$SbC{l}_3$ and $BiC{l}_3$ on hydrolysis gives 

${\mathrm{SbCl}}_3+{\mathrm{H}}_2\mathrm{O}\to \mathrm{SbOCl}+2\mathrm{HCl}$

$BiC{l}_3+H_{2}O⟶BiOCl+2HCl$

(Partial Hydrolysis)

The halides of $\mathrm{Sb}$ and $\mathrm{Bi}$generate a partially hydrolyzed product as ionic character rises with falling group 15 reactivity.

Hence correct option C