$(B) \,I_3^-$ undergoes $sp^3 d$ hybridisation and linear shape so its bond angle is $180^{\circ}$.

$(C) \,NO _2^{-}$ undergoes $s p^2$ hybridisation and So, its bond angle is nearly $120^{\circ}$

$(D) \,PH _3$ undergoes $sp ^3$ hybridisation and contains one lone pair, so its bond angle is mearly $90^{\circ}$

$\therefore$ Among all the options, $PH _3$ has least angle around the central atom.