The magnetic moment (\( \mu \)) of a transition metal ion is calculated using the formula:\[\mu = \sqrt{n(n+2)} \, \text{B.M.}\]Here, \( n \) represents the number of unpaired electrons.For the Mn\(^{2+}\) ion, which has an electronic configuration of \( [Ar] 3d^5 \), there are 5 unpaired electrons. Plugging \( n = 5 \) into the magnetic moment formula yields:\[\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.}\]Consequently, Mn\(^{2+}\) exhibits a magnetic moment of approximately 5.92 B.M.
