Question

Hybridisation of Al atom changes in :
$Al{H}_3$ changes to $Al{H}_4^{-}$

• A. $s{p}^2$ to $s{p}^3$
• B. $s{p}^3$ to $s{p}^2$
• C. $s{p}^{3}d$ to $s{p}^3$
• D. both have same hybridisation

Answer 1

The correct option is A $s{p}^2$ to $s{p}^3$

Steric Number $\left(x\right)$ = $\frac{1}{2}\left(M+V-C+A\right)$.
where, M = number of monovalent surrounding atoms
V = number of valence electrons on the central atom
C = charge (with sign)
A=charge on anion
$Al{H}_3$
$\text{Steric Number (x)}=\frac{1}{2}(3+3-0+0)$.
$\text{Steric Number (x)}=3$.
Hybridisation =$s{p}^2$

$Al{H}_4^{-}$
$\text{Steric Number (x)}=\frac{1}{2}(3+4-0+1)$.
$\text{Steric Number (x)}=4$.
Hybridisation =$s{p}^3$