In compounds of type $$EC{l_3},$$  where $$E = B,P,$$   $$As$$  or $$Bi,\,$$ the angles $$Cl - E - Cl$$    for different $$E$$  are in the order

Solution :
$$BC{l_3},\,\,H = \frac{1}{2}\left( {3 + 3 + 0 - 0} \right) = 3;\,s{p^2}\,{\text{hybridization}}$$
$$\left( {{\text{bond}}\,{\text{angle}} = {{120}^ \circ }} \right)$$    similarly $$PC{l_3}\,AsC{l_3}$$   and $$BiC{l_3}$$  are found to have $$s{p^3}$$ hybridized central atom with one lone pair of electrons on the central atom. The bond angle \[\le {{109}^{\circ }}28',\]   since the central atoms belong to the same group, the bond angle of the chlorides decreases as we go down the group. Thus the order of bond angle is, $$BC{l_3} > PC{l_3} > AsC{l_3} > BiC{l_3}.\,$$