# What are the bond angles of PF_3Cl_2?

And so for such a $E {X}_{5}$ structure there are $3 \times \angle X - E - X = {120}^{\circ}$, i.e. for the equatorial $X$ ligands, and $1 \times \angle X - E - X = {180}^{\circ}$.
Of course, an $E {X}_{5}$ structure, is conformationally mobile, and the axial and equatorial ligands undergo facile exchange.