The d-electron configuration of [Ru(en)3]Cl2 and [Fe(H2O)6]Cl2 respectively are:

🧠 Identify Each Metal's d-Count and Spin State

Complex 1: $[\mathrm{Ru(en)_3}]\mathrm{Cl_2}$.

• Outer Cl₂ → complex is +2 → Ru = +2.
• Ru(II) is 4d⁶.
• en is strong-field; Ru is 4d (always LS regardless) → low-spin.
• Configuration: $t_{2g}^6 e_g^0$.

Complex 2: $[\mathrm{Fe(H_2O)_6}]\mathrm{Cl_2}$.

• Outer Cl₂ → complex is +2 → Fe = +2.
• Fe(II) is 3d⁶.
• H₂O is weak-field; Fe is 3d → high-spin.
• Configuration: $t_{2g}^4 e_g^2$.

🗺️ Why Ru Is Always LS

Going from 3d → 4d → 5d, $\Delta_o$ scales up sharply (50% jump 3d→4d, ~25% more 4d→5d). For 4d/5d metals, $\Delta_o$ exceeds pairing energy for virtually any ligand → second- and third-row metals are nearly always low-spin.

So $[\mathrm{Ru(en)_3}]^{2+}$ is unambiguously LS → $t_{2g}^6 e_g^0$, diamagnetic.

⚡ Two d⁶ Ions, Two Different Configurations

Both Ru(II) and Fe(II) are d⁶, but they sit on different sides of the HS/LS boundary:

| Ion | Period | Field needed for LS | This complex | |---|---|---|---| | Fe(II) 3d⁶ | 3d | strong (CN⁻, CO, en) | weak H₂O → HS | | Ru(II) 4d⁶ | 4d | always LS | en → LS |

⚠️ Don't Carry Over 3d Reasoning to 4d/5d

A common slip: assuming the ligand-strength criterion that works for 3d metals also works for 4d/5d. It doesn't — Δ_o is so large for 4d/5d that any ligand pushes them to LS.

$\boxed{\text{Answer: (3) } t_{2g}^6 e_g^0 \text{ and } t_{2g}^4 e_g^2}$