Question

Hybridization of central atom in $N{F_3}$ is:
$A.$ $s{p^3}$
$B.$ $sp$
$C.$ $s{p^2}$
$D.$ $ds{p^2}$

Answer 1

Hint: Before solving the question we need to have some idea about hybridization. Hybridization is defined as the intermixing of two atomic orbits having the similar energy to produce a new different orbital, a new orbital which is produced known as hybrid orbitals. This process is called hybridization.

Complete step by step answer:
In Nitrogen trifluoride $\left( {N{F_3}} \right)$ the central metal atom is nitrogen atom which is attached to three fluorine atoms. Nitrogen has five valence electrons in its valence shell; to complete its octet it requires three more electrons. So it forms three sigma bonds with fluorine atoms.
The hybridization of nitrogen atoms in $N{F_3}$ molecules can be determined by valence shell electron pair repulsion theory $\left( {VSEPR} \right)$ . The Lewis structure of $N{F_3}$ molecules shows that the nitrogen atom uses only its three electrons to complete the octet. It has five valence electrons in which three are used in the formation of \[3\sigma \] with fluorine atom and remaining two electrons on nitrogen atom, they remain of nitrogen atom as non- bonding electron is now an lone pair.
Hence, in the $N{F_3}$ molecule there are three bond pairs and one lone pair. Sum of bond pairs and lone pairs is four. Nitrogen atoms are $N{F_3}$ molecules and are supposed to be \[s{p^3}\] hybridization.
                           

As we discussed above, the hybridisation of nitrogen atoms in $N{F_3}$ molecules. So, the correct option is $\left( A \right)$ .

Note:
Valence Shell Electron Pair Repulsion Theory $\left( {VSEPR} \right)$ .This is a very useful theory to predict the geometry or shape of a number of polyatomic molecules or ions on a non-transition element. This theory says that shapes of a species and hybridization of a molecule depend on the number of and nature of electron pairs surrounding the central atom of a species.