Arrange the following coordination compounds in the increasing order of magnetic moments (Atomic numbers: Mn=25,…

🧠 Walk Each, Tally Unpaired

| Complex | Metal | d-count | Field | Unpaired | |---|---|---|---|---| | A. $[\mathrm{FeF_6}]^{3-}$ | Fe(III) | d⁵ | weak F⁻ → high-spin | 5 | | B. $[\mathrm{Fe(CN)_6}]^{3-}$ | Fe(III) | d⁵ | strong CN⁻ → low-spin | 1 | | C. $[\mathrm{MnCl_6}]^{3-}$ (HS) | Mn(III) | d⁴ | weak Cl⁻ → high-spin | 4 | | D. $[\mathrm{Mn(CN)_6}]^{3-}$ | Mn(III) | d⁴ | strong CN⁻ → low-spin | 2 |

Order of μ (= order of unpaired count): $B (1) < D (2) < C (4) < A (5)$ → option (2).

🗺️ The d⁴ vs d⁵ Spin-Switching Behaviour

Both d⁴ and d⁵ can split between high-spin and low-spin depending on ligand:

• d⁴ HS: 4 unpaired ($\mathrm{t_{2g}^3 e_g^1}$); LS: 2 unpaired ($\mathrm{t_{2g}^4}$).
• d⁵ HS: 5 unpaired; LS: 1 unpaired ($\mathrm{t_{2g}^5}$).

Spread of 3–4 unpaired between HS and LS makes these the most ligand-sensitive d-counts.

⚡ CN⁻ vs F⁻: Always Opposite Spin States

CN⁻ is at the strong-field end; F⁻ is at the weak-field end. For the same metal in the same OS, you can drop unpaired count from 5 to 1 (Fe(III)) or 4 to 2 (Mn(III)) just by swapping ligand.

⚠️ Don't Forget d⁴ Has Two Spin States Too

A common slip: students remember CN⁻ vs F⁻ for Fe(III) (d⁵) but forget the same logic applies to Mn(III) (d⁴). Always check d-count and ligand together.

$\boxed{\text{Answer: (2) B < D < C < A}}$