Hybridisation 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 $s p^{3}$
Number of hybrid orbitals $=$ number of $σ$ bonds $+$ number of lone pairs
$\Rightarrow 3 σ$ bonds $+ 1$ lone pair $= 4$
$∴$ Hybridisation of $N$ is ${s p}^{3}$ and geometry of the molecule is pyramidal.

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