Amongst the following,the lowest degree of pa | vedclass

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(B) To determine the degree of paramagnetism,we calculate the number of unpaired electrons $(n)$ in the metal ion of each compound:$1$. In $MnSO_4 \cd

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$CuSO_4 \cdot 5H_2O$

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(B) To determine the degree of paramagnetism,we calculate the number of unpaired electrons $(n)$ in the metal ion of each compound:$1$. In $MnSO_4 \cdot 4H_2O$,the metal ion is $Mn^{2+}$ $(3d^5)$,which has $n = 5$ unpaired electrons.$2$. In $CuSO_4 \cdot 5H_2O$,the metal ion is $Cu^{2+}$ $(3d^9)$,which has $n = 1$ unpaired electron.$3$. In $FeSO_4 \cdot 6H_2O$ and $FeSO_4 \cdot 5H_2O$,the metal ion is $Fe^{2+}$ $(3d^6)$,which has $n = 4$ unpaired electrons.Since the degree of paramagnetism is directly proportional to the number of unpaired electrons,$CuSO_4 \cdot 5H_2O$ shows the lowest degree of paramagnetism.

Amongst the following,the lowest degree of paramagnetism per mole of the compound at $298 \ K$ will be shown by:

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