Which element from the following in +2 oxidat | vedclass

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(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Fe^{2

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$Mn$

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(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Fe^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ BM$.$2$. For $Cr^{2+}$ $(3d^4)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ BM$.$3$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ BM$.$4$. For $Ni^{2+}$ $(3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.Comparing the values,$Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,and therefore exhibits the highest magnetic moment.

Which element from the following in $+2$ oxidation state exhibits the highest magnetic moment?

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