JEE Main 2024 question paper pdf with solutions- Download April 4 Shift 1 Chemistry Question Paper pdf

Step 1: Understanding the question.

The question asks for the number of complexes that have an even number of unpaired electrons in their d-orbitals. These complexes contain transition metal ions with different oxidation states, and the electron configuration depends on the metal and its charge.


Step 2: Analyze the complexes.

For each of the given complexes, we calculate the number of unpaired electrons in the d-orbitals:


(Cu(H₂O)₆)²⁺: The copper ion in this complex is in the +2 oxidation state, with a d⁹ electron configuration. This gives 1 unpaired electron.

(Fe(H₂O)₆)³⁺: The iron ion is in the +3 oxidation state, with a d⁵ electron configuration. This gives 5 unpaired electrons.

(Cr(H₂O)₆)²⁺: The chromium ion is in the +2 oxidation state, with a d⁴ electron configuration. This gives 4 unpaired electrons.

(Ni(H₂O)₆)²⁺: The nickel ion is in the +2 oxidation state, with a d⁸ electron configuration. This gives 2 unpaired electrons.


Step 3: Conclusion.

The complexes that have an even number of unpaired electrons are [Cu(H₂O)₆]²⁺, [Cr(H₂O)₆]²⁺, and [Ni(H₂O)₆]²⁺, all having even numbers of unpaired electrons (1, 5, and 4 respectively). Therefore, the correct answer is (02.00).
Quick Tip: In transition metal complexes, the number of unpaired electrons can be calculated based on the electron configuration of the metal ion in a given oxidation state. An even number of unpaired electrons corresponds to a stable electron configuration in the context of d-orbitals.

Step 1: Understanding the relationship.

The equation for an ideal gas is given by: \[ P = \frac{R \cdot P}{M} \cdot T \]
where \( P \) is the pressure, \( M \) is the molar mass, and \( T \) is the temperature.


Step 2: Analyzing the graph.

From the graph, we observe that as the density increases, the pressure increases at the same temperature. Thus, the slope of the graph will be greater for higher densities.


Step 3: Conclusion.

The graph shows that \( \rho_1 > \rho_2 > \rho_3 \), which indicates that the density of the gas decreases from \( \rho_1 \) to \( \rho_3 \). Therefore, the correct answer is (2) \( \rho_1 > \rho_2 > \rho_3 \).
Quick Tip: In pressure vs temperature graphs for ideal gases, the slope can provide information about the gas's density and other properties. Higher density results in steeper slopes at the same temperature.

Step 1: Understanding ionisation enthalpy.

Ionisation enthalpy is the energy required to remove an electron from an atom in the gaseous state. The trend in ionisation enthalpy is affected by atomic size and the effective nuclear charge.


Step 2: Analyzing the electronic configurations.

The electronic configurations of the elements are:
- B: 2s² 2p¹
- Be: 2s²
- O: 2s² 2p⁴
- F: 2s² 2p⁵
- N: 2s² 2p³

Step 3: Explanation of the trend.

The correct order of ionisation enthalpy is (2) because:
- B < Be: B has a higher atomic radius and lower ionisation enthalpy than Be.
- Be < O: O has a higher effective nuclear charge and thus a higher ionisation enthalpy than Be.
- O < F: F has a higher ionisation enthalpy than O because of its higher nuclear charge.
- F < N: N has a half-filled stable configuration, making it harder to ionise than F.


Step 4: Conclusion.

The correct order of 1st ionisation enthalpy is (2) B < Be < O < F < N.
Quick Tip: The ionisation enthalpy increases with an increase in effective nuclear charge and decreases with an increase in atomic radius. Elements with half-filled or fully-filled orbitals have higher ionisation enthalpies.

Step 1: Understanding the equation.

We are given the equation for \( E_{overall} \): \[ E_{overall} = E_{a1} + E_{a2} - E_{a3} \]

Step 2: Applying the values.

Substituting the given values of \( E_{a1} = 400, E_{a2} = 300, E_{a3} = 200 \), we get: \[ E_{overall} = 400 + 300 - 200 = 500 \]

Step 3: Conclusion.

Therefore, the correct answer is (3) 500.
Quick Tip: The overall activation energy for a reaction is calculated by summing up the individual activation energies, considering the energy barriers for each step of the reaction.

Step 1: Writing the balanced reaction.

The balanced reaction is: \[ MnO_4^- + C_2O_4^{2-} \rightarrow Mn^{2+} + CO_2 \]

Step 2: Using the stoichiometric relationships.

From the reaction, we know the stoichiometric coefficients:
- \( n_f \) for MnO₄ = 5
- \( n_f \) for \(C_2O_4^{2-}\) = 2

The number of moles of KMnO₄ is calculated using the formula: \[ moles of KMnO_4 = M \times V \] \[ moles of KMnO_4 = 2 \, M \times 2 \, mL = 4 \, mL \times 2 = 8 \, moles \]

Step 3: Conclusion.

The molarity of H₂C₂O₄ is 0.5 M. Therefore, the correct answer is (00.50).
Quick Tip: Use stoichiometry to calculate the molarity of a substance by balancing the chemical reaction and considering the relationship between the species involved.

Step 1: Understanding basic strength.

Basic strength refers to the ability of a species to accept a proton (H⁺). The stronger the base, the more readily it accepts a proton.


Step 2: Analyzing the species.

- H^- is a very strong base, it is a hydride ion.
- OR^- is the alkoxide ion, which is also a strong base.
- OH^- is a strong base but weaker than H^- and OR^-.
- CH₃COO^- is an acetate ion, which is weaker than OH^- due to resonance stabilization.
- HCOO^- is a formate ion, which is weaker than acetate due to its smaller resonance effect.


Step 3: Conclusion.

The order of basic strength is: \[ H^- > OR^- > OH^- > CH_3COO^- > HCOO^- \]
Thus, the correct answer is (2) II > V > I > III > IV.
Quick Tip: To determine basic strength, consider the ability of the species to donate electrons, the size of the ion, and resonance stabilization. The more electron-donating the species, the stronger the base.

Step 1: Identifying the valence electrons of each element.

The number of valence electrons for each element is as follows: \[ O = 6, \quad S = 6, \quad F = 7, \quad N = 5, \quad Al = 3, \quad C = 4, \quad Si = 4 \]

Step 2: Analyzing which elements do not use all of their valence electrons in bonding.

- Oxygen (O), Sulfur (S), and Fluorine (F) tend to form bonds using all their valence electrons.
- Nitrogen (N) does not use all of its valence electrons in bonding as it forms 3 bonds, leaving 1 lone pair.
- Aluminum (Al) does not use all of its valence electrons in bonding as it forms 3 bonds, leaving 0 lone pairs.
- Carbon (C) and Silicon (Si) form bonds using all their valence electrons (4 bonds each).

Step 3: Conclusion.

The elements that do not use all of their valence electrons in bonding are N, Al, and Si. Therefore, the correct answer is (03.00).
Quick Tip: To determine whether an element uses all its valence electrons in bonding, check the number of bonds it forms. If the number of bonds is less than the number of valence electrons, the element does not use all of its electrons.

Step 1: Understanding the reaction.

The reaction shown involves the conversion of a halogenated aromatic compound into another halogenated aromatic compound. The typical reagents for such reactions are used in nucleophilic substitution or elimination reactions.


Step 2: Identifying the reagents.

- Alcoholic NaOH/80°C is commonly used for elimination reactions, especially to form alkenes.
- HBr/CH₃COOH is used for electrophilic substitution, typically in the presence of an aromatic ring.


Step 3: Conclusion.

The correct reagents for this reaction are Alcoholic NaOH/80°C and HBr/CH₃COOH. Therefore, the correct answer is (3).



Quick Tip: Alcoholic NaOH is often used in elimination reactions to form alkenes, and HBr is used for electrophilic substitution in aromatic compounds.

Step 1: Understanding ligand strength.

Ligand strength refers to the ability of a ligand to donate electrons to a central metal ion. The more electron-donating the ligand, the stronger it is. Ligands like \( CO \) are strong field ligands (SFL), while ligands like \( F^- \) are weak field ligands (WFL).


Step 2: Analyzing the order of ligand strength.

- \( CO \) is a strong field ligand and is the strongest ligand in this list.
- \( OH^- \) is a weaker ligand compared to \( CO \) but stronger than \( F^- \).
- \( SCN^- \) is a weaker ligand compared to \( OH^- \) and \( F^- \).
Thus, the correct order of ligand strength is \( CO > OH^- > F^- > SCN^- \).


Step 3: Conclusion.

Therefore, the correct answer is (1) \( CO > OH^- > F^- > SCN^- \).
Quick Tip: In ligand field theory, strong field ligands like \( CO \) are better electron donors compared to weak field ligands like \( F^- \).

Step 1: Understanding the reaction.

The given reaction shows the conversion of \( K_2MnO_4 \) in alkaline medium to \( KMnO_4 \) and \( MnO_2 \). We need to find the sum of the spin-only magnetic moment of the central metal ion in both products.


Step 2: Analyze the oxidation states of manganese.

In \( KMnO_4 \), manganese has an oxidation state of +7 (Mn⁷⁺), and in \( MnO_2 \), manganese has an oxidation state of +4 (Mn⁴⁺).


Step 3: Calculating the magnetic moments.

For \( KMnO_4 \) (Mn⁷⁺):
- Mn⁷⁺ has a \( d^0 \) configuration, so its magnetic moment is zero.

For \( MnO_2 \) (Mn⁴⁺):
- Mn⁴⁺ has a \( d^3 \) configuration, and the magnetic moment is calculated using the formula: \[ \mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \]

Step 4: Conclusion.

The sum of the spin-only magnetic moments of the two central metal ions is \( 0 + 3.87 = 3.87 \), which rounds to the nearest integer of 4. Therefore, the correct answer is (4) 4.00.
Quick Tip: Spin-only magnetic moment is calculated using the formula \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons.

Step 1: Understanding the purpose of adding NH₄Cl with NH₄OH.

In qualitative analysis, NH₄OH (ammonium hydroxide) increases the concentration of OH^- ions in the solution, which can precipitate group IV elements as their hydroxides. Adding NH₄Cl (ammonium chloride) helps to reduce the concentration of OH^- ions by forming the buffer system, thereby controlling the precipitation of other elements.


Step 2: Conclusion.

Therefore, NH₄Cl is added with NH₄OH to decrease the concentration of OH^- ions in order to avoid the precipitation of further group elements. Thus, the correct answer is (2).
Quick Tip: In qualitative analysis, NH₄Cl is often used with NH₄OH to control the concentration of OH^- ions and prevent the precipitation of undesired compounds.

Step 1: Understanding the reaction.

The reaction is a Clemmensen reduction, which is used to reduce aldehydes and ketones into their respective alkanes. In the given reaction, the compound undergoes reduction with zinc and mercury in hydrochloric acid (Zn–Hg/HCl), which will reduce the carbonyl group to a CH₂ group.


Step 2: Identifying the product.

The reduction of the carbonyl group (C=O) in the given aromatic aldehyde results in the conversion of the aldehyde group (CHO) into a methyl group (CH₂). Thus, the correct product is the alkane derived from the original aldehyde. The structure shown in option (1) represents this reduction product.


Step 3: Conclusion.

Therefore, the correct answer is (1).


Quick Tip: The Clemmensen reduction is commonly used for the reduction of carbonyl groups in aldehydes and ketones, converting them into corresponding alkanes without affecting other functional groups.