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

(N/A) $1$. Hybridization: In $PCl_{5}$,the central phosphorus atom $(P)$ undergoes $sp^{3}d$ hybridization. In the excited state,one $3s$ electron is promoted to the $3d$ orbital,resulting in five half-filled orbitals $(3s, 3p_{x}, 3p_{y}, 3p_{z}, 3d_{z^{2}})$ which hybridize to form five equivalent $sp^{3}d$ hybrid orbitals.
$2$. Geometry: These five $sp^{3}d$ hybrid orbitals are directed towards the corners of a trigonal bipyramid,forming five $P-Cl$ $\sigma$-bonds.
$3$. Bond Lengths: The axial $P-Cl$ bonds are longer than the equatorial $P-Cl$ bonds because the three equatorial $Cl$ atoms exert greater repulsive forces on the two axial $Cl$ atoms. To minimize this repulsion,the axial bonds are pushed slightly away,resulting in a longer bond length compared to the equatorial bonds.