Describe hybridization in the case of $PCl_5$ and $SF_6$. The axial bonds are longer as compared to equatorial bonds in $PCl_5$ whereas in $SF_6$ both axial bonds and equatorial bonds have the same bond length. Explain.

(N/A) Formation of $PCl_5$: In $PCl_5$,phosphorus is $sp^3d$ hybridized to produce a set of five $sp^3d$ hybrid orbitals which are directed towards the five corners of a trigonal bipyramid. These five $sp^3d$ hybrid orbitals overlap with singly occupied $p$-orbitals of $Cl$ atoms to form five $P-Cl$ sigma bonds.
Three $P-Cl$ bonds are in one plane and make an angle of $120^{\circ}$ with each other. These bonds are called equatorial bonds.
The remaining two $P-Cl$ bonds,one lying above and the other lying below the plane,make an angle of $90^{\circ}$ with the equatorial plane. These bonds are called axial bonds. Axial bonds are slightly longer than equatorial bonds because axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.
Formation of $SF_6$: In $SF_6$,sulfur is $sp^3d^2$ hybridized to produce a set of six $sp^3d^2$ hybrid orbitals which are directed towards the six corners of a regular octahedron. These six $sp^3d^2$ hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six $S-F$ sigma bonds. Thus,the $SF_6$ molecule has a regular octahedral geometry and all $S-F$ bonds have the same bond length.