JEE Main 2024 question paper pdf with solutions- Download April 9 Shift 1 Chemistry Question Paper pdf

Step 1: Concept of paramagnetism.

A molecule is paramagnetic if it contains at least one unpaired electron. Molecular orbital (MO) theory is used to determine the presence of unpaired electrons.


Step 2: Electron configuration analysis.

O\(_2\) has two unpaired electrons → paramagnetic.

O\(_2^-\) has one unpaired electron → paramagnetic.

O\(_2^{2-}\) has all electrons paired → diamagnetic.

C\(_2^-\) has all electrons paired → diamagnetic.

CN\(^-\) is isoelectronic with CO (all paired) → diamagnetic.

N\(_2\) has all electrons paired → diamagnetic.


Step 3: Conclusion.

Only O\(_2\) and O\(_2^-\) show paramagnetism due to unpaired electrons.
Quick Tip: In MO theory, species with odd electron count or partially filled antibonding orbitals are usually paramagnetic.

Step 1: Applying the n+\(\ell\) rule.

Lower (n+\(\ell\)) means lower energy. If (n+\(\ell\)) is equal, the orbital with lower n has lower energy.


Step 2: Calculate values.

(i) 4s → n+\(\ell\) = 4+0 = 4

(ii) 4d → 4+2 = 6

(iii) 4p → 4+1 = 5

(iv) 3d → 3+2 = 5

(v) 3p → 3+1 = 4


Step 3: Ordering.

Smallest (n+\(\ell\) = 4): v (3p), i (4s)

Next (n+\(\ell\) = 5): iii (4p), iv (3d); lower n first → iii < iv

Highest (n+\(\ell\) = 6): ii (4d)


Thus: v < i < iii < iv < ii
Quick Tip: Use the n+\(\ell\) rule: lower value means lower energy; if equal, choose smaller n.

Step 1: Understanding ambident ligands.

Ambident ligands can coordinate through two different donor atoms.


Step 2: Check each ligand.

NO\(_2^-\) → can bind via N (nitro) or O (nitrito).

CN\(^-\) → can bind via C (cyano) or N (isocyano).

SCN\(^-\) → can bind via S (thiocyanato-S) or N (thiocyanato-N).

H\(_2\)O → only O → not ambident.

NH\(_3\) → only N → not ambident.

C\(_2\)O\(_4^{2-}\) → bidentate, not ambident.


Step 3: Conclusion.

The ligands showing ambidentate behaviour are NO\(_2^-\), CN\(^-\) and SCN\(^-\).
Quick Tip: Ambident ligands attach through two different atoms, unlike bidentate ligands that attach through two atoms simultaneously.

Step 1: Understand the reaction of PbS with dilute HNO\(_3\).

Dilute nitric acid oxidises sulphide (S\(^{2-}\)) to elemental sulphur (S). Lead sulphide dissolves forming Pb(NO\(_3\))\(_2\). Nitric acid (dilute) reduces mostly to NO gas.


Step 2: Identify products formed.

PbS + 2HNO\(_3\) → Pb(NO\(_3\))\(_2\) + S + 2H\(_2\)O + NO


Step 3: Identify which is NOT formed.

Products actually formed: S, NO, Pb(NO\(_3\))\(_2\)

N\(_2\)O is never produced from dilute nitric acid in this reaction.


Step 4: Conclusion.

Since N\(_2\)O is not formed, option (2) is correct.
Quick Tip: Dilute HNO\(_3\) reduces mainly to NO, not to N\(_2\)O or NO\(_2\).

Step 1: Understanding geometry.

[Co(en)\(_2\)Cl\(_2\)]\(^{+}\) contains Co(III) with 6 ligands (4 donor atoms from 2 en + 2 Cl). Therefore, complex is octahedral. So Statement–II is true.


Step 2: Number of geometrical isomers.

In an octahedral complex of the type MA\(_2\)B\(_2\)C\(_2\) (here en = A\(_2\), Cl = B), three geometrical isomers exist: cis–cis, cis–trans, and trans–cis. Therefore, 3 GI are possible. Statement–I is true.


Step 3: Conclusion.

Both statements are correct → option (3).
Quick Tip: Complexes of type A\(_2\)B\(_2\) often show 3 geometrical isomers in octahedral geometry.

Step 1: Concept of colour in lanthanides.

Lanthanide ions show colour due to unpaired 4f electrons. The ions with 4f\(^0\) or 4f\(^{14}\) configuration are colourless.


Step 2: Check each species.

La\(^{3+}\) → 4f\(^0\) → colourless

Lu\(^{3+}\) → 4f\(^{14}\) → colourless

Eu\(^{3+}\) → 4f\(^6\) → coloured

Sm\(^{3+}\) → 4f\(^5\) → coloured

Nd\(^{2+}\) → 4f\(^4\) → coloured


Step 3: Conclusion.

Only La\(^{3+}\) and Lu\(^{3+}\) are colourless due to fully empty or fully filled 4f subshell.
Quick Tip: Lanthanides are colourless only when 4f shell is either empty (La\(^{3+}\)) or completely filled (Lu\(^{3+}\)).

Step 1: Identify hybridisation of each species.

BF\(_3\): Boron has 3 bond pairs and 0 lone pairs → trigonal planar → sp\(^2\).

NO\(_2^-\): Nitrogen has 2 bond pairs + 1 lone pair → bent shape but steric number = 3 → sp\(^2\).

NH\(_2^-\): N has 2 bond pairs + 2 lone pairs → steric number = 4 → sp\(^3\).

H\(_2\)O: 2 bond pairs + 2 lone pairs → steric number = 4 → sp\(^3\).


Step 2: Choose the correct sp\(^2\) pair.

Only BF\(_3\) and NO\(_2^-\) both have sp\(^2\) hybridisation.


Step 3: Conclusion.

Hence option (1) is the correct pair of sp\(^2\) species.
Quick Tip: Hybridisation = steric number; if steric number is 3, hybridisation is sp\(^2\).

Step 1: Checking Statement–I.

Sulphur exists as S\(_8\) molecules in both monoclinic and rhombic forms. Oxygen does not exist as S\(_8\)-type rings; it exists as O\(_2\) molecules. So Statement–I is true.


Step 2: Checking Statement–II.

The statement claims that O\(_2\) has a \(\pi n\)–\(\pi n\) bond and sulphur does not. But O\(_2\) has a \(\pi\)–\(\pi\) bond (p–p overlap), not \(\pi n\)–\(\pi n\).

Sulphur does not form O\(_2\)-type double bonds due to large atomic size and prefers S–S single bonds forming S\(_8\) rings. So Statement–II is false.


Step 3: Conclusion.

Statement–I is correct but Statement–II is incorrect → option (4).
Quick Tip: Oxygen forms O=O double bonds, but sulphur prefers single S–S bonds forming cyclic S\(_8\) rings.

Step 1: Understanding effect on acidity.

Lower pKa = stronger acid.
Electron-withdrawing groups increase acidity → decrease pKa.
Electron-donating groups decrease acidity → increase pKa.


Step 2: Analyse each group.

Compound IV (p-OCH\(_3\)): +M donating → least acidic → highest pKa.

Compound III (m-NO\(_2\)): NO\(_2\) at meta has –I effect → moderately acidic → moderate pKa.

Compound I (phenol): baseline acidity.

Compound II (p-NO\(_2\)): strong –M \& –I group → most acidic → lowest pKa.


Step 3: Arrange in decreasing pKa.

Highest pKa → lowest acidic: IV > III > I > II.
Quick Tip: Electron-withdrawing groups decrease pKa of phenols; donating groups increase pKa.

Step 1: List of essential amino acids.

Essential amino acids include: Histidine, Lysine, Valine, Phenylalanine, Tryptophan, Leucine, Isoleucine, Threonine, Methionine.


Step 2: Match with given list.

Histidine → essential

Arginine → semi-essential (not counted)

Lysine → essential

Valine → essential

Proline → non-essential

Glutamic acid → non-essential

Phenylalanine → essential

Tryptophan → essential

Glycine → non-essential


Step 3: Conclusion.

Total essential amino acids present = 5.
Quick Tip: Remember the mnemonic: PVT TIM HALL → all essential amino acids.

Step 1: Identify tooth enamel composition.

Tooth enamel primarily contains fluorapatite, a fluoride-containing hydroxyapatite derivative.


Step 2: Standard formula.

The chemical formula of fluorapatite is Ca\(_{10}\)(PO\(_4\))\(_6\)F\(_2\).


Step 3: Conclusion.

Thus, the correct option is (1).
Quick Tip: Fluoride replaces OH\(^-\) in hydroxyapatite, strengthening tooth enamel.

Step 1: Analyse electronic configuration.

Cu\(^{+}\) has 3d\(^{10}\) → stable but less hydrated in water.

Cu\(^{2+}\) has 3d\(^9\) → strong hydration due to high charge density.


Step 2: Hydration enthalpy effect.

Cu\(^{2+}\) has very high hydration energy, stabilising it strongly in aqueous medium.

Cu\(^{+}\) readily undergoes disproportionation:

2Cu\(^+\) → Cu\(^{2+}\) + Cu (s).


Step 3: Conclusion.

Cu\(^{2+}\) is more stable in aqueous solutions.
Quick Tip: High hydration enthalpy stabilises higher oxidation states of transition metals in water.

Step 1: Compare electron availability for protonation.

(i) Cyclohexylamine: Aliphatic amines are strongly basic → highest basicity.

(ii) Aniline: Lone pair delocalised into benzene ring → least basic.

(iii) Pyridine: Lone pair is not part of aromatic sextet → moderately basic.


Step 2: Arrange in decreasing basic strength.

Aliphatic amine > heterocyclic amine > aromatic amine


Step 3: Conclusion.

Correct basicity order: i > iii > ii.
Quick Tip: Aromatic amines are least basic due to delocalisation of lone pair.

Step 1: Write the thermochemical relation.

CuSO\(_4\)(s) + 5H\(_2\)O → CuSO\(_4\cdot\)5H\(_2\)O(s)

The heat of hydration = \(\Delta\)H(soln, hydrated) – \(\Delta\)H(soln, anhydrous).


Step 2: Substitute values.

Heat of solution:

CuSO\(_4\)(s) = –72 kJ/mol

CuSO\(_4\cdot\)5H\(_2\)O(s) = +12 kJ/mol


Heat of hydration = 12 – (–72) = 84 kJ/mol.


Step 3: Conclusion.

Thus, the molar heat of hydration is 84 kJ/mol.
Quick Tip: Hydration enthalpy = (heat of solution of hydrated salt) – (heat of solution of anhydrous salt).

Step 1: Compare reduction potential.

Cr\(_2\)O\(_7^{2-}\)/Cr\(^{3+}\) has E\(^\circ\) = +1.33 V → strong oxidising agent.
Any metal with lower reduction potential than 1.33 V gets oxidised.


Step 2: Standard potentials (approx values).

Fe\(^{2+}\)/Fe = –0.44 V → oxidised

Ni\(^{2+}\)/Ni = –0.25 V → oxidised

Cr\(^{3+}\)/Cr = –0.74 V → oxidised

Cu\(^{2+}\)/Cu = +0.34 V → not oxidised

Ag\(^{+}\)/Ag = +0.80 V → not oxidised

Au\(^{3+}\)/Au = +1.50 V → not oxidised


Step 3: Conclusion.

Only Fe, Ni and Cr will oxidise → total = 3.
Quick Tip: A metal is oxidised if its E\(^\circ\) value is lower than the oxidising agent’s E\(^\circ\).

Step 1: Identify the reaction.

CO + HCl + AlCl\(_3\) on benzene produces benzaldehyde.
This is the Gattermann–Koch reaction.


Step 2: Mechanism overview.

CO + HCl + AlCl\(_3\) generate formyl chloride (HCOCl) in situ.
This formyl chloride performs electrophilic substitution on benzene forming benzaldehyde.


Step 3: Conclusion.

The product is benzaldehyde → option (1).
Quick Tip: Gattermann–Koch reaction always converts benzene → benzaldehyde using CO/HCl/AlCl\(_3\).

Step 1: First step – Friedel–Crafts acylation.

Benzene reacts with phthalic anhydride in the presence of AlCl\(_3\) forming benzoyl-benzoic acid derivative (P).


Step 2: Second step – Clemmensen reduction (Zn-Hg/HCl).

The carbonyl group is reduced to CH\(_2\) forming ortho-phenyl toluene derivative (Q).


Step 3: Third step – Intramolecular cyclisation with conc. H\(_2\)SO\(_4\).

The –CH\(_2\) group and the benzene ring undergo intramolecular Friedel–Crafts cyclisation producing a fused bicyclic hydrocarbons system.
This gives tetralin-type compound → option (2).


Step 4: Conclusion.

Therefore, the correct final product is structure (2).
Quick Tip: Phthalic anhydride + benzene + AlCl\(_3\) often leads to fused-ring hydrocarbons after reduction and cyclisation.

Step 1: Analyse curve I.

Curve I shows nearly constant molar conductivity with decrease in concentration.
This is characteristic of strong electrolytes, which dissociate completely.


Step 2: Analyse curve II.

Curve II shows a steep increase of molar conductivity at low concentration.
This is characteristic of weak electrolytes, whose ionisation increases sharply at low concentration.


Step 3: Conclusion.

I = Strong electrolyte; II = Weak electrolyte.
Quick Tip: Strong electrolytes show slight change in molar conductivity; weak electrolytes show a steep rise at low concentration.

Step 1: Product P from acid-catalysed hydration.

CH\(_3\)–CH=CH\(_2\) (propene) reacts with H\(_2\)O/H\(^+\) via Markovnikov addition.

OH attaches to the more substituted carbon → forms 2-propanol (CH\(_3\)–CH(OH)–CH\(_3\)).


Step 2: Product Q from hydroboration–oxidation.

Hydroboration–oxidation gives anti-Markovnikov alcohol.

Thus OH attaches to terminal carbon → forms 1-propanol (CH\(_3\)–CH\(_2\)–CH\(_2\)OH).


Step 3: Conclusion.

P = 2-propanol, Q = 1-propanol → matches option (2).
Quick Tip: Hydration gives Markovnikov alcohol; hydroboration–oxidation always gives anti-Markovnikov alcohol.