JEE Main 2023 Chemistry April 11 Shift 1 Question Paper with Solution PDF- Download Here

The reaction of L-tetrose with HNO\(_3\) results in the formation of a compound with two chiral centers, and on reduction with NaBH\(_4\), a compound (B) is formed which is optically active.

Quick Tip: For carbohydrates, reactions such as Schiff’s test and acetylation help identify the presence of aldehyde groups and determine the stereochemistry of the compound.

- A. KCl: KCl is used in thermoluminescent reactions, making it related to the light-emitting properties of certain compounds.
- B. KCl: KCl is also a fertilizer as it contains potassium and chlorine.
- C. KOH: KOH is essential for the sodium-potassium pump, which is crucial for cell function and active transport.
- D. KOH: KOH is an absorber of CO\(_2\) because of its basic nature and ability to neutralize CO\(_2\).

Thus, the correct matching is A-IV, B-III, C-I, D-II. Quick Tip: KCl is often used as a fertilizer because potassium is an essential nutrient for plant growth, and KOH’s role in the sodium-potassium pump is vital for maintaining cellular function.

- Ga forms bond with Cl: GaCl\(_3\) shows ionic bonding, and the bond between Ga and Cl is ionic.
- Ga is coordinated with Cl: Ga in GaCl\(_3\) is in the +3 oxidation state, which results in covalent bonding with Cl.
- Ga is more electronegative than Cl: This statement is incorrect as Ga has lower electronegativity than chlorine.
- Oxidation state of Ga: In GaCl\(_3\), Ga is in the +3 oxidation state.

Therefore, the correct answer is the oxidation state of Ga in GaCl\(_3\) is +3. Quick Tip: In GaCl\(_3\), Ga is in the +3 oxidation state, and this compound forms a covalent bond with chlorine.

According to the observation, A is more mobile and interacts with the mobile phase more than C, and C is more than B.

This suggests that A is more mobile, C is less mobile, and B has the least mobility. Hence the correct order is A > C > B. Quick Tip: Thin layer chromatography helps in understanding the relative mobility of different substances based on their interaction with the mobile and stationary phases.

- When a mixture containing the acetates reacts with ferric chloride (FeCl\(_3\)), it forms a complex and the characteristic brown red precipitate and coloration is formed, which then disappears on boiling.
- The reaction suggests the presence of CH\(_3\)COO\(^-\) and FeCl\(_3\) in the mixture, leading to the formation of blood-red and brown colours.

Thus, the correct answer is (3), CH\(_3\)COO\(^-\) & FeCl\(_3\). Quick Tip: When testing for specific ions in inorganic salts, observe changes in colour and precipitate formation which can give clues about the compounds involved.

- Low density polyethylene (LDPE) is made by free radical polymerization of ethylene. It is prepared under high pressure and with a catalyst like triethylaluminium and titanium tetrachloride.
- LDPE has a high degree of branching, making it less dense and more flexible.

Thus, the correct answer is (3), Low density polyethylene. Quick Tip: Polymerization under high pressure and specific catalysts leads to the formation of low-density polymers with high flexibility.

- A. H\(_2\)O: Water has a bent shape due to the lone pairs on oxygen, making the geometry bent.
- B. Acetylene: Acetylene is a linear molecule with a triple bond between the carbons.
- C. NH\(_3\): Ammonia has a trigonal pyramidal shape due to the lone pair on nitrogen.
- D. ClO\(_2\): Chlorine dioxide is a bent molecule due to the lone pairs on chlorine.

Thus, the correct matching is A-I, B-II, C-III, D-IV. Quick Tip: Remember that the geometry of molecules is determined by the number of bonding pairs and lone pairs on the central atom, according to the VSEPR theory.

- Statement I is correct. When methane reacts with steam over a nickel catalyst, hydrogen gas is produced in the reaction:
\[ CH_4(g) + H_2O(g) \xrightarrow{Ni} CO(g) + 3H_2(g) \]
- Statement II is correct. Sodium nitrate reacts with ammonium chloride to give sodium chloride and water in the reaction:
\[ NaNO_3(aq) + NH_4Cl(aq) \rightarrow NaCl(aq) + NaNO_3(aq) + H_2O \]

Thus, the correct answer is (4), Both the statements I and II are correct. Quick Tip: The reaction of methane and steam over a nickel catalyst is a classic example of a reforming reaction, where hydrogen gas is generated.

- Benzene becomes more reactive towards electrophilic aromatic substitution (EAS) when an electron-donating group (-OH, -NH\(_2\)) is attached to the benzene ring.


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- Benzene with electron-withdrawing groups (-NO\(_2\), -COOH) decreases the reactivity, as it deactivates the ring towards electrophilic attack.

Thus, the order is: \[ (a) < (b) < (c) \] Quick Tip: Remember that groups like -OH and -NH\(_2\) activate the aromatic ring by donating electron density, whereas -NO\(_2\) and -COOH deactivates it by withdrawing electron density.

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When o-phenylenediamine is treated with nitric acid, nitration occurs at the position ortho to the amino group. The nitro group (\(NO_2\)) replaces the hydrogen attached to the benzene ring, forming a nitrated product. In this case, the major product formed is an imine (\(C = N\)) structure, which is the result of the reaction between the amino group and the nitro group.

Thus, the correct product is a structure with a carbon-nitrogen double bond (C = N). Quick Tip: When performing reactions involving amines, the functional group often reacts with the reagent to form imines or other derivatives based on the substitution pattern of the aromatic ring.

In the extraction of copper, when copper oxide (\(CuO\)) is reduced by hydrogen gas (\(H_2\)), the copper is reduced to its metallic form and water vapor is produced as a byproduct. This process produces what is referred to as reduced copper.

Blister copper, on the other hand, refers to copper that is in a nearly pure state but still contains small amounts of other elements such as sulfur, which can create small bubbles (blisters) on its surface during the smelting process.

Thus, the correct product is reduced copper, which is the result of the reduction of \(CuO\) with \(H_2\). Quick Tip: The reduction of metal oxides with hydrogen gas results in the formation of pure metals, while the process in smelting where impurities cause bubbling forms blister copper.

On adding \( AgNO_3 \) to KI, \( AgI \) will form, and the solubility of \( AgI \) is very low. As a result, only \( Ag⁺ \) will remain in very small quantities in the solution. The majority of the \( I⁻ \) ions will precipitate out as \( AgI \), leaving very few ions in solution.

Thus, the correct answer is \( Ag⁺ \) ions only. Quick Tip: When mixing solutions of ionic compounds, check the solubility products (Ksp) to predict whether the ions will precipitate or remain in solution.

- Statement I: A BOD value of 4 ppm and a dissolved oxygen level of 8 ppm indicate that the water has a good quality because the dissolved oxygen should be higher than the BOD to support aquatic life.
- Statement II: If the concentration of zinc and nitrate is 5 ppm, this is within permissible limits for drinking water, indicating good quality.

Both statements are correct as they provide accurate interpretations of water quality standards. Quick Tip: When evaluating water quality, check for parameters like BOD, dissolved oxygen, and concentrations of hazardous elements such as zinc and nitrates.

The reaction is: \[ Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe \]

Now calculate the heat evolved:
\[ \Delta H = (\Delta H_f^{\circ} (Al_2O_3)) - (\Delta H_f^{\circ} (Fe_2O_3)) \]

Substitute the given values:
\[ \Delta H = (-1700) - (-840) = -860 \, KJ/mol \]

Moles of \(Fe_2O_3\) and Al are in the ratio 1 : 2. So, 1 mole of \(Fe_2O_3\) weighs \(112 \, g\), and 2 moles of Al weigh \(54 \, g\).

Total mass of the mixture is:
\[ 160 + 54 = 214 \, g \]

Now, the heat evolved per gram of the mixture is:
\[ \frac{-860 \, KJ}{214} \approx -4.01 \, KJ/g \]

Thus, the heat evolved per gram is approximately 4 KJ. Hence, the correct answer is (4). Quick Tip: To calculate the heat evolved per gram, divide the total heat by the total mass of the mixture.

Rate of reaction is given by the change in concentration of FeSO₄:
\[ \frac{-\Delta [FeSO₄]}{\Delta t} \]

Substitute the given values:
\[ \frac{-10 + 8.8}{30 \times 60} = \frac{1.2}{1800} = 6.67 \times 10^{-4} \]

From the given reaction, the rate of production of Fe₂(SO₄)₃ is related to the rate of FeSO₄:
\[ \frac{1}{6} \times \frac{-\Delta [FeSO₄]}{\Delta t} \]

Substitute the value of \(\frac{-\Delta [FeSO₄]}{\Delta t}\):
\[ Rate of production of Fe_2(SO₄)_3 = \frac{3}{6} \times 6.67 \times 10^{-4} = 333.33 \times 10^{-6} \]

Thus, the rate of production of Fe₂(SO₄)₃ is \(333 \times 10^{-6}\) mol L⁻¹ s⁻¹. Hence, the correct answer is (1). Quick Tip: The rate of production of a product in a reaction can be calculated by determining the rate of change of concentration of a reactant and using stoichiometric ratios.

For isotonic solution: \[ i(glucose) = i(K_2SO_4) \] \[ 0.01 = i(K_2SO_4) \times 0.004 \] \[ i(K_2SO_4) = \frac{0.01}{0.004} = 2.5 \]

Now, for \( K_2SO_4 \): \[ i = 1 + (n-1) \] \[ 2.5 = 1 + (n-1) \] \[ n = 3 for K_2SO_4 \]

Percentage dissociation: \[ \alpha = \frac{3}{2} = 75% \]

Thus, the percentage dissociation of K_2SO_4 is 75%. The correct answer is (2). Quick Tip: For isotonic solutions, use the equation \( i = 1 + (n-1) \) to find the number of particles and dissociation percentage.

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In the given reaction, the carbocation formed after the removal of \( H^+ \) can undergo several hyperconjugation structures. We see that the carbocation can shift, and the hyperconjugation structures involve the bonding electrons of the adjacent C-H bonds.

After shifting, the carbocation is stabilized by the hyperconjugation effect, and counting all the hyperconjugation structures (including the shift), there are 7 such structures.

Thus, the correct number of hyperconjugation structures is 7. Therefore, the correct answer is (2). Quick Tip: Hyperconjugation involves the delocalization of electrons from adjacent C-H or C-C bonds to stabilize carbocations.

From the given equation, the change in concentrations is: \[ At equilibrium \quad 1 - 0.4 = 0.6 \quad 0.4 \quad 0.4 \] \[ K_c = \frac{0.4 \times 0.4}{0.6 \times 0.6} = \frac{0.16}{0.36} = 0.444 \approx 44 \]

Thus, the equilibrium constant is \( 44 \times 10^7 \). Therefore, the correct answer is (1). Quick Tip: To calculate the equilibrium constant, use the relationship between the concentrations of the products and reactants at equilibrium.

Given:
- Molar mass \( M = 12 \) g mol\(^{-1}\)
- Density \( d = 3.0 \) g mL\(^{-1}\)
- Edge length \( a = 300 \) pm \( = 300 \times 10^{-12} \) m
- Avogadro's number \( N_A = 6.02 \times 10^{23} \) mol\(^{-1}\)

Formula for number of atoms in one unit cell: \[ Z = \frac{N_A \times M}{d \times a^3} \]
Substitute values: \[ Z = \frac{6.02 \times 10^{23} \times 12}{3.0 \times (300 \times 10^{-12})^3} \] \[ Z = 40.635 \times 10^{21} = 6 \]

Thus, the number of atoms present in one unit cell is 6. The correct answer is (4). Quick Tip: The number of atoms in a unit cell can be calculated using the formula \( Z = \frac{N_A \times M}{d \times a^3} \).

The given reaction is: \[ Pb^{2+} + 2e^- \rightarrow Pb \] \[ E^{\circ} = m \quad and \quad \Delta G^{\circ} = -2Fm \] \[ Pb^{4+} + 4e^- \rightarrow Pb^{2+} \] \[ E^{\circ} = n \quad and \quad \Delta G^{\circ} = -4Fn \]
Now, \[ \Delta G^{\circ} = \Delta G^{\circ}_1 - \Delta G^{\circ}_2 \] \[ -2Fm = -4Fn \] \[ 2FE = 2Fm + 4Fn \quad \Rightarrow \quad E^{\circ} = m - 2n \]

Thus, the value of \( x \) is 2. The correct answer is (2). Quick Tip: When dealing with electrochemical reactions, the relationship between the potentials and the standard Gibbs free energies can help calculate the overall cell potential.