The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal…

🧠 Two Clues, One Identity

Two pieces of data — CFSE and magnetic moment — and we need a single ion that satisfies both. Strategy:

• $\mu = 3.87$ BM → $n(n+2) = 15 \Rightarrow n = 3$ unpaired electrons.
• CFSE $= -0.8\Delta_o$ for an octahedral aqua complex.

Both must point to the same ion.

🗺️ Match Each Option

| Ion | d-count | High-spin filling (octahedral aqua) | Unpaired | CFSE (units of $\Delta_o$) | |---|---|---|---|---| | $\mathrm{V^{3+}}$ | $\mathrm{d^2}$ | $\mathrm{t_{2g}^2}$ | 2 (μ = 2.83) | $-2(0.4) = -0.8$ | | $\mathrm{Co^{2+}}$ | $\mathrm{d^7}$ | $\mathrm{t_{2g}^5\,e_g^2}$ | 3 (μ = 3.87) | $-5(0.4) + 2(0.6) = -2.0 + 1.2 = -0.8$ | | $\mathrm{Cr^{3+}}$ | $\mathrm{d^3}$ | $\mathrm{t_{2g}^3}$ | 3 (μ = 3.87) | $-3(0.4) = -1.2$ | | $\mathrm{Mn^{4+}}$ | $\mathrm{d^3}$ | $\mathrm{t_{2g}^3}$ | 3 (μ = 3.87) | $-3(0.4) = -1.2$ |

Only $\mathrm{Co^{2+}}$ satisfies both μ = 3.87 BM and CFSE = $-0.8\Delta_o$.

⚡ The CFSE Quick-Compute

For octahedral high-spin: each $\mathrm{t_{2g}}$ electron contributes $-0.4\Delta_o$, each $\mathrm{e_g}$ electron contributes $+0.6\Delta_o$. So CFSE $= -0.4\,n_{t_{2g}} + 0.6\,n_{e_g}$ (in units of $\Delta_o$). Memorise this single line; CFSE questions become trivial.

⚠️ The "μ Alone" Trap

V³⁺, Cr³⁺, Mn⁴⁺ all can give μ = 3.87 BM individually, but only one of them matches CFSE = −0.8Δo too. Always cross-verify both data points — never identify a metal from a single property when multiple ions match.

$\boxed{\text{Answer: (2) } \mathrm{Co^{2+}}}$