Low solubility of LiF in water is due to its small hydration enthalpy.
\(KO _2\) is paramagnetic.
Solution of sodium in liquid ammonia is conducting in nature.
Sodium metal has higher density than
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The Correct Option isA
Solution and Explanation
To determine the incorrect statement among the given options, let's evaluate each statement based on chemical properties and principles:
Low solubility of LiF in water is due to its small hydration enthalpy.
Lithium fluoride (LiF) exhibits low solubility in water, but this is primarily due to its very high lattice energy rather than small hydration enthalpy. The lattice energy is so large because LiF involves a strong ionic bond between Li+ and F- ions, making it difficult to break apart.
This statement incorrectly attributes its low solubility to small hydration enthalpy, making it the incorrect statement.
\(KO_2\) is paramagnetic.
Potassium superoxide (\(KO_2\)) is indeed paramagnetic. This can be attributed to the presence of unpaired electrons in the molecular orbital of \(O_2^-\) (superoxide) ions.
Therefore, this statement is correct.
Solution of sodium in liquid ammonia is conducting in nature.
Sodium dissolves in liquid ammonia to form a deep blue solution that conducts electricity. The conduction is due to the presence of solvated electrons, which are free to move and carry charge.
This makes the statement correct.
Sodium metal has higher density than potassium metal.
Sodium has a density of approximately 0.97 g/cm3, while potassium's density is about 0.86 g/cm3. Thus, sodium is indeed denser than potassium.
This statement is correct.
Based on the analysis, the incorrect statement is: "Low solubility of LiF in water is due to its small hydration enthalpy."
