JEE Main 2025 April 4 Shift 2 Chemistry Question Paper, Exam Analysis, and Answer Keys (Available)

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According to Bredt's rule, \( R \) has a more localized lone pair, making it the most basic, followed by \( Q \), which has cross conjugation, and finally \( P \) which is the least basic.

Thus, the correct order is \( R > Q > P \). Quick Tip: Bredt's rule helps explain basicity based on the localization of lone pairs.

The incorrect relationship is between \( Fe^{2+} \) and \( Fe^{3+} \). According to ionisation enthalpies, \( Mn^{2+} \) has more ionisation energy than \( Fe^{2+} \), and \( Mn^{3+} \) has more ionisation energy than \( Fe^{3+} \).

Thus, the correct answer is (4). Quick Tip: Ionisation enthalpy depends on the electron configuration, stability, and half-filled stability of orbitals.

When alcohols are reacted with NaOI/NaOH, a yellow solid is produced due to the formation of iodine complexes. Upon examining the compounds:

- (A) \( CH_3 CH_2 C_2 H_5 \) (an alcohol) reacts with NaOI/NaOH, producing a yellow solid.
- (B) \( CH_3 CH_2 C_2 H_2 OH \) (an alcohol) reacts with NaOI/NaOH, producing a yellow solid.
- (C) \( CH_3 C_2 H_5 \) (an aldehyde) does not react with NaOI/NaOH to produce a yellow solid.
- (D) \( CH_3 OH \) does not react with NaOI/NaOH to form a yellow solid.
- (E) \( CH_3 CH_2 H \) does not react with NaOI/NaOH to form a yellow solid.

Thus, the correct answer is (2). Quick Tip: The reaction of alcohols with NaOI/NaOH often leads to yellow solid formation due to iodine complexation.

Let’s break down the key information provided:

1. Hydrolysis of “x”:
- The dipeptide "x" undergoes complete hydrolysis to produce two amino acids: y and z.
- y is a compound that, when treated with aqueous HNO\(_2\), produces lactic acid. This strongly suggests that y is glycine, as glycine reacts with nitrous acid to form lactic acid. Therefore, glycine must be one of the products after hydrolysis.

2. Heating of “z”:
- z on heating forms a cyclic molecule. This strongly indicates that z is proline, as proline is an amino acid that can form a cyclic structure under heating conditions.

3. Identifying the Dipeptide:
- The dipeptide must be one that hydrolyzes to give glycine (which produces lactic acid upon treatment with HNO\(_2\)) and proline (which forms a cyclic structure upon heating).
- The only dipeptide in the given options that fits this pattern is alanine-glycine (option 2), as alanine can undergo cyclization to form proline under heat.

Thus, the correct dipeptide x is alanine-glycine.

Quick Tip: The key to solving this question lies in recognizing that glycine reacts with HNO\(_2\) to produce lactic acid, and proline (formed from alanine) can form a cyclic structure when heated.

In the pair (A), the ion with the methoxy group (\( OMe \)) is more stable than the one with the methyl group (\( Me \)) because the oxygen in the methoxy group can donate electron density via resonance, making it more stable.

In pair (B), the nitro group (\( NO_2 \)) is an electron-withdrawing group, making the ion less stable than the one with the oxygen-nitrogen double bond (\( O,N \)), which stabilizes the ion through resonance.

Thus, the correct answer is (2). Quick Tip: Resonance and inductive effects play a key role in the stability of ions, with electron-donating groups increasing stability and electron-withdrawing groups decreasing it.

Statement (I) is incorrect because alkyl chlorides react with aqueous potassium hydroxide to undergo substitution (SN) reactions, not elimination. Alcohols are typically formed through substitution reactions, not elimination.

Statement (II) is correct because in alcoholic potassium hydroxide, alkyl chlorides undergo elimination (E2) reactions, forming alkenes by abstracting the hydrogen from the \( \beta \)-carbon.

Thus, the correct answer is (2). Quick Tip: Elimination and substitution reactions are key processes for forming alkenes and alcohols, respectively, based on the conditions (e.g., solvent, temperature).

1. Statement I: The formula for the molal depression constant is correctly given by:
\[ k_f = \frac{M_1 R T_f}{\Delta S_{fus}} \]
Here, \( M_1 \) represents the molality of the solution, \( R \) is the gas constant, \( T_f \) is the freezing point depression, and \( \Delta S_{fus} \) is the enthalpy of fusion. Therefore, Statement I is correct.

2. Statement II: The molal depression constant \( k_f \) for benzene is greater than for water, not less. Specifically, \( k_f \) for water is 1.86 °C/molal and for benzene is 5.12 °C/molal. Hence, Statement II is incorrect.

Thus, the correct answer is (4). Quick Tip: The molal depression constant depends on the solvent. Water has a lower \( k_f \) than benzene, which is why Statement II is incorrect.

The structure contains a hydroxyl group at position 4, an alkene at position 1, and an alkyne at position 6. Based on the IUPAC naming conventions, the correct name for this compound is Hept-1-en-6-yn-4-ol.

Thus, the correct answer is (4). Quick Tip: When naming organic compounds, identify and number the substituents, functional groups, and multiple bonds to derive the correct IUPAC name.

- (A) Aniline from aniline-water mixture: This is typically done using steam distillation (IV).
- (B) Glycerol from spent-lye in soap industry: This is done using fractional distillation (II).
- (C) Different fractions of crude oil: This is done using distillation at reduced pressure (III).
- (D) Chloroform-Aniline mixture: This is separated using simple distillation (I).

Thus, the correct answer is (2). Quick Tip: The method of distillation is chosen based on the boiling points and volatility of the components involved.

- A is methanol because it reacts with NaCN in an acidic medium to form formaldehyde (B).
- B is formaldehyde, which is then converted to methyl chloride (C) by dibromane.
- C (methyl chloride) reacts with oleum at 140°C to form chloroform (D), which is an anesthetic.

Thus, the correct answer is (3). Quick Tip: Organic compound transformations depend on reactions like nucleophilic substitution and addition of reagents like dibromane and oleum.

- Electronic configuration of each complex:
- \( [FeF_6]^{3-} \): \( [Ar] 3d^5 4s^0 \). There are 5 unpaired electrons in the \( 3d \)-orbitals.
- \( [CoF_6]^{3-} \): \( [Ar] 3d^6 4s^0 \). There are 4 unpaired electrons in the \( 3d \)-orbitals.
- \( [Ni(CO)_4] \): \( [Ar] 3d^8 4s^2 \). The \( CO \) ligand is a strong field ligand, so the pairing of electrons leads to 0 unpaired electrons.
- \( [Ni(CN)_4]^{2-} \): \( [Ar] 3d^8 4s^0 \). The \( CN \) ligand is a strong field ligand, leading to the pairing of all electrons, so there are 0 unpaired electrons.

Thus, the order of the unpaired electrons is: \[ [FeF_6]^{3-} > [CoF_6]^{3-} > [Ni(CN)_4]^{2-} = [Ni(CO)_4] \] Quick Tip: The number of unpaired electrons in transition metal complexes can be determined based on the electronic configuration of the metal ion and the nature of the ligands.

In the given data, we have two reactions:

1. Reaction (a): The dissolution of HCl in 10 moles of water gives a heat of solution of \( \Delta H = -69.01 \, kJ/mol \).
2. Reaction (b): The dissolution of HCl in 40 moles of water gives a heat of solution of \( \Delta H = -72.79 \, kJ/mol \).

The negative values for both enthalpy changes indicate that both processes are exothermic, meaning heat is released when HCl dissolves in water. This contradicts the statement (1), which suggests that the dissolution is endothermic.

Step 1: Calculation of the Heat of Dilution
Now, to understand how the heat of solution changes with the amount of solvent, we subtract the enthalpy change of reaction (a) from that of reaction (b):
\[ \Delta H = -72.79 \, kJ/mol - (-69.01 \, kJ/mol) = -3.78 \, kJ/mol \]

This value represents the difference in heat of solution when the amount of solvent changes from 10 moles of water to 40 moles of water. This shows that the heat of solution depends on the amount of solvent used, confirming that statement (2) is correct.

Step 2: Why Statement (3) is Incorrect
Statement (3) suggests that the heat of dilution for the HCl (HCl.10H\(_2\)O to HCl.40H\(_2\)O) is 3.78 kJ/mol. However, the value we calculated is actually the difference in heat of solution, not the heat of dilution itself. Therefore, statement (3) is incorrect. Quick Tip: The heat of solution for a substance can depend on the amount of solvent used. A larger amount of solvent typically reduces the heat released during dissolution, as seen in the negative change in \( \Delta H \) between reactions (a) and (b).

For \( Z = 24 \), the electron configuration is \( [Ar] 3d^6 4s^2 \).
- \( l = 1 \) corresponds to the p-orbital, which can hold 6 electrons.
- \( l = 2 \) corresponds to the d-orbital, which can hold 10 electrons.

Thus, there are 12 electrons in the \( l = 1 \) shell and 5 electrons in the \( l = 2 \) shell.

Therefore, the correct answer is (3). Quick Tip: The number of electrons in orbitals depends on the quantum numbers, with the \( p \)-orbitals having a maximum of 6 electrons and \( d \)-orbitals having a maximum of 10.

- Statement I: The first ionisation enthalpy of group 14 elements is indeed higher than that of group 13 elements, as the number of valence electrons increases from group 13 to 14, leading to a higher ionisation energy. Thus, Statement I is correct.

- Statement II: This statement is incorrect. The melting points and boiling points of group 13 elements are generally lower than those of group 14 elements due to the stronger metallic bonding in group 14 elements. Thus, Statement II is incorrect.

Therefore, the correct answer is (1). Quick Tip: Ionisation enthalpy increases across a period as the nuclear charge increases, but the melting and boiling points of the elements generally follow the trends in atomic size and bonding strength.

The activation energy \( E_a \) of a reaction is related to the slope of the plot of \( \log k \) vs \( \frac{1}{T} \) by the Arrhenius equation: \[ \log k = -\frac{E_a}{2.303R} \times \frac{1}{T} + constant \]
where:
- \( \log k \) is the log of the rate constant,
- \( T \) is the temperature,
- \( R \) is the gas constant.

The steeper the slope, the higher the activation energy. In this plot:
- Reaction 1 (the line with the steepest slope) corresponds to the highest activation energy, i.e., \( E_{a1} \).
- Reaction 2 (the line with a less steep slope) corresponds to the middle activation energy, i.e., \( E_{a2} \).
- Reaction 3 (the line with the least steep slope) corresponds to the lowest activation energy, i.e., \( E_{a3} \).

Thus, the correct order of activation energies is \( E_{a2} > E_{a1} > E_{a3} \), which corresponds to option (1). Quick Tip: The slope of the Arrhenius plot is directly related to the activation energy. A steeper slope corresponds to a higher activation energy.

1. Identifying the most stable complex ion:

The stability of the complex depends on factors like ligand field strength and the charge on the metal ion. Among the given complexes, the most stable complex is \( [Fe(C_2O_4)_3]^{3-} \) because oxalate (C\(_2\)O\(_4\)) is a bidentate ligand and provides a strong ligand field, stabilizing the iron(III) ion effectively.

2. Electron Configuration of Iron in \( [Fe(C_2O_4)_3]^{3-} \):

- Iron in the \( [Fe(C_2O_4)_3]^{3-} \) complex is in the \( +3 \) oxidation state, so its electron configuration is \( [Ar] 3d^5 \). This means it has 5 electrons in its \( 3d \)-orbitals.
- For the octahedral \( [Fe(C_2O_4)_3]^{3-} \) complex, these 5 \( d \)-electrons will occupy the \( t_{2g} \) orbitals, as these orbitals are lower in energy in an octahedral field.

Thus, the number of electrons in the \( t_{2g} \) orbitals is \( \mathbf{5} \), and X = 5.

3. The nature of \( V_2O_x \):

- Vanadium oxides, such as \( V_2O_5 \), exhibit amphoteric properties. This means they can act as both acids and bases depending on the reaction conditions. Therefore, the correct answer for the nature of vanadium oxide is amphoteric.

Thus, the correct answer is (4). Quick Tip: The stability of a complex ion depends on the ligand field and the metal ion's charge. The presence of strong ligands, like oxalate, increases the stability of the complex.

- The ionisation enthalpy decreases as we move down a group. Boron (B) has the highest ionisation enthalpy in Group 13, while Gallium (Ga) has the lowest due to the larger atomic radius and shielding effect as we move down the group.

Thus, the correct answer is (1). Quick Tip: Ionisation enthalpy decreases down the group because of increasing atomic size and electron shielding.

The structure of \( X \) is the molecule shown in Option 2. Upon the addition of \( H^+ \), a tertiary carbocation is formed, which is stable and leads to the formation of the major product, which involves the substitution of the bromine atom at the most stable position. Therefore, the structure of \( X \) corresponds to the molecule in Option 2. Quick Tip: In reactions involving carbocation formation, the stability of the carbocation determines the major product. Tertiary carbocations are more stable than secondary and primary ones.

- Statement I is correct. The structure involves all three possible resonance structures where the fluorine atoms are positioned at different bond angles with respect to the central atom. The lone pairs on fluorine atoms may vary depending on the electron distribution.

- Statement II is incorrect. In \( sp^3d \) hybridization, the lone pairs occupy equatorial positions, not axial, due to minimizing \( \ell p - \beta p \) repulsion. Therefore, the statement about lone pairs occupying axial positions in structure III is incorrect.

Thus, the correct answer is (2). Quick Tip: In reactions involving lone pairs, remember that in \( sp^3d \) hybridization, lone pairs prefer equatorial positions to minimize \( \ell p - \beta p \) repulsion, not axial positions.

For zero-order reaction:

The half-life is given by: \[ Half life = \frac{A_0}{2k} \]
Given, \( half-life = 1 \) hour and the initial concentration \( A_0 = 2.0 \, mol/L \), we can write: \[ 60 \, min = \frac{2}{2k} \]
Solving for \( k \): \[ k = \frac{1}{60} \, M/min \]

Now, using the formula for zero-order reaction: \[ A_t = A_0 - kt \] \[ t = \frac{A_0 - A_t}{k} \]
Substitute the values: \[ t = \frac{0.5 - 0.25}{\frac{1}{60}} = 0.25 \times 60 = 15 \, min \]

Thus, the time required to decrease the concentration of A from 0.50 to 0.25 mol L\(^{-1}\) is 15 minutes. Quick Tip: For zero-order reactions, the concentration of reactant decreases linearly with time, and the rate constant \( k \) is directly related to the half-life.

We are given that sea water is a 6 molar (6 M) solution of NaCl and has a density of 2 g mL\(^{-1}\). We also know that the concentration of dissolved oxygen (O\(_2\)) in sea water is 5.8 ppm.

First, we need to calculate the concentration of O\(_2\) in mol/L using the given data:

1. Molar mass of NaCl = 58.5 g/mol
- Sea water has a molarity of 6 M, meaning each liter of sea water contains 6 moles of NaCl.
- Therefore, in 1 liter of sea water, the mass of NaCl is:
\[ mass of NaCl = 6 \times 58.5 = 351 grams \]

2. Given that the density of the solution is 2 g/mL, the mass of 1000 mL (1 L) of sea water is:
\[ mass of solution = 2 \times 1000 = 2000 grams \]

3. Now, we can calculate the mass of O\(_2\) dissolved in 1 liter of sea water using the given ppm value of 5.8 ppm:
\[ ppm of O_2 = \frac{mass of O_2}{mass of solution} \times 10^6 \]
\[ mass of O_2 = 5.8 \times 10^{-3} g \]
\[ mass of O_2 = 5.8 mg = 5.8 \times 10^{-3} grams \]

4. To calculate the molarity (mol/L) of O\(_2\) in the solution, we use the molar mass of O\(_2\):
\[ molality for O_2 = \frac{5.8 \times 10^{-3}}{32} = 1.81 \times 10^{-4} moles \]
This corresponds to a concentration of O\(_2\) as:
\[ = 2.19 \times 10^{-4} mol/L \]

Therefore, the concentration of O\(_2\) in sea water is \( \mathbf{2.19 \times 10^{-4}} \) mol/L, which is approximately \( 2.19 \times 10^{-6} \). Quick Tip: When dealing with ppm and molarity, remember that ppm represents parts per million, and it can be converted to mass per volume. Once you have the mass of a solute, converting it to moles using the molar mass will give you the molarity.

Given that: \[ CaCO_3 \rightarrow CaO + CO_2 \]
We start by calculating the mass of CaCO\(_3\): \[ mass of CaCO_3 = \frac{150 \times 75}{100} = 112.5 \, kg \]

Next, calculate the moles of CaCO\(_3\): \[ n_{CaCO_3} = \frac{mass}{molar mass of CaCO_3} = \frac{1125000}{100} = 1125 \, moles \]

Since each mole of CaCO\(_3\) produces 1 mole of CaO, the moles of CaO formed will be the same: \[ n_{CaO} = 1125 \, moles \]

Now, we calculate the mass of CaO: \[ mass of CaO = n_{CaO} \times molar mass of CaO = 1125 \times 56 = 63000 \, grams = 63 \, kg \]

Thus, the amount of calcium oxide produced is 63 kg. Quick Tip: When calculating mass from moles, always use the correct molar masses and conversion factors (grams to kilograms) to ensure accurate results.

The complex MC\(\ell_4\)3NH\(_3\) undergoes sp\(^3\)d\(^2\) hybridisation. The central atom of BrF\(_5\) has 1 lone pair, so x = 1.

The complex MC\(\ell_4\)3NH\(_3\) has octahedral geometry, leading to 2 possible geometrical isomers: fac and mer.

Thus, the number of geometrical isomers is 2. Quick Tip: When dealing with octahedral complexes, the number of geometrical isomers is determined by how the ligands are arranged. For complexes with three identical ligands and three other different ligands, two possible isomers (fac and mer) exist.

For ammonium chloride, the molar conductance at infinite dilution is: \[ \lambda^0_{NH_4Cl} = 185 \, S cm^{-1} mol^{-1} \]
The ionic conductance of hydroxyl and chloride ions are given as 170 and 70, respectively. The dissociation of ammonium hydroxide is represented by: \[ \lambda^0_{NH_4OH} = \lambda^0_{NH_4^+} + \lambda^0_{OH^-} \]
The molar conductance of the 0.02 M solution of ammonium hydroxide is 85.5 S cm\(^{-1}\) mol\(^{-1}\), and we use the following relation to find the degree of dissociation \( \alpha \): \[ \lambda = \alpha \cdot \lambda^0 \]
where \( \alpha \) is the degree of dissociation, so \[ 85.5 = 0.02 \times (170 + 70) \times \alpha \]
Solving for \( \alpha \): \[ \alpha = \frac{85.5}{0.02 \times 240} = 0.177 \quad or \quad x = 3 \]

Thus, the degree of dissociation is \( x = 3 \times 10^{-1} \). Quick Tip: To calculate degree of dissociation, use the formula \( \lambda = \alpha \cdot \lambda^0 \), where \( \lambda^0 \) is the molar conductance at infinite dilution and \( \lambda \) is the observed molar conductance.

Given:

- pH = 10.0

- pOH = 4.0

- [OH\(^-\)] = \(10^{-4}\) M


The concentration of OH\(^-\) is \(10^{-4}\) M, and since Mg(OH)\(_2\) dissociates completely in water, the number of moles of OH\(^-\) will be equal to the number of moles of Mg(OH)\(_2\).

Step-by-step solution:

1. Number of moles of OH\(^-\) = \(10^{-4}\) moles (from the concentration of OH\(^-\)).
2. Number of moles of Mg(OH)\(_2\) = \(\frac{10^{-4}}{2}\) = \(5 \times 10^{-5}\) moles, because one mole of Mg(OH)\(_2\) gives two moles of OH\(^-\).
3. The mass of Mg(OH)\(_2\) is then calculated as:
\[ mass of Mg(OH)\(_2\) = 5 \times 10^{-5} \times 58 \times 10^3 \, mg \]
\[ = 2.9 \, mg \]

Thus, the value of x is 2.9 mg, which is approximately 3 mg. Quick Tip: When calculating the mass of a compound from its concentration, remember that dissociation of salts like Mg(OH)\(_2\) can provide multiple moles of ions for each mole of the compound. The number of moles of the compound is related to the ion concentration through the dissociation stoichiometry.