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

Explanation

P in `PH_(3)` is `sp^(3)`-hybridized. 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, the tetraheldral 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)^(+)` ionn 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`. This explains why the bond angle in `PH_(4)^(+)` is higher than in `PH_(3)`.