# Bond angle in CH3NCO

I was doing past year papers of A levels Chemistry, when I came across this problem:

I answered (A), but the correct answer is (C) according to the marking scheme. I thought it was a bent-shaped, because the lone pair of $\ce{e-}$ would repel the bond pairs of electron (similarly to $\ce{H2O}$ or $\ce{SO2}$)

My current assumption is that the $\ce{e-}$ in $\ce{N}$ in $\ce{CH3NCO}$ forms $\mathrm{sp^2}$ hybridization (and therefore should forms a shape like ethene).

I'm not sure if my thinking is correct, so hopefully you can help me understand why (C) is correct and why (A) is impossible.

The answer is C because double bonds act as only one pair of electrons around the central atom (the Nitrogen) therefore in this case the nitrogen has 2 bonding pairs (one from the double bond and one from the single bond) and one lone pair. As the nitrogen is surrounded by 3 pairs it will take a triangular planar structure around the atom, meaning that is extra repulsion from the lone pair are ignored the best angle approximation for the bond is 120 degrees.

I think that to answer this question, you need to do the process of elimination (A Level multiple choice questions are like this sometimes). Not to mention that the question you supplied looks like it's from an OCR new specification exam (I could be wrong), and hybridisation isn't on the OCR specification so I highly doubt they'd expect students to think about that. My following method of thinking is purely from being taught the OCR specification:

• It can't be A, since to achieve a 104° bond angle, you require 2 bonding pairs and 2 lone pairs,
• It can't be B, since to acheive a 109° bond angle, you require 4 bonding pairs and 0 lone pairs, and,
• It can't be D, since to acheive a 180° bond angle, you require to have a few bond areas and 0 lone pairs (for example, the 2 double bonds in CO2 count as 2 bond areas).

The bonds around nitrogen have 2 bond areas and 1 lone pair, which isn't covered by the OCR specification; this leaves C as the answer due to the fact we eliminated the other answers.

I would vote for C, the C-N-C angle will be slightly smaller than 120 degrees. This is becuase the lone pair is slightly closer to the nucleus than the bonded pairs in sigma bonds. As a result the lone pair occupies a slightly larger solid angle than a bonded pair.

As a result the H-N-H angle of ammonia and the C-N-C angle of methyl isocyanate are slightly smaller than the values which would be predicted by the VSEPR theroy if it is assumed that all electron pairs in sigma bonds / lone pairs are equal.