JEE Main · 2012 · 29 Apr 2012 · Offline · d and f Block Elements

Comparing the two oxidation states, the lower one is more basic and more ionic, so $\mathrm{FeO}$ is more basic than $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$, and ferrous compounds are more ionic and, having higher lattice energy, less volatile than ferric ones — all true. Hydrolysis, however, is driven by charge density, and the smaller, more highly charged $\mathrm{F}{\mathrm{e}}^{3+}$ polarises water far more strongly than $\mathrm{F}{\mathrm{e}}^{2+}$, so ferric compounds hydrolyse more readily. The claim that ferrous compounds are more easily hydrolysed than ferric is therefore the incorrect statement.