Which of the following arrangements is correct in respect of solubility in water?
- $\mathrm{CaSO}_{4}>\mathrm{BaSO}_{4}>\mathrm{BeSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{SrSO}_{4}$
- $\mathrm{BeSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{CaSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4}$
- $\mathrm{BaSO}_{4}>\mathrm{Sr} \mathrm{SO}_{4}>\mathrm{CaSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{BeSO}_{4}$
- $\mathrm{BeSO}_{4}>\mathrm{CaSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4}$
Solution
$$
\mathrm{BeSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{CaSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4}
$$
Solublity of 2nd group sulphates decreases as we down the group due to less release of hydration energy. $\mathrm{Be}^{2+} < \mathrm{Mg}^{2+} < \mathrm{Ca}^{2+} < \mathrm{Sr}^{2+} < \mathrm{Ba}^{24}$ (lonic Size)
As hydration energy decreases more rapidly than latice energy, the solubility decreases down the group. Hydration Energy $=\frac{\text { Charge }}{\text { Size }}$
While lathice energy almost remains constant. Hence, solubility decreases.
Asked in: JEE-TOPICTESTS-CHEMISTRY