Question

Describe the hybridisation in case of $PC{l}_5$. Why are the axial bonds longer as compared to.

Answer 1

The ground state and the excited state outer electronic configuration of phosphorus $(Z=15)$ are $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$ and $1s^{2}2s^{2}2p^{6}3s^{1}3p^{3}3d^1$ respectively.

Now the five orbitals are available for hybridisation to yield a set of five $s{p}^{3}d$ hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal. It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent . In $PC{l}_5$, the five $s{p}^{3}d$ orbitals of phosphorus overlap with the singly occupied $p$ orbitals of chlorine atoms to from five $P-Cl$ sigma bond. Three $P-Cl$ bond lie in one plane and make an angle of ${120}^o$ with each other, these bond are termed as equatorial bonds. The remaining two $P-Cl$ bonds one lying about and the other lying below the equatorial plane, make an angle of ${90}^o$ with the plane. These bonds are called axial bonds. s axial bonds suffer more repulsion than equatorial bond, these are found to be slightly longer and hence slightly weaker than equatorial bonds.