JEE Main 2024 Jan 29 Shift 2 Chemistry question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 29 Shift 2 exam from 3 PM to 6 PM. The Chemistry question paper for JEE Main 2024 Jan 29 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 Jan 29 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Jan 29 Shift 2 Chemistry Question Paper PDF Download
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JEE Main 29 Jan Shift 2 2024 Chemistry Questions with Solutions
| Question | Answer | Detailed Solution |
|---|---|---|
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61: The ascending acidity order of the following H atoms is: |
(1) C < D < B < A | The acidity of H atoms depends on the stability of the conjugate base after deprotonation. The order of stability is: C < D < B < A. |
|
62: Match List I with List II: |
(4) A-II, B-III, C-I, D-IV | The correct matching is based on the composition of bio-polymers: Starch (α-glucose), Cellulose (β-glucose), Nucleic acid (nucleotide), and Protein (α-amino acid). |
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63: Match List I with List II: |
(4) A-II, B-I, C-IV, D-III | The pKa values correspond as follows: Ethanol (15.9), Phenol (10.0), m-Nitrophenol (8.3), p-Nitrophenol (7.1). |
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64: Which of the following reaction is correct? |
(2) Correct reaction | The addition of HI occurs such that iodine attaches to the carbon with fewer hydrogen atoms, following Markovnikov’s rule. |
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65: According to IUPAC system, the compound is named as: |
(4) Cyclohex-2-en-1-ol | The compound contains a double bond and an alcohol group. Numbering starts from the alcohol group to give it the lowest locant. |
|
66: The correct IUPAC name of K₂MnO₄ is: |
(2) Potassium tetraoxidomanganate (VI) | The oxidation state of manganese in K₂MnO₄ is +6. Based on IUPAC rules, the correct name is Potassium tetraoxidomanganate (VI). |
|
67: A reagent which gives a brilliant red precipitate with Nickel ions in a basic medium is: |
(4) Dimethyl glyoxime | Dimethyl glyoxime reacts with Nickel ions (Ni²⁺) in a basic medium to form a characteristic red precipitate of [Ni(dmg)₂]. |
|
68: Phenol treated with chloroform in the presence of sodium hydroxide, followed by hydrolysis with acid, results in: |
(4) 2-Hydroxybenzaldehyde | This reaction is known as the Reimer-Tiemann reaction. It produces 2-Hydroxybenzaldehyde (salicylaldehyde) as the major product. |
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69: Match List I with List II: |
(3) A-II, B-IV, C-III, D-I | The Lyman series lies in the UV region, Balmer in the visible region, and Paschen and Pfund in the infrared region. |
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70: On passing a gas, ‘X,’ through Nessler’s reagent, a brown precipitate is obtained. The gas ‘X’ is: |
(3) NH₃ | Nessler’s reagent reacts with ammonia (NH₃) to form a brown precipitate of Millon’s base. |
|
71: The product A formed in the following reaction is: |
(3) Chlorobenzene | The reaction involves the formation of a diazonium salt followed by the Sandmeyer reaction. The intermediate diazonium salt reacts with Cu₂Cl₂ to produce Chlorobenzene as the final product. |
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72: Identify the reagents used for the following conversion: |
(4) A = DIBAL-H, B = NaOH(alc), C = Zn/HCl | DIBAL-H selectively reduces esters to aldehydes. NaOH(alc) facilitates aldol condensation, and Zn/HCl performs Clemmensen reduction to produce hydrocarbons. |
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73: Which of the following acts as a strong reducing agent? (Atomic numbers: Ce = 58, Eu = 63, Gd = 64, Lu = 71) |
(3) Eu²⁺ | Eu²⁺ acts as a strong reducing agent as it readily undergoes oxidation to Eu³⁺, achieving a stable half-filled 4f⁶ configuration. |
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74: Chromatographic technique(s) based on the principle of differential adsorption is/are: |
(3) A & B only | Column chromatography and Thin Layer Chromatography (TLC) operate on differential adsorption. Paper chromatography, however, relies on partition rather than adsorption. |
|
75: Which of the following statements are correct about Zn, Cd, and Hg? |
(1) B, D only | Statements B and D are correct: Zn and Cd exhibit fixed oxidation states (+2), while Hg can show +1 and +2. All three are classified as soft metals. |
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76: The element having the highest first ionization enthalpy is: |
(3) N | The ionization enthalpy follows the trend: Al < Si < C < N. Nitrogen has the highest ionization enthalpy due to its half-filled 2p³ configuration, which is particularly stable. |
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77: Alkyl halide is converted into alkyl isocyanide by reaction with: |
(4) AgCN | Alkyl halides react with AgCN to form alkyl isocyanides (R-NC) due to the covalent nature of AgCN, which favors bonding through nitrogen. |
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78: Which one of the following will show geometrical isomerism? |
(3) CH₃CH=CHBr | Geometrical isomerism arises due to restricted rotation around a double bond. CH₃CH=CHBr has different groups attached to each carbon of the double bond, enabling geometrical isomerism. |
|
79: Given below are two statements: |
(4) Statement I is false but Statement II is true | Statement I is false because chlorine, not fluorine, has the most negative electron gain enthalpy in Group 17 due to lower repulsion in larger orbitals. Statement II is true because oxygen has the least negative electron gain enthalpy in Group 16 due to electron repulsion in its compact size. |
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80: Anomalous behavior of oxygen is due to its: |
(3) Small size and high electronegativity | Oxygen exhibits anomalous behavior in Group 16 due to its small atomic size and high electronegativity, leading to unique properties like hydrogen bonding and oxide formation. |
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81: The total number of antibonding molecular orbitals formed from 2s and 2p atomic orbitals in a diatomic molecule is: |
4 | The antibonding molecular orbitals in diatomic molecules include 1 from 2s and 3 from 2p. Therefore, the total number of antibonding orbitals is 4. |
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82: The oxidation number of iron in the compound formed during the brown ring test for NO₃⁻ ion is: |
+1 | In the compound [Fe(H₂O)₅(NO)]²⁺ formed during the brown ring test, the oxidation number of iron is +1, calculated as x + 0 + (+1) = +2. |
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83: The equilibrium constant for the formation of NH₃ from N₂ and H₂, given [N₂] = 2×10⁻² M, [H₂] = 3×10⁻² M, [NH₃] = 1.5×10⁻² M at 500 K, is: |
417 | Using the formula \( K_c = \frac{[NH₃]²}{[N₂][H₂]³} \) and substituting the given values, the equilibrium constant \( K_c \) is calculated as 417. |
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84: The molality of a 0.8 M H₂SO₄ solution with density 1.06 g/cm³ is: |
815 × 10⁻³ m | Given the density and molarity, the mass of 1L solution is 1060 g, and the mass of solute is 78.4 g. Using the molality formula, the calculated molality is 815 × 10⁻³ m. |
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85: The amount of NaOH in 50 mL of a solution neutralized by 50 mL of 0.5 M oxalic acid is: |
4 g | Using the equation \( M_1V_1 = M_2V_2 \) and given that oxalic acid neutralizes NaOH with an equivalent weight calculation, the mass of NaOH is 4 g. |
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86: The total number of sigma (σ) and pi (π) bonds in 2-formylhex-4-enoic acid is: |
22 | The structure of 2-formylhex-4-enoic acid has 16 sigma (σ) bonds and 6 pi (π) bonds. Summing these, the total number of bonds is 22. |
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87: The half-life of radioisotopic bromine-82 is 36 hours. The fraction that remains after one day is: |
0.63 | Using the first-order decay equation, the fraction remaining is calculated as 0.63 after 24 hours, given a half-life of 36 hours. |
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88: Standard enthalpy of vaporization for CCl₄ is 30.5 kJ/mol. Heat required for vaporization of 284 g of CCl₄ at constant temperature is: |
56 kJ | The molar mass of CCl₄ is 154 g/mol. Calculating the moles and multiplying by the enthalpy of vaporization, the total heat required is 56 kJ. |
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89: A constant current was passed through a solution of AuCl₄⁻ ions between gold electrodes. After 10 minutes, the increase in the cathode’s mass was 1.314 g. The total charge passed through the solution is: |
2 F | Using Faraday’s laws of electrolysis, the charge is calculated as 2 Faraday, based on the given mass of deposited gold and its equivalent weight. |
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90: The total number of molecules with zero dipole moment among CH₄, BF₃, H₂O, HF, NH₃, CO₂, and SO₂ is: |
3 | Molecules with zero dipole moment are CH₄, BF₃, and CO₂, as they are symmetrical and have dipole moments that cancel out. Other molecules are asymmetrical and have non-zero dipole moments. |
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 Jan 29 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 | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 Jan 29 Shift 2 Chemistry Paper Analysis
JEE Main 2024 Jan 29 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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