JEE Main 2024 8 April Shift 2 Chemistry question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 8 April Shift 2 exam from 3 PM to 6 PM. The Chemistry question paper for JEE Main 2024 8 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 8 April Shift 2 exam is available for download using the link below.
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JEE Main 2024 8 April Shift 2 Chemistry Question Paper PDF Download
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JEE Main 8 Apr Shift 2 2024 Chemistry Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 61. In qualitative tests for the identification of the presence of phosphorus, the compound is heated with an oxidizing agent. It is then treated with nitric acid and ammonium molybdate, respectively. The yellow-colored precipitate obtained is: (1) Na₃PO₄ · 12MoO₃ (2) (NH₄)₃PO₄ · 12(NH₄)₂MoO₄ (3) (NH₄)₃PO₄ · 12MoO₃ (4) MoPO₄ · 21NH₄NO₃ |
(3) (NH₄)₃PO₄ · 12MoO₃ | In the qualitative analysis of phosphorus, the reaction with ammonium molybdate in the presence of nitric acid results in the formation of the yellow precipitate ammonium phosphomolybdate, (NH₄)₃PO₄ · 12MoO₃. |
| 62. For a reaction A → B → C, if the rate of formation of B is set to zero, then the concentration of B is given by: (1) k₁k₂[A] (2) (k₁ − k₂)[A] (3) (k₁ + k₂)[A] (4) (k₁/k₂)[A] |
(4) (k₁/k₂)[A] | The rate of formation of B is given by the equation d[B]/dt = k₁[A] − k₂[B]. Setting the rate to zero and solving for [B], we get the concentration of B as (k₁/k₂)[A]. |
| 63. When ψₐ and ψᵦ are the wave functions of atomic orbitals, then σ∗ is represented by: (1) ψₐ − 2ψᵦ (2) ψₐ − ψᵦ (3) ψₐ + 2ψᵦ (4) ψₐ + ψᵦ |
(2) ψₐ − ψᵦ | The anti-bonding molecular orbital (σ*) is formed by the destructive interference of atomic orbitals, represented by ψₐ − ψᵦ. |
| 64. Which one of the following compounds will readily react with dilute NaOH? (1) C₆H₅CH₂OH (2) C₂H₅OH (3) (CH₃)₃COH (4) C₆H₅OH |
(4) C₆H₅OH | Phenol (C₆H₅OH) reacts readily with dilute NaOH due to its higher acidity, forming the phenoxide ion (C₆H₅O⁻). |
| 65. The shape of a carbocation is: (1) Trigonal planar (2) Diagonal pyramidal (3) Tetrahedral (4) Diagonal |
(1) Trigonal planar | A carbocation has a central carbon with three bonded groups and an empty p-orbital, resulting in a trigonal planar geometry due to sp² hybridization. |
| 66. Given below are two statements: Statement I: SN2 reactions are ‘stereospecific,’ indicating that they result in the formation of only one stereoisomer as the product. Statement II: SN1 reactions generally result in the formation of products as racemic mixtures. In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is true but Statement II is false (2) Statement I is false but Statement II is true (3) Both Statement I and Statement II are true (4) Both Statement I and Statement II are false |
(3) Both Statement I and Statement II are true | SN2 reactions are stereospecific, leading to an inversion of configuration due to a backside attack. SN1 reactions proceed via a planar carbocation intermediate, resulting in racemic mixtures. Both statements are correct. |
| 67. Match List-I with List-II: List-I (Reaction) | List-II (Product) (A) Diazotization Reaction | (I) Diazonium Salt (B) Oxidation with Na2Cr2O7 | (II) Carboxylic Acid (C) Grignard Reaction | (III) Alcohol (D) Friedel-Crafts Alkylation | (IV) Alkyl Benzene Choose the correct answer from the options given below: (1) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (3) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) (4) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) |
(4) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) | The correct matches are: (A) Diazotization forms diazonium salts. (B) Oxidation with Na2Cr2O7 produces carboxylic acids. (C) Grignard reagents react to form alcohols. (D) Friedel-Crafts Alkylation produces alkyl benzenes. |
| 68. Match List-I with List-II: List-I (Test) | List-II (Identification) (A) Bayer’s Test | (I) Unsaturation (B) Ceric Ammonium Nitrate Test | (II) Alcoholic -OH group (C) Phthalein Dye Test | (III) Phenol (D) Schiff’s Test | (IV) Aldehyde Choose the correct answer from the options given below: (1) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (2) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (3) (A)-(IV), (B)-(I), (C)-(II), (D)-(III) (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) |
(4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) | The correct matches are: (A) Bayer’s Test identifies unsaturation. (B) Ceric Ammonium Nitrate Test identifies alcohols. (C) Phthalein Dye Test confirms phenols. (D) Schiff’s Test detects aldehydes. |
| 69. Identify the incorrect statements about Group 15 elements: (A) Dinitrogen is a diatomic gas that acts like an inert gas at room temperature. (B) The common oxidation states of these elements are -3, +3, and +5. (C) Nitrogen has a unique ability to form pπ-pπ multiple bonds. (D) The stability of +5 oxidation state increases down the group. (E) Nitrogen shows a maximum covalency of 6. Choose the correct answer from the options given below: (1) (A), (B), (D) only (2) (A), (C), (E) only (3) (B), (D), (E) only (4) (D) and (E) only |
(4) (D) and (E) only | The stability of the +5 oxidation state decreases down the group due to the inert pair effect. Nitrogen can have a maximum covalency of 4, not 6, as it cannot expand its octet. |
| 70. The IUPAC name of the following hydrocarbon is: (1) 2-Ethyl-3,6-dimethylheptane (2) 2-Ethyl-2,6-diethylheptane (3) 2,5,6-Trimethyloctane (4) 3,4,7-Trimethyloctane |
(3) 2,5,6-Trimethyloctane | The correct IUPAC name is 2,5,6-Trimethyloctane, as the substituents are given the lowest locants, and the parent chain contains eight carbons (octane). |
| 71. The equilibrium Cr2O72- ⇀↽ 2CrO42- is shifted to the right in: (1) an acidic medium (2) a basic medium (3) a weakly acidic medium (4) a neutral medium |
(2) a basic medium | In a basic medium, the hydroxide ions (OH-) consume H+, decreasing the H+ concentration, which shifts the equilibrium to the right, producing more CrO42-. |
| 72. Given below are two statements: Statement I: A buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities. Statement II: Blood is a naturally occurring buffer solution whose pH is maintained by H2CO3/HCO3- concentrations. Choose the correct answer from the options given below: (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 but Statement II is true | Statement I is incorrect because a buffer solution specifically requires a weak acid and its conjugate base (or a weak base and its conjugate acid). Statement II is correct as blood maintains its pH through the H2CO3/HCO3- buffer system. |
| 73. The correct sequence of acidic strength of the following aliphatic acids in their decreasing order is: CH3CH2COOH, CH3COOH, CH3CH2CH2COOH, HCOOH (1) HCOOH > CH3COOH > CH3CH2COOH > CH3CH2CH2COOH (2) HCOOH > CH3CH2CH2COOH > CH3CH2COOH > CH3COOH (3) CH3CH2CH2COOH > CH3CH2COOH > CH3COOH > HCOOH (4) CH3COOH > CH3CH2COOH > CH3CH2CH2COOH > HCOOH |
(1) HCOOH > CH3COOH > CH3CH2COOH > CH3CH2CH2COOH | Formic acid (HCOOH) is the strongest acid due to the absence of an electron-donating group. Acetic acid (CH3COOH) is weaker due to the methyl group, and the acidity decreases as the alkyl chain length increases. |
| 74. Given below are two statements: Statement I: Fusion of MnO2 with KOH and an oxidizing agent gives dark green K2MnO4. Statement II: Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion. Choose the correct answer from the options given below: (1) Both Statement I and Statement II are true (2) Both Statement I and Statement II are false (3) Statement I is true but Statement II is false (4) Statement I is false but Statement II is true |
(1) Both Statement I and Statement II are true | In alkaline medium, fusion of MnO2 with KOH and an oxidizing agent gives K2MnO4, which is dark green. The manganate ion (MnO42-) undergoes oxidation to permanganate (MnO4-) in further oxidation reactions. |
| 75. The emf of the cell: Tl |Tl+ (0.001M) ||Cu2+ (0.01M) |Cu is 0.83V at 298K. It could be increased by: (1) increasing concentration of Tl+ ions (2) increasing concentration of both Tl+ and Cu2+ ions (3) decreasing concentration of both Tl+ and Cu2+ ions (4) increasing concentration of Cu2+ ions |
(4) increasing concentration of Cu2+ ions | According to the Nernst equation, the emf can be increased by increasing the concentration of Cu2+ ions, as it shifts the reaction to the right and increases the emf. |
| 76. The correct statements about p-block elements and their compounds: (A) Non-metals have higher electronegativity than metals. (B) Non-metals have lower ionization enthalpy than metals. (C) Compounds formed between highly reactive non-metals and highly reactive metals are generally ionic. (D) The non-metal oxides are generally basic in nature. (E) The metal oxides are generally acidic or neutral in nature. Choose the correct answer: (1) (D) and (E) only (2) (A) and (C) only (3) (B) and (E) only (4) (B) and (D) only |
(2) (A) and (C) only | Statement (A) is true as non-metals are more electronegative. Statement (C) is true as compounds formed between highly reactive non-metals (like halogens) and metals are typically ionic. The other statements are false. |
| 77. Given below are two statements: Statement I: Kjeldahl method is applicable to estimate nitrogen in pyridine. Statement II: The nitrogen present in pyridine can easily be converted into ammonium sulfate in Kjeldahl method. In the light of the above statements, choose the correct answer from the options given below: (1) Both Statement I and Statement II are false (2) Statement I is false but Statement II is true (3) Both Statement I and Statement II are true (4) Statement I is true but Statement II is false |
(1) Both Statement I and Statement II are false | The Kjeldahl method cannot be used to estimate nitrogen in pyridine as it contains an aromatic nitrogen that is not easily converted to ammonium sulfate in this method. |
| 78. The reaction: 1/2 H2(g) + AgCl(s) → H+(aq) + Cl-(aq) + Ag(s) occurs in which of the following galvanic cells? (1) Pt|H2(g)|HCl(soln.)|AgCl(s)|Ag (2) Pt|H2(g)|HCl(soln.)|AgNO3(aq)|Ag (3) Pt|H2(g)|KCl(soln.)|AgCl(s)|Ag (4) Ag|AgCl(s)|KCl(soln.)|AgNO3(aq)|Ag |
(3) Pt|H2(g)|KCl(soln.)|AgCl(s)|Ag | The galvanic cell involving hydrogen gas (H2) at the anode and AgCl at the cathode, with KCl as the electrolyte, correctly represents the given reaction. |
| 79. Given below are two statements: Statement I: Fusion of MnO2 with KOH and an oxidizing agent gives dark green K2MnO4. Statement II: Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion. Choose the correct answer from the options given below: (1) Both Statement I and Statement II are true (2) Both Statement I and Statement II are false (3) Statement I is true but Statement II is false (4) Statement I is false but Statement II is true |
(1) Both Statement I and Statement II are true | Both statements are true. Fusion of MnO2 with KOH and an oxidizing agent forms K2MnO4 (dark green), and manganate ions are oxidized to permanganate ions under electrolytic conditions. |
| 80. Match List-I with List-II: List-I (Complex Ion) | List-II (Spin-only magnetic moment in B.M.) (A) [Cr(NH3)6]3+ | (I) 4.90 (B) [NiCl4]2- | (II) 3.87 (C) [CoF6]3- | (III) 0.0 (D) [Ni(CN)4]2- | (IV) 2.83 Choose the correct answer from the options given below: (1) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) (2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (4) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) |
(3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) | The correct matches are: (A) [Cr(NH3)6]3+ has a magnetic moment of 3.87 B.M. (II). (B) [NiCl4]2- has a magnetic moment of 2.83 B.M. (IV). (C) [CoF6]3- has a magnetic moment of 4.90 B.M. (I). (D) [Ni(CN)4]2- is diamagnetic with 0.0 B.M. (III). |
| 81. ∆vapH° for water is +40.49 kJ/mol at 1 bar and 100°C. The change in internal energy for this vaporization under the same condition is: | (1) 38 kJ/mol | The change in internal energy (∆vapU°) can be found using the equation ∆vapH° = ∆vapU° + ∆nRT. For water, ∆vapU° = 38 kJ/mol at 100°C. |
| 82. Number of molecules having bond order 2 from the following molecules is: C2, O2, Be2, Li2, Ne2, N2, H2 |
(2) 3 | C2, O2, and N2 have bond order 2. Bond order is calculated using the molecular orbital theory based on the number of bonding and anti-bonding electrons. |
| 83. Total number of optically active compounds from the following is: CH3CH(OH)CH(OH)CH3>, CH3CH2CH2OH, CH3CH2CH(Cl)CH3 |
(1) 1 | Among the given compounds, only CH3CH(OH)CH(OH)CH3 is optically active due to the presence of two chiral centers. |
| 84. The total number of carbon atoms present in tyrosine, an amino acid, is: | (1) 9 | Tyrosine contains a benzene ring (6 carbon atoms), a side chain (-CH2OH, 2 carbons), and the amino group (-NH2) attached to the alpha carbon (1 carbon), giving a total of 9 carbon atoms. |
| 85. Two moles of benzaldehyde and one mole of acetone under alkaline conditions using aqueous NaOH after heating gives x as the major product. The number of π-bonds in the product x is: | (2) 9 | The aldol condensation of benzaldehyde and acetone results in a conjugated system, which includes 3 π-bonds in the benzene ring, 2 π-bonds formed during the aldol reaction, and 1 π-bond from the carbonyl group, giving a total of 9 π-bonds. |
| 86. Total number of aromatic compounds among the following compounds is: | (1) C6H6, C7H8, C8H10, C6H5CH3 | The compounds C6H6 (benzene), C7H8 (toluene), C8H10 (ethylbenzene), and C6H5CH3 (toluene) are all aromatic, as they contain benzene rings. |
| 87. Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in the solution is x × 10-3. Value of x is: | 74 | Molality (m) = 4.44 m, which means 4.44 moles of urea in 1 kg of water. The mole fraction of urea is calculated using the formula: mole fraction = moles of urea / (moles of urea + moles of water). After calculation, the value of x is 74. |
| 88. Total number of unpaired electrons in the complex ion [Co(NH3)6]3+ and [NiCl4]2− is: | (2) 2 | In [Co(NH3)6]3+, Co has 6 d-electrons, and in [NiCl4]2−, Ni has 8 d-electrons. Both ions have 2 unpaired electrons in their respective electronic configurations. |
| 89. Wavenumber for a radiation having 5800 Å wavelength is x × 10-2 cm-1. The value of x is: | 1724 | Wavenumber (ν) is the reciprocal of the wavelength. Using the formula ν = 1/λ, where λ = 5800 Å = 5800 × 10-10 m, the wavenumber is calculated to be 1724 × 10-2 cm-1. |
| 90. A solution is prepared by adding 1 mole of ethyl alcohol in 9 moles of water. The mass percent of solute in the solution is: (Given: Molar mass in g mol-1: Ethyl alcohol = 46, Water = 18) |
22 | Mass of solute (ethyl alcohol) = 1 × 46 = 46 g Mass of solvent (water) = 9 × 18 = 162 g Total mass of the solution = 46 + 162 = 208 g Mass percent = (46 / 208) × 100 = 22% |
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 8 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 8 April Shift 2 Chemistry Paper Analysis
JEE Main 2024 8 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:
- JEE Main 2024 question paper pattern and marking scheme
- Most important chapters in JEE Mains 2024, Check chapter wise weightage here
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