Meaning of negative deviation from Raoult’s law

A solution shows negative deviation when the observed vapour pressure is less than that predicted by Raoult’s law.

This happens when unlike molecules attract each other more strongly than like molecules.

So, for components $A$ and $B$: $A-B \, \text{interaction} > A-A \text{ and } B-B \text{ interactions}$

Thermodynamic consequences of negative deviation

Because $A-B$ attractions are stronger, mixing is accompanied by:

• decrease in volume due to closer packing: $\Delta V_{mix} < 0$
• release of heat: $\Delta H_{mix} < 0$

Therefore, any option claiming $\Delta V_{mix} > 0$ or $\Delta V_{mix} = 0$ is not characteristic of negative deviation.

Check each option

Option A: $\Delta V_{mix} > 0, \, \Delta S_{mix} > 0$

• Positive volume change suggests weaker interactions.
• Not correct for negative deviation.

Option B: $\Delta V_{mix} = 0, \, \Delta S_{mix} > 0$

• Ideal-like volume behavior does not indicate negative deviation.
• Not correct.

Option C: $A-B$ interaction is weaker than $A-A$ and $B-B$ interaction

• This corresponds to positive deviation, not negative deviation.
• Not correct.

Option D: $A-B$ interaction is stronger than $A-A$ and $B-B$ interaction

• This is exactly the condition for negative deviation.
• Correct.

Final answer

The correct option is: $\boxed{D}$