Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?

For an ionic solid to dissolve in water, we compare two energy terms. First, lattice enthalpy $$\Delta H_{latt}$$ is the energy released when gaseous ions come together to form one mole of the solid. Second, hydration enthalpy $$\Delta H_{hyd}$$ is the energy released when the gaseous ions become surrounded by water molecules. A salt will be appreciably soluble, or at least its ions will prefer to separate, when the magnitude of $$\Delta H_{hyd}$$ exceeds that of $$\Delta H_{latt}$$, that is when $$|\Delta H_{hyd}| \gt |\Delta H_{latt}|$$.

Both of these enthalpies depend strongly on ionic size. The general relationships are:

$$|\Delta H_{latt}| \propto \dfrac{z^{+}z^{-}}{r^{+}+r^{-}}$$ where $$r^{+}$$ and $$r^{-}$$ are the ionic radii of the cation and anion, and $$z^{+},z^{-}$$ are their charges.

$$|\Delta H_{hyd}| \propto \dfrac{z^{2}}{r}$$ for each individual ion, so the smaller the ion, the larger the hydration enthalpy.

Inside the alkaline-earth family $$Be^{2+},\, Mg^{2+},\, Ca^{2+},\, Sr^{2+},\, Ba^{2+}$$ the ionic radius increases steadily as we go down the group:

$$r(Be^{2+}) \lt r(Mg^{2+}) \lt r(Ca^{2+}) \lt r(Sr^{2+}) \lt r(Ba^{2+}).$$

Because the radius of $$Be^{2+}$$ is the smallest, the hydration enthalpy $$|\Delta H_{hyd}|$$ drops very rapidly from $$Be^{2+}$$ to $$Ba^{2+}$$. On the other hand, lattice enthalpy also decreases down the group, but this decrease is slower, since the radius of the large anion $$SO_{4}^{2-}$$ dominates the denominator $$r^{+}+r^{-}$$ in the lattice enthalpy expression. Symbolically we may write

$$|\Delta H_{hyd}(Be^{2+})| \gg |\Delta H_{hyd}(Ba^{2+})|$$ while $$|\Delta H_{latt}(BeSO_{4})| \approx |\Delta H_{latt}(BaSO_{4})|$$ but with only a moderate downward trend.

Therefore, for the smallest cation $$Be^{2+}$$, the drop in hydration enthalpy is not matched by an equal drop in lattice enthalpy. The inequality

$$|\Delta H_{hyd}(Be^{2+})| \gt |\Delta H_{latt}(BeSO_{4})|$$

remains valid, whereas for the larger cations $$Ca^{2+}, Sr^{2+}, Ba^{2+}$$ we reach

$$|\Delta H_{hyd}| \lt |\Delta H_{latt}|,$$

making their sulphates sparingly soluble and ensuring that their lattice enthalpy is the greater quantity.

Consequently, among the given choices, only $$BeSO_{4}$$ possesses a hydration enthalpy greater than its lattice enthalpy.

Hence, the correct answer is Option C.