Given below are two statements:one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: LiF is sparingly soluble in water.
Reason R: The ionic radius of $$Li^+$$ ion is smallest among its group members, hence has least hydration enthalpy.
In the light of the above statements, choose the most appropriate answer from the options given below.

Analyzing Assertion A:

LiF is indeed sparingly soluble in water. This is because both $$Li^+$$ and $$F^-$$ are very small ions, resulting in a very high lattice energy. The high lattice energy makes it difficult for water molecules to overcome the strong electrostatic attraction between the ions, leading to low solubility. This statement is true.

Analyzing Reason R:

The reason states that $$Li^+$$ has the smallest ionic radius among its group members and hence has the least hydration enthalpy. This is false. According to the relationship between ionic size and hydration enthalpy, a smaller ion has a higher charge density, which attracts water molecules more strongly. Therefore, $$Li^+$$ actually has the highest (most negative) hydration enthalpy among the alkali metal cations, not the least.

The order of hydration enthalpy (magnitude) is:

$$Li^+ > Na^+ > K^+ > Rb^+ > Cs^+$$

The actual reason LiF is sparingly soluble is that its very high lattice energy (due to both small ions) outweighs the hydration enthalpy, making dissolution energetically unfavorable.

Since Assertion A is true but Reason R is false, the correct answer is Option C.