Question:
Calculated magnetic moment value for \( \text{Fe}^{2+} \) ion in Bohr Magnetons (BM) is:
Calculated magnetic moment value for \( \text{Fe}^{2+} \) ion in Bohr Magnetons (BM) is:
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Use \( \mu = \sqrt{n(n+2)} \) BM, where \( n \) = number of unpaired electrons.
Updated On: Mar 18, 2026
- 3.87
- 4.90
- 2.84
- 1.73
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The Correct Option is B
Solution and Explanation
Electronic configuration of \( \text{Fe}^{2+} \) is:
\[
[\text{Ar}] \, 3d^6
\]
Number of unpaired electrons \( n = 4 \)
Magnetic moment \( \mu = \sqrt{n(n+2)} \)
\[
\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM}
\]
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