
JEE Main 2024 Apr 5 Shift 2 Question Paper with Solution pdf is available for download here. Students found Physics easy and Mathematics hard. Chemistry carried the highest weightage and overall difficulty level was moderate.
| JEE Main 2024 Chemistry Question Paper with Answer Key April 5 Shift 2 | Check Solution |
Match List-I with List-II:
List-I:
(A) ICl
(B) ICl₃
(C) ClF₅
(D) IF₇
List-II:
(I) T-Shape
(II) Square pyramidal
(III) Pentagonal bipyramidal
(IV) Linear
The molecular geometry of the compounds is determined by VSEPR theory:
- (A) ICl: Linear (IV).
- (B) ICl₃: T-Shape (I).
- (C) ClF₅: Square pyramidal (II).
- (D) IF₇: Pentagonal bipyramidal (III).
Geometry is based on the steric number and lone pairs around the central atom.
While preparing crystals of Mohr’s salt, dilute H₂SO₄ is added to a mixture of ferrous sulfate and ammonium sulfate. Before dissolving this mixture in water, dilute H₂SO₄ is added here to:
Dilute H₂SO₄ is added to prevent the hydrolysis of ferrous sulfate. Without the acidic medium, ferrous sulfate would hydrolyze to form ferric hydroxide, which interferes with crystal formation.
The acidic medium ensures the stability of Fe²⁺ ions, preventing oxidation or hydrolysis.
Identify the major product in the following reaction:
The reaction involves elimination via the E2 mechanism, where the base removes a β-hydrogen and the leaving group departs, forming a double bond. The major product follows Zaitsev’s rule, favoring the more substituted alkene: cyclopentene.
Zaitsev’s rule predicts that elimination will form the more stable, highly substituted alkene.
The nomenclature for the following compound is:
The parent chain has 7 carbons with a double bond at position 6. Functional groups include a formyl group at position 2, a hydroxyl group at position 4, and a carboxylic acid at the terminal position. The correct name is 2-Formyl-4-hydroxyhept-6-enoic acid.
Follow IUPAC rules to assign positions and prioritize functional groups in naming.
Given below are two statements:
Assertion (A): NH₃ and NF₃ molecules have a pyramidal shape with a lone pair of electrons on the nitrogen atom. The resultant dipole moment of NH₃ is greater than that of NF₃.
Reason (R): In NH₃, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the N-H bonds. F is the most electronegative element.
NH₃ has a higher dipole moment because the lone pair dipole aligns with the N-H bond dipoles. In NF₃, the lone pair dipole opposes the N-F bond dipoles. Both statements are true, and the reason explains the assertion.
Analyze the direction and relative magnitude of bond and lone pair dipoles in NH₃ and NF₃.
Given below are two statements:
Statement I: On passing HCl(g) through a saturated solution of BaCl₂ at room temperature, white turbidity appears.
Statement II: When HCl(g) is passed through a saturated solution of NaCl, sodium chloride is precipitated due to the common ion effect.
HCl(g) increases Cl⁻ concentration, reducing the solubility of BaCl₂ and forming white turbidity due to the common ion effect. However, NaCl is highly soluble, and its solubility is not significantly affected by HCl. Hence, Statement I is correct, but Statement II is incorrect.
The solubility product (Ksp) of BaCl₂ decreases in the presence of additional Cl⁻ ions from HCl.
The metal atom present in the complex MABXL (where A, B, X, and L are unidentate ligands and M is a metal) involves sp³ hybridization. The number of geometrical isomers exhibited by the complex is:
In sp³ hybridization, the complex adopts a tetrahedral geometry. Geometrical isomerism is not possible in tetrahedral complexes because all positions are equivalent. Thus, the number of geometrical isomers is 0.
Tetrahedral complexes lack cis and trans arrangements, as all ligands are symmetrically distributed.
Match List-I with List-II:
List-I (Pair of Compounds):
(A) n-Propanol and isopropanol
(B) Methoxypropane and ethoxyethane
(C) Propanone and propanal
(D) Neopentane and isopentane
List-II (Type of Isomerism):
(I) Metamerism
(II) Chain Isomerism
(III) Position Isomerism
(IV) Functional Isomerism
(A) n-Propanol and isopropanol exhibit functional isomerism (IV).
(B) Methoxypropane and ethoxyethane exhibit metamerism (I).
(C) Propanone and propanal exhibit position isomerism (III).
(D) Neopentane and isopentane exhibit chain isomerism (II).
The type of isomerism is determined by differences in structure, functional groups, or connectivity of atoms.
The quantity of silver deposited when one coulomb of charge is passed through AgNO₃ solution is:
The amount of a substance deposited during electrolysis is proportional to its electrochemical equivalent. For one coulomb of charge, the quantity of silver deposited is equal to its electrochemical equivalent.
Use Faraday’s first law of electrolysis: m = ZQ, where Z is the electrochemical equivalent.
Which one of the following reactions is NOT possible?
Phenol does not react with HCl to form chlorobenzene because the −OH group is strongly bonded to the benzene ring. Thus, Reaction (2) is not possible, while the other reactions are feasible under appropriate conditions.
The −OH group in phenol is not replaced by Cl⁻ under normal acidic conditions due to resonance stabilization.
Given below are two statements:
Statement I: The metallic radius of Na is 1.86 Å, and the ionic radius of Na⁺ is lesser than 1.86 Å.
Statement II: Ions are always smaller in size than the corresponding elements.
The metallic radius of Na is 1.86 Å, and its ionic radius decreases due to the loss of an electron, which reduces electron-electron repulsion. Statement II is false because anions (e.g., Cl⁻) are larger than their corresponding neutral atoms due to increased electron repulsion.
For cations, the size decreases due to fewer electrons, while for anions, additional electrons increase repulsion and size.
Consider the above reaction sequence and identify the major product P:
The reaction sequence involves oxidation of ethanol to acetic acid, further oxidation to carbon dioxide, and subsequent decarboxylation with soda lime to yield methane as the final product. Thus, the major product is methane (CH₄).
Decarboxylation is a reaction that removes a carboxyl group, releasing carbon dioxide.
Consider the given chemical reaction. Product 'A' is:
Cyclohexane undergoes oxidation in the presence of KMnO₄ and H₂SO₄, resulting in the cleavage of CH₂ groups to carboxylic acid groups. This leads to the formation of adipic acid (HOOC-(CH₂)₄-COOH).
Adipic acid is a dicarboxylic acid commonly produced through the oxidation of cyclohexane derivatives.
For the electrochemical cell M—M²⁺||X—X²⁻, if E°(M²⁺/M) = 0.46V and E°(X/X²⁻) = 0.34V, which of the following is correct?
The standard cell potential E°cell is calculated as E°cathode - E°anode. Here, E°cell = 0.34V - 0.46V = -0.12V. Since E°cell is negative, the reverse reaction M²⁺ + X²⁻ → M + X is spontaneous.
A negative E°cell indicates that the reverse reaction is thermodynamically favored.
The number of moles of methane required to produce 11 g of CO₂ after complete combustion is:
From the combustion reaction CH₄ + 2O₂ → CO₂ + 2H₂O, one mole of methane produces one mole of CO₂. Given the molar mass of CO₂ is 44 g/mol, 11 g corresponds to 0.25 moles of CO₂. Therefore, 0.25 moles of methane are required.
The stoichiometric ratio between CH₄ and CO₂ in combustion is 1:1.
The number of complexes from the following with no electrons in the t2 orbital is:
TiCl4, [MnO4]-, [FeO4]2-, [FeCl4]-, [CoCl4]2-
The number of complexes with no electrons in the t2 orbital is 3: TiCl4, [MnO4]-, and [FeO4]2-. Each has a 3d0 electronic configuration.
Complexes with no electrons in the t2 orbital have their d-electrons fully paired or absent in the orbital.
The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is:
Ti2+, Cr2+, V2+
All three ions (Ti2+, Cr2+, and V2+) are strong reducing agents capable of reacting with H+ ions to liberate H2 gas. Hence, the count is 3.
Reducing agents donate electrons to H+, producing H2.
Identify A and B in the given chemical reaction sequence:
Friedel-Crafts acylation followed by Clemmensen reduction
Friedel-Crafts acylation forms acetophenone (A), which undergoes Clemmensen reduction to produce ethylbenzene (B).
Friedel-Crafts acylation introduces an acyl group, and Clemmensen reduction converts the carbonyl group to a CH2 group.
The correct decreasing order of atomic radii of Group 13 elements is:
The correct order is Tl > In > Ga > Al > B, due to increasing nuclear charge and varying shielding effects down the group.
The atomic radii generally increase down a group, but poor shielding by d and f electrons causes deviations in the trend.
The number of ways the set S = {2, 4, 8, ..., 512} can be partitioned into three subsets of equal size is:
The set has 9 elements. To partition it into three subsets of equal size, use the formula for combinations and calculate the number of ways as 1680.
The partitioning involves distributing the elements into subsets such that no two subsets overlap, and each subset has an equal number of elements.
Combustion of 1 mole of benzene is expressed as:
C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)
The standard enthalpy of combustion of 2 moles of benzene is -x kJ. Calculate the value of x given the following data:
- Standard enthalpy of formation of C₆H₆(l): 48.5 kJ/mol
- Standard enthalpy of formation of CO₂(g): -393.5 kJ/mol
- Standard enthalpy of formation of H₂O(l): -286 kJ/mol
Using Hess's Law:
ΔH = ΣΔHf(products) − ΣΔHf(reactants).
ΔH = [(6 × -393.5) + (3 × -286)] − [2 × 48.5].
For 2 moles of benzene, ΔH = -6535 kJ.
The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products A and B along with the evolution of CO₂. The sum of spin-only magnetic moment values of A and B is ___ B.M. (Nearest integer).
Given atomic numbers: C = 6, Na = 11, O = 8, Fe = 26, Cr = 24
Chromite ore reacts with sodium carbonate to form Na₂CrO₄ and Fe₂O₃:
- Spin-only magnetic moment for Na₂CrO₄ (Cr⁶⁺) = 0 B.M. (no unpaired electrons).
- For Fe₃⁺ in Fe₂O₃: Magnetic moment = √35 ≈ 5.92 B.M. (nearest integer is 6).
In an atom, the total number of electrons having quantum numbers n = 4, |ml| = 1, and ms = -1/2 is:
For n = 4, |ml| = 1 corresponds to 3 orbitals (p, d, or f). Each orbital can hold one electron with ms = -1/2. Thus, the total number of electrons is 6.
Using the given figure, the ratio of Rf values of sample A and sample C is x × 10⁻². Value of x is:
Figure: Solvent front = 12.5 cm, Sample A = 5.0 cm, Sample C = 10.0 cm.
The Rf value is the ratio of the distance traveled by the sample to the distance traveled by the solvent front:
- Rf(A) = 5.0 / 12.5 = 0.4.
- Rf(C) = 10.0 / 12.5 = 0.8.
Ratio Rf(A) / Rf(C) = 0.4 / 0.8 = 0.5 = 50 × 10⁻².
In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is ________ g (nearest integer).
In the Claisen-Schmidt reaction:
1 mole of acetone reacts with 2 moles of benzaldehyde. Molar masses: Acetone = 58 g/mol, Benzaldehyde = 106 g/mol.
87 g of acetone corresponds to 1.5 moles. Benzaldehyde required = 1.5 × 2 × 106 = 318 g.
In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is ________ g (nearest integer).
In the Claisen-Schmidt reaction, 1 mole of acetone reacts with 2 moles of benzaldehyde. Given:
87 g of acetone corresponds to 1.5 moles. The required amount of benzaldehyde is:
1.5 × 2 × 106 = 318 g.
Consider the following single-step reaction in the gas phase at constant temperature:
2A(g) + B(g) → C(g)
The initial rate of the reaction is r₁ when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate r₂ is recorded when the pressure of C becomes 0.5 atm. The ratio r₁ : r₂ is ________ × 10⁻¹ (nearest integer).
Using the rate law for the reaction:
Rate ∝ [A]²[B].
Initial pressures: [A] = 1.5 atm, [B] = 0.7 atm.
After C reaches 0.5 atm, pressures are updated:
Calculating the ratio:
r₁ : r₂ = (1.5² × 0.7) : (0.5² × 0.2) = 315 × 10⁻¹.
The product C in the following sequence of reactions has ________ π bonds:
Reaction sequence: KMnO₄–KOH, ∆ → A; H₃O⁺ → B; Br₂/FeBr₃ → C
The sequence involves oxidation of the alkyl chain in a benzene derivative:
- Product C is para-bromobenzoic acid.
- It has 3 π bonds in the benzene ring and 1 π bond in the carboxyl group.
Total = 4 π bonds.
Considering acetic acid dissociates in water, its dissociation constant is 6.25 × 10⁻⁵. If 5 mL of acetic acid is dissolved in 1 liter of water, the solution will freeze at −x × 10⁻² °C, provided pure water freezes at 0 °C. x = _________. (Nearest integer).
Given: Kf(water) = 1.86 K·kg·mol⁻¹, density of acetic acid = 1.2 g·mL⁻¹, molar mass of acetic acid = 60 g·mol⁻¹, density of water = 1 g·cm⁻³.
Molality (m) of acetic acid:
Mass of acetic acid = 5 × 1.2 = 6 g, moles = 6 / 60 = 0.1 mol.
Molality = 0.1 / 1 = 0.1 m.
Considering dissociation:
Effective molality = 0.1 × (1 + α), where α = degree of dissociation.
∆Tf = i × Kf × m, substituting values gives x = 19.
Number of compounds from the following with zero dipole moment is ___________. HF, H₂, H₂S, CO₂, NH₃, BF₃, CH₄, CHCl₃, SiF₄, H₂O, BeF₂
Compounds with symmetrical geometry and no net dipole moment are:
- H₂, CO₂, BF₃, CH₄, SiF₄, BeF₂.
Total = 6 compounds.
*The article might have information for the previous academic years, please refer the official website of the exam.