Need help with a past paper question

from 9701_w11_qp_11 (international a level):
Q4.Methyl isocyanate, CH3NCO, is a toxic liquid which is used in the manufacture of somepesticides.In the methyl isocyanate molecule, the sequence of atoms is H3C—N C O.What is the approximate angle between the bonds formed by the N atom?A104°B109°C120°D180°

MS says it's C, however i don't quite understand.
I figured that Nitrogen would have 2 lone pairs left over, therefore it will have a tetrahedral shape (repulsion between the 2 C's and the 2 lone pairs), resulting in 104 degrees.

Original post by TeachChemistry

The angle wouldn't be 120° though and the shape wouldn't be trigonal planar. The lone pair would repel the electrons in the bonding pairs more and so the angle would be around 117/118° and the shape would be considered bent.

Reply 3

username986184

3

Original post by Jpw1097

The angle wouldn't be 120° though and the shape wouldn't be trigonal planar. The lone pair would repel the electrons in the bonding pairs more and so the angle would be around 117/118° and the shape would be considered bent.

Since 117 is not an option in the given answers your reasoning is, although correct, irrelevant.

Reply 4

Pigster

20

Original post by Jpw1097

The angle wouldn't be 120° though and the shape wouldn't be trigonal planar. The lone pair would repel the electrons in the bonding pairs more and so the angle would be around 117/118° and the shape would be considered bent.

Your suggestion seems sensible, but it turns out that angle is actually 125^o

Int VSEPR fab?

Reply 5

Jpw1097

19

Original post by TeachChemistry

Since 117 is not an option in the given answers your reasoning is, although correct, irrelevant.

I realised...

Reply 6

Jpw1097

19

Original post by Pigster

Your suggestion seems sensible, but it turns out that angle is actually 125^o

Int VSEPR fab?

Perhaps that's due to the high e^- density in the N=C bond? The electrons in the double bond might repel the electrons in the lone pair and the bonding pair, resulting in an angle of 125°.

Original post by Jpw1097

Perhaps that's due to the high e^- density in the N=C bond? The electrons in the double bond might repel the electrons in the lone pair and the bonding pair, resulting in an angle of 125°.

Clearly the inter-domain repulsion depends on the relative electron densities of the domains and their relative distance from the nucleus.

This makes applying VSEPR a bit of a hand wavy exercise, particularly coupled with the fact that it falls apart in period 3 (and above) when the domains used in bonding are now much further from the nuclear center AND there is often contribution from 'd' orbitals.

.. is all.

The equivalent molecule: SiH3NCO has a bond angle of 180^o.

The lone pair on the N becomes a second bond to the Si, in a p(pi)d(pi) style.

H₃Si^-=N⁺=C=O

(edited 8 years ago)

Reply 10

username986184

3

Original post by Pigster

The equivalent molecule: SiH3NCO has a bond angle of 180^o.

The lone pair on the N becomes a second bond to the Si, in a p(pi)d(pi) style.

H₃Si^-=N⁺=C=O

interesting. :smile:

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