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

Explanation

In `PH_(3)`, P is `sp_(3)` 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 `sp_(3)` bonding is changed to pyramidal. `PH_(3)` combines with a proton to form in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in `PH_(3)`