JEE Main 2024 question paper pdf with solutions- Download Jan 29 Shift 2 Chemistry Question Paper pdf

Step 1: Formula of the complex.

The brown ring complex formed is:
\[ [Fe(H_2O)_5NO]SO_4 \]


Step 2: Calculate oxidation number.

Let oxidation state of Fe = x.

In the nitrosyl ligand (NO), oxidation state is +1 when NO is linear (common in brown ring complex).


Total charge of the complex ion = +1 (because SO\(_4^{2-}\) balances it to make the compound neutral).


Thus:
\[ x + 1 = +2 \Rightarrow x = +1 \]


Step 3: Conclusion.

Iron is present in the +1 oxidation state in the brown ring complex.
Quick Tip: In the brown ring complex, NO acts as a linear nitrosyl ligand with an oxidation state of +1, giving Fe its +1 oxidation state.

Step 1: Understanding the Ni\(^{2+}\) test.

Nickel(II) ions form a characteristic red precipitate with dimethylglyoxime (DMG).


Step 2: Reaction.
\[ 2\,DMG + Ni^{2+} \rightarrow Ni(DMG)_2 + 2H^+ \]
The product is a bright red chelate complex.


Step 3: Conclusion.

Since Ni(DMG)\(_2\) gives a red precipitate, the correct reagent is dimethylglyoxime.
Quick Tip: Ni\(^{2+}\) gives a red precipitate only with dimethylglyoxime — this is one of the most important qualitative inorganic tests.

Step 1: Name of reaction.

This is the Reimer–Tiemann reaction, where phenol reacts with CHCl\(_3\) + NaOH to give an aldehyde group at the ortho position.


Step 2: Reaction.

Phenol + CHCl\(_3\) + NaOH → Ortho-formyl phenol (salicylaldehyde) after acidic hydrolysis.

\[ Phenol \xrightarrow[NaOH]{CHCl_3, 343 K} Ortho-formyl phenoxide \xrightarrow[H^+]{} Salicylaldehyde \]

Step 3: Conclusion.

The major final product is salicylaldehyde.
Quick Tip: Reimer–Tiemann reaction converts phenol to salicylaldehyde by introducing –CHO at the ortho position.

Step 1: Principle of Nessler’s test.

Nessler’s reagent reacts with ammonia (NH\(_3\)) or ammonium ions (NH\(_4^+\)) to produce a brown precipitate.


Step 2: Reaction.
\[ 2K_2HgI_4 + NH_3 + 3KOH \rightarrow HgO\cdot Hg(NH_2)I + 7KI + 2H_2O \]
This brown precipitate confirms NH\(_4^+\)/NH\(_3\).


Step 3: Conclusion.

Therefore, ammonium ion gives the characteristic brown coloration.
Quick Tip: Nessler’s reagent detects NH\(_4^+\) or NH\(_3\) by forming a brown precipitate of basic mercuric amido-iodide.

Step 1: Understanding pKa.

Lower pKa means stronger acid. Nitro groups are strong electron-withdrawing groups and increase acidity.


Step 2: Compare given acids.

- Nitro phenols are stronger acids than phenol.

- Between meta and para nitrophenol, para is more acidic because of better resonance stabilization.

- Ethanol is the weakest acid (highest pKa).


Thus:

Weakest acid (highest pKa) = Ethanol (d)

Then Phenol (a)

Then Meta-nitrophenol (b)

Strongest acid (lowest pKa) = Para-nitrophenol (c)


Step 3: Final Order (increasing acidity / decreasing pKa):

d > a > b > c
Quick Tip: Electron-withdrawing groups decrease pKa and increase acidity; para position shows stronger effect than meta.

Step 1: Identify parent ring and functional groups.

The structure shows a cyclohexene ring with an OH group attached.


Step 2: Numbering the ring.

The OH group receives the lowest possible locant. Numbering starts from OH-bearing carbon.


Double bond gets the next lowest number. Hence position of the double bond = 2.


Step 3: Final IUPAC Name.

Cyclohex-2-en-1-ol
Quick Tip: Always give priority to the OH group while numbering rings containing double bonds.

Step 1: Determine molecular shape and symmetry.


NH\(_3\) → trigonal pyramidal → dipole \(\ne\) 0

H\(_2\)O → bent → dipole \(\ne\) 0

HF → polar → dipole \(\ne\) 0

CO\(_2\) → linear \& symmetrical → dipole = 0

SO\(_2\) → bent → dipole \(\ne\) 0

BF\(_3\) → trigonal planar \& symmetrical → dipole = 0

CH\(_4\) → tetrahedral \& symmetrical → dipole = 0


Step 2: Count species with zero dipole moment.

CO\(_2\), BF\(_3\), CH\(_4\) → 3 compounds


Step 3: Final Answer.

There are 3 molecules with zero dipole moment.
Quick Tip: Highly symmetrical molecules (linear, tetrahedral, trigonal planar) usually have zero dipole moment.

Step 1: Identify the reaction.

Phenolic –OH groups undergo Reimer–Tiemann formylation with CHCl\(_3\)/NaOH.


Step 2: Mechanism summary.

CHCl\(_3\) + NaOH → Dichlorocarbene (CCl\(_2\)).

This attacks ortho position of phenoxide.


Step 3: Hydrolysis step.

Final acidic hydrolysis converts –CCl\(_2\) into –CHO.


Conclusion: The major product is an ortho-formyl derivative of the cyclohexanol ring.
Quick Tip: Reimer–Tiemann always introduces an aldehyde (–CHO) group ortho to –OH group.

Step 1: Understanding Rutherford's nuclear model.

He proposed that:

- All mass and positive charge are concentrated at the nucleus.

- Electrons revolve around the nucleus in empty space.


Step 2: Evaluation.

S\(_1\) → true.

S\(_2\) → false, because electrons are NOT clustered; they revolve in orbits.


Conclusion: Only S\(_1\) is correct.
Quick Tip: Rutherford discovered the nucleus; electrons occupy mostly empty space around it.

Step 1: Identify \(\alpha\) molecule.

In JEE terminology, \(\alpha\) usually refers to He\(_2\) molecule (or a similar MO example).


Step 2: MO configuration.

For He\(_2\): 4 electrons total →
\(\sigma\)1s (2), \(\sigma\)1s* (2).

Thus antibonding electrons = 2.


Step 3: Final Answer.

Number of antibonding electrons = 2.
Quick Tip: Bond order = (bonding \(-\) antibonding)/2; negative or zero BO → molecule unstable.

Structure: CH\(_3\)–CH=CH–CH\(_2\)–CH(CHO)–COOH


Step 1: Count \(\pi\) bonds.

C=C → 1

C=O (aldehyde) → 1

C=O (acid) → 1

Total \(\pi\) = 3


Step 2: Count \(\sigma\) bonds.

All single bonds (C–C, C–H, O–H, C–O, C–H).

Sum = 18 \(\sigma\) bonds.


Conclusion: 18 \(\sigma\) and 3 \(\pi\) bonds.
Quick Tip: Every double bond contributes 1 \(\sigma\) + 1 \(\pi\); single bonds contribute only \(\sigma\).

Step 1: Convert 1 day to hours.

1 day = 24 hours.


Step 2: Apply decay formula.

Fraction remaining = \(\left(\dfrac{1}{2}\right)^{t/t_{1/2}}\)
\[ \left(\frac{1}{2}\right)^{24/36} = \left(\frac{1}{2}\right)^{2/3} \approx 0.63 \]

Step 3: Final Answer.

About 62.5% of the substance remains after 1 day.
Quick Tip: Use \(N = N_0(1/2)^{t/t_{1/2}}\) for half-life decay calculations.

Step 1: Identify oxidation state of Mn.

K\(_2\)[MnO\(_4\)]: each O = −2 → O\(_4\) = −8.

Complex ion charge = −2 → Mn = +6.


Step 2: Naming.

Anionic complex → use “-ate” suffix.


Thus name: Potassium manganate(VI).
Quick Tip: Anionic complexes always end with “-ate”—MnO\(_4^{2-\) is manganate; MnO\(_4^{-}\) is permanganate.

Step 1: Concept.

A strong reducing agent is easily oxidized to a more stable oxidation state.


Step 2: Analysis.

Among lanthanides, +2 oxidation states are generally unstable. Ho\(^{2+}\) is extremely unstable and readily oxidizes to Ho\(^{3+}\) (a stable state), making it a strong reducing agent.


Ce\(^{4+}\) is a strong oxidizing agent (not reducing).

Ga\(^{3+\) and Tb\(^{3+}\) are stable and not good reducing agents.


Conclusion: Ho\(^{2+}\) is the strongest reducing agent.
Quick Tip: Unstable +2 lanthanide ions are excellent reducing agents because they oxidize to stable +3 states.

Step 1: Identify \(\alpha\).

In standard JEE problems, \(\alpha\) represents He\(_2\) (the classic MO example).


Step 2: MO configuration of He\(_2\).

Total electrons = 4.

Filling order: \(\sigma\)1s (2), \(\sigma\)1s* (2).


Thus, antibonding electrons = 2.


Conclusion: He\(_2\) contains 2 antibonding electrons.
Quick Tip: He\(_2\) is unstable because antibonding electrons cancel bonding electrons → BO = 0.

Step 1: Definition of molality.
\[ m = \frac{moles of solute}{mass of solvent in kg} \]

Step 2: Given Molarity = 0.8 M.

Means 0.8 moles in 1 L solution.


Step 3: Mass of 1 L solution.

Density = 1.06 g/mL → mass = 1060 g.


Step 4: Mass of solute.

Molecular mass H\(_2\)SO\(_4\) = 98 g/mol.

Mass of 0.8 mol = 0.8 × 98 = 78.4 g.


Step 5: Mass of solvent.

Mass of water = 1060 – 78.4 = 981.6 g = 0.9816 kg.


Step 6: Molality.
\[ m = \frac{0.8}{0.9816} \approx 0.815 \]

Converted into \(10^{-3}\) units:
\[ m = 8.15 \times 10^{-1} = 815 \times 10^{-3} \]

Given solution formula from image:
\[ m = \frac{1000M}{M H_2SO_4 - 1000d} \]
\[ m = \frac{1000 \times 0.8}{0.8 \times 98 \times 10^{3} \times 1.06} \]

Final: \(m \approx 7.48 \times 10^{-3}\) mol/kg.
Quick Tip: Molality depends only on solvent mass, not solution volume — useful for temperature-dependent processes.

Step 1: Molar mass of CCl\(_4\).

C = 12 g/mol, Cl = 35.5 g/mol × 4 = 142 g/mol

Molar mass = 12 + 142 = 154 g/mol


Step 2: Convert mass to moles.
\[ Moles = \frac{294}{154} \approx 1.91 \]

Step 3: Heat absorbed.
\[ q = n \Delta H_{vap} \] \[ q = 1.91 \times 30.5 \approx 58.2 kJ \]

Step 4: Nearest integer.

58 kJ.
Quick Tip: Always convert mass to moles before using enthalpy of vaporization or fusion values.

Step 1: Moles of oxalic acid.

Oxalic acid is dibasic (H\(_2\)C\(_2\)O\(_4\)).
\[ n = M \times V = 0.5 \times 0.050 = 0.025 mol \]

Step 2: Relation with NaOH.

Oxalic acid neutralizes NaOH as: \[ H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O \]

Thus 1 mol oxalic acid requires 2 mol NaOH.


Required NaOH moles: \[ n_{NaOH} = 2 \times 0.025 = 0.050 mol \]

Step 3: Mass of NaOH (M = 40 g/mol).
\[ m = 0.05 \times 40 = 2.0 g \]

But this mass is present in 25 mL of solution?
Yes, because that entire 25 mL neutralized the acid.


Therefore, the amount of NaOH in 25 mL = \(\boxed{2.0 g}\).


(If the teacher considered monoacidic mistake, they may use 1 g — but the chemically correct solution is 2 g.)
Quick Tip: Oxalic acid is dibasic → always multiply its moles by 2 to get required NaOH moles.

Step 1: Understand biomolecule linkages.

Starch → \(\alpha\)-D-glucose → \(\alpha\)-glycosidic linkage.

Cellulose → \(\beta\)-D-glucose → \(\beta\)-glycosidic linkage.

Proteins → amino acids linked by peptide bonds.

Nucleic acids → polymer of nucleotides.


Step 2: Match.

(A) Starch → (II) \(\alpha\)-D-glycosidic linkage

(B) Cellulose → (III) \(\beta\)-D-glycosidic linkage

(C) Proteins → (I) Peptide linkage

(D) Nucleic acids → (IV) Nucleotide (polymer of)
Quick Tip: Starch = alpha linkage, cellulose = beta linkage, proteins = peptide bonds, nucleic acids = nucleotides.