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Q. Hybridisation of central atom in $NF_{3}$ is

WBJEEWBJEE 2009Chemical Bonding and Molecular Structure

Solution:

Number of hybrid orbitals = number of $\sigma$ bonds + number of lone pairs
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$\Rightarrow 3 \sigma$ bonds $+ 1$ lone pair $= 4$
$\therefore $ Hybridisation of $N$ is $sp^3$ and geometry of the molecule is pyramidal.