Question

the electron cloud of hf is smaller than that of f₂, however, hf has a much higher boiling point than f₂ has. which of the following explains how the dispersion - force model of intermolecular attraction does not account for the unusually high boiling point of hf?
a f₂ is soluble in water, whereas hf is insoluble in water.
b the f₂ molecule has a greater mass than the hf molecule has.
c liquid f₂ has weak dispersion force attractions between its molecules, whereas liquid hf has strong ionic interactions between h⁺ and f⁻ ions.
d liquid f₂ has weak dispersion force attractions between its molecules, whereas liquid hf has both weak dispersion force attractions and hydrogen bonding interactions between its molecules.

Explanation:

• Option A: Solubility in water is not related to boiling point differences.
• Option B: \(F_2\) has a smaller mass than \(HF\) (molar mass of \(F_2 = 38\space g/mol\), molar mass of \(HF=20\space g/mol\)), so this is incorrect.
• Option C: \(HF\) is a covalent compound, not ionic. There are no \(H^+\) and \(F^-\) ions in liquid \(HF\) in the sense of ionic interactions.
• Option D: \(F_2\) has only weak dispersion (London) forces. \(HF\) has both weak dispersion forces (due to its electrons) and strong hydrogen - bonding (because of the \(H - F\) bond where \(F\) is highly electronegative). Hydrogen bonding is a much stronger intermolecular force than dispersion forces. A substance with stronger intermolecular forces (like \(HF\) with hydrogen bonding) will have a higher boiling point as more energy is required to break these intermolecular attractions.

Answer:

D. Liquid \(F_2\) has weak dispersion force attractions between its molecules, whereas liquid \(HF\) has both weak dispersion force attractions and hydrogen bonding interactions between its molecules.