JEE Main 2024 question paper pdf with solutions- Download April 5 Shift 2 Chemistry Question Paper pdf

Step 1: Identifying functional groups.

The compound contains a formyl group (-CHO) and a hydroxyl group (-OH) attached to a heptene backbone. The formyl group is at position 2, and the hydroxyl group is at position 4. The correct IUPAC name is based on this arrangement.


Step 2: Conclusion.

The IUPAC name of the compound is 2-formyl-4-hydroxy hept-6-enoic acid, which corresponds to option (1).
Quick Tip: When naming compounds, always identify and prioritize functional groups such as formyl and hydroxyl, and follow the numbering based on the longest carbon chain.

Step 1: Understanding square planar complexes.

A square planar complex typically has cis-trans isomerism, meaning it can have two different configurations — one where two similar ligands are adjacent (cis) and one where they are opposite (trans). In some cases, additional isomerism may arise from different types of ligands.


Step 2: Conclusion.

For a square planar complex like Mabcd, there are three possible isomers: two geometric isomers (cis and trans) and one optical isomer if the ligands are asymmetric. The correct answer is (2) 3.
Quick Tip: In square planar complexes, the key source of isomerism is the arrangement of ligands, which can lead to cis-trans and optical isomerism.

Step 1: Understanding quantum numbers.

The quantum numbers are as follows:
- Principal quantum number \( n = 4 \)
- Orbital angular momentum quantum number \( l = 1 \) (which corresponds to a p-orbital)
- Magnetic quantum number \( m \) can have values from \( -l \) to \( +l \), i.e., \( m = -1, 0, +1 \)
- Spin quantum number \( ms = \frac{1}{2} \), so each orbital can accommodate 2 electrons with opposite spins.

Step 2: Calculating maximum number of electrons.

For \( l = 1 \), there are 3 orbitals (\( m = -1, 0, +1 \)), and each orbital can hold 2 electrons due to the spin quantum number. Hence, the total maximum number of electrons is: \[ 3 \, orbitals \times 2 \, electrons per orbital = 6 \, electrons. \]

Step 3: Conclusion.

The maximum number of electrons is \( \boxed{6} \).
Quick Tip: The maximum number of electrons in an orbital set is given by \( 2(2l + 1) \), where \( l \) is the orbital angular momentum quantum number.

Step 1: Understanding the reactions.

The first reaction is the oxidation of propene (CH\(_2\)=CH\(_2\)-CH\(_3\)) using KMnO\(_4\)/KOH, which produces a carboxylic acid (COOH). The second step involves bromination (Br\(_2\)) in the presence of FeBr\(_3\), which results in the addition of a bromine atom to the benzene ring. The product is a substituted benzene with a carboxyl group (COOH) and a bromine atom (Br) attached.




Step 2: Identifying the \(\pi\) bonds in product B.

In the product B, the benzene ring still contains 3 \(\pi\) bonds from its aromatic system. The carboxyl group (COOH) contributes no additional \(\pi\) bonds, and the bromine atom does not affect the \(\pi\) bond count. The total number of \(\pi\) bonds is 4, including the 3 from the benzene ring and 1 from the double bond in the carboxyl group.

Step 3: Conclusion.

The total number of \(\pi\) bonds in product B is 4.
Quick Tip: In aromatic compounds, the number of \(\pi\) bonds is determined by the number of bonds in the conjugated system, such as the benzene ring.

Step 1: Understanding electronic configuration.

- Ti\(^{4+}\) has the electron configuration of [Ar], which means it does not have any electrons in the \(t_{2g}\) orbital. Therefore, Ti\(^{4+}\) has a \(d^0\) electronic configuration.

- For Mn\(^{2-}\), Fe\(^{2-}\), and Co\(^{2-}\), the electronic configuration would involve electrons in the \(t_{2g}\) orbitals.


Step 2: Conclusion.

The compound \([TiCl_4]\) corresponds to Ti\(^{4+}\), which has a \(d^0\) configuration and therefore does not have electrons in the \(t_{2g}\) orbital. The correct answer is (1) [TiCl\(_4\)].
Quick Tip: A \(d^0\) configuration indicates no electrons in the \(t_{2g}\) orbital. This is typical for metal ions with a high oxidation state like Ti\(^{4+}\).

Step 1: Analyzing the dipole moments.

NF\(_3\) has less dipole moment than NH\(_3\). This is because the electronegativity of F is higher than that of N, which makes the bond moment of N-H in NH\(_3\) more significant.


Step 2: Explanation of R.

In NH\(_3\), the bond moment of N-H is in the same direction as the lone pair, creating a stronger dipole moment. However, in NF\(_3\), the lone pair of electrons on nitrogen does not align with the N-F bonds, leading to a smaller dipole moment. This correctly explains why NF\(_3\) has less dipole moment than NH\(_3\).


Step 3: Conclusion.

Both statements A and R are correct, and R is the correct explanation of A. The correct answer is (1).
Quick Tip: Dipole moment depends on the electronegativity difference and the geometry of the molecule, as well as the alignment of the lone pair.

Step 1: Understanding the reactions.

- The first step involves the oxidation of ethanol to acetic acid (CH\(_3\)COOH) using Jones reagent (CrO\(_3\)).
- The second step uses KMnO\(_4\), which further oxidizes acetic acid, but does not affect the outcome for this reaction sequence.
- The third step involves decarboxylation, where the carboxyl group is removed from acetic acid, resulting in methane (CH\(_4\)).


Step 2: Conclusion.

The major product of the given sequence is methane (CH\(_4\)), formed after the decarboxylation of acetic acid. The correct answer is (4).
Quick Tip: In a decarboxylation reaction, the removal of the carboxyl group (COOH) from a compound typically results in the formation of methane or other alkanes.

Step 1: Identifying the shapes.

- IF has a linear geometry due to the bonding of I to F atoms in a straight line.
- IF\(_7\) has a pentagonal bipyramidal structure because it involves seven bonding groups around the central atom.
- ICl\(_3\) has a T-shape structure, which is typical for molecules with three bonding atoms and one lone pair.
- BF\(_3\) has a trigonal planar geometry due to its three bonds and no lone pairs on the central atom.


Step 2: Conclusion.

The correct match is P\(\rightarrow\)1; Q\(\rightarrow\)2; R\(\rightarrow\)3; S\(\rightarrow\)4.
Quick Tip: To identify molecular shapes, refer to the number of bonding pairs and lone pairs on the central atom, and use VSEPR theory for predictions.

Step 1: Identifying oxidation states.

- A = Fe\(_2\)O\(_3\) \(\rightarrow\) Fe\(^{3+}\), which has a \(d^5\) configuration, and the magnetic moment is 5.87 BM.
- B = Na\(_2\)CrO\(_4\) \(\rightarrow\) Cr\(^{6+}\), which has a \(d^0\) configuration, and the magnetic moment is 0 BM.


Step 2: Conclusion.

The sum of the magnetic moments is \(5.87 + 0 = 5.87\), which rounds to 6. The correct answer is (6).
Quick Tip: Magnetic moments are related to the number of unpaired electrons in a species. A \(d^0\) configuration means no unpaired electrons, resulting in zero magnetic moment.

Step 1: Calculating the cell potential.

The cell potential is calculated as: \[ E^\circ_{cell} = E^\circ_{M^{2+}/M} + E^\circ_{X^{2-}/X} = 0.34V + (-0.46V) = -0.12V \]

Step 2: Analyzing the reaction.

- For the reaction to be spontaneous, the cell potential must be positive.
- Since the calculated potential is negative, the reverse reaction is spontaneous:
M\(^{2+}\) + 2X\(^{2-}\) \(\rightarrow\) M + X\(_2\).


Step 3: Conclusion.

The correct answer is (4) because the cell reaction M\(^{2+}\) + 2X\(^{2-}\) \(\rightarrow\) M + X\(_2\) is spontaneous.
Quick Tip: A negative cell potential indicates a non-spontaneous reaction in the forward direction, but spontaneous in the reverse direction.

Step 1: Analyzing the reactions.

- Ti\(^{2+}\), Cr\(^{2+}\), and V\(^{2+}\) are all capable of reacting with dilute mineral acids such as HCl. These reactions typically lead to the liberation of H\(_2\) gas due to the reduction of the metal ion and the oxidation of hydrogen.


Step 2: Conclusion.

All of the metal ions (Ti\(^{2+}\), Cr\(^{2+}\), and V\(^{2+}\)) will liberate H\(_2\) gas when treated with dilute mineral acid. The correct answer is (4).
Quick Tip: Transition metal ions in lower oxidation states can often react with dilute acids to liberate hydrogen gas, due to the ease of reduction.

Step 1: Understanding periodic trends.

- The atomic size increases as you move down a group in the periodic table.
- The atomic size also decreases from left to right across a period due to increasing effective nuclear charge.


Step 2: Analyzing the elements in the 13\(^th\) group.

- Tl (Thallium) is at the bottom of the group, so it has the largest atomic size.
- In (Indium) follows Tl, so it has a smaller atomic size.
- Al (Aluminum) is smaller than In.
- Ga (Gallium) is smaller than Al due to the additional d-electrons in the transition metals that contract the size.
- B (Boron) is the smallest due to its position in the group and periodic table.


Step 3: Conclusion.

The correct order of increasing atomic size is Tl \(>\) In \(>\) Al \(>\) Ga \(>\) B, which corresponds to option (1).
Quick Tip: Atomic size generally increases down a group and decreases across a period due to the increasing nuclear charge.

Step 1: Formula for electrochemical equivalent.

The work done (W) when passing a charge \(q\) through a solution is given by: \[ W = Z \times q \]
where \(Z\) is the electrochemical equivalent and \(q\) is the charge passed.

Step 2: Given values.

Here, \(q = 1\) C (1 Coulomb of charge) and \(Z\) is the electrochemical equivalent of Ag.

Step 3: Conclusion.

Thus, 1 electrochemical equivalent of Ag is deposited. The correct answer is (2).
Quick Tip: The electrochemical equivalent of an element is the mass of the element deposited when 1 Coulomb of charge passes through its solution.

Step 1: Determining the van't Hoff factor.

Since acetic acid dimerizes, the van't Hoff factor \( i \) is 1 (for dimerization) minus \( \frac{1}{2} \) for each molecule of acetic acid that forms a dimer, giving \( i = \frac{1}{2} \).


Step 2: Applying the formula for freezing point depression.

The formula for freezing point depression is: \[ \Delta T_f = i \times K_f \times \frac{m}{M} \]
where:
- \( i = \frac{1}{2} \) (since the acid dimerizes),
- \( K_f = 5 \, K \cdot kg/mol \),
- \( m = 60 \, g \) of acetic acid,
- \( M = 60 \, g/mol \) (molar mass of acetic acid),
- The mass of benzene is \( 500 \, g = 0.5 \, kg \).


Step 3: Calculating the freezing point depression.

Substitute the values into the equation: \[ \Delta T_f = \frac{1}{2} \times 5 \times \frac{60}{60 \times 0.5} = 5 \, K. \]

Step 4: Conclusion.

Thus, the freezing point depression \( \Delta T_f = 5 \, K \). The correct answer is (5).
Quick Tip: The van't Hoff factor \( i \) accounts for the number of particles that result from dissociation or association in the solution. For dimerization, \( i = \frac{1}{2} \).

Step 1: Enthalpy of formation equation.

The enthalpy change for the combustion reaction is: \[ \Delta H_{comb} = \left[ \sum \Delta H_f (products) \right] - \left[ \sum \Delta H_f (reactants) \right] \]
Here, the combustion of 1 mole of benzene is given. The formation of 2 moles of benzene will involve multiplying the result by 2.

Step 2: Calculate the enthalpy change for the reaction.

We know the combustion reaction is: \[ C_6H_6 + O_2 \rightarrow 6 CO_2 + 3 H_2O \]
Thus, the enthalpy change is: \[ \Delta H_{comb} = \left[ 6(-393.5) + 3(-285.83) \right] - \left[ 1(0) + 1(0) \right] = -3267 \, kJ/mol \]

Step 3: Conclusion.

The enthalpy of formation for benzene is \(\boxed{97.02 \, kJ/mol}\).
Quick Tip: Standard enthalpy of formation for any element in its standard state is zero, and it is important to balance the stoichiometry of the reaction while calculating enthalpies.

Step 1: Setup for reaction rates.

At time \( t = 0 \), [A] = 1.5, [B] = 0.7

Given at time \( t = 0.5 \), [C] = 0.5 and [A] = 0.5, [B] = 0.2.


The rate equation for the reaction is: \[ R = k[A]^2[B] \]

Step 2: Calculating \(\frac{R_1}{R_2}\).

For \( R_1 \) (at \( t = 0 \)): \[ R_1 = k[A]^2[B] = k(1.5)^2(0.7) \]
For \( R_2 \) (at \( t = 0.5 \)): \[ R_2 = k[A]^2[B] = k(0.5)^2(0.2) \]

Step 3: Calculating the ratio.
\[ \frac{R_1}{R_2} = \frac{k(1.5)^2(0.7)}{k(0.5)^2(0.2)} = \frac{1.5^2 \times 0.7}{0.5^2 \times 0.2} = \frac{63}{2} = 31.5 \]

Step 4: Conclusion.

The value of \(\frac{R_1}{R_2} = 31.5\). The correct answer is (5).
Quick Tip: The rate of a reaction can be expressed in terms of concentrations of reactants raised to appropriate powers.

Step 1: Understanding the substitution in phenol.

The substitution of the \(-OH\) group in phenol is not possible in the presence of HCl because the lone pair of electrons on oxygen in the \(-OH\) group is delocalized into the aromatic ring. This delocalization prevents nucleophilic attack by the chloride ion. Therefore, the reaction with HCl is incorrect.


Step 2: Correct reactions.

- (2) Substitution with NaOH \(\Delta\) leads to phenoxide, which is a common reaction for phenol.
- (3) Substitution with HBr leads to the formation of bromo derivatives.
- (4) Chlorination with Cl\(_2\)/AlCl\(_3\) leads to substitution of the hydroxyl group with chlorine.


Step 3: Conclusion.

The substitution of \(-OH\) with HCl is incorrect due to delocalization of its lone pair in phenol. The correct answer is (1).
Quick Tip: In phenol, the delocalization of the lone pair on oxygen reduces its nucleophilicity, making it less reactive toward electrophiles like HCl.

Step 1: Understanding the reaction.

The given reaction starts with benzene (C\(_6\)H\(_6\)) undergoing oxidation by KMnO\(_4\) in the presence of KOH. This oxidizes the aromatic ring to a straight-chain dicarboxylic acid, resulting in the formation of adipic acid, which has the structure COOH-(CH\(_2\))\(_4\)-COOH.


Step 2: Conclusion.

The product formed is adipic acid, which is a six-carbon dicarboxylic acid. The correct answer is (1).
Quick Tip: KMnO\(_4\) in alkaline conditions (like KOH) typically oxidizes aromatic compounds to carboxylic acids, and when followed by acid hydrolysis (H\(_3\)O\(^+\)), it forms dicarboxylic acids.