JEE Main 2024 question paper pdf with solutions- Download Jan 30 Shift 1 Chemistry Question Paper pdf

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JEE Main 2024 Jan 30 Shift 1 Chemistry question paper with solutions and answers pdf is available here . NTA conducted JEE Main 2024 Jan 30 Shift 1 exam from 9 AM to 12 PM. The Chemistry question paper for JEE Main 2024 Jan 30 Shift 1 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 1 exam is available for download using the link below.

JEE Main 2024 Jan 30 Shift 1 Chemistry Question Paper PDF Download

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JEE Main 2024 Jan 30 Shift 1 Chemistry Questions with Solutions

Question	Answer	Solution
61. Given below are two statements: Statement-I: The gas liberated on warming a salt with dilute H₂SO₄, turns a piece of paper dipped in lead acetate into black; it is a confirmatory test for sulphide ion. Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphide. Choose the most appropriate answer from the options below: (1) Both Statement-I and Statement-II are false (2) Statement-I is false but Statement-II is true (3) Statement-I is true but Statement-II is false (4) Both Statement-I and Statement-II are true	(3) Statement-I is true but Statement-II is false	The gas liberated is H₂S, which reacts with lead acetate paper to form black PbS. Statement-I is true. However, the blackening is due to lead sulphide (PbS), not lead sulphite, making Statement-II false.
62. This reduction reaction is known as: (1) Rosenmund reduction (2) Wolff-Kishner reduction (3) Stephen reduction (4) Etard reduction	(1) Rosenmund reduction	The reduction of acid chlorides to aldehydes in the presence of hydrogen and palladium on barium sulfate is called the Rosenmund reduction.
63. Sugar which does not give reddish brown precipitate with Fehling’s reagent is: (1) Sucrose (2) Lactose (3) Glucose (4) Maltose	(1) Sucrose	Sucrose is a non-reducing sugar as it lacks a free aldehyde or ketone group. It does not react with Fehling’s reagent, unlike other sugars that give a reddish-brown precipitate.
64. Given below are the two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A): There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. Reason (R): Covalent and ionic radii in a particular oxidation state increase down the group. Choose the most appropriate answer from the options below: (1) (A) is false but (R) is true (2) Both (A) and (R) are true but (R) is not the correct explanation of (A) (3) (A) is true but (R) is false (4) Both (A) and (R) are true and (R) is the correct explanation of (A)	(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)	Both statements are true, but the small increase in covalent radius from As to Bi is due to poor shielding by d- and f-electrons, not the general trend described in Reason (R).
65. Which of the following molecule/species is most stable? (1) Aromatic compound (2) Anti-aromatic compound (3) Non-aromatic compound (4) All are equally stable	(1) Aromatic compound	Aromatic compounds are the most stable due to their delocalized π-electrons, which follow Hückel’s rule (4n+2 π-electrons).
66. Diamagnetic Lanthanoid ions are: (1) Nd³⁺ and Eu³⁺ (2) La³⁺ and Ce⁴⁺ (3) Nd³⁺ and Ce⁴⁺ (4) Lu³⁺ and Eu³⁺	(2) La³⁺ and Ce⁴⁺	Diamagnetic ions have all electrons paired. La³⁺ and Ce⁴⁺ have no unpaired electrons, making them diamagnetic.
67. Aluminium chloride in acidified aqueous solution forms an ion having geometry: (1) Octahedral (2) Square Planar (3) Tetrahedral (4) Trigonal bipyramidal	(1) Octahedral	In acidified aqueous solution, AlCl₃ forms the complex [Al(H₂O)₆]³⁺, which has an octahedral geometry.
68. Given below are two statements: Statement-I: The orbitals having the same energy are called as degenerate orbitals. Statement-II: In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals. Choose the most appropriate answer from the options 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	(1) Statement-I is true but Statement-II is false	In a hydrogen atom, all orbitals with the same principal quantum number (such as 3s, 3p, and 3d) are degenerate. Therefore, Statement-I is correct and Statement-II is incorrect.
69. Example of vinylic halide is: (1) CH₂=CH–Cl (2) CH₃–CH₂–CH₂Cl (3) C₆H₅–CH₂Cl (4) CH₃–C≡CCl	(1) CH₂=CH–Cl	A vinylic halide has a halogen attached to a carbon atom that is part of a double bond (sp² hybridized). Option (1) is a vinylic halide because the halogen is attached to a sp² carbon in a C=C double bond.
70. Structure of 4-Methylpent-2-enal is: (1) H₂C = C - CH₂ - C = C - H (2) CH₃ - CH₂ - C = CH - C = CH (3) CH₃ - CH₂ - CH = C - CH₃ (4) CH₃ - CH = CH - C = C - H	(4) CH₃ - CH = CH - C = C - H	The IUPAC name "4-Methylpent-2-enal" indicates a five-carbon chain, a double bond at position 2, an aldehyde group at position 1, and a methyl group at position 4. Option (4) matches this structure.
71. Match List-I with List-II: List-I (Molecule) \| List-II (Shape) (A) BrF₅ \| (I) T-shape (B) H₂O \| (II) See-saw (C) ClF₃ \| (III) Bent (D) SF₄ \| (IV) Square pyramidal Choose the correct answer from the options below: (1) (A) - I, (B) - III, (C) - IV, (D) - II (2) (A) - II, (B) - I, (C) - III, (D) - IV (3) (A) - III, (B) - IV, (C) - I, (D) - II (4) (A) - IV, (B) - III, (C) - I, (D) - II	(4) (A) - IV, (B) - III, (C) - I, (D) - II	Using VSEPR theory, the molecular shapes are determined based on the number of bonding and lone pairs of electrons around the central atom. - BrF₅: Square pyramidal (IV) - H₂O: Bent (III) - ClF₃: T-shape (I) - SF₄: See-saw (II)
72. The final product A, formed in the following multistep reaction sequence is: CH₃ - C≡CH + Na → A B → CH₃ - C≡C - CH₂ - CH₂ - CH₃ + NaBr Choose the correct answer from the options below: (1) A = CH₃–C≡CNa, B = CH₃–CH₂–CH₂–Br (2) A = CH₃–CH₂–CH₂Br, B = CH₃–C≡C–CH₃ (3) A = CH₃–C≡CNa, B = CH₃–C≡CH (4) A = CH₃–C≡CNa, B = CH₃–CH₂–CH₃	(1) A = CH₃–C≡CNa, B = CH₃–CH₂–CH₂–Br	In the first step, sodium acetylide (CH₃–C≡CNa) is formed. Then, it reacts with 1-bromopropane (CH₃–CH₂–CH₂Br) to give the final product CH₃–C≡C–CH₂–CH₂–CH₃.
73. In the given reactions, identify the reagent A and reagent B: CH₃ - C≡CH + Na → A B → CH₃ - C≡C - CH₂ - CH₂ - CH₃ + NaBr (1) A-CrO₃, B-CrO₃ (2) A-CrO₃, B-CrO₂Cl₂ (3) A-CrO₂Cl₂, B-CrO₂Cl₂ (4) A-CrO₂Cl₂, B-CrO₃	(2) A-CrO₃, B-CrO₂Cl₂	In the first step, sodium acetylide (CH₃–C≡CNa) is formed when the alkyne reacts with Na. In the second step, CrO₃ (Chromium trioxide) is used to oxidize the acetylide to an aldehyde. In the second reaction, CrO₂Cl₂ is used for further functionalization, leading to the final product CH₃–C≡C–CH₂–CH₂–CH₃.
74. Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A): CH₂=CH−CH₂−Cl is an example of allyl halide. Reason (R): Allyl halides are the compounds in which the halogen atom is attached to an sp² hybridised carbon atom. Choose the most appropriate answer from the options below: (1) (A) is true but (R) is false (2) Both (A) and (R) are true but (R) is not the correct explanation of (A) (3) (A) is false but (R) is true (4) Both (A) and (R) are true and (R) is the correct explanation of (A)	(1) (A) is true but (R) is false	CH₂=CH−CH₂−Cl is an allyl halide because the halogen is attached to the carbon adjacent to the C=C double bond (sp³ hybridized). However, the definition in Reason (R) is incorrect because allyl halides have the halogen attached to an sp³ carbon, not sp².
75. What happens to the freezing point of benzene when a small quantity of naphthalene is added to benzene? (1) Increases (2) Remains unchanged (3) First decreases and then increases (4) Decreases	(4) Decreases	When a non-volatile solute like naphthalene is added to benzene, it causes a depression in the freezing point due to the lowering of the vapor pressure.
76. Match List-I with List-II: List-I (Species) \| List-II (Electronic Distribution) (A) Cr²⁺ \| (I) 3d⁸ (B) Mn⁺ \| (II) 3d⁵4s¹ (C) Ni²⁺ \| (III) 3d⁴ (D) V⁺ \| (IV) 3d³4s¹ Choose the correct answer from the options below: (1) (A) - I, (B) - II, (C) - III, (D) - IV (2) (A) - III, (B) - IV, (C) - I, (D) - II (3) (A) - IV, (B) - III, (C) - I, (D) - II (4) (A) - II, (B) - I, (C) - IV, (D) - III	(1) (A) - I, (B) - II, (C) - III, (D) - IV	The electronic configurations of the ions are as follows: - Cr²⁺: 3d⁴ (I) - Mn⁺: 3d⁵4s¹ (II) - Ni²⁺: 3d⁸ (III) - V⁺: 3d³4s¹ (IV)
77. Compound A formed in the following reaction reacts with B, giving the product C. CH₃−C≡CH + Na → A B → CH₃−C≡C−CH₂−CH₂−CH₃ + NaBr Find out A and B. (1) A = CH3–C≡CNa, B = CH3–CH2–CH2–Br (2) A = CH3–CH2–CH2Br, B = CH3–C≡C–CH3 (3) A = CH3–C≡CNa, B = CH3–C≡CH (4) A = CH3–C≡CNa, B = CH3–CH2–CH3	A = CH₃–C≡CNa, B = CH₃–CH₂–CH₂–Br	In the first reaction, sodium acetylide (CH₃–C≡CNa) is formed when the alkyne reacts with sodium metal. In the second reaction, the acetylide reacts with 1-bromopropane (CH₃–CH₂–CH₂Br) to give the final product, CH₃–C≡C–CH₂–CH₂–CH₃.
78. Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B). A: NaNO₂ + HCl, 0–5°C B: Phenol Find out A and B.	A = NaNO₂ + HCl, B = Phenol	This is the diazotization reaction, where primary aromatic amines react with sodium nitrite (NaNO₂) in the presence of HCl to form diazonium salts. The diazonium salt then couples with phenol to form a red azo dye.
79. The Lassaigne’s extract is boiled with dilute HNO₃ before testing for halogens because: (1) AgCN is soluble in HNO₃ (2) Silver halides are soluble in HNO₃ (3) Ag₂S is soluble in HNO₃ (4) Na₂S and NaCN are decomposed by HNO₃	(4) Na₂S and NaCN are decomposed by HNO₃	The Lassaigne’s test is used for detecting halogens, nitrogen, and sulfur. The extract is treated with dilute nitric acid (HNO₃) to decompose sodium cyanide (NaCN) and sodium sulfide (Na₂S), which might otherwise interfere with the halogen test.
80. Choose the correct statements from the following: (A) Ethane-1,2-diamine is a chelating ligand. (B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite. (C) Cyanide ion is used as ligand for leaching of silver. (D) Phosphine acts as a ligand in Wilkinson catalyst. (E) The stability constants of Ca²⁺ and Mg²⁺ are similar with EDTA complexes. Choose the correct answer from the options below: (1) (B), (C), (E) only (2) (A), (B), (C) only (3) (B), (C), (D), (E) only (4) (A), (B), (C), (D), (E)	(3) (B), (C), (D), (E) only	- (B) Metallic aluminium is produced by electrolyzing alumina in the presence of cryolite. - (C) Cyanide ion forms complexes with silver and is used in the leaching process. - (D) Phosphine acts as a ligand in the Wilkinson catalyst (used in hydrogenation reactions). - (E) The stability constants of Ca²⁺ and Mg²⁺ with EDTA are quite similar due to their similar ionic sizes.
81. The rate of a first-order reaction is 0.04 mol·L⁻¹·s⁻¹ at 10 minutes and 0.03 mol·L⁻¹·s⁻¹ at 20 minutes after initiation. The half-life of the reaction is:	24 minutes	For a first-order reaction, using the integrated rate law and the given rate data at two different times, we can find the rate constant k. Then, the half-life (t_1/2) is calculated using t_1/2 = 0.693/k. Substituting the values, the half-life is found to be 24 minutes.
82. The pH at which Mg(OH)₂ [Ksp = 1 × 10⁻¹¹] begins to precipitate from a solution containing 0.10 M Mg²⁺ ions is:	9	Precipitation occurs when the ion product (Q) exceeds the solubility product (Ksp). Using the solubility product expression for Mg(OH)₂, we calculate the OH⁻ concentration required for precipitation. From this, the pH is calculated as 9.
83. An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A → B → C → A as shown in the diagram. The total work done in the process is:	200 J	The total work done by the gas in a cyclic process is equal to the area enclosed by the path on the P-V diagram. The area is calculated to be 200 J.
84. If the IUPAC name of an element is "Unununnium", then the element belongs to nth group of the periodic table. The value of n is:	11	"Unununnium" corresponds to element 111, which belongs to Group 11 of the periodic table.
85. The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule is:	8	From the 2s and 2p atomic orbitals, we form 8 molecular orbitals in total: 2 from 2s (σ2s, σ2s) and 6 from 2p (σ2pz, σ2pz, π2px, π2py, π2px, π2py).
86. On a thin layer chromatographic plate, an organic compound moved by 3.5 cm, while the solvent moved by 5 cm. The retardation factor of the organic compound is ×10⁻¹:	7	The retardation factor (R_f) is calculated as the ratio of the distance traveled by the compound to the distance traveled by the solvent: R_f = (3.5 cm) / (5 cm) = 0.7 = 7 × 10⁻¹.
87. The compound formed by the reaction of ethanol with semicarbazide contains number of nitrogen atoms.	3	The reaction between ethanol and semicarbazide forms an imine (semicarbazone), which contains three nitrogen atoms: two from the semicarbazide group and one from the imine bond.
88. A 0.05 cm thick coating of silver is deposited on a plate of 0.05 m² area. The number of silver atoms deposited on the plate is ×10²³ (At. mass Ag = 108, d = 7.9 g/cm³):	11	Area into cm²: 0.05 m² = 0.05 × (100 cm × 100 cm) = 0.05 × 10,000 cm² = 500 cm². Volume = Area × Thickness = 500 cm² × 0.05 cm = 25 cm³. Mass = Density × Volume = 7.9 g/cm³ × 25 cm³ = 197.5 g. Moles = Mass / Atomic mass = 197.5 g / 108 g/mol ≈ 1.83 mol. Number of atoms = Moles × Avogadro's number ≈ 1.83 × 6.022 × 10²³ ≈ 1.1 × 10²⁴ atoms. This can be expressed as 11 × 10²³ atoms.
89. 2MnO₄⁻ + 6I⁻ + 4H₂O → 3I₂ + 2MnO₂ + 8OH⁻. If the above equation is balanced with integer coefficients, the value of z is:	8	The equation is balanced by comparing the reduction and oxidation half-reactions. The value of z, representing the number of hydroxide ions produced, is 8.
90. The mass of sodium acetate (CH₃COONa) required to prepare 250 mL of 0.35 M aqueous solution is (in grams). (Molar mass of CH₃COONa is 82.02 g/mol)	7	First, convert the volume to liters: 250 mL = 0.250 L. Moles required = Molarity × Volume = 0.35 mol/L × 0.250 L = 0.0875 mol. Mass = Moles × Molar mass = 0.0875 mol × 82.02 g/mol ≈ 7.18 g. Rounded off, this is approximately 7 g.

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JEE Main 2024 Jan 30 Shift 1 Chemistry Paper Analysis

JEE Main 2024 Jan 30 Shift 1 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.

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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 2024 question paper pdf with solutions- Download Jan 30 Shift 1 Chemistry Question Paper pdf

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JEE Main Questions

1.
Match List - I with List - II:

2.
Match List - I with List - II:
Choose the correct answer from the options given below:

3.
In the following substitution reaction:

5.
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?

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JEE Main 2024 Jan 30 Shift 1 Chemistry Question Paper PDF Download

JEE Main 2024 Jan 30 Shift 1 Chemistry Questions with Solutions

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JEE Main Previous Year Question Paper

JEE Main Questions

1.Match List - I with List - II:

2.Match List - I with List - II: Choose the correct answer from the options given below:

3.In the following substitution reaction:

5.The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?

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1.
Match List - I with List - II:

2.
Match List - I with List - II:
Choose the correct answer from the options given below:

3.
In the following substitution reaction:

5.
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?