Magnetic Properties
The spin-only magnetic moment of a d5 high-spin ion is closest to:
Select the correct option:
Solution
5.92 BM
For spin-only magnetic moment, mu = sqrt(n(n+2)) BM, where n is number of unpaired electrons. For a high-spin d5 ion, n = 5. Hence mu = sqrt(5 x 7) = sqrt(35) = 5.92 BM (approximately). This value is consistent with strong paramagnetism expected for five unpaired electrons.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More magnetic properties Practice Questions
The magnetic moment (spin-only) of [Fe(CN)6]4− is approximately:
The magnetic moment (spin-only) of [Fe(CN)6]4− is approximately:
Paramagnetism in transition metal ions is due to:
Paramagnetism in transition metal ions is due to:
[Fe(CN)6]4− magnetic character is best described as:
[Fe(CN)6]4− magnetic character is best described as:
The magnetic moment of [Fe(CN)₆]⁴⁻ ion is:
The magnetic moment of [Fe(CN)₆]⁴⁻ ion is:
About This Question
- Subject
- chemistry
- Chapter
- d- and f-block elements
- Topic
- magnetic properties
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5.92 BM
For spin-only magnetic moment, mu = sqrt(n(n+2)) BM, where n is number of unpaired electrons. For a high-spin d5 ion, n = 5. Hence mu = sqrt(5 x 7) = sqrt(35) = 5.92 BM (approximately). This value is consistent with strong paramagnetism expected for five unpaired electrons.
This medium difficulty chemistry question is from the chapter d- and f-block elements, covering the topic of magnetic properties. It appeared in the 2025 exam.
Looking for more practice? Explore all chemistry questions or browse d- and f-block elements questions on RankGuru.