JEE Main 2024 question paper pdf with solutions- Download Jan 29 Shift 2 Chemistry Question Paper pdf

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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 29 Jan Shift 2 2024 Chemistry Questions with Solutions

Question Answer Detailed Solution

61: The ascending acidity order of the following H atoms is:

(1) C < D < B < A
(2) A < B < C < D
(3) A < B < D < C
(4) D < C < B < A

(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:

List I (Bio Polymer) | List II (Monomer)
A. Starch   |   I. Nucleotide
B. Cellulose   |   II. α-glucose
C. Nucleic acid   |   III. β-glucose
D. Protein   |   IV. α-amino acid

(1) A-II, B-I, C-III, D-IV
(2) A-IV, B-II, C-I, D-III
(3) A-I, B-III, C-IV, D-II
(4) A-II, B-III, C-I, D-IV

(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).

63: Match List I with List II:

List I (Compound) | List II (pKa value)
A. Ethanol   |   II. 15.9
B. Phenol   |   I. 10.0
C. m-Nitrophenol   |   IV. 8.3
D. p-Nitrophenol   |   III. 7.1

(1) A-I, B-II, C-III, D-IV
(2) A-IV, B-I, C-II, D-III
(3) A-III, B-IV, C-I, D-II
(4) A-II, B-I, C-IV, D-III

(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).

64: Which of the following reaction is correct?

(1) Incorrect reaction
(2) Correct reaction
(3) Partially correct reaction
(4) Not a valid reaction

(2) Correct reaction The addition of HI occurs such that iodine attaches to the carbon with fewer hydrogen atoms, following Markovnikov’s rule.

65: According to IUPAC system, the compound is named as:

(1) Cyclohex-1-en-2-ol
(2) 1-Hydroxyhex-2-ene
(3) Cyclohex-1-en-3-ol
(4) Cyclohex-2-en-1-ol

(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:

(1) Potassium tetraoxopermanganate (VI)
(2) Potassium tetraoxidomanganate (VI)
(3) Dipotassium tetraoxidomanganate (VII)
(4) Potassium tetraoxidomanganese (VI)

(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:

(1) Sodium nitroprusside
(2) Neutral FeCl₃
(3) Meta-dinitrobenzene
(4) Dimethyl glyoxime

(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:

(1) Salicylic acid
(2) Benzene-1,2-diol
(3) Benzene-1,3-diol
(4) 2-Hydroxybenzaldehyde

(4) 2-Hydroxybenzaldehyde This reaction is known as the Reimer-Tiemann reaction. It produces 2-Hydroxybenzaldehyde (salicylaldehyde) as the major product.

69: Match List I with List II:

List I (Spectral Series for Hydrogen) | List II (Spectral Region)
A. Lyman   |   I. Infrared region
B. Balmer   |   II. UV region
C. Paschen   |   III. Infrared region
D. Pfund   |   IV. Visible region

(1) A-II, B-III, C-I, D-IV
(2) A-I, B-III, C-II, D-IV
(3) A-II, B-IV, C-III, D-I
(4) A-I, B-II, C-III, D-IV

(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.

70: On passing a gas, ‘X,’ through Nessler’s reagent, a brown precipitate is obtained. The gas ‘X’ is:

(1) H₂S
(2) CO₂
(3) NH₃
(4) Cl₂

(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.

72: Identify the reagents used for the following conversion:

(1) A = LiAlH₄, B = NaOH(aq), C = NH₂NH₂/KOH
(2) A = LiAlH₄, B = NaOH(alc), C = Zn/HCl
(3) A = DIBAL-H, B = NaOH(aq), C = NH₂NH₂/KOH
(4) A = DIBAL-H, B = NaOH(alc), C = Zn/HCl

(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.

73: Which of the following acts as a strong reducing agent? (Atomic numbers: Ce = 58, Eu = 63, Gd = 64, Lu = 71)

(1) Lu³⁺
(2) Gd³⁺
(3) Eu²⁺
(4) Ce⁴⁺

(3) Eu²⁺ Eu²⁺ acts as a strong reducing agent as it readily undergoes oxidation to Eu³⁺, achieving a stable half-filled 4f⁶ configuration.

74: Chromatographic technique(s) based on the principle of differential adsorption is/are:
A. Column chromatography
B. Thin layer chromatography
C. Paper chromatography


(1) B only
(2) A only
(3) A & B only
(4) C only

(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?
A. They exhibit high enthalpy of atomization as the d-subshell is full.
B. Zn and Cd do not show variable oxidation state while Hg shows +1 and +2.
C. Compounds of Zn, Cd, and Hg are paramagnetic in nature.
D. Zn, Cd, and Hg are called soft metals.


(1) B, D only
(2) B, C only
(3) A, D only
(4) C, D only

(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.

76: The element having the highest first ionization enthalpy is:

(1) Si
(2) Al
(3) N
(4) C

(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.

77: Alkyl halide is converted into alkyl isocyanide by reaction with:

(1) NaCN
(2) NH₄CN
(3) KCN
(4) AgCN

(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.

78: Which one of the following will show geometrical isomerism?

(1) CH₃CH=CHCH₃
(2) CH₂=CBr₂
(3) CH₃CH=CHBr
(4) Cyclohexane

(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:
Statement I: Fluorine has the most negative electron gain enthalpy in its group.
Statement II: Oxygen has the least negative electron gain enthalpy in its group.
In the light of the above statements, choose the most appropriate option from the options given below:


(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true

(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.

80: Anomalous behavior of oxygen is due to its:

(1) Large size and high electronegativity
(2) Small size and low electronegativity
(3) Small size and high electronegativity
(4) Large size and low electronegativity

(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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.


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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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JEE Main Previous Year Question Paper

JEE Main Questions

  • 1.
    A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is ________________________.


      • 2.
        When sec-butylcyclohexane reacts with bromine in the presence of sunlight, the major product is:

          • sec-butylcyclohexane
          • sec-butylcyclohexane
          • sec-butylcyclohexane
          • sec-butylcyclohexane

        • 3.
          The total number of hydrogen bonds of a DNA-double Helix strand whose one strand has the following sequence of bases is _______ . $5^{\prime}-\mathrm{G}-\mathrm{G}-\mathrm{C}-\mathrm{A}-\mathrm{A}-\mathrm{A}-\mathrm{T}-\mathrm{C}-\mathrm{G}-\mathrm{G}-\mathrm{C}-\mathrm{T}-\mathrm{A}-3^{\prime}$


            • 4.
              The reaction \( {A}_2 + {B}_2 \to 2{AB} \) follows the mechanism: \[ {A}_2 \xrightarrow{k_1} {A} + {A} \ ({fast}) \quad {A} + {B}_2 \xrightarrow{k_2} {AB} + {B} \ ({slow}) \quad {A} + {B} \to {AB} \ ({fast}) \] The overall order of the reaction is:

                • 3
                • 1.5
                • 2.5
                • 2

              • 5.
                Total number of nucleophiles from the following is: $NH_3$, $PhSH$, $(H_3C_2S)_2$, $H_2C=CH_2$, $OH^-$, $H_3O^+$, $(CH_3)_2CO$, $NCH_3$

                  • 6
                  • 5
                  • 4
                  • 7

                • 6.

                  Given below are two statements: 

                  Statement I: In the oxalic acid vs KMnO$_4$ (in the presence of dil H$_2$SO$_4$) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO$_4$ titration (in the presence of dil H$_2$SO$_4$). 

                  Statement II: In oxalic acid vs KMnO$_4$ titration, the initial formation of MnSO$_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMnO$_4$, heating oxidizes Fe$^{2+}$ into Fe$^{3+}$ by oxygen of air and error may be introduced in the experiment. 

                  In the light of the above statements, choose the correct answer from the options given below:

                    • Both Statement I and Statement II are true
                    • Both Statement I and Statement II are false
                    • Statement I is false but Statement II is true
                    • Statement I is true but Statement II is false

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