JEE Main 2024 question paper pdf with solutions- Download April 5 Shift 1 Chemistry Question Paper pdf

Step 1: Understanding Wolff-Kishner reduction.

Wolff-Kishner reduction is a reaction that reduces aldehydes and ketones to the corresponding alkanes by using hydrazine (N\(_2\)H\(_4\)) and a base such as KOH in glycol.


Step 2: Analysis of options.

(1) Product 1: This is the correct product as it corresponds to the reduction of a carbonyl group to a methylene group.

(2) Product 2: This is incorrect because it suggests a different type of product, possibly a hydroxylated product.

(3) Product 3: This is incorrect as it suggests an alcohol, which is not the expected outcome of a Wolff-Kishner reduction.

(4) Product 4: This is incorrect because it does not correspond to the reduction of the carbonyl group.


Step 3: Conclusion.

The correct answer is (1) Product 1, as it correctly shows the reduction of a carbonyl group to an alkane.
Quick Tip: The Wolff-Kishner reduction is a strong reduction method for converting carbonyl compounds (like aldehydes and ketones) into alkanes, using hydrazine and a base.

Step 1: Understanding maximum oxidation states.

The maximum oxidation state of an element corresponds to the highest number of electrons it can lose in a compound.


Step 2: Analysis of options.

(1) Mn: Manganese has a maximum oxidation state of +7, as seen in compounds like KMnO\(_4\) (potassium permanganate).

(2) Ti: Titanium can have a maximum oxidation state of +4, as seen in TiO\(_2\).

(3) Co: Cobalt typically has a maximum oxidation state of +3 in most of its compounds.

(4) Na: Sodium has a maximum oxidation state of +1, as it typically loses one electron to form Na\(^+\).


Step 3: Conclusion.

The correct answer is (1) Mn, as manganese can show a maximum oxidation state of +7.
Quick Tip: In transition metals, the maximum oxidation state is often observed in the highest oxidation state compound, which can vary depending on the element's electron configuration.

Step 1: Understanding paramagnetic character.

Paramagnetism is caused by the presence of unpaired electrons in an atom or ion. The more unpaired electrons, the higher the paramagnetic character.


Step 2: Electron configurations in the +2 oxidation state.

Fe\(^{2+}\): The electron configuration for Fe\(^{2+}\) is \(d^6\), with 4 unpaired electrons.

Co\(^{2+}\): The electron configuration for Co\(^{2+}\) is \(d^7\), with 3 unpaired electrons.

Ni\(^{2+}\): The electron configuration for Ni\(^{2+}\) is \(d^8\), with 2 unpaired electrons.

Mn\(^{2+}\): The electron configuration for Mn\(^{2+}\) is \(d^5\), with 5 unpaired electrons.


Step 3: Conclusion.

The ion with the least number of unpaired electrons is Ni\(^{2+}\), which has the lowest paramagnetic character. Hence, the correct answer is (3) Ni.
Quick Tip: In paramagnetism, the number of unpaired electrons determines the strength of the magnetic character. The fewer unpaired electrons, the weaker the paramagnetic character.

Step 1: Understanding the bonding in ethylene.

Ethylene (\(C_2H_4\)) consists of a double bond between two carbon atoms. A double bond consists of one \(\sigma\) bond and one \(\pi\) bond. Each carbon-hydrogen single bond is a \(\sigma\) bond.


Step 2: Counting the bonds.

- The \(\sigma\) bonds: There are 5 \(\sigma\) bonds — 4 between carbon and hydrogen, and 1 between the two carbon atoms.

- The \(\pi\) bond: There is 1 \(\pi\) bond between the two carbon atoms.


Step 3: Conclusion.

The total number of \(\sigma\) bonds is 5, and the total number of \(\pi\) bonds is 1. Hence, the correct answer is (2) 5,1.
Quick Tip: In a double bond, there is always one \(\sigma\) bond and one \(\pi\) bond. \(\sigma\) bonds are formed from head-on overlap, while \(\pi\) bonds arise from sideways overlap.

Step 1: Analyzing the species.

The species provided are O\(^{2-}\), F\(^-\), Na\(^+\), and Mg\(^{2+}\). These species all have the same number of electrons (10 electrons) and are isoelectronic. Isoelectronic species have the same number of electrons, but their sizes vary due to differing nuclear charges.


Step 2: Evaluating the statements.

- (a) O\(^{2-}\) is the largest in size because it has the fewest protons (8 protons) compared to the others, leading to the least effective nuclear charge.

- (b) Mg\(^{2+}\) is the smallest in size because it has the highest effective nuclear charge (12 protons), pulling its electrons closer.

- (c) All have the same effective nuclear charge: This is incorrect, as the effective nuclear charge varies with the number of protons in each species.

- (d) All are isoelectronic: This is correct, as all species have 10 electrons.


Step 3: Conclusion.

The correct statements are (a), (b), and (d). Hence, the correct answer is (2) a, b and d.
Quick Tip: Isoelectronic species have the same number of electrons but different sizes due to differences in nuclear charge. The higher the nuclear charge, the smaller the size.

Step 1: Understanding the reaction.

The given reaction involves the transformation of a phenol (C\(_6\)H\(_5\)OH) with concentrated sulfuric acid (H\(_2\)SO\(_4\)) followed by treatment with concentrated nitric acid (HNO\(_3\)). This forms two compounds, A and B.


Step 2: Identifying compound A.

When phenol is treated with concentrated sulfuric acid (H\(_2\)SO\(_4\)), it undergoes sulfonation to form compound A, which is 2-hydroxybenzenesulfonic acid (C\(_6\)H\(_4\)(OH)SO\(_3\)H). In this compound, there are 7 oxygen atoms in total: 2 from the hydroxyl group (OH), 1 from the sulfonic acid group (SO\(_3\)H), and 4 from the sulfonate group.


Step 3: Identifying compound B.

When compound A is treated with concentrated nitric acid (HNO\(_3\)), it undergoes nitration to form compound B, which is 2-hydroxy-5,6-dinitrobenzenesulfonic acid (C\(_6\)H\(_3\)(OH)(NO\(_2\))\(_2\)SO\(_3\)H). In this compound, there are also 7 oxygen atoms: 1 from the hydroxyl group (OH), 2 from the nitro groups (NO\(_2\)), 1 from the sulfonic acid group (SO\(_3\)H), and 3 from the nitro groups attached to the benzene ring.


Step 4: Conclusion.

The total number of oxygen atoms in both compounds A and B is 7 in A and 7 in B, adding up to a total of 14 oxygen atoms. Therefore, the correct answer is (3) 14.
Quick Tip: In reactions involving aromatic compounds, common reactions like sulfonation and nitration involve adding functional groups that contain oxygen. Always count the oxygen atoms from each functional group.

Step 1: Understanding the reaction.

The reaction involves chlorobenzene, which undergoes a nucleophilic substitution with NaOH at high temperature (620°C), followed by nitration using a mixture of HNO\(_3\) and H\(_2\)SO\(_4\). This forms a nitro group at the ortho position with respect to the hydroxyl group, resulting in ortho-nitrophenol.


Step 2: Conclusion.

The major product formed is ortho-nitrophenol due to the strong activating effect of the hydroxyl group at the ortho position. Hence, the correct answer is (1) Orthonitrophenol.


Quick Tip: In aromatic substitution reactions, electron-donating groups like OH activate the ring, making the ortho and para positions more reactive toward electrophilic substitution.

Step 1: Understanding the boiling points.

The boiling point of a compound depends on its intermolecular forces. Compounds with stronger intermolecular forces will have higher boiling points.


Step 2: Intermolecular forces and their effects on boiling points.

- (P) Diethyl ether: Diethyl ether has a dipole moment but is not involved in hydrogen bonding. Its boiling point is relatively lower than compounds with hydrogen bonding.

- (Q) n-butanol: n-Butanol has hydrogen bonding, which significantly raises its boiling point.

- (R) n-butane: n-Butane has only London dispersion forces, which are weak, leading to the lowest boiling point.

- (S) Ethylmethyl ketone: Ethylmethyl ketone has dipole-dipole interactions, which are stronger than London dispersion forces but weaker than hydrogen bonding.


Step 3: Conclusion.

The correct order of boiling points is (Q > S > P > R). Hence, the correct answer is (2) Q > S > P > R.
Quick Tip: When comparing boiling points, remember that hydrogen bonding increases boiling points, followed by dipole-dipole interactions and London dispersion forces.

Step 1: Understanding Le Chatelier's principle.

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions (concentration, pressure, or temperature), the system will shift in a direction that opposes the change.


Step 2: Evaluating the options.

- (1) Removal of CO\(_2\): This would shift the equilibrium to the right to replace the removed CO\(_2\).

- (2) Addition of CO\(_2\): This would shift the equilibrium to the left to counteract the increase in CO\(_2\).

- (3) Removal of CO: This would shift the equilibrium to the left to replace the removed CO.

- (4) Addition of Fe\(_2\)O\(_3\): This would shift the equilibrium to the right, as it provides more reactant for the reaction to proceed. This is consistent with Le Chatelier's principle.


Step 3: Conclusion.

The correct answer is (4) Addition of Fe\(_2\)O\(_3\), as it will shift the equilibrium to the right according to Le Chatelier's principle.
Quick Tip: When a system at equilibrium is disturbed, the equilibrium will shift to counteract the change. Adding more reactant shifts the equilibrium to the right.

Step 1: Understanding ligand strength.

Ligand strength is determined by the ability of the ligand to donate electron pairs to the metal center. The order of strength is influenced by factors such as the charge and size of the ligand.


Step 2: Evaluating the ligands.

- Br\(^-\) is a weak ligand due to its larger size and lower charge density.

- F\(^-\) is stronger than Br\(^-\) due to its smaller size and higher charge density.

- H\(_2\)O is a neutral molecule and is a stronger ligand than halides due to its ability to form hydrogen bonds and donate electron pairs.

- NH\(_3\) is a strong ligand due to its lone pair of electrons and small size.


Step 3: Conclusion.

The correct order of strength is Br\(^-\) < F\(^-\) < H\(_2\)O < NH\(_3\). Hence, the correct answer is (3).
Quick Tip: In general, smaller and highly charged ligands (like NH\(_3\)) are stronger than larger, less charged ones (like halides).

Step 1: Rate law.

The rate law is given by: \[ Rate = k [A]^x [B]^y \]
Where \(x\) and \(y\) are the orders of reaction with respect to A and B, respectively.

Step 2: Using the given data to find \(x\) (order with respect to A).

Using the first two trials, we compare the rates with respect to the concentration of A, keeping B constant: \[ \frac{6 \times 10^{-3}}{12 \times 10^{-3}} = \frac{[0.1]^x}{[0.4]^x} \]
Simplifying the equation: \[ \frac{1}{2} = \left(\frac{0.1}{0.4}\right)^x = \left(\frac{1}{4}\right)^x \]
Taking the logarithm: \[ \frac{1}{2} = \left(\frac{1}{4}\right)^x \quad \Rightarrow \quad x = 1/2 \]

Step 3: Using the given data to find \(y\) (order with respect to B).

Using the first and third trials, we compare the rates with respect to the concentration of B, keeping A constant: \[ \frac{12 \times 10^{-3}}{48 \times 10^{-3}} = \frac{[0.4]^x [0.1]^y}{[0.4]^x [0.2]^y} \]
Simplifying the equation: \[ \frac{1}{4} = \left(\frac{0.1}{0.2}\right)^y \quad \Rightarrow \quad y = 2 \]

Step 4: Conclusion.

Thus, the order with respect to A is 0.5 and the order with respect to B is 2. Hence, the overall order of reaction is \(0.5 + 2 = 2.5\). Therefore, the correct answer is (2.5).
Quick Tip: When determining the overall order of reaction, combine the individual orders with respect to each reactant. The order can be fractional if it doesn't follow simple integer values.

Step 1: Understanding the spin-only magnetic moment formula.

The spin-only magnetic moment (\(\mu\)) is given by the formula: \[ \mu = \sqrt{n(n + 2)} \]
Where \(n\) is the number of unpaired electrons. The strongest oxidizing agent will be the one with the highest number of unpaired electrons, as it will have the highest magnetic moment.

Step 2: Evaluating each species.

- Ti\(^{2+}\): \(Ti^{2+}\) has 1 unpaired electron. Thus, \[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73 \, \mu_B \]
- V\(^{2+}\): \(V^{2+}\) has 3 unpaired electrons. Thus, \[ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \, \mu_B \]
- Mn\(^{2+}\): \(Mn^{2+}\) has 5 unpaired electrons. Thus, \[ \mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \, \mu_B \]
- Co\(^{3+}\): \(Co^{3+}\) has 6 unpaired electrons. Thus, \[ \mu = \sqrt{6(6 + 2)} = \sqrt{48} \approx 6.93 \, \mu_B \]

Step 3: Conclusion.

Co\(^{3+}\) has the highest magnetic moment, making it the strongest oxidizing agent. Hence, the correct answer is (4) Co\(^{3+}\).
Quick Tip: The higher the number of unpaired electrons, the stronger the magnetic moment, and typically, the stronger the oxidizing agent.

Step 1: Understanding the stoichiometry of the reaction.

The reaction is between \(NH_2\) and Br\(_2\) in the presence of H\(_2\)O to form a white ppt. The molar masses of the compounds involved are used to determine the number of moles.


Step 2: Calculating moles.

- Moles of \(NH_2\) = \(\frac{9.3}{93} = 0.1\) mol (Molar mass of \(NH_2\) = 93 g/mol)

- Moles of white ppt = \(\frac{24.6}{348} = 0.07\) mol (Molar mass of white ppt = 348 g/mol)


Step 3: Calculating the percentage yield.

The theoretical yield would be 0.1 mol of white ppt, and the actual yield is 0.07 mol. Thus, the percentage yield is: \[ Percentage yield = \frac{0.07}{0.1} \times 100 = 70% \]

Step 4: Conclusion.

The percentage yield of the white ppt product is 70%. Hence, the correct answer is (2) 70%.
Quick Tip: To calculate percentage yield, divide the actual yield by the theoretical yield and multiply by 100.

Step 1: Understanding the borax bead test.

In the borax bead test, certain metal cations produce characteristic colors when heated in a reducing flame. The colors depend on the metal ion and its oxidation state.


Step 2: Analyzing the options.

- (1) Iron: Iron ions produce a green color in the flame due to the formation of iron borate in the borax bead test.

- (2) Cobalt: Cobalt typically gives a blue color in the borax bead test.

- (3) Manganese: Manganese gives a pale brown or pink color.

- (4) Nickel: Nickel produces a green color, but it is not typically the same shade as that produced by iron.


Step 3: Conclusion.

The correct answer is (1) Iron, as iron gives a green color in the reducing flame in the borax bead test.
Quick Tip: In the borax bead test, metal ions like iron (Fe\(^2+\)) give a characteristic green color in a reducing flame.

Step 1: Dalton's Atomic Theory.

Dalton’s atomic theory was based on several postulates, the first of which stated that all matter is made up of indivisible atoms. Another postulate states that atoms of the same element are identical in mass and properties. However, we know today that this statement is incorrect, as atoms of the same element can have different masses (due to isotopes).


Step 2: Evaluating the postulates.

- (1) Matter consists of indivisible atoms: This is largely true, but later discoveries like subatomic particles (electrons, protons, neutrons) showed that atoms are divisible. However, this does not invalidate Dalton’s basic theory.

- (2) All atoms of a given element have identical properties but different masses: This is false, as isotopes of the same element have different masses.

- (3) Compounds are formed when atoms of different elements combine in a fixed ratio: This is correct.

- (4) Chemical reactions involve the rearrangement of atoms: This is also correct.


Step 3: Conclusion.

The incorrect postulate is (2), which states that all atoms of a given element have identical properties but different masses.
Quick Tip: Isotopes of the same element have identical chemical properties but different masses due to differing numbers of neutrons.