Step 1: {Central Atom Hybridisation}
In sp\(^3\) d\(^2\) hybridisation, one s, three p, and two d orbitals of the central atom combine to form six hybrid orbitals. This occurs when the central atom is bonded to six other atoms.
Step 2: {Compound Analysis}
In BrF\(_5\), bromine exhibits sp\(^3\) d\(^2\) hybridisation because it is bonded to five fluorine atoms.
In SF\(_6\), sulfur exhibits sp\(^3\) d\(^2\) hybridisation because it is bonded to six fluorine atoms.
In [CrF\(_6\)]\(^{3-}\), chromium exhibits sp\(^3\) d\(^2\) hybridisation due to bonding with six fluoride ions.
In PF\(_5\), phosphorus exhibits sp\(^3\) hybridisation as it is bonded to five fluorine atoms, not six.
Therefore, the correct option is (D).
Hybridisation and geometry of [Ni(CN)$_4$]$^{2-}$ are