Question

Solubility of an ionic compound in water is mainly dependent on:
a. Lattice enthalpy
b. Hydration enthalpy
Both these factors oppose each other, and the resultant of these factors determines the solubility of an ionic compound in water. If the lattice enthalpy has a larger value, the compound is less soluble. If the hydration enthalpy has larger value, the compound is highly soluble in water.

$Be{F}_2$ is soluble in water while fluorides of other alkaline earth metals are insoluble because of:

• A. Covalent nature of $Be{F}_2$
• B. Ionic nature of $Be{F}_2$
• C. Greater hydration enthalpy of $B{e}^{2+}$ ion
• D. Greater lattice enthalpy of $B{e}^{2+}$ ion

Answer 1

The correct option is C Greater hydration enthalpy of $B{e}^{2+}$ ion

In case of $Be{F}_2$, $B{e}^{2+}$ has high hydration enthalpy which can overcome the lattice enthalpy which makes it soluble. However in case of other alkaline earth metals, hydration enthalpy is not that high.