JEE Main 2024 4 April Shift 2 Chemistry question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 4 April Shift 2 exam from 3 PM to 6 PM. The Chemistry question paper for JEE Main 2024 4 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 4 April Shift 2 exam is available for download using the link below.
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JEE Main 2024 4 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 4 April Shift 2 Chemistry Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 61: The equilibrium constant for the reaction: SO3(g) ⇀↽ SO2(g) + (1/2)O2(g) is Kc = 4.9 × 10-2. The value of Kc for the reaction given below is: 2SO2(g) + O2(g) ⇀↽ 2SO3(g) (1) 4.9 (2) 41.6 (3) 49 (4) 416 |
(4) 416 | The given reaction is the reverse of the initial reaction, and its stoichiometric coefficients are doubled. The equilibrium constant for the reverse reaction is 1/Kc, and doubling the stoichiometric coefficients squares this value. Kc for the new reaction = (1/Kc)² = (1/(4.9 × 10-2))² = 416. |
| 62: Find out the major product formed from the following reaction. [Me: −CH3] (1) Product A (2) Product B (3) Product C (4) Product D |
(3) Product C | The reaction proceeds via an SN2 mechanism, where the bromine atoms are sequentially replaced by Me2N groups. The final product contains two Me2N groups attached to the cyclopentane ring, the same for both cis and trans isomers. |
| 63: When MnO2 and H2SO4 are added to a salt (A), the greenish-yellow gas liberated indicates that salt (A) is: (1) NaBr (2) CaI2 (3) KNO3 (4) NH4Cl |
(4) NH4Cl | The reaction is: 2NH4Cl + MnO2 + 2H2SO4 → MnSO4 + (NH4)2SO4 + 2H2O + Cl2. Chlorine gas (Cl2), a greenish-yellow gas, is liberated. |
| 64: The correct statements about Hydrogen bonding are: A. Hydrogen bonding exists when H is covalently bonded to a highly electronegative atom. B. Intermolecular H bonding is present in o-nitro phenol. C. Intramolecular H bonding is present in HF. D. The magnitude of H bonding depends on the physical state of the compound. E. H bonding has a powerful effect on the structure and properties of compounds. Choose the correct answer: (1) A only (2) A, D, E only (3) A, B, D only (4) A, B, C only |
(2) A, D, E only | A: True, as H bonding occurs with electronegative atoms like F, O, N. D: True, as the magnitude of H bonding varies with the compound's physical state. E: True, as H bonding affects properties like boiling point and structure. B and C are incorrect in the given context. |
| 65: In the above chemical reaction sequence, "A" and "B" respectively are: (1) O3, Zn/H2O and NaOH (alc)/I2 (2) H2O, H+ and NaOH (alc)/I2 (3) H2O, H+ and KMnO4 (4) O3, Zn/H2O and KMnO4 |
(1) | Step 1: Ozonolysis (O3, Zn/H2O) cleaves the double bond, forming compound A with aldehyde groups. Step 2: Haloform reaction (NaOH (alc), I2) converts compound A to compound B containing a carboxylate ion and a secondary alcohol. |
| 66: Common name of Benzene-1,2-diol is: (1) Quinol (2) Resorcinol (3) Catechol (4) o-Cresol |
(3) Catechol | The structure of Benzene-1,2-diol corresponds to catechol. It features hydroxyl groups at adjacent positions on a benzene ring. - 1,2-diol is catechol - 1,3-diol is resorcinol - 1,4-diol is quinol |
| 67: Consider the above reactions, identify product B and product C: (1) B = C = 2-Propanol (2) B = 2-Propanol, C = 1-Propanol (3) B = 1-Propanol, C = 2-Propanol (4) B = C = 1-Propanol |
(2) B = 2-Propanol, C = 1-Propanol | The reactions involve oxidation and reduction steps. Product B (2-Propanol) is formed as a secondary alcohol. Product C (1-Propanol) retains the primary alcohol structure. |
| 68: The adsorbent used in adsorption chromatography is/are: (1) Silica gel (2) Alumina (3) Quick lime (4) Magnesia |
(3) A and B only | Silica gel and alumina are common adsorbents used in chromatography. Silica gel is acidic and polar, while alumina is basic and polar, making them suitable for different types of adsorption. |
| 69: The reaction shown below leads to the formation of a major product "P": (1) Product A (2) Product B (3) Product C (4) Product D |
(2) Product B | The reaction involves elimination using alcoholic KOH. The most substituted alkene is formed as the major product through an E2 elimination mechanism. |
| 70: Correct order of stability of carbanions is shown below: (1) c > b > d > a (2) a > b > c > d (3) d > a > c > b (4) d > c > b > a |
(4) d > c > b > a | Stability order is determined by resonance and aromaticity. Compound (d) is aromatic, giving it maximum stability. Compound (a) is anti-aromatic, making it the least stable. |
| 71: The correct order of the first ionization enthalpy is: (1) Al > Ga > Tl (2) Ga > Al > B (3) B > Al > Ga (4) Tl > Ga > Al |
(4) Tl > Ga > Al | Due to the lanthanide contraction, Tl has the highest ionization enthalpy among these elements. Ga has a higher ionization enthalpy than Al due to scandide contraction, which increases the effective nuclear charge. |
| 72: If an iron (III) complex with the formula [Fe(NH3)x(CN)y]3+ has no electron in its eg orbital, then the value of x + y is: (1) 5 (2) 6 (3) 3 (4) 4 |
(2) 6 | In [Fe(NH3)x(CN)y]3+, Fe3+ (d5) is low-spin due to strong field ligands like CN. This results in all t2g orbitals being filled and eg orbitals empty. Here, x = 2 and y = 4, giving x + y = 6. |
| 73: A fuel cell, using hydrogen and oxygen as fuels: A. Has been used in spaceships B. Has an efficiency of 40% to produce electricity C. Uses aluminum as a catalyst D. Is eco-friendly E. Is a type of galvanic cell (1) A, B, C only (2) A, B, D only (3) A, B, D, E only (4) A, D, E only |
(4) A, D, E only | Fuel cells are used in spaceships and are eco-friendly, producing only water as a by-product. They are classified as galvanic cells. Aluminum is not used as a catalyst; instead, noble metals like platinum are commonly used. |
| 74: Choose the incorrect statement about Dalton’s Atomic Theory: (1) Compounds are formed when atoms of different elements combine in any ratio. (2) All the atoms of a given element have identical properties including identical mass. (3) Matter consists of indivisible atoms. (4) Chemical reactions involve reorganization of atoms. |
(1) Compounds are formed when atoms of different elements combine in any ratio | Dalton's Atomic Theory states that compounds are formed when atoms of different elements combine in fixed, simple, whole-number ratios, not in any ratio. This aligns with the law of definite proportions. |
| 75: Match List I with List II: List I: A. α-Glucose and α-Galactose B. α-Glucose and β-Glucose C. α-Glucose and α-Fructose D. α-Glucose and α-Ribose List II: I. Functional isomers II. Homologous III. Anomers IV. Epimers Choose the correct answer: (1) A-III, B-IV, C-II, D-I (2) A-III, B-IV, C-I, D-II (3) A-IV, B-III, C-I, D-II (4) A-IV, B-III, C-II, D-I |
(3) A-IV, B-III, C-I, D-II | - A: α-Glucose and α-Galactose differ in configuration at one carbon atom, making them epimers. - B: α-Glucose and β-Glucose differ at the anomeric carbon, making them anomers. - C: α-Glucose and α-Fructose have different functional groups, making them functional isomers. - D: α-Glucose and α-Ribose belong to different homologous series. |
| 76: Given below are two statements: Statement I: The correct order of first ionization enthalpy values of Li, Na, F, and Cl is Na < Li < Cl < F. Statement II: The correct order of negative electron gain enthalpy values of Li, Na, F, and Cl is Na < Li < F < Cl. Choose the correct answer: (1) Both Statement I and Statement II are true. (2) Both Statement I and Statement II are false. (3) Statement I is false but Statement II is true. (4) Statement I is true but Statement II is false. |
(1) Both Statement I and Statement II are true. | - Statement I is correct as F has the highest ionization energy followed by Cl, Li, and Na due to periodic trends. - Statement II is correct as Cl has a more negative electron gain enthalpy than F due to less inter-electronic repulsion. |
| 77: For a strong electrolyte, a plot of molar conductivity against √(concentration) is a straight line with a negative slope. The correct unit for the slope is: (1) S cm2 mol−3/2 L1/2 (2) S cm2 mol−1 L1/2 (3) S cm2 mol−3/2 L (4) S cm2 mol−3/2 L−1/2 |
(1) S cm2 mol−3/2 L1/2 | For molar conductivity (Λm) = Λ0m − A√C, the slope has units consistent with molar conductivity divided by √concentration, yielding S cm2 mol−3/2 L1/2. |
| 78: A first-row transition metal in its +2 oxidation state has a spin-only magnetic moment value of 3.86 BM. The atomic number of the metal is: (1) 25 (2) 26 (3) 22 (4) 23 |
(4) 23 | The spin-only magnetic moment (µs) is given by µs = √n(n + 2). For µs = 3.86 BM, solving gives n = 3 unpaired electrons. This corresponds to V (Vanadium, Z = 23) in its +2 state with configuration [Ar]3d3. |
| 79: The number of unpaired d-electrons in [Co(H2O)6]3+ is: (1) 4 (2) 2 (3) 0 (4) 1 |
(3) 0 | Co3+ has a d6 configuration. In [Co(H2O)6]3+, the octahedral ligand field of water induces pairing of all d-electrons in the t2g orbitals, leaving no unpaired electrons. |
| 80: The number of species from the following that have pyramidal geometry around the central atom is: S2O32−, SO42−, SO32−, S2O72− (1) 4 (2) 3 (3) 1 (4) 2 |
(3) 1 | Only S2O32− exhibits pyramidal geometry around one of the sulfur atoms due to the presence of a lone pair. Other species like SO42− and S2O72− have tetrahedral or planar structures. |
| 81: The maximum number of orbitals which can be identified with n = 4 and ml = 0 is: | 4 | For n = 4, the subshells are 4s, 4p, 4d, and 4f. Each subshell has one orbital with ml = 0, giving a total of 4 orbitals with ml = 0. |
| 82: The number of compounds/species from the following with non-zero dipole moment is: BeCl2, BCl3, NF3, XeF4, CCl4, H2O, H2S, HBr, CO2, H2, HCl |
5 | Compounds with non-zero dipole moments: NF3, H2O, H2S, HBr, HCl. Non-polar molecules: BeCl2, BCl3, XeF4, CCl4, CO2, H2. |
| 83: Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using a constant pressure of 5 atm. Heat exchange Q for the compression is: | 200 Lit. atm | For isothermal processes, Q = −W. Work done is W = P(V2 − V1). Substituting P = 5 atm, V1 = 60 L, and V2 = 20 L gives W = −200 Lit. atm. Hence, Q = 200 Lit. atm. |
| 84: From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ___ ×10−1 g. | 95 | From stoichiometry, 93 g of aniline gives 135 g of acetanilide. For 6.55 g of aniline: Amount of acetanilide = (135/93) × 6.55 = 9.5 g = 95 × 10−1 g. |
| 85: A first-order reaction has a rate constant (k) of 4.6×10−2 s−1. The time taken for the concentration of reactant to drop from 1 M to 0.1 M is: | 50 s | Using the first-order reaction formula t = (2.303/k) log([A]0/[A]): Substituting k = 4.6×10−2, [A]0 = 1 M, [A] = 0.1 M gives t = 50 s. |
| 86: Phthalimide is made to undergo the following sequence of reactions: (a) KOH (b) Benzyl chloride Phthalimide → Product 'P' Total number of π-bonds present in product 'P' is/are: |
6 | The reaction involves the conversion of phthalimide to its N-benzyl derivative using KOH and benzyl chloride. In product 'P', the aromatic rings and the double bond in the imide group account for a total of 6 π-bonds. |
| 87: The total number of 'sigma' and 'pi' bonds in 2-oxohex-4-ynoic acid is: | 18 sigma, 2 pi | 2-Oxohex-4-ynoic acid contains: - 6 single bonds (C-H) = 6 sigma bonds. - 5 single bonds (C-C, C-O, O-H) = 5 sigma bonds. - 1 triple bond (C≡C) = 1 sigma + 2 pi bonds. - 1 double bond (C=O) = 1 sigma + 1 pi bond. - Total = 18 sigma and 2 pi bonds. |
| 88: A first-row transition metal with the highest enthalpy of atomisation, upon reaction with oxygen at high temperature, forms oxides of formula M2On (where n = 3, 4, 5). The "spin-only" magnetic moment value of the amphoteric oxide from the above oxides is: | 5.92 BM | The metal with the highest enthalpy of atomisation is Chromium (Cr). It forms amphoteric oxide Cr2O3. Chromium in Cr2O3 is in the +3 oxidation state with a d3 configuration. The spin-only magnetic moment is calculated as µ = √n(n+2), where n = 3 unpaired electrons. Substituting n = 3 gives µ = 5.92 BM. |
| 89: 2.7 kg of each of water and acetic acid are mixed. The freezing point of the solution will be −x °C. Consider the acetic acid does not dimerise in water, nor dissociates in water. Given: - Molar mass of water = 18 g mol−1 - Molar mass of acetic acid = 60 g mol−1 - Kf for H2O = 1.86 K kg mol−1 - Freezing point of H2O = 273 K Calculate x. |
x = 8.37°C | The molality of acetic acid is: m = (2700 g / 60 g mol−1) / 2.7 kg = 15 mol/kg. The freezing point depression is: ΔT = Kf × m = 1.86 × 15 = 27.9 K. The freezing point of the solution is: T = 273 K − 27.9 K = −8.37°C. |
| 90: The vanillin compound obtained from vanilla beans has a total sum of oxygen atoms and π electrons of: | 16 | The molecular structure of vanillin contains: - 3 oxygen atoms. - 13 π electrons from aromatic rings and C=O bonds. The total sum is 3 + 13 = 16. |
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:
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| JEE Main 2024 Cutoff | JEE Main 2024 Marks vs Rank |
JEE Main 2024 4 April Shift 2 Chemistry Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Vedantu | Download PDF |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 4 April Shift 2 Chemistry Paper Analysis
JEE Main 2024 4 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 |
Read More:
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