Question

Describe the hybridisation in case of $PCl_5$. Why are the axial bonds longer as compared to equatorial bonds?

Answer 1

The ground state and excited state outer electronic configurations of phosphorus $(Z = 15)$ are:
Ground state:

Excited state:

Phosphorus atom is $sp ^3d$ hybridized in the excited state. These orbitals are filled by the electron pairs donated by five $Cl$ atoms as:
$PCl_5$

The five $sp ^3d$ hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of $PCl_5$ can be represented as:

There are five $P-Cl$ sigma bonds in $PCl_5$. Three $P-Cl$ bonds lie in one plane and make an angle of $120\degree$ with each other. These bonds are called equatorial bonds.
The remaining two $P-Cl$ bonds lie above and below the equatorial plane and make an angle of $90\degree$ with the plane. These bonds are called axial bonds.
As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.