For a d⁴ metal ion in an octahedral field, the correct electronic configuration is:

🧠 d⁴ Octahedral: HS vs LS

For d⁴ in octahedral:

• High-spin ($\Delta_o < P$): electrons spread; fill $t_{2g}^3 e_g^1$ → 4 unpaired.
• Low-spin ($\Delta_o > P$): electrons pair within $t_{2g}$; fill $t_{2g}^4 e_g^0$ → 2 unpaired.

So when $\Delta_o < P$: $t_{2g}^3 e_g^1$. ✓ matches option 1.

🗺️ Walk the Other Options

| Option | Config | Field condition | Verdict | |---|---|---|---| | (1) $t_{2g}^3 e_g^1$ when $\Delta_o < P$ | HS | $\Delta_o < P$ ✓ | Correct ✓ | | (2) $t_{2g}^3 e_g^1$ when $\Delta_o > P$ | HS but condition is LS | mismatch | ✗ | | (3) $t_{2g}^4 e_g^0$ when $\Delta_o < P$ | LS but condition is HS | mismatch | ✗ | | (4) $e_g^2 t_{2g}^2$ when $\Delta_o < P$ | impossible (eg above t₂g, can't fill eg before t₂g) | — | ✗ |

⚡ Memorise the d⁴ HS↔LS Pair

| Spin | Config | Unpaired | μ (BM) | |---|---|---|---| | HS | $t_{2g}^3 e_g^1$ | 4 | 4.9 | | LS | $t_{2g}^4 e_g^0$ | 2 | 2.83 |

These correspond to Cr(II), Mn(III), and the "spin-crossover" region of d-counts.

⚠️ Filling Order Always Goes $t_{2g}$ First

In octahedral, $t_{2g}$ is always lower than $e_g$. Even in HS, you don't put electrons in $e_g$ before $t_{2g}$ has at least 3 (one per orbital). Option (4) violates this.

$\boxed{\text{Answer: (1) } t_{2g}^3 e_g^1 \text{ when } \Delta_O < P}$