Which one of the following complexes will have 0 = 0 and = 5.96 B.M.? (a) [Fe(CN)6]4- (b) [Co(NH3)6]3+ (c) [FeF6]4- (d)…

🧠 μ = 5.96 BM → 5 Unpaired

$\mu = \sqrt{n(n+2)} = 5.96 \Rightarrow n(n+2) = 35.5 \Rightarrow n = 5$.

So we need a complex with 5 unpaired electrons AND $\Delta_o$ (CFSE) = 0.

🗺️ Evaluate Each

| Complex | Metal/d-config | Spin | Configuration | Unpaired | CFSE | |---|---|---|---|---|---| | $[\mathrm{Fe(CN)_6}]^{4-}$ | Fe(II) d⁶ LS | $t_{2g}^6 e_g^0$ | 0 | $-2.4\Delta_o$ | | $[\mathrm{Co(NH_3)_6}]^{3+}$ | Co(III) d⁶ LS | $t_{2g}^6 e_g^0$ | 0 | $-2.4\Delta_o$ | | $[\mathrm{FeF_6}]^{4-}$ | Fe(II) d⁶ HS | $t_{2g}^4 e_g^2$ | 4 | $-0.4\Delta_o$ | | $[\mathrm{Mn(SCN)_6}]^{4-}$ | Mn(II) d⁵ HS | $t_{2g}^3 e_g^2$ | 5 | 0 |

For Mn(II) HS d⁵: $\mathrm{CFSE} = 3 \times (-0.4\Delta_o) + 2 \times (+0.6\Delta_o) = -1.2\Delta_o + 1.2\Delta_o = 0$.

🗺️ Both Conditions Satisfied

$[\mathrm{Mn(SCN)_6}]^{4-}$: 5 unpaired (μ ≈ 5.92 BM) AND CFSE = 0. ✓

⚡ Mn(II) HS d⁵ is Special

It's the only common configuration with zero CFSE despite a non-zero $\Delta_o$. Hence Mn(II) HS complexes are commonly described as having "Δ-independent" stability — they exist due to electrostatic and not crystal-field effects.

⚠️ CFSE = 0 ≠ Δₒ = 0

The question phrases "$\Delta_o = 0$" loosely; what's meant is CFSE = 0. The crystal field still splits orbitals, but the half-filled d⁵ HS configuration sits exactly at the unsplit average energy.

$\boxed{\text{Answer: (4) } [\mathrm{Mn(SCN)_6}]^{4-}}$