Magnetic Moment (µ)=√n(n+2)BM
When n=1,
µ=√1(1+2)=√3=1.73 BM≈2
When n=2,
µ=√2(2+2)=√8=2.83 BM≈3
When n=3,
µ=√3(3+2)=√15=3.87BM≈4
When n≈4,
µ=√4(4+2)=√24=4.90BM=5
When n=5,
µ=√5(5+2)=√35=5.92BM≈6
K4[Mn(CN)6]: Oxidation state of Mn is +2, Mn‌2+ has an electronicconfiguration of [Ar]3d‌5.
As µ = 2.2 BM which corresponds to one unpaired electron. The orbitaldiagramfor Mn‌2+ is[Fe(H2O)6]2+:µ = 5.3 BM corresponds to 4 unpaired electrons. Theconfiguration of Fe‌2+ ions is [Ar]3d‌6and orbital diagram isK2[MnCl4],: µ = 5.9 BM, Mn2+ ion has 5 unpaired electrons. Its orbitaldiagram is