Question

Hybridisation of the central atom changes in:

• A. $Al{H}_3\text{to}Al{H}_4^{-}$
• B. $H_{2}O\text{to}{H}_3{O}^{+}$
• C. $N{H}_3\text{to}N{H}_4^{+}$
• D. All of these

Answer 1

The correct option is A $Al{H}_3\text{to}Al{H}_4^{-}$

The hybridisation of $Al$ in $Al{H}_3$ is $s{p}^2$ (3 bond pair + 0 lone pair ; trigonal planar) whereas in $Al{H}_4^{+}$ the hybridisation on $Al$ is $s{p}^3$ (4 bond pairs + 0 lone pair ; tetrahedral).
The hybridisation on $O$ in both $H_{2}O$ and $H_3{O}^{+}$ is same, i.e., $s{p}^3$. But the shape of $H_{2}O$ is bent whereas that of $H_3{O}^{+}$ is pyramidal.
The hybridisation on $N$ in both $N{H}_3$ and $N{H}_4^{+}$ is same, i.e., $s{p}^3$. But the shape of $N{H}_3$ is pyramidal whereas that of $N{H}_4^{+}$ is tetrahedral .