The complex ion that will lose its crystal field stabilization energy upon oxidation of metal to +3 state is: (phen =…

🧠 Find the Complex Where M(III) Has Less CFSE Than M(II)

For each, compute CFSE in M(II) and M(III) states (with strong-field phen), then check which loses CFSE on oxidation.

🗺️ Evaluate Each

| Complex (M²⁺) | d-config | Oct + strong field config | CFSE | M³⁺ d-config | M³⁺ CFSE | |---|---|---|---|---|---| | $[\mathrm{Ni(phen)_3}]^{2+}$ | Ni²⁺ d⁸ | $t_{2g}^6 e_g^2$ | $-1.2\Delta_o$ | Ni³⁺ d⁷ LS: $t_{2g}^6 e_g^1$ | $-1.8\Delta_o$ (more stable!) | | $[\mathrm{Fe(phen)_3}]^{2+}$ | Fe²⁺ d⁶ LS | $t_{2g}^6 e_g^0$ | $-2.4\Delta_o$ | Fe³⁺ d⁵ LS: $t_{2g}^5 e_g^0$ | $-2.0\Delta_o$ (less!) | | $[\mathrm{Co(phen)_3}]^{2+}$ | Co²⁺ d⁷ LS | $t_{2g}^6 e_g^1$ | $-1.8\Delta_o$ | Co³⁺ d⁶ LS: $t_{2g}^6 e_g^0$ | $-2.4\Delta_o$ (more stable!) | | $[\mathrm{Zn(phen)_3}]^{2+}$ | Zn²⁺ d¹⁰ | $t_{2g}^6 e_g^4$ | $0$ | Zn³⁺ d⁹: $t_{2g}^6 e_g^3$ | $-0.6\Delta_o$ (more!) |

So $[\mathrm{Fe(phen)_3}]^{2+}$ goes from $-2.4\Delta_o$ (LS d⁶, max CFSE) to $-2.0\Delta_o$ (LS d⁵) on oxidation → loses 0.4Δ₀ of CFSE.

🗺️ Why Fe(II) → Fe(III) Loses CFSE

LS d⁶ has the maximum possible CFSE for an octahedral configuration ($t_{2g}^6 e_g^0$ = $6 \times -0.4\Delta_o = -2.4\Delta_o$). Removing one electron from $t_{2g}$ (going to LS d⁵) reduces CFSE to $5 \times -0.4\Delta_o = -2.0\Delta_o$.

⚡ Maximum CFSE Configurations

For octahedral LS:

• d⁶ LS: $-2.4\Delta_o$ (max, all 6 in $t_{2g}$)
• d⁵ LS: $-2.0\Delta_o$
• d⁴ LS: $-1.6\Delta_o$

⚠️ Phen is Strong Field

phen, like bpy and CN⁻, is strong-field. Forces LS configuration for first-row 3d M²⁺ and M³⁺ ions with ≥4 d-electrons.

$\boxed{\text{Answer: (2) } [\mathrm{Fe(phen)_3}]^{2+}}$