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(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $M

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$Mn^{2+}$

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(A) The spin-only magnetic moment 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 $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.$2$. For $Ti^{3+}$ $(Z=22)$: Electronic configuration is $[Ar] 3d^1$. Number of unpaired electrons $(n)$ = $1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$3$. For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$4$. For $Ni^{2+}$ $(Z=28)$: Electronic configuration is $[Ar] 3d^8$. Number of unpaired electrons $(n)$ = $2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.Comparing the values,$Mn^{2+}$ has the highest number of unpaired electrons and thus the highest magnetic moment.

Which of the following elements in their respective oxidation states develops the highest spin-only magnetic moment?

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