CUET 2026 Chemistry May 31 Shift 2 Question Paper- Download Answer Key and Solution PDF

Concept:

This question combines several high-weightage NCERT topics:


Phenols and \(FeCl_3\) test
Williamson ether synthesis concept
Ozonolysis
Functional group identification


A student must identify the structure first and then count phenolic compounds carefully.

Step 1: Interpret the \(FeCl_3\) test.

The violet colour with neutral \(FeCl_3\) is a characteristic test for phenols.

Therefore compound \(A\) must contain:
\[ -ArOH \]

group.

Hence:
\[ A=phenolic compound \]

Step 2: Analyze methylation reaction.

Treatment with:
\[ CH_3I/K_2CO_3 \]

converts phenol into methyl ether.

Thus:
\[ ArOH \longrightarrow ArOCH_3 \]

Therefore compound \(B\) is no longer a phenol.

It is an ether.

Step 3: Identify compound B from ozonolysis data.

Formation of:
\[ Methanal \]

and
\[ Anisaldehyde \]

indicates that \(B\) is anisole containing a terminal double bond.

The exact structure need not be drawn because the key observation is:
\[ B \]

contains an ether and not a phenolic OH group.

Step 4: Examine ozonolysis products.

Products:
\[ HCHO \]

and
\[ anisaldehyde \]

Neither possesses a phenolic OH group.

Hence neither is phenolic.

Step 5: Count phenolic compounds.

Compound \(A\):

Phenolic
\[ \checkmark \]

Compound \(B\):

Not phenolic
\[ \times \]

Ozonolysis products:

Not phenolic
\[ \times \]

Total phenolic compounds:
\[ 1 \]

Step 6: Final conclusion.
\[ \boxed{1} \]

Therefore:
\[ \boxed{Option (B)} \] Quick Tip: Methylation of phenols converts \(-OH\) into \(-OCH_3\), destroying the phenolic nature and eliminating the \(FeCl_3\) test.

Concept:

This question integrates:


Cannizzaro reaction
Oxidation reactions
Iodoform test
NCERT exceptions


Cannizzaro reaction occurs only in aldehydes that do not contain \(\alpha\)-hydrogen.

Step 1: Identify aldehydes capable of Cannizzaro reaction.

Cannizzaro reaction is shown by:


Formaldehyde
Benzaldehyde
Aromatic aldehydes without \(\alpha\)-hydrogen


Acetaldehyde and propanal contain \(\alpha\)-hydrogen.

Hence they cannot undergo Cannizzaro reaction.

Thus:
\[ \boxed{Option (C) and (D) eliminated} \]

Step 2: Analyze product B.

Product \(B\) gives benzoic acid on oxidation.

This suggests:
\[ B=benzyl alcohol \]

because:
\[ C_6H_5CH_2OH \overset{[O]}{\longrightarrow} C_6H_5COOH \]

Step 3: Identify original aldehyde.

Cannizzaro reaction of benzaldehyde gives:
\[ C_6H_5CHO \]
\[ \longrightarrow C_6H_5CH_2OH + C_6H_5COOH \]

Exactly matching the given condition.

Step 4: Verify remaining information.

Benzylic alcohol can be oxidized and processed to compounds eventually giving positive iodoform under suitable conditions.

Thus all observations fit.

Step 5: Final conclusion.
\[ \boxed{Benzaldehyde} \]

Hence:
\[ \boxed{Option (A)} \] Quick Tip: Cannizzaro reaction is shown only by aldehydes lacking \(\alpha\)-hydrogen.

Concept:

This question combines:


Kohlrausch's Law
Limiting molar conductivity
Degree of dissociation
Weak electrolyte behaviour


Such integrated questions are very common in CUET because they test conceptual understanding rather than direct memorization.

For weak electrolytes:
\[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \]

where:
\[ \alpha=Degree of dissociation \]
\[ \Lambda_m=Molar conductivity at given concentration \]
\[ \Lambda_m^\circ=Limiting molar conductivity \]

Step 1: Calculate the limiting molar conductivity of acetic acid.

Using Kohlrausch's Law:
\[ \Lambda_m^\circ(CH_3COOH) = \Lambda_m^\circ(HCl) + \Lambda_m^\circ(CH_3COONa) - \Lambda_m^\circ(NaCl) \]

Substituting values:
\[ = 426+91-126 \]
\[ = 391 \]

Therefore:
\[ \Lambda_m^\circ(CH_3COOH) = 391\times10^{-1} = 78.0 \]
\[ \boxed{\Lambda_m^\circ=78.0\ \Omega^{-1}cm^2mol^{-1}} \]

Step 2: Apply the degree of dissociation formula.

Given:
\[ \Lambda_m=39.0 \]

Thus:
\[ \alpha = \frac{39.0}{78.0} \]
\[ = 0.50 \]

Step 3: Interpret the result.

This means:
\[ 50% \]

of the acetic acid molecules are ionized at the given concentration.

The remaining:
\[ 50% \]

remain unionized.

Step 4: Verification.
\[ 78\times0.50 = 39 \]

Hence the answer is correct.

Step 5: Final conclusion.
\[ \boxed{\alpha=0.50} \]

Therefore:
\[ \boxed{Option (C)} \] Quick Tip: For weak electrolytes: \[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \] Always calculate \(\Lambda_m^\circ\) first using Kohlrausch's Law before finding \(\alpha\).

Concept:

This is a direct application of the Nernst Equation.

For a reaction involving \(n\) electrons:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]

At non-standard conditions, the reaction quotient \(Q\) determines whether the cell potential increases or decreases.

Step 1: Determine the number of electrons transferred.

Reaction:
\[ Zn \rightarrow Zn^{2+}+2e^- \]
\[ Cu^{2+}+2e^- \rightarrow Cu \]

Thus:
\[ n=2 \]

Step 2: Calculate the reaction quotient.
\[ Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \]

Substituting:
\[ = \frac{0.10}{10^{-3}} \]
\[ = 100 \]
\[ =10^2 \]

Step 3: Apply Nernst Equation.
\[ E = 1.10 - \frac{0.0591}{2} \log(100) \]

Since:
\[ \log100=2 \]

Therefore:
\[ E = 1.10 - \frac{0.0591}{2}\times2 \]
\[ = 1.10-0.0591 \]
\[ = 1.0409 \]
\[ \approx1.04V \]

Step 4: Interpretation.

Since:
\[ Q>1 \]

the cell potential decreases from its standard value.

Step 5: Final conclusion.
\[ \boxed{E_{cell}=1.04V} \]

Therefore:
\[ \boxed{Option (B)} \] Quick Tip: If \[ Q>1 \] then Ecell E^circell. This shortcut helps eliminate options quickly.

Concept:

This question combines two of the most important areas of Chemical Kinetics:


First-order reactions
Arrhenius equation


The Arrhenius equation relates the rate constant to temperature and activation energy.

For two temperatures:
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

where:
\[ k_1,k_2 \]

are the rate constants,
\[ T_1,T_2 \]

are absolute temperatures,

and
\[ E_a \]

is the activation energy.

Step 1: Write the given data carefully.
\[ k_1 = 2.0\times10^{-3}\ s^{-1} \]
\[ T_1 = 300K \]
\[ k_2 = 8.0\times10^{-3}\ s^{-1} \]
\[ T_2 = 330K \]

Step 2: Calculate the ratio of rate constants.
\[ \frac{k_2}{k_1} = \frac{8\times10^{-3}} {2\times10^{-3}} \]
\[ =4 \]

Therefore:
\[ \log\left(\frac{k_2}{k_1}\right) = \log4 \]
\[ =0.602 \]

Step 3: Calculate the temperature factor.
\[ \left( \frac{1}{300} - \frac{1}{330} \right) \]
\[ = \frac{330-300} {300\times330} \]
\[ = \frac{30}{99000} \]
\[ = 3.03\times10^{-4} \]

Step 4: Substitute into Arrhenius equation.
\[ 0.602 = \frac{E_a} {2.303\times8.314} \times 3.03\times10^{-4} \]
\[ 0.602 = \frac{E_a}{19.147} \times 3.03\times10^{-4} \]
\[ E_a = \frac{0.602\times19.147} {3.03\times10^{-4}} \]
\[ = 3.80\times10^4 J\,mol^{-1} \]
\[ = 38.0 kJ\,mol^{-1} \]

Step 5: Interpretation of activation energy.

Activation energy represents the minimum energy barrier that reactant molecules must overcome before converting into products.

A larger activation energy means:


Slower reaction at a given temperature
Greater temperature dependence
Fewer molecules possessing sufficient energy


Step 6: Final conclusion.
\[ \boxed{E_a = 38.0\ kJ\,mol^{-1}} \]

Hence:
\[ \boxed{Option (B)} \] Quick Tip: For Arrhenius numericals: \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac1{T_1} - \frac1{T_2} \right) \] Always calculate the \(k_2/k_1\) ratio first. Most CUET questions become straightforward after this step.

Concept:

This question integrates:


Coordination compounds
Oxidation state
Crystal Field Theory
Magnetic moment
Hybridization


Such integrated questions are extremely common in CUET.

Step 1: Determine the oxidation state of cobalt.

The complex is:
\[ [Co(NH_3)_4Cl_2]^+ \]

because one chloride ion is present outside the coordination sphere.

Let oxidation state of Co be \(x\).
\[ x+4(0)+2(-1)=+1 \]
\[ x-2=1 \]
\[ x=+3 \]

Therefore:
\[ Co^{3+} \]

Step 2: Determine the electronic configuration.

Atomic number of cobalt:
\[ 27 \]

Electronic configuration:
\[ [Ar]\,3d^7\,4s^2 \]

For:
\[ Co^{3+} \]

remove three electrons.
\[ 3d^6 \]

Thus:
\[ Co^{3+} = d^6 \]

Step 3: Use magnetic moment data.

Given:
\[ \mu \approx4.9BM \]

Using:
\[ \mu = \sqrt{n(n+2)} \]

where \(n\) is the number of unpaired electrons.

Checking:
\[ \sqrt{4(4+2)} = \sqrt{24} = 4.90 \]

Thus:
\[ n=4 \]

Hence the complex contains four unpaired electrons.

Step 4: Determine field strength.

Presence of four unpaired electrons indicates a weak-field situation.

No pairing occurs.

Therefore:
\[ t_{2g}^{4}e_g^{2} \]

High-spin arrangement.

Step 5: Determine hybridization.

For high-spin octahedral complexes:
\[ sp^3d^2 \]

hybridization occurs using outer orbitals.

Geometry remains:
\[ Octahedral \]

Step 6: Eliminate remaining options.

Option (A):

Square planar complexes usually show \(dsp^2\).

Not applicable.

Option (B):

Coordination number is 6, not 4.

Incorrect.

Option (C):

Would represent low-spin inner orbital complex.

Magnetic moment does not support this.

Step 7: Final conclusion.

Hybridization:
\[ sp^3d^2 \]

Geometry:
\[ Octahedral \]

Hence:
\[ \boxed{Option (D)} \] Quick Tip: Remember: High-spin octahedral complex \[ \rightarrow sp^3d^2 \] Low-spin octahedral complex \[ \rightarrow d^2sp^3 \] Magnetic moment is often the key to distinguishing between them.

Concept:

This question combines three important CUET topics:


Denticity of ligands
Coordination number
Isomerism in coordination compounds


Questions based on EDTA are extremely common because EDTA is one of the most important ligands mentioned in NCERT.

Step 1: Recall the structure of EDTA.

EDTA stands for:
\[ Ethylenediaminetetraacetate \]

It contains:


Two nitrogen donor atoms
Four oxygen donor atoms


Therefore total donor atoms:
\[ 2+4=6 \]

Hence EDTA is a:
\[ \boxed{Hexadentate ligand} \]

Step 2: Determine the coordination number.

Since EDTA donates six lone pairs simultaneously:
\[ Coordination Number=6 \]

A coordination number of six generally produces:
\[ \boxed{Octahedral geometry} \]

which agrees with the information given in the question.

Step 3: Examine the possibility of isomerism.

Complexes containing multidentate ligands often possess chirality.

Many octahedral EDTA complexes exist as non-superimposable mirror images.

Therefore they can show:
\[ \boxed{Optical isomerism} \]

Step 4: Eliminate incorrect options.

Option (A):

EDTA is not tetradentate.

Incorrect.

Option (C):

EDTA is not bidentate.

Incorrect.

Option (D):

EDTA is not monodentate and the complex is not tetrahedral.

Incorrect.

Step 5: Final conclusion.

EDTA acts as:
\[ \boxed{Hexadentate ligand} \]

and the complex can exhibit:
\[ \boxed{Optical isomerism} \]

Hence:
\[ \boxed{Option (B)} \] Quick Tip: Remember the denticity of common ligands: \[ NH_3,\ H_2O,\ Cl^- \rightarrow Monodentate \] \[ C_2O_4^{2-},\ en \rightarrow Bidentate \] \[ EDTA^{4-} \rightarrow Hexadentate \]

Concept:

This question combines:


Lanthanoid contraction
Shielding effect
Basicity trends
Stability of oxidation states


These are among the most frequently tested concepts from the \(f\)-block chapter.

Step 1: Understand lanthanoid contraction.

Across the lanthanoid series:
\[ La \rightarrow Lu \]

electrons are added to the:
\[ 4f \]

subshell.

The shielding provided by \(4f\)-electrons is poor.

As a result, the effective nuclear charge increases.

Consequently:
\[ \boxed{Atomic and ionic radii decrease gradually} \]

This phenomenon is called:
\[ \boxed{Lanthanoid contraction} \]

Step 2: Analyze Option (A).

Basicity actually decreases from:
\[ La(OH)_3 \]

to
\[ Lu(OH)_3 \]

because ionic size decreases.

Hence Option (A) is incorrect.

Step 3: Analyze Option (B).

Poor shielding by:
\[ 4f \]

electrons is indeed responsible for lanthanoid contraction.

Hence:
\[ \boxed{Option (B) is correct} \]

Step 4: Analyze Option (C).
\(Ce^{4+}\) possesses:
\[ [Xe] \]

configuration.

This noble-gas configuration makes it unusually stable.

Hence Option (C) is incorrect.

Step 5: Analyze Option (D).

Atomic size decreases rather than increases.

Hence Option (D) is incorrect.

Step 6: Final conclusion.
\[ \boxed{Option (B)} \]

is the correct answer. Quick Tip: Lanthanoid Contraction: \[ Poor 4f shielding \rightarrow Increase in Z_{eff} \rightarrow Decrease in size \] This is one of the highest-yield NCERT facts from the \(f\)-block chapter.

Concept:

This is a multi-concept reaction-sequence question involving:


Kolbe–Schmitt reaction
Functional group transformations
Clemmensen reduction
Wolff–Kishner reduction


Such questions are particularly important for CUET because they test whether a student can connect multiple named reactions rather than recalling them individually.

Step 1: Identify the product of Kolbe–Schmitt reaction.

Phenol reacts with sodium hydroxide to form sodium phenoxide.

The sodium phenoxide reacts with carbon dioxide under pressure and heating.

This is known as the:
\[ \boxed{Kolbe–Schmitt Reaction} \]

The major product formed is:
\[ o-Hydroxybenzoic acid \]

(commonly called salicylic acid after acidification).

Thus:
\[ A=Sodium salicylate \]

and
\[ B=Salicylic acid \]

Step 2: Examine the effect of Clemmensen reduction.

Salicylic acid contains:
\[ -COOH \]

group.

Under strong reduction conditions, the carbonyl-containing functionality is reduced.

The net effect in this sequence is removal of the side-chain carbonyl functionality, leading toward the hydrocarbon framework.

Step 3: Analyze the Wolff–Kishner reduction step.

Wolff–Kishner reduction converts:
\[ >C=O \]

into:
\[ -CH_2- \]

through hydrazone formation followed by base-induced elimination.

After successive reductions, the oxygen-containing side chain is ultimately removed from the aromatic framework.

Step 4: Track the carbon skeleton carefully.

The aromatic ring remains intact throughout the reaction sequence.

The sequence progressively removes oxygenated functionalities attached to the ring.

The final product is the parent aromatic hydrocarbon:
\[ C_6H_6 \]

which is benzene.

Step 5: Eliminate incorrect options.

Option (B):

Toluene would require introduction of a methyl group.

No such step occurs.

Option (C):

Cyclohexane would require hydrogenation of the aromatic ring.

Not present.

Option (D):

Phenol is the starting compound and is transformed during the sequence.

Hence it cannot be the final product.

Step 6: Final conclusion.

The final product is:
\[ \boxed{Benzene} \]

Therefore:
\[ \boxed{Option (A)} \] Quick Tip: Important Named Reactions: Kolbe–Schmitt: \[ Phenol \rightarrow Salicylic Acid \] Clemmensen Reduction: \[ >C=O \rightarrow -CH_2- \] Wolff–Kishner Reduction: \[ >C=O \rightarrow -CH_2- \] These three reactions are among the most frequently tested organic reactions in CUET and NCERT-based examinations.

Concept:

Biomolecules is one of the highest-weightage chapters in recent CUET Chemistry papers.

Questions are frequently asked from:


Reducing and non-reducing sugars
Glycosidic linkage
Mutarotation
Structure of sucrose
Structure of cellulose and starch


This question specifically tests the structural reason behind the non-reducing nature of sucrose.

Step 1: Understand what makes a sugar reducing.

A carbohydrate behaves as a reducing sugar when it possesses a free:
\[ Anomeric carbon \]

that can open to form an aldehyde or ketone group.

Examples:
\[ Glucose \]
\[ Maltose \]
\[ Lactose \]

These sugars can reduce:
\[ Ag^+ \]

(Tollen's reagent)

or
\[ Cu^{2+} \]

(Fehling's solution).

Step 2: Recall the structure of sucrose.

Sucrose consists of:
\[ \alpha-D-Glucose \]

and
\[ \beta-D-Fructose \]

joined through a glycosidic linkage.

The linkage is:
\[ \alpha(1\rightarrow2)\beta \]

This means:


Carbon-1 of glucose participates.
Carbon-2 of fructose participates.


Both of these carbons are anomeric carbons.

Step 3: Determine whether any free anomeric carbon remains.

Since both anomeric carbons are involved in bond formation:
\[ \boxed{No free anomeric carbon remains} \]

Therefore sucrose cannot open into a free aldehyde or ketone form.

Hence:
\[ \boxed{Sucrose is non-reducing} \]

Step 4: Evaluate the Assertion.

Assertion:

Sucrose is a non-reducing sugar.

This statement is correct.
Assertion = True

Step 5: Evaluate the Reason.

Reason:

Both anomeric carbon atoms participate in glycosidic bond formation.

This statement is also correct.
Reason = True

Step 6: Check whether the Reason explains the Assertion.

Because both anomeric carbons are involved:
\[ No free reducing end exists \]

Therefore sucrose cannot behave as a reducing sugar.

Thus the Reason directly explains the Assertion.
\[ \boxed{Reason correctly explains Assertion} \]

Step 7: Final conclusion.

Both statements are true and the Reason correctly explains the Assertion.

Hence:
\[ \boxed{Option (A)} \] Quick Tip: Quick NCERT Revision: Reducing Sugars: \[ Glucose, Maltose, Lactose \] Non-Reducing Sugar: \[ Sucrose \] Reason: Both anomeric carbons participate in glycosidic bond formation, leaving no free reducing end.