The homoleptic and octahedral complex of Co ^ 2 + and H _ 2 O has...

The electronic configuration of ${Co}^{2 +}$ is $3{d}^{7}$. In an octahedral complex of ${Co}^{2 +}$ with ${H}_{2}O$ ligands, the d orbitals split into two sets of energy levels: ${t}_{2g}$ (lower energy)$and {e}_{g}$ (higher energy).

In this case, since the complex is octahedral, we have six ligands surrounding the central cobalt ion. Each ligand donates a pair of electrons to the cobalt ion, occupying the available d orbitals.

So, ${\left[Co{\left({H}_{2}O\right)}_{6}\right]}^{2 +} = {d}^{7}\left(Highspin\right) = {t}_{2g}^{5}{e}_{g}^{2}$

In the ${t}_{2g}$set of orbitals, there will be $1$ unpaired electron