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Methyl isocyanate, $\ce{HC3NCO}$, is a toxic liquid which is used in the manufacture of some pesticides. What is the approximate angle between the bonds formed by the nitrogen atom in a molecule of methyl isocyanate?

(A) $104^\circ$ (B) $109^\circ$ (C) $120^\circ$ (D) $180^\circ$

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.

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This is not a good exercise question, because none of the options is actually correct. It basically asks you to blindly apply a model which cannot yield the correct result.

Before continuing any deeper into the matter, let's have a look at the actual structure. I was not able to find a crystal structure, so I ran a quick optimisation on the RI-BP86/def2-SVP level of theory. For this purpose, this is accurate enough.

structure of methyl isocyanate

The bond angle $\angle(\ce{CNC}) = 144^\circ$ is right in between $120^\circ \dots 180^\circ$.

The (pure) VSEPR model must fail for this molecule, as it is only really working for central atoms with distinct and non-delocalised ligands. For simple cases it performs very well in describing the approximate shape of a molecule. Once you have gotten this down, you can refine predictions with other tools. Please note, the lone pairs take up more space argument is actually a post-rationalisation and should not be used within the framework of VSEPR theory. More reliable predictions can be made with Bent's rule, which says that more electronegative ligands will concentrate p-orbital contributions towards them, decreasing the bond angle (this is a very, very brief summary).

As far as the VSEPR model goes, one should concentrate on simple coordination patterns:

  1. linear; the bond angles will be close to $180^\circ$
  2. trigonal; the bond angles will be close to $120^\circ$
  3. tetrahedral; the bond angles will be close to $109^\circ$
  4. higher coordinations like trigonal bipyramidal, octahedral, etc.

(TL;DR) From that we see that the central nitrogen has three ligands, so the general coordination around it is trigonal and the bond angle should be close to $120^\circ$. This is the answer, which your book will accept as correct, because it blindly applies the VSEPR model.

Applying a bit of Bent's rule won't help much. Carbon is slightly less electronegative than nitrogen, hence one would expect a slight opening up of the angle (more s-character directed towards carbon), but that's not enough to push it to $144^\circ$.

More importantly, the $\ce{-NCO}$ group is highly delocalised. Another very important concept comes into play: resonance. Ignoring the underlying structure for a moment, these are the most important contributors: $$\ce{Me-N=C=O <-> Me-N^+\bond{3}C-O^-}\tag1\label{eq:resonance}$$ Within the concept of resonance, the positions of the nuclei will not change, and the structures do not exist on its own. A single structure of a set is therefore always an incomplete description. Please read What is resonance, and are resonance structures real?

The left structure in \eqref{eq:resonance} is basically what we have described above with VSEPR. From the depiction on the right, we would expect a $\angle(\ce{CNC}$ of about $180^\circ$, because triple bonds are coordinated linear. Since these are electronic descriptions of the same structure, we can expect the actual (underlying) structure to be somewhere in between these hypothetical forms.

The observed molecular structure is always the result of a delicate balance of effects. So repulsion of nuclei, electronic repulsion, electronic attraction, dispersion, etc. will all have an effect. Many of these interact quite complicated with each other. Sometimes our used models will yield good approximations, sometimes they don't.
In this particular case, we see that VSEPR alone fails; including resonance will refine the prediction. However, without a more rigorous approach, it is impossible to predict the bond angle correctly, i.e. we do not know the contribution of the right structure without doing some more maths. I used a quantum chemical approach, which is obviously nothing you could do in an exam. This is one of the reasons why I dislike this question.


A quick word on hybridisation:

The concept of hybridisation is independent of VSEPR. Many books will throw them together, but it is actually incorrect. You can make a prediction of the shape of a molecule with VSEPR. After that, you can use the molecular structure to choose the closest matching hybridisation model (if any at all). As such, hybridisation is always a result from a particular structure, never a cause. Therefore, your initial assertion is reverse:

My current assumption is that the $\ce{e-}$ in nitrogen in $\ce{CH3NCO}$ forms sp2 hybridization (and therefore should forms a shape like ethene).

Correct is, because the structure around nitrogen is trigonal, the electronic structure can be approximated with sp2 orbitals.


Why $\angle\ce{CNC} \approx 104^\circ$ cannot be correct

In order for this angle to be achieved, the nitrogen first needs to be surrounded by four ligands (or electron pairs), giving us an angle of about $109^\circ$ to start with. Then more p-character needs to go into the bonds, reducing the angle. Hence the ligands must be more electronegative than the other ligands. A popular example is dichloromethane, or even water (but that is a bit more complicated). See the Wikipedia article on Bent's rule, our Q&A Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?, and multiple other questions with regard to Bent's rule.

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  • $\begingroup$ Hnmmm. I did not expect 144° at all so TIL... Did you manage to look at hybridisation at N? $\endgroup$ – orthocresol Jul 8 at 16:22
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    $\begingroup$ @orthocresol It is indeed more extreme than I expected, too, (Spoiler: even more in ~thio~). I don't have NBO or similar program currently available, but with Coulson's theorem, you could approximate this. For $\mathrm{sp}^x$, $x = \sqrt{\left(\frac{1}{-\cos(\angle)}\right)^2}$, ergo $\mathrm{sp}^{1.2}$. $\endgroup$ – Martin - マーチン Jul 8 at 16:32
  • $\begingroup$ would an x value less than 1.5 indicate that the resonance structure with sp contributes more than sp2? $\endgroup$ – Safdar Jul 8 at 16:47
  • $\begingroup$ @Safdar these are quite a crude approximation, so I wouldn't read too much into it. A more involved investigation would be necessary. At first glance though, it appears like you say. However, I don't feel confident enough in these calculations to make that statement. Also, there are probably more effects at play than electronics. $\endgroup$ – Martin - マーチン Jul 8 at 17:13
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    $\begingroup$ @Martin-マーチン I did not expect to find a crystal structure for methyl isocyanate (after all, a mp of −45 °C contributes to some difficulties [yet orderly crystallization at T lower than RT in a capillary mounted on a diffractometer is possible]), but fond some $\ce{R-N=C=O}$ in the COD to compare with: crystallography.net/cod/4025444.html, crystallography.net/cod/1504501.html, and crystallography.net/cod/1502171.html. $\endgroup$ – Buttonwood Jul 8 at 20:51
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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.

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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.

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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.

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