The spin only magnetic moment of Fe\(^{3+}\) ion (in BM) is approximately.
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The spin-only magnetic moment formula applies when there is no contribution from orbital motion of electrons, which is true for transition metal ions like Fe\(^{3+}\).
Step 1: {Electronic Configuration of Fe}
The electronic configuration of Fe is \( {Fe} = [{Ar}] 3d^6 4s^2 \). For Fe\(^{3+}\), it loses three electrons, resulting in the configuration:
\[
{Fe}^{3+} = [{Ar}] 3d^5
\]
Step 2: {Number of Unpaired Electrons}
For Fe\(^{3+}\), the number of unpaired electrons, \( n \), is 5 (since it has 5 electrons in the 3d orbitals).
Step 3: {Spin Only Magnetic Moment Formula}
The formula for calculating the spin-only magnetic moment is:
\[
\mu_s = \sqrt{n(n+2)} { BM}
\]
where \(n = 5\). Therefore:
\[
\mu_s = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} { BM} \approx 6 { BM}
\]
Thus, the correct answer is (C).