Explain the geometry of $SF_{6}$.

(N/A) The electronic configuration of sulphur ($S$,$Z=16$) is $[Ne] 3s^{2} 3p^{4}$.
In the excited state,one electron from $3s$ and one from $3p$ are promoted to the $3d$ orbitals,resulting in six half-filled orbitals ($3s$,$3p_{x}$,$3p_{y}$,$3p_{z}$,$3d_{x^{2}-y^{2}}$,and $3d_{z^{2}}$).
These six orbitals undergo $sp^{3}d^{2}$ hybridization to form six equivalent $sp^{3}d^{2}$ hybrid orbitals.
These hybrid orbitals are directed towards the corners of an octahedron,resulting in an octahedral geometry with bond angles of $90^{\circ}$.
In $SF_{6}$,each of the six fluorine atoms ($F$,$[He] 2s^{2} 2p^{5}$) uses its half-filled $2p$ orbital to overlap axially with one of the $sp^{3}d^{2}$ hybrid orbitals of sulphur,forming six $S-F$ sigma bonds.