An octahedral complex of ${\mathrm{Co}}^{3+}$ is diamagnetic. The hybridisation involved in the formation of the complex is:

The outer electronic configuration of ${ }_{27}{\mathrm{Co}}^{3+}$ is $(3\mathrm{d}{)}^6$, i.e. free ${ }_{27}{\mathrm{Co}}^{3+}\text{ ion }$;

Since the octahedral complex of ${\mathrm{Co}}^{3+}$ is diamagnetic, it does not possess any unpaired electron. The configuration of ${\mathrm{Co}}^{3+}$ ion in the complex will be

Thus, ${\mathrm{Co}}^{3+}$ion in the complex involves ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridization.