Question

Hybridization of central atom in $N{F}_3$ is:

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

Answer 1

The correct option is A

${sp}^3$

We can approach this problem by two ways. Let's take a look at electronic configuration of Nitrogen.

Now why Nitrogen?

Because nitrogen is our central atom.

$\to$ 3 Flourine will overlap with 3 ${sp}^3$ hybrid orbitals.

$\to$ In one ${sp}^3$ hybrid oribital, a lone pair will be present.

So, ${sp}^3$ hybridization

OR

By VSEPR, let's calculate electron pairs.

ep = $\frac{V+L+A-C}{2}$ = $\frac{5+3}{2}$ = 4
4 ⇒ tetrahedral geometry ⇒ sp³ hybridization