Question

The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to:
(A.) ionic nature of lithium fluoride
(B.) high lattice enthalpy
(C.) high hydration enthalpy for lithium-ion
(D.) low ionization enthalpy of the lithium atom

Answer 1

Hint: Here is one tip for you, the small size of Li cation and F anion has an important role in solving this question. Now try to answer this accordingly.

Complete step by step solution:
First, Let’s try to understand some important terms given in the question.

Lattice enthalpy - The amount of energy required to completely separate one mole of the solid ionic compound into constituent gaseous ions is known as lattice enthalpy.

Hydration enthalpy - The hydration enthalpy is the enthalpy change when 1 mole of gaseous ions dissolves in sufficient water to give an infinitely dilute solution.

Ionization enthalpy - The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so to convert it into gaseous cation is called ionization enthalpy.

Solubilities of alkali metal halides in water can be explained in terms of lattice enthalpy and hydration enthalpies favor dissolution. Among fluorides, the order of solubility is LiF < NaF < KF < RbF < CsF.
The low solubility of LiF is due to very high lattice energy. Except for LiF, other halides of lithium are highly soluble in water.

Therefore, we can conclude that the correct answer to this question is option B.

Note: We should also know that on moving down in the group LiF to CsF, solubility is increasing because lattice energy is decreasing.