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

Explanation

P is `sp^(3)` -hybridised both in `PH_(3)` and `PH_(4)^(+)cdot PH_(4)^(+)` contains all four bond pairs with bond angle `109^(@)28'` , as expected in case of tetrahedral geometry. However, `PH_(3)` contains three bond pairs and one lone pair. Since lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, the bond angle reduces to `93.6^(@)` from `109^(@)28'`. Hence, the bond angle in `PH_(4)^(+)` is higher than `PH_(3)`.