Question

Arrange the following in the increasing order of deviation from normal tetrahedral angle.

• A. $P_4<P{H}_3<H_{2}O$
• B. $P{H}_3<H_{2}O<P_4$
• C. $P_4<H_{2}O<P{H}_3$
• D. $H_{2}O<P{H}_3<P_4$

Answer 1

The correct option is D $H_{2}O<P{H}_3<P_4$

All the molecules have $s{p}^3$ hybridization. $P_4$ has one central $P$ atom has one lone pair, surrounded by $3P$ ligands having one lone pair on all of them and are bigger in size. $P{H}_3$ also has one lone pair on central $P$ atom. But the $H$ atoms are smaller in size thus will have less repulsion from lone pair. Similarly in $H_{2}O$ central $O$ atom has two lone pairs. All these molecules have deviation because of the presence of lone pair. The order of Deviation from the tetrahedral angle $H_{2}O<P{H}_3<P_4$