JEE Main 2024 5 April Shift 2 Chemistry question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 5 April Shift 2 exam from 3 PM to 6 PM. The Chemistry question paper for JEE Main 2024 5 April Shift 2 includes 30 questions divided into 2 sections, Section 1 with 20 MCQs and Section 2 with 10 numerical questions. Candidates must attempt any 5 numerical questions out of 10. The memory-based JEE Main 2024 question paper pdf for the 5 April Shift 2 exam is available for download using the link below.
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JEE Main 2024 5 April Shift 2 Chemistry Question Paper PDF Download
| JEE Main 2024 Chemistry Question Paper | JEE Main 2024 Chemistry Answer Key | JEE Main 2024 Chemistry Solution |
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JEE Main 2024 5 April Shift 2 Chemistry Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 61: Match List-I with List-II: List-I: (A) ICl (B) ICl₃ (C) ClF₅ (D) IF₇ List-II: (I) T-Shape (II) Square pyramidal (III) Pentagonal bipyramidal (IV) Linear (1) (A)-(I), (B)-(IV), (C)-(III), (D)-(II) (2) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (3) (A)-(IV), (B)-(I), (C)-(II), (D)-(III) (4) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) |
(3) | The molecular geometry of the compounds is determined by VSEPR theory: - (A) ICl: Linear (IV). - (B) ICl₃: T-Shape (I). - (C) ClF₅: Square pyramidal (II). - (D) IF₇: Pentagonal bipyramidal (III). |
| 62: While preparing crystals of Mohr’s salt, dilute H₂SO₄ is added to a mixture of ferrous sulfate and ammonium sulfate. Before dissolving this mixture in water, dilute H₂SO₄ is added here to: (1) Prevent the hydrolysis of ferrous sulfate (2) Prevent the hydrolysis of ammonium sulfate (3) Make the medium strongly acidic (4) Increase the rate of formation of crystals |
(1) | Dilute H₂SO₄ is added to prevent the hydrolysis of ferrous sulfate. Without the acidic medium, ferrous sulfate would hydrolyze to form ferric hydroxide, which interferes with crystal formation. |
| 63: Identify the major product in the following reaction: (1) Cyclopentane (2) Cyclopentanol (3) Cyclopentene (4) Cyclopentanone |
(3) | The reaction involves elimination via the E2 mechanism, where the base removes a β-hydrogen and the leaving group departs, forming a double bond. The major product follows Zaitsev’s rule, favoring the more substituted alkene: cyclopentene. |
| 64: The nomenclature for the following compound is: (1) 2-Carboxy-4-hydroxyhept-6-enal (2) 2-Carboxy-4-hydroxyhept-7-enal (3) 2-Formyl-4-hydroxyhept-6-enoic acid (4) 2-Formyl-4-hydroxyhept-7-enoic acid |
(3) | The parent chain has 7 carbons with a double bond at position 6. Functional groups include a formyl group at position 2, a hydroxyl group at position 4, and a carboxylic acid at the terminal position. The correct name is 2-Formyl-4-hydroxyhept-6-enoic acid. |
| 65: Given below are two statements: Assertion (A): NH₃ and NF₃ molecules have a pyramidal shape with a lone pair of electrons on the nitrogen atom. The resultant dipole moment of NH₃ is greater than that of NF₃. Reason (R): In NH₃, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the N-H bonds. F is the most electronegative element. (1) Both (A) and (R) are true, and (R) is the correct explanation of (A). (2) (A) is true, but (R) is false. (3) (A) is false, but (R) is true. (4) Both (A) and (R) are true, but (R) is not the correct explanation of (A). |
(1) | NH₃ has a higher dipole moment because the lone pair dipole aligns with the N-H bond dipoles. In NF₃, the lone pair dipole opposes the N-F bond dipoles. Both statements are true, and the reason explains the assertion. |
| 66: Given below are two statements: Statement I: On passing HCl(g) through a saturated solution of BaCl₂ at room temperature, white turbidity appears. Statement II: When HCl(g) is passed through a saturated solution of NaCl, sodium chloride is precipitated due to the common ion effect. (1) Statement I is correct, but Statement II is incorrect. (2) Both Statement I and Statement II are correct. (3) Statement I is incorrect, but Statement II is correct. (4) Both Statement I and Statement II are incorrect. |
(1) | HCl(g) increases Cl⁻ concentration, reducing the solubility of BaCl₂ and forming white turbidity due to the common ion effect. However, NaCl is highly soluble, and its solubility is not significantly affected by HCl. Hence, Statement I is correct, but Statement II is incorrect. |
| 67: The metal atom present in the complex MABXL (where A, B, X, and L are unidentate ligands and M is a metal) involves sp³ hybridization. The number of geometrical isomers exhibited by the complex is: (1) 4 (2) 0 (3) 2 (4) 3 |
(2) | In sp³ hybridization, the complex adopts a tetrahedral geometry. Geometrical isomerism is not possible in tetrahedral complexes because all positions are equivalent. Thus, the number of geometrical isomers is 0. |
| 68: Match List-I with List-II: List-I (Pair of Compounds): (A) n-Propanol and isopropanol (B) Methoxypropane and ethoxyethane (C) Propanone and propanal (D) Neopentane and isopentane List-II (Type of Isomerism): (I) Metamerism (II) Chain Isomerism (III) Position Isomerism (IV) Functional Isomerism (1) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (2) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (3) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) (4) (A)-(IV), (B)-(I), (C)-(III), (D)-(II) |
(4) | - (A) n-Propanol and isopropanol exhibit functional isomerism (IV). - (B) Methoxypropane and ethoxyethane exhibit metamerism (I). - (C) Propanone and propanal exhibit position isomerism (III). - (D) Neopentane and isopentane exhibit chain isomerism (II). |
| 69: The quantity of silver deposited when one coulomb of charge is passed through AgNO₃ solution is: (1) 0.1 g atom of silver (2) 1 chemical equivalent of silver (3) 1 g of silver (4) 1 electrochemical equivalent of silver |
(4) | The amount of a substance deposited during electrolysis is proportional to its electrochemical equivalent. For one coulomb of charge, the quantity of silver deposited is equal to its electrochemical equivalent. |
| 70: Which one of the following reactions is NOT possible? (1) Hydrolysis of an ether to form phenol (2) Conversion of phenol to chlorobenzene using HCl (3) Hydrolysis of chlorobenzene to phenol (4) Electrophilic substitution of anisole with chlorine |
(2) | Phenol does not react with HCl to form chlorobenzene because the −OH group is strongly bonded to the benzene ring. Thus, Reaction (2) is not possible, while the other reactions are feasible under appropriate conditions. |
| 71: Given below are two statements: Statement I: The metallic radius of Na is 1.86 Å, and the ionic radius of Na⁺ is lesser than 1.86 Å. Statement II: Ions are always smaller in size than the corresponding elements. (1) Statement I is correct, but Statement II is false. (2) Both Statement I and Statement II are true. (3) Both Statement I and Statement II are false. (4) Statement I is incorrect, but Statement II is true. |
(1) | The metallic radius of Na is 1.86 Å, and its ionic radius decreases due to the loss of an electron, which reduces electron-electron repulsion. Statement II is false because anions (e.g., Cl⁻) are larger than their corresponding neutral atoms due to increased electron repulsion. |
| 72: Consider the above reaction sequence and identify the major product P: (1) Methane (2) Methanal (3) Methoxymethane (4) Methanoic acid |
(1) | The reaction sequence involves oxidation of ethanol to acetic acid, further oxidation to carbon dioxide, and subsequent decarboxylation with soda lime to yield methane as the final product. Thus, the major product is methane (CH₄). |
| 73: Consider the given chemical reaction. Product 'A' is: (1) Picric acid (2) Oxalic acid (3) Acetic acid (4) Adipic acid |
(4) | Cyclohexane undergoes oxidation in the presence of KMnO₄ and H₂SO₄, resulting in the cleavage of CH₂ groups to carboxylic acid groups. This leads to the formation of adipic acid (HOOC-(CH₂)₄-COOH). |
| 74: For the electrochemical cell M—M²⁺||X—X²⁻, if E°(M²⁺/M) = 0.46V and E°(X/X²⁻) = 0.34V, which of the following is correct? (1) Ecell = -0.80V (2) M + X²⁻ → M²⁺ + X²⁻ is a spontaneous reaction (3) M²⁺ + X²⁻ → M + X is a spontaneous reaction (4) Ecell = 0.80V |
(3) | The standard cell potential E°cell is calculated as E°cathode - E°anode. Here, E°cell = 0.34V - 0.46V = -0.12V. Since E°cell is negative, the reverse reaction M²⁺ + X²⁻ → M + X is spontaneous. |
| 75: The number of moles of methane required to produce 11 g of CO₂ after complete combustion is: (1) 0.75 (2) 0.25 (3) 0.35 (4) 0.50 |
(2) | From the combustion reaction CH₄ + 2O₂ → CO₂ + 2H₂O, one mole of methane produces one mole of CO₂. Given the molar mass of CO₂ is 44 g/mol, 11 g corresponds to 0.25 moles of CO₂. Therefore, 0.25 moles of methane are required. |
| 76: The number of complexes from the following with no electrons in the t2 orbital is: TiCl4, [MnO4]-, [FeO4]2-, [FeCl4]-, [CoCl4]2- (1) 3 (2) 1 (3) 4 (4) 2 |
(1) | The number of complexes with no electrons in the t2 orbital is 3: TiCl4, [MnO4]-, and [FeO4]2-. Each has a 3d0 electronic configuration. |
| 77: The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is: Ti2+, Cr2+, V2+ (1) 0 (2) 2 (3) 3 (4) 1 |
(3) | All three ions (Ti2+, Cr2+, and V2+) are strong reducing agents capable of reacting with H+ ions to liberate H2 gas. Hence, the count is 3. |
| 78: Identify A and B in the given chemical reaction sequence: Friedel-Crafts acylation followed by Clemmensen reduction (1) A = Benzaldehyde, B = Benzyl alcohol (2) A = Acetophenone, B = Ethylbenzene (3) A = Benzene, B = Phenol (4) A = Phenol, B = Benzene |
(2) | Friedel-Crafts acylation forms acetophenone (A), which undergoes Clemmensen reduction to produce ethylbenzene (B). |
| 79: The correct decreasing order of atomic radii of Group 13 elements is: (A) Tl > In > Ga > Al > B (B) Tl > In > Ga > B > Al (C) Al > Ga > In > Tl > B (D) In > Tl > Ga > Al > B (1) A (2) B (3) C (4) D |
(1) | The correct order is (A): Tl > In > Ga > Al > B, due to increasing nuclear charge and varying shielding effects down the group. |
| 80: The number of ways the set S = {2, 4, 8, ..., 512} can be partitioned into three subsets of equal size is: (1) 1680 (2) 1720 (3) 1640 (4) 1600 |
(1) | The set has 9 elements. To partition it into three subsets of equal size, use the formula for combinations and calculate the number of ways as 1680. |
| 81: Combustion of 1 mole of benzene is expressed as: C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l) The standard enthalpy of combustion of 2 moles of benzene is −x kJ. Calculate the value of x given the following data: - Standard enthalpy of formation of C₆H₆(l): 48.5 kJ/mol - Standard enthalpy of formation of CO₂(g): −393.5 kJ/mol - Standard enthalpy of formation of H₂O(l): −286 kJ/mol |
(1) | Using Hess's Law, the enthalpy change for combustion is calculated as ∆H = Σ∆Hf(products) − Σ∆Hf(reactants). For 2 moles, ∆H = −6535 kJ. |
| 82.The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products A and B along with the evolution of CO₂. The sum of spin-only magnetic moment values of A and B is ___ B.M. (Nearest integer). (Given atomic numbers: C = 6, Na = 11, O = 8, Fe = 26, Cr = 24) |
6 | During the reaction, chromite ore (FeCr₂O₄) reacts with sodium carbonate in the presence of air to form sodium chromate (Na₂CrO₄) and iron oxide (Fe₂O₃) along with CO₂. The spin-only magnetic moment for sodium chromate (Cr⁶⁺, no unpaired electrons) is 0 B.M., while Fe₃⁺ in Fe₂O₃ has a magnetic moment of √35 ≈ 5.92 B.M. (nearest integer is 6). |
| 83: In an atom, the total number of electrons having quantum numbers n=4, |ml|=1, and ms=−1/2 is: | (2) | The total number of electrons is 6, based on the possible values of l and corresponding orbitals, each holding one electron with ms = −1/2. |
| 84: In an atom, the total number of electrons having quantum numbers n = 4, |ml| = 1, and ms = −1/2 is: | 6 | For n = 4, |ml| = 1 corresponds to three possible orbitals (p, d, or f). Each orbital can hold one electron with ms = −1/2. Therefore, the total number of electrons is 6. |
| 85: Using the given figure, the ratio of Rf values of sample A and sample C is x × 10⁻². Value of x is: Figure: Solvent front = 12.5 cm, Sample A = 5.0 cm, Sample C = 10.0 cm |
50 | The Rf value is the distance traveled by the spot divided by the distance traveled by the solvent front. Rf(A) = 5.0/12.5, Rf(C) = 10.0/12.5. The ratio Rf(A)/Rf(C) = 50 × 10⁻². |
| 86: In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is ________ g (nearest integer). | 318 | The reaction stoichiometry shows that 1 mole of acetone reacts with 2 moles of benzaldehyde. Using the molar masses, the required amount of benzaldehyde is calculated as 318 g. |
| 87: Consider the following single-step reaction in gas phase at constant temperature: 2A(g) + B(g) → C(g) The initial rate of the reaction is r₁ when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate r₂ is recorded when the pressure of C becomes 0.5 atm. The ratio r₁ : r₂ is ________ × 10⁻¹ (nearest integer). |
315 | The rate law is determined using the partial pressures of A and B. Substituting the values for initial and intermediate conditions, the ratio r₁ : r₂ is calculated as 315 × 10⁻¹. |
| 88: The product C in the following sequence of reactions has ________ π bonds. Reaction sequence: KMnO₄–KOH, ∆ → A; H₃O⁺ → B; Br₂/FeBr₃ → C |
4 | The product C, para-bromobenzoic acid, has 3 π bonds in the benzene ring and 1 π bond in the carboxyl group, making a total of 4 π bonds. |
| 89: Considering acetic acid dissociates in water, its dissociation constant is 6.25 × 10⁻⁵. If 5 mL of acetic acid is dissolved in 1 liter of water, the solution will freeze at −x × 10⁻² °C, provided pure water freezes at 0 °C. x = _________. (Nearest integer). Given: Kf(water) = 1.86 K·kg·mol⁻¹, density of acetic acid = 1.2 g·mL⁻¹, molar mass of acetic acid = 60 g·mol⁻¹, density of water = 1 g·cm⁻³. |
19 | The molality of acetic acid is calculated using its mass and dissociation constant. Considering partial dissociation, the effective molality is used in ∆Tf = i·Kf·m. Substituting the values gives x = 19. |
| 90: Number of compounds from the following with zero dipole moment is ___________. HF, H₂, H₂S, CO₂, NH₃, BF₃, CH₄, CHCl₃, SiF₄, H₂O, BeF₂ | 6 | Compounds with symmetrical geometry and no net dipole moment are H₂, CO₂, BF₃, CH₄, SiF₄, and BeF₂. Therefore, the total number of compounds with zero dipole moment is 6. |
JEE Main 2024 Question Paper Session 1 (January)
Those appearing for JEE Main 2024 can use the links below to practice and keep track of their exam preparation level by attempting the shift-wise JEE Main 2024 question paper provided below.
| Exam Date and Shift | Chemistry Question Paper |
|---|---|
| JEE Main 2024 Feb 1 Shift 2 Chemistry Question Paper | Check Here |
| JEE Main 2024 Feb 1 Shift 1 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 31 Shift 2 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 31 Shift 1 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 30 Shift 2 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 30 Shift 1 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 29 Shift 2 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 29 Shift 1 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 27 Shift 2 Chemistry Question Paper | Check Here |
| JEE Main 2024 Jan 27 Shift 1 Chemistry Question Paper | Check Here |
Also Check:
| JEE Main 2024 Paper Analysis | JEE Main 2024 Answer Key |
| JEE Main 2024 Cutoff | JEE Main 2024 Marks vs Rank |
JEE Main 2024 5 April Shift 2 Chemistry Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Vedantu | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 5 April Shift 2 Chemistry Paper Analysis
JEE Main 2024 5 April Shift 2 Chemistry paper analysis is updated here with details on the difficulty level of the exam, topics with the highest weightage in the exam, section-wise difficulty level, etc.
JEE Main 2024 Chemistry Question Paper Pattern
| Feature | Question Paper Pattern |
|---|---|
| Examination Mode | Computer-based Test |
| Exam Language | 13 languages (English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu, and Urdu) |
| Sectional Time Duration | None |
| Total Marks | 100 marks |
| Total Number of Questions Asked | 30 Questions |
| Total Number of Questions to be Answered | 25 questions |
| Type of Questions | MCQs and Numerical Answer Type Questions |
| Section-wise Number of Questions | 20 MCQs and 10 numerical type, |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
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