The homoleptic and octahedral complex of Co²⁺ and H₂O has ____unpaired electrons in the t_2g set of orbitals.

When the water ligands coordinate to the Co²⁺ ion in ${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$, the d-orbitals of the Co²⁺ ion splits into two sets of orbitals: ${\mathrm{t}}_{2\mathrm{g}}$ set, which consists of the ${\mathrm{d}}_{\mathrm{xy}},{\mathrm{d}}_{\mathrm{xz}}, \mathrm{and} {\mathrm{d}}_{\mathrm{yz}}$ orbitals, and the e_g set, which consists of the ${\mathrm{d}}_{{\mathrm{z}}^2} \mathrm{and} {\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ orbitals, and the e_g set, which consists of the ${\mathrm{d}}_{{\mathrm{z}}^2} \mathrm{and} {\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ orbitals.

The ${\mathrm{t}}_{2\mathrm{g}}$ orbitals are lower in energy than the e_g orbitals.

The Co²⁺ ion has d⁷ electronic configuration, with 5 electrons in the ${\mathrm{t}}_{2\mathrm{g}}$ set and 2 electrons in the e_g set. In an octahedral crystal field, the three ${\mathrm{t}}_{2\mathrm{g}}$ orbitals will be lower in energy than the two e_g orbitals. Therefore, the 5 electrons in the ${\mathrm{t}}_{2\mathrm{g}}$ set will occupy all three ${\mathrm{t}}_{2\mathrm{g}}$ orbitals and the 2 electrons in the e_g set will occupy the higher energy e_g orbitals. As a result, there is only 1 unpaired electron in the ${\mathrm{t}}_{2\mathrm{g}}$ set of orbitals.