JEE Main Chemistry question paper with solutions

Sample questions with solutions

Five questions from the bank – try them before you read the answer.

1. Q1.Elevation of boiling point · Fresh

   The following aqueous solutions are heated at 1 atm: 0.12 m glucose, 0.050 m NaCl, 0.030 m Al₂(SO₄)₃ and 0.040 m K₃[Fe(CN)₆]. All electrolytes are completely dissociated. K_b (water) = 0.52 K kg mol⁻¹. The elevation of boiling point (in K) of the solution that boils at the highest temperature is

   1. (1)0.052
   2. (2)0.062
   3. (3)0.083
   4. (4)0.078

   Answer (3)

   Particles per kg (i × m): glucose 0.12, NaCl 2 × 0.050 = 0.10, Al₂(SO₄)₃ 5 × 0.030 = 0.15, K₃[Fe(CN)₆] 4 × 0.040 = 0.16.

   The largest i m boils highest: ΔT_b = 0.52 × 0.16 = 0.0832 ≈ 0.083 K. (0.078 K is the trap – it belongs to Al₂(SO₄)₃.)

2. Q2.KMnO₄ and K₂Cr₂O₇ · Fresh

   Orange aqueous K₂Cr₂O₇ turns yellow on adding NaOH and returns to orange on adding H₂SO₄. Which option correctly gives the yellow species and the nature of the change?

   1. (1)CrO₄²⁻ (Cr +6); a redox change in which Cr(VI) is reduced
   2. (2)CrO₄²⁻ (Cr +6); a pH-dependent equilibrium, with no change in oxidation state
   3. (3)Cr(OH)₄⁻ (Cr +3); a pH-dependent equilibrium
   4. (4)CrO₅ (Cr +6); formation of a peroxide on adding alkali

   Answer (2)

   2CrO₄²⁻ (yellow) + 2H⁺ ⇌ Cr₂O₇²⁻ (orange) + H₂O. Alkali removes H⁺ and gives yellow chromate; acid brings back orange dichromate.

   Cr is +6 on both sides, so it is an acid–base equilibrium, not a redox change.

3. Q3.Gabriel phthalimide synthesis · Fresh

   Which one of the following amines is a primary amine and cannot be prepared in good yield by the Gabriel phthalimide synthesis?

   1. (1)2-Methylpropan-2-amine, (CH₃)₃C–NH₂
   2. (2)N-Methylbenzylamine, C₆H₅CH₂NHCH₃
   3. (3)Benzylamine, C₆H₅CH₂NH₂
   4. (4)Butan-1-amine, CH₃CH₂CH₂CH₂NH₂

   Answer (1)

   N-Methylbenzylamine is secondary, so it fails the first condition.

   Gabriel synthesis needs S_N2 attack of the phthalimide anion on an alkyl halide. (CH₃)₃C–Br is tertiary – no S_N2, only elimination – so (CH₃)₃C–NH₂ cannot be made this way.

4. Q4.Mole concept · Fresh

   Which of the following samples contains the same number of molecules and the same number of atoms as 8 g of O₂? (H = 1, C = 12, N = 14, O = 16)

   1. (1)11 g of CO₂
   2. (2)4 g of CH₄
   3. (3)4.5 g of H₂O
   4. (4)7.5 g of NO

   Answer (4)

   8 g O₂ = 0.25 mol molecules and 0.50 mol atoms.

   Every option is 0.25 mol of molecules (11/44, 4/16, 4.5/18, 7.5/30), but atoms are 0.75, 1.25, 0.75 and 0.50. Only diatomic NO matches both.

5. Q5.Isomerism in complexes · JEE Main PYQ · 8 Apr 2025 · Shift 2

   Given below are two statements :
   Statement-I : A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism
   Statement-II : cis and trans platin are heteroleptic complexes of Pd
   In the light of the above statements, choose the correct answer from the options given below

   1. (1)Both Statement I and Statement II are true
   2. (2)Statement I is true but Statement II is false
   3. (3)Both Statement I and Statement II are false
   4. (4)Statement I is false but Statement II is true

   Answer (2)

   Statement I: [MA₆] has six equivalent positions – no geometrical or optical isomer. True.

   Statement II: cis- and trans-platin, [PtCl₂(NH₃)₂], are heteroleptic, but the metal is Pt, not Pd. False.

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