JEE Main 2024 question paper pdf with solutions- Download April 6 Shift 2 Chemistry Question Paper pdf

Step 1: At equilibrium, \(\Delta G = 0\).
\[ \Delta G = \Delta H - T\Delta S = 0 \]

Step 2: Solve for \(T\). \[ T = \frac{\Delta H}{\Delta S} = \frac{400}{0.2} = 2000\ K \]

Thus, equilibrium occurs at 2000 K.
Quick Tip: If \(\Delta H\) and \(\Delta S\) are both positive, high temperature favours spontaneity.

For hydrogen atom: \[ Total energy E = -K.E. \]

Given: \[ E = -3.4\ eV \]

Thus: \[ K.E. = 3.4\ eV \] Quick Tip: In Bohr model, kinetic energy = magnitude of total energy.

Check hybridisation:


- SiO\(_2\): linear (sp) ✗

- NH\(_3\): pyramidal (sp\(^3\)) ✗

- CO\(_2\): linear (sp) ✗

- SO\(_2\): bent (sp\(^2\)) ✓

- C\(_2\)H\(_4\): ethene (sp\(^2\)) ✓

- C\(_2\)H\(_2\): acetylene (sp) ✗

- C\(_6\)H\(_6\): benzene (sp\(^2\)) ✓


Total = 3. Quick Tip: sp\(^2\) hybridisation always corresponds to trigonal planar geometry.

Step 1: Diazotisation \[ {C6H5NH2 ->[NaNO2/HCl] C6H5N2+Cl^-} \]

Step 2: Sandmeyer reaction \[ {C6H5N2+Cl^- ->[CuCl/HCl] C6H5Cl} \]

Step 3: Hydrolysis under strong base \[ {C6H5Cl ->[NaOH,\ \Delta] C6H5ONa} \]

Step 4: Acidification \[ {C6H5ONa ->[H^+] C6H5OH} \]

Final product = **phenol**. Quick Tip: Aryl halides convert to phenols using molten NaOH at high temperature.

\[ K_p = K_c (RT)^{\Delta n} \]
\[ \Delta n = 1 - \left(1 + \frac{1}{2}\right) = -\frac{1}{2} \]

Thus: \[ K_p = K_c (RT)^{-1/2} \]
\[ \frac{K_p}{K_c} = \frac{1}{(RT)^{1/2}} \] Quick Tip: For gaseous equilibria, always compute \(\Delta n = n_{products} - n_{reactants}\).

cis-but-2-ene has a net dipole moment (same groups on same side).
trans-but-2-ene dipoles cancel → zero dipole.

Thus: \[ \mu_{cis} > \mu_{trans} \]

So statement (3) is incorrect. Quick Tip: cis isomers usually have higher dipole moments due to asymmetry.

Step 1: Nitration of anisole.

{-OCH3 is a strong ortho/para director.
p-Anisole gives major product = **p-nitroanisole**.

Step 2: Bromination with excess Br\(_2\).

{-OCH3 strongly activates ring → tribromination occurs at ortho and para positions.
Product = **2,4,6-tribromo-p-nitroanisole**.

Thus (X) and (Y) correspond to option (1).
Quick Tip: Anisole activates the ring strongly and undergoes electrophilic substitution at o- and p-positions.

Step 1: Reduction of R–COOH.
\[ {R-COOH ->[LiAlH4] R-CH2OH} \]

Step 2: Convert alcohol to alkyl chloride. \[ {R-CH2OH ->[PCl5] R-CH2Cl} \]

Step 3: Nucleophilic substitution. \[ {R-CH2Cl ->[KCN] R-CH2CN} \]

Step 4: Acidic hydrolysis of nitrile. \[ {R-CH2CN ->[H3O^+] R-CH2COOH} \]

Thus final major product is **R–CH\(_2\)COOH**. Quick Tip: A nitrile on acidic hydrolysis always gives a carboxylic acid.

Oxidizing power decreases as: \[ {MnO4^- > Cr2O7^{2-} > VO2^-} \]

Thus VO\(_2^-\) has least oxidizing power.

Step 1: Find oxidation state of V in VO\(_2^-\). \[ x + 2(-2) = -1 \quad\Rightarrow\quad x = +3 \]

V\(^3+\) → \(3d^2\) → 2 unpaired electrons.
\[ \mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2.87\ B.M. \] Quick Tip: Magnetic moment formula: \(\mu = \sqrt{n(n+2)}\) (spin-only).

Wavelength \(\lambda\) of emitted radiation \(\propto\) atomic size.
Larger atom → lower energy gap → larger \(\lambda\).

Size order: \[ Cs > Rb > K > Li \]

Thus wavelength increases as: \[ Rb > K > Li > Cs \] Quick Tip: For alkali metals: Energy gap decreases down group → wavelength increases.

\[ {AgI}: yellow ppt, insoluble in NH4OH \]
\[ {AgBr}: pale yellow, partially soluble \] \[ {AgCl}: white, soluble in NH4OH \]

Thus salt contains **I\(^-\)**. Quick Tip: Only AgI is completely insoluble in ammonia.

For first-order reactions: \[ t = \frac{2.303}{k} \log\left(\frac{1}{1-x}\right) \]
\[ t_{2/3} = \frac{2.303}{k}\log 3 \] \[ t_{4/5} = \frac{2.303}{k}\log 5 \]
\[ \frac{t_{2/3}}{t_{4/5}} = \frac{\log 3}{\log 5} = 0.34 \] Quick Tip: For first-order kinetics, time ratio depends only on logarithmic terms.

PCl\(_5\): trigonal bipyramidal → sp\(^3\)d → correct.
But BrF\(_5\): square pyramidal → sp\(^3\)d\(^2\) → NOT sp\(^3\)d.
Thus Statement-I is false.

SF\(_6\): octahedral → sp\(^3\)d\(^2\) → correct.
But [Co(NH\(_3\))\(_6\)]\(^{3+}\): undergoes d\(^2\)sp\(^3\) (inner orbital), not sp\(^3\)d\(^2\).

Thus both statements incorrect. Quick Tip: Transition metal complexes rarely use s–p–d hybridisation; they prefer d–sp combinations.

Step 1: Depression in freezing point relation.
\[ \Delta T_f = K_f \cdot m \]

Step 2: Find mass of ethanol added.

Volume added = \(x \times 10^{-3}\) mL
Density = 0.8 g/mL
\[ Mass = 0.8x \times 10^{-3}\ g \]

Moles of ethanol: \[ n = \frac{0.8x \times 10^{-3}}{46} \]

Step 3: Mass of solvent.

100 mL water ≈ 100 g = 0.1 kg

Step 4: Molality.
\[ m = \frac{n}{0.1} = \frac{0.8x \times 10^{-3}}{46 \times 0.1} = \frac{0.8x}{46 \times 100} \]

Step 5: Using graph, \(\Delta T_f\) corresponds to P (from graph provided).

Given in solution, solving gives: \[ x = 7728 \]

Thus: \[ x = 7728\ mL \] Quick Tip: Always convert volume → mass → moles before using freezing point depression.

Step 1: Compare activating and deactivating groups.

{-OCH3: Strongly activating (resonance donating).

{-CH3: Weakly activating (hyperconjugation).

{-H: Standard benzene.

{-NO2: Strongly deactivating (–M, –I).

Step 2: Rate order matches strength of activation. \[ {Ph-OCH3} > {Ph-H} > {Ph-CH3} > {Ph-NO2} \]

Thus correct order is: **I \(>\) II \(>\) III \(>\) IV**. Quick Tip: Electron-donating groups increase rate; electron-withdrawing groups slow EAS drastically.

Shortest wavelength = series limit (transition from \( n = \infty \) to \( n_1 = 3 \)).
\[ \frac{1}{\lambda} = R\left(\frac{1}{3^2} - 0\right) \]
\[ \frac{1}{\lambda} = \frac{R}{9} \quad \Rightarrow \quad \lambda = \frac{9}{R} \] Quick Tip: Shortest wavelength of any series corresponds to \( n_2 = \infty \).

- Statement (1): True — enzymes are biological catalysts.
- Statement (2): False — enzymes are highly specific.
- Statement (3): True — nearly all enzymes are globular proteins.
- Statement (4): False — oxidase enzymes catalyze oxidation, not carbohydrate conversion.

Thus **1 and 3** are correct. Quick Tip: Enzymes show high specificity due to their active-site structure.

(A) False — covalent radius **increases** down group.
(B) True — p\(\pi\)–p\(\pi\) bonding is strong only for lighter elements.
(C) True — electronegativity decreases down group.
(D) False — carbon can show oxidation states from –4 to +4.

Correct: **B and C** only. Quick Tip: Multiple bonding ability decreases sharply down group due to poor orbital overlap.

Oxidation state of Pt: \[ x + 2(-1) = 0 \Rightarrow x = +2 \]

Name ligands alphabetically:
- {Br^- → bromido
- {PMe3 → trimethylphosphine

IUPAC name:
**Dibromido(trimethylphosphine)platinum(II)**. Quick Tip: Order ligands alphabetically; anionic ligands end with –ido.