Consider the following iodides :

$\begin{array}{ccc}{\mathrm{PI}}_3& {\mathrm{AsI}}_3& {\mathrm{SbI}}_3\\ {102}^{\circ }& {100.2}^{\circ }& {99}^{\circ }\end{array}$

The bond angle is maximum in ${\mathrm{PI}}_3$, which is :

The molecules ${\mathrm{PF}}_3$ and ${\mathrm{OSF}}_2$ have the shortest angles as a result of having the most electronegative exterior atoms.

The halogen exerts a higher pull on the electron pairs it shares with the core atom as its electronegativity rises. 

As a result, there are fewer electrons close to the central atom, which lessens the attraction between bonding pairs there and gives the lone pair more power to compress the halogen central atom halogen angles. 

Hence the bond angle is maximum in ${\mathrm{PI}}_3$,