Another query is from June 2016 /MCQ. I am afraid that this
paper is not available in any site on the web.
Methyl isocyanate , CH3NCO, is a toxic liquid.
What is the approximate angle between the bonds formed by the N
atoms in a a molecule of methyl isocyanate?
H3C- N=C=O 120 is the answer.
The electron pairs around N are 3 ; there are 2 bond pairsaround
N; one is lone pairs. 321....so the bond angle must deviate from
the ideal situation 330...120. It would be less than 120, around
109.
Excellent question. There are actually 5 levels to this
at-1st-glance straightforward question.
Methyl
isocyanate
At the simplest level, the N atom has 3 electron regions or
charge clouds : a single bond, a double bond, and a lone pair.
Hence electron geometry is expected to be trigonal planar, ie. bond
angle close to 120 degrees.
On a deeper level, the lone pair is expected to have slightly
more repelling power, hence the bond angle may now be expected to
decrease slightly below 120 degrees.
On an even deeper level, the double bond is expected to have
slightly more repelling power, hence the bond angle may now be
expected to increase slightly back to approximately 120
degrees.
On a way deeper level, O has the greatest electronegativity,
hence the O atom withdraws strongly by induction (ie. through the
sigma bond) as well as by resonance (ie. through the pi bond) from
the C atom, which consequently itself becomes somewhat
electron-withdrawing vis-à-vis the N atom, despite N's greater
electronegativity vis-à-vis the C atom. The effect of which is to
reduce the electron density of the N=C double bond, hence the bond
angle may now be expected to decrease back to slightly less than
120 degrees.
On a way, way, way deeper level, you should be able to figure
out that there is a minor resonance contributor in which the lone
pair on the N atom forms a 2nd pi bond with the C atom, and the pi
bond between the C atom and the O atom becomes a lone pair on the O
atom, ie. +ve formal charge on the N atom, -ve formal charge on the
O atom. This is a minor resonance contributor, valid only because
of the high electronegativity of O. As such, the lone pair resides
in an orbital which is mostly sp2, but has a little unhybridized
'p' character, in order for the slight overlap of unhybridized p
orbitals (of N and C) to occur and for slight delocalization of the
lone pair to form a partial 2nd pi bond with the C atom, ie. the
N=C bond has partial triple bond character in the resonance hybrid.
As such, the bond angle may now be expected to increase to above
120 degrees, ie. the N atom is mostly sp2 hybridized, with a little
sp hybridized character.
Indeed, experimental evidence has proven the C-N-C bond angle to
be approximately 125 degrees, thus concurring with my 5 points of
consideration above.
Of course, Cambridge is aware that A level students are not able
to consider all these factors to elucidate as deeply as I just did,
and hence if it were a non-MCQ question, Cambridge will be
reasonable and accept reasonable answers on condition of reasonable
rationales. For a MCQ question, Cambridge will (as they did in this
MCQ) only provide clear-cut options such as A) 120 degrees B) 109.5
degrees C) 90 degrees D) 180 degrees, in which the required answer
is obvious, because the A level student is expected to choose the
option which gives the closest answer.