Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes?

Solution

Chemistry ma’am · 3 steps

Look for the pair whose molecules like each other more than themselves: phenol and aniline.

1. Link deviation to azeotrope. A positive deviation from Raoult's law means A–B attractions are weaker than A–A and B–B. The vapour pressure rises, so the mixture boils at a minimum temperature. A negative deviation means A–B attractions are stronger. The vapour pressure falls, so the azeotrope boils at a maximum temperature.

   Mixture	What mixing does	Deviation	Azeotrope
   CH₃OH + CHCl₃	Chloroform molecules break up the H-bonds between methanol molecules	positive	minimum boiling
   C₆H₅OH + C₆H₅NH₂	Acidic O–H of phenol H-bonds strongly to the basic N of aniline	negative	maximum boiling
   H₂O + CH₃COC₂H₅	The ketone breaks up the extensive H-bonding in water	positive	minimum boiling
   CS₂ + CH₃COCH₃	Non-polar CS₂ weakens the dipole–dipole attraction between acetone molecules	positive	minimum boiling

2. Why phenol and aniline. Phenol is weakly acidic and aniline is basic. The new O–H···N hydrogen bond formed between them is stronger than what each liquid had on its own. Fewer molecules escape into the vapour.
3. So. Phenol + aniline shows negative deviation and forms a maximum boiling azeotrope. It is the one pair that does not form a minimum boiling azeotrope, which is option (b).

The idea to keep. Weaker A–B attraction gives positive deviation and a minimum boiling azeotrope. Stronger A–B attraction, such as an acid–base H-bond, gives negative deviation and a maximum boiling azeotrope.

Answer from the NTA final answer key. Solution by PrepWiser.