Question

Bond angle in $PH_{3}$ is closer to $90^\circ $ while that in $NH_{3}$ is $104.5^\circ $ . Which of the following best explains this structural feature?

• A. Due to larger size of the lone pair electron cloud, there is larger lone pair - bond pair repulsion in $PH_{3}$ compared to $NH_{3}$ .
• B. Higher electronegativity of nitrogen concentrates the bond pair electron cloud near the central atom which increases the bond pair - bond pair repulsion which in turn decreases the bond angle in $NH_{3}$ .
• C. Energy difference between $3s$ and $3p$ orbitals is quite high and hence the lone pair on phosphorous prefers to occupy unhybridized $s-$ orbital rather than hybridized $s p^{3}$ hybridized orbital which causes its $s-$ orbital energy to increase.
• D. Phosphorous forms $p\pi -d\pi $ bonds while nitrogen does not.

Answer 1

Answer: (C)

In the hydrides of group $15$ and group $16$ (except $NH_{3}$ and $H_{2}O$ ) the energy difference between $3s$ and $3p$ orbital’s is quite high. Hybridization increases the energy of $3s$ orbital so much that lone pair rather prefers to occupy unhybridized s orbital. For example, in $PH_{3}$ , $600kJmol^{- 1}$ of energy is required to hybridize the central atom. So, to avoid such energy demanding hybridization P forms bonds with unhybridized p orbitals leaving the lone pair in the spherical s orbital which leads to a bond angle close to $90^\circ $ .