A solution of two liquids A and B shows a negative deviation from Raoult’s law when:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The A–B interactions are stronger than the A–A and B–B interactions
A solution shows a negative deviation from Raoult’s law when its vapour pressure is lower than Raoult’s law predicts.
If the attraction between unlike molecules (A–B) is stronger than that between like molecules (A–A and B–B), the molecules escape less easily into the vapour, and the vapour pressure falls.
Phenol–aniline and chloroform–acetone, where A–B hydrogen bonds form, behave in this way.
Why the other options are incorrect
Option (2) The A–B interactions are weaker than the A–A and B–B interactions — Incorrect. Weaker A–B interactions make escape easier and raise the vapour pressure: that is a positive deviation.
Option (3) ΔmixV > 0 and ΔmixS > 0 — Incorrect. Stronger A–B attractions draw the molecules closer, so the solution does not expand on mixing; an increase in volume goes with weaker A–B attractions.
Option (4) ΔmixV = 0 and ΔmixS > 0 — Incorrect. A zero volume of mixing is a property of an ideal solution, which shows no deviation.
NCERT (2026-27 edition):
1.5.2 Non-ideal Solutions – “… In case of negative deviations from Raoult’s law, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B and leads to decrease in vapour pressure resulting in negative deviations.”