JEE Main 2024 question paper pdf with solutions- Download Jan 27 Shift 1 Chemistry Question Paper pdf

Step 1: Understanding magnetic moment.

Magnetic moment depends on the number of unpaired electrons. Higher the number of unpaired electrons, higher the magnetic moment.


Step 2: Determine unpaired electrons.

For 3d orbitals:

- \(3d^3\) → 3 unpaired electrons

- \(3d^6\) → 4 unpaired electrons (Hund's rule)

- \(3d^7\) → 3 unpaired electrons


Step 3: Conclusion.

Since \(3d^6\) has the highest number of unpaired electrons (4), it has maximum magnetic moment.
Quick Tip: Magnetic moment increases with the number of unpaired electrons: \(\mu = \sqrt{n(n+1)}\).

Step 1: Check truth of Assertion.

Boron indeed has a very high melting point (2453 K). So the assertion is true.


Step 2: Check truth of Reason.

Boron has a strong 3D crystalline network made of rigid covalent bonds. So the reason is also true.


Step 3: Check explanation.

A strong crystalline lattice directly leads to a very high melting point. Hence the reason correctly explains the assertion.
Quick Tip: If strong covalent bonding or network lattice is present, melting point is always high.

Step 1: Find oxidation states.

- BaSO\(_4\): S = +6

- SOCl\(_2\): S = +4

- SF\(_4\): S = +4

- H\(_2\)SO\(_3\): S = +4

- H\(_2\)S\(_2\)O\(_7\): S = +6 (average)

- SO\(_3\): S = +6


Step 2: Count species with +4 oxidation state.

SOCl\(_2\), SF\(_4\), H\(_2\)SO\(_3\) → 3 compounds.


Step 3: Conclusion.

Total = 3 compounds.
Quick Tip: Always assign oxygen as –2 and hydrogen as +1 while calculating oxidation states.

Step 1: Understand the reaction.

PbCrO\(_4\) reacts with excess hot NaOH to form a soluble chromate complex. Lead(II) precipitate dissolves in strong base, producing a tetra-anionic chromate species.


Step 2: Coordination concept.

Chromate ion (CrO\(_4^{2-}\)) has coordination number 4 because chromium is surrounded by four oxygen atoms in a tetrahedral structure.


Step 3: Conclusion.

Thus, the product formed is a tetra-anion with CN = 4.
Quick Tip: Chromate (CrO\(_4^{2-}\)) is always tetrahedral with coordination number 4.

Step 1: Understanding negative deviation.

Negative deviation occurs when intermolecular forces between different components (A–B) are stronger than between like components (A–A and B–B).


Step 2: Effect on vapour pressure.

Stronger interactions lower vapour pressure because fewer molecules escape into vapour phase.


Step 3: Effect on boiling point.

Lower vapour pressure means the solution requires more heating to boil → boiling point increases.


Step 4: Conclusion.

Therefore, in negative deviation: BP increases and VP decreases.
Quick Tip: Stronger intermolecular attraction → low vapour pressure → high boiling point.

Step 1: Identify compound (A).

K\(_2\)Cr\(_2\)O\(_7\) + NaCl + H\(_2\)SO\(_4\) produces a red volatile compound CrO\(_2\)Cl\(_2\) (chromyl chloride).


Step 2: Hydrolysis of (A).
\[ CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O \]
This gives yellow sodium chromate (B).


Step 3: Conclusion.

Thus, (A) = CrO\(_2\)Cl\(_2\), (B) = Na\(_2\)CrO\(_4\).
Quick Tip: Red fumes in dichromate + chloride + acid test always indicate chromyl chloride formation.

Step 1: Check Statement-1.

(NH\(_4\))\(_2\)CO\(_3\) contains NH\(_4^+\) (weak acid) and CO\(_3^{2-}\) (strong base conjugate), giving an overall basic salt. Hence S\(_1\) is true.


Step 2: Check Statement-2.

For salts of weak acid and weak base, pH depends on relative strengths (K\(_a\) vs K\(_b\)). Hence S\(_2\) is true.


Step 3: Conclusion.

Both statements are true.
Quick Tip: Weak acid + weak base → pH depends on K\(_a\) and K\(_b\).

Step 1: Write oxidation state equation for sulphur in H\(_2\)SO\(_3\).

H = +1, O = -2.
\[ 2(+1) + x + 3(-2) = 0 \]

Step 2: Solve for x.
\[ 2 + x - 6 = 0 \] \[ x = +4 \]

Step 3: Conclusion.

Sulphur in H\(_2\)SO\(_3\) is in oxidation state +4.
Quick Tip: In oxyacids, oxidation state of the central atom is found using:
\(H = +1\), \(O = -2\), and overall charge = 0.

Step 1: General behaviour of halogens.

Cl, Br and I show multiple oxidation states such as –1, +1, +3, +5, +7.


Step 2: Behaviour of fluorine.

Fluorine is the most electronegative element and can only gain electrons.

Therefore, it shows oxidation state –1 only.


Step 3: Conclusion.

Fluorine does not show variable oxidation states.
Quick Tip: Fluorine is always –1 in all its compounds because it is the strongest oxidizing agent.

Step 1: Identify stabilizing effects.

Acidity increases if the conjugate base is resonance-stabilized or has electron-withdrawing groups.


Step 2: Analyse structure (3).

The compound in option (3) contains two carbonyl groups adjacent to the acidic hydrogen.
This allows strong resonance stabilization of the conjugate base.


Step 3: Compare with others.

Other structures have either one carbonyl group or less stabilization.
Thus, (3) is most acidic.


Step 4: Conclusion.

Compound (3) is the most acidic due to maximum resonance stabilization.
Quick Tip: More carbonyl groups near a hydrogen → more acidic due to resonance stabilization of conjugate base.

Step 1: S\(_N1\) mechanism.

S\(_N1\) involves formation of a planar carbocation intermediate.
Nucleophile can attack from both sides → mixture of 50% inversion and 50% retention → racemisation.


Step 2: S\(_N2\) mechanism.

S\(_N2\) is a one-step backside attack mechanism.
Backside attack always causes inversion of configuration (Walden inversion).


Step 3: Conclusion.

S\(_N1\) → racemisation, S\(_N2\) → inversion.
Quick Tip: S\(_N1\) = planar carbocation → racemisation. S\(_N2\) = backside attack → inversion.

Step 1: Understand enolisation.

Enol content depends on stability of the enol form.
More resonance or intramolecular H-bonding → greater enol content.


Step 2: Analyse option (1).

The compound has two carbonyl groups positioned to allow strong intramolecular hydrogen bonding in its enol form.
This stabilizes the enol tautomer significantly.


Step 3: Compare with other options.

Other compounds have less resonance or no intramolecular H-bonding → less enol content.


Step 4: Conclusion.

Thus, option (1) has the maximum enol content.
Quick Tip: Enol content increases with intramolecular hydrogen bonding and conjugation.

Step 1: Identify electron-withdrawing and donating groups.

- NO\(_2\) is strong \(-M\), \(-I\) → increases acidity.

- OCH\(_3\) is strong \(+M\) → decreases acidity.


Step 2: Analyse compounds.

(II) has two NO\(_2\) groups → strongest acid.

(I) has one NO\(_2\) group → next strongest.

(III) phenol → moderately acidic.

(V) methoxy phenol → electron donation reduces acidity.

(IV) alcohol → least acidic.


Step 3: Conclusion.

Hence the order: II \(>\) I \(>\) III \(>\) V \(>\) IV.
Quick Tip: Electron-withdrawing groups increase acidity; electron-donating groups decrease acidity.

Step 1: Examine structure.

Cyclohexene is a cyclic compound (ring) but not aromatic.


Step 2: Classification rule.

A cyclic compound that is not aromatic is called an alicyclic compound.


Step 3: Conclusion.

Cyclohexene is an alicyclic compound.
Quick Tip: Alicyclic = cyclic but not aromatic.

Step 1: Check polarity.

CHCl\(_3\) (chloroform) is polar because of C–Cl bond dipoles that do not fully cancel.


Step 2: Compare with others.

- CCl\(_4\): symmetrical tetrahedral → nonpolar.

- CH\(_2\)=CH\(_2\): alkene, nonpolar.

- CO\(_2\): linear and symmetrical → nonpolar.


Step 3: Conclusion.

Only CHCl\(_3\) is polar.
Quick Tip: Symmetry cancels dipole moments → molecule becomes nonpolar.

Step 1: Understand nucleotide structure.

A nucleotide contains a sugar, phosphate, and nitrogen base.


Step 2: How two nucleotides link.

The phosphate group of one nucleotide links to the 3’–OH of another nucleotide.
This bond is called a phosphodiester bond.


Step 3: Conclusion.

Therefore, nucleotides form dimers through phosphodiester linkage.
Quick Tip: DNA and RNA strands grow via 3’–5’ phosphodiester bonds.

Step 1: Identify the nature of substituents.

Meta-directing groups are electron-withdrawing groups (–M or strong –I) that deactivate the ring.


Step 2: Check each substituent:

(1) \({-NHCOCH3}\): Electron-donating (amide) → ortho/para directing.

(2) OH: +M group → ortho/para directing.

(3) COR (acyl): Strong –M → Meta directing

(4) Cl: –I but +M → overall ortho/para directing.

(5) COOH: Strong –M, –I → Meta directing

(6) OCH\(_3\): +M → ortho/para directing.

(7) NO\(_2\): Very strong –M, –I → Meta directing

(8) CN: Strong –M, –I → Meta directing

(9) CH\(_3\): +I → ortho/para directing.


Step 3: Count meta directors.

Meta directing groups = (3), (5), (7), (8) = 4 groups.


Step 4: Conclusion.

Total meta-directing groups = 4.
Quick Tip: All strong electron-withdrawing groups (–NO\(_2\), –CN, –COOH, –COR, –SO\(_3\)H) are ALWAYS meta-directing.

Step 1: Understand the reaction mechanism.

HBr + Peroxide → Anti-Markovnikov addition (free radical mechanism).
Br adds to the less substituted carbon of the double bond.


Step 2: Identify the two new chiral centers.

The product formed has two stereocenters (marked *): \[ {R-CH(Br)-CH2-CH(R)-R} \]
Two stereocenters → possible stereoisomers = 2² = 4.


Step 3: List types of stereoisomers.

Four stereoisomers include:
- RR
- SS
- RS
- SR

Step 4: Conclusion.

Thus, total products including stereoisomers = 4.
Quick Tip: Anti-Markovnikov addition of HBr in the presence of peroxides follows a radical pathway and often creates new chiral centers → multiple stereoisomers.