Statement - I : BeSO₄ and MgSO₄ are readily soluble in water

Statement - II : The greater hydration enthalpy of Be²⁺ and Mg²⁺ ions overcome the lattice enthalpy of BeSO₄ and MgSO₄

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Both statement - I and II are incorrect

Statement I and II are correct and II does not explain I

Statement I and II are correct and II explains I

Statement - I is correct and Statement - II is incorrect

answer is A.

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Detailed Solution

\large Solubility\;:\;\xrightarrow[{As\;hydration\;energy\;decreases\;solubility\;decreases}]{{BeS{O_4}\; > \;MgS{O_4}\; > \;CaS{O_4}\; > \;SrS{O_4}\; > \;BaS{O_4}}}

→ If anion size is large $[{SO}_{4}^{- 2}, {CO}_{3}^{- 2}]$ then with small change in size of cation there is no considerable change in lattice energy, in such cases the solubility depends upon the hydration energy. As the size of metal ion increases hydration energy decreases, Therefore solubility decreases and hence BeSO₄ , MgSO₄ & BeCO₃ ; MgCO₃ are highly soluble, where as BaSO₄, BaCO₃ almost all insoluble
Hydration energy $α$ Solubility
TRICK - I :-
In general in all IA group compounds except fluorides & hydroxides solubility down the group decreases.
But the two exceptions are carbonates & bicarbonates of IA group where solubility increases down the group
TRICK - II :-
More the difference between the sizes of cation and anion then more will be the solubility