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

Explanation

Phosphorous in both `PH_(3) and PH_(4)^(+) is sp^(3)` hybridised. Due to the absence of lone pair-bond pair repulsion and presence of four identical bond pair - bond pair interactions. `PH_(4)^(+)` assume tetrahedral geometry with a bond angle of `109^(@)28.`
But `PH_(3)` has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion. the tetrahedral angle decreases from `109^(@)28. to 93.6^(@)`. As a result PH, is pyramidal.
`PH_(3)` - yramidal with bond angle of `93.^(@)`
`PH_(4)^(+)` - Tetrahedral with bond angle of `109^(@)28..`