The correct order of the spin-only magnetic moments of the following complexes is: (I) [Cr(H2O)6]Br2 (II) Na4[Fe(CN)6]…

🧠 Walk Each, Count Unpaired

| # | Complex | Metal | d-count | Field | Unpaired | |---|---|---|---|---|---| | (I) | $[\mathrm{Cr(H_2O)_6}]\mathrm{Br_2}$ | Cr²⁺ | d⁴ | weak (HS) | 4 | | (II) | $\mathrm{Na_4[Fe(CN)_6]}$ | Fe²⁺ | d⁶ | strong (LS) | 0 | | (III) | $\mathrm{Na_3[Fe(C_2O_4)_3]}$, $\Delta_o > P$ | Fe³⁺ | d⁵ | LS | 1 | | (IV) | $(\mathrm{Et_4N})_2[\mathrm{CoCl_4}]$ | Co²⁺ | d⁷ | Td (weak) | 3 |

Order: (I) 4 > (IV) 3 > (III) 1 > (II) 0.

🗺️ The Subtle One: III

$\mathrm{C_2O_4^{2-}}$ alone is borderline; for Fe(III), oxalate is usually weak-field → d⁵ HS, n=5. But the question explicitly states $\Delta_o > P$, forcing low-spin d⁵: $\mathrm{t_{2g}^5}$ → 1 unpaired. Always read the parenthetical condition — it overrides the default.

⚡ Td Co²⁺ Trick (IV)

In tetrahedral geometry, $\Delta_t$ is always small → high-spin by default. Co²⁺ d⁷ Td: $\mathrm{e^4 t_2^3}$ → 3 unpaired. This is the standard result; you don't need to check the ligand.

⚠️ Don't Miss the Override

If the question gives an explicit field-strength condition like "$\Delta_o > P$", trust it over your spectrochemical-series instinct.

$\boxed{\text{Answer: (3) (I) > (IV) > (III) > (II)}}$