Question

The sum of $p$ orbitals involved in the hybridisation of $N{H}_3$ and $B{F}_3$ molecules are:

• A. $6$
• B. $5$
• C. $4$
• D. $3$

Answer 1

The correct option is B $5$

In $N{H}_3$, $N$ is bonded to three $H$ atoms and also has a lone pair. Thus, it is $s{p}^3$ hybridised.

In $B{F}_3$, $B$ is bonded to three $F$ atoms and has no lone pair. Thus, it is $s{p}^2$ hybridised.

The hybridisation of $N{H}_3$ and $B{F}_3$ are $s{p}^3$ and $s{p}^2$ respectively.

Hence, a total of $5$ $p$ orbitals are involved in the hybridisation of both the molecules.