Question

Which of the following statement is incorrect?

• A. Low solubility of LiF in water is due to its small hydration enthalpy.
• B. \(KO_2\) is paramagnetic.
• C. Solution of sodium in liquid ammonia is conducting in nature.
• D. Sodium metal has higher density than potassium metal.

Answer 1

The Correct Option is A

To determine the incorrect statement, let us examine each option provided:

Low solubility of LiF in water is due to its small hydration enthalpy.

Explanation: LiF is only sparingly soluble in water primarily because of its high lattice enthalpy, not because of low hydration enthalpy. The small ionic size of lithium contributes to this high lattice enthalpy, which is not sufficiently compensated by hydration enthalpy. Therefore, this statement is incorrect.

\(KO_2\) is paramagnetic.

Explanation: \(KO_2\) is a superoxide. Superoxide ions \((\text{O}_2^-)\) have an odd number of electrons (in this case, 33), resulting in unpaired electrons. Therefore, \(KO_2\) exhibits paramagnetism. The statement is correct.

Solution of sodium in liquid ammonia is conducting in nature.

Explanation: When sodium dissolves in liquid ammonia, it forms solvated electrons which are free to move, making the solution an excellent conductor of electricity. This statement is correct.

Sodium metal has a higher density than potassium metal.

Explanation: Sodium has a lower atomic mass than potassium, but its atomic radius is also smaller, leading to a higher packing density. Therefore, sodium metal is denser than potassium metal. This statement is correct.

Conclusion: Based on the explanations above, the incorrect statement is "Low solubility of LiF in water is due to its small hydration enthalpy." The correct reason for the low solubility is its high lattice enthalpy.