Explain $sp^3d^2$ hybridization with a suitable example.

(N/A) Example: $SF_6$ molecule.
$1$. Electronic configuration of Sulphur $(S)$: Atomic number $Z=16$,configuration is $[Ne] 3s^2 3p^4$.
$2$. Excited state: In the excited state,electrons from $3s$ and $3p$ orbitals are promoted to $3d$ orbitals to provide six half-filled orbitals: one $3s$,three $3p$,and two $3d$ orbitals.
$3$. $sp^3d^2$ hybridization: These six half-filled orbitals undergo hybridization to form six equivalent $sp^3d^2$ hybrid orbitals. These orbitals are directed towards the corners of an octahedron with bond angles of $90^{\circ}$.
$4$. Bond formation: Each of the six $sp^3d^2$ hybrid orbitals of $S$ overlaps with the half-filled $2p$ orbital of a Fluorine $(F)$ atom $([He] 2s^2 2p^5)$ to form six $S-F$ sigma bonds.