Question

The correct order of hybridisation of the central atom in the species $N{F}_3,B{F}_3,P{F}_5$ and $[Si{F}_6{]}^{2-}$ is respectively:

• A. $s{p}^2,s{p}^3,s{p}^3{d}^2,s{p}^{3}d$
• B. $s{p}^3,s{p}^2,s{p}^{3}d,s{p}^3{d}^2$
• C. $s{p}^2,s{p}^3,s{p}^{3}d,s{p}^3{d}^2$
• D. $s{p}^3,s{p}^2,s{p}^3{d}^2,s{p}^3{d}^3$

Answer 1

The correct option is B $s{p}^3,s{p}^2,s{p}^{3}d,s{p}^3{d}^2$

The hybridisation of $N$ in $N{F}_3$ is $s{p}^3$ as there are 3 bond pairs and one lone pair on N atom.

The hybridisation of $B$ in $B{F}_3$ is $s{p}^2$ as there are 3 bond pairs.

The hybridisation of $P$ in $P{F}_5$ is $s{p}^{3}d$ as there are 5 bond pairs.

The hybridisation of $Si$ in $[Si{F}_6{]}^{2-}$ is $s{p}^3{d}^2$ as there are 6 bond pairs.

Hence, the correct order is given by option B.