JEE Advanced 2025 — Paper 2 — CHEM-III solution

JEE Advanced2025Paper 2CHEM-III

The sum of the spin only magnetic moment values (in B.M.) of [Mn(Br) $_{6}$ ] $^{3 -}$ and [Mn(CN) $_{6}$ ] $^{3 -}$ is ______.

Your answer

Solution

Mn in both complexes is in oxidation state +3, electronic configuration $d^{4}$ .

[Mn(Br) $_{6}$ ] $^{3 -}$ : Br $^{-}$ is a weak-field ligand, so the complex is high-spin. The four $d$ electrons occupy $t_{2 g}^{3} e_{g}^{1}$ , giving 4 unpaired electrons.

$μ = n (n + 2) = 4 \times 6 = 24 \approx 4.90$ B.M.

[Mn(CN) $_{6}$ ] $^{3 -}$ : CN $^{-}$ is a strong-field ligand, so the complex is low-spin. The four $d$ electrons occupy $t_{2 g}^{4}$ (with one pair), giving 2 unpaired electrons.

$μ = 2 \times 4 = 8 \approx 2.83$ B.M.

Sum $= 4.90 + 2.83 = 7.73$ B.M.

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