For an octahedral field, the five $d$-orbitals split into: $t_{2g} \text{ (lower energy)} \quad \text{and} \quad e_g \text{ (higher energy)}$

For a $d^4$ metal ion, we need to place 4 electrons in these orbitals. The arrangement depends on the comparison between:

• $\Delta_o$ = octahedral crystal field splitting energy
• $P$ = pairing energy

Case 1: $\Delta_o < P$

• It is energetically cheaper to put the 4th electron into the higher $e_g$ orbital rather than pair up in $t_{2g}$.
• So the configuration is: $t_{2g}^3 e_g^1$
• This is the high-spin case.

Case 2: $\Delta_o > P$

• It is energetically cheaper to pair electrons in the lower $t_{2g}$ set.
• So the configuration is: $t_{2g}^4 e_g^0$
• This is the low-spin case.

Now evaluate the options:

• A: $t_{2g}^4e_g^0$ when $\Delta_o < P$
  Wrong, because $t_{2g}^4e_g^0$ occurs when $\Delta_o > P$.

• B: $t_{2g}^3e_g^1$ when $\Delta_o > P$
  Wrong, because $t_{2g}^3e_g^1$ occurs when $\Delta_o < P$.

• C: $e_g^2t_{2g}^2$ when $\Delta_o < P$
  Wrong, electrons fill lower-energy $t_{2g}$ first, so this arrangement is not correct.

• D: $t_{2g}^3e_g^1$ when $\Delta_o < P$
  Correct.

Therefore, the correct answer is: $\boxed{D}$