Match List-I with List-II List-I Components of binary solution List-I…

The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

List-I Components of binary solution	List-II Deviation from Raoult's Law
(A) Ethanol and water	(IV) Large positive
(B) Chloroform and acetone	(III) Negative
(C) n-Hexane and n-heptane	(II) No deviation
(D) Carbon disulphide and acetone	(I) Positive

(A)-(IV) Pure ethanol forms strong $\text{H}$-bonds. When water is added, the original $\text{H}$-bonds in ethanol are broken, and new $\text{H}$-bonds between ethanol and water are weaker than the original ones. This leads to large positive deviation.

(B)-(III) In the mixture, new, strong intermolecular $\text{H}$-bonds form between the $\text{H}$ of chloroform ($\text{CHCl}_3$) and the $\text{O}$ of acetone ($\text{CH}_3\text{COCH}_3$). This increased attraction lowers the escaping tendency of molecules, causing a negative deviation.

(C)-(II)These two are non-polar organic liquids with very similar molecular size and weak Van der Waals forces. The $\text{A-A}$ and $\text{B-B}$ interactions are nearly identical to the $\text{A-B}$ interactions. This forms an ideal solution with no deviation.

(D)-(I)Both components have weak intermolecular forces. When mixed, the new $\text{A-B}$ interactions are weaker than the strong dipole-dipole interactions in pure acetone. This makes molecules escape more easily, resulting in a positive deviation.