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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.

List-$I$ (Element)	List-$II$ (Block)
$(A)$ $Ra$	$(I)$ $p$-block
$(B)$ $Uuq$	$(II)$ $s$-block
$(C)$ $Ds$	$(III)$ $f$-block
$(D)$ $Fm$	$(IV)$ $d$-block

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

Similar Questions

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Reason : Transition metals have a large energy difference between the $ns^2$ and $(n-1)d$ electrons.

In the $3d$ transition series,as the atomic number increases,the paramagnetic character:

The period number and group number of Tantalum $(Z = 73)$ are respectively:

Which one of the following ions will be colourless in an aqueous solution? (Atomic numbers: $Sc = 21, Fe = 26, Ti = 22, Mn = 25$)

Match the following:
List-$I$ (Element) List-$II$ (Block)
$(A)$ $Ra$ $(I)$ $p$-block
$(B)$ $Uuq$ $(II)$ $s$-block
$(C)$ $Ds$ $(III)$ $f$-block
$(D)$ $Fm$ $(IV)$ $d$-block

The correct answer is

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