[Borek]

In 1966, while researching reactions between NO and a simple substance A, Neil Bartlett found two products – the main one, B, and an interesting side product C. Mass spectrum of the compound C contained peaks listed in the attachment.

C survived heating up to 260 °C and was stable in the presence of concentrated alkaline solutions. Mixed with hydrogen it exploded at 170 °C. Reaction with iodides produced 1.5 mmol I₂ for every mmol of C (NO evolved during this reaction). C reacted with the hydrofluoric acid substantially increasing conductivity of the solution, and after further addition of BF₃ two crystalline adducts were produced – one (D) contained 6.98% boron and the other (E) 9.71% boron.

Give formulas of all substances A to E.

(Note: generally speaking it is not possible to deduct formula for B from the information given, so if you have no idea  what it can be - don't worry.)

[DrCMS]

I think Jack Passmore knows all the answers.

[Schrödinger]

I'm just going to take a guess at A and C, having done nothing other than look at the mass spec

N = syhbevar
P = ABS2 ?

[Borek]

You are on the right track, but not entirely correct.

Actually you felt in the same trap I did when first approaching this problem.

[Borek]

Extending the deadline for another week...

[DrCMS]

I found Jack Passmore's PhD thesis that describes this work but he only describes the characterisation of one product from C with HF + BF₃ and I get a slightly different %B (9.26) for the material he found?

[Borek]

Hard to comment - I am relying on the question prepared by someone else (the answer given looks reasonable). I have checked that the numbers I posted reflect the formulas expected, so there is no simple mistake or typo in what I posted.

[DrCMS]

A = F
B = NOF or NO₂
C = NOF₃
D = ??
E = NOBF₄ but I calculate it to be 9.26%B

[Borek]

Answer key gives B as NOF.

NOBF₄ would be made from NOF (B) and not NOF₃ (C).

[DrCMS]

Answer key gives B as NOF.

His PhD thesis (https://circle.ubc.ca/bitstream/handle/2429/37267/UBC_1968_A1%20P375.pdf?sequence=1) lists both NOF and NO₂ as the major products from NO + F₂

NOBF₄ would be made from NOF (B) and not NOF₃ (C).

No his PhD quite clearly states the reaction of NOF₃ with HF and BF₃ was done to try to make HNOF₃BF₄ but they could find no evidence it was formed but that the IR spectra of the solid was identical to NOBF₄

[Borek]

No his PhD quite clearly states the reaction of NOF₃ with HF and BF₃ was done to try to make HNOF₃BF₄ but they could find no evidence it was formed but that the IR spectra of the solid was identical to NOBF₄

Sadly, you are right. I had troubles getting the document, but eventually I managed to download it and have it here. Page 77:

2.3.3.(d) HF + BF₃ + ONF₃

It has been shown that boron trifluoride does not react with ONF₃ at ambient temperatures and pressures, but it was possible that the acid HBF₄ might protonate ONF₃ and form the salt HONF₃BF₄.

Equal quantities of BF₃ and ONF₃ were condensed into an infrared cell and the excess anhydrous HF was added. The infrared spectrum was taken and all three components to be present.

The experiment was repeated in a a Kel-F trap. The gases were added to the trap and some solid was seen to form. The volatile gases were removed into the infrared cell and a spectrum showed them to be HF, ONF₃ and BF₃. The X-ray diffraction pattern of the solid was complex and did not correspond to NOBF₄ or to NO₂BF₄. The solid partly dissolved in anhydrous HF, the ¹⁹F n.m.r. of which revealed the presence of BF₄^- ion. The infrared spectrum of the solid was identical to NOBF₄.

So it looks like someone "invented" the chemistry behind the question. Sigh.

[DrCMS]

So it looks like someone "invented" the chemistry behind the question. Sigh.

It might be that later work done by another student of Neil Barlett got some different results using different conditions to Jack Passmore?

I found the info I did based on the date and name etc but I do not have access to the published papers.

What were the "correct" answers you had?

[Borek]

A - F₂
B - NOF
C - NOF₃
D - NOF₃·BF₃
E - NOF₃·2BF₃