Q28 / 125 · BITSAT 2012 · Chemistry

Among the following the lowest degree of paramagnetism per mole of the compound at 298 $\mathrm{K}$ will be shown by

  • A
    $\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}$
  • Correct answer
    B
    $\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}$
  • C
    $\mathrm{FeSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$
  • D
    $\mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$

Solution

$\begin{array}{lllll}\text { Ion } & \mathrm{Mn}^{2+} & \mathrm{Cu}^{2+} & \mathrm{Fe}^{2+} & \mathrm{Ni}^{2+} \\ \text { EC } & 3 d^{5} & 3 d^{9} & 3 d^{6} & 3 d^{8} \\ \text { Number of } & 5 & 1 & 4 & 2\end{array}$ unpaired electron Hence lowest paramagnetism is shown by $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$

Hence, the correct answer is option: 'B'.

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