The sum of the spin only magnetic moment values (in B.M.) of and is ____.
Answer: 7.71
Solution
For [Mn(Br)₆]³⁻:
For [Mn(CN)₆]³⁻:
Sum = 4.89 + 2.83 = 7.72 B.M.
(Acceptable range: 7.70 to 7.73 B.M.)
- Mn is in +3 oxidation state: [Ar]3d⁴
- Br⁻ is a weak field ligand → high spin complex
- Electronic configuration: t₂g³ eg¹ (4 unpaired electrons)
- μ = √[n(n+2)] = √[4(6)] = √24 = 4.89 B.M.
For [Mn(CN)₆]³⁻:
- Mn is in +3 oxidation state: [Ar]3d⁴
- CN⁻ is a strong field ligand → low spin complex
- Electronic configuration: t₂g⁴ eg⁰ (2 unpaired electrons)
- μ = √[n(n+2)] = √[2(4)] = √8 = 2.83 B.M.
Sum = 4.89 + 2.83 = 7.72 B.M.
(Acceptable range: 7.70 to 7.73 B.M.)
JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer