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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 $Mn^

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$Mn^{2+} > Cr^{2+} > V^{2+}$

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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 $Mn^{2+}$ $(Z=25)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.$2$. For $Cr^{2+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.$3$. For $V^{2+}$ $(Z=23)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.Therefore,the order of magnetic moment is $Mn^{2+} > Cr^{2+} > V^{2+}$.

What is the correct order of the theoretical magnetic moment for $Mn^{2+}$,$Cr^{2+}$,and $V^{2+}$?

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