Question

Bond angle in ${\displaystyle P{H}_4^{+}}$ is higher than that in PH₃. Why?

Answer 1

In PH₃, P is sp³ hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp³ bonding is changed to pyramidal. PH₃ combines with a proton to form ${\displaystyle P{H}_4^{+}}$ in which the lone pair is absent. Due to the absence of lone pair in ${\displaystyle P{H}_4^{+}}$, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in PH₃

Answer 2

P in PH₃ is sp³-hybridized with 3 bond pairs and one lone pair around P. Due to stronger lp-bp repulsions than bp-bp repulsions, tetrahedral angle decreases from 109°28′ to 93.6°. As a result, PH₃ is pyramidal. In PH₄⁺, there are 4 bp’s and no lone pair. As a result, there are only identical bp-bp repulsions so that PH₄⁺ assumes tetrahedral geometry and the bond angle is 109°28′.Hence, bond angle of PH₄⁺ > bond angle of PH₃