Question

Among the following molecules, N-Si bond length is shortest in:
(a) $ N{(Si{H_3})_3} $
(b) $ N{(Si{H_3})_2} $
(c) $ N{H_2}(Si{H_3}) $
(d) All have equal N-Si bond length

Answer 1

Hint: In Chemistry, Bond length is defined as the average distance between the centers or the nuclei of two bonded atoms. In molecules having N-Si bonds, back bonding occurs. Back bonding is the type of bonding that occurs between two atoms which are adjacent to each other; one having a lone pair of electrons while the other has a vacant orbital. The stronger the back bond, the smaller the bond length.

Complete answer:
In back bonding, A weaker $ \pi - {\text{bond}} $ is formed, since the two atoms already have a sigma bond $ \left( {\sigma - {\text{bond}}} \right). $ The $ \pi - {\text{bond}} $ is formed due to sideways overlap between the filled orbital of one atom and the vacant orbital of the other atom. $ \pi - {\text{character}} $ is exhibited by the bond due to its formation after $ \sigma - {\text{bond}}. $
In the case of molecules having N-Si bonds, N atom has an electron pair in $ 2p\pi $ orbital while the Si atom has a vacant orbital $ 2d\pi . $ Hence, $ \pi - {\text{back bonding}} $ occurs in the N-Si bond. When the number of Si atoms bonded to a N atom increases, the distribution of the electron pair occurs, due to which back bonding occurs to a lesser effect. Hence, the $ \pi - {\text{bond}} $ formed is not strong, and correspondingly the $ \pi - {\text{characteristic}} $ decreases. A direct effect of decrease in $ \pi - {\text{characteristic}} $ is the increase in the bond-length.
Accordingly, the decreasing order of bond length will be $ N{(Si{H_3})_3} > N{(Si{H_3})_2} > N{H_2}(Si{H_3}). $ Therefore, among the given molecules, N-Si bond length is shortest in (c) $ N{H_2}(Si{H_3}). $

Note:
 Back bonding is considered as intermolecular Lewis acid-base interaction being a $ \pi - {\text{bond}}{\text{.}} $ It is seen that back bonding is most effective between the following orbital pairs: $ \;2p - 2p,{\text{ }}2p - 3p,{\text{ }}2p - 3d\; $ and the extent of overlapping is in the order: $ 2p - 2p > {\text{ }}2p - 3d{\text{ }} > 2p - 3p.\;\; $ It has also been observed that back bonding does not occur between a certain pair of orbitals, the pair being $ \;{d_{{x^2}}}{\text{ and }}\;{d_{{x^2} - {y^2}}}. $