JEE Main 2025 April 4 Shift 1 Chemistry Question Paper, Exam Analysis, and Answer Keys (Available)

1. Normal osmosis occurs from chamber 2 to chamber 1.
2. For reverse osmosis from chamber 1 to chamber 2, the pressure \(\mathrm{p}_{1}\) must be greater than the osmotic pressure \(\pi\).
3. Therefore, the correct conditions are A and C.

Therefore, the correct answer is (3) A and C only. Quick Tip: Reverse osmosis requires the pressure in the chamber with higher concentration to be greater than the osmotic pressure.

1. For a reaction to be spontaneous, \(\Delta \mathrm{G} < 0\).
\[ \Delta \mathrm{G} = \Delta \mathrm{H} - \mathrm{T} \Delta \mathrm{S} \]
2. Given that both \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are positive:
\[ \Delta \mathrm{G} = \Delta \mathrm{H} - \mathrm{T} \Delta \mathrm{S} < 0 \]
\[ \mathrm{T} > \frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} = \mathrm{T}_{\mathrm{e}} \]

Therefore, the correct answer is (3) \(\mathrm{T}>\mathrm{T}_{\mathrm{e}}\). Quick Tip: A reaction is spontaneous when the Gibbs free energy change is negative.

1. Number of unpaired electrons:
- (A) \(\mathrm{O}_{2}\): 2
- (B) \(\mathrm{N}_{2}\): 0
- (C) \(\mathrm{F}_{2}\): 0
- (D) \(\mathrm{S}_{2}\): 2
- (E) \(\mathrm{Cl}_{2}\): 0

2. Paramagnetic behavior:
- Molecules with unpaired electrons exhibit paramagnetic behavior.
- Therefore, \(\mathrm{O}_{2}\) and \(\mathrm{S}_{2}\) are paramagnetic.

Therefore, the correct answer is (4) A \& D only. Quick Tip: Molecules with unpaired electrons are paramagnetic.

1. Intramolecular aldol condensation products:
- (1), (2), and (3) are products of intramolecular aldol condensation.
- (4) is a product of intermolecular aldol condensation.

Therefore, the correct answer is (4). Quick Tip: Intramolecular aldol condensation involves the reaction within the same molecule, while intermolecular aldol condensation involves the reaction between two different molecules.

1. Given:
- Isothermal expansion from \(10 \mathrm{dm}^{3}\) to \(20 \mathrm{dm}^{3}\) at \(300 \mathrm{~K}\).
- \(\mathrm{R} = 8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

2. Calculate the work done (w):
\[ w = -nRT \ln \frac{V_2}{V_1} \]
\[ w = -8.3 \times 300 \times \ln \left( \frac{20}{10} \right) \]
\[ w = -1.718 \mathrm{~kJ} \]

3. Calculate the heat transferred (q):
\[ q = -w = 1.718 \mathrm{~kJ} \]

4. Calculate the change in internal energy (\(\Delta U\)):
\[ \Delta U = 0 \quad (since \Delta T = 0) \]

Therefore, the correct answer is (4) \(0,178 \mathrm{~kJ},-1.718 \mathrm{~kJ}\). Quick Tip: For an isothermal process, the change in internal energy is zero.

1. \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\):
- \(\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}\)
- \(\mathrm{CN}^{-}\) is a strong field ligand.
- \(\mu = 0\)

2. \(\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\):
- \(\mathrm{Co}^{3+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}\)
- \(\mathrm{NH}_{3}\) is a strong field ligand.
- \(\mu = 0\)

3. \(\left[\mathrm{FeF}_{6}\right]^{4}\):
- \(\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}\)
- \(\mathrm{F}^{-}\) is a weak field ligand.
- \(\mu = 0\)

4. \(\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4}\):
- \(\mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}\)
- \(\mathrm{SCN}^{-}\) is a weak field ligand.
- \(\mu = \sqrt{35} \mathrm{~BM} = 5.96 \mathrm{~BM}\)
- \(\Delta_{0} = 0\)

Therefore, the correct answer is (4) \(\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4}\). Quick Tip: The magnetic moment and crystal field stabilization energy depend on the ligand field strength.

1. Given:
- \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}\)
- \(\mathrm{E}_{\mathrm{f}} = 180 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- \(\mathrm{E}_{\mathrm{b}} = 200 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

2. Calculate the enthalpy change (\(\Delta \mathrm{H}\)):
\[ \Delta \mathrm{H} = \mathrm{E}_{\mathrm{f}} - \mathrm{E}_{\mathrm{b}} = 180 \mathrm{~kJ} \mathrm{~mol}^{-1} - 200 \mathrm{~kJ} \mathrm{~mol}^{-1} = -20 \mathrm{~kJ} \mathrm{~mol}^{-1} \]

3. Effect of catalyst:
- Catalyst lowers the activation energy but does not change the Gibbs free energy change (\(\Delta \mathrm{G}\)) or the enthalpy change (\(\Delta \mathrm{H}\)) of the reaction.

Therefore, the correct answer is (1) Catalyst does not alter the Gibbs energy change of a reaction. Quick Tip: Catalysts lower the activation energy but do not change the thermodynamic properties of the reaction.

1. Initial rate law:
\[ \mathrm{r}_{1} = \mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} \]

2. New concentrations:
- Concentration of A is doubled: \(2[\mathrm{A}]\)
- Concentration of B is halved: \(\frac{[\mathrm{B}]}{2}\)

3. New rate law:
\[ \mathrm{r}_{2} = \mathrm{k}(2[\mathrm{A}])^{\mathrm{n}} \left( \frac{[\mathrm{B}]}{2} \right)^{\mathrm{m}} \]
\[ \mathrm{r}_{2} = \mathrm{k} \cdot 2^{\mathrm{n}} [\mathrm{A}]^{\mathrm{n}} \cdot \frac{[\mathrm{B}]^{\mathrm{m}}}{2^{\mathrm{m}}} \]

4. Ratio of new rate to initial rate:
\[ \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}} = \frac{\mathrm{k} \cdot 2^{\mathrm{n}} [\mathrm{A}]^{\mathrm{n}} \cdot \frac{[\mathrm{B}]^{\mathrm{m}}}{2^{\mathrm{m}}}}{\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}} = 2^{\mathrm{n}} \cdot \frac{1}{2^{\mathrm{m}}} = 2^{(\mathrm{n}-\mathrm{m})} \]

Therefore, the correct answer is (1) \(2^{(\mathrm{n}-\mathrm{m})}\). Quick Tip: The rate law depends on the concentrations of the reactants raised to their respective orders.

1. \(\left[\mathrm{CrCl}_{3}(\mathrm{py})_{3}\right]\):
- Facial and meridional isomers are possible.
- Total stereoisomers = 2.

2. \(\left[\mathrm{CrCl}_{2}(\mathrm{ox})_{2}\right]^{3-}\):
- Geometrical isomers: cis and trans.
- Optical isomers for cis: 2.
- Optical isomers for trans: 1.
- Total stereoisomers = 3.

Therefore, the correct answer is (3) \(2 \& 3\). Quick Tip: Stereoisomers include geometrical and optical isomers.

1. Reaction sequence:
- The major product formed is (2).

Therefore, the correct answer is (2). Quick Tip: Follow the reaction sequence to determine the major product.

1. Anode reaction:
- \(\mathrm{PbSO}_{4}\) is reduced to \(\mathrm{Pb}\).
- \(\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{0}\)
- \(\mathrm{x}_{1} = +2\), \(\mathrm{y}_{1} = 0\)

2. Cathode reaction:
- \(\mathrm{PbSO}_{4}\) is oxidized to \(\mathrm{PbO}_{2}\).
- \(\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}\)
- \(\mathrm{x}_{2} = +2\), \(\mathrm{y}_{2} = +4\)

Therefore, the correct answer is (2) \(+2,0,+2,+4\). Quick Tip: The oxidation states change during the charging of a lead-acid battery.

1. Statement I:
- Nitrogen can form oxides with oxidation states from +1 to +5 due to the formation of \(\mathrm{p} \pi-\mathrm{p} \pi\) bonds with oxygen.
- This statement is true.

2. Statement II:
- Nitrogen does not form halides with a +5 oxidation state due to the absence of d-orbitals.
- This statement is true.

Therefore, the correct answer is (4) Both Statement I and Statement II are true. Quick Tip: Nitrogen's ability to form oxides and halides depends on its electronic configuration and the availability of d-orbitals.

1. Reaction sequence:
- Benzene treated with oleum produces benzene sulfonic acid (X).
- Heating with molten sodium hydroxide followed by acidification produces phenol (Y).
- Treatment with zinc metal reduces phenol to cyclohexanol (Z).

Therefore, the correct answer is (2). Quick Tip: Follow the reaction sequence to identify the final product.

1. Reactants:
- \(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}\) (Glycine chloride)
- \(\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}\) (Alanine)

2. Reaction:
- The reaction between these reactants with the elimination of HCl will produce the dipeptide Gly-Ala.

Therefore, the correct answer is (1) \(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COCl}\) and \(\mathrm{NH}_{2}-\mathrm{CH}-\mathrm{COOH}\). Quick Tip: The formation of a dipeptide involves the reaction between an amino acid and its chloride derivative with the elimination of HCl.

1. Incorrect pair:
- \(\mathrm{Al}<\mathrm{Ga}\)

2. Ionic radius comparison:
- \(\mathrm{Al}^{3+}<\mathrm{Ga}^{3+}\)
- The atomic number of \(\mathrm{Ga}\) is 31.

Therefore, the correct answer is (1) 31. Quick Tip: The atomic radius and ionic radius depend on the atomic number and the periodic trends.

1. Compound (X):
- \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (Acetone)

2. Reduction to (Y):
- \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) (Isopropyl alcohol)

3. Reaction with HBr to form (Z):
- \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\) (2-Bromopropane)

4. Grignard reagent and reaction with (X):
- The Grignard reagent formed from (Z) reacts with acetone to form 2,3-dimethylbutan-2-ol after hydrolysis.

Therefore, the correct answer is (2) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\). Quick Tip: Follow the reaction sequence to identify the compounds involved.

% Option
(A) Step 1: Bromination (Br\(_2\)/hv)
\[\ce{CH3 ->[Br2/hv] CH2Br}\]

% Option
(B) Step 2: Elimination (Alcoholic KOH)
\[\ce{CH2Br ->[Alc. KOH][\Delta] CH2=}\]

% Option
(C) Step 3: Anti-Markovnikov addition (HBr/ROOR)
\[\ce{CH2= ->[HBr/ROOR][hv] Br-CH3}\]


Mechanistic Explanation:

Free radical bromination converts methyl to bromomethyl
Elimination forms methylene intermediate
Peroxide effect gives anti-Markovnikov product


Therefore, the correct answer is \boxed{(2) \ce{Br-CH3. Quick Tip: Key Points: Radical bromination prefers allylic position Alcoholic KOH causes elimination ROOR reverses normal addition orientation

1. Analysis of Statement I:

For \chemfig{CH_3-CH=CH-CH=O: \(\mu = q \times d\)
Conjugated system creates greater charge separation
More distance between charges than in saturated compound
Therefore, greater dipole moment
Statement I is TRUE


2. Analysis of Statement II:

In \chemfig{CH_3-CH=CH-CH=O, \(C_1-C_2\) has partial double bond character
Double bond character means shorter bond length
Compared to pure single bond in \chemfig{CH_3-CH_2-CH_2-CH=O
Statement II is FALSE (actual bond length is shorter) Quick Tip: Key concepts: Dipole moment depends on charge magnitude and separation distance Conjugation affects both electronic distribution and bond lengths Partial double bond character decreases bond length

1. \(\mathrm{V}^{2+}\):
- \(\mathrm{V}^{2+} \Rightarrow 3 \mathrm{~d}^{3} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 3

2. \(\mathrm{Co}^{2+}\):
- \(\mathrm{Co}^{2+} \Rightarrow 3 \mathrm{~d}^{7} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 3

3. \(\mathrm{Ti}^{2+}\):
- \(\mathrm{Ti}^{2+} \Rightarrow 3 \mathrm{~d}^{2} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 2

4. \(\mathrm{Fe}^{3+}\):
- \(\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 5

5. \(\mathrm{Cr}^{2+}\):
- \(\mathrm{Cr}^{2+} \Rightarrow 3 \mathrm{~d}^{4} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 4

6. \(\mathrm{Ti}^{3+}\):
- \(\mathrm{Ti}^{3+} \Rightarrow 3 \mathrm{~d}^{1} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 1

7. \(\mathrm{Mn}^{2+}\):
- \(\mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}\)
- Number of unpaired electrons = 5

Therefore, the correct answer is (1) \(\mathrm{V}^{2+}, \mathrm{Co}^{2+}\). Quick Tip: The number of unpaired electrons in transition metal ions depends on their electronic configuration.

1. Probability density:
- The probability density of finding the electron is maximum at the nucleus.

2. Distance from the nucleus:
- The electron can be found at a distance \(2 \mathrm{a}_{0}\) from the nucleus.

3. Spherical symmetry:
- The 1s orbital is spherically symmetrical.

4. Total energy:
- The total energy of the electron is maximum when it is at a distance \(\mathrm{a}_{0}\) from the nucleus. This statement is incorrect.

Therefore, the correct answer is (4). Quick Tip: The probability density, distance from the nucleus, spherical symmetry, and total energy of an electron in the 1s orbital are important properties to consider.

1. Calculate the partial pressure of \(\mathrm{N}_{2}\):
\[ \mathrm{p}_{\mathrm{N}_2} = 900 \mathrm{~mm~Hg} - 15 \mathrm{~mm~Hg} = 885 \mathrm{~mm~Hg} \]

2. Calculate the moles of \(\mathrm{N}_{2}\):
\[ Moles of \mathrm{N}_2 = \frac{885 \mathrm{~mm~Hg} \times 0.15 \mathrm{~L}}{0.0821 \mathrm{~L~atm/mol} \times 300 \mathrm{~K}} = 0.0071 \mathrm{~mol} \]

3. Calculate the percentage of nitrogen:
\[ Percentage of nitrogen = \frac{0.0071 \mathrm{~mol} \times 28 \mathrm{~g/mol}}{1 \mathrm{~g}} \times 100 = 19.85% \approx 20% \]

Therefore, the correct answer is (20). Quick Tip: Use the ideal gas law to calculate the moles of nitrogen and then determine the percentage composition.

1. Oxidation states of Mn:
- Reactant: \(\mathrm{Mn}^{7+}\)
- Product: \(\mathrm{Mn}^{2+}\)
- Difference in oxidation states: \(X = 7 - 2 = 5\)

2. Brown red precipitate:
- The brown red precipitate is \(\mathrm{Fe}(\mathrm{OH})_2(\mathrm{CH}_3\mathrm{COO})_n\).
- \(\mathrm{Fe}^{3+}\) has 5 d-electrons.
- Therefore, \(Y = 5\).

3. Calculate \(\mathrm{X}+\mathrm{Y}\):
\[ \mathrm{X} + \mathrm{Y} = 5 + 5 = 10 \]

Therefore, the correct answer is (10). Quick Tip: Determine the oxidation states and the number of d-electrons to find the values of X and Y.

1. Calculate the mass of iron required:
\[ Mass of iron = \frac{12 \mathrm{~ppm} \times 150 \mathrm{~kg}}{10^6} = 1.8 \mathrm{~g} \]

2. Calculate the moles of iron:
\[ Moles of iron = \frac{1.8 \mathrm{~g}}{56 \mathrm{~g/mol}} = 0.0321 \mathrm{~mol} \]

3. Calculate the moles of \(\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}\):
\[ Moles of \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \]

4. Calculate the mass of \(\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}\):
\[ Molar mass of \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 56 + 32 + 7 \times 18 = 277 \mathrm{~g/mol} \]
\[ Mass of \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \times 277 \mathrm{~g/mol} = 8.8935 \mathrm{~g} \approx 9 \mathrm{~g} \]

Therefore, the correct answer is (9). Quick Tip: Use the molar mass and stoichiometry to calculate the mass of the compound required.

1. Initial pH calculation:
\[ \mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \quad \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10} \]
\[ 0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad Not justified \]
\[ \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \]

2. Calculate \(\mathrm{K}_{\mathrm{a}}\):
\[ \mathrm{K}_{\mathrm{a}}=0.01 \alpha^{2}=10^{-8} \]

3. Data given is inconsistent \& contradictory. This should be bonus. Quick Tip: Check the consistency of the given data and ensure the calculations are justified.

1. Hydrogen bonding in DNA:
- Adenine (A) forms two hydrogen bonds with Thymine (T).
- Guanine (G) forms three hydrogen bonds with Cytosine (C).

2. Count the hydrogen bonds:
- Number of G-C pairs: 4
- Number of A-T pairs: 4

3. Total number of hydrogen bonds:
\[ Total hydrogen bonds = 4 \times 3 + 4 \times 2 = 12 + 8 = 20 \]

Therefore, the correct answer is (33). Quick Tip: Count the number of hydrogen bonds formed by each base pair in the DNA strand.