JEE Advanced 2025 Paper 2 · Q15 · Crystal Field Theory

Correct answer:7.7

Solution

$Mn^{3+}$ is $d^4$ in both complexes ($Mn$ = $[Ar] 3d^5 4s^2$; $Mn^{3+}$ = $[Ar] 3d^4$). $[Mn(Br)_6]^{3-}$: $Br^-$ is a weak-field ligand -> high-spin; $t_{2g}^3 e_g^1$ -> 4 unpaired electrons -> $\mu = \sqrt{n(n+2)} = \sqrt{4 \times 6} = \sqrt{24} = 4.90$ BM. $[Mn(CN)_6]^{3-}$: $CN^-$ is a strong-field ligand -> low-spin; $t_{2g}^4 e_g^0$ -> 2 unpaired electrons -> $\mu = \sqrt{2 \times 4} = \sqrt{8} = 2.83$ BM. Sum = $4.90 + 2.83 = 7.73 \approx 7.7$ BM.

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