Describe the hybridisation in case of $PCl_{5}$. Why are the axial bonds longer as compared to equatorial bonds?

(N/A) The ground state and excited state outer electronic configurations of phosphorus $(Z=15)$ are:
Ground state: $3s^{2} 3p^{3}$
Excited state: $3s^{1} 3p^{3} 3d^{1}$
Phosphorus atom is $sp^{3}d$ hybridized in the excited state. These orbitals are filled by the electron pairs donated by five $Cl$ atoms to form $PCl_{5}$.
The five $sp^{3}d$ hybrid orbitals are directed towards the five corners of a trigonal bipyramid.
There are five $P-Cl$ sigma bonds in $PCl_{5}$. Three $P-Cl$ bonds lie in one plane and make an angle of $120^{\circ}$ with each other. These are called equatorial bonds.
The remaining two $P-Cl$ bonds lie above and below the equatorial plane and make an angle of $90^{\circ}$ with the plane. These are called axial bonds.
As the axial bond pairs suffer more repulsion from the three equatorial bond pairs,the axial bonds are slightly longer than the equatorial bonds.