JEE Advanced 2025 · Paper 2, Chemistry question 15

Both complexes contain Mn(III), a $d^4$ ion.

• $\ce{[MnBr6]^3-}$: $\ce{Br-}$ is a weak-field ligand, so the complex is high spin with 4 unpaired electrons. $\mu=\sqrt{24}=4.90$ BM.
• $\ce{[Mn(CN)6]^3-}$: $\ce{CN-}$ is a strong-field ligand, so the complex is low spin ($t_{2g}^4$) with 2 unpaired electrons. $\mu=\sqrt8=2.83$ BM.

Sum: $4.90+2.83\approx7.73$ BM.