Identify the following species in which d²sp³ hybridisation is shown by central atom.

The correct option is [Co(NH₃)₆]³⁺.

To determine the hybridization of the central atom in each of the given species, we can analyze their chemical bonding and electronic configurations:

1. ${\mathrm{BrF}}_5$ (Bromine pentafluoride):
The central bromine (Br) atom has 7 valence electrons. It forms 5 covalent bonds with fluorine atoms and has 1 lone pair of electrons. The steric number is 5 + 1 = 6, which corresponds to ${\mathrm{sp}}^3{d}^2$ hybridization (outer orbital octahedral geometry).

2. ${\mathrm{SF}}_6$ (Sulfur hexafluoride):
The central sulfur (S) atom has 6 valence electrons and forms 6 single covalent bonds with fluorine atoms. The steric number is 6, which corresponds to ${\mathrm{sp}}^3{d}^2$ hybridization.

3. ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$ (Hexaamminecobalt(III) ion):
In this complex, the central cobalt atom is in the +3 oxidation state (${\mathrm{Co}}^{3+}$).
The electronic configuration of neutral Cobalt (atomic number = 27) is [Ar] ${\mathrm{3d}}^7{\mathrm{4s}}^2$.
The electronic configuration of ${\mathrm{Co}}^{3+}$ is [Ar] ${\mathrm{3d}}^6$.
Since ammonia (${\mathrm{NH}}_3$) is a strong-field ligand, it causes the pairing of the electrons in the 3d subshell, which leaves two 3d orbitals vacant.
These two vacant 3d orbitals, along with one 4s orbital and three 4p orbitals, undergo hybridization to form six vacant $d^2{\mathrm{sp}}^3$ hybrid orbitals. These orbitals accept lone pairs of electrons from the six ammonia ligands, resulting in an inner orbital octahedral complex with $d^2{\mathrm{sp}}^3$ hybridization.

4. ${\left[{\mathrm{PtCl}}_4\right]}^{\mathrm{2-}}$ (Tetrachloroplatinate(II) ion):
Platinum is in the +2 oxidation state with a coordination number of 4. It forms a square planar complex that involves ${\mathrm{dsp}}^2$ hybridization.