JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
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.
- (a) Both A and R are true and R is the correct explanation of A.
- (b) Both A and R are true but R is NOT the correct explanation of A.
- (c) A is true but R is false.
- (d) A is false but R is true.
Answer and solution
Answer (c) A is true but R is false.
Solution
Chemistry ma’am · 3 steps
The smallest ion pulls water hardest, so Li⁺ has the highest hydration enthalpy, not the least.
- Check Assertion A. LiF is only sparingly soluble in water. Both Li⁺ and F⁻ are very small, so they pack very close together and the lattice enthalpy is very high. Water cannot supply enough energy to break this lattice, so A is true.
- Check Reason R. Li⁺ is indeed the smallest ion in group 1. A smaller ion has a higher charge density, so it attracts water molecules more strongly. That gives Li⁺ the highest hydration enthalpy in the group, not the least, so R is false.
Ion Size Hydration enthalpy Li⁺ smallest highest Na⁺ larger lower K⁺ larger still lower still - So. A is true and R is false, so the answer is (c). LiF stays undissolved because its lattice enthalpy is so high, even though Li⁺ is strongly hydrated.
The idea to keep. Small ion means high charge density, which means high hydration enthalpy. Solubility is a contest between lattice enthalpy and hydration enthalpy, and for LiF the lattice wins.
Answer from the NTA final answer key. Solution by PrepWiser.