For a $$d^4$$ metal ion in an octahedral field, the correct electronic configuration is:

In an octahedral crystal field, the metal’s five $$d$$-orbitals split into two sets having different energies. By crystal-field theory we have

$$d\;(5) \;\longrightarrow\; t_{2g}\;(d_{xy},d_{yz},d_{zx})\; \text{(lower energy)} \;+\; e_g\;(d_{x^2-y^2},d_{z^2})\; \text{(higher energy)}$$

The energy gap between these two sets is called the octahedral crystal-field splitting energy and is denoted by $$\Delta_O$$. In addition, whenever two electrons occupy the same orbital we must supply the pairing energy $$P$$. Whether an electron prefers to jump to the higher $$e_g$$ level or to pair up in the lower $$t_{2g}$$ level depends on the relative magnitudes of $$\Delta_O$$ and $$P$$.

The rule is stated as follows: “If $$\Delta_O < P$$, promotion to the higher $$e_g$$ level costs less energy than pairing, so electrons remain unpaired as long as possible (high-spin case). If $$\Delta_O > P$$, pairing costs less energy than promotion, so electrons pair up in the lower level first (low-spin case).”

Now we place four $$d$$ electrons ($$d^4$$) in the octahedral field.

We first fill the three $$t_{2g}$$ orbitals singly according to Hund’s rule:

$$t_{2g}: \uparrow \;\; \uparrow \;\; \uparrow \qquad e_g:$$

Three electrons are now accommodated. One electron is still left to be placed.

Case 1: $$\Delta_O < P$$ (weak-field, high spin)

Because the energy required to pair ($$P$$) exceeds the energy required to promote ($$\Delta_O$$), the fourth electron will not pair in $$t_{2g}$$. Instead it will occupy the empty but higher $$e_g$$ orbital :

$$t_{2g}: \uparrow \;\; \uparrow \;\; \uparrow \qquad e_g: \uparrow$$

Therefore the electronic configuration is

$$t_{2g}^3 \, e_g^1$$

Case 2: $$\Delta_O > P$$ (strong-field, low spin)

Here pairing costs less energy than promotion, so the fourth electron pairs up in one of the $$t_{2g}$$ orbitals :

$$t_{2g}: \uparrow\downarrow \;\; \uparrow \;\; \uparrow \qquad e_g:$$

giving the configuration

$$t_{2g}^4 \, e_g^0$$

Comparing these results with the options, we see that the high-spin configuration $$t_{2g}^3 e_g^1$$ is obtained under the condition $$\Delta_O < P$$, which exactly matches Option A.

Hence, the correct answer is Option A.