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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)
$A$. $Cd$	$I$. $f$-block
$B$. $Eu$	$II$. $s$-block
$C$. $Se$	$III$. $d$-block
$D$. $Ba$	$IV$. $p$-block

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

Similar Questions

The general electronic configuration of transition elements is

Match the following:
List-$I$ (Element) List-$II$ (Block)
$A$. $Cd$ $I$. $f$-block
$B$. $Eu$ $II$. $s$-block
$C$. $Se$ $III$. $d$-block
$D$. $Ba$ $IV$. $p$-block

The correct answer is:

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

The common oxidation state of scandium $(Sc)$,a transition element,is:

Which of the following is a soft metal?

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