When the water ligands coordinate to the $\mathrm{Co}^{2+}$ ion in $\mathrm{[Co(H_2O)6]^{2+}}$, the $d$ orbitals of the $\mathrm{Co}^{2+}$ ion split into two sets of orbitals: the $t{2g}$ set, which consists of the $d_{xy}$, $d_{xz}$, and $d_{yz}$ orbitals, and the $e_g$ set, which consists of the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals.

The $t_{2g}$ orbitals are lower in energy than the $e_g$ orbitals.

The $\mathrm{Co}^{2+}$ ion has a d$^7$ electron configuration, with three electrons in the $t_{2g}$ set and two electrons in the $e_g$ set. In an octahedral crystal field, the three $t_{2g}$ orbitals will be lower in energy than the two $e_g$ orbitals. Therefore, the three electrons in the $t_{2g}$ set will occupy all three $t_{2g}$ orbitals, and the two electrons in the $e_g$ set will occupy the higher energy $e_g$ orbitals. As a result, there is only $\boxed{1}$ unpaired electron in the $t_{2g}$ set of orbitals.