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JEE Main 2024 Question Paper with Answer Key PDF for Chemistry (April 6 - Shift 1)

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JEE Main 2024 Apr 6 Shift 1 Question Paper with Solution pdf is available for download here. Students found Chemistry easy and Mathematics hard. Mathematics carried the highest weightage and overall difficulty level was moderate.

JEE Main 2024 Chemistry Question Paper with Answer Key PDF

JEE Main 2024 Chemistry Question Paper with Answer Key April 6 Shift 1	Download	Check Solution

JEE Main 2024 6 April Shift 1 Chemistry Questions with Solution

Question 1:

Functional group present in sulphonic acid is:

SO₃H
SO₃H
−S−OH
−SO₂

Correct Answer: (2) SO₃H

Solution: The sulphonic acid group is characterized by the functional group SO₃H.

The functional group in sulphonic acid consists of a sulfur atom double-bonded to two oxygen atoms and a hydroxyl group (-OH).
This group is responsible for the acidic properties of sulphonic acids.

Question 2:

Match List I with List II:

List I (Molecule / Species)	List II (Property / Shape)
A. SO₂Cl₂	I. Paramagnetic
B. NO	II. Diamagnetic
C. NO₂^-	III. Tetrahedral
D. I₃^-	IV. Linear

Choose the correct answer from the options given below:

A-IV, B-I, C-III, D-II
A-III, B-I, C-II, D-IV
A-III, B-III, C-I, D-IV
A-III, B-IV, C-II, D-I

Correct Answer: (2) A-III, B-I, C-II, D-IV

Solution: Match the properties and shapes of the molecules and ions.

A: SO₂Cl₂ is tetrahedral in shape (III).
B: NO is paramagnetic due to an unpaired electron (I).
C: NO₂^- is diamagnetic and has a bent shape (II).
D: I₃^- is linear due to its molecular geometry (IV).

Question 3:

Given below are two statements:

Statement I: Picric acid is 2,4,6-trinitrotoluene.
Statement II: Phenol-2,4-disulphuric acid is treated with conc. HNO₃ to get picric acid.

Choose the most appropriate answer from the options given below:

Statement I is incorrect but Statement II is correct.
Both Statement I and Statement II are incorrect.
Statement I is correct but Statement II is incorrect.
Both Statement I and Statement II are correct.

Correct Answer: (1) Statement I is incorrect but Statement II is correct

Solution: Picric acid is not trinitrotoluene; it is 2,4,6-trinitrophenol.

Statement I is incorrect as picric acid is 2,4,6-trinitrophenol, not trinitrotoluene.
Statement II is correct because phenol-2,4-disulphuric acid reacts with concentrated HNO₃ to form picric acid.

Question 4:

Which of the following is a metamer of the given compound (X)?

Option 1
Option 2
Option 3
Option 4

Correct Answer: (4) Option 4

Solution: Metamers have the same molecular formula and functional group but differ in the alkyl groups on either side of the functional group.

Analyze the alkyl groups around the functional group of the given compound.
The correct metamer has a different arrangement of alkyl groups but retains the same molecular formula and functional group.

Question 5:

DNA molecule contains 4 bases whose structures are shown below. One of the structures is not correct. Identify the incorrect base structure.

Option 1
Option 2
Option 3
Option 4

Correct Answer: (3) Option 3

Solution: DNA contains adenine, guanine, cytosine, and thymine. The incorrect structure does not match these bases.

Examine the structure of the four bases.
Adenine and guanine are purines, while cytosine and thymine are pyrimidines.
The incorrect structure is one that does not correspond to any of these bases.

Question 6:

Match List I with List II:

List I (Hybridization)	List II (Orientation in Space)
A. sp³	III. Tetrahedral
B. dsp²	IV. Square planar
C. sp³d	I. Trigonal bipyramidal
D. sp³d²	II. Octahedral

Choose the correct answer from the options below:

A-III, B-I, C-IV, D-II
A-III, B-I, C-II, D-IV
A-IV, B-III, C-I, D-II
A-III, B-IV, C-I, D-II

Correct Answer: (4) A-III, B-IV, C-I, D-II

Solution: Match the hybridization with its spatial orientation.

A (sp³): Tetrahedral structure with bond angles of approximately 109.5°.
B (dsp²): Square planar geometry, seen in coordination complexes like [Ni(CN)₄]^2-.
C (sp³d): Trigonal bipyramidal geometry with axial and equatorial positions.
D (sp³d²): Octahedral geometry, typical for complexes like [SF₆].

Question 7:

Given below are two statements:

Statement I: Gallium is used in the manufacturing of thermometers.
Statement II: A thermometer containing gallium is useful for measuring the freezing point (256 K) of brine solution.

Choose the correct answer from the options below:

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

Correct Answer: (4) Statement I is true but Statement II is false

Solution: Gallium thermometers are designed for high-temperature measurements and are unsuitable for freezing points like 256 K.

Gallium's melting point is approximately 303 K, making it liquid at room temperature, ideal for high-temperature applications.
However, it cannot be used to measure low temperatures like the freezing point of brine solution (256 K).

Question 8:

Which of the following statements are correct?

A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point.
B. Aniline can be purified by steam distillation because it decomposes at its normal boiling point.
C. Ethanol can be separated from ethanol-water mixture by azeotropic distillation because it forms azeotrope.
D. An organic compound is pure if mixed melting point remains unchanged.

Choose the correct answer from the options below:

A, B, C only
A, C, D only
B, C, D only
A, B, D only

Correct Answer: (2) A, C, D only

Solution: Statements A, C, and D are correct as per the properties of the compounds and methods of purification.

A: Glycerol decomposes at its normal boiling point, hence vacuum distillation is used.
C: Ethanol and water form an azeotrope, separated by azeotropic distillation.
D: A pure organic compound retains its mixed melting point.
B is incorrect as aniline is immiscible in water but does not decompose at its boiling point.

Question 9:

Match List I with List II:

List I (Compound/Species)	List II (Shape/Geometry)
A. SF₄	III. See-saw
B. BrF₃	IV. Bent T-shape
C. BrO₃^-	II. Pyramidal
D. NH₄⁺	I. Tetrahedral

Choose the correct answer from the options below:

A-II, B-III, C-I, D-IV
A-III, B-IV, C-II, D-I
A-II, B-IV, C-III, D-I
A-III, B-II, C-IV, D-I

Correct Answer: (2) A-III, B-IV, C-II, D-I

Solution: Match the compounds with their shapes based on VSEPR theory.

A: SF₄: See-saw due to one lone pair.
B: BrF₃: T-shape (Bent) due to two lone pairs.
C: BrO₃^-: Pyramidal due to one lone pair.
D: NH₄⁺: Tetrahedral as all valence electrons are bonded.

Question 10:

In Reimer-Tiemann reaction, phenol is converted into salicylaldehyde through an intermediate. The structure of the intermediate is:

Option 1
Option 2
Option 3
Option 4

Correct Answer: (4) Option 4

Solution: The intermediate is the sodium salt of dichlorophenol, formed via the attack of dichlorocarbene (:CCl₂) on phenol.

The Reimer-Tiemann reaction introduces a -CHO group ortho to the phenolic group.
The dichlorocarbene reacts with phenol to form the intermediate, which upon hydrolysis gives salicylaldehyde.

Question 11:

Which of the following material is not a semiconductor?

Germanium
Graphite
Silicon
Copper oxide

Correct Answer: (2) Graphite

Solution: Graphite is a good conductor of electricity, not a semiconductor.

Semiconductors like silicon and germanium have a specific band gap that allows them to conduct electricity under certain conditions.
Graphite, on the other hand, conducts electricity due to its delocalized π-electrons and does not have a band gap like semiconductors.

Question 12:

Consider the following complexes:

(A) [CoCl(NH₃)₅]²⁺
(B) [Co(CN)₆]^3-
(C) [Co(NH₃)₅(H₂O)]³⁺
(D) [Cu(H₂O)₄]²⁺

The correct order of A, B, C, and D in terms of wavenumber of light absorbed is:

C < D < A < B
D < A < C < B
A < C < B < D
B < C < A < D

Correct Answer: (2) D < A < C < B

Solution: The ligand field strength determines the wavenumber of absorption. Strong field ligands like CN^- cause higher energy transitions.

[CoCl(NH₃)₅]²⁺ has weaker field ligands compared to [Co(CN)₆]^3-, leading to a lower wavenumber absorption.
[Cu(H₂O)₄]²⁺ has the weakest field ligands, hence the lowest wavenumber.

Question 13:

Match List I with List II:

List I (Precipitating reagent and conditions)	List II (Cation)
A. NH₄Cl + NH₄OH	III. Al³⁺
B. NH₄OH + Na₂CO₃	IV. Sr²⁺
C. NH₄OH + NH₄Cl + H₂S gas	II. Pb²⁺
D. Dilute HCl	I. Mn²⁺

Choose the correct answer from the options below:

A-IV, B-III, C-II, D-I
A-IV, B-II, C-I, D-II
A-III, B-IV, C-II, D-I
A-III, B-IV, C-II, D-I

Correct Answer: (3) A-III, B-IV, C-II, D-I

Solution: Match the reagents with the cations based on their precipitation reactions.

A: NH₄Cl + NH₄OH precipitates Al(OH)₃ (III).
B: NH₄OH + Na₂CO₃ precipitates SrCO₃ (IV).
C: NH₄OH + NH₄Cl + H₂S precipitates PbS (II).
D: Dilute HCl precipitates MnCl₂ (I).

Question 14:

The electron affinity values are negative for:

A. Be → Be^-
B. N → N^-
C. O → O^-
D. Na → Na^-
E. Al → Al^-

Choose the most appropriate answer from the options below:

D and E only
A, B, D and E only
A and D only
A, B and C only

Correct Answer: (1) D and E only

Solution: Electron affinity is negative for elements that do not readily accept an electron to form a stable anion.

D: Sodium (Na) does not form a stable anion, leading to negative electron affinity.
E: Aluminum (Al) has a similar property, resulting in a negative electron affinity.

Question 15:

The number of elements from the following that do not belong to lanthanoids is:

Eu, Cm, Er, Tb, Yb, and Lu

Correct Answer: (3) 1

Solution: Cm (Curium) is an actinide, while the rest belong to the lanthanoid series.

Lanthanoids include Eu, Er, Tb, Yb, and Lu.
Cm (Curium) is an actinide and does not belong to the lanthanoid series.

Question 16:

The density of 'x' M solution ('x' molar) of NaOH is 1.12 g/mL, while in molality, the concentration of the solution is 3m (3 molal). Then x is:

Correct Answer: (2) 3.0

Solution: The relationship between molarity and molality is used with the given density, resulting in x = 3.0 M.

Molarity (M) = (Molality × Density × 1000) / (1000 + (Molality × Molar Mass of Solute)). Substituting values, we find x = 3.0.

Question 17:

Which among the following aldehydes is most reactive towards nucleophilic addition reactions?

HCHO
C₂H₅CHO
CH₃CHO
C₃H₇CHO

Correct Answer: (1) HCHO

Solution: Formaldehyde (HCHO) is the most reactive due to the absence of alkyl groups, leading to reduced steric hindrance and a higher partial positive charge on the carbonyl carbon.

As alkyl groups increase, steric hindrance and electron donation to the carbonyl carbon reduce reactivity towards nucleophiles.

Question 18:

At −20°C and 1 atm pressure, a cylinder is filled with an equal number of H₂, I₂, and HI molecules for the reaction:

H₂(g) + I₂(g) ⇌ 2HI(g)

K_p for the process is x × 10⁻¹. x =?

2
1
10
0.01

Correct Answer: (3) 10

Solution: Using the equilibrium constant expression, K_p = (P_HI)² / (P_H2 × P_I2), and substituting mole fractions, x = 10.

The partial pressures of the gases and the stoichiometry of the reaction allow for the calculation of K_p.

Question 19:

Match List I with List II:

List I (Compound)	List II (Uses)
A. Iodoform	III. Antiseptic
B. Carbon tetrachloride	I. Fire extinguisher
C. CFC	IV. Refrigerants
D. DDT	II. Insecticide

Choose the correct answer from the options below:

A-I, B-II, C-III, D-IV
A-III, B-II, C-IV, D-I
A-III, B-I, C-IV, D-II
A-II, B-IV, C-I, D-III

Correct Answer: (3) A-III, B-I, C-IV, D-II

Solution: A-III: Iodoform is an antiseptic; B-I: Carbon tetrachloride is a fire extinguisher; C-IV: CFCs are refrigerants; D-II: DDT is an insecticide.

Iodoform has antiseptic properties due to its iodine content.
Carbon tetrachloride is a non-flammable liquid used in fire extinguishers.
CFCs (chlorofluorocarbons) are widely used in refrigeration systems.
DDT is an effective pesticide but is banned in many countries due to its environmental impact.

Question 20:

A conductivity cell with two electrodes (dark side) is half filled with an infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will:

Increase sharply
Remain same or cannot be measured accurately
Decrease sharply
Depend upon type of electrolyte

Correct Answer: (2) Remain same or cannot be measured accurately

Solution: For infinitely dilute solutions, molar conductivity becomes constant, and further dilution does not change it.

At infinite dilution, ion interactions become negligible, and the limiting molar conductivity remains unaffected by additional water.

Question 21:

Consider the dissociation of the weak acid HX as given below:

HX(aq) ⇌ H⁺(aq) + X⁻(aq), K_a = 1.2×10⁻⁵

The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is ___ ×10⁻² bar (nearest integer).

Given: R = 0.083 L·bar·mol⁻¹·K⁻¹

Correct Answer: 76

Solution: The osmotic pressure is calculated using the formula:

Π = iCRT, where i is the van 't Hoff factor considering dissociation.

Considering the degree of dissociation and using the formula for i, the total concentration is adjusted to account for dissociation, leading to an osmotic pressure of approximately 76 × 10⁻² bar.

Question 22:

The difference in the ‘spin-only’ magnetic moment values of KMnO₄ and the manganese product formed during titration of KMnO₄ against oxalic acid in acidic medium is ___ BM (nearest integer).

Correct Answer: 6

Solution: In KMnO₄, Mn is in the +7 oxidation state (no unpaired electrons, μ = 0 BM). In the reduced product (Mn²⁺), there are 5 unpaired electrons, giving a spin-only magnetic moment of 6 BM.

The difference is due to the change in the oxidation state of Mn from +7 (KMnO₄) to +2 (Mn²⁺), where the latter has unpaired electrons contributing to its magnetic moment.

Question 23:

Time required for 99.9% completion of a first-order reaction is ___ times the time required for completion of 90% reaction (nearest integer).

Correct Answer: 3

Solution: The time for a given percentage completion is proportional to ln(remaining concentration). For 99.9% completion:

t_99.9% = (ln(1000) / ln(10)) × t_90%, which simplifies to 3 times t_90%.

The relationship is derived from the first-order reaction equation: t = (1/k) ln([A]₀ / [A]).

Question 24:

Number of molecules from the following which can exhibit hydrogen bonding is ___ (nearest integer):

CH₃OH, H₂O, C₆H₆, C₆H₅NO₂, HF, NH₃

Correct Answer: 5

Solution: Hydrogen bonding is exhibited by CH₃OH, H₂O, HF, NH₃, and C₆H₅NO₂, due to the presence of N-H, O-H, or F-H bonds.

C₆H₆ (benzene) cannot form hydrogen bonds as it lacks the necessary functional groups for hydrogen bonding.

Question 25:

9.3 g of pure aniline upon diazotization followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _____ g (nearest integer).

Correct Answer: 20

Solution: The molecular weight of aniline is 93 g/mol. Given 9.3 g of aniline, this corresponds to 0.1 mol. Each mole of aniline produces 1 mole of dye, and the molecular weight of the dye is 200 g/mol, so:

Mass of dye = 0.1 × 200 = 20 g.

The reaction is a 1:1 stoichiometric conversion, ensuring all aniline converts to dye in 100% yield.

Question 26:

The major product of the following reaction is P:

CH₃C≡C-CH₃
(i) Na/liq. NH₃ → (ii) dil. KMnO₄, 273 K → P
The number of oxygen atoms present in product P is _____ (nearest integer).

Correct Answer: 2

Solution: The reaction produces a vicinal diol as the final product, with two hydroxyl groups attached to the adjacent carbon atoms, contributing two oxygen atoms.

Na in liquid NH₃ reduces the alkyne to a trans-alkene, and KMnO₄ hydroxylates the double bond, forming a vicinal diol.

Question 27:

The frequency of the de-Broglie wave of an electron in Bohr’s first orbit of the hydrogen atom is ___ ×10¹³ Hz (nearest integer).

Given: RH (Rydberg constant) = 2.18×10⁻¹⁸ J, h (Planck’s constant) = 6.6×10⁻³⁴ J·s.

Correct Answer: 661

Solution: The frequency is calculated using the relation ν = E/h, where E is the kinetic energy of the electron in the first orbit.

The energy of the first orbit is derived from E = RH, and substituting values gives ν ≈ 661 ×10¹³ Hz.

Question 28:

The major products from the following reaction sequence are product A and product B:

B (i) Br₂ → (ii) alc. KOH (3 eq.) → A
B (i) Br₂ → (ii) Na⁺/O⁻ (1.0 eq.) → B
The total sum of π electrons in product A and product B are ____ (nearest integer).

Correct Answer: 8

Solution: Product A is benzene (6 π electrons), and product B is an alkene (2 π electrons). The total sum is 8 π electrons.

The reaction sequence involves dehydrohalogenation and elimination, leading to aromatic and alkenic structures as products.

Question 29:

Among CrO, Cr₂O₃, and CrO₃, the sum of spin-only magnetic moment values of basic and amphoteric oxides is ___ ×10⁻² BM (nearest integer).

Given: Atomic number of Cr is 24.

Correct Answer: 877

Solution: CrO (Cr²⁺) has 4 unpaired electrons, Cr₂O₃ (Cr³⁺) has 3 unpaired electrons, and CrO₃ (Cr⁶⁺) has no unpaired electrons.

The spin-only magnetic moment is μ = √n(n + 2), where n is the number of unpaired electrons. Adding the moments for CrO and Cr₂O₃ gives 8.77 BM (or 877 ×10⁻² BM).

Question 30:

An ideal gas, C_V = 5/2 R, is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively, then the final temperature is ____ K (nearest integer).

Correct Answer: 274

Solution: Using the adiabatic relation PV^γ = constant, where γ = C_P/C_V, the final temperature is calculated.

For γ = 1.4 (diatomic gas), the relationship T₁V₁^γ−1 = T₂V₂^γ−1 is applied, leading to T₂ ≈ 274 K.