Sp³d² hybridization is present in [Co(NH₃)₆³⁺], find its geometry

The correct answer/option is octahedral geometry.

To understand why $[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6{]}^{3+}$ possesses octahedral geometry under ${\mathrm{sp}}^3{d}^2$ (or outer orbital octahedral coordination) hybridization, let us break down the bonding and coordination details step-by-step:

1. Coordination Number and Geometry:
In the complex ion $[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6{]}^{3+}$, the cobalt central metal ion is bonded to six ammine (${\mathrm{NH}}_3$) ligands. This means the coordination number of the central metal is 6.
A coordination number of 6 corresponds directly to an octahedral arrangement of ligands around the central metal atom/ion to minimize steric repulsion.

2. Hybridization State:
An ${\mathrm{sp}}^3{d}^2$ hybridization involves the mixing of one s-orbital, three p-orbitals, and two d-orbitals from the valence shell.
This combination yields six equivalent hybrid orbitals oriented toward the vertices of a regular octahedron. Thus, any metal complex exhibiting ${\mathrm{sp}}^3{d}^2$ hybridization will have an octahedral geometry.