Animated Solution for Chemistry - s and p-Block Elements: Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?

Hydration vs. Lattice Enthalpy

Imagine you are a tiny Beryllium ion, ${\text{Be}}^{2+}$, dropped into a vast ocean of water molecules. What decides whether you will break free from your crystal lattice and dissolve, or remain stubbornly locked in a solid state? The answer lies in an epic tug-of-war between two powerful thermodynamic forces: Lattice Enthalpy and Hydration Enthalpy.

Meet the Contenders

In this problem, we are looking at the sulphates of alkaline earth metals: ${\text{BeSO}}_4$, ${\text{CaSO}}_4$, ${\text{SrSO}}_4$, and ${\text{BaSO}}_4$.

As we move down Group 2 of the periodic table, the size of the metal cation increases significantly: ${\text{Be}}^{2+}<{\text{Ca}}^{2+}<{\text{Sr}}^{2+}<{\text{Ba}}^{2+}$. On the other hand, the sulphate anion, ${\text{SO}}_4^{2-}$, is a massive, bulky polyatomic ion that remains constant across all these compounds.

Lattice Enthalpy

Lattice enthalpy ($\Delta H_{\text{lattice}}$) is the energy required to completely separate one mole of a solid ionic compound into its gaseous ions. It depends heavily on how well the ions pack together in the crystal lattice.

When a tiny ${\text{Be}}^{2+}$ ion tries to pack with a giant ${\text{SO}}_4^{2-}$ ion, there is a severe size mismatch. They fit together awkwardly, leaving empty spaces. Because of this poor packing, the lattice energy of ${\text{BeSO}}_4$ is relatively low. Conversely, a large ${\text{Ba}}^{2+}$ ion matches perfectly with the large ${\text{SO}}_4^{2-}$ ion, resulting in excellent packing and a very high lattice energy.

Hydration Enthalpy

Hydration enthalpy ($\Delta H_{\text{hydration}}$) is the energy released when gaseous ions are surrounded by water molecules. This energy is directly proportional to the charge density of the ion (Charge/Radius).

Because ${\text{Be}}^{2+}$ is incredibly small but carries a full $+2$ charge, its charge density is off the charts. It acts like a powerful magnet, intensely attracting the polar water molecules. This intense interaction releases a massive amount of hydration energy. As we move down the group to ${\text{Ba}}^{2+}$, the size increases, the charge density drops, and the hydration energy plummets.

For an ionic compound to dissolve in water, the energy released by hydration must overcome the energy holding the lattice together:

In the case of ${\text{BeSO}}_4$, the lattice energy is low (due to size mismatch), and the hydration energy is exceptionally high (due to the tiny size of ${\text{Be}}^{2+}$). Therefore, its hydration enthalpy easily exceeds its lattice enthalpy, making it highly soluble in water.

For the other sulphates like ${\text{BaSO}}_4$, the high lattice energy and low hydration energy mean they remain insoluble. Thus, ${\text{BeSO}}_4$ is the undisputed winner of this thermodynamic battle!