JEE Main 2025 April 3 Shift 1 Chemistry Question Paper, Exam Analysis, and Answer Keys (Available)

The electron in a H-atom's stationary state moves in a spherical path. Quick Tip: Bohr's model assumes electrons move in circular orbits, while quantum mechanics describes electrons in terms of probability distributions (orbitals).

N-N single bond weaker than P-P due to more lp-lp repulsion.
Bond length \( d_{N-N} > d_{P-P} \) (size↑, B.L.↑)
In group 15 elements only N and P show disproportionation in +3 oxidation state. As, Sb and Bi have almost inert for disproportionation in +3 oxidation state.
So both statements are false. Quick Tip: Consider the bond strengths and lengths in nitrogen and phosphorus. Recall the trend of disproportionation reactions in group 15 elements.

A catalyst can change equilibrium composition if it is added at constant pressure, but it can not change equilibrium constant. Quick Tip: A catalyst affects the rate of a reaction, but not the equilibrium constant. It can alter the equilibrium composition by changing the forward and reverse reaction rates equally.

Given magnetic moment = 4.9 B.M.


We know, M.M = \( \sqrt{n(n+2)} \) B.M.

Where, \( n = \) Number of unpaired electrons (\( e^- \))

\( 4.9 = \sqrt{n(n+2)} \)

We get \( n = 4 \)


(A) \( \mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4 \) (4 unpaired \( e^- \))

(B) \( \mathrm{Fe}^{2+} = [\mathrm{Ar}]\,3d^6 \) (4 unpaired \( e^- \))

(C) \( \mathrm{Fe}^{3+} = [\mathrm{Ar}]\,3d^5 \) (5 unpaired \( e^- \))

(D) \( \mathrm{Co}^{2+} = [\mathrm{Ar}]\,3d^7 \) (3 unpaired \( e^- \))

(E) \( \mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5 \) (5 unpaired \( e^- \))
Quick Tip: Use the spin-only magnetic moment formula to calculate the number of unpaired electrons. Then, determine the electronic configurations of the given metal ions and identify those with 4 unpaired electrons.

\[ Given: [A]_0 = 8[B]_0 \] \[ t_{1/2(A)} = 10 min \] \[ t_{1/2(B)} = 40 min \] \[ 1^{st} order kinetics \] \[ t = ? \] \[ [A] = [B] \] \[ -k_A \times t = \ln \frac{[A]}{[A]_0} \] \[ [A] = [A]_0 e^{-k_A t} \] \[ [B] = [B]_0 e^{-k_B t} \] \[ [A] = [B] \] \[ [A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \] \[ 8[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \] \[ 8 = e^{(k_A - k_B)t} \] \[ \ln 8 = (k_A - k_B)t \] \[ t = \frac{\ln 8}{k_A - k_B} \] \[ t = \frac{\ln 8}{\frac{\ln 2}{10} - \frac{\ln 2}{40}} \] \[ t = \frac{\ln 2^3}{\frac{\ln 2}{10} - \frac{\ln 2}{40}} \] \[ t = \frac{3 \ln 2}{\ln 2 \left( \frac{1}{10} - \frac{1}{40} \right)} \] \[ t = \frac{3}{\frac{4-1}{40}} = \frac{3}{\frac{3}{40}} \] \[ t = 40 min \] Quick Tip: Use the first-order rate law and the given half-lives to find the time when the concentrations of A and B are equal.

Quick Tip: Remember the structure of L-fructose, including the position of the hydroxyl groups and the orientation of the chiral centers.

In option C are momologues to each - other and option D are only organic molecule not isomers. Quick Tip: Remember the definitions of metamers, functional isomers, position isomers, and homologous series. Identify the relationships between the given pairs of compounds based on their structures and functional groups.

\[ No. of atoms = \frac{Mass in g}{Molar Mass (g/mol)} \times N_A \]
Therefore for the same Mass element having the least Molar mass will have the higher no. of atoms. \[ M_{Pb} = 209 \] \[ M_{Pr} = 141 \] \[ M_{Po} = 207 \] \[ M_{Pt} = 195 \] Quick Tip: The number of atoms in a given mass is inversely proportional to the molar mass of the element.

[(A)] The process of adding an \( e^- \) to a neutral gaseous atom is not always exothermic; it may be exothermic or endothermic.

[(B)] Be: 1s²2s²

B: 1s²2s²2p¹

In Be, the 2s subshell is fully filled.

So, high energy is needed to remove an \( e^- \) as compared to B.

[(D)] In \( CCl_4 \), due to partially positive charge, \( Z_{eff} \uparrow \)

So, EN of C: \( CCl_4 > CH_4 \)

[(E)] Cs is most electropositive Quick Tip: Review the definitions of electron affinity, ionization energy, electronegativity, and electropositivity. Consider the electronic configurations and trends in the periodic table.

Sol. Both solutions are having same composition, which is 1 mole of 'x' in 1L \( H_2O \), so all the intensive properties will remain same, but as total amount is greater in solution '1' compared to solution '2'. So extensive properties will be different hence Gibbs free energy will be different. Quick Tip: Distinguish between intensive and extensive properties. Intensive properties do not depend on the amount of substance, while extensive properties do.

Following will not give iodoform reaction/test.
(1) Butanal

(2) 2-Pentanone

(3) Pentanal

(4) 3-Pentanol Quick Tip: Remember the structural requirements for the iodoform test. The compound must have a \( CH_3CO- \) or \( CH_3CH(OH)- \) group.

Quick Tip: Recognize the reactions involved: diazotization, Sandmeyer reaction, and coupling reaction.

Not correct Quick Tip: Recognize the reactions involved: Sandmeyer reaction, hydrolysis, and nucleophilic substitution.

\[ We know E = h\nu = \frac{hC}{\lambda} \] \[ E \propto \frac{1}{\lambda} \]
Here all Co in +3 oxidation state.
So, as the ligand field strength ↑, CFSE ↑
Order of field strength of ligand : \[ CN^- > NH_3 > H_2O > Cl^- \]
CFSE order : C \(>\) B \(>\) A \(>\) D
Wavelength order : D \(>\) A \(>\) B \(>\) C Quick Tip: Remember the spectrochemical series and how it relates to the strength of ligands and the energy of absorbed light.

On addition of inert gas at constant P and T, reaction moves in the direction of greater no. of moles so it will shift in forward direction, so \( [PCl_5] \) decrease and \( [PCl_3] \) and \( [Cl_2] \) will increase. Quick Tip: Adding an inert gas at constant pressure and temperature increases the volume of the container. This causes the equilibrium to shift towards the side with more moles of gas.

\[ \Delta T_b = i_1 m_1 k_b + i_2 m_2 k_b \] \[ \Delta T_b = 1 \times \frac{2}{0.5} \times 0.52 + 1 \times \frac{2}{0.5} \times 0.52 = 4.16 \] \[ (T_b)_{solution} = 373.16 + 4.16 = 377.3 K \] Quick Tip: Use the formula for boiling point elevation. Remember to account for the van't Hoff factor (i) for each solute and the total molality of the solution.

3-Methyl-6-ketoheptanal Quick Tip: Remember that ozonolysis of alkenes cleaves the double bond and forms carbonyl compounds.

\begin{align*
&\( PF_5 \): \quad 5\sigma + 0 \, lone pair \Rightarrow sp^3d \, \text{hybridisation

&\( SF_6 \): \quad 6\sigma + 0 \, \text{lone pair \Rightarrow sp^3d^2 \, \text{hybridisation

&\( Ni(CO)_4 \): \quad \text{Ni oxidation state = 0

&\text{In presence of ligand field:

&\quad \text{Ni(0): [Ar] \quad \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, _ _ _ _

&\quad \text{Orbitals: \quad 3d \quad \quad \quad \quad \quad \quad 4s \quad 4p

&\quad \Rightarrow sp^3 \, \text{hybridisation

&\( [PtCl_4]^{2- \): \quad \text{Pt oxidation state = +2

&\text{In presence of ligand field:

&\quad \text{Pt^{2+ \text{: [Kr] \quad \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, _ _ _

&\quad \text{Orbitals: \quad 5d \quad \quad \quad \quad 6s \quad 6p

&\quad \Rightarrow dsp^2 \, \text{hybridisation
\end{align* Quick Tip: Determine the hybridization of the central atom in each molecule/ion by counting the number of sigma bonds and lone pairs.

C.B. of terminal alkyne will be sp hybridisation and localised. In other C.B. will be resonance stabilised. Quick Tip: Acidity depends on the stability of the conjugate base. Consider the hybridization of the carbon atom from which the proton is removed and the possibility of resonance stabilization.

Limiting Molar Conductivities of ions : \[ \lambda^0_{H^+} : 349.8 Sem^2 mol^{-1} \] \[ \lambda^0_{Na^+} : 50.11 Sem^2 mol^{-1} \] \[ \lambda^0_{K^+} : 73.52 Sem^2 mol^{-1} \] \[ \lambda^0_{Ca^{2+}} : 119 Sem^2 mol^{-1} \] \[ \lambda^0_{Mg^{2+}} : 106.12 Sem^2 mol^{-1} \]
Therefore correct order of limiting molar conductivity of cations will be - \[ H^+ > Ca^{2+} > Mg^{2+} > K^+ > Na^+ \] Quick Tip: The limiting molar conductivity depends on the size and charge of the ion. Smaller and highly charged ions are more hydrated, leading to lower mobility and thus lower conductivity. However, \(H^+\) has exceptionally high conductivity due to its movement through a proton hopping mechanism.

M.wt of given compound = 86

Applying POAC on 'N' \( n_N \times 2 = n_{N_2} \times 2 \) \( n_N = n_{N_2} \)
Moles of N in 0.42 g compound = \( \frac{0.42}{86} \times \frac{14 \times 1}{14} = \frac{0.42}{86} \)
Moles of \( N_2 \) formed = \( \frac{0.42}{86} \)
Volume \( (N_2) \) at STP = \( \frac{0.42}{86} \times 22.4 L \)
Volume \( (N_2) \) at STP = \( 0.1108 L = 110.8 mL \approx 111 mL \) Quick Tip: In Dumas' method, all the nitrogen in the organic compound is converted to \( N_2 \) gas. Use the molar ratio of nitrogen in the compound to the moles of \( N_2 \) produced and then calculate the volume at STP.

Organic Compound \(\rightarrow\) CO\(_2\)

Applying POAC on ‘C’

(mole) of ‘C’ in compound \hspace{0.2cm = \hspace{0.2cm \(n_{CO_2} \times 1\)
[10pt]
So mass of ‘C’ in compound
[5pt]
\hspace*{1cm = \(\dfrac{1.46}{44} \times 12\)
[10pt]
So, % of ‘C’ in compound = \[ \dfrac{1.46}{44} \times \dfrac{12}{0.5} \times 100 \] \[ = 79.63 \] Quick Tip: During combustion of an organic compound, all the carbon present in the compound is converted to \( CO_2 \). Use the molar mass ratio to find the mass of carbon in the \( CO_2 \) produced and then calculate the percentage of carbon in the organic compound.

\[ FeCl_3 + KOH + H_2C_2O_4 \rightarrow K_3[Fe(C_2O_4)_3] \]
(A) \[ [Fe(C_2O_4)_3]^{3-} is [M(AA)_3] type complex. \]
So total optical isomers = 2 Quick Tip: For \( [M(AA)_3] \) type complexes, where AA is a symmetrical bidentate ligand, the complex is chiral and exists as two optical isomers (d and l forms).

\[ [\Delta H_f^0]_{C_2H_4(g)} = (2 \times 710) + (2 \times 436) - 611 - 4 \times 414 \] \[ [\Delta H_f^0]_{C_2H_4(g)} = 1420 + 872 - 611 - 1656 \] \[ [\Delta H_f^0]_{C_2H_4(g)} = 2292 - 2267 = 25 kJ mol^{-1} \]

Final Answer: The final answer is \(\boxed{25}\) Quick Tip: The enthalpy of formation of a compound can be related to the enthalpies of formation of its constituent atoms and the bond enthalpies of the bonds within the molecule. Set up an equation relating the enthalpy of formation of ethylene to the sublimation enthalpy of carbon, the bond enthalpy of hydrogen, and the bond enthalpies of C=C and C-H bonds, then solve for the C=C bond enthalpy.

\[ Cr_2O_7^{2-} + HCl + H_2SO_4 \rightarrow CrO_2Cl_2 \] \[ CrO_2Cl_2(vapour) + NaOH \rightarrow Na_2CrO_4 + NaCl + H_2O \]
\[ Na_2CrO_4 + H^+ \rightarrow Na_2Cr_2O_7 + H_2O \]
\[ 2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O \] \[ CrO_4^{2-} \xrightarrow{H^+} Cr_2O_7^{2-} \]

\includegraphics{75.png

No of terminal "O" = 6 Quick Tip: Identify the chromium-containing species formed in each reaction. Chromyl chloride (\( CrO_2Cl_2 \)) reacts with NaOH to form chromate ions (\( CrO_4^{2-} \)), which in acidic solution convert to dichromate ions (\( Cr_2O_7^{2-} \)). Draw the structure of the dichromate ion to count the terminal oxygen atoms.