Animated Solution for Chemistry - Coordination Compounds: For a $d^4$ metal ion in an octahedral field, the correct electronic configuration is

Imagine you are a tiny electron, the fourth one to arrive at a transition metal ion that has just been surrounded by six ligands in a perfect octahedral geometry. The first three electrons had it easy. They settled comfortably into the lower energy $t_{2g}$ orbitals, following Hund's rule of maximum multiplicity. But for you, the fourth electron, a critical decision awaits.

The arrival of the ligands has shattered the degeneracy of the five $d$-orbitals. They are no longer equal in energy. The repulsion from the ligands has pushed two of the orbitals ($d_{x^2-y^2}$ and $d_{z^2}$) to a higher energy level, forming the $e_g$ set. The remaining three ($d_{xy}$, $d_{yz}$, and $d_{zx}$) sit at a lower energy, forming the $t_{2g}$ set. The energy gap between these two sets is known as the Crystal Field Splitting Energy, denoted by ${\Delta }_0$.

As the fourth electron, you face a crossroads. You can either: 1. Jump the gap: Move up to the higher energy $e_g$ orbital. This requires an energy equal to ${\Delta }_0$. 2. Pair up: Stay in the lower energy $t_{2g}$ orbital and pair up with an existing electron. This requires Pairing Energy ($P$), which is the energy needed to overcome the electrostatic repulsion of sharing an orbital.

Your choice depends entirely on which path requires less energy. Nature always favors the path of least resistance!

The Weak Field (High Spin)

If the ligands surrounding the metal ion are "weak field" ligands (like halogens or water), they don't interact very strongly with the metal's $d$-orbitals. As a result, the splitting energy ${\Delta }_0$ is relatively small.

In this scenario, ${\Delta }_0<P$.

Since the gap is small, it takes less energy for you to jump up to the $e_g$ level than to endure the repulsion of pairing up in the $t_{2g}$ level. You make the jump! The electronic configuration becomes $t_{2g}^3{e}_g^1$. Because all four electrons are unpaired, this is called a high spin complex.

The Strong Field (Low Spin)

Now, imagine the ligands are "strong field" ligands (like cyanide or carbon monoxide). They interact fiercely with the metal, causing a massive split between the energy levels.

In this scenario, ${\Delta }_0>P$.

The gap is now a towering wall. It takes far more energy to jump to the $e_g$ level than to simply pair up in the $t_{2g}$ level. You choose to stay down and pair up. The electronic configuration becomes $t_{2g}^4{e}_g^0$. Because there are fewer unpaired electrons, this is called a low spin complex.

Returning to our original problem, we are asked for the configuration when ${\Delta }_0<P$. As we've just discovered, when the splitting energy is less than the pairing energy, the fourth electron jumps to the higher level. This results in the configuration $t_{2g}^3{e}_g^1$. Therefore, the correct option is indeed (a).