Question:
Calculated magnetic moment value for \(\mathrm{Fe}^{2+}\) ion in BM is:
Calculated magnetic moment value for \(\mathrm{Fe}^{2+}\) ion in BM is:
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For spin-only magnetic moment, use:
\[
\mu=\sqrt{n(n+2)}
\]
where \(n\) is the number of unpaired electrons.
Updated On: Jun 24, 2026
- \(3.87\)
- \(4.90\)
- \(2.84\)
- \(1.73\)
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The Correct Option is B
Solution and Explanation
Step 1: Find the electronic configuration of \(\mathrm{Fe}^{2+}\).
Atomic number of iron is \[ 26 \] Electronic configuration of neutral iron is \[ \mathrm{Fe}: [Ar]\,3d^6\,4s^2 \] For \(\mathrm{Fe}^{2+}\), two electrons are removed from \(4s\) orbital.
Therefore, \[ \mathrm{Fe}^{2+}: [Ar]\,3d^6 \]
Step 2: Find the number of unpaired electrons.
For \(3d^6\) configuration in high spin state, the number of unpaired electrons is \[ n=4 \]
Step 3: Use the spin-only magnetic moment formula.
The magnetic moment is \[ \mu=\sqrt{n(n+2)} \] Substitute \[ n=4 \] \[ \mu=\sqrt{4(4+2)} \] \[ \mu=\sqrt{24} \] \[ \mu=4.90\,\text{BM} \]
Step 4: Final conclusion.
Hence, the calculated magnetic moment of \(\mathrm{Fe}^{2+}\) is \[ \boxed{4.90\,\text{BM}} \]
Atomic number of iron is \[ 26 \] Electronic configuration of neutral iron is \[ \mathrm{Fe}: [Ar]\,3d^6\,4s^2 \] For \(\mathrm{Fe}^{2+}\), two electrons are removed from \(4s\) orbital.
Therefore, \[ \mathrm{Fe}^{2+}: [Ar]\,3d^6 \]
Step 2: Find the number of unpaired electrons.
For \(3d^6\) configuration in high spin state, the number of unpaired electrons is \[ n=4 \]
Step 3: Use the spin-only magnetic moment formula.
The magnetic moment is \[ \mu=\sqrt{n(n+2)} \] Substitute \[ n=4 \] \[ \mu=\sqrt{4(4+2)} \] \[ \mu=\sqrt{24} \] \[ \mu=4.90\,\text{BM} \]
Step 4: Final conclusion.
Hence, the calculated magnetic moment of \(\mathrm{Fe}^{2+}\) is \[ \boxed{4.90\,\text{BM}} \]
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