Question

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

• A. $CaSO_4$
• B. $BeSO_4$
• C. $BaSO_4$
• D. $SrSO_4$

Answer 1

The Correct Option is B

To determine which alkaline earth metal sulfate has a hydration enthalpy greater than its lattice enthalpy, we must understand the relationship between these two thermodynamic quantities.

The lattice enthalpy of a compound is the energy required to break one mole of a solid ionic compound into its gaseous ions. It is always positive since energy is required to overcome the forces holding the lattice together.

The hydration enthalpy is the energy released when gaseous ions dissolve in water to form a hydrated solution. It is usually negative since energy is released during this process.

For a salt to be soluble, the hydration enthalpy must compensate for the lattice enthalpy. If the hydration enthalpy (absorbance) is greater in magnitude than the lattice enthalpy, the salt is more likely to dissolve in water.

The sizes of the cations affect these enthalpies. Smaller ions generally have higher hydration enthalpies due to their higher charge density, which attracts water molecules more strongly. Conversely, for lattice enthalpy, larger ions often result in lower values due to lower ionic interactions.

In the given options, beryllium sulfate, BeSO_4, involves the smallest cation, Be^{2+}, among the alkaline earth metals listed:

• Be^{2+} (Beryllium)
• Ca^{2+} (Calcium)
• Sr^{2+} (Strontium)
• Ba^{2+} (Barium)

Because of its smaller size and higher charge density, Be^{2+} should have a much higher hydration enthalpy. Consequently, the hydration enthalpy of BeSO_4 is greater than its lattice enthalpy, leading to better solubility in water compared to others in the group.

Therefore, the correct option is BeSO_4, as it has a hydration enthalpy that is greater than its lattice enthalpy.