Explain $sp^{3}d$ hybridization by a suitable example.

(N/A) The $sp^{3}d$ hybridization involves the mixing of one $s$,three $p$,and one $d$ orbital to form five equivalent $sp^{3}d$ hybrid orbitals.
Example: Formation of $PCl_{5}$ molecule.
$1$. Electronic configuration of Phosphorus ($P$,atomic number $15$): Ground state: $[Ne] 3s^{2} 3p^{3} 3d^{0}$.
$2$. In the excited state,one electron from the $3s$ orbital is promoted to the $3d$ orbital: Excited state: $[Ne] 3s^{1} 3p^{3} 3d^{1}$.
$3$. These five orbitals $(3s, 3p_{x}, 3p_{y}, 3p_{z}, 3d_{z^{2}})$ undergo $sp^{3}d$ hybridization to form five equivalent $sp^{3}d$ hybrid orbitals.
$4$. These hybrid orbitals are directed towards the corners of a trigonal bipyramid,resulting in a trigonal bipyramidal geometry.
$5$. The bond angles are $120^{\circ}$ (equatorial) and $90^{\circ}$ (axial).
$6$. Each of these five $sp^{3}d$ hybrid orbitals overlaps with the half-filled $p$-orbital of a chlorine atom to form five $P-Cl$ $\sigma$-bonds.