JEE Main 2023 — Coordination Compounds Question with Solution

When the water ligands coordinate to the ${\mathrm{Co}}^{2+}$ ion in $[\mathrm{Co}({\mathrm{H}}_2\mathrm{O})6{]}^{2+}$, the $d$ orbitals of the ${\mathrm{Co}}^{2+}$ ion split into two sets of orbitals: the $t2g$ 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.