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

Step 1: Understanding the statements.

- Statement-I suggests that PH3 has a lower boiling point than NH3.

- Statement-II discusses van der Waals forces and hydrogen bonding, which are key factors affecting boiling points.


Step 2: Analyzing the statements.

Statement-I: This is correct. PH3 has a lower boiling point than NH3 because NH3 exhibits strong hydrogen bonding, which is absent in PH3. The boiling point of NH3 is 239.7 K, whereas PH3 boils at 185.5 K.

Statement-II: This is incorrect. While NH3 exhibits hydrogen bonding, PH3 only has weaker van der Waals forces due to the absence of hydrogen bonding.


Step 3: Conclusion.

The correct answer is (3) Statement-I is true but Statement-II is false.
Quick Tip: Boiling points of molecules depend significantly on intermolecular forces like hydrogen bonding. Molecules with hydrogen bonding (like NH3) have higher boiling points than those without (like PH3).

Step 1: Identifying the trigonal bipyramidal geometry.

Trigonal bipyramidal geometry occurs when there are five substituents around a central atom, resulting in two axial and three equatorial positions.


Step 2: Analyzing the given compounds.

- PF5, PBr5, PCl5: These molecules are known to exhibit trigonal bipyramidal geometry due to five bonding atoms around the central phosphorus atom.

- [PtCl4]^2-: This complex ion also has trigonal bipyramidal geometry.

- SF6: This has an octahedral geometry, not trigonal bipyramidal.

- [Fe(CO)5]: This complex shows a trigonal bipyramidal geometry with five CO ligands.


Step 3: Conclusion.

The correct answer is (1) PF5, PBr5, [PtCl4]^2-, SF6, BrF5, PCl5, [Fe(CO)5] only as these compounds exhibit trigonal bipyramidal geometry.
Quick Tip: Trigonal bipyramidal geometry typically occurs with five substituents around a central atom, while SF6 has octahedral geometry.

Step 1: Understanding the de Broglie relation.

The de Broglie wavelength \( \lambda \) is related to the momentum \( p \) of a particle by the equation: \[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. This shows that \( \lambda \) is inversely proportional to \( p \). As \( p \) increases, \( \lambda \) decreases.

Step 2: Analyzing the plot.

- Option (1): Correct. Since \( \lambda \) is inversely proportional to \( p \), the plot between \( \lambda \) and \( p \) will show a curve that decreases as \( p \) increases. This matches the expected relationship.

- Option (2): Incorrect. A linear increasing relationship does not match the inverse proportionality of \( \lambda \) and \( p \).

- Option (3): Incorrect. A linear decreasing relationship also does not fit the inverse relation between \( \lambda \) and \( p \).

- Option (4): Incorrect. The plot with a minimum at \( \frac{1}{p} \) does not match the inverse proportionality.


Step 3: Conclusion.

The correct answer is (1) Curve decreasing exponentially, as the relationship between \( \lambda \) and \( p \) follows an inverse proportionality.
Quick Tip: The de Broglie wavelength decreases as the momentum of the particle increases. This inverse relationship is crucial in understanding the wave-like properties of matter.

Step 1: Identifying the reactions.

The given reaction involves the chlorination of methyl chloride (\( CH_3Cl \)) with chlorine in the presence of light, leading to the formation of a product A. The second step involves treatment with NaOH, which would lead to the formation of a carboxylate ion (forming B after hydrolysis). The second reaction is most likely a reduction to an aldehyde.


Step 2: Conclusion.

Thus, the correct answer is (3) (A) \(CH_3Cl\), (B) \(CHO\).
Quick Tip: When halogenating alkyl compounds and following with NaOH treatment, oxidation and reduction processes can occur, leading to aldehydes or carboxylic acids.

Step 1: Formula for molar heat capacity.

The molar heat capacity of a mixture of gases is the weighted average of the heat capacities of the individual gases: \[ C_{v,m} = \frac{2 \times \left(\frac{3R}{2}\right) + 6 \times \left(\frac{5R}{2}\right)}{2 + 6} \]

Step 2: Calculation.

Substituting the values: \[ C_{v,m} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9R}{4} \]

Step 3: Conclusion.

The correct answer is (1) \( \frac{9R}{4} \).
Quick Tip: When calculating the heat capacity of a mixture, consider the mole fractions and the individual heat capacities of each component.

Step 1: Analyzing Statement-I.

The –NH\(_2\) group in aniline is an electron-donating group, which increases the electron density on the benzene ring, making it more reactive towards electrophilic aromatic substitution (EAS) reactions. Therefore, the statement that it is a strong deactivating group for all ESR is incorrect.


Step 2: Analyzing Statement-II.

Aniline does not show Friedel-Craft alkylation reaction because the –NH\(_2\) group is too basic and forms a complex with the catalyst (AlCl\(_3\)), deactivating the ring for this reaction. Therefore, Statement-II is correct.


Step 3: Conclusion.

Since Statement-I is incorrect and Statement-II is correct, the correct answer is (4) Statement-I is incorrect and statement-II is correct.
Quick Tip: For electrophilic aromatic substitution reactions, the presence of electron-donating groups like –NH\(_2\) makes the ring more reactive. However, in Friedel-Craft reactions, such groups can interfere by coordinating with the catalyst, preventing the reaction.

Step 1: Understanding Bond Cleavage.

In ionic reactions, heterolytic bond cleavage occurs where ions are formed. This is characteristic of reactions in which one atom retains both electrons of the bond, resulting in the formation of cations and anions.


Step 2: Analyzing the options.

(1) Heterolytic cleavage: Correct. In heterolytic cleavage, ions are formed, making it the correct process for ionic reactions in organic compounds.

(2) Homolytic cleavage: This involves the splitting of a bond where each atom retains one electron, forming free radicals. It is not relevant in ionic reactions.

(3) Free radical: This is not a type of bond cleavage but refers to a species with unpaired electrons, typically formed during homolytic cleavage.

(4) No cleavage of bond: This option is incorrect because cleavage is required in ionic reactions.


Step 3: Conclusion.

The correct answer is (1) Heterolytic cleavage, as this is the bond cleavage type in ionic reactions.
Quick Tip: In ionic reactions, look for heterolytic cleavage where ions are formed, not homolytic cleavage or no cleavage.

Step 1: Understanding Ka values.

The \(Ka\) value represents the acid dissociation constant, which indicates the strength of an acid. A higher \(Ka\) value means a stronger acid.


Step 2: Analyzing the options.

The given \(Ka\) values for the acids are:

- Acid A: \(10^{-3}\) (strongest)

- Acid B: \(5 \times 10^{-9}\)

- Acid C: \(9 \times 10^{-11}\) (weakest)


Therefore, the order of acidic strength is A > B > C based on the higher \(Ka\) values.


Step 3: Conclusion.

The correct answer is (1) A > B > C, as acid A has the highest \(Ka\) value, indicating the strongest acid.
Quick Tip: Remember: A higher \(Ka\) value means a stronger acid. Compare the \(Ka\) values to determine the strength of the acids.

Step 1: Understanding Disproportionation Reactions.

A disproportionation reaction is a reaction in which a substance (element) is simultaneously oxidised and reduced.


Step 2: Analyzing the options.

(A) \( Cu^+ \longrightarrow Cu^{2+} + Cu \): This is a disproportionation reaction because \( Cu^+ \) is both oxidized to \( Cu^{2+} \) and reduced to \( Cu \).

(B) \( MnO_4^{2-} \longrightarrow MnO_4^+ + MnO_2 \): This is also a disproportionation reaction. The \( MnO_4^{2-} \) ion is oxidized to \( MnO_4^+ \) and reduced to \( MnO_2 \).

(C) \( H_2O_2 \longrightarrow O_2 + H_2O \): This is a disproportionation reaction as \( H_2O_2 \) is both oxidized to \( O_2 \) and reduced to \( H_2O \).

(D) \( Cr_2O_7^{2-} \longrightarrow Cr^{3+} + H_2O \): This is not a disproportionation reaction because the chromium is only reduced, not oxidized.


Step 3: Conclusion.

The correct answer is (4) A, B and C only because reactions A, B, and C all involve a substance being both oxidized and reduced.
Quick Tip: Disproportionation reactions involve the simultaneous oxidation and reduction of the same element. Always look for reactions where the same element undergoes both processes.

We will use the Nernst equation for the calculation of \( E_{cell} \).

\[ E_{cell} = 0 - \frac{0.06}{2} \log 2 \]

Step 1: Calculation.
\[ E_{cell} = -\frac{0.06}{2} \log 2 \] \[ = -0.03(0.3) \] \[ = -0.009 \] \[ E_{cell} = -9 \times 10^{-3} \, V \]

Step 2: Conclusion.

Thus, \( x = 9 \).
Quick Tip: Use the Nernst equation to calculate the cell potential for non-standard conditions, and remember to adjust for the number of electrons transferred.

Step 1: Understanding Amphoteric Oxides.

Amphoteric oxides are those that can react with both acids and bases.


Step 2: Analyzing the oxides.

- \( SnO_2 \), \( PbO_2 \), and \( Al_2O_3 \) are amphoteric oxides. They can react with both acids and bases.

- \( SiO_2 \), \( P_2O_5 \), and \( CO_2 \) are acidic oxides. They only react with bases.

- \( CO \), \( NO \), and \( N_2O \) are neutral oxides. They do not react significantly with acids or bases.


Step 3: Conclusion.

Therefore, the total number of amphoteric oxides is 3: \( SnO_2 \), \( PbO_2 \), and \( Al_2O_3 \).
Quick Tip: Amphoteric oxides can react with both acids and bases. Keep in mind that acidic oxides react with bases, and neutral oxides have minimal reactivity with both.

Step 1: Understanding the equation.

The given formula for carbon dating is: \[ \left( \frac{{C^{14}}}{{C^{12}}} \right)_t = \left( \frac{{C^{14}}}{{C^{12}}} \right)_0 \left( \frac{1}{2} \right)^n \]
where \( n \) is the number of half-lives, \( t \) is the time, and the half-life of \( C^{14} \) is given as 1580 years.


Step 2: Calculating the time.

We are told that \( n = 3 \), so the time \( t \) is: \[ t = 3 \times 1580 = 4740 \, years \]

Step 3: Conclusion.

Thus, the life of the wood sample is 4740 years.
Quick Tip: In carbon dating, the number of half-lives is used to estimate the age of the sample. The formula for half-life decay is \( \left( \frac{1}{2} \right)^n \), where \( n \) is the number of half-lives.

Step 1: Understanding the formula for energy levels.

The energy of an electron in a hydrogen atom at a particular energy level is given by the formula: \[ \Delta E = 13.6 \, eV \times \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right) \]
where \( n_1 = 1 \) (ground state) and \( n_2 = 2 \) (1st excited state).


Step 2: Calculating the energy.
\[ \Delta E = 13.6 \times \left( \frac{1}{{1}^2} - \frac{1}{{2}^2} \right) \] \[ \Delta E = 13.6 \times (1 - 0.25) \] \[ \Delta E = 13.6 \times 0.75 = 10.05 \, eV \]

Step 3: Conclusion.

Thus, the minimum energy required for the electron to excite from the ground state to the 1st excited state is 10.05 eV.
Quick Tip: The energy required for an electron to move between energy levels in a hydrogen atom can be calculated using the formula \( \Delta E = 13.6 \times \left( \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right) \) where \( n_1 \) and \( n_2 \) are the principal quantum numbers.

The given reaction is: \[ PbCl_2 + (NH_4)_2SO_4 \longrightarrow PbSO_4 \downarrow + 2 \, NH_4Cl \]

Step 1: Use the mole ratio.

From the balanced equation, we see that 1 mole of \( PbCl_2 \) reacts with 1 mole of \( (NH_4)_2SO_4 \) to form 1 mole of \( PbSO_4 \).


Step 2: Identify the limiting reactant.

We are given 72 moles of \( PbCl_2 \) and 50 moles of \( (NH_4)_2SO_4 \). According to the reaction, both reactants are in a 1:1 molar ratio. The limiting reactant here is \( (NH_4)_2SO_4 \), as there are fewer moles of it (50 moles).


Step 3: Calculate moles of precipitate.

Since 1 mole of \( (NH_4)_2SO_4 \) reacts with 1 mole of \( PbCl_2 \) to form 1 mole of \( PbSO_4 \), 50 moles of \( (NH_4)_2SO_4 \) will form 50 moles of \( PbSO_4 \).


Step 4: Conclusion.

Thus, 50 moles of \( PbSO_4 \) are formed as the precipitate.
Quick Tip: In reactions, the limiting reactant determines the amount of product formed. Use the mole ratio to identify the limiting reactant and calculate the moles of the product.