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(D) 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 $Cu^

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$Mn^{2+}$

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(D) 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 $Cu^{2+}$ $([Ar] 3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$2$. For $Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$3$. For $Ni^{2+}$ $([Ar] 3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.$4$. For $Mn^{2+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Since $Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,it exhibits the highest magnetic moment.

Which one of the following ions exhibits the highest magnetic moment?

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