Which one of the following ions exhibits the | vedclass

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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. $1$. $Cu^{2+

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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. $1$. $Cu^{2+}$ $([Ar]3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \ BM$ $2$. $Ti^{3+}$ $([Ar]3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \ BM$ $3$. $Ni^{2+}$ $([Ar]3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.82 \ BM$ $4$. $Mn^{2+}$ $([Ar]3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \ BM$ 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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