Question

Arrange the following in order of decreasing bond angles, giving reasons
(i) $C{H}_4,N{H}_3,H_{2}O,B{F}_3,C_2{H}_2$ (ii) $N{H}_3,N{H}_2^{-},N{H}_4^{+}$.

Answer 1

(i) $C_2{H}_2\left({180}^{\circ }\right)>C{H}_4\left({109.28}^{\circ }\right)>B{F}_3\left({120}^{\circ }\right)>N{H}_3\left({107}^{\circ }\right)>H_{2}O\left({104.5}^{\circ }\right)$
(ii) $N{H}_4^{+}>N{H}_3>N{H}_2^{-}$
This is because all of them involve $s{p}^3$ hybridization. The number of lone pair of electrons present on N-atom are 0, 1 and 2 respectively. Greater the number of lone pairs, greater are the repulsion on the bond pairs and hence smaller is the angle.