Step 1: {Electronic Configuration of Fe}
The electronic configuration of Fe is \( {Fe} = [{Ar}] 3d^6 4s^2 \). For Fe\(^{3+}\), three electrons are lost, yielding the configuration: \[{Fe}^{3+} = [{Ar}] 3d^5\]Step 2: {Number of Unpaired Electrons}
For Fe\(^{3+}\), the count of unpaired electrons, \( n \), is 5 (due to 5 electrons in the 3d orbitals).Step 3: {Spin Only Magnetic Moment Formula}
The formula for the spin-only magnetic moment is:\[\mu_s = \sqrt{n(n+2)} { BM}\]with \(n = 5\). Consequently:\[\mu_s = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} { BM} \approx 6 { BM}\]Therefore, the correct option is (C).