Question

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

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

Answer 1

Answer: (C)

$CuS{O}_4.5H_{2}O$

Option (C) is correct. The lowest degree of paramagnetism per mole of the compound at 298 K will be shown by $CuS{O}_4.5H_{2}O$

In the presence of water, which is a weak field ligand, the configurations of metal ions in hydrated compounds reflects those in isolated gaseous ions i.e., no pairing of electrons is possible as the interaction with water molecules is weak.

All the metal ions in above compounds are divalent and their outer shell electronic configurations are shown above.

The paramagnetic nature of a compound is proportional to the number of unpaired electrons in it. $M{n}^{2+}$ ion has more number of unpaired electrons. Hence $MnS{O}_4.4H_{2}O$ shows greater paramagnetic nature. In contrast, there is only one unpaired electron in $C{u}^{2+}$ and hence $CuS{O}_4.5H_{2}O$ shows lowest degree of paramagnetism.