Hybridization of central atom in $N F_{3}$ is:

${s p}^{3}$

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$s p$

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${s p}^{2}$

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${d s p}^{2}$

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Solution

The correct option is A
${s p}^{3}$

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 ${s p}^{3}$ hybrid orbitals.

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

So, ${s p}^{3}$ 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

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Hybridization of central atom in $N F_{3}$ is:

In which of the following the central atom does not use $s p^{3}$ hybrid orbitals in its bonding

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