JEE Advanced 2026 — Paper 1 — CHEM-IV solution

JEE Advanced2026Paper 1CHEM-IV

List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy ( $Δ$ H) and entropy ( $Δ$ S). Match each entry in List-I to appropriate entry in List-II, and choose the correct option.

List-I	List-II
(P) Physisorption	(1) $Δ H > 0$ and $Δ S > 0$
(Q) Diamond $\to$ Graphite	(2) $Δ H < 0$ and $Δ S < 0$
(R) Denaturation of protein	(3) $Δ H < 0$ and $Δ S = 0$
(S) Propene $\to$ Cyclopropane	(4) $Δ H > 0$ and $Δ S < 0$
	(5) $Δ H < 0$ and $Δ S > 0$

A
P $\to$ 2; Q $\to$ 3; R $\to$ 5; S $\to$ 4
B
P $\to$ 4; Q $\to$ 3; R $\to$ 5; S $\to$ 1
C
P $\to$ 2; Q $\to$ 5; R $\to$ 1; S $\to$ 4
D
P $\to$ 2; Q $\to$ 5; R $\to$ 1; S $\to$ 3

Solution

(P) Physisorption. A gas-phase molecule loses translational freedom on becoming adsorbed on a surface ( $Δ S < 0$ ); the process is exothermic because weak van der Waals attractions release energy ( $Δ H < 0$ ). P $\to$ 2.

(Q) Diamond $\to$ Graphite. Graphite is thermodynamically more stable than diamond at standard conditions, so $Δ H < 0$ . Graphite's layered structure with weak interlayer forces affords more positional/configurational disorder than rigid diamond, so $Δ S > 0$ . Q $\to$ 5.

(R) Denaturation of protein. Hydrogen bonds and other tertiary-structure interactions are broken, requiring heat input ( $Δ H > 0$ ); the unfolded polypeptide chain has many more accessible conformations than the native folded state ( $Δ S > 0$ ). R $\to$ 1.

(S) Propene $\to$ Cyclopropane. Cyclopropane has substantial ring (angle) strain, so the cyclisation is endothermic ( $Δ H > 0$ ). The open-chain propene has more rotational/conformational freedom than the rigid cyclic isomer, so $Δ S < 0$ . S $\to$ 4.

Mapping P $\to$ 2, Q $\to$ 5, R $\to$ 1, S $\to$ 4 corresponds to option (C).

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