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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?

Similar Questions

Predict which of the following will be coloured in aqueous solution: $Ti^{3+}, V^{3+}, Cu^{+}, Sc^{3+}, Mn^{2+}, Fe^{3+},$ and $Co^{2+}$. Give reasons for each.

Which of the following magnetic moment values corresponds to the highest ionization energy for $Mn$?

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