# Cl-P-Cl and Cl-S-Cl Bond angles in respective PCl3 and SCl2 compounds

Both $$\ce{PCl3}$$ and $$\ce{SCl2}$$ are $$\mathrm{sp^3}$$ hybrid. $$\ce{PCl3}$$ has 1 lone pair whereas $$\ce{SCl2}$$ has 2 lone pairs, so due to more lp-lp repulsion, $$\ce{SCl2}$$ should have smaller bond angle than $$\ce{PCl3}$$. But practically, $$\ce{SCl2}$$ has got larger bond angle than $$\ce{PCl3}$$. Why is it so?