I and modify it. Let's add Chlorine looking into the rings, which should push the triple bonds apart. That does not change anything with regard to the first molecule ... TBNBL V has $\pu{143.4 pm}$ with a BDE of $\pu{555 kJ/mol}$.

E(RB97D3) =  -2531.12566838
C         -0.00000       -0.00000        0.71686
C          0.00000       -0.00000       -0.71686
C         -1.25995        0.00000        1.35010
C          1.25995       -0.00000        1.35010
C         -1.25995        0.00000       -1.35010
C          1.25995        0.00000       -1.35010
C         -2.47385        0.00000        1.43772
C         -2.47385        0.00000       -1.43772
C          2.47385        0.00000       -1.43772
C          2.47385       -0.00000        1.43772
C         -3.88167        0.00000        1.11523
C          3.88167       -0.00000        1.11523
C          4.23226        1.05834        0.00000
C          3.88167        0.00000       -1.11523
C          4.23226       -1.05834       -0.00000
C         -3.88167        0.00000       -1.11523
C         -4.23226        1.05834       -0.00000
C         -4.23226       -1.05834        0.00000
H         -4.54203        0.00000       -1.98574
H         -5.30546       -1.25818        0.00000
H         -5.30546        1.25818        0.00000
H         -4.54203        0.00000        1.98574
H          4.54203       -0.00000       -1.98574
H          5.30546        1.25818       -0.00000
H          5.30546       -1.25818        0.00000
H          4.54203        0.00000        1.98574
Cl        -3.41634       -2.62778        0.00000
Cl        -3.41634        2.62778        0.00000
Cl         3.41634        2.62778       -0.00000
Cl         3.41634       -2.62778        0.00000

What about adding more chlorine atoms? Seems to work; TBNBL VI: $\pu{144.1 pm}$ with a BDE of $\pu{542 kJ/mol}$.

E(RB97D3) =  -4369.66038511
C          0.00000        0.00000        0.72068
C          0.00000        0.00000       -0.72068
C         -1.25395        0.00000        1.35718
C          1.25395       -0.00000        1.35718
C         -1.25395        0.00000       -1.35718
C          1.25395        0.00000       -1.35718
C         -2.46546        0.00000        1.48386
C         -2.46546        0.00000       -1.48386
C          2.46546        0.00000       -1.48386
C          2.46546       -0.00000        1.48386
C         -3.85845        0.00000        1.12299
C          3.85845       -0.00000        1.12299
C          4.13677        1.09997        0.00000
C          3.85845        0.00000       -1.12299
C          4.13677       -1.09997       -0.00000
C         -3.85845        0.00000       -1.12299
C         -4.13677        1.09997       -0.00000
C         -4.13677       -1.09997        0.00000
H         -4.57107        0.00000       -1.94770
H         -4.57107        0.00000        1.94770
H          4.57107       -0.00000       -1.94770
H          4.57107        0.00000        1.94770
Cl        -3.09602       -2.53375        0.00000
Cl        -3.09602        2.53375        0.00000
Cl         3.09602        2.53375        0.00000
Cl         3.09602       -2.53375        0.00000
Cl        -5.83899        1.64902        0.00000
Cl        -5.83899       -1.64902       -0.00000
Cl         5.83899       -1.64902        0.00000
Cl         5.83899        1.64902        0.00000

As Geoff and I were curious to see what happens, when Bromine is used instead of Chlorine, I did the calculations (only geometry optimization so far) and they indeed yield a slightly longer bond but nothing is dramatically changed.

$144.3$, $144.3$ and $\pu{144.4 pm}$.
Iodine also adds a little to the bond length.

I asked a colleague and he said, what about about extending the lever (see law of the lever)? Indeed, it adds a little length.

His other thought was towards a BN triple bond. Not better than the CC triple bond in TBNBL I.

... and this is where I stop (again), because someone needs to beat my awesome DFT value of a super realistic molecule of 1445 pm now.
 EXPERIMENTALISTS THINK SILICON IS REALLY FUN TO USE
     ITS PLACE IN NOVEL COMPOUNDS IS CERTAIN TO AMUSE
 THEY SIT ALL DAY IN LABORATORIES MAKING ALL THIS SLUDGE
     "LOADED WITH THE SILICON THEY SAY", TO ME IT LOOKS LIKE FUDGE.
 FOR HAPPY THOUGH THEY BE WITH CRUD, I'D LIKE TO KNOW A LITTLE
      ABOUT THE PI BONDS ON THE EDGE AND SIGMAS IN THE MIDDLE.

 SO LETS DERIVE A WAVEFUNCTION.....6-31G*
     USE AN OPTIMAL GEOMETRY AND SEE WHERE ELECTRONS ARE.
 BUT WHAT OF CORRELATION?  ASKS THE WIRY LITTLE SKEPTIC.
     WE'LL THROW IN PERTURBATION AS AN ELECTRON ANTISEPTIC.
 AND WHEN THE PROGRAM GIVES US ANSWERS IN THEM WE CAN TRUST
     SINCE NOBODY CAN MAKE THE STUFF, WE HAVE NO CHOICE, WE MUST.

 SO THEORY GUYS HAVE GOT IT MADE, IN ROOMS FREE OF POLLUTION.
     INSTEAD OF PROBLEMS WITH THE REFLUX, THEY HAVE ONLY SOLUTIONS.
 AND WHEN THE FEDS ANNOUNCE THE LIST OF CARCINOGENIC TERRORS,
     THE THEORISTS SIT SAFELY AT THEIR TERMINALS FIXING ERRORS.
 IN OTHER WORDS, EXPERIMENTALISTS WILL LIKELY DIE OF CANCER
     FROM WORKING HARD YET FRUITLESSLY...TILL THEORY GIVES THE ANSWER.
          -- THOMAS A. HOLME, 1983

[1] Andreas A. Zavitsas, J. Phys. Chem. A, 2003, 107 (6), pp 897–898
[2] Images where created using CYLview, 1.0.564 BETA; Legault, C. Y., Université de Sherbrooke, 2009 (http://www.cylview.org)

I didn't check yesterday, but: TBNBL II "has an RHF->UHF instability". The rest "is stable under the perturbations considered." I was asked, if it is really a double bond or maybe a radical: HOMO and LUMO, pi and pi* of the DB, are separated by more than 2 eV (except for TBNBL II, where it is 0.23 eV). I'll leave TBNBL II in the answer, because it shows my working process, but I guess it is actually no DB.

Out of curiosity.. what if you change the Cl to Br? Seems like the added radius would help the "ring push" aspect?

I thought about investigating push-pull alkenes. The explosive 1,1–diamino—2,2—dinitroethylene also known as FOX-7 has a double bond with the length 145.6 pm determined by X-ray crystallography:

The evidence is in this publication and CIF.

Calculations on the B97D3/def2-TZVPP level of theory (Gaussian 09 Rev. E.01) reveal the bond length in the gas phase to be 142.4 pm, which is in quite good agreement of the crystal structure.

E(RB97D3/def2-TZVPP)       =       -598.257208193
C          0.04916        0.00000       -0.00001
C         -1.37453        0.00031       -0.00000
N          0.80303       -1.22125        0.11237
O          0.24490       -2.28958       -0.23784
O          1.92916       -1.18705        0.58174
N         -2.07141       -1.13253       -0.20314
H         -3.03559       -1.17626        0.08434
H         -1.51303       -1.97791       -0.31069
N         -2.07092        1.13344        0.20315
H         -3.03510        1.17757       -0.08431
H         -1.51218        1.97858        0.31060
N          0.80359        1.22090       -0.11241
O          0.24599        2.28948        0.23787
O          1.92972        1.18619       -0.58172

@Martin-マーチン strictly speaking, this crystal structure has been measured at -100 °C. Even though it shouldn't affect the bond distances much, but this is one thing to keep in mind when replicating the geometry with computational methods. Also, there is one weak spot in this crystal structure: G00F is way off (1.655 vs 1, ideally).

I think this is the way more impressive find than the desperate attempt I came up with... you'll get my bounty tomorrow ;)

@Martin-マーチン Too bad that no one else participated in this.

I hope that by providing one data-point I might get the ball rolling. Most of my attempts have been unfruitful, and I suspect that stretching a carbon-carbon double bond is difficult in general.
That is due to the fact that if introducing bulky substituents in most of the cases I looked at the planarity is lost to avoid stress, i.e. the methylene groups rotate and form an approximately 90 degree angle. Restricting this movement is more challenging than I previously thought.
In other cases the double bond is still typically around 134 pm, as the sum of their covalent radii would suggest.[1]
Jumping off of that the only somewhat stretched bond I could find is within the following molecule, which I am not even attempting to name. The key features here are the alkynyl links to get the bulky diamantyl moieties far enough away from the double bond to simultaneously allow them to have positive dispersion interactions. Even with these rather extreme measures, I was only able to stretch the bond to 139.7 pm at the B97D3/def2-TZVPP level of theory, calculated with Gaussian 09 Rev. E.01. (The image is from the previous B97D3/def2-SVP optimisation and hence shows 140.8 pm bond distance. The overall change is too small to justify re-uploading a new image.)

I have yet to increase the level of theory and obtain a frequency calculation, but since this is quite the sizeable molecule it'll take some more time.

Pekka Pyykkö and Michiko Atsumi published covalent (single-, double-, triple-) bond radii for (almost) the entire periodic table in Chem. Eur. J. 2009, 15, 12770-12779. It is summarised in figure 3. These are obtained from experimental or theoretical data.
The figure itself can be found on Pyykkö's homepage as a pdf-file.

xyz coordinates

142
symmetry c2h
C        0.000000000      0.000000000      0.703991000
C        0.000000000      0.000000000     -0.703991000
C       -0.000036000      1.228392000     -1.422011000
C       -0.000036000      1.228392000      1.422011000
C        0.000036000     -1.228392000      1.422011000
C        0.000036000     -1.228392000     -1.422011000
C       -0.000039000      2.307702000     -2.004101000
C       -0.000039000      2.307702000      2.004101000
C        0.000039000     -2.307702000      2.004101000
C        0.000039000     -2.307702000     -2.004101000
C        0.000007000      3.632781000      2.620363000
C       -0.000321000      3.543471000      4.170911000
H       -0.890097000      2.977508000      4.502792000
H        0.889220000      2.977332000      4.503111000
C       -1.260688000      4.427875000      2.173610000
H       -2.166884000      3.872289000      2.476464000
H       -1.273063000      4.492650000      1.070965000
C        1.261096000      4.427671000      2.174190000
H        2.167022000      3.871915000      2.477502000
H        1.273950000      4.492403000      1.071573000
C        0.000007000      3.632781000     -2.620363000
C        1.261096000      4.427671000     -2.174190000
H        1.273950000      4.492403000     -1.071573000
H        2.167022000      3.871915000     -2.477502000
C       -1.260688000      4.427875000     -2.173610000
H       -2.166884000      3.872289000     -2.476464000
H       -1.273063000      4.492650000     -1.070965000
C       -0.000321000      3.543471000     -4.170911000
H        0.889220000      2.977332000     -4.503111000
H       -0.890097000      2.977508000     -4.502792000
C       -0.000007000     -3.632781000     -2.620363000
C       -1.261096000     -4.427671000     -2.174190000
H       -1.273950000     -4.492403000     -1.071573000
H       -2.167022000     -3.871915000     -2.477502000
C        1.260688000     -4.427875000     -2.173610000
H        2.166884000     -3.872289000     -2.476464000
H        1.273063000     -4.492650000     -1.070965000
C        0.000321000     -3.543471000     -4.170911000
H       -0.889220000     -2.977332000     -4.503111000
H        0.890097000     -2.977508000     -4.502792000
C       -0.000007000     -3.632781000      2.620363000
C        0.000321000     -3.543471000      4.170911000
H        0.890097000     -2.977508000      4.502792000
H       -0.889220000     -2.977332000      4.503111000
C        1.260688000     -4.427875000      2.173610000
H        2.166884000     -3.872289000      2.476464000
H        1.273063000     -4.492650000      1.070965000
C       -1.261096000     -4.427671000      2.174190000
H       -2.167022000     -3.871915000      2.477502000
H       -1.273950000     -4.492403000      1.071573000
C       -1.256835000      5.742791000     -4.340720000
C       -1.254127000      5.830583000     -2.794873000
C        0.000330000      6.620635000     -2.348365000
C        1.254467000      5.830385000     -2.795414000
C        1.256518000      5.742596000     -4.341263000
C       -0.000316000      4.952467000     -4.782788000
H       -2.160732000      5.192099000     -4.666973000
C       -1.255695000      7.150419000     -4.960009000
H       -2.159549000      6.376170000     -2.463344000
H        0.000576000      6.699088000     -1.243023000
C        0.000320000      8.027542000     -2.968740000
H        2.160119000      6.375816000     -2.464258000
H        2.160191000      5.191761000     -4.667890000
C        1.255319000      7.150218000     -4.960570000
H       -0.000550000      4.866220000     -5.887095000
H       -1.277972000      7.076821000     -6.064351000
H       -2.168257000      7.697354000     -4.653970000
H        0.890296000      8.590343000     -2.626973000
H       -0.889406000      8.590507000     -2.626592000
H        1.277086000      7.076611000     -6.064921000
C       -0.000024000      7.918685000     -4.505704000
H        2.168108000      7.697008000     -4.654946000
H       -0.000037000      8.931342000     -4.951455000
C        1.256518000      5.742596000      4.341263000
C        1.254467000      5.830385000      2.795414000
C        0.000330000      6.620635000      2.348365000
C       -1.254127000      5.830583000      2.794873000
C       -1.256835000      5.742791000      4.340720000
C       -0.000316000      4.952467000      4.782788000
H        2.160191000      5.191761000      4.667890000
C        1.255319000      7.150218000      4.960570000
H        2.160119000      6.375816000      2.464258000
H        0.000576000      6.699088000      1.243023000
C        0.000320000      8.027542000      2.968740000
H       -2.159549000      6.376170000      2.463344000
H       -2.160732000      5.192099000      4.666973000
C       -1.255695000      7.150419000      4.960009000
H       -0.000550000      4.866220000      5.887095000
H        1.277086000      7.076611000      6.064921000
H        2.168108000      7.697008000      4.654946000
H       -0.889406000      8.590507000      2.626592000
H        0.890296000      8.590343000      2.626973000
H       -1.277972000      7.076821000      6.064351000
C       -0.000024000      7.918685000      4.505704000
H       -2.168257000      7.697354000      4.653970000
H       -0.000037000      8.931342000      4.951455000
C       -0.000330000     -6.620635000      2.348365000
C        1.254127000     -5.830583000      2.794873000
C        1.256835000     -5.742791000      4.340720000
C        0.000316000     -4.952467000      4.782788000
C       -1.256518000     -5.742596000      4.341263000
C       -1.254467000     -5.830385000      2.795414000
H       -0.000576000     -6.699088000      1.243023000
C       -0.000320000     -8.027542000      2.968740000
H        2.159549000     -6.376170000      2.463344000
H        2.160732000     -5.192099000      4.666973000
C        1.255695000     -7.150419000      4.960009000
H        0.000550000     -4.866220000      5.887095000
H       -2.160191000     -5.191761000      4.667890000
C       -1.255319000     -7.150218000      4.960570000
H       -2.160119000     -6.375816000      2.464258000
H       -0.890296000     -8.590343000      2.626973000
H        0.889406000     -8.590507000      2.626592000
H        1.277972000     -7.076821000      6.064351000
H        2.168257000     -7.697354000      4.653970000
H       -2.168108000     -7.697008000      4.654946000
C        0.000024000     -7.918685000      4.505704000
H       -1.277086000     -7.076611000      6.064921000
H        0.000037000     -8.931342000      4.951455000
C        1.256835000     -5.742791000     -4.340720000
C        1.254127000     -5.830583000     -2.794873000
C       -0.000330000     -6.620635000     -2.348365000
C       -1.254467000     -5.830385000     -2.795414000
C       -1.256518000     -5.742596000     -4.341263000
C        0.000316000     -4.952467000     -4.782788000
H        2.160732000     -5.192099000     -4.666973000
C        1.255695000     -7.150419000     -4.960009000
H        2.159549000     -6.376170000     -2.463344000
H       -0.000576000     -6.699088000     -1.243023000
C       -0.000320000     -8.027542000     -2.968740000
H       -2.160119000     -6.375816000     -2.464258000
H       -2.160191000     -5.191761000     -4.667890000
C       -1.255319000     -7.150218000     -4.960570000
H        0.000550000     -4.866220000     -5.887095000
H        1.277972000     -7.076821000     -6.064351000
H        2.168257000     -7.697354000     -4.653970000
H       -0.890296000     -8.590343000     -2.626973000
H        0.889406000     -8.590507000     -2.626592000
H       -1.277086000     -7.076611000     -6.064921000
C        0.000024000     -7.918685000     -4.505704000
H       -2.168108000     -7.697008000     -4.654946000
H        0.000037000     -8.931342000     -4.951455000

Great insightful answer! I'm wondering is there a good reason that the bond is shown with a single stick?

"Rotated" double bonds (i.e., so-called "tictoid" $\pi$ conjugation) is a known, accepted way to increase bond length in conjugated systems. ;-)

@andselisk Yes, there is: I prefer it that way. In all my pictures you'll never see me use a double or even triple rod.

@Martin-マーチン I can tell:) But why is it so, if I may ask? Is it because the bond order is often fractional and the integer values denoted via multiple-stick bonds bother you?

@andselisk yes.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Is there a point of defining the reaction rate as the rate of disappearing of a reactant **over its coefficent**?
$$aA+bB \rightarrow cC + dD$$
The reaction rate is defined as:
\begin{align}
Rate = (\frac{1}{a})(\frac{-\Delta[A]}{\Delta t}) = (\frac{1}{b})(\frac{-\Delta[B]}{\Delta t}) = (\frac{1}{c}) (\frac{\Delta[C]}{\Delta t}) = (\frac{1}{d})(\frac{\Delta[D]}{\Delta t}) \\
\end{align}
Is there a point of defining the reaction rate as the rate of disappearance of a reactant over its coefficent, or the rate of appearance of a product over its coefficent? Why not just saying the rate of disappearance of a reactant or the rate of appearance of a product?

Because then, as your equations show, we would have 4 different rates for a reaction like this. Its much simpler to divide each by the coefficients so that there is one common rate of reaction.

The question was just marked as a duplicate because it has been answered elsewhere on the site

There is no defining.
When you write down the multiplication table, you don't define that $2\times4=8$, as if it could have been anything else. Instead, you derive it according to the common rules. Same thing here.
The rate of anything is just how many times a second something happens. If we are interested in the rate of reaction, then our "something" stands for the reaction, just as written in the equation above, i.e., the act of consuming $a$ moles of $A$, $b$ moles of $B$, etc.
So it goes.


text:
Why doesn't a catalyst shift the position of equlibrium, when temperature does?
Maybe this is a silly question, but it's not one I've been able to answer to my own satisfaction. Increasing the temperature of a reaction favours the endothermic direction, because it makes it relatively more likely for particles to collide with enough energy to make the activation energy. Yes?
But doesn't lowering the activation energy have almost the same effect on the relative likelihood of particles colliding at least that energetically? Someone on a related question about whether catalysts shift the position of equilibrium said:

The energy difference between the reactants and products is unchanged
  by catalysis

But isn't that equally true for increasing the temperature?

Your presumption in the first paragraph is incorrect. The fact that more particles collide with enough energy to make the activation energy with an increase in temperature means that the rate of reaction increases as temperature goes up, but that doesn't affect the equilibrium of the reaction.
Instead, the effect on temperature has to do with the comparative free energy of the two sides of the equilibrium. The controlling factor in where the equilibrium lies is the difference in free energy between the reactants and products (the ΔG). In fact, the equilibrium constant has a very simple relationship to the free energy: $\ce{\Delta G = -RT ln K }$ ... or $\ce{K = e^{-\frac{\Delta G}{RT} }}$. The ΔG is made up of two components - the enthalpy and the entropy: $\ce{\Delta G = \Delta H -T\Delta S}$, where T is the (absolute) temperature. The side of the reaction which is favored is the one with the lowest (most negative) free energy.
So when we substitute, we get the following relationship for the temperature dependence of the equilibrium constant, in terms of entropy and enthalpy:
$$ K = e^{-\frac{\Delta H}{RT} } \cdot e^{\frac{\Delta S}{R} } $$
The temperature dependence is only in the enthalpy ($\ce{\Delta H}$). Enthalpy, by the way, is what we normally think of as endothermic/exothermic. Roughly, reactions with a positive $\ce{\Delta H}$ absorb heat and are thus endothermic, and those with a negative $\ce{\Delta H}$ give off heat and are thus exothermic.
So imagine we have an endothermic reaction (positive $\ce{\Delta H}$), which means that the exponent for enthalpy in the above equation is negative. When we increase the temperature, that exponent gets smaller in magnitude (closer to zero), which means it becomes less negative. This causes an increase in $e^x$, which means that K gets bigger, which means the equilibrium shifts in the forward (endothermic) reaction. Likewise, if you have the reaction written in the exothermic direction, $\ce{\Delta H}$ is negative, the exponent is positive, an increase in temperature causes the exponent to become less positive, causing a decrease in the $e^x$, which results in a decrease of K -- the equilibrium now shifts in the reverse (endothermic) direction.
So while the statement that increasing temperatures means that the equilibrium shifts in the endothermic direction is true, it has nothing to do with the activation energy of the reaction. The arguments above are from a purely thermodynamic standpoint, and don't involve any consideration about the mechanism of reaction or the activation energy.


text:
Why doesn't the carbocation form on the carbon that is attached to the ethyl group in this reaction?
Why doesn't the carbocation form on the carbon that is attached to the ethyl group in this reaction?
Subsequently, why doesn't the $\ce{Cl-}$ attack the carbon that is attached to the ethyl group? Is a tertiary carbocation less stable than a secondary allylic carbocation?
Reaction:

1,2 -hydride shifting takes place after the formation of the tertiary carbocation. After the hydride shifting, the carbocation becomes resonance stabilised. And the chlorine adds up there(the first product that you have shown). Also, as I said the carbocation was resonance stabilised. So you can draw resonance structure of the carbocation intermediate and the positive charge goes to 4 position and then chlorine adds up there (the second product that you have shown) and thus we get the second product. Hope this helps.

Thanks.  This helps a lot.  I never thought to look for potential hydride shifts when I already have a tertiary carbocation.  So this suggests that an allylic carbocation is more stable than a tertiary carbocation, yes?

Yes resonance stabilisation is preferred over +I stabilisation. So, I hope it's clear. Do tick mark my answer as this question's answer in Chemistry.SE if my answer helped you.

Have labeling studies been done that confirm your proposed mechanism?  If, as you say, an allyl carbocation is more stable than a tertiary carbocation, then why not protonate at the ethyl end of the double bond to directly produce the allyl carbocation which would then directly lead to the observed products? I'm a bit surprised  at the reported products.  I would have thought that protonation would occur at the other double bond to give a substituted allyl carbocation which would then lead to 2 different products.

@ron First, we should protonate at the second position following Markovnikov's rule then think of carbocation rearrangement. Also, we should add the H+ ar the pi bond at the 1 position because there the electron density increases due to +I of ethyl group and hence the electrophile attacks there. Please reply if you agree and feel free to point out any shortcoming that my answer might contain.


text:
Theoretical limits of (NIR) spectroscopy for quantitative analysis
Edit (months later): In hindsight, I was asking about the theoretical limits of multivariate curve resolution (see Lawton & Sylvestre: Self-modeling curve resolution) in chemometric applications.

NIR spectra are commonly used to estimate analyte concentrations. The common approach is to measure the spectra of different samples with known concentration (training data) and then estimate the unknown concentrations by regression.
My question is whether we could determine analyte concentrations of any substance just from the following:
1) Infinitely precise spectra of all elements.
2) Infinitely precise spectrum of the sample.
If not, would it help if we relaxed on the NIR range assumption?
If still not, would it help if we knew the spectra of all possible molecules with infinite precision?
The first step in determining the answer is probably determining whether there are two different atoms identical in their spectra. I don't know the answer. Unfortunately my knowledge of physics and chemistry is very limited.

"Infinitely precise spectrum of the sample" ? Precise in what? BTW I find this question unclear and rather strange.

@Mithoron precise as in free from measurement errors.

Firstly, atoms do not have an IR spectrum. You mean molecules, or any kind of chemical substance. Anyway. Then, by "infinite precision", you mean noise-free. Yes of course, no two molecules have the same vibrational modes, except if it's the identical substance. But this is completely pointless, because there is no such thing as a noise-free measurement, and also because you can't just split a single absorption spectrum into the individual spectra of an arbitrary number of possibe components. This approach is 100% wilful nonsense, imo.

to continue comments by @karl, molecular spectra have transitions that have a finite spectral width and because of this spectra from one molecule will always overlap that of another to a greater of lesser extent. You only have to look at an IR spectrum to see this. This means that it can be impossible to properly measure different species as some lines will be hidden under those from different species. Having noise free signals will not help much as you will have to fit data anyway and this will in itself introduce uncertainty is assignments.

Thank you for your feedback! I will do some more reading and possibly refine my question. I'll add an analogy to mass spectroscopy, where it is feasible to generate a database of all possible peptide spectra (de novo approach) and then search for the measured spectrum in this database e.g. by the cosine distance metric. I don't hope to build such database for NIR spectra. The goal of this question is just to understand the data at hand 'from bottom up'.

The problem is the same for MS. For pure substances, you can make a database for practically every spectroscopic method, and perfectly identify everything. For mixtures, you have to put in a lot of thinking and additional constraints (limit range of possible constituents etc) before you can quantify anything.

@Karl "you can't just split a single absorption spectrum into the individual spectra of an arbitrary number of possibe components" - but isn't that what the Beer-Lambert law assumes in its $A = \sum_{i = 1}^N A_i = \sum_{i = 1}^N \varepsilon_i \int_0^\ell c_i(z),\mathrm{d}z$ (equation from Wikipedia)?

@Josef Just because you have an equation for sth. doesn't mean it has a single, unique solution. Further spectra e.g. look slightly different depending on concentration and solvent identity. Lambert-Beer has it's limits. There are far, far too many imponderables in your problem.

@Karl thank you for your reply. Of course, I'm just making sure I understand what the Lambert-Beer law is about. Regarding physical effects, this paper claims it can eliminate them, albeit under the strong assumption that the mixture is made of just two species: https://www.ncbi.nlm.nih.gov/m/pubmed/12585463/

@Josef Not "physical effects", just scattering in turbid samples. There are a dozen more effects that creep in when your sample is not perfectly dilute. The spectrum of a 1M solution of sth. does NOT necessarily look identical to the 1µM solution. Surely never perfectly identical. Even if you try to separate solvent and solute spectra and scale them accordingly.


text:
How to Dissolve Glyphosate?
I am a microbiologist forming a growth media which has glyphosate in it. Unfortunately, I am having troubles dissolving it in water. At a .2M solution in 200ml of water, it is not completely dissolving. Does anyone know why this is chemically happening? Even upon adjusting pH and raising temperatures, I am unable to dissolve it.
https://pubchem.ncbi.nlm.nih.gov/compound/glyphosate
If you have any ideas please let me know.

As written, this is a combination of "too broad" and borderline homework - in the link you give, it says the pH of a 1% solution is 2.5. You might consider narrowing the scope of the question and/or adding your thoughts to avoid closure.

Will do. Though, in all serious, this is not a homework problem. The biologist at the lab I am working at do not know why this is.

So you've got 0.2 M, which is about 34 g/L. Solubility - from your link - is in the range of 1 to 12 g/L. Your solution is supersaturated.

This structure has a phosphonic and a carboxylic acid moieties with a minimum number of carbon atoms. As such, it should be soluble in basic water to deprotonate at least one site

1, this is the correct answer, but you might consider elaborating.  What exactly what do you mean by "basic water" (sodium hydroxide?  ammonium hydroxide?  some other base?)  Would recommend focusing on the molarity of the hydroxide added or the pH of the resulting solution?


text:
Reaction of amines with diethyl oxalate (Hofmann amine separation method)
Why does only one molecule of secondary amine react with diethyl oxalate and two molecules of primary amine react with diethyl oxalate?
I feel secondary amine also should have reacted similar to primary amine.
From JEE-Main 2015:

Reaction with diethyl oxalate
1°, 2°, and 3° amines can be distinguished by their reactions with diethyl
  oxalate. Primary (1°) amines react with diethyl oxalate forming N,N-oxamide, which is a solid.

Secondary (2°) amines react with diethyl oxalate forming oxamic ester,
  which is a liquid.

Tertiary (3°) amines do not react with diethyl oxalate.

Probably a steric issue. Seems reasonable.

You may want to look at the Hinsberg test for distinguishing between amines.

@Zhe: From this site, I looked at structutr of oxamic ester, which was different from the one given in this post. As, the -OR and -NR2 groups are on opposite sides, hence I think steric reason aren't reasonable.

The second reaction has a typo: it should be $\ce{CONR2}$.

Tetra-substituted oxamide or N,N,N',N'-tetraalkyloxamide can be prepared by the action of oxalyl chloride on secondary amines.(see here)

When a 1° amine attacks to a carboxyl group, The structure below (left) is generated which lowers the energy level of the intermediate such that the second amine can easily react with the other carboxyl group.
It is clear that 2° amines (right) don't have such proton for hydrogen bonding with the oxygen. Why the second 2° amine doesn't attack to the other carboxyl at all is maybe due to the decreased partial positive charge of the amide (less electronegativity of nitrogen) toward the carboxyl.
Finally, 3° amines don't have even a leaving proton to substitute with the ethoxide.

The tetrahedral intermediate in the primary amine attack case would have an intramolecular hydrogen bond to stabilize.  In the secondary amine case, no such H-bond is available.

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


text:
Won't gas thermometers break on using fixed volume of gas?
From the ideal gas equation, PV is proportional to T. Now if V is kept constant, we'll get Gay-Lussac's law.
Now my question is:
According to observations, we know that substances undergo a change in dimensions or they expand on exposure to heat. So wouldn't that mean the fixed volume of gas must expand on exposure to heat? But if this happens the pressure must also increase, and wouldn't that break the glass thermometer?

There is no "must". The gas will want to expand, that's right. But whether we let that happen or not is another question.

When a basketball is inflated, pressure  increases and it must be accompanied by volume decrease right?. The volume is the space occupied by the molecules of the gas and it's increasing as the no.of molecules of gas pumped into the body increases?

If the temperature is high enough, yes, the glass thermometer will indeed break.
However, the glass (or other material) used to construct a thermometer should be strong enough to withstand the internal pressure in the temperature range for which it is intended. Else the thermometer would not be safe to use.
According to wikipedia, glass typically has a tensile strength (ability to withstand being pulled apart) of 7 MPa (approximately 70 atm), which should be sufficient for most applications.

Or the inability of the glass to deform ensures that the gas has a constant volume?

You are aware that a tensile strength of 7 MPa does not mean that a given  equipment geometry can survive a pressure of 70 atm, correct?  There is a big difference between the radial pressure within a cylinder, and the tensile stress in the cylindrical shell.

Yes I am aware! But the number serves as an indication.


text:
If melting points and freezing points are the same, what state is pure equilibrium?

If melting points and freezing points are the same, what state is pure equilibrium?

I understand melting points and freezing points are the same; adding or removing temperature of melting points for specific variables, like water at $\pu{0^\circ C}$ will melt or freeze them. Then, I have to ask the question, what is in between freezing and melting? What is that state? What is the behavior of the atoms if you somehow managed a perfect system where there wasn't extra heat being added nor removed and equilibrium was exact.

The state between freezing and melting is the state when some part of water is frozen and the other part is still liquid.

equilibrium of two states: liquid-solid.

At the melting point there is a change of state. This means that solid is transformed in liquid or viceversa. For this process you need energy which is supplied or withdrawn in the form of heat. The whole process occurs at constant temperature, which is precisely the melting point.
Adding or removing temperature has no sense. It is heat what  is added or removed. The temperature is just a result of the state. When you have a mixture of solid and liquid in equilibrium, this temperature is defined, is the same in the whole system and is the melting point.
Since it is state change, both solid and liquid are present and the temperature will not change while both are there in any quantity and if the system is in equilibrium. The quantities of solid and liquid will not vary until there is some heat flow. However, on the microscopic level, the molecules are still moving and some will migrate from the solid to the liquid while others (the same amount) will go from the liquid to the solid. This results in that, even if the system is in equilibrium, the small crystals will dissolve while the bigger ones will grow bigger.

Yes, I understand what melting point is. I apologize for the erroneous reference of temperature as heat. I wrote this in a rush trying to get my thoughts out and didn't adjust this error. Didn't really answer my question though. You essentially defined melting point and provided an example to an exception. My question is, what state are the atoms in, or behavior when at equilibrium. Ivan made an interesting comment though, "state between freezing and melting is the state when some part of water is frozen and the other part is still liquid".  Does this mean it continuously melts and solidifies?

@KingHarambe I have edited the answer to include the microscopic situation.

The other answer is right but I just wish to clarify a point because I see this error all over the place.
There is a strong misconception that melting point and freezing point happen at the same temperature. It all practicality they do, in almost all practical experimentation they should be treated as such, and 99.9999% of the time its the same.
However, when we get away from bulk chemistry and get to the small scale, there can be variations, see this paper for example.
Also, I did read about one super cooled bulk material that had a difference in intermediate crystal structures going from liquid to solid vs solid to liquid that actually resulted in a difference in the two temperatures. I can not find that article now, but it's out there somewhere.
Again, these are extreme examples, but to truly say melting and freezing point are the same is technically wrong. I just needed to put this out there now.
It's also an important distinction because there is a directional component to the energy vector by saying either melting point or freezing point. (endothermic vs exothermic), although I guess "point" implies otherwise, but I find differentiating the two helps when understanding which state you started from.


text:
Acidity of cyclopentadiene when fused to benzene rings
Why does fusion to a benzene ring reduce the acidity of a cyclopentadiene ring?
For example, acidity decreases going from cyclopenta-1,3-diene to indene and to fluorene:

Clar's rule states that (Solà, 2013):

The Kekulé resonance structure with the largest number of disjoint aromatic π-sextets, i.e., benzene-like moieties, is the most important for characterization of properties of polycyclic aromatic hydrocarbons (PAHs).

Applying Clar's rule to the the question, we would realise that in the indenyl and fluorenyl anions, the resonance structures where there are 6 π-electrons delocalised exclusively within the fused benzene rings would contribute more to the actual structure than the resonance structures where the π-electrons are delocalised along the 5-membered ring. In the major resonance structure for the fluorenyl anion, we would have the negative charge concentrated on the carbon atom, without delocalisation.

In the cyclopentadienyl anion, the negative charge is completely delocalised, as shown below in the resonance structures. Thus, it is the most stable among them. In the indenyl and fluorenyl anions, the negative charge is not as delocalised and thus, these are less stable.

Due to this, the π-electrons in the 5-membered ring experiences the most delocalisation and achieves most aromatic stabilisation in the cyclopentadienyl anion, and this gradually decreases going to the indenyl and fluorenyl anions.
Reference
Solà, M. (2013). Forty years of Clar's aromatic π-sextet rule. Frontiers in Chemistry, 1(22). doi:10.3389/fchem.2013.00022

Now-obsolete discussions on previous revisions of this answer have been moved to chat.

When you deprotonate cyclopentadiene you make the ring aromatic, bringing all five carbon atoms and six pi electrons into an aromatic cycle.  When you deprotonate indene, you again make the five-atom ring aromatic, but ... in this case part of the ring had already been aromatic by being incorporated into the benzenoid ring.  You don't add as much to the aromatic conjugation and thus the impact of aromaticity on the deprotonation is less.  Add another fused benzenoid ring in fluorene and you get another decrement in the net effect of aromaticity.

Your explanation does not explain why the conjugate base upon deprotonation is least stable for fluorene and most stable for cyclopentadiene.

In this paradigm the bases do not differ so much in stability as the conjugate acids do.  The acids are different because of the aromatization of the benzenoid rings partially overlapping into the cyclopentadienoid rings.


text:
Reduction of nitrobenzene with LiAlH4 and NaBH4
After searching a lot on the internet I've found that nitrobenzene is reduced by LiAlH4 to the following compound whose naming I couldn't do.

I couldn't find the mechanism for this reaction anywhere.
Can someone tell what is the mechanism for it?

The structure you provided is known as azobenzene . It is a parent compound in a well-known family of aromatic azo compounds. The compounds usually are not very thermally stable, but often have intense color and are used as light duty dies (or as pH or other indicators) Aliphatic azo compounds are not stable at all and very prone to fragmentation with formation of molecular nitrogen and free radicals and some of them are used exactly for that.

I couldn't find the mechanism for this reaction anywhere

The reduction of the nitro-group might procede through attack of hydride onto the nitrogen atom or electron transfer onto the group with subsequent reaction of the anion with proton (the exact mechanism depends on conditions and reagents). It can produce a wide amount of monomeric products, in particular: nitrosocompounds, hydroxilamines and amines. It is known, that amines can react with nitrosocompounds forming the $\ce{N=N}$ bond. It looks like the first step in the reaction is a nucleoplilic attack of the amine nitrogen onto nitroso nitrogene with subsequent elimination of water.

+1 for providing reference to the mechanism. I saw the mechanism at many places but was not able to find a satisfying reference. You have one.


text:
Thermochemistry - zero heat energy
We were just wondering in our chemistry class: is there any reaction which is neither exothermic nor endothermic?

I suppose you mean single-step / elementary reactions? If so, it seems unlikely; since if there's no energy gain or loss in the reactant species at the end of a reaction, why bother reacting in the first place? In other words: If there's no energy change, then there are no atoms rearranged or charges created...meaning there'd be no reaction. (Not a chemist, but this is the way I see it) :-)

@paracetamol You think right, but for the wrong reason.

@Ivan Could you clarify? :-)

Zero energy gain does not mean the reaction will not happen (there is also entropy to consider). Also, zero energy change does not mean that there are no atoms rearranged. It could just have happened that the new arrangement has the same energy as the old one, yet still they differ. True, this is unlikely, but not impossible.

@Ivan Aha! I stand corrected! :-)

many reactions such as  electron transfer $\ce{  Fe^{3+} + Fe^{2+}  \to Fe^{2+} + Fe^{3+}  } $ and atom transfer $ \ce{H2 + H}$ reactions for example. In each case there is an activation energy barrier between reactants and products so that they are 'real' reactions in the proper sense of the word.

You could just as well have wondered is there any number which is neither negative nor positive. Yes, there is such number, and it is called zero. It is rarely encountered among the real-life physical values, though, unless enforced by some law (which is not the case here). After all, there are immensely many positive numbers, and equally many negative numbers, and only one zero. You can hardly expect to hit it by choosing numbers at random.
Then again, reaction enthalpies are subject to variation with temperature, so if we pick any reaction with low enthalpy, it is quite probable that there is a certain temperature at which the said enthalpy is exactly 0. That would be the answer to your question. It is for entropy to decide whether the reaction will occur spontaneously at this temperature. Chances are it will.
On a side note, there is a huge number of "void reactions" like $\ce{H2 + H -> H + H2}$ with guaranteed zero enthalpy at any temperature, but many would object against calling these things reactions in the first place.
So it goes.

Now it becomes the matter of controlling the temperature to arbitrary precision ...

Precision similar to that of enthalpy measurements will suffice.

As it happens $\ce{H2 + H}$ is a very interesting reaction when studied in molecular beams as the transient $\ce{H3}$ species formed has proved very difficult to understand. There is much information to be obtained about the potential energy surfaces when the amount of energy in rotational and vibrational levels is observed under conditions of different collision energies .

@porphyrin Good to know you are not one of those many who would object.


text:
What is the logic behind the division into groups of anions and cations in qualitative inorganic analysis?
I find it very confusing and hard to memorize.
Is there a way to deduce it, like understanding how it is derived, or is it just one of a lot of ways you can do it but is the standard one?
Do people use to test every possible reagent or combination of reagents and see if something happens or is it deducible?

"What is the logic behind the division into groups of anions and cations in qualitative inorganic analysis?" Well then, what's the logic behind discussing the constitution of various substances in terms of  elements (and compounds thereof)? Why don't we just accept that everything is made of "something" that occupies space and has mass and just forget this whole chemistry business? <--- Food for thought

Well the resolution of substances into compounds and compounds into elements seems meaningful because elements are characterized by differing quantum numbers like baryon number, election number and so on. It makes sense that they would have different properties but the division of groups in inorganic analysis seems rather arbitrary and imprecise if there are more than one way to do it.

I can deduce, in some sense, the structure of the atom, elements, ideas of electrovalent and covalent bonds and then the categorization of elements with the postulates of quantum mechanics and some ideas of electrostatics. What I want to know is if there is some idea behind the categorization for qualitative analysis.

@paracetamol The logic behind discussing the constitution of compounds in terms of elements is that an element has a simple physical model and using many elements in conjunction, compounds form whose properties can be explained on the basis of the elements that form it, from the same physical model. By physical I mean, approximately physical, statistical, whatever you'd like to consider

What is the logic behind the division into groups of anions and cations in qualitative inorganic analysis?

There is a somewhat established classic procedure of qualitative and quantitative chemical analysis (also known as "systematic analysis") based on treatment of the solution by specific set of reagents in specific order. Cations are divided into groups based on the stage they are detected in the procedure. There is NO other logic behind this division, nor it has any meaningful chemical significance.

Also, how does one get at the various tests, like, did people use to test every possible reagent or combination of reagents and see if something happens or is it deducible?

Somewhat deducible. Solubility product constant can be used to check if the solution is prone to precipitation of particular salt barring cases of redox reaction. For redox reactions standard electrode potentials and Nernst equation might be used for a quick check. Still, knowledge of these particular constants is mostly (if not completely) empirical.

Is there a good source to get a nice grip in the entire procedure without having to conduct the experiments on my own?

@RishiRaj Search old (<90s) analytic chemistry books with chapters like "systematic analysis" that contain use of H2S. In recent years the method is virtually never used, there are X-ray spectroscopic methods that are reasonably affordable and billion times safer. Otherwise wikipedia gives reasonable qualititve description of the procedure https://en.wikipedia.org/wiki/Qualitative_inorganic_analysis


text:
What is three dimensional aromaticity?
I have recently come across a statement which states:

The relative stability of the 1,3-dehydro-5,7-adamantanediyl dication is ascribed to its three-dimensional aromaticity. [1]

My understanding of aromatic nature is, that the molecule must be cyclic and planar, every atom in the ring must be $\ce{sp^2}$-hybridized and the molecule must have [4n+2] $\pi$ electrons, where $n\in\mathbb{N}$ (Hückel’s rule). The molecular orbitals also have to have the correct orientation and phase for aromaticity (2).
In my understanding 1,3-dehydro-5,7-adamantanediyl dication does not follow these rules for Y-aromaticity (3).
So my question is: What is three-dimensional aromaticity and how does it stabilise the 1,3-dehydro-5,7-adamantanediyl dication?
References

Firme, C. L.; Antunes, O. A. C.; Esteves, P. M. Electronic nature of the aromatic adamantanediyl ions and its analogues. J. Braz. Chem. Soc. 2008, 19 (1), 140–149. DOI: 10.1590/S0103-50532008000100020.
LibreTexts
What is Y-aromaticity? Is the trinitromethanide anion aromatic?

https://en.wikipedia.org/wiki/Adamantane#Adamantane_cations

Two more contributing structures, for a total of six.  There is one with a bond on each edge of the "aromatic" tetrahedron.

This 3d-aromaticity has nothing to do with regular Hückel-aromaticity. The analogy is that thoses structures are mesomeric, that is they do not describe actual, discernible molecular structures. The inner bond of that three-ring, like the double bonds in benzene, does not exist as is, but is delocalised over all possible positions.

Aromaticity in Organic chemistry textbooks covers the Huckel version, which abide by the rules you have correctly listed above. However, these rules are mainly a tool to identify an organic molecule which has a high delocalization of electrons across a ring structure (a valuable tool in identifying a molecule's stability/reactivity/etc...).
So, for 3-dimensional aromaticity, the molecule must reach a certain threshold of delocalization in the ring structure (which, in 3D, becomes more of a sphere). Rules such as trigonal planar hybridization in 3D are no longer useful, and thus different approaches are taken to measure the resonance state of a system, and thus its level of aromaticity.
Nozawa et al. (2016, cited below) discusses use of "[t]he bond length alternation (BLA) in cyclic compounds" as "a good indicator of aromaticity, as it allows an evaluation of the degree of effective $\pi$-electron delocalization." These values, using a "harmonic oscillator model of aromaticity (HOMA)," are close to 1 for aromatic molecules by definition, and Nozawa obtains these from a variety of experimental and computational techniques to compare aromaticity in normally anti-aromatic structures.
For your above structure, as some comments pointed out as well, they are mesomeric, or resonance structures of the molecule illustrating the ability for the electrons to delocalize to an aromatic state. Firme et al's article mentions "delocalization indexes among bridged atoms," what I assume to be yet another similar measure of aromaticity related to the BLA using HOMA quantification.
Hope this helps! Please let me know if there is any confusion, as this response is a bit packed with summary detail.
References:
1. Nozawa, R. et al. Stacked antiaromatic porphyrins. Nature Communications 7, 13620 (2016).

By the way, are you a journalist or translator or so? The use of square brackets to enclose very minor text change in the citation seems rather confusing than useful here, in my opinion.

@mykhal, nope. Chemist and avid writer. It's simply the appropriate form for in-text citations when slightly modifying the punctuation and/or word choice to fit the sentence flow. Note is appreciated, but I don't believe there is any loss of meaning to be had.


text:
Do molecules like NH2 cation exist?
I saw that the book said that nitrogen cations must have four bonds and no lone pairs. I wasn't sure why you couldn't have a lone pair and 2 bonds.

If the nitrogen atom has one lone pair, it must have THREE bonds. So can you revise your question?

Nitrogen must satisfy the octet rule; having one lone pair and two bonds would be too unstable.

How could it have a lone pair and three bonds if the nitrogen has a positive formal charge? @SteffX

On a similar note, why is it that oxygen with a negative formal charge may have 1 bond and 3 lone pairs? I'm struggling with this concept to determine what the appropriate amount of bonds and lone pairs are. Why couldn't a negative oxygen have 3 bonds and 2 lone pairs? Wouldn't you still have 7 electrons? I feel like I'm missing a pretty important concept. @a-cyclohexane-molecule

Oxygen must also satisfy the octet rule: there must be eight electrons associated with each oxygen atom, either through lone pairs or bonds. With respect to bonds, both electrons in a bond count toward the eight electrons, as opposed to a formal-charge calculation, where only one contributes.

Such cations are called nitrenium ions and are very reactive.

Please cite your source.

The radical NH$2\cdot$ is also known. It  has C${2v}$ symmetry and has been studied in the gas phase. It has an absorption spectrum from 900 to 400 nm.

As @Mithoron said, $\ce{NH2+}$ ion is called parent nitrenium ion. The following text is an abstract from a paper1 which discussed its generation and various trapping pathways:

[...] (It) has been generated in various solutions through photolysis of 1-amino-2,4,6-triphenylpyridinium perchlorate. It was
found that the hydrocarbons, toluene and triphenylmethane trap
$\ce{NH2+}$ through two independent pathways. (1) free radical pathway
which leads to bibenzyl (in the case of toluene) or the trityl radical
(in the case of triphenylmethane). (2) hydride transfer
pathway which leads to benzyltoluene (in the case of toluene) or
triphenylmethanol (in the case of triphenylmethane). Results from dilution, triplet sensitization, and EPR experiments lead to the assignment of the hydride transfer pathway  to the singlet state of $\ce{NH2+}$ and the radical pathway to the triplet state of $\ce{NH2+}$ [...]
$$\ce{PhCH2+ + NH3 <-[PhCH3] ^{1}NH2+ -> ^{3}NH2+ ->[PhCH3] NH3+^{.} + PhCH2. }$$

There is also another paper2 which discuss about the reaction of $\ce{NH+, NH2+}$, and $\ce{NH3+}$ ions with $\ce{H2}$ at low temperatures which leads to formation of ammonium ions (the general pathway of $\ce{NH4+}$ generation in interstellar environment):
$$\ce{N+ ->[H2] NH+ ->[H2] NH2+ ->[H2] NH3+ ->[H2] NH4+}$$
This reaction scheme is very much temperature dependent as it can lead to different pathways. For instance, reaction between $\ce{NH+}$ and $\ce{H2}$ has two pathways:
$$
\begin{align}
\ce{NH+ + H2 &->[k^{a}_{NH^{+}}] NH2+ + H} &\quad \Delta H^{\circ} = \pu{-1.9 eV}  & \tag{1}\\
\ce{NH+ + H2 &->[k^{b}_{NH^{+}}] H3+ + N} &\quad \Delta H^{\circ} = \pu{-0.8 eV} & \tag{2}\\
\end{align}
$$
The former pathway occurs at 300 K while the latter one occurs at 15 K.
Reference

On the Solution Chemistry of Parent Nitrenium Ion NH2+:  The Role of
the Singlet and Triplet States in Its Reactions with Water,
Methanol, and Hydrocarbons Sanjay Srivastava, Michael Kercher, and
Daniel E. Falvey The Journal of Organic Chemistry 1999 64 (16),
5853-5857 DOI: 10.1021/jo990324+
Reaction of NH+, NH2+, and NH3+ ions with H2 at low temperatures
The pathway to ammonia production in the interstellar medium
S. Rednyk, Š. Roučka, A. Kovalenko, T. D. Tran, P. Dohnal, R. Plašil and J. Glosík, A&A
Volume 625, 2019, DOI: 10.1051/0004-6361/201834149


text:
Why is there a comparative lack of variety in positive counterions in drugs?
Pharmaceutical salts are important in the process of drug development. Using different chemical species to neutralise the parent drug can produce a diverse series of compounds, and this process is traditionally used to improve drug solubility and drug dissolution rates.
For negatively charged drugs which are paired with cationic counterions, there seems to be a general lack of variety in the counterion choice, which is usually a metal ion such as sodium. Other counterions such as ammonium cations (e.g choline and respective cations of benzathine, diethanolamine, diethylamine, meglumine, piperazine, procaine) or silver have not been used over the past 25 years.
On the other hand, there seem to be many more anionic counterions, most notably chloride, but also mesylate (e.g imatinib mesylate), besylate (amlodipine besylate), hydrocortisone succinate, and so on.
How can such a trend be explained in drug development process or chemistry of substances thereof? Could it simply be because there are fewer negatively charged drugs, or are there other explanations?

Lacking a background in medicinal chemistry, I cannot really give an answer. I would think that counterions are mostly governed by toxicity and solubility, in that aspect, $\ce{Na+}$ seems a reasonable choice.

Could this just generally be an availability issue? The only stable cations I can think of are metal cations, while there's a whole bunch of inorganic oxoanions (chlorate, chloride, perchlorate, chlorite, and hypochlorite are all "chlorine anions") and a lot more organic anions (e.g. conjugates to carboxylic acids)

Why do you think they weren't used? For example salt of diclofenak and diethylamine is used.

Here is what I managed to find:

In various researches, cationic counterions were associated with a high prevalence of unexpected incompatibilities in drug development processes. Owing to their low solubilities, (typically  <0.1% ) cationic counterions are infrequently used in many pharmaceutical preparations, however may be preferred to provide prolonged slow absorption while achieving effective plasma concentrations for some therapeutic uses, which is the case for benzathine and procaine penicillin G intramuscular injections.1 Furthermore cationic counterions, to a greater extent are susceptible to salting. Salting out results when highly hydrated inorganic ions (e.g $\ce{K+}$, $\ce{Na+}$) deprive organic ions and molecules of adequate water molecules to remain dissolved.
In addition another important factor influencing pharmaceutical salt selection is compatibility with excipients (diluents, binders, disintergrants, glidants, preservatives etc). Most amine derived counterions are incompatible with excipients. For example, the counterion of tromethamine (a primary amine) should not be used with reducing sugars as excipients because of a possible Maillard reaction, consequently limiting the available entities in the cationic counterions' pool .2
The trend can also be explained as a consequence of the changes in research techniques employed by the pharmaceutical industry. It can also be attributed to small absolute numbers of approved drug products containing salts formed from acidic entities.3
Overally, anionic counterions are favoured in drug development due to the wide availability of  weakly  basic pharmaceutically  active principles (approximately 75% of drugs are estimated to be weak bases) 4, consequently leading to correspondingly low availability of cationic counterions for acidic drugs which comprise only about 20%, with remaining percentage being neutral.

References

Newton D, W. Drug Incompatibility chemistry (American Journal of Health-System Pharmacy) Available from: http://www.medscape.com/viewarticle/590261.
Wu, Y., Levons, J., Narang, A. S., Raghavan, K., & Rao, V. M. (2011). Reactive impurities in excipients: Profiling, identification and mitigation of drug-excipient incompatibility. AAPS PharmSciTech, 12(4), 1248-1263.
Paulekuhn, G. S., Dressman, J. B., & Saal, C. (2007). Trends in active pharmaceutical ingredient salt selection based on analysis of the orange book database. Journal of Medicinal Chemistry, 50(26), 6665-6672.
Manallack, D. T. (2007). The pKa Distribution of Drugs: Application to Drug Discovery. Perspectives in Medicinal Chemistry 1( )25–38.


text:
Assigning the 13C NMR spectrum of bromobenzene and 4-bromobenzophenone
I've been trying to assign the peaks of 4-bromobenzophenone (I have the peaks, just need to assign them to the correct carbons), but my assignments and what it should be don't seem to agree. As part of this, I tried (and failed) to correctly assign the NMR of bromobenzene.
Bromobenzene's spectrum:
https://www.chemicalbook.com/SpectrumEN_108-86-1_13CNMR.htm
Bromobenzene, my labelling of the carbon environments: https://i.stack.imgur.com/2XPQ9.jpg
Here's what I've thought so far for bromobenzene:
The halogen atom is a deactivating substituent for the cyclic ring. The two effects of the halogen are the inductive and conjugative effect.
1) The inductive effect falls away with distance. Therefore, carbon environment 2 experiences a stronger inductive effect than carbon environment 3, and likewise for carbon environment 3 compared to 4. Therefore, due to the inductive effect only, carbon 2 has a more positive, and so higher shift than carbon 3, which has a higher one than 4 i.e. 2>3>4.
2) The conjugative effect acts mainly on carbon 2 and 4 by resonance arrows (as bromobenzene is ortho/para directing). Therefore, carbon environment 3 has a more positive environment, and therefore higher shift, than 2 and 4 i.e. 3>2=4
However, when it comes to carbon 1, it is directly attached to the bromine. The inductive effect is strongest for the carbon 1, therefore, and the conjugative effect isn't the same strength here as for carbons 2/4 and so carbon 1 should be the highest shift?
According to my rationalisations, carbon 1 should have the highest shift, followed by 2/3 (order is hard to determine) and then 4. This is not seen in the spectra.
Where am I going wrong then? I don't see how you can (qualitatively) predict that:
1) carbon environment 1 actually has the lowest shift
2) carbon environment 2 has a higher shift than carbon environment 3 i.e. the increased inductive effect outweighs the increased conjugative effect for carbon environment 2.

Good question. For 13C the shifts are never as intuitive as that for 1H, because the factors dictating shifts can be different (see linked question in next comment). I find that this is especially so for aromatics. It is sensible to give a range in which you expect to find aromatic peaks, but trying to assign them based on 1H NMR intuition/heuristics will nearly always lead one down the wrong path. In general I find that a HSQC (plus HMBC if there's more than one substituent) is needed to confidently assign aromatic 13C peaks.

Related: Nitrobenzene and other monosubstituted benzenes 1H vs 13C NMR assignments

Thanks for the reply! I'm assigning this NMR as part of my course, however, so unfortunately I only have the 1H NMR and 13C NMR spectra seperately.
With regards to the range, what range width would you be considering? For example, in my 13C spectrum of 4-bromobenzophenone, the aromatic peaks range from 132.7 ppm to 127.4 ppm. Would it be possible to separate the 132.7 ppm peak from the 127.4 ppm, or is this simply not feasible?

Bromine is an example of what is sometimes described as the 'heavy atom effect', where chemical shift follows an opposite trend than would be expected due to electronegativity. These types of atoms (Br to a lesser extent than for I and Te) have a large electron cloud distribution. The electron cloud has a significant effect on shielding of the ipso carbon, effectively immersing the carbon within a cloud of electrons. This contributes to the diamagnetic shielding around the carbon nucleus and results in an upfield shift. As an aside, because of their size, Br and I often have significant γ effects also, which are usually steric in origin.
For calculating 13C shifts at C1, the trend between halogen-substituted benzene series follows what is expected based on electronegativities. So, qualitatively, one might expect shifts to be in the order:
F  >  Cl  >  Br  >  I  >  H

However, you just need to be aware of the heavy atom effect for Br and I, and that these will result in an upfield shift for the ispo. Hence, one can expect, qualitatively, that the progression of shifts would look like:
F  >  Cl  >  H  >  Br  >  I

This heavy atom effect is also well demonstrated in the substituted methane series:
    CH4     CH3X    CH2X2   CHX3    CX4
Cl  -2.5    25.6    53.5    77.4    96.3
Br  -2.5    9.5     21.7    12.4    -29.4
I   -2.5    -24.0   -53.8   -139.7  -292.2

So, you can qualitatively predict shifts for C1, as long as you are aware of this contribution for Br. Predicting the order for C2/C3 is not so easy. Qualitatively, it is a reasonable assumption to say that the effect at C3 is negligible, that is, without any other knowledge, δ shift will be ~0. So you just need to look at the β carbon, or C2. Again, a reasonable guestimate might predict an upfield shift for those substituents with strong electronegativities. And, this holds for many cases. However, there are many exceptions to this, and the case becomes very complicated as soon as you start introducing multiple substituents. In this case, one should always assign peaks using 2D methods rather than just predictions.

Thank you for your reply, it has been very helpful!
However, regarding your comment that for C2, we may predict an upfield shift for substituents with strong electronegativities - shouldn't these substituents provide a downfield shift, conjugation of the halogen's electron pair notwithstanding?

Additionally, with regard to multiple substituents: seeing as my ultimate task is to assign the spectrum of 4-bromobenzophenone, would I be correct in the following assumptions:
For the ring with both Br and the ketone group: I may be able to rank the relative shifts of the ipso and para carbons, but not the ortho/meta positions.
Additionally, I cannot compare each ring's carbons ortho/meta carbons with each other, nor can I compare the carbons of each ring connected to the ketone group?


text:
Calculating mass from just density?
So I just started chem 101 and I'm not sure how to solve this problem. This is a problem from the dimensional analysis section of our book. It seems like the question wants me to get mass purely from density? The book never talked about that. It feels like I need more information to solve this problem.
Anyway here's the problem:

A substance known as aerogel has one of the lowest known densities of any solid. Around 2004, NASA used aerogel to obtain interstellar dust from the tail of a comet. How many pounds of aerogel are in a cubic foot? The density of aerogel is 3.0 mg/cc. Do not enter “pounds” as part of your answer. Do not use scientific notation. Report your answer to two significant figures.

I'm not looking to be given the answer. I want to know how to solve this.

Hmm, I thought NASA already used metric system at that time.

NASA did, but the book doesn't.

The question is poorly-worded: if you are asked to report your answer as "There are X pounds of aerogel in a cubic foot of aerogel," you necessarily have to use the word "pounds."
In any event, density is mass per unit volume; thus, you need to convert milligrams to pounds and cubic centimeters to cubic feet, and that's your answer (spoiler alert):

$$3.0\,{\mathrm{mg}\over\mathrm{cm}^3}\times\left({2.20462\times10^{-6}\,\mathrm{lb}\over\mathrm{mg}}\right)\times\left({1\,\mathrm{cm}^3\over 3.53147\times 10^{-5}\,\mathrm{ft}^3}\right) = 0.19\,{\mathrm{lb}\over\mathrm{ft}^3}$$

or

"There are 0.19 pounds of aerogel in one cubic foot of aerogel."

Thank you! And this is a question on the web that will only accept numbers, so the answer I have to give is just "0.19".


text:
What is the atomic radius of chlorine?
I'm trying to study for chemistry and there's a question in my book about identifying an element based on its atomic radius. But when I looked up the value for chlorine, I found different answers from different sources:

79 pm
175 pm (from the Princeton Review)
0.97 Å
175 pm (van der Waals radius?)
79 pm

What is the correct value?

According to the environmentalchemistry website, the unit is angstroms (the value 0.97 is equivalent to 97 pm)

There is no such thing as unified atomic radius. An atomic radius is a class consisting of van der Waals radii $R_\mathrm{vdW}$ (steric interactions), covalent radii $R_\mathrm{cov}$, and ionic radii $R_\mathrm{i}$ (and some other as well).
From the recent edition of CRC Handbook [1, p. 9-57]:

$$
\begin{array}{llrr}
\hline
\text{Element} & \text{Symbol} & R_\mathrm{vdW}~(Å) & R_\mathrm{cov}~(Å)\\
\hline
\text{Chlorine} & \ce{Cl} & 1.75 & 1.00\\
\hline
\end{array}
$$

Ionic radii are given for the crystallographic data separately [1, p. 12-12]:

$$
\begin{array}{llr}
\hline
\text{Ion} & \text{C.N.} & R_\mathrm{i}~(Å)\\
\hline
\ce{Cl^{+5}} & 3~\text{(pyramidal)} & 0.12\\
\ce{Cl^{+7}} & 4 & 0.08\\
\hline
\end{array}
$$

Reference

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.

Related: Pyykkö's summary of covalent radii, see my answer here.

@Martin-マーチン - Agreed. I think I've linked to Pyykkö's radii in other questions too.


text:
Polymers that aren't brittle at the cryogenic temperatures found on Titan?
For entertainment purposes, I'm imagining a colony on a non-terraformed Titan, and wondering what is achievable. Hydrocarbons are incredibly plentiful there, while metals and even silicon would be extremely rare and precious, which makes me think that polymers would be used to an even greater extent than on Earth in nearly every form of engineering.
On Titan, the pressure is around $\pu{1.5 atmospheres}$ and the surface temperature is cryogenic at $\pu{76 K}$ on average. This makes me imagine that they would want to use a lot of polymers with extremely low glass transition temperatures (below and around $\pu{76 K}$ at $\pu{1.5 atmospheres}$), so as to have rubbery and non-brittle materials that work outdoors.
Do we know of such materials, and if not, are they physically plausible? I didn't see any examples in tables of glass transition temperatures that I found on the internet, but then again, such polymers would have a lot less application on Earth, if they even exist as solids here.
Of course, these cryogenic temperatures only apply to things that go outside, like suits, vehicles, extraction and construction equipment, buildings, and the necessary pumps and other equipment needed for life support.

Unrelated, but for entertainment purposes you can read Stanislaw Lem's Fiasco. The first chapter "Birnam Wood" gives quite a colorful sci-fi interpretation of Titan's colonization.

Azotosomes, maybe?

related Titan questions How quickly might a Titan rover get covered in oil and dirt? Will it need windshield-wipers? as well as Any proposed missions to explore the black liquid on Titan? Technical challenges? and also What would a “water” landing on Titan be like? How viscous is the liquid?

Considering liquid nitrogen has a bp of 77 K, I highly doubt we have developed polymers that work at such low temps. This Book might help you out. Also see this article.

According to NASA1 the main factor inhibiting the widespread use of polymer materials in space environments is oxygen erosion, the materials tested so far to be proven effective in space are  Teflon, Mylar, Kapton and Tedlar.
Further experimentation is being conducted on board ISS according to their website2.
I do not believe there are polymers currently capable of handing the cold, pressure, and oxygen erosion of Titan, as such, the probe3 Huygens sent to Titan was constructed of aluminum coated Kapton, which seems to be the industry standard for space environments.
References:

https://www.nasa.gov/centers/glenn/shuttlestation/station/misse.html
http://www.spacestationresearch.com/tag/polymer-erosion-and-contamination-experiment-polymers/
https://en.wikipedia.org/wiki/Huygens_(spacecraft)


text:
Does "silica colloidal anhydrous" contain aluminium?
Some silicas such as sodium aluminosilicate contain aluminium. Is silica colloidal anhydrous one of these substances?

If it's prepared as commonly defined, it should be pure silica without an aluminum component.

It will probably have some aluminium in it, likely to be less than 0.1%, and most probably less than 0.01%. Not only that, also expect Ti and Zr in those quantities.
This level of impurities does not matter for most practical applications. If you do require exceptionally pure silica, you can buy it for a premium.
A note about “anhydrous”: when exposed to air, powdered silica will absorb moisture. Particularly if it’s amorphous. If you want it to be really anhydrous, dry the thing before using. Above 100 C will remove most of it, but completely removing all traces of water will require firing at ~1000 C in a noble metal crucible such as platinum.


text:
Briggs-Rauscher Reaction Temperature vs Oscillation Period
I am investigating the effect of temperature on oscillation period in the Briggs-Rauscher Reaction. By period I mean the average of the time the reaction to take one cycle (orange --> blue).
I have attempted to make a connection to the Arrhenius Equation with my data. However, I have no information about the rate constant and I am curious about how my period data connects to the rate constant.
Below is the graph I produced after assuming that the rate constant is linearly related to the period. Any insight on this problem would be very appreciated.

What period are you referring to exactly? The length of time between each oscillation gets shorter as the reaction progresses and eventually reaches equilibrium. You need to be careful to measure periods that are actually comparable. Also, what are the axis labels?

@jheindel Clarified the question.