JEE Main 2025 April 2 Shift 1 Chemistry Question Paper, Exam Analysis, and Answer Keys (Available)

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Aromatic compounds (follow Huckel's rule):

(a) Cyclic, planar, conjugated with 6\(\pi\) electrons (4n+2 where n=1)
(c) Cyclic, planar, conjugated with 6\(\pi\) electrons
(d) Cyclic, planar, conjugated with 6\(\pi\) electrons
(e) Cyclic, planar, conjugated with 6\(\pi\) electrons
(h) Cyclic, planar, conjugated with 6\(\pi\) electrons


Non-aromatic compounds:

(b) Not fully conjugated (sp\(^3\) hybridized carbon breaks conjugation)
(f) Not planar (twisted structure prevents conjugation)
(g) Has 4\(\pi\) electrons (doesn't satisfy 4n+2 rule) Quick Tip: \textbf{Huckel's Rule for Aromaticity:} Cyclic, planar molecule Fully conjugated \(\pi\)-system Contains 4n+2 \(\pi\) electrons (n=0,1,2...)

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The reaction sequence is as follows:

(A) Reacts with KOH to form alkene [B] (Elimination of HBr).
(B) Reacts with Br\(_2\) to give dibromide [C].
(C) Dibromide reacts with NaNH\(_2\) to form [D] (Alkyne formation).
(D) The alkyne [D] undergoes hydration with mercuric sulfate to give [E]. Quick Tip: For alkene to alkyne conversions, always check for the necessary reagents such as NaNH\(_2\) and the correct conditions for hydration (HgSO\(_4\)).

The order of first four elements of group-17 are as follows.

F < Cl < Br < I (Covalent radius)
Cl > F > Br > I (Electron affinity)
F < Cl < Br < I (Ionic radius)
F > Cl > Br > I (First ionization energy)


Electron affinity order is irregular. Quick Tip: For group-17 elements, remember that the irregularities mainly arise due to the electron configuration and atomic size trends.

As temperature increases, solubility first decreases then increases, hence \( K_H \) first increases, then decreases. At moderate temperature, the value of \( K_H \) follows the order: \[ He > N_2 > CH_4 \]


% Diagram for solution of question 54 Quick Tip: For gases in water, the temperature dependence of Henry’s Law constant shows an initial decrease followed by an increase in solubility at higher temperatures. This behavior varies across different gases.

We know that for Bohr’s model: \[ r \propto n^2 \]
Where \( n \) is the principal quantum number. Hence, we have: \[ \frac{r_3}{r_1} = \left(\frac{3}{1}\right)^2 = 9, \quad \frac{r_8}{r_4} = \left(\frac{8}{4}\right)^2 = 4, \quad \frac{r_6}{r_4} = \left(\frac{6}{4}\right)^2 = 2.25, \quad \frac{r_4}{r_2} = \left(\frac{4}{2}\right)^2 = 4 \]
Thus, the incorrect statement is option (3).
Quick Tip: For Bohr's model, the radius of each orbit increases with the square of the principal quantum number \( n \). Thus, radius comparisons can be calculated using the formula \( r \propto n^2 \).

Since there is no change in temperature after the stopcock is opened, it is a free expansion, implying: \[ w = 0, \quad q = 0, \quad \Delta U = 0 \]
Thus, the correct answer is option (4). The pressure in vessel B before opening the stopcock is zero.
Quick Tip: In a free expansion, there is no work done, no heat transfer, and no change in internal energy, as the process is adiabatic and occurs without a temperature change.

The relation between vapor pressures is given by Raoult's law: \[ P_A = P_A^0 \cdot X_A, \quad P_B = P_B^0 \cdot X_B \]
Given: \[ P_A + P_B = 500 \, mm Hg, \quad P_A^0 = 200 \, mm Hg, \quad P_B^0 = 600 \, mm Hg \]
Using the mole fractions and Raoult's law: \[ P_A = 200 \times \frac{1}{4}, \quad P_B = 600 \times \frac{3}{4} \]
Thus, the answer is \( P_A = 600 \, mm Hg, P_B = 600 \, mm Hg \).
Quick Tip: Raoult's Law is useful for determining the vapor pressure of components in an ideal solution. It states that the vapor pressure is proportional to the mole fraction of the component in the solution.

Using stoichiometry, the moles of \( CaCO_3 \) are calculated: \[ Moles of CaCO_3 = \frac{1000}{100} = 10 \, mol \]
Now, for the reaction: \[ CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O \]
Hence, the moles of \( CaCl_2 \) formed will be \( 10 \, mol \). Finally: \[ Mass of CaCl_2 = 10.545 \, g \] Quick Tip: In stoichiometry, make sure the balanced chemical equation is used to relate the moles of reactants and products. Converting moles to grams requires using the molar mass.

When equal volumes are mixed, the molarity of each component will be halved. Let’s calculate the value of \( Q \) for precipitation and compare it with \( K_{sp} \).

For precipitation, we use the equation: \[ Q = [A^{1-}][Y^{1-}] \]

Let’s calculate \( Q \) for each option:
\[ Q = (1.8 \times 10^{-7}) \left( \frac{5 \times 10^{-4}}{2} \right)^2 \]
This will give: \[ Q = (1.8 \times 10^{-7}) \times (2.5 \times 10^{-4})^2 = 1.8 \times 10^{-7} \times 6.25 \times 10^{-8} = 1.125 \times 10^{-14} \]
We can see that the value of \( Q \) for this combination is smaller than \( K_{sp} \), indicating that a precipitate will form.

Now let’s check the other combinations for comparison:
\[ Q = (10^{-7}) \left( \frac{0.4 \times 10^{-3}}{2} \right)^2 \] \[ Q = (10^{-7}) \times (0.2 \times 10^{-3})^2 = 10^{-7} \times 4 \times 10^{-8} = 4 \times 10^{-15} \]
Again, this \( Q \) value is smaller than \( K_{sp} \), indicating no precipitate. Quick Tip: When dealing with solubility products, remember that the concentration of ions is key. Use the product of ion concentrations and compare it with \( K_{sp} \) to determine if a precipitate will form.

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Among the given molecules, SF₆ has a non-zero dipole moment and lone-pair electrons on the central atom, thus the hybridization of SF₆ is \( sp^3d \).
Quick Tip: When determining hybridization, count the number of bonding and lone pairs of electrons on the central atom. The hybridization depends on this count (e.g., \( sp^3d \) for SF₆).

Vanillin does not give self-aldol reaction due to the lack of acidic H for condensation. It will react with NaOH and Tollen’s reagent, thus making Statement I correct. Statement II is incorrect because vanillin does not undergo self-aldol condensation easily.


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% Replace with the correct image path for the solution part Quick Tip: Aromatic aldehydes generally do not undergo self-aldol condensation due to the lack of an active hydrogen atom, which is required for the reaction.

- Glycine is optically inactive.
- Aspartic acid also contains COOH group at the side chain.
- \(\alpha\)-Amino acids have a chiral carbon except for glycine.
- Cysteine undergoes dimerization due to the presence of free –SH group, but the dimerization process is not as common.
Quick Tip: For amino acids, remember that glycine is the only amino acid without a chiral centre. Most amino acids, except for glycine, have one chiral carbon.

Basic strength of amines depends on hydrogen bonding and electronic inductive effect. Thus, the correct order is: \[ NH_3 > NH_2 > CH_3 NH_2 > CH_3 CH_2 NH_2 > (CH_3 CH_2)_2 NH \] Quick Tip: The presence of alkyl groups enhances basic strength due to the electron-donating effect, but the bulkier groups like \( (CH_3 CH_2)_2 NH \) decrease the basicity due to steric hindrance and reduced availability of the lone pair.

The metallic radius of Al is less than Ga, which is correct. The ionic radius of Al\(^{3+}\) is more than Ga\(^{3+}\), making Statement II incorrect.
Quick Tip: The metallic radius of an element generally increases down a group. The ionic radius decreases with an increase in positive charge on the ion.

High spin complexes have low values of \( \Delta_o \), whereas low spin complexes are formed when \( \Delta_o \) is large. Thus, Statement I is incorrect, and Statement II is correct.
Quick Tip: High spin complexes generally form in weak field ligands, while low spin complexes form in strong field ligands due to the splitting of d-orbitals.

N/A

Moles of 'C' = \( n_{CO_2} = \frac{1.46}{44} = 0.033 \)

Moles of 'C' = \( W_c = 0.033 \times 12 = 0.396 \)

Moles of 'H' = \( 2 \times n_{H_2O} = 2 \times \frac{0.567}{18} = 0.063 \)

Mass of 'H' = \( 0.0063 \)

Mass of Oxygen \( O \) = \( 1 - (W_c + W_h) = 1 - (0.033 \times 12 + 0.063 \times 1) = 0.541 \)

Moles of 'O' = \( \frac{0.541}{16} = 0.033 \)


Empirical formula = \( CH_2O \)

Empirical formula mass = 30
Quick Tip: When calculating the empirical formula, first determine the moles of each element from the given combustion data. Then, divide each element’s mass by its atomic mass to obtain the ratio of atoms.

The most stable carbon radical due to stabilization by adjacent alkyl groups is III, while the least stable radical is I as it lacks such stabilization.



% Replace with correct image path for structural formula Quick Tip: Radical stability increases with the number of alkyl groups attached to the carbon carrying the radical. The more alkyl groups, the more stable the radical.

The rate of hydrolysis is influenced by the availability of the leaving group, and the stability of the leaving group. In this case, the order is determined based on the ease of displacement of chloride ions.



% Replace with correct image path for structural formula Quick Tip: For nucleophilic substitution reactions, the stability of the leaving group is key. Chlorine, being a good leaving group, affects the rate of hydrolysis.

The molecule \( A X_2 Y_2 \) follows the square pyramidal structure based on the given criteria. The electronegativity and ionization energy of element A explain its rarest behavior.



% Replace with correct image path for structural formula Quick Tip: In molecules with elements from the p-block, consider the electron configuration and the electronegativity values to determine the most likely molecular geometry.

Co has the highest standard electrode potential among Mn, Cr, Co, and Fe. The complex is \([ Co(CN)_6 ]^{3-}\), and its splitting is as follows:

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The number of electrons in the \( e \)-orbital of the complex is 6. Quick Tip: In coordination chemistry, the number of electrons in the \( e \)-orbital of a complex depends on its d-orbital splitting. This splitting occurs when ligands interact with the central metal ion, influencing the number of available electrons in the \( e \)-orbital.

The cell reaction is: \[ QH_2 + 2Ag^+ \rightarrow 2Ag + Q + 2H^+ \]
The equation for this reaction is: \[ E = E^\circ - \frac{0.06}{2} \log [H^+]^2 \] \[ E = E^\circ - 0.06 \log [H^+] \]
Now, using the given data and solving for pH: \[ pH = - \log [H^+] = \frac{E - E^\circ}{0.06} = \frac{0.4 - 0.1}{0.06} = 5 \]

Now, we consider the ammonium halide salt (NH₄X) with the relation: \[ pH + NH_4X = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log C \]
Substituting values: \[ 5 = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log (10^{-3}) \] \[ 5 = 7 - \frac{1}{2} pK_a + \frac{1}{2} \times 3 \] \[ 5 = 7 - \frac{1}{2} pK_a + 1.5 \] \[ \Rightarrow pK_a = 6 \] Quick Tip: When dealing with electrochemical cells, use the Nernst equation to calculate potential at non-standard conditions. Additionally, in pH calculations, use the relationship between pH, pKa, and concentration to determine the equilibrium.

Molar mass is given as 372 g/mol for compound (P). Hence, for 0.1 mole, the mass will be: \[ Mass = Molar mass \times Number of moles = 372 \times 0.1 = 37.2 \, g \]


% Replace with the correct image path for the diagram in the solution Quick Tip: When calculating the weight of a compound, simply multiply the molar mass by the number of moles to get the total mass in grams.

The reaction is: \[ CO(g) + H_2(g) \rightleftharpoons CH_3OH(g) \]

At time \( t = 0 \), moles of CO = 0.1 mol and moles of H\(_2\) = 0.1 mol. At equilibrium, the number of moles is: \[ CO(g) = 0.1 - x \quad H_2(g) = 0.1 - 2x \quad CH_3OH(g) = x = 0.04 \, mol \]
Substituting values: \[ x = 0.04 \quad CO(g) = 0.06 \quad H_2(g) = 0.08 \]

Given: \[ V = 2 \, L \quad T = 500 \, K \quad P_{total} = 5 \, bar \]

Using the ideal gas law: \[ n_{total} = 0.25 \, mol \] \[ P_{total} = \frac{n_{total} \times R \times T}{V} \] \[ P_{total} = \frac{(0.06 + 0.08 + 0.04) \times 0.08 \times 500}{2} = 5 \, bar \]

Thus, \( K_p = 74 \).

Now continuing the calculation: \[ K_p = \frac{X_{CH_3OH} \times X_{CO} \times X_{H_2}}{P_{total}^2} \] \[ K_p = \frac{0.04}{(0.06)(0.15)^2} = \frac{4}{6 \times 0.15 \times 16} \times \frac{1}{25} \] \[ K_p = \frac{100 \times 100}{24 \times 225 \times 25} = 0.074 \quad \Rightarrow K_p = 74 \times 10^7 \] Quick Tip: When calculating equilibrium constants, use the ideal gas law to determine the total moles and pressure. For equilibrium calculations, ensure that you account for the changes in the concentration of reactants and products.

From the graph, we know that \( t_{1/2} \) is proportional to \( [A] \). The slope is given as 76.92. Thus, using the equation for zero-order reaction:
\[ t_{1/2} = \frac{A_0}{2K} \quad where \quad slope = \frac{1}{2K} = 76.92 \]

Thus, \[ K = \frac{1}{2 \times 76.92} = \frac{1}{153.84} \]

Now, applying the formula for zero-order reaction: \[ [A] = - Kt + A_0 \] \[ [A] = - \frac{1}{2 \times 76.92} \times 10 + 2.5 = 2.435 \, mol/L \]

Thus, the concentration of A at 10 minutes is \( 2435 \times 10^{-3} \, mol/L \). Quick Tip: In a zero-order reaction, the rate of reaction is constant and the concentration of reactant decreases linearly with time. The equation \( [A] = -Kt + A_0 \) is used to calculate the concentration at any given time.