Quantum Numbers
Spin-only magnetic moment (μ) in Bohr magnetons for a high-spin d5 ion (e.g., Mn2+) is approximately:
Select the correct option:
Solution
5.92 BM
- Formula: μspin=n(n+2) Bohr Magnetons (BM), where n is the number of unpaired electrons.
- Configuration: A high-spin d5 ion has all five d orbitals singly occupied (dxy1,dyz1,dzx1,dx2−y21,dz21).
- Unpaired Electrons: n=5.
- Calculation:
- μ=5(5+2)=35
- μ≈5.916 BM.
- Conclusion: Rounded to two decimal places, the value is 5.92 BM.
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- quantum numbers
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5.92 BM
- Formula: μspin=n(n+2) Bohr Magnetons (BM), where n is the number of unpaired electrons.
- Configuration: A high-spin d5 ion has all five d orbitals singly occupied (dxy1,dyz1,dzx1,dx2−y21,dz21).
- Unpaired Electrons: n=5.
- Calculation:
- μ=5(5+2)=35
- μ≈5.916 BM.
- Conclusion: Rounded to two decimal places, the value is 5.92 BM.
This medium difficulty chemistry question is from the chapter atomic structure, covering the topic of quantum numbers. It appeared in the 2025 exam.
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