The correct order of spin only magnetic moments for the following complex ions is:

🧠 Compute Unpaired for Each, Then Order

| Complex | M | d-count | Field | Unpaired | |---|---|---|---|---| | $[\mathrm{Fe(CN)_6}]^{3-}$ | Fe(III) | d⁵ | strong CN⁻ → low-spin | 1 | | $[\mathrm{Mn(CN)_6}]^{3-}$ | Mn(III) | d⁴ | strong CN⁻ → low-spin | 2 | | $[\mathrm{CoF_6}]^{3-}$ | Co(III) | d⁶ | weak F⁻ → high-spin | 4 | | $[\mathrm{MnBr_4}]^{2-}$ | Mn(II) | d⁵ | tetrahedral, high-spin | 5 |

Order: $[\mathrm{Fe(CN)_6}]^{3-} \,(1) < [\mathrm{Mn(CN)_6}]^{3-}\,(2) < [\mathrm{CoF_6}]^{3-}\,(4) < [\mathrm{MnBr_4}]^{2-}\,(5)$.

🗺️ Why Each Spin State

• CN⁻ at strong-field end: forces low-spin for d⁴, d⁵, d⁶ on 3d metals.
• F⁻ at weak end: high-spin for the same d-counts.
• Tetrahedral complexes are always high-spin (smaller $\Delta_t \approx 4/9 \Delta_o$ < pairing energy).

⚡ The "More Unpaired = Higher μ" Direct Mapping

For ranking μ, just rank unpaired counts. $\mu = \sqrt{n(n+2)}$ is monotonic in $n$.

⚠️ Tetrahedral = Always High-Spin

The tetrahedral splitting ($\Delta_t$) is too small to ever exceed pairing energy. So tetrahedral complexes are universally high-spin — even with strong-field ligands.

$\boxed{\text{Answer: (3) } [\mathrm{Fe(CN)_6}]^{3-} < [\mathrm{Mn(CN)_6}]^{3-} < [\mathrm{CoF_6}]^{3-} < [\mathrm{MnBr_4}]^{2-}}$