JEE Main 2024 Jan 30 Shift 2 Chemistry question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 30 Shift 2 exam from 3 PM to 6 PM. The Chemistry question paper for JEE Main 2024 Jan 30 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 30 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Jan 30 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 30 Jan Shift 2 2024 Chemistry Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 61: Which among the following purification methods is based on the principle of “Solubility” in two different solvents? (1) Column Chromatography (2) Sublimation (3) Distillation (4) Differential Extraction |
(4) Differential Extraction | Differential extraction is based on the varying solubility of a compound in two immiscible solvents, allowing separation of components from a mixture. |
| 62: Salicylaldehyde is synthesized from phenol, when reacted with: (1) HCl, NaOH (2) CO₂, NaOH (3) CCl₃, NaOH (4) HCl, NaOH |
(1) HCl, NaOH | Salicylaldehyde is synthesized through the Reimer-Tiemann reaction where phenol reacts with chloroform (HCl) and NaOH to form the aldehyde group on the benzene ring. |
| 63: Given below are two statements: Statement I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow SN2 mechanism. Statement II: A secondary alkyl halide when treated with a large excess of ethanol follows SN1 mechanism. In the light of the above statements, choose the most appropriate 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 false. (4) Both Statement I and Statement II are true. |
(4) Both Statement I and Statement II are true | Statement I is true because SN2 mechanisms are favored by strong nucleophiles and unhindered secondary alkyl halides. Statement II is also true as SN1 mechanisms occur in polar protic solvents like ethanol, which stabilize the carbocation intermediate. |
| 64: m-Chlorobenzaldehyde on treatment with 50% KOH solution yields: (1) Benzyl alcohol (2) m-Chlorobenzoate and m-Chlorobenzyl alcohol (3) Benzoic acid (4) m-Chlorobenzoic acid |
(2) m-Chlorobenzoate and m-Chlorobenzyl alcohol | When m-Chlorobenzaldehyde is treated with 50% KOH, it undergoes a Cannizzaro reaction. This reaction involves disproportionation of the aldehyde, forming m-Chlorobenzoate ion and m-Chlorobenzyl alcohol as products. |
| 65: Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: H₂Te is more acidic than H₂S. Reason R: Bond dissociation enthalpy of H₂Te is lower than H₂S. In light of the above statements, choose the most appropriate from the options given below: (1) Both A and R are true but R is NOT the correct explanation of A. (2) Both A and R are true and R is the correct explanation of A. (3) A is false but R is true. (4) A is true but R is false. |
(2) Both A and R are true and R is the correct explanation of A. | H₂Te is more acidic than H₂S because the bond dissociation enthalpy of H₂Te is lower, making it easier to release H⁺ ions. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion. |
| 66: Product A and B formed in the following set of reactions are: (1) A = CH₄, B = CH₃OH (2) A = CH₃, B = CH₂OH (3) A = C₂H₆, B = C₂H₅OH (4) A = CH₂, B = CH₃OH |
(2) A = CH₃, B = CH₂OH | B₄H₆ undergoes hydrolysis and reduction in the presence of water and a base. The reaction leads to the cleavage of B-H bonds, forming an alcohol (CH₂OH) as product B and a simple hydrocarbon radical (CH₃) as product A. |
| 67: IUPAC name of the following compound is: CH₃—CH—CH₂—CN NH₂ (1) 2-Aminopentanitrile (2) 2-Aminobutanitrile (3) 3-Aminobutanenitrile (4) 3-Aminopropanenitrile |
(3) 3-Aminobutanenitrile | The compound consists of a four-carbon chain with a cyano group (-CN) at the terminal position and an amino group (-NH₂) at the third carbon. The correct IUPAC name is 3-Aminobutanenitrile. |
| 68: The products A and B formed in the following reaction scheme are respectively: 1. Benzene + HNO₃ + H₂SO₄ → A 2. A + Sn/HCl → B (1) A = C₆H₅NO₂, B = C₆H₅NH₂ (2) A = C₆H₆, B = C₆H₅NH₂ (3) A = C₆H₅NO₂, B = C₆H₆ (4) A = C₆H₅NH₂, B = C₆H₅NO₂ |
(1) A = C₆H₅NO₂, B = C₆H₅NH₂ | Benzene reacts with concentrated HNO₃ and H₂SO₄ to undergo nitration, forming nitrobenzene (A = C₆H₅NO₂). Reduction of nitrobenzene with Sn and HCl converts it to aniline (B = C₆H₅NH₂). |
| 69: The molecule/ion with square pyramidal shape is: (1) [Ni(CN)₆]²⁻ (2) PCl₅ (3) BrF₅ (4) PF₅ |
(3) BrF₅ | BrF₅ exhibits a square pyramidal geometry due to five bonding pairs and one lone pair around the central bromine atom, as predicted by the VSEPR theory. |
| 70: The orange color of K₂Cr₂O₇ and purple color of KMnO₄ is due to: (1) Charge transfer transition in both. (2) d → d transition in KMnO₄ and charge transfer transitions in K₂Cr₂O₇. (3) d → d transition in K₂Cr₂O₇ and charge transfer transitions in KMnO₄. (4) d → d transition in both. |
(1) Charge transfer transition in both | Both K₂Cr₂O₇ and KMnO₄ exhibit their respective colors due to charge transfer transitions involving the movement of electrons between metal and ligand orbitals. |
| 71: Alkaline oxidative fusion of MnO₂ gives “A” which on electrolytic oxidation in alkaline solution produces B. A and B respectively are: (1) Mn₂O₇ and MnO₄⁻ (2) MnO₂ and MnO₄⁻ (3) Mn₂O₇ and MnO₂⁻ (4) MnO₂⁻ and Mn₂O₇ |
(2) MnO₂ and MnO₄⁻ | Alkaline oxidative fusion of MnO₂ forms manganate ion (MnO₂⁻), and subsequent electrolytic oxidation produces permanganate ion (MnO₄⁻). This sequence explains the products A and B. |
| 72: If a substance ‘A’ dissolves in a solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as nₐ, nᵦ, and n????, the mole fraction of C in the solution is: (1) Option 1 (2) Option 2 (3) Option 3 (4) Option 4 |
(1) Option 1 | Mole fraction is defined as the number of moles of a component divided by the total number of moles in the solution. |
| 73: Given below are two statements: Statement I: Along the period, the chemical reactivity of the element gradually increases from group 1 to group 18. Statement II: The nature of oxides formed by group 1 elements is basic, while that of group 17 elements is acidic. In the light of the above statements, choose the most appropriate from the questions given below: (1) Both Statement I and Statement II are true. (2) Statement I is true but Statement II is false. (3) Statement I is false but Statement II is true. (4) Both Statement I and Statement II are false. |
(2) Statement I is true but Statement II is false | Statement I is true as chemical reactivity generally increases along the period due to the increasing non-metallic nature. Statement II is false because oxides of group 17 elements are not acidic; they are often neutral or slightly basic. |
| 74: The coordination geometry around the manganese in decacarbonylmanganese(0) is: (1) Octahedral (2) Trigonal bipyramidal (3) Square pyramidal (4) Square planar |
(1) Octahedral | The geometry of decacarbonylmanganese(0) is octahedral due to the symmetrical arrangement of ligands around the manganese atom, minimizing repulsion and maximizing stability. |
| 75: Given below are two statements: Statement I: Since fluorine is more electronegative than nitrogen, the net dipole moment of NF₃ is greater than NH₃. Statement II: In NH₃, the orbital dipole due to lone pair and the dipole moment of NH bonds are in opposite directions, but in NF₃, the orbital dipole due to lone pair and dipole moments of N − F bonds are in the same direction. In light of the above statements, choose the most appropriate from the options given below: (1) Statement I is true but Statement II is false. (2) Both Statement I and Statement II are true. (3) Both Statement I and Statement II are false. (4) Statement I is false but Statement II is true. |
(2) Both Statement I and Statement II are true | Both statements are true. Fluorine's higher electronegativity increases NF₃'s dipole moment compared to NH₃. The directions of dipoles in NF₃ and NH₃ align as described, leading to their respective dipole moment magnitudes. |
| 76: The correct stability order of carbocations is: (1) C₃⁺ > CH₃⁺ > (CH₂)₂CH⁺ > (CH₃)₂CH₂⁺ (2) CH₃⁺ > (CH₂)₂CH⁺ > (CH₃)₂CH₂⁺ > C₃⁺ (3) (CH₃)₂CH⁺ > (CH₂)₂CH⁺ > C₃⁺ > (CH₃)₂CH⁺ (4) (CH₃)₂CH₂⁺ > (CH₂)₂CH⁺ > CH₃⁺ > C₃⁺ |
(3) (CH₃)₂CH⁺ > (CH₂)₂CH⁺ > C₃⁺ > (CH₃)₂CH⁺ | Carbocation stability depends on hyperconjugation and inductive effects. Tertiary carbocations are more stable than secondary or primary carbocations due to greater alkyl group stabilization. |
| 77: The solution from the following with the highest depression in freezing point/lowest freezing point is: (1) 180 g of acetic acid dissolved in water (2) 180 g of acetic acid dissolved in benzene (3) 180 g of benzoic acid dissolved in benzene (4) 180 g of glucose dissolved in water |
(1) 180 g of acetic acid dissolved in water | Acetic acid undergoes partial ionization in water, leading to an increased number of particles in solution and a higher depression in freezing point compared to its dissolution in benzene or other solutes like glucose. |
| 78: A and B formed in the following reactions are: Cr₂O₇²⁻ + 4NaOH → Na₂Cr₂O₄ + 2NaCl + 2H₂O A + 2Cl₂ + 2H₂O → B + 3H₂O (1) A = Na₂Cr₂O₄, B = CrO₃ (2) A = Na₂Cr₂O₄, B = Cr₂O₇ (3) A = Na₂Cr₂O₄, B = NaCrO₄ (4) A = Na₂Cr₂O₄, B = Cr₃O₈ |
(1) A = Na₂Cr₂O₄, B = CrO₃ | Na₂Cr₂O₄ is formed by the reaction of Cr₂O₇²⁻ with NaOH. When A reacts with Cl₂ and H₂O, it forms CrO₃, which is the oxide of chromium. Thus, A = Na₂Cr₂O₄ and B = CrO₃. |
| 79: Choose the correct statements about the hydrides of group 15 elements: 1. A: The stability of the hydrides decreases in the order NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃. 2. B: The reducing ability of the hydrides increases in the order NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃. 3. C: Among the hydrides, NH₃ is a strong reducing agent while BiH₃ is a mild reducing agent. 4. D: The basicity of the hydrides increases in the order NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃. Choose the most appropriate from the options given below: (1) B and C only (2) C and D only (3) A and B only (4) A and D only |
(1) B and C only | The reducing ability of hydrides increases as we move down the group due to decreasing bond strength (B). NH₃ is a strong reducing agent due to its lone pair, while BiH₃ is mild (C). Statements A and D are incorrect. |
| 80: Reduction potential of ions are given below: ClO₃⁻: E° = 1.19 V IO₃⁻: E° = 1.65 V BrO₃⁻: E° = 1.74 V The correct order of their oxidising power is: (1) ClO₃⁻ > IO₃⁻ > BrO₃⁻ (2) BrO₃⁻ > IO₃⁻ > ClO₃⁻ (3) IO₃⁻ > ClO₃⁻ > BrO₃⁻ (4) IO₃⁻ > BrO₃⁻ > ClO₃⁻ |
(2) BrO₃⁻ > IO₃⁻ > ClO₃⁻ | Oxidising power is directly proportional to the reduction potential. The given potentials place BrO₃⁻ > IO₃⁻ > ClO₃⁻ in oxidising ability based on their standard reduction values. |
| 81: Number of complexes which show optical isomerism is: (1) cis-[Cr(ox)₂Cl₂]³⁻ (2) [Co(en)₃]³⁺ (3) cis-[Pt(en)₂Cl₂]²⁺ (4) trans-[Cr(ox)₂Cl₂]³⁻ |
(2) [Co(en)₃]³⁺ | [Co(en)₃]³⁺ exhibits optical isomerism due to its chiral nature. Complexes with cis geometry and bidentate ligands, like ethylenediamine, often show such isomerism. |
| 82: NO₂, required for a reaction, is produced by decomposition of N₂O₄ in CCl₄, as per the equation: 2N₂O₄ ⇌ 4NO₂ + O₂ The initial concentration of N₂O₄ is 3 mol L⁻¹ and it is 2.75 mol L⁻¹ after 30 minutes. The rate of formation of NO₂ is x × 10⁻³ mol L⁻¹ min⁻¹. The value of x is: |
17 | The change in concentration of N₂O₄ is 0.25 mol L⁻¹ over 30 minutes. Since 1 mol of N₂O₄ produces 2 mol of NO₂, the rate of formation is (2 × 0.25) / 30 = 0.0167 mol L⁻¹ min⁻¹, or 17 × 10⁻³. |
| 83: Two reactions are given below: 2Fe³⁺ + 3O₂(g) → Fe₂O₃(s), ∆Hf = −822 kJ/mol C(g) + ½O₂(g) → CO(g), ∆Hf = −110 kJ/mol Enthalpy change for the reaction: 3C(g) + Fe₂O₃(s) → 2Fe(g) + 3CO(g), ∆H = ? |
492 | Using Hess’s law, calculate ∆H: Reverse the first reaction and multiply the second reaction by 3. Adding these gives ∆H = 492 kJ/mol for the target reaction. |
| 84: The total number of correct statements regarding the nucleic acids is: A. RNA is regarded as the reserve of genetic information. B. DNA molecule self-duplicates during cell division. C. DNA synthesizes proteins in the cell. D. The message for the synthesis of particular proteins is present in DNA. E. Identical DNA strands are transferred to daughter cells. |
3 | Statements B, D, and E are correct. RNA is not the genetic reserve (A is false), and DNA does not directly synthesize proteins (C is false). Thus, there are 3 correct statements. |
| 85: The pH of an aqueous solution containing 1M benzoic acid (pKa = 4.20) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is mL. | 100 | Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). Substituting values, pH = 4.20 + log(200/100). The volume of benzoic acid is 100 mL. |
| 86: Number of geometrical isomers possible for the given structure is/are: | 4 | The structure supports four distinct geometrical isomers due to the presence of multiple double bonds or coordination centers, each allowing cis-trans configurations. |
| 87: Total number of species from the following which can undergo disproportionation reaction: H₂O₂, ClO₃⁻, P₄, Cl₂, Ag⁺, F₂, NO₂, K |
6 | Species like H₂O₂, Cl₂, NO₂, and others undergo disproportionation due to their ability to be both oxidized and reduced. A total of 6 species exhibit this property. |
| 88: Number of metal ions characterized by flame test among the following: Sr²⁺, Ba²⁺, Ca²⁺, Cu²⁺, Zn²⁺, Co²⁺, Fe²⁺ |
4 | Flame tests are characteristic for Sr²⁺, Ba²⁺, Ca²⁺, and Cu²⁺, as these ions produce distinct colors in a flame. Other ions like Zn²⁺ and Fe²⁺ do not exhibit this property. |
| 89: 2-chlorobutane + Cl₂ → C₄H₇Cl₃ (isomers) Total number of optically active isomers shown by C₄H₇Cl₃, obtained in the above reaction is: |
6 | The reaction generates multiple chlorinated products. Among these, six are optically active due to the presence of chiral centers formed during substitution. |
| 90: Number of spectral lines obtained in He⁺ spectra, when an electron makes a transition from the fifth excited state to the first excited state is: | 10 | The number of spectral lines is given by the formula n(n − 1)/2, where n is the initial principal quantum number. For n = 5, the number of lines is 5(5 − 1)/2 = 10. |
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JEE Main 2024 Jan 30 Shift 2 Chemistry Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
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JEE Main 2024 Jan 30 Shift 2 Chemistry Paper Analysis
JEE Main 2024 Jan 30 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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