Question

Bond angle in $P{H}_3$ is closer to $90^{\circ }$ while that in $N{H}_3$ 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 $P{H}_3$ compared to $N{H}_3$ .
• 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 $N{H}_3$ .
• 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 $N{H}_3$ 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 $P{H}_3$ , $600kJmo{l}^{-1}$ 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 $90^{\circ }$ .