Question

Hybridisation of central atom in $N{F}_3$ is

• A. $s{p}^3$
• B. $sp$
• C. $s{p}^2$
• D. $ds{p}^2$

Answer 1

Answer: (A)

Number of hybrid orbitals = number of $\sigma$ bonds + number of lone pairs

$\Rightarrow 3\sigma$ bonds $+1$ lone pair $=4$
$\therefore$ Hybridisation of $N$ is $s{p}^3$ and geometry of the molecule is pyramidal.