The increasing order of magnetism of (I) MnSO | vedclass

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(B) The central metal ions are $Mn^{2+}$,$Fe^{2+}$,$Ni^{2+}$,and $Cu^{2+}$.Electronic configurations are $3d^5$,$3d^6$,$3d^8$,and $3d^9$ respectively.

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$IV < III < II < I$

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(B) The central metal ions are $Mn^{2+}$,$Fe^{2+}$,$Ni^{2+}$,and $Cu^{2+}$.Electronic configurations are $3d^5$,$3d^6$,$3d^8$,and $3d^9$ respectively.The number of unpaired electrons $(n)$ for each is: $n=5$ for $Mn^{2+}$,$n=4$ for $Fe^{2+}$,$n=2$ for $Ni^{2+}$,and $n=1$ for $Cu^{2+}$.Magnetic moment $\mu$ is directly proportional to the number of unpaired electrons $(n)$.Therefore,the decreasing order of magnetism is $I > II > III > IV$.The increasing order of magnetism is $IV < III < II < I$.

The increasing order of magnetism of $(I)$ $MnSO_4 \cdot 4H_2O$,$(II)$ $FeSO_4 \cdot 7H_2O$,$(III)$ $NiSO_4 \cdot 6H_2O$,and $(IV)$ $CuSO_4 \cdot 5H_2O$ is:

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