Question

Hybridisation of central atom in NF$_3$ is

• A. $sp^3$
• B. sp
• C. $sp^2$
• D. $dsp^2$

Answer 1

Answer: (A)

$sp^3$

Answer 2

$H=\frac{1}{2}(V+M-C+A)$ 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$^3$