Question

The lowest degree of paramagnetism per mole of the compound at 298 K will be shown by :

• A. $FeS{O}_4.6H_{2}O$
  
• B. $NiS{O}_4.6H_{2}O$
• C. $MnS{O}_4.4H_{2}O$
• D. $CuS{O}_4.5H_{2}O$

Answer 1

The correct option is D $CuS{O}_4.5H_{2}O$

$F{e}^{2+}$, $N{i}^{2+}$, $M{n}^{2+}$ and $C{u}^{2+}$ ions have outer electronic configuration of $3d^6$, $3d^8$, $3d^5$ and $3d^9$ respectively. They contain 4, 2, 5 and 1 unpaired electron respectively.

Hence, the lowest degree of paramagnetism per mole of the compound at 298 K will be shown by, $CuS{O}_4.5H_{2}O$

Hence the correct option is D.