Spin-only Magnetic Moment
Spin-only magnetic moment (μ) for high-spin d5 (Mn2+) is approximately (μ = √(n(n+2)) BM):
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About This Question
- Subject
- chemistry
- Chapter
- d- and f-block elements
- Topic
- spin-only magnetic moment
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5.92 BM
- Configuration: Mn2+ [Ar]3d5.
- Unpaired Electrons (n): In a high-spin octahedral or tetrahedral field, d5 has 5 unpaired electrons.
- Apply Formula: μ=n(n+2) Bohr Magnetons (BM).
- Calculation:
- μ=5(5+2)=35.
- 25=5 and 36=6.
- 35≈5.92 BM.
- Note: As a shortcut, the magnetic moment starting digit is usually equal to n (e.g., 5 unpaired → 5.xx BM).
This medium difficulty chemistry question is from the chapter d- and f-block elements, covering the topic of spin-only magnetic moment. It appeared in the 2025 exam.
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