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Can reaction of gastric acid with swallowed things be dangerously exothermic?
Okay, okay. I know that swallowing a large enough amount of any substance would be considered dangerous. That isn't the point of this question, however.
As you probably know, the stomach has hydrochloric acid in it to help digest food, and kill off some microorganisms. I suspect the swallowing aluminum foil (or a decent-sized ingot if aluminum) would be dangerous, because it reacts with hydrochloric acid exothermically to form aluminum chloride and hydrogen gas.
$$\ce{2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)}$$
Here is a video of aluminum's exothermic and vigorous reaction with hydrochloric acid:
https://youtu.be/LaaORZ0ofCY?t=1m12s

Now imagine that going on inside your stomach. I believe that the hydrogen gas may be able to escape through the mouth, but the intense heat might evaporate water (thus creating more gas that must escape), and would burn and kill cells inside the stomach. Plus, there's always the possibility that the hydrogen gas that is escaping could ignite, and burn through the mouth (where it escaped from) and into the body.
So why isn't everyone worried about accidentally swallowing aluminum foil? Wouldn't it be considered very dangerous to do so? Is there something I'm missing or otherwise not thinking of?

!! SAFETY !! - Swallowing any kind of metal is STUPID. If the metal pierces your intestines then you are facing a life threatening medical emergency.   // Assuming that this is strictly hypothetical, then @andselisk gave you the answer in his first comment. The hydrogen gas produced won't be dangerous.

I think its significance depends on the health conditions and biometrics of the given individual. For the majority there shouldn't be any severe consequences -- yes, transition metals are dangerous in general, but having certain decease may amplify the effect of how digestive system reacts to heat and AlCl3. I'm afraid I cannot provide a full answer beyond average calculated amount of Al that would dissolve (and, therefore, amount of heat released). If this is fine, I can post an answer, but it won't cover the question entirely. I retracted my vote, but it all seems for me too broad anyway.

I think the real question here is "Does Al react with gastric juice in similar way as it does with concentrated HCl solution" and the answer is "No it doesn't, you could even not notice any reaction if you used such diluted acid in lab.

For an average healthy adult person maximum concentration of hydrochloric acid
in gastric juice is $C = \pu{0.16 M}$, and on average there is $V = \pu{2 L}$ of gastric acid in the stomach, assuming it's daily secretion of roughly $\pu{1.5 L}$ and some leftovers. Considering that the acid hasn't been used for the purpose yet, and the following reaction
$$\ce{Al + 3HCl -> AlCl3 +1.5H2},\tag{1}$$
one can make an assumption that the mass of aluminium metal that can possibly be digested is
$$m_\ce{Al} = \frac{1}{3} CVM_\ce{Al} = \frac{1}{3} \times \pu{0.16 mol L^{-1}} \times \pu{2 L} \times \pu{26.98 g mol^{-1}} = \pu{2.85 g}.$$
In reality the outcome depends on the health conditions and biometrics of the given individual. People suffering from acid deficiency won't be able to digest even this amount. Larger or overweight specimen can have larger/stretched stomach, so the amount of acid and its pH may vary quite a lot.
On the other hand, having certain decease such as gastric ulce may amplify the effect of digestive system reacting to the produced $\ce{AlCl3}$ and heat. Also, aluminium(III) chloride is considered to be a neurotoxin. Acute oral
toxicity ($\ce{LD_{50}}$, mouse) of $\ce{AlCl3}$ is $\pu{3805 mg/kg}$. Reaction (1) produces approx. $\pu{14 g}$ of $\ce{AlCl3}$, which is not all that bad.
As @MaxW noticed in the comments, swallowing any metal is potentially dangerous and may or may not cause unwanted effects.

It might also be worth pointing out that a square centimetre of relatively thick domestic foil weighs just about .06g so you would need to eat a sample of foil about 7cm square to get that much Al into your stomach. If my calculations are right that is a lot to ingest, especially accidentally.

@matt_black yep, it seems about right, I just tried to make a spherical object out of 7x7 cm foil piece and I got a ball approx. 1 cm in diameter, which is not that hard to put into the stomach. Though I would like not to include my childhood experience to the answer  as is not something objectively reliable:)

2L of gastric acid in the stomach? Are you sure about that? At that amount your entire stomach (and then some) would be filled with acid.

I don't understand this answer. Would the reaction caused by swallowing 3g of aluminium be dangerous or not?

Things that enter the stomach are typically coated with saliva and mucus, which protects it from unwanted and uncontrolled reactions. Remember, your body isn’t a glass filled with HCl, there are other “things” there, which play a protective role. If you ever saw an animal swallowing a Christmas decoration (also made out foil) and then look how it comes out from the other end, you’d notice that it’s absolutely unchanged, which means our body has mechanisms in place to protect it from unwanted events that can throw homeostasis out of the window and cause death. Every single reaction is tightly controlled by enzymes and whenever something does not look right, it’s neutralized. Simple inorganic chemistry does not explain biological processes in full and does not give you a true picture of what is happening in the body. Whatever you learn in inorganic chemistry you’d have to relearn when you get to biochemistry as the game changes drastically. So the short answer is: if you swallow a piece of foil, it will likely not react with the stomach acid and will exit the body unchanged.

The problem is not that one must unlearn inorganic chemistry but that one did not learn enough inorganic chemistry in the first place to be useful in biochemistry. Enzymes require delicate balances of ionic concentrations and electrical potentials to do their regulating.


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Why in solution glucose not convert into galactose or mannose?
In glucose solution glucose is present in ring form but 1% can be in open chain form. As carbon carbon bond (c-c) roation is possible. Then why does Glucose can't be converted in galactose or mannose in solution?

Optical isomers are quite distinct from rotamers.

Yes, carbon-carbon bond rotation can happen in the open chain form of these carbohydrates. But you are confusing rotation with stereochemical configuration. This can only be altered by breaking bonds. Consider the following Fischer projection of D-glucose.

Compare this to the Fischer projection of D-galactose.

These two molecules differ only in the orientation of the hydroxyl, OH, and hydrogen, H groups. If you want to turn one into the other, you will have to break bonds and reattach them. This does not spontaneously happen in water and is frankly difficult even in the laboratory. You cannot merely turn one into the other through rotations around a bond.
It is a common mistake that chemistry students make thinking that they can simply rotate around one of the carbons to change the stereochemistry. If you build a model, you will tangibly see why this cannot happen. Models are great true to life representations of molecules (the better ones at least). You will find out that you cannot turn glucose into galactose without breaking several bonds.
And remember, rotations do not change stereochemistry. Only bond breaking and reforming does.

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Synthesis Golf IV: Plaunotol
A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

This fourth round of golf concerns the synthesis of plaunotol, a peptic ulcer treatment:
InChI=1S/C20H34O2/c1-17(2)8-5-9-18(3)10-6-12-20(16-22)13-7-11-19(4)14-15-21/h8,10,13-14,21-22H,5-7,9,11-12,15-16H2,1-4H3/b18-10+,19-14+,20-13-

In a departure from previous rounds of synthesis-golf, no stereocentres to worry about (hopefully easier for even more people to get involved), though several challenging alkenes for which the geometry must be controlled.
You may start from anything commercially available in the Sigma Aldrich catalogue, but to make this 'simple' challenge a bit harder, you must make each of the C=C bonds (i.e. no buying in the alkenes that form the product, but you may use an alkene if it is not preserved into the product).

Sorry its a bit early, I'll be away when this was meant to happen (though actually a month seems to be little long anyway, all of the action happens in the first fortnight or so...).

For clarification, may we start with alkenes so long as the alkene we start with is not preserved, for example as a precursor to one of the alcohols through hydroboration or something similar?

Are we allowed to use alkynes if they are transformed into an alkene of the final product?

Yep!  Though how you'll reduce them to give the right thing looks like it could cause a headache

Thinking about it I've definitely learnt more ways to control chiral centres than ways to make trisubstituted alkenes stereoselectively...

Polyketiiiiiiiiides....

@Zhe You were taught by Evans, no? Did he not impart his love of polyketides upon you ...

I took chem 206 with Evans. But my advisers were Jacobsen and Fu, so I'm more of an asymmetric catalysis acolyte. Dave still owes me lunch though. I need to cash that in... @NotEvans.

16 steps total
Longest linear sequence of 9 steps
Synthesis of Precursor 1:

The hydroformylation of allylamine, Rhodium catalysis (drew the product wrong, it has one more carbon please ignore it for now, I drew the right structure in the one-pot step)
Diazotization of product, occurs immediately before addition to other precursors in one-pot step later on (produces precursor 1)

Synthesis of Precursors 2 & 3

Hydroformylation of methylacrolein, Rhodium catalysis (produces precursor 2)
Protection of aldehyde via hemithioacetal
Protection of ketone through dioxolane
Deprotection of aldehyde by base (produces precursor 3)

Formation of grignard reagent by reaction with magensium of bromoethanal diethyl acetal

One pot reaction, diazo thermolysis resulting in carbene substitution followed by aldol
Weak acid dehydration to form alpha-beta unsaturated ketone
Wittig reaction to establish alkene
Dehydration using Rhenium Catalyst with good selectivity, click here for the paper
Reduction of aldehyde (special reducing agent used just in case free ketone exists)

Deprotection of the ketone (loss of ethane-1,2-diol)
Grignard reaction followed by dehydration to form the ketone of the product
Reduction with lithium aluminum hydride (not shown) to get product, Plaunotol

You are going to need to protect your aldehyde before you attempt to form the Grignard in step 2. Why would you expect to get the selectivity you show on the weak acid dehydration when you have a tertiary alcohol present as well as the dioxolane, both of which can potentially react?

Meant to do the weak acid dehydration in the presence of some Lewis acid to chelate the beta hydroxy ketone to achieve selectivity in that elimination.  And regarding your first comment, I probably would have to, but the ring closing is disfavored so depending on how fast the ether solution is added to the solution containing dicarbonyls, it may not be needed.

Inter- molecular reaction would still be a problem with the Grignard formation

That is true, protecting the aldehyde will likely result in failure of the one-pot step, but the compound is still accessible via a diazo route.  Start with acrolein, react with hydrazine, hydroformylation, then oxidation to get the diazo derivative of the Grignard, which will be suitable for the one-pot step.

Doesn't acrolein plus hydrazine give pyrazine? Easier to start from 4-Br-butanol

Apparently it gives pyrazole, but regardless I'm just saying that you can go via the diazo route.

Please add precedence sources for the one pot reaction at the beginning of your coupling sequence and for the Wittig reaction occuring at the saturated aldehyde preferably (with respect to the unsaturated aldehyde).

(You might actually want to change the order of steps; iirc, a Luche reduction should reduce the unsaturated aldehyde while leaving the saturated one untouched.)

That would eliminate that problem in the wittig step and the one pot step is more theoretical, I just thought that the carbene would add to most substituted carbon before the aldol, then the aldol could occur with one of the aldehydes, there are only three aldehydes in which the aldol could occur, two of which are basically equivalent (apart from a methyl group), so this would likely be one of the products.

Preface
As some might have guessed from the question I asked in the comments and others may have guessed based on the group I’m now working in, I decided to rely strongly on alkyne starting materials to build up the double bonds. Also, each double bond is built up in a different method. Of course, this just ‘works out’ but I also tried to make it happen this way. But enough pre-talk, let’s get to the topic.
Retrosynthesis

Scheme 1: retrosynthetic analysis of plaunotol (1).
The leftmost double bond can be built up from terminal alkyne 3 using a carboalumination with $\ce{Me3Al}$ and trapping the aluminate with formaldehyde (2). This terminal aldehyde moiety can be introduced in a Breit $\ce{C}_\mathrm{sp^3}\ce{-C}_\mathrm{sp^3}$ cross coupling using propargyl bromide (4) and allyl Grignard 5. To access this intermediate, I envisioned coming from a protected allylic alcohol 6, whose double bond would be synthesised the corresponding alkyne 8 in a silylformylation.
The rightmost double bond can then be built up in a standard Wittig reaction using acetone (10) as the ketone functionality. A Wittig reaction of Wittig salt 9 suggests itself here since this double bond is symmetric so no diastereoselectivity issues are encountered. The corresponding bromide can be accessed in a Suzuki-type cross-coupling using boronate 11 and 1,3-dibromopropane (12). Finally, trans-boronate 11 can be accessed from commercially available (i.e. in Sigma Aldrich’s catalogue) dialkyne 13 in a trans-selective hydroboration.
I chose this order of functionalisation for a number of reasons. Firstly, protecting the primary alcohol of 13 with a bulky silyl protecting group should allow differentiation of the two internal alkynes sterically. While I had originally considered carrying the boronate along the synthesis, I noticed that no harm is done in immediately elaborating the right-hand terminus.
The silylformylation reaction attaches the carbonyl functionality on the less hindered side, meaning that a simple alkyl chain should direct the carbonyl to its side rather than the bulky silyl protecting group. Also, it was necessary to install the silyl before deprotection of the primary alcohol so both deprotections can take place in one step. This leaves only the elaboration of the final, terminal alkyne. It makes sense to have this last since trimethyl aluminium is a very small reactant that could, in principle also attack any remaining internal alkynes. By having only one alkyne functionality present in the molecule, this is no longer a problem.
Forward synthesis

Scheme 2: proposed forward synthesis. Abbreviations: $\ce{Cp^*}$: 1,2,3,4,5-pentamethylcyclopentadienyl; $\ce{L}$: generic ligand, e.g. acetonitrile; $\ce{dba}$: dibenzylideneacetone; $\ce{TBAF}$: tetrabutylammonium fluoride; $\ce{Cp}$: cyclopentadienyl.
As mentioned earlier, the initial step is the protection of the primary alcohol by the Corey procedure to give silyl 15. Subsequently, Fürstner’s ruthenium-catalysed method of trans-selective hydroboration is used to install the required trans-boronate 11.[1] As mentioned, the selectivity should arise from the bulky silyl group; if necessary, more bulky groups like $\ce{TBDPS}$ can be used instead. The subsequent $\ce{C}_\mathrm{sp^2}\ce{-C}_\mathrm{sp^3}$ cross-coupling with dibromide 12 is performed with $\ce{[Pd2(dba)3]}$ as published by Nishihara et. al. to give bromide 16.[2] This can be transformed into Wittig salt 9 by standard procedures and reacted with acetone to give diene 8 with practically any base.
With the first two double bonds installed now is the time for the silylformylation employing the methodology of Matsuda et al.[3] Thus, tetrarhodium octacarbonyl catalyses the addition of methyldiphenylsilane under positive $\ce{CO}$ pressure to give aldehyde 7. This is reduced under Luche conditions and the resulting alcohol protected with $\ce{MOM}$. Disilane 6 can then be deprotected using $\ce{TBAF}$; I chose to buffer with acetic acid to prevent undesired isomerisations or double bond migrations due to the highly acidic conditions. Potentially, less reactive fluoride sources will also do the trick.
After deprotection, the primary alcohol is converted to bromide 17 in an Appel reaction ($\ce{PPh3, CBr4}$. This bromide 17 is transformed into the corresponding Grignard and coupled in the presence of lithium bromide and $\ce{CuBr}$ with propargyl bromide (4) to give the terminal alkyne 3.[4] Our final remaining double bond is installed under the action of trimethylaluminium and capturing the resulting aluminate with formaldehyde as reported by Patel et al.[5] This only leaves the final acidic deprotection of the $\ce{MOM}$ ether with practically any acid.
Step count and remarks
Unfortunately, this reaction is linear. It requires 13 steps — formation of the Grignard and subsequent Breit cross-coupling is a one-step two-stage procedure.
Carbon sources are dialkyne 13, dibromide 12, acetone, carbon monoxide, propargyl bromide, trimethyl aluminium and formaldehyde. I confirmed that 13, 12 and 4 are available at Sigma Aldrich and I have no doubt about the availability of the remaining.
References
[1]: B. Sundararaju, A. Fürstner, Angew. Chem. Int. Ed. 2013, 52, 14050–14054. DOI: 10.1002/anie.201307584.
[2]: Y. Nishihara, Y. Okada, J. Jiao, M. Suetsugu, M.-T. Lan, M. Kinoshita, M. Iwasaki, K. Takagi, Angew. Chem. Int. Ed. 2011, 50, 8660–8664. DOI: 10.1002/anie.201103601.
[3]: I. Matsuda, A. Ogiso, S. Sato, Y. Izumi, J. Am. Chem. Soc. 1989, 111, 2332–2333. DOI: 10.1021/ja00188a074.
[4]: K. Xu, N. Thieme, B. Breit, Angew. Chem. Int. Ed. 2014, 53, 2162–2165. DOI: 10.1002/anie.201309126.
[5]: H. H. Patel, M. S. Sigman, J. Am. Chem. Soc. 2016, 138, 14226–14229. DOI: 10.1021/jacs.6b09649.

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I have a question regarding chemical bonds
can NaCl bond be called polar covalent bond? I understand that it's ionic bond
but Na is still sharing its electron with Cl. I am confused with Ionic and polar covalent bond.

There should be some duplicate around here somewhere that explains how to call a bond (if ionic or polar covalent is more acceptable). Unfortunately, I'm too lazy to look for it now.

The sodium and chlorine ions do not actually share their electrons. Instead, chlorine "takes" an electron from sodium. They are kept together in a crystal lattice by electrostatic interactions between the net +1 charge of Na and net -1 charge of Cl, thusly making the bond ionic. If they were truly sharing their electrons, then yes, it would be considered polar covalent. But if that also were true, then dissolving NaCl in water would be much more difficult as the sodium and chloride cannot be separated.
An example of a polar covalent bond will be between carbon and oxygen, say carbon monoxide CO. These two atoms actually share their electrons and make it more difficult to separate the two atoms.

You're wrong and OP's right... All "ionic" bonds are partially covalent.

https://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound/17072?s=3|0.2164#17072

@Mithron As this is the OP's first post, and because they are having difficulty differentiating b/w covalent and non-covalent, I assumed they are a student of general chemistry. It is best not to confuse these students with % ionic character so early in their education. But yes, you are right in that there exists partial covalency.

Well, denying what he already knows would be rather more confusing... Edit your post.


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Is my electrolysis emitting compounds from additional reactions?
I am separating water into hydrogen and oxygen via electrolysis.
The electrolyte solution is sodium hydroxide (NaOH) in deionized water.
Electrodes are 99.96% pure nickel.
All gases and liquids are perfectly clear.  Even after running for hours, the electrolyte solution is crystal clear and colorless.
However, the reaction seems to be emitting something that burns the nose and throat of everyone in the room.
Is that some sort of gas from a side reaction at one of the electrodes, or is it simply bits of NaOH being carried into the air by the splashing bubbles?

What is the NaOH concentration of your solution?

I used about 250 grams, in 6-8 litres of water, so maybe 4-5% ?

I think your guess of the NaOH forming aerosols from the bubbling is pretty likely. I don't know what your apparatus looks like, but it may be that just covering it, or the bubbling portion of it, could resolve your problem. Of course protect your eyes when working with this solution (I don't know your level of chemistry training either, so have to mention it ;). Maybe someone will come up with something else, but I suspect that is it.

@MadScientist. From your description it smells like Ozone. Perhaps you are at some overpotential.

TL;DR: Your suggestion that the problem is "simply bits of $\ce{NaOH}$ being carried into the air by the splashing bubbles" is very likely correct.
You have a very simple system, meaning a minimal number of variables to consider in order to eliminate other possibilities. For example, if you were using $\ce{NaCl}$ as the electrolyte, then chlorine gas could be the irritant. But with $\ce{NaOH}$ as the electrolyte there really is no side reaction that could produce such an irritant.
On the other hand, $\ce{NaOH}$ itself can irritate sensitive tissues like mucous membranes, and even skin for that matter. According to one of your comments, you are using a $\ce{NaOH}$ strength of a bit over 1 molar. Droplets of this solution will eventually be neutralized by $\ce{CO2}$ in the air, but at that strength it should stay around long enough to elicit the symptoms you described.

Thanks for that.  Can these aerosol droplets be filtered using something like glass wool?  I'm pumping out the hydrogen and don't want to contaminate the output with NaOH.

It's hard to say what would work best without knowing what your apparatus looks like. You can probably just cover the area where the bubbles are coming out of solution with a plastic lid just over the surface. No need to be air-tight or anything, you just need to stop the splattering from freely going into the air. Again, not having seen your apparatus, that's kind of a shot in the dark answer.

Dependending on the potential you apply, side reactions occur. One taking place at moderate overpotential is
O2 + 2 OH-  giving O3 + H2O and 2 electrons. .
Considering the above in conjunction with

the alcaline pH you are working with,
that the "smell" spread to the room in spite of CO2 neutralising the hypothetical NaOH droplets

I really think that what irritates your nose is O3.

Thanks for that.  Any idea how what voltage/pH I can have in order to prevent this?

I think it might be easy and safer to work with NaCl but it is also easy to get Cl2. Why you do not try some sulphate, e.g. Na2SO4? Moreover try to control the potential, just enough to keep your stuff bubbling.

You should get water electrolysis with about 1.5 V. The point is that overpotential (for water splitting as well as for secondary reactions) will depend on your specific setup. Are you demonstrating or somehow producing? :)

It's not for demonstrating purposes, I'm producing H2 at a large scale (inflating large lighter-than-air aircraft), and I produce much more with higher voltages.  (Yes, I know, Hindenberg..)

Then it is your source. You need current rather than voltage. Is really matter of setup. Anyway give a try to sodium sulphate


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Why do polarizable nucleophiles favor SN2 over E2?
Weakly basic ions and polarizable bases favor $\mathrm{S_{N}2}$ over $\mathrm{E2}$. Why? The argument used that polarizable nucleophiles form bonds earlier, stabilizing the transition state, can be applied to either $\mathrm{S_{N}2}$ or $\mathrm{E2}$.

Polarisable Nucleophiles increase the rate of an Sn2-type reaction but not an E2 reaction because the more polarisable a base is, lower the electronegativity and charge density (conversely, higher ionic radius) it has.
Now, we know that the basic strength depends on these factors( in a period, we check the electronegativity and in a group, the ionic radius).
As, the Rate Determining Step of an E2 reaction involves the attack of a base, lower basic strength would mean that the reaction would be slower. Hence, more polatisable bases means that E2 would be slower.

But why doesn't the stronger bonding in the TS for E2 that would result from using a more polarizable nucleophile matter

the base is only used for removing H+ ion from the carbon in the rds.

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What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride?
What is the hybridisation of $\ce{F}$ in $\ce{BF3}$ or $\ce{CH3F}$ or $\ce{PF5}$ or $\ce{SF6}$? My book states that the p orbitals of fluorine overlaps, but I feel that sp3 hybridisation occurs in $\ce{F}$, so its sp3 hybrid orbital overlaps with the empty p orbital of boron.
According to Ron's answer,  all atoms with p or d orbitals can hybridise.

It's unhybridized.

@xasthor please see Ron's answer in the link!

Just because it’s Ron doesn’t mean it must be correct ;)

@Jan Ok I understand that there are no general rules in chemistry! I mostly refer to his answer because he explains things in very elementary manner :)

I was once very puzzled with this as well. When I consulted a Chemistry expert the other day, what I learnt was that hybridisation is a concept that is used to explain features of a molecule, such as geometry. Thus, determining hybridisation states for terminal atoms would not be useful and would be unnecessary. What implications would there be if it were sp or sp3? There would likely be not a lot of difference in the consequences. Thus, I would like to argue that it is unnecessary to determine the hybridisation states of terminal atoms. Please feel free to correct me if I'm wrong.

TL;DR: As a rule of thumb, terminal (heavy) atoms are almost always best described as having sp hybrid orbitals (at the most).
For more on this, I refer you to my answer on What is the hybridization of chlorine in vinyl chloride?
Here are the relevant visualisations for $\ce{BF3}$ at the DF=BP86/def2-SVP level of theory. First I'll start with the canonical orbitals obtained from a molecular orbital point of view. The molecule has $D_\mathrm{3h}$ symmetry, therefore some orbitals are degenerated.

With appropriate transformations, we can interpret these orbitals in a localised fashion, which will give us hybrid orbitals. In this view the σ bond orbitals, and the lone pairs are triply degenerated. Note that there is a significant amount of delocalisation of the fluorine lone pairs into the empty boron p orbital.

Here is the representative section from the natural bond orbital analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(core and symmetry equivalent orbitals skipped)
   5. (1.99345) LP ( 1) F  2            s( 63.79%)p 0.57( 36.19%)d 0.00(  0.01%)
   6. (1.94124) LP ( 2) F  2            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
   7. (1.86896) LP ( 3) F  2            s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
  14. (1.99778) BD ( 1) B  1- F  2
               ( 17.29%)   0.4158* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 82.71%)   0.9094* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.38551) LV ( 1) B  1            s(  0.00%)p 1.00(100.00%)d 0.00(  0.00%)
  18. (0.05374) BD*( 1) B  1- F  2
               ( 82.71%)   0.9094* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 17.29%)  -0.4158* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)

Here you can clearly see the approximate two sp orbitals and the remaining p lone pair orbitals.
The above remains true for (almost) all terminal atoms, because of the local linear symmetry (only one bonding partner, negligible external field), although with lesser extend for heavier atoms (like bromine). This is because hybridisation in general becomes a less reliable description due to the increase in the s-p gap.

Thanks! Is it true for all Halogen atoms like $F,Cl$ etc in such compounds like $PCl_3$ or $CH_3Cl$ ?

@samjoe it is true for all terminal atoms. To a lesser extent for very heavy elements because of the general lack of hybridisation due to a large s p gap.

Hehe, I started writing up an answer but then I had to go back to the wet lab. You obviously beat me to it ;)

@Jan I hope you would have been saying the same, and I also hope I didn't ruin too much of your work. Knowing that, I'm glad I hit send before I went to the seminar then, it would most likely have been the other way around ;)

@Martin-マーチン What is DF-BP86/def2-SVP theory?

@samjoe It is a computational method to calculate approximations for the wave function. In this case it is a density functional with a specific basis set. This level usually give reasonable results for not much computational cost. The details of the methodology are only interesting for those who are regularly working with it. For the intents of this answer, you can ignore it. It's just a more accurate way to generate orbitals than guessing.

It would have ended up more or less at the same conclusion except without the cool orbital calculations and more based on thought experiments ;)


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How does barium chloride identify sulfate ions?
What is it about barium chloride specifically that makes it able to form a precipitate with sulfate ions.
What is the chemistry behind this?

Not necessarily about $\ce{BaCl2}$ itself, but when treated with a more reactive compound (containing $\ce{SO4^2-}$) ions such as sulfuric acid, the ($\ce{Cl^-}$) ions in   $\ce{BaCl2}$ are displaced with the  $\ce{SO4^2-}$ essentially forming an insoluble precipitate:
$$\ce{BaCl2(aq) + [SO4]^2-(aq) -> BaSO4(s) +2Cl^-(aq)}$$
$\ce{BaSO4}$ is a sparingly soluble salt ($K_{\mathrm{sp}} = 1.07 \times 10^{-10}$).
(The solubility of the ionic salt is extremely small, such that the saturated solution has very few ions)


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Are acyl chlorides acidic?
Are acyl chlorides such as $\ce{CH3COCl}$
 acidic in water or any other solvent?
Are they in any context referred to as acid chlorides? Or is that just a mistake?

Acyl chlorides don't even exist in water. They would hydrolyze instantly and produce acids.

Of course. I forgot. So is that why they are considered acidic? Beacause they produce carboxyllic acids in reaction with water.

@IvanNeretin They don't hydrolyze that fast. You can do kinetics at zero degrees C: http://pubs.acs.org/doi/abs/10.1021/jo9609844?journalCode=joceah

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Why are p-type solar cells more prone to degradation in space than the n-type ones?
Why are p-type (boron doped substrate) solar cells more prone to degradation in space than n-type (phosphorus doped substrate)?
All the sources I've looked at simply state it was found that p-type degrades slower, hence they were used for space application.

What is the main damaging radiation from the sun?
Why is it more damaging to p-type than n-type?

In your second paragraph you mean n type

The space radiation environment is a quite complex subject. Generally speaking one needs to be careful about just where in space you are concerned about - near earth orbit (within the Van Allen Belts), outside the belts, regularly passing through the belts, interplanetary, etc.
That being said, the three main areas of concern are protons from the solar wind, electrons from the solar wind, and cosmic rays from the rest of the universe. For testing purposes the standards for the solar wind are 10MeV protons and 1MeV electrons. I will focus on these for the purposes of this answer.
As the protons or electrons hit a solar cell, the will go in some ways and lose energy through interactions with electrons and the nuclei (electronic and nuclear stopping, respectively). Electronic excitations at best make a little more current in the depletion region. The interactions with nuclei are more problematic - either particle can transfer enough energy to a nucleus to displace it off a crystalline lattice site (protons are about 1000 times more effective at it then electrons because their mass is much higher). So, as the cell accumulates dose, more and more atoms are off their sites.
OK, what does that do? Well, point defects in semiconductors often create mid-gap electronic states that allow carriers to recombine - your cell is now less efficient at making current since some gets lost. Further, point defects (vacancies and interstitials) can combine with either other point defects or impurities to form more complex defects, and these may be even better recombination centers.
Of particular relevance for your question is the impact of boron. To quote from
T. Hisamatsu et al. Radiation degradation of large fluence irradiated space silicon solar cells (1998):

Radiation-induced defects associated with boron atoms are said to form
  recombination centers for minority carrier electron and this phenomenon is said to
  cause the degradation in low-fluence regions. Therefore, using highly boron-doped
  substrates in order to avoid the sudden drop of cell performance should not be
  a proper means to provide cells with higher radiation tolerance.


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What is smell??How come 2 things have the same smell
I have been wondering how something smells.
How come something smells the same as other thing? What is similar in them?
Like what is the chemistry behind a particular smell of a substance?

For example compounds with SH (thiol) smell awful and similar, so what is similar between them? I know SH, but I am asking why can't they smell different.

You aren't surprised that two things may have the same color, are you?

I'm afraid unless you provide a certain example you question is way too broad.

This is probably more of a biology question, no? Smell is a property of the interaction between a molecule and the receptors in the nose. Although compounds with a similar smell may have the same functional groups, they may be very different. On the other hand, there is D/L- limonene that has a different smell for each of the two isomers.

The bit in your post that's suggestive of being off-topic(ish): "smell awful and similar". Smell is human perception, not a physical quantity, and hence doesn't really pertain to "pure" Chemistry. Sure, in Chemistry we talk about "smell" a lot, but that's really just limited to the laboratory ( "Phenolic" smell, "Pungent" smell, etc), but wanting to know how something has a particular odor is really more of a human physiology question. I'm not saying there's no Chem. involved...but this is way better off on the Biology.SE. As such, I vote to keep this closed.

Question could be migrated, but premise isn't true, I'm pretty sure one can make thiols with distinct smell. It's not that simple.

http://blogs.sciencemag.org/pipeline/archives/2006/04/20/sulfurous_stenches_a_connisseurs_guide

Smell is not one sense, or three like with colours, but a few hundred. Thiols strongly trigger one of them. And there are of course nuances in thiols. Stick your nose in there until the thiol sense is saturated, and you will start noticing. Wait, DON'T, thiols are not only disgusting but many of them also toxic.


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Correlation between electron counting and structure for metal borides
When it comes to metal borides, the electron counting and its correspondence to structure seems weird. For instance, $\ce{MgB2}$ and $\ce{AlB2}$ are isostructural, which is surprising.
Further consider the example of metal hexaborides. $\ce{MB6}$ structures are isostructural for  $\ce{M}= \{\ce{Ca}, \ce{La}, \ce{Th}, \ce{Ce}, \ce{K}\}$. How does the bonding in in these metal diborides, hexaborides, and so on occur?

You need to stop thinking of molecules and start thinking of crystal structures. These compounds (and many more) can all be thought of as a metal structure with the B (and often C, N, and/or O) sitting in particular interstitial sites within the metal structure. The fact that many metal-(B,N,C,O) compounds have a broad range of mutual solid solubility is a hint that the bonding is not what you think.


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Why can't Schiff's reagent distinguish between glucose and fructose?
I have heard that Schiff's base only gives a positive test for aldehydes. Glucose is an aldehyde. However, I was told that glucose doesn't give a positive Schiff's base test. Is this correct? If so, how?

Hello, and welcome to ChemSE. I have completely modified your question, but I retained it's meaning, for it was difficult for other users to understand your question. I hope your question will get reopened soon, and you will get an answer. Good luck!

My bad...My doubt wasn't framed right..

Related: https://chemistry.stackexchange.com/questions/64363/why-do-alpha-hydroxy-ketones-respond-positively-to-tollens-fehlings-benedict https://chemistry.stackexchange.com/questions/40127/why-does-fructose-reduce-tollens-reagent-and-fehlings-solution

Schiff test is a colorimetric method for determination of aldehyde groups. Both glucose and fructose are monosaccharides and exhibit oxo-cyclo tautomerism. Glucose is an aldose and has aldehyde group in linear form. Whereas fructose is a ketose and doesn't give a positive Schiff test.


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Glycoside Formation Reaction
Whenever we add $\ce{H+}$/ Ethanol to glucose in its hemiacetal form, why doesn't pinacone/pinacolone rearrangement take place in place of nucleophillic substitution. The product would lead to formation of a carbonyl bond and so, should be the favoured product.
The actual pathway is:
But, why can't the reaction occur like this with ethanol as a polar solvent?:

Not on topic, but I feel like it would be a disservice to you to not point this out: your curly arrows for the first step (protonation and loss of water) aren't correct. The protonation and loss of water should also be two separate steps. I would also prefer to depict carbocations adjacent to oxygen with the alternative resonance form on the right: $$\ce{>C^{+}-OH <-> >C=O^+H}$$

First of all, there's no pinacole in the first place, so pinacole/pinacolone rearrangement isn't possible. Your rearrangement reaction/hydride shift wouldn't be called pinacole/pinacolone rearrangement.
Second, we have to think about why this would happen. Such rearrangements happen because we produce a more stable carbocation. In case of pinacole/pinacolone, if we do the rearrangement we end up with:

Which is well stabilized and after deprotonation a stable compound.
In case of sugars it's a little bit different, since it's not an alcohol in 1-position, it's a hemiacetale. This is extremly important since it has quite different reactivities.

We can cleave of water under much weaker acidic and mild conditions than in case of alcohols. The produced carbocation is quite stable since we got an oxygen stabilizing the cation, therefore no rearrangement is happening.


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How to measure energy density of 1 gramm of fat
I'm an absolute beginner in chemistry.
How do I measure the energy density of fat?
As a specific example, I'd like to verify that 1 gramm of fat has 37 kilojoules.
I guess that measuring the energy density involves burning that fat.

What type of food are talking ab . . . Actually, that doesn't matter. Since food isn't just fat, this gets way too broad, even for just a specific type of food. You'd never get accurate enough.

Burn it! Calorimetry.

Thanks for your comments! @M.A.R. If we never get accurate enough, where do the figures on the nutrition labels come from?

Hi @MatthiasBraun, the determination of caloric content, in the US at least, is briefly discussed in the article below. The accuracy of the measurement is dependent on the sensitivity and and quality of the instruments used. If you perform this measurement with common items, the accuracy will suffer due to sources of error.
https://www.scientificamerican.com/article/how-do-food-manufacturers/

You are correct that you would need to burn the sample, in your case a fat. The basic approach is to burn the sample in an apparatus that will cause the heat generated in the reaction to warm up a known quantity of water. The temperature change of the water is measured, and with a known mass of water and the water's heat capacity, you can calculate the amount of heat energy that was generated from burning your fat sample.
Here is an example procedure of how to conduct a basic calorimetry experiment to determine the energy content of a sample of food and the equation that is used to calculate the energy content. Note, however, that this method is very rudimentary and will not guarantee a high degree of accuracy.


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What's the most trusted source for bond lengths?
I'm doing some computational work and am manually inputting Z-matrices for the molecule (no GUI support for my program). For this, I need bond lengths:

I tried Herzberg's book
I tried a basic optimize in Avogadro and then asked it to generate a Z matrix.
I checked Scifinder for structural properties

However, bond lengths vary quite a bit from these sources (upto 0.4 Angstroms). I understand that these sources are a few decades apart from each other.
Should I just stick to the newest source? Or should I stick to Herzberg because he's like the established standard?
(I'm not sure what are the best tags for this question, please feel free to suggest more/edit them)
Thanks!

Your (2) is not a source at all. Also, some bond lengths vary quite a bit in real life, so even the newest sources would not remove the discrepancy.

I'd immediately answer single-crystal analysis data of a given compound with well-established ESDs for atom coordinates, but it's only for the solid state.

for small molecules look at NIST databases.


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Nucleophilic aromatic substitution on pyrimidines: C2 vs C4 selectivity
[There are many similar questions on Chemistry.SE (e.g. 1, 2) and a quick search will reveal all of them, but nothing on this in particular.]
The reaction of 2,4-halopyrimidines with nucleophiles generally leads to selective displacement of the halide at the 4-position:1

In their textbook, Joule and Mills don't explain why, though, simply saying that it "is the normal situation for nucleophilic – 4-chloro > 2-chloro".
I don't see any obvious difference in the stability of the anionic intermediates, either (the so-called Meisenheimer complexes).
What can explain the regioselectivity of the above reaction?

Reference

Joule, J. A.; Mills, K. Heterocyclic Chemistry, 5th ed.; Wiley: Chichester, U.K., 2010, p 257.

This is a simple enough system that it might be amenable to computation...

It's maybe not as obvious as you might think, @Zhe. If you go via just the electron density (or even just the NMR shifts/how deshielded) it kinda looks like it should go the other way round.

My question is linked anyways. I've also synthesized and calculated 2- and 4pyrimidyl compounds and in ym case the 4-pyrimidyl residue is more electron withdrawing than the 2.

@notevans I specifically chose not to make assumptions about electronics or orbital overlap factors. If the computed preference agrees with experiment, you can then proceed to rationalize based on the results.

What kinda mechanism does this reaction undergo ?

An explanation can be provided by frontier molecular orbital theory.  The LUMO coefficient is higher on the C4 vs C2.  This is a general effect meaning that an EWG para to the leaving group is more reactive than ortho.  See 4-halopyridines vs 2-halo and 4-nitrobenznes vs 2-nitro.  There could also be an effect from the repulsion between the nitrogen lone pairs and the approaching electron rich nucleophile, which would be greater in proximity to 2 nitrogens rather than 1.

Try drawing the resonance hybrides out. You'll soon realise that the intermediate formed when substitution accours at the c4 position can be stabilised by three resonance hybrids!
:)

Unless I’m gravely mistaken, the same can be said of the intermediate for C2 substitution.


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Deposition of elemental sulfur
I want to create a thin film of elemental sulfur. Based on significant searching, I can find that elemental sulfur deposition is a significant problem in oil and gas. However, I cannot find any specifications for depositing elemental sulfur via CVD, ALD, or other methods.
Can anyone suggest a method for creating a thin film of elemental sulfur? Thanks!

Evaporation would seem to be a direct way. The melting point is so low that it should be pretty easy in either a boat or e-gun system. Watch for sulfur-metal interactions on the boat. It may also react upon opening the system up.

Can you explain a little more exactly what you mean? I have previously tried sublimation, as mentioned in the paper "A binder-free sulfur/carbon composite electrode prepared by a sulfur sublimation method for Li–S
batteries" by Cui et al. However, I found that very little sulfur actually deposited and it was not possible to obtain a full film. However, my processing may have been imperfect.

Reaction of copper sulfate and sodium thiosulfate in a 1 to 1 ratio.


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Why polymers have a range over which they melt instead of having a specific melting temperature?
Does it have something to do with the fact that all chains are not of the same length?

The uniformity of the chain matters as well. For example: https://en.wikipedia.org/wiki/Tacticity

Many polymers do not have a melting point at all.

what sort of polymers, polyethylene, proteins, dna?

No. Although we need to be careful in defining what is and isn't a "polymer". The nature of (nearly all) polymers is that they are (covalently) bonded (in at least one dimension) over large (relative to atomic diameters) distances. The nature of a liquid is that is exhibits no long range order, no long range "connectivity". There are, as far as I know, no polymer crystals, in their most ordered configurations they are quasi- or semi-crystalline, having disordered zones as well as often but not always having zones of some regularity/order. Often the degree of crystallinity of a polymer is determined by its processing history (temperature, pressure, and stresses applied to it), but even the most crystalline of polymers is not fully crystalline. What this means is that the transition between solid and liquid phases is not sharp; the disengagement of the polymer chains from one another takes place gradually with no sharp/distinct difference between the solid phase and the liquid phase. Perhaps a better way of saying that is that the polymer becomes more and more amorphous but continues to have some long-range order as each chain's motion relative to its neighbors increases. see https://en.wikipedia.org/wiki/Reptation and https://en.wikipedia.org/wiki/Amorphous_solid. Note that the presence of polymer  chains (assuming linear or nearly linear polymers) of varying MW increases the gradual transition from amorphous solid to polymer melt. If the amount of oligiomer is large enough, it may act as a solvent for the larger chains. Note also that the theory of glass to liquid and glass to solid phase transitions is an active area (not yet "settled science").


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Can sulphur from low-flying planes be cause of silver tarnishing rapidly?
My silver dishes, tea set, etc. that previously needed polishing once or twice a year are now turning black in just a few weeks. This area has low-flying planes.  Could they be releasing sulphur which causes the blackening of silver?

Related: Abundance of sulfur containing compounds in the air

Planes normally don't release all that much sulphur.

In general, yes, this theoretically might be one of the reasons as sulfur is imminently present in the majority of jet fuels (e.g. International Standard for Jet-A1 fuel allows up to 0.30 w/w% S (mercaptanes mostly)), but before suing the airline companies I'd check out local analysis reports on sulfur dioxide air content.

There are 3 main culprits for the silver tarnishing reaction: hydrogen sulfide, water, and oxygen. Please note that sulfur dioxide doesn't (directly) contribute to this. Airplane exhausts are principally CO2, H2O, and NOx with small amounts of ozone, sulfur dioxide, and particulates. Note: no H2S (hydrogen sulfide). Increased humidity is one obvious reason why you might see an increase in the rate of tarnishing - have you changed the heating/air conditioning that your silver is exposed to? Is the weather wetter now than in the past? It's also been established that NOx accelerates tarnishing - and anything that burns (in air, which is 78% N2 and 21% O2 and about 1% H2O) makes NOx. (NOx is shorthand for a family of nitrogen oxygen compounds). It gets worse: the more you polish silver the faster it tarnishes. So, you'd expect the longer you own the silver that you'd find an increasing need to polish it if you wanted to maintain its silvery appearance (and didn't use it much). There are other environmental sources of H2S: volcanoes, gas wells (frakking has increased dramatically in the last 2 decades), chemical plants, sewers, water treatment pools, landfills. So, while it's possible that increased emissions from near-by internal combustion engines (cars and planes and boats) could be making the problem worse, I'd not expect aircraft emissions to have much of an effect. I've read (but have no personal experience with) anti-tarnish bags, cloths and strips do wonders. Also, you should consider a well sealed cabinet perhaps with a desiccant (remember humidity accelerates tarnishing) if you display your silver. The experts use activated carbon and zinc oxide in their cabinets to absorb the H2S, but that seems excessive for home use. I'd try the 3M strips and as small an air volume as possible. I'd also be willing to try large desiccant packs but have some doubts that they'd work for long enough to be worth while (unless you kept the cabinet or bags really well sealed and almost never opened it). Just my 2 cents, HTH. I should also mention that kitchens (and compost) can be a significant source of H2S.

great explanation, just a small suggestion: Would you mind editing your posts to make them easier to read? Like spacing, formatting mathjax blockquotes etc. Thanks for sharing great knowledge!


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Geometry of the pentachloridoindate(III) anion
According to Brown et al.,1 the crystal structure of $\ce{[NEt4]2[InCl5]}$ reveals that the pentachloridoindate(III) anion, $\ce{[InCl5]^2-}$, adopts a square pyramidal $C_\mathrm{4v}$ geometry (VSEPR theory would predict trigonal bipyramidal $D_\mathrm{3h}$):

According to Greenwood and Earnshaw,2

It will be noted that $\ce{[InCl5]^2-}$ is not isostructural with the isoelectronic species $\ce{[SnCl5]-}$ and $\ce{SbCl5}$, which have the more common $D_\mathrm{3h}$ symmetry. […] $\ce{[NEt4]2[TlCl5]}$ is isomorphous with $\ce{[NEt4]2[InCl5]}$ and presumably has a similar structure for the anion.

However, the structure of $\ce{[PPh4]2[InCl5].MeCN}$ was reported to have the usual trigonal bipyramidal structure for the anion,3 which throws a spanner in the works. I can't understand much of the paper as it's in German. Since it's a report of a crystal structure, though, I don't really expect it to say much about the chemistry of the compound.
Can the reported differences in molecular geometry here be rationalised?

References

Brown, D. S.; Einstein, F. W. B.; Tuck, D. G. Tetragonal-pyramidal indium(III) species. Crystal structure of tetraethylammonium pentachloroindate(III). Inorg. Chem. 1969, 8 (1), 14–18. DOI: 10.1021/ic50071a004.
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth–Heinemann: Oxford, U.K., 1997, p 238.
Bubeinheim, W.; Frenzen, G.; Müller, U. Die Chloroindate $\ce{[PPh4]2[In2Cl6]}$ und $\ce{[PPh4]2[InCl5].CH3CN}$. Acta Crystallogr., Sect. C: Cryst. Struct. Commun. 1995, 51, 1120–1124. DOI: 10.1107/S0108270194011789.

Do you have a CIF-file from [Brown1969]by any chance? In addition to [Bubenheim1995] I've just found three more recent crystal structures with nearly perfect trigonal bipyramidal $D_\mathrm{3h}$ geometry of $\ce{[InCl5]^{2-}}$ and it looks like the Brown's work is the only featuring square pyramid. I wonder whether it has something to do with the experimental data or refinement at that time -- $R$-factor of nearly 10% seems quite lousy for such a simple small-molecule structure.

@andselisk no cif, sorry. I guess that might be a possibility, although I don't know much about crystallography to begin with - so I'm happy to hear any thoughts on the matter.

Well, having a look at the structure with checkcif.iucr.org is where I would start. As I said, $R$-factor might be a bit off due to the wrong initial symmetry and it might cause the wrong structure, especially if there is a proposed disorder (according to the paper). I cannot find the data in CCDC either, though the entry exists. Weird, maybe it's just me doing something wrong.

Related: Use of geometry index for the determination of coordination environment

TL;DR Initially published crystal structure of $\ce{[NEt4]2[InCl5]}$ [1] according to the further investigations [3], is not valid. The $\ce{InCl5^2-}$ ion does not have $C_\mathrm{4v}$ symmetry, and VSEPR theory pretty much explains formation of numerous slightly distorted trigonal bipyramidal $\ce{InCl5^2-}$-containing complexes according to the most recent single-crystal structure experiments.
From the other side, the bond lengths and geometry index refer to the dominating character of square pyramidal environment, which is in this case can be dictated by the crystal packing, towards which $\ce{[InCl5]^2-}$ is more sensible than $\ce{[SnCl5]-}$ due to the noticeable size difference ($\ce{[InCl5]^2-}$ is bulkier). Either way, it looks like another crystallographic experiment (preferably at lower temperatures to decrease the size of thermal ellipsoids) is needed to determine the angles more precisely.
As a general advice, always pay attention to the $R_1$ and $R_2$ values of the crystal structure. A rule of thumb: a good structure refinement of small molecules should lead to $R_1 < 0.05$ and $R_2 < 0.12$.

Crystal structure of $\ce{[NEt4]2[InCl5]}$, determined in 1969 [1], has been subsequently criticized in several publications, mainly involving additional symmetry vibration analysis of Raman spectra for $\ce{MX5^2-}$ anions [2]. It's been established that $\ce{InCl5^2-}$ ion does not have $C_\mathrm{4v}$ symmetry (but retains $C_\mathrm{2v}$), and that the crystal may not be centrosymmetric.
The results are summarized in [3]:

Deconvolution of the single-crystal data was again performed with the aid
  of the low temperature data, revealing $11$ $\ce{InCl5^2-}$ bands between $300$ and $\pu{100 cm-1}$. Two overlapping A stretches are clearly evident at $292$ and $\pu{286 cm-1}$, but no B stretch is apparent. The two E symmetry stretches expected for the anion of a $C_2$ site, but not for a $C_4$ site, are evident at $281$ and $\pu{271 cm-1}$. [...] With the aid of low-temperature data, the four E modes predicted for $\ce{InCl5^2-}$ at $C_2$ sites are found at $144$, $136$, $122$, and $\pu{103 cm-1}$ in the single-crystal data. With this reinterpretation of the spectrum it is no longer necessary to assume arbitrarily the presence of a lattice mode in this region.
In summary, the vibrational data for the tetraethylammonium salts of $\ce{InCl5^2-}$ and $\ce{TlCl5^2-}$ indicate that these complexes reside on sites lacking full $C_4$ symmetry. The most straightforward conclusion from the vibrational data is that the $\ce{InCl5^2-}$ ions have local $C_2$ symmetry, which would be inconsistent with the previous structure determination [1]. [...]
The structure of $\ce{[(C2H5)4N]2[InCl5]}$, as previously reported, consisted of centrosymmetrically related $\ce{InCl5^2-}$ ions situated about the
  Wyckoff c positions of $C_4$ symmetry in space group $C_\mathrm{4h}-P4/n$.
  Coaxial with the $\ce{InCl5^2-}$ ions are disordered $\ce{(C2H5)4N+}$ ions
  also situated about the Wyckoff c positions. The N atoms of the two remaining $\ce{(C2H5)4N+}$ ions in the unit cell occupy the centrosymmetrically related Wyckoff b sites of $S_4$ symmetry, the cation as a whole again being disordered. There are two disturbing features of this solution and its subsequent refinement: the disorder of the $\ce{(C2H5)4N+}$ ions and the significant residual electron density in the region of the basal Cl atoms of the tetragonal-pyramidal anion.

It is demonstrated further that for the ordered model the most satisfying crystallographic interpretation coexists with $C_2$ rather than $C_4$ symmetry. A correlation of allowed point symmetries implies switching from $P4/n$ to $P\bar{4}$ space group, thus reducing $R$-factor ($R_1$) from $9.5\%$ to $6.7\%$.

During the refinement in $P\bar{4}$, small but significant changes
  occur in the basal plane of the $\ce{InCl5^2-}$ ion. The basal Cl atoms
  have moved by $\pu{-0.5 A}$ from their fourfold symmetric positions in $P4/n$, leading to variations in the bond distances and angles about the In atom (Figure 3) that are qualitatively sufficient to explain the vibrational results.

Geometry index $\tau_5$ for $\ce{[InCl5]}$ fragment can be used to determine formal coordination environment as follows:
$$\tau_5 = \frac{\beta - \alpha}{60^\circ},$$
where $\alpha$, $\beta$ - two greatest valence angles of the coordination center ($\angle \ce{Cl - In - Cl}$, $\alpha < \beta$).
\begin{align}
\begin{cases}
\tau_5 &= 0 \qquad &\text{square pyramidal geometry} \\
\tau_5 &= 1 \qquad &\text{trigonal bipyramidal geometry}
\end{cases}
\end{align}
For the original $\ce{[NEt4]2[InCl5]}$ structure refined in [3] I've taken angular values from their supplementary materials. It turned out that changing space group to $P\bar{4}$ still results in dominating character of square pyramidal geometry of $\ce{[InCl5]^2-}$, though one must remember that the rotation axis of highest order is $C_2$, not $C_4$:
$$\alpha = (150.6 \pm 5.0)^\circ, \beta = (154.1 \pm 6.0)^\circ, \bar{\tau_5} = \frac{154.1^\circ - 150.6^\circ}{60^\circ} = 0.06; \tau_5 \in [0.04; 0.24]$$
Paper [4] mentions that for isoelectronic $\ce{[SnCl5]-}$-contining structures with various cations only trigonal-bipyramidal geometries were reported, and suggests higher influence of crystal packing on $\ce{[InCl5]^2-}$ geometry. This can also be explained by higher flexibility of $\ce{MX5}$ fragment in $\ce{[PPh4]2[InCl5]}$ in comparison with $\ce{[PPh4][SnCl5]}$ [5] due to greater average bond length: $d(\ce{In - Cl}) = \pu{2.5 A}$, $d(\ce{Sn - Cl}) = \pu{2.3 A}$.
As for auxiliary experimental data, all crystal structures including $\ce{[InCl5]^2-}$ fragment determined in the past two decade completely support VSEPR theory that predicts trigonal bipyramidal $D_\mathrm{3h}$ symmetry (with one intermediate exception of $\tau_5 \approx 50\%$).

bis(Tetraphenylphosphonium) pentachloro-indate(III) [4], trigonal bipyramidal geometry:
$$\alpha = 124.24^\circ, \beta = 174.74^\circ, \tau_5 = \frac{174.74^\circ - 124.24^\circ}{60^\circ} = 0.84$$
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

bis(Diphenyldichlorophosphonium) pentachloro-indate(III) [6], trigonal bipyramidal geometry:
$$\alpha = 120.93^\circ, \beta = 179.29^\circ, \tau_5 = \frac{179.29^\circ - 120.93^\circ}{60^\circ} = 0.97$$
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

tetrakis($\mu_3$-Selenido)-tetrakis($\mu_2$-1,5-bis(diphenylphosphino)pentane)-deca-gold(I) pentachloro-indate(III) [7], intermediate geometry:
$$\alpha = 138.45^\circ, \beta = 166.50^\circ, \tau_5 = \frac{166.50^\circ - 138.45^\circ}{60^\circ} = 0.47$$
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#FFA100}{\Large\bullet}~\ce{Se}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$;
$\color{#FFD123}{\Large\bullet}~\ce{Au}$.

tetrakis($\mu_2$-chloro)-chloro-tetrakis(triphenylphosphine-P)-di-copper(I)-indium(III) tetrahydrofuran solvate [8], trigonal bipyramidal geometry (though authors refer to it as to "quasi square-pyramidal" coordination based on a single slightly shorter $\ce{In - Cl}$ bond distance $(d(\ce{In-Cl_\mathrm{ap}}) = \pu{2.36 A})$, shortest $d(\ce{In-Cl_\mathrm{eq}}) = \pu{2.42 A})$:
$$\alpha = 123.90^\circ, \beta = 172.89^\circ, \tau_5 = \frac{172.89^\circ - 123.90^\circ}{60^\circ} = 0.82$$
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#C88033}{\Large\bullet}~\ce{Cu}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

Bibliography

Brown, D. S.; Einstein, F. W. B.; Tuck, D. G. Inorganic Chemistry 1969, 8 (1), 14–18. DOI 10.1021/ic50071a004.
Adams, D. M.; Smardzewski, R. R. Journal of the Chemical Society A: Inorganic, Physical, Theoretical 1971, 714. DOI 10.1039/j19710000714.
Joy, G.; Gaughan, A. P.; Wharf, I.; Shriver, D. F.; Dougherty, J. P. Inorganic Chemistry 1975, 14 (8), 1795–1801. DOI 10.1021/ic50150a011.
Bubenheim, W.; Frenzen, G.; Müller, U. Acta Crystallographica Section C 1995, 51 (6), 1120–1124. DOI 10.1107/S0108270194011789.
Müller, U.; Siekmann, J. F. Acta Cryst C 1996, 52 (2), 330–333. DOI 10.1107/S0108270195011073.
Taraba, J.; Zak, Z. Inorganic Chemistry 2003, 42 (11), 3591–3594. DOI 10.1021/ic034091n.
Olkowska-Oetzel, J.; Sevillano, P.; Eichhöfer, A.; Fenske, D. European Journal of Inorganic Chemistry, 2004 (5), 1100–1106. DOI 10.1002/ejic.200300774.
Zhang, X.-Z.; Song, Y.-W.; HuiWu, F.; Zhang, Q.-F. Zeitschrift für Naturforschung B 2007, 62 (6). DOI 10.1515/znb-2007-0605.

Certainly not the answer I expected - but excellent nonetheless. Thank you!

@orthocresol I don't know what was wrong with me last night, but I calculated $\tau_5$ for your complex wrongly (improper $\alpha$ from SI), my sincere apologies. It actually supports square pyramidal geometry even for the structure refined for the second time, though due to lowered symmetry one cannot attribute $C_\mathrm{4v}$. I edited the answer correspondingly adding a few more details. Probably you might want to unmark it as accepted (or even downvote), as new experimental data or QC-calculations could be more significant that the idea of crystal packing effect proposed here.

No problem. Well, generally, if new information comes in, the accepted answer can always be changed, so IMO it's not much of a big deal. Downvoting would be cruel, no? ;)

@orthocresol Well, it still was my fault and the faults should be punished/ignored, not rewarded. But thank you for the nice attitude!


text:
Kinetic vs thermodynamic control of HBr to 1,3-Butadiene

The author claims that the transition state leading to the formation of the 1,2 product from the allylic carbocation is lower in energy since the carbon (which is attached to Br in the 1,2 product) is more electrophilic.
This makes sense to me, however, doesn't the transition state represent the energies of both, the reactants, and the products? In which case the thermodynamically favored product (1,4 product) should also have a lower energy transition state.

Please add the citation to which the context is referring, i.e. who is the author with that claim? For addition reactions there are a multitude of different paths, it can be concerted or step wise, aided by the solvent or not, etc. In any case, the specific atomic arrangements in the transition states will certainly be of different energies, none of which can be predicted from the reactants by any easy measure. It is not uncommon, that the more stable thermodynamic product has a higher activation barrier than a different path.

I'm not sure if it's an outright duplicate, but I covered this topic in a previous question: Addition of hydrogen bromide to 1,3-butadiene: thermodynamic and kinetic control

Without any other information than in your question it would seem that the 1,2- has the lower barrier as this is preferentially formed at very low temperatures.  Is there not  then a second reaction (the rearrangement you show after the low temperature one) and so there is a also a  barrier for rearrangement between species 1,2- to 1,4- that is observable at the higher temperature only since the ratio of products changes? Direct reaction at 40 C would follow this scheme also, low barrier to 1,2- (80% product) then rearrangement.

@orthocrestol that answer was helpful, but I want to know why the accepted answer before Nordlander et al makes sense. Although I understand why a more electrophilic carbon will stabilize the transition state, doesn't the transition state represent the product as well as the reactant? In which case the thermodynamically more stable product should lower the energy of the transition state leading to its formation.

@martin LG Wade. But that's exactly what I don't understand. The transition state represents the products and the reactants. So why wouldn't a more stable product imply a more stable transition state?

Pinging @orthocresol since Xasthor made a typo in pinging you.

Sometimes there is a correlation between the energy of the TS and the energy of the product, and it can be reflected in a linear free energy relationship, e.g. Evans–Polanyi principle. However, it is important to realise that this is not always the case. In general there is not necessarily a link between the kinetics of a reaction (energy of a TS) and the thermodynamics of a reaction (energy of a product). [Thanks, @Pritt.]

It is always worth bearing in mind that the transition state is a hypothetical idea and although they must exist,  almost no direct experimental evidence about their properties, such as geometry, vibrational frequencies, exists. Furthermore in solution the 'transition state' will be some thermal average depending on the prevailing conditions.

A transition state does not ‘represent the energies of […] reactants and […] products’. A transition state is a certain arrangement of atoms somewhere between the most stable atom arrangement the reactants can have and the most stable atom arrangement the products can have. The key characteristic of a transition state is that it is directly en route from reactant to product and that it is the arrangement with the highest internal energy.
For any arrangement of atoms you can calculate a single point energy: reactants, products and the transition state. The energy difference between reactants and the transition state is colloquially known as activiation energy although that may not always be correct. It is, however, important to realise that each arrangement has its own energy that cannot a priori be deduced from any other arrangement. Thus, a transition state’s energy is not directly related to the energies of reactants or products (with the exception that by definition it must be higher in energy than both).
While it is a general rule that the transition state leading to the thermodynamically more stable products has a lower energy than any other transition state, this is not true for all cases. Indeed, for any reaction in which one can discuss kinetic (lower energy transition state) versus thermodynamic (lower energy products) control, the transition state energies must be ‘reversed’ with respect to the product energies.


text:
SCF (Roothan-Hall) Method, why Density Matrix?
I've been aksing myself, when I try to variationally minimize a Hartree-Fock energy by using the Roothan equations, what I do is to set a Fock-Matrix, then transform it, then diagonalize it so I can read the energy from the diagonal elements and then I test for convergence. But why is usually the density matrix compared to the previous run and not just the sum of the diagonal elements? I know in the end they both have to be evaluated but just for testing for any changes, pretty much any parameter, which changes with the Fock-Matrix and has some relationship with the energy could be used as a criteria for convergence. Or where do I see something wrong here?

In your implementation, you may do whatever you wish. I have (successfully) used the change in total energy to detect convergence in my very first Hartree-Fock SCF.
In more involved projects, the root mean square deviation (RMSD, compared to the previous iteration) or other measures that take the whole density matrix into account are often used because these quantities are easily accessible or even necessary for the convergence speedup procedures that are in place, such as extrapolating the density matrix, direct inversion of iterative space (DIIS) and its variants etc.

When doing the SCF calculation taking the RMSD of the density matrix compared to taking the RMSD of the diagonal of the density matrix takes about the same with respect to the total calculation time. The comparrison of the density matrices is not the time consuming step in the SCF calcultion. When you take the RMSD of the total density matrix you also "keep an eye" on the off diagonal elements, this is a more rigid test for convergence.


text:
Why is the rate of a reaction proportional to the concentrations of reactants raised to their stoichiometric coefficients?
Consider a gaseous state elementary reaction $$\ce{aA(g) + bB(g)} \overset{k_\mathrm{f}}{\underset{k_{\mathrm{b}}}{\ce{<=>}}}\ce{ cC(g) + dD(g)}$$
I know that for this reaction, $$\Delta G = \Delta G^{\circ} + RT \ln \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) \tag1$$
Now, if $$K = \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) ,$$
then $\Delta G = \Delta G^{\circ} + RT \ln K$.
Now, how can we say that $$K = \frac{k_{\mathrm{b}}}{k_{\mathrm{f}}}$$
and $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$  and $\, r_\mathrm{b} = k_{\mathrm{f}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$
$\,$where $r_{\mathrm{f}}$ and $r_{\mathrm{b}}$ are rates of forward and backward reaction as assumed in this answer.
I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$  and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

You've neglected the big assumption in the post you quoted---that this reaction is assumed elementary.

@cyclohexane-molecule Edited question. But even though the problem remains.

Khan Academy related video - https://youtu.be/psLX080RQR8

I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$  and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

I might try to give you some intuition to back up that, for a given (elementary) reaction $\ce{A ->[k] B}$, the reaction rate $r$ can be written as
$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} \propto P_\ce{A}\text{.}$$
(Observe that the second equality above is true due to $P_\ce{A} + P_\ce{B} = \text{constant}$.)
First, for an ideal gas, $P_\ce{A} = \frac{n_\ce{A}}{V} RT$.
This means that $P_\ce{A} \propto n$ for fixed temperature and volume.
Let's say the reaction happens as a random process.
That is to say that, for every time interval $\Delta t$, we have a probability per unit time $p$ of having a single molecule $\ce{A}$ turning into $\ce{B}$.
If we wait longer, proportionally more molecules will turn.
We'll thus have, for initially $n_\ce{A}$ molecules of $\ce{A}$, after $\Delta t$ seconds,
$$\Delta n_\ce{B} = p n_\ce{A} \Delta t\text{.}$$
This means that, in the time interval $\Delta t$, the population of $\ce{B}$ goes from 0 to $\Delta n_\ce{B}$ (assuming no $\ce{B}$ initially).
From the stoichiometry of the reaction, $\Delta n_\ce{B} = -\Delta n_\ce{A}$ (i.e., there's conservation of moles).
Thus,
$$\frac{\Delta n_\ce{B}}{\Delta t} = -\frac{\Delta n_\ce{A}}{\Delta t} = p n_\ce{A}\text{.}$$
After taking the limit of $\Delta t \rightarrow 0$ and multiplying by $\frac{1}{V}$ (i.e., dividing by volume), we get
$$\frac{d \frac{n_\ce{B}}{V}}{dt} = \frac{d [\ce{B}]}{dt} = p \frac{n_\ce{A}}{V} = p [\ce{A}]\text{,}$$
where $[\ce{A}]$ and $[\ce{B}]$ stand for the concentrations of $\ce{A}$ and $\ce{B}$, respectively.
Since, again for an ideal gas, $[\ce{A}] = \frac{P_\ce{A}}{RT}$, we get
$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} = p P_\ce{A}\text{,}$$
where the proportionality constant is the probability of reaction per unit time.

A late answer...
In a molecular approch to chemical reaction rates we asume that the molecules move randomly in solution and have a velocity randomly assigned according to the Boltzmann distribution and that the chance of being in one position in solution is the same as being at any other, i.e. the molecules are in thermal equilibrium at some temperature T.
The reaction rate for the reaction $\mathrm {A+B \rightarrow}$ can be found by considering the number of collisions per second and then this number reduced by the chance that a reaction occurs on collision. To arrive at a rate the chance that a randomly chosen combination of molecules of type A and B at time t will react in the next infinitesimal time interval $t+\delta t$ and has not reacted up to t has to be calculated. As a result the total number of collisions /second between $N_A$ molecules of type A and $N_B$ of type B is found to be $\sim N_AN_B\sigma_{AB}$, where $\sigma_{AB}$ is the cross sectional area for a collision, and this expression produces the rate  when  multiplied by the probability $p_{AB}$ of A and B reacting, i.e. rate $\sim N_AN_B\sigma_{AB}p_{AB}$. The probability of reaction $p_{AB}$ depends critically on the activation energy.
The important point is, however, that the  rate is proportional to the product of the numbers of molecules of type A and B which in constant volume is the same as being proportional to the product of concentrations. That it also is proportional to the collision cross section and the chance of reacting is also physically realistic. The rate constant is proportional just to the probability of reaction and the cross section for collision.
Thus the origin of the product of concentrations in the rate expression is found in the chance that a molecule of type A will collide with that of type B in a small time interval $t, t +\delta t$.  Thus if we ignore the fact that there is variation due to changes in experimental conditions, such as temperature, the electronics of the detectors used to measure the reaction etc. perfect kinetics would be measured, i.e $dA/dt$ or $ dB/dt$ vs. time would be noise free.
If we now consider the random nature of the collisions between molecules a slightly different approach has to be taken. If a small volume of the solution is considered and the rate calculated in that region, because the number of species entering and leaving in any given time period varies, the rate expression is also potentially subject to variation. This happens simply because of the random nature of the motion of molecules in a solution or gas. In this case the rate of change of a species X should be written in terms of its average (or mean) value represented as $\langle X(t)\rangle$;
$$ \frac{d\langle X(t)\rangle}{dt} = \langle f\times X(t)\rangle $$
where $\langle f\times X(t)\rangle$ represents some rate expression such as $k_1\langle X(t)\rangle + k_2\langle X(t)\rangle^2$ etc. except that the number of  $X$ are represented as their average values. Clearly the $X$ are always positive values. These  values have a variance (square of standard deviation) and so we would expect that different experiments give slightly different results because of this. (This is the approach taken by Gillespie, see 'Markov Processes' publ. Academic Press 1992)
In a bimolecular reaction such as $X + X \rightarrow $ with rate constant k the rate expression is
$$ \frac{d\langle X(t)\rangle}{dt} = k\langle \;X(t)(X(t)-1)\;\rangle/2 $$
The term $\langle X(t)(X(t)-1)\rangle /2$ arises becaus of the number of distinct X-X pairs in the reaction volume at time t. The question then arises as how to deal with this. Two approximations can be made; the first is to assume that X is large compared to $1$ and so $\langle X(t)(X(t)-1)\rangle \rightarrow \langle X^2(t)\rangle $, which seems to be a reasonably sensible approximation, the next step is more problematic and is $\langle X^2(t)\rangle \sim \langle X(t)\rangle^2$. This is a drastic step and assumes that the variance in the number of molecules is zero, i.e. that $X(t)=\langle X(t)\rangle$ which means that $X$ is a deterministic process, so that the randon nature of the numbers of $X$ entering into the reaction volume is always the same, which is not what was assumed at the outset and seems to fly in the face of what we know about how molecules move in solution, i.e. their random diffusional nature.
However, in any normal kinetic experiment the number of molecules we measure is vast, $\langle X(t)\rangle \sim 10^{18}$ would not be untypical, so that the fluctuations in the  number of molecules is tiny and the approximation made above is valid.
$$ \frac{\sigma_X}{\langle X \rangle} \sim \frac{\sqrt{\langle X \rangle}}{\langle X \rangle} \sim \frac{1}{\sqrt{\langle X \rangle}}\sim 10^{-9}$$
Even if there are as few as 1000 molecules the realtive fluctuation  is not too drastic but in experiments using small numbers of molecules, as in the case of many 'single molecule' experiments then clearly the rate equation approach needs to be modified.
Notes :
(A)  To see that the variance is zero use the definition $\sigma^2_X = \langle X^2 \rangle -\langle X \rangle^2$ where $\sigma$ is the standard deviation)
(B) There are other situations where the rate equation approach breaks down in macroscopic systems and that is when they are sufficiently non-linear, one example is the Schlogl system of equations, which for certain values of concentrations and rate constants becomes bi-stable;
$$B_1+2X \rightleftharpoons 3X; \;\; B_2 \rightleftharpoons X$$

I understood around half of the answer. I don't know the maths of variance and deviation. But still, this answer could be useful to people who know that or to me in the future.

:)   This web page explains variance and standard deviation. http://www.mathsisfun.com/data/standard-deviation.html


text:
Molecular Hamiltonian of Helium hydride ion
Currently, I am going through Quantum Simulation of Helium Hydride Cation in a Solid-State Spin Register. I am not a chemist but rather a computer scientist hence having trouble following the paper.
I am using this easy to access note on Hartree-Fock  approximation.
So, I am not sure how the authors created the molecular Hamiltonian in Eq. 2 of the supplementary material.
$$
H = \begin{pmatrix}
-2.85404 & 0.130671 & 0.0 \\
0.130671 & -0.760916 & -0.323568\\
0.0 & -0.323568 & -1.91238
\end{pmatrix}
$$
From the note, I understand that the Born-Oppenheimer approximation (Eq. 3) is the starting point.
$$
H = -\sum^{N}_{i = 1} \frac{1}{2} \nabla^2_i - \sum^{N}_{i = 1} \sum^{M}_{A = 1} \frac{Z_A}{r_i A} + \sum^{N}_{i = 1} \sum^{N}_{j > 1} \frac{1}{r_{ij}}
$$
What should be my following steps to reach at the molecular Hamiltonian.
I am extremely sorry if I am asking for a really long answer.
UPDATE 1:
The molecular Hamiltonian is:
\begin{align}
\hat{H} &= \underbrace{-\frac{\hbar^2}{2 M_{He}} \nabla^2_{He} - \frac{\hbar^2}{2 M_H}}_{\text{Nuclear kinetic energy}} \underbrace{- \frac{\hbar^2}{2 m_e} \nabla_1^2 - \frac{\hbar^2}{2 m_e} \nabla^2_2}_{e^-\text{'s kinetic energy}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 1\text{\textsuperscript{st} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 2\text{\textsuperscript{nd} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{r_2}|}
 }_{e^-\text{ - }  e^- \text{ repulsion}}
 \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{R_{He}} - \overrightarrow{R_H}|}
 }_{\text{Nuclear - Nuclear repulsion}}
\end{align}
After Born–Oppenheimer approximation:
\begin{align}
\hat{H} &=  \underbrace{- \frac{\hbar^2}{2 m_e} \nabla_1^2 - \frac{\hbar^2}{2 m_e} \nabla^2_2}_{e^-\text{'s kinetic energy}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 1\text{\textsuperscript{st} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 2\text{\textsuperscript{nd} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{r_2}|}
 }_{e^-\text{ - }  e^- \text{ repulsion}}
 \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{R_{He}} - \overrightarrow{R_H}|}
 }_{\text{Now a constant since nuclei positions are fixed}}
\end{align}

Your provided note is about the HF Hamiltonian in the one-electron basis set (atomic orbitals). Since a minimal STO-3G basis set is used here, the corresponding matrix representation of this Hamiltonian is a $2\times 2$ matrix.
The $3\times 3$ Hamiltonian from the SI is the FCI Hamiltonian in the many-electron basis set. These basis functions are Configuration State Functions, arising from linear combinations of (excited) Slater determinants (electron configurations), which in turn are built from the HF molecular orbitals.
Configuration Interaction calculations (like FCI) are typically done on top of a HF calculation, to correct for missing electron correlation effects. Your linked note is only about HF. For details about CI, I can recommend the book by Szabo and Ostlund which is also cited in your note.


text:
Why is this intermediate borane-hydroperoxide adduct unstable?

Boron has an empty p-orbital and is highly electron deficient because of an incomplete octet.
So shouldn't this intermediate be more stable than $\ce{BR3}$ (whose electron deficiency is only somewhat fulfilled through hyperconjugation) since boron's electron deficiency is satisfied by the increased negative charge (although the majority of negative charge is still on oxygen due to its high electronegativity).

Sure, this thing feels somewhat more satisfied than BR3, but it has a suitable decomposition pathway, while BR3 has none.

But why would it decompose into something less stable? Could it be because of the favorable entropy change?

Who said "less stable"? The ultimate product is even more satisfied (and hence more stable).

Why are you comparing a borate and a borane? Those don't even have the same number of atoms, let along the same number of electrons.

Basically an intermediate form, that fulfills the octet rule, would be more stable than the trialkylborane, but only if we suppose that there are no other effects! And if there were no other ffects, it wouldn't be an intermediate form but an isolable compound.
In your reaction scheme you see that $\ce{OH-}$ will leave. You maybe know that $\ce{OH-}$ is - under basic conditions - a good leaving group. Moreover a peroxide is mostly pretty unstable, so it's happy to react and let $\ce{OH-}$ leave. The hydroboration–oxidation reaction gives you the tetrahydroxyborane-anion ($\ce{B(OH)4-}$) and three alcohols ($\ce{ROH}$) as final product. So you'll get an anion that fulfills the octet rule and is more stable than your reactant.

Edit: More details on the hydroxide as leaving group
The reaction finds place under strongly basic conditions. Usually you take $\ce{NaOH}$ to deprotonate the $\ce{H2O2}$ and form $\ce{OOH-}$, so in your solution there are plenty of $\ce{OH-}$ ions. The intermediate also splits off an hydroxide ion which is a good leaving group under this conditions. This means that the ion is stabilized in solution and is not less stable as the intermediate. Even if you can't really compare the stability of those two things because they are not the same... So let's only say, that the hydroxide ion is pretty stable in a basic solution.
As I also said in my original answer above, peroxides are pretty unstable. This is because the two oxygens are right next to each other. As you may know, oxygen is one of the most electronegative elements of the periodic table, so it "fights to get electrons". If you connect two oxygens like in a peroxide, they both fight to get the electron of the other one, so it resembles rope pulling. This isn't the best case scenario for a stable bond.
If you look at hydrogen peroxide, especially in concentrated solutions, it tends to decompose without any external influence.
$\ce{2H2O2 -> 2H2O + O2}$
This reaction releases 98,2 kJ/mol [1] of energy. This means that the products (oxygen and water) are way more stable than the hydrogen peroxide itself. This is mainly due to the very unstable rope pulling-bond between the two oxygens in $\ce{H2O2}$: $\ce{RO-OR}$.
This should explain why the intermediate group is so unstable.

I understand that the final product is thermodynamically favored compared to the reactants. What I don't understand is why this intermediate is less stable than the first two reactants. Not only is boron's electron deficiency fulfilled but the reactive negative charge on oxygen is gone in the intermediate.

Why would the intermediate decompose into two parts which are less stable than itself(one part having a deficient boron and the other part having a reactive negative charge)

@xasthor Please consider my edit.


text:
How can the effect of a pulse sequence on a strongly coupled spin system be analysed?
In introductory NMR theory there are two models which are commonly used to analyse the effect of a pulse sequence: namely, these are the vector model and the product operator formalism.
The vector model serves as a useful pictorial representation, but is generally considered too simplistic to be used when the system contains coupled spins. Usually at this point the product operator approach is introduced, based on the Liouville–von Neumann equation
$$\frac{\mathrm d\hat{\rho}}{\mathrm dt} = -\mathrm{i}[\hat{H}, \hat{\rho}]$$
and its solution $\hat{\rho}(t) = \exp(-\mathrm{i}\hat{H}t)\hat{\rho}(0)\exp(\mathrm{i}\hat{H}t)$. Then with a knowledge of the initial density matrix $\hat{\rho}(0)$ and the relevant Hamiltonians $\hat{H}$, the density matrix after a pulse sequence $\hat{\rho}(t)$ can be calculated.
Yet, it is also stated that the product operator approach "only applies to weakly coupled spin systems" (Keeler, Understanding NMR Spectroscopy 2nd ed., section 7.1.3). "Weakly coupled", of course, means that the coupling constant $J_\mathrm{IS}$ is much smaller than the difference in resonance frequencies $|\nu_\mathrm I - \nu_\mathrm S|$. This raises two questions:

Why does the product operator formalism not work for strongly coupled systems?
How can the effect of a pulse sequence on such systems be analysed?

Full disclosure: the answer is described in Hore, NMR: The Toolkit, How Pulse Sequences Work, 2nd ed., Chpt 10. But I haven't gotten around to reading it fully yet and I just thought I'd post the question here anyway.

The product operator formalism still applies to strongly coupled systems.  The simple "magic formula" for calculating a closed-form solution to the step-wise time evolution only applies when the terms of the Hamiltonian commute with each other, which is not the case in a strongly-coupled system.  Maybe more accurately stated: in a weakly-coupled system, the non-commuting terms can be neglected.  The other assumption that allows the removal of non-commuting terms is that the pulses are very short compared to other evolution times, so during pulse periods the free-evolution terms are omitted from the Hamiltonian.
The effects of a pulse sequence without these on such a system can still be found by computing the exponential of the matrix rather than using the shortcut.
Now a shameless plug: the SpinDrops tool (http://spindrops.org) can be used to simulate and visualize the effects of pulse sequences with arbitrary Hamiltonians. It shows an exact 1-1 mapping of coupled-spin operators to the visualization which can not be done with the vector model.  The setting 'Pure Pulses' can be disabled to include the free-evolution Hamiltonian during pulses, but incorporating strong coupling would currently need to be done by defining the Hamiltonian explicitly.


text:
Conceptual tussle with kinetics of equilibrium and non-equilibrium processes
Prologue:
Recently started with Chemical Kinetics at class. Came across a couple of points in Levine's Physical Chemistry (the book was recommended on Chem.SE; so I thought I'd give it a read :D) that I'm having serious issues wrapping my head around. This query is about the kinetics of equilibrium (and non-equilibrium) processes, and the fact that this has puzzled me for quite a while... I attribute to my (seriously questionable) understanding of equilibrium processes (courtesy: High-school education). It would be prudent to keep my shortcomings in Kinetics (newbie) and Equilibrium (possibly flawed understanding) in mind while addressing my query in your answer or the comments section. Thanks!

According to Physical Chemistry (Levine,I.N), Chapter 15, "Kinetics":

Processes in systems in equilibrium are reversible and are comparatively easy to treat. This chapter and the next deal with non-equilibrium processes, which are irreversible and hard to treat. The rate of a reversible process is infinitesimal. Irreversible processes occur at nonzero rates.

Which, if I understood correctly, can be paraphrased as:

Equilibrium processes are reversible and have infinitesimal (practically "non-existent") rates.
Non-equilibrium process are irreversible and have non-zero ("finite" or "significant"?) rates.

From my "understanding" of Equilibrium; chemical equilibrium is an example of dynamic equilibrium. When a chemical reaction, $$\ce{A + B -> C + D}$$
has proceeded to equilibrium. The "forward" rate of reaction (rate at which $C$ + $D$ is spit out) must equal the "backward" rate of reaction (rate at which $A$ + $B$ is spit out). There is no net formation of products/reactants, but both the forward and backward rates are positive and finite.

My question?
Why does Levine say that equilibrium processes have "infinitesimal" (almost non-existent) reaction rates? I thought the forward and backward reaction rates in a reaction at equilibrium are finite...

I'm currently trying to wrap my head around @Curt's answer here [ https://chemistry.stackexchange.com/questions/47262/is-every-chemical-reaction-in-equilibrium ], since it might help here. It sounds amazing (and correct)...but it contradicts everything I've learnt at school so far. :'(

I cannot find the corresponding passage in the 6th edition of Levine, so I can only guess, but my assumption is that the author is referring to the net rate of reaction being infinitesimal. After all, the rate of reaction is defined to be proportional to $\mathrm d[\ce{A}]/\mathrm dt$ and if the system is at equilibrium, the concentrations of all components are constant (or at most changing infinitesimally due to some reversible change in external conditions) and hence the rate is zero (or at most infinitesimal).

@ortho It's on page 474 (section 15.1), and yeah...she could be referring to the net reaction rate. Bear with me here, but...wouldn't the reaction then no longer be in equilibrium? I mean, the net rate ought to be zero, right? Thanks! (re-reading your comment...didn't see the edit)

OK, I think I am correct about this. If not then I suppose somebody will come along and correct me anyway.
The author here is referring to the net rate being infinitesimal. To understand where the infinitesimal comes from, we need to figure out what is being meant by an "equilibrium process", or a "reversible process". On page 43, Levine describes a reversible process as

one where the system is always infinitesimally close to equilibrium, and an infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.

Often these are exemplified using thermodynamics processes, e.g. expansion/compression of a gas, but since we're talking about chemical kinetics, we could use an example where the temperature of the system is being increased in a reversible fashion. This means that the temperature is being increased in infinitesimal steps of $\mathrm{d}T$. The total change in temperature of the process is calculated via integration:
$$\Delta T = \int_{T_1}^{T_2}\mathrm{d}T = T_2 - T_1$$
and likewise for other properties. If something (e.g. the pressure $p$) depends on temperature, i.e. $p = p(T)$, then we can find the overall change in pressure
$$\Delta p = \int_{p(T_1)}^{p(T_2)} p(T)\,\mathrm{d}T$$
If the system is at equilibrium ($Q = K$), then an infinitesimal increase in temperature leads to an infinitesimal change in the equilibrium constant $K$, and hence an infinitesimal change in $Q$ (essentially this is due to Le Chatelier's principle, but more properly it is explained by the system moving in the direction of decreasing Gibbs free energy). This leads to an infinitesimal change in the concentrations of the reacting species, and hence the overall rate of reaction is infinitesimal.
If nothing is changing, the system would still be considered as being at equilibrium. The net rate in this case is, of course, zero. However, such a system is not particularly interesting, as there is no process actually occurring.
So, the bottom line is that for a system in equilibrium, its properties must be changing at most infinitesimally.

Yup, this works, thanks! (I'll be keeping an eye out for other answers too...but I guess I can safely "accept" this for now :D)

Quick question: On page 516, Levine says the rate conversion rate $J$ (and hence the reaction rate $r$) becomes zero at equilibrium...which would indicate she's talking about the net rate. However, given the way the conversion rate $J$ is defined (equation), I find this contradictory. Could you take a look at it?

I don't have Levine, but I'd assume using the term "infinitesimal" justifies an assumption of continuity and hence allows taking the derivative, you know: dX/dy.
I could be wrong (and anyway did you really mean to suggest that we would be able to tell you what Levine was thinking when that was written? (I forgot my crystal ball...)). By definition the net reaction "at equilibrium" is zero. For equilibrium processes, the net reaction (rate) is infinitesimal (because what we mean by "at equilibrium" for changes in the (macroscopic) system is "near equilibrium", yeah we're totally sloppy but we don't have much choice because the concept is really hard to pin down into a simple definition.)
As far as what Levine means by Non-equilibrium processes being irreversible is probably (just my guess here) about the path that the process takes. Most meaningful Thermodynamic values are path independent (for beginning students). Path independence means that no matter how you get from state A to state B, the Thermodynamic result is the same. In real world (irreversible) processes, the path you take matters. Consider a simple physical system, a ball and a hill. The potential energy difference between the ball at the bottom of the hill and the ball at the top is easy to calculate (P.E.= mg*(h2-h1)). The actual energy you use to move the ball up depends on the actual path you take. The actual path is irreversible, the ideal difference is reversible (although I didn't explain why - it has to do with it being an extremal value - a maximum or minimum).

Alphonse I can concur with your answer (though the last bit wasn't really necessary, I guess?), yes, improved formatting would help. You didn't have to "delete" it :/

There is "delete" button there for you. Answer is not a comment.