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According to VSEPR theory, for $\mathrm{sp^2}$ hybridisation the bond angle is approximately $120^\circ$. In $\ce{H3CNCS}$, the nitrogen has one lone pair. As lone-pair-bond-pair repulsion is greater than bond-pair-bond-pair repulsion, the bond angle of $\ce{CNC}$ should be less than $120^\circ$, but it is actually greater. How can this be rationalised?

orthocresol
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Shiva
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1 Answers1

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I have just answered What is the CNC bond angle in methyl isocyanate? Pretty much everything written there also applies to methyl isothiocyanate, just a bit more extreme.

Here is the structure calculated on the RI-BP86/def2-SVP level of theory:

molecular structure methyl isothiocyanate

The bond angle is $\angle(\ce{CNC}) = 156^\circ$, so we expect a much larger contribution of the right side in the following resonance description: $$\ce{H3C-N=C=S <-> H3C-N^+\bond{3}C-S^-}$$

Reasons for this are not trivial, but that would exceed the scope of this answer.

TL;DR From VSEPR we would expect a trigonal coordination and a $120^\circ$ angle. Refining with resonance, we'll expect a widening of the bond angle. Everything beyond that needs much more involved models.

Martin - マーチン
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  • Similar to an other comment, not about methyl isothiocyanates, but $\ce{R-N=C=S}$ there are at least these 15 entries in the COD with coordinates to compare with: http://www.crystallography.net/cod/7116329.html and (just substitute the numbers), COD7116328, 7116327, 7116326, 7116325, 7103098, 4117347, 4108942, 4108356, 4075734, 4024672, 4024671, 2234166, 2233483, and 2214501. – Buttonwood Jul 08 '20 at 21:01