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Among the following the lowest degree of paramagnetism per mole of the compound at 298 $\mathrm{K}$ will be shown by
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Verified Answer
The correct answer is:
$\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}$
$\begin{array}{lllll}\text { Ion } & \mathrm{Mn}^{2+} & \mathrm{Cu}^{2+} & \mathrm{Fe}^{2+} & \mathrm{Ni}^{2+} \\ \text { EC } & 3 d^{5} & 3 d^{9} & 3 d^{6} & 3 d^{8} \\ \text { Number of } & 5 & 1 & 4 & 2\end{array}$
unpaired electron
Hence lowest paramagnetism is shown by $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$
unpaired electron
Hence lowest paramagnetism is shown by $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$
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