Question

Hybridisation of central atom in NF${}_3$ is

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

Answer 1

Answer: (A)

$H=\frac{1}{2}(V+M-C+A)$
where, V = number of electrons in valence shell
M = number of monovalent atoms
C = charge on cation
A = charge on anion
$H=\frac{1}{2}(5+3)=4$
$\therefore$ State of hybridisation = sp${}^3$