The homoleptic and octahedral complex of Co 2 + and H 2 O has ______ unpaired electron(s) in the t 2 g set…

The electronic configuration of ${\mathrm{Co}}^{2+}$ is $3{\mathrm{d}}^7$. In an octahedral complex of ${\mathrm{Co}}^{2+}$ with ${\mathrm{H}}_2\mathrm{O}$ ligands, the d orbitals split into two sets of energy levels: ${\mathrm{t}}_{2\mathrm{g}}$ (lower energy) and ${\mathrm{e}}_{\mathrm{g}}$ (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, ${\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}={\mathrm{d}}^7\left(\mathrm{High} \mathrm{spin}\right)={\mathrm{t}}_{2\mathrm{g}}^5 {\mathrm{e}}_{\mathrm{g}}^2$

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