JEE Advance - Chemistry (2025 - Paper 2 Online - No. 15)
The sum of the spin only magnetic moment values (in B.M.) of $\left[\mathrm{Mn}(\mathrm{Br})_6\right]^{3-}$ and $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$ is _________.
Answer
7.73
Explanation
First, note that in both complexes Mn is in the +3 state (Br– and CN– are monovalent ligands), so Mn3+ is a d4 ion.
[Mn(Br)₆]³⁻
Br⁻ is a weak‐field ligand ⇒ high‐spin d⁴ ⇒ 4 unpaired electrons
Spin‐only moment:
$$\mu_1=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}\approx4.90\;\mathrm{B.M.}$$
[Mn(CN)₆]³⁻
CN⁻ is a strong‐field ligand ⇒ low‐spin d⁴ ⇒ 2 unpaired electrons
Spin‐only moment:
$$\mu_2=\sqrt{2(2+2)}=\sqrt{8}\approx2.83\;\mathrm{B.M.}$$
Sum of the two moments:
$$4.90+2.83\approx7.73\;\mathrm{B.M.}$$
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