Bond angle in `PH_(4)^(+)` is higher than that in `PH_(3)`. Why?

Explanation

P in `PH_(3)`, is `sp^(3)` hybridised. It has three bond pairs and one lone pair around P. Due to stronger lone pair-bond pair repulsions than bond pair-bond pair repulsions, thetetrahedral angle decreases from `109^(@)`28. to `93.6^(@)`. As a result, `PH_(3)` , is pyramidal.However, when it reacts with a proton, it forms `PH_(4)^(+)` ion which has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsions and presence of four identical bond pair-bond pair interactions, `PH_(4)^(+)` assumes tetrahedral geometry with a bond angle of `109^(@)` 28.. Hence, bond angle in `PH_(4)^(+)` is higher than `PH_(3)`.