JEE Main 2024 Feb 1 Shift 2 Chemistry question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Feb 1 Shift 2 exam from 3 PM to 6 PM. The Chemistry question paper for JEE Main 2024 Feb 1 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 Feb 1 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Feb 1 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 1 Feb Shift 2 2024 Chemistry Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 61: The transition metal having the highest 3rd ionisation enthalpy is: (1) Cr (2) Mn (3) V (4) Fe |
(2) Mn | The 3rd ionisation energy of Mn is the highest due to the stable half-filled electronic configuration (3d⁵4s²). This stability increases the energy required for ionisation, making Mn the correct choice. |
| 62: Given below are two statements: Statement I: A π-bonding MO has lower electron density above and below the inter-nuclear axis. Statement II: The π-antibonding MO has a node between the nuclei. (1) Both Statement I and Statement II are false (2) Both Statement I and Statement II are true (3) Statement I is false, but Statement II is true (4) Statement I is true, but Statement II is false |
(3) | Statement I is false because π-bonding molecular orbitals have higher electron density above and below the inter-nuclear axis. Statement II is true as π-antibonding molecular orbitals indeed have a node between the nuclei. |
| 63: Assertion (A): In aqueous solutions, Cr2+ is reducing while Mn3+ is oxidising in nature. Reason (R): Extra stability of half-filled electronic configuration is observed than incompletely filled configurations. (1) Both (A) and (R) are true, and (R) is the correct explanation of (A). (2) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (3) (A) is false, but (R) is true. (4) (A) is true, but (R) is false |
(1) | Cr2+ gets oxidised to Cr3+, achieving stability due to a t2g3 configuration, while Mn3+ gets reduced to Mn2+, which is a stable d⁵ configuration. Both statements are true, and (R) explains (A). |
| 64: Match List-I with List-II: List-I (Reactants) (A) Phenol, Zn/∆ (B) Phenol, CHCl3, NaOH, HCl (C) Phenol, CO2, NaOH, HCl (D) Phenol, Conc. HNO3 List-II (Products) (I) Salicylaldehyde (II) Salicylic acid (III) Benzene (IV) Picric acid (1) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) (2) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) (3) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) |
(2) | - (A) Phenol reduced with Zn forms benzene. - (B) Reimer-Tiemann reaction produces salicylaldehyde. - (C) Kolbe-Schmitt reaction produces salicylic acid. - (D) Nitration produces picric acid. |
| 65: Given below are two statements: Statement I: Both metal and non-metal exist in p- and d-block elements. Statement II: Non-metals have higher ionisation enthalpy and higher electronegativity than metals. (1) Both Statement I and Statement II are false (2) Statement I is false, but Statement II is true (3) Statement I is true, but Statement II is false (4) Both Statement I and Statement II are true |
(2) | Statement I is false as d-block elements contain only metals, while p-block contains both metals and non-metals. Statement II is true since non-metals exhibit higher ionisation enthalpy and electronegativity due to their smaller atomic size and high nuclear charge. |
| 66: The strongest reducing agent among the following is: (1) NH3 (2) SbH3 (3) BiH3 (4) PH3 |
(3) BiH3 | BiH3 is the strongest reducing agent because of its low bond dissociation energy. The reducing ability increases down the group as the bond strength decreases. |
| 67: Which of the following compounds shows colour due to d-d transition? (1) CuSO4.5H2O (2) K2Cr2O7 (3) K2CrO4 (4) KMnO4 |
(1) CuSO4.5H2O | CuSO4.5H2O exhibits colour due to the presence of Cu2+ ions with an unpaired electron in the d-orbital undergoing a d-d transition. Other compounds exhibit colour due to charge transfer. |
| 68: The set of meta-directing functional groups from the following sets is: (1) −CN,−NH2,−NHR,−OCH3 (2) −NO2,−NH2,−COOH,−COOR (3) −NO2,−CHO,−SO3H,−COR (4) −CN,−CHO,−NHCOCH3,−COOR |
(3) −NO2,−CHO,−SO3H,−COR | Meta-directing groups withdraw electron density through resonance or inductive effects. Groups like −NO2,−CHO,−SO3H,−COR are strong electron-withdrawing groups and direct substituents to the meta position. |
| 69: Select the compound from the following that will show intramolecular hydrogen bonding: (1) H2O (2) NH3 (3) C2H5OH (4) A compound containing -OH and adjacent -CHO group |
(4) | The compound with an -OH group adjacent to a carbonyl (-CHO) group exhibits intramolecular hydrogen bonding due to proximity and structure. |
| 70: Lassaigne’s test is used for the detection of: (1) Nitrogen and Sulphur only (2) Nitrogen, Sulphur, and Phosphorus only (3) Phosphorus and Halogens only (4) Nitrogen, Sulphur, Phosphorus, and Halogens |
(4) Nitrogen, Sulphur, Phosphorus, and Halogens | Lassaigne’s test detects elements like nitrogen, sulphur, phosphorus, and halogens in organic compounds by converting them into ionic forms for easy detection. |
| 71: Which among the following has the highest boiling point? (1) CH3CH2CH3 (2) CH3CH2CH2OH (3) CH3CH2CHO (4) CH3C(=O)−CH2CH3 |
(2) CH3CH2CH2OH | CH3CH2CH2OH has the highest boiling point because of strong hydrogen bonding between alcohol molecules, which increases the boiling point significantly compared to aldehydes, ketones, or alkanes. |
| 72: In the given reactions, identify A and B: CH3C≡CH + H2 (Pd/C) → A A + Na/Liquid NH3 → B (1) A: 2-Pentyne, B: trans-2-butene (2) A: n-Pentane, B: trans-2-butene (3) A: 2-Pentyne, B: cis-2-butene (4) A: n-Pentane, B: cis-2-butene |
(1) A: 2-Pentyne, B: trans-2-butene | Partial hydrogenation of CH3C≡CH with Pd/C produces A (2-Pentyne). Further reduction with sodium in liquid ammonia converts A into trans-2-butene (B). |
| 73: The number of radial nodes for a 3p orbital is: (1) 1 (2) 2 (3) 3 (4) 4 |
(1) 1 | The formula for radial nodes is n−l−1. For a 3p orbital, n = 3 and l = 1, resulting in 1 radial node. |
| 74: Match List-I with List-II: List-I (Compound) (A) Carbon tetrachloride (B) Methylene chloride (C) DDT (D) Freons List-II (Use) (I) Paint remover (II) Refrigerators and air conditioners (III) Fire extinguisher (IV) Non-biodegradable insecticide (1) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) (2) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (3) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) |
(2) | (A) Carbon tetrachloride is used in fire extinguishers. (B) Methylene chloride is a paint remover. (C) DDT is a non-biodegradable insecticide. (D) Freons are used in refrigerators and air conditioners. |
| 75: The functional group that shows negative resonance effect is: (1) −NH2 (2) −OH (3) −COOH (4) −OR |
(3) −COOH | The −COOH group exhibits a negative resonance effect due to the electron-withdrawing nature of the carbonyl group, reducing electron density on the conjugated system. |
| 76: [Co(NH3)6]3+ and [CoF6]3− are respectively known as: (1) Spin free Complex, Spin paired Complex (2) Spin paired Complex, Spin free Complex (3) Outer orbital Complex, Inner orbital Complex (4) Inner orbital Complex, Spin paired Complex |
(2) | [Co(NH3)6]3+ forms a spin-paired (low-spin) complex due to strong-field NH3. [CoF6]3− forms a spin-free (high-spin) complex due to weak-field F−. |
| 77: Given below are two statements: Statement I: SiO2 and GeO2 are acidic, while SnO and PbO are amphoteric. Statement II: Allotropic forms of carbon arise due to catenation and pπ-pπ bonding. (1) Both Statement I and Statement II are false (2) Both Statement I and Statement II are true (3) Statement I is true, but Statement II is false (4) Statement I is false, but Statement II is true |
(3) | Statement I is true as SiO2 and GeO2 are acidic, whereas SnO and PbO are amphoteric. Statement II is false because allotropic forms of carbon are due to catenation, not pπ-pπ bonding. |
| 78: Acid D formed in the reaction is: (1) Gluconic acid (2) Succinic acid (3) Oxalic acid (4) Malonic acid |
(2) Succinic acid | The reaction produces Succinic acid (HOOC-CH2-CH2-COOH) through the following steps: alkene bromination, cyanide addition, and hydrolysis. |
| 79: Solubility of calcium phosphate (molecular mass, M) in water is W g per 100 mL at 25°C. Its solubility product at 25°C will be approximately: (1) 107(W/M)3 (2) 107(W/M)5 (3) 107(W/M)7 (4) 107(W/M)4 |
(2) 107(W/M)5 | The dissociation of calcium phosphate, Ca3(PO4)2, gives 3Ca2+ and 2PO43−. Solubility product Ksp is proportional to the fifth power of solubility, leading to the formula 107(W/M)5. |
| 80: Given below are two statements: Statement I: Dimethyl glyoxime forms a six-membered covalent chelate when treated with NiCl2 solution in the presence of NH4OH. Statement II: Prussian blue precipitate contains iron in both +2 and +3 oxidation states. (1) Statement I is false, but Statement II is true (2) Both Statement I and Statement II are true (3) Both Statement I and Statement II are false (4) Statement I is true, but Statement II is false |
(1) | Statement I is false because dimethyl glyoxime forms a five-membered chelate, not a six-membered one. Statement II is true as Prussian blue contains Fe2+ and Fe3+. |
| 81: Total number of isomeric compounds (including stereoisomers) formed by monochlorination of 2-methylbutane is: | 6 | Monochlorination of 2-methylbutane produces six isomers due to substitution at different carbon atoms and the presence of chiral centers, leading to structural and stereoisomers. |
| 82: The following data were obtained during the first-order thermal decomposition of a gas A at constant volume: A(g) → 2B(g) + C(g) Time (s): 0, 115 Total Pressure (atm): 0.1, 0.28 The rate constant of the reaction is ×10−2 s−1 (nearest integer): |
2 | Using the first-order integrated rate law, k = (2.303/t) log (Pfinal − Pinitial) / Pinitial, the rate constant is calculated as approximately 2 × 10−2 s−1. |
| 83: The number of tripeptides formed by three different amino acids using each amino acid once is: | 6 | The total number of tripeptides is determined by permutations of three distinct amino acids. Using 3! = 6, there are six possible arrangements. |
| 84: Number of compounds which give reaction with Hinsberg’s reagent is: | 5 | Hinsberg’s reagent reacts with primary and secondary amines to form derivatives. Among the given compounds, five amines react with the reagent, confirming their classification. |
| 85: Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at −24°C is: | 14.88 kg | Using the formula for depression in freezing point, ∆Tf = Kf × molality, the required mass of ethylene glycol is calculated as 14.88 kg to achieve the desired freezing point protection. |
| 86: Following Kjeldahl’s method, 1g of organic compound released ammonia that neutralized 10 mL of 2M H2SO4. The percentage of nitrogen in the compound is: | 56 | The moles of NH3 are calculated as 10 × 2 × 10−3 = 0.02 mol. The weight of nitrogen is 0.02 × 14 = 0.28 g. Percentage nitrogen is (0.28 / 1) × 100 = 56%. |
| 87: The amount of electricity in Coulombs required for the oxidation of 1 mol of H2O to O2 is ×105 C: | 2 | For 1 mol of H2O, 4 mol of electrons are required. Total charge = 4 × 96500 = 3.86 × 105 C. Dividing by 2 gives 2 × 105 C. |
| 88: For a certain reaction at 300 K, K = 10. Then ∆G° for the same reaction is ×10−1 kJ/mol: | 57 | Using the formula ∆G° = −RT lnK, with R = 8.314 J/mol·K, T = 300 K, and ln(10) = 2.303, ∆G° = −8.314 × 300 × 2.303 = −5744.14 J/mol = −57.4 × 10−1 kJ/mol. |
| 89: Consider the redox reaction: MnO4− + H+ + H2C2O4 ⇌ Mn2+ + H2O + CO2. If the equilibrium constant of the above reaction is Keq = 10x, then the value of x is (nearest integer): |
338 | Using E°cell = 1.51 − (−0.49) = 2.00 V and the formula logK = (nE°cell) / 0.0591, with n = 5, logK = (5 × 2.00) / 0.0591 ≈ 338. |
| 90: 10 mL of gaseous hydrocarbon on combustion gives 40 mL of CO2 and 50 mL of water vapor. Total number of carbon and hydrogen atoms in the hydrocarbon is: | 14 | The hydrocarbon is C4H10 (Butane). Based on combustion stoichiometry, x = 4 and y = 10. Total atoms = x + y = 4 + 10 = 14. |
Also Check:
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JEE Main 2024 Feb 1 Shift 2 Chemistry Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | To be updated |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Vedantu | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 Feb 1 Shift 2 Chemistry Paper Analysis
JEE Main 2024 Feb 1 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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