Question

Which one of the following shows a change in the type of hybridisation?

• A. $N{H}_3$ combines with $H$
• B. $Al{H}_3$ combines with $H$
• C. In both the cases
• D. In none of the case

Answer 1

The correct option is B $Al{H}_3$ combines with $H$

In $Al{H}_3$ (Refer IMAGE $01$)

$SN=\frac{1}{2}[3+3]=3\Rightarrow s{p}^3\to$ Trigonal planar.

$Al{H}_3+H^{-}\to Al{H}_4^{-}$

Hybridization of $Al$ shifts from $s{p}^2$ to $s{p}^3$ shape is tetrahedral.

In$N{H}_3$ (Refer IMAGE $02$)

$SN=\frac{1}{2}[5+3]=4\Rightarrow s{p}^3\to$ Tetrahedral

$N{H}_3+H^{+}\to N{H}_4^{+}$

The hybridization of $N$ remain same $s{p}^3$. Geometry remain same, tetrahedral. Only shape changes from trigonal pyramidal to tetrahedral.