Magnetic Moment $(\mu)=\sqrt{n(n+2)} \text{BM}$ When $n=1,$ $\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73 \text{BM}\approx 2$ When $n=2,$ $\mu=\sqrt{2(2+2)}=\sqrt{8}=2.83 \text{BM} \approx 3$ When $n=3,$ $\mu=\sqrt{3(3+2)}=\sqrt{15}=3.87 \text{BM} \approx 4$ When $n\approx 4,$ $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\text{BM}=5$ When $n=5,$ $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92 \text{BM}\approx 6$ $\mathrm{K}_4[\mathrm{Mn}(\mathrm{CN})_6] :$ Oxidation state of Mn is +2, Mn$^{2+}$ has an electronicconfiguration of [Ar]3d$^{5}$. As $\mu$ = 2.2 BM which corresponds to one unpaired electron. The orbitaldiagramfor Mn$^{2+}$ is$[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+} :$$\mu$ = 5.3 BM corresponds to 4 unpaired electrons. Theconfiguration of Fe$^{2+}$ ions is [Ar]3d$^{6}$and orbital diagram is$\mathrm{K}_2[\mathrm{MnCl}_4], :$ $\mu$ = 5.9 BM, Mn2+ ion has 5 unpaired electrons. Its orbitaldiagram is