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${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{+3}-{\mathrm{d}}^2{\mathrm{sp}}^3$ hybridization

In ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$, $\mathrm{Co}$ is in $+3$ oxidation state with the configuration $3{\mathrm{d}}^6$. In the presence of ${\mathrm{NH}}_3$ a strong ligand, the $3\mathrm{d}$ electrons pair up leaving two d-orbitals empty. Hence, the hybridization is ${\mathrm{d}}^2{\mathrm{sp}}^3$forming an inner orbital octahedral complex. 

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