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(A) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n 2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. In $MnSO_4$,the

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$MnSO_4$

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(A) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n 2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. In $MnSO_4$,the ion is $Mn^{2 }$. The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. Thus,$Mn^{2 }$ is $[Ar] 3d^5$. It has $n=5$ unpaired electrons.$2$. In $Cr_2(SO_4)_3$,the ion is $Cr^{3 }$. The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$. Thus,$Cr^{3 }$ is $[Ar] 3d^3$. It has $n=3$ unpaired electrons.$3$. In $FeSO_4$,the ion is $Fe^{2 }$. The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. Thus,$Fe^{2 }$ is $[Ar] 3d^6$. It has $n=4$ unpaired electrons.$4$. In $CuSO_4$,the ion is $Cu^{2 }$. The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$. Thus,$Cu^{2 }$ is $[Ar] 3d^9$. It has $n=1$ unpaired electron.Since $Mn^{2 }$ has the highest number of unpaired electrons $(n=5)$,it will have the highest magnetic moment.

Which of the following substances will have the highest magnetic moment for its transition metal ion?

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