Question

The correct order of hybridization of the central atom in the following species is:

${NH}_3,{\left[Pt{Cl}_4\right]}^{2-},P{Cl}_5$ and $B{Cl}_3$

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

Answer 1

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

${NH}_3,{\left[Pt{Cl}_4\right]}^{2-},P{Cl}_5$ and $B{Cl}_3$ have ${sp}^3,d{sp}^2,{sp}^{3}d$ and ${sp}^2$-hybridization respectively. Note that hybridization of $P$ in $P{Cl}_5$ is wrongly reported in the problem.
Hybridization in ${NH}_3$ :It is is $s{p}^3$ because $N$ is bonded to 3 hydrogen groups and contains two unpaired electrons. For these three bonds and unpaired electron, the $s$ orbital and three $p$ orbitals hybridize forming $s{p}^3$ hybridized orbitals.