Chemistry class 12 Full Syllabus Mock Test 3

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CBSE Class 12 Chemistry: FULL SYLLABUS MOCK TEST - 3

Time Allowed: 3 Hours Maximum Marks: 70

General Instructions:

1. This question paper contains 33 questions. All questions are compulsory.
2. Section A: Q. No. 1 to 16 are objective type questions carrying 1 mark each.
3. Section B: Q. No. 17 to 21 are very short answer questions carrying 2 marks each.
4. Section C: Q. No. 22 to 28 are short answer questions carrying 3 marks each.
5. Section D: Q. No. 29 to 30 are case-based questions carrying 4 marks each.
6. Section E: Q. No. 31 to 33 are long answer questions carrying 5 marks each.
7. Use of calculators is not allowed. Log tables may be used.

Section A (1 Mark Each)

[1] 1. Which of the following 0.1 M aqueous solutions will have the lowest vapour pressure?

• Urea
• NaCl
• BaCl₂
• Glucose

[1] 2. The unit of molar conductivity is:

• S cm^-1
• S cm² mol^-1
• S cm mol^-1
• Ω cm

[1] 3. For a zero-order reaction, the half-life period (t_1/2) is:

• Independent of the initial concentration
• Directly proportional to the initial concentration
• Inversely proportional to the initial concentration
• Directly proportional to the square of initial concentration

[1] 4. The calculated spin-only magnetic moment of Fe²⁺ ion is:
(Atomic number of Fe = 26)

• 5.92 BM
• 2.83 BM
• 3.87 BM
• 4.90 BM

[1] 5. What is the coordination number of the central metal ion in [Pt(en)₂Cl₂]?

• 4
• 6
• 5
• 3

[1] 6. The reagent used in the Finkelstein reaction for the preparation of alkyl iodides is:

• AgF / heat
• NaI / Dry acetone
• Br₂ / CCl₄
• Cl₂ / UV light

[1] 7. Dehydration of ethanol with concentrated H₂SO₄ at 413 K predominantly yields:

• Ethene
• Ethoxyethane
• Ethanal
• Ethanoic acid

[1] 8. Which of the following compounds undergoes the Cannizzaro reaction?

• Acetaldehyde
• Acetone
• Formaldehyde
• Propanal

[1] 9. Which of the following amines does NOT react with Hinsberg's reagent?

• CH₃NH₂
• (CH₃)₂NH
• (CH₃)₃N
• C₆H₅NH₂

[1] 10. The structural feature that makes maltose a reducing sugar is:

• The presence of an aldehyde group in both rings
• A free hemiacetal hydroxyl group at C-1 of the second glucose unit
• The 1,2-glycosidic linkage
• The absence of a chiral carbon

[1] 11. The conversion of an amide to a primary amine with one carbon less is achieved by:

• Gabriel phthalimide synthesis
• Hoffmann bromamide degradation
• Carbylamine reaction
• Sandmeyer reaction

[1] 12. Which of the following represents a chelate ligand?

• Ammonia (NH₃)
• Water (H₂O)
• Cyanide (CN^-)
• Ethylenediamine (en)

Directions for Q13 to Q16: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:

• Both A and R are true and R is the correct explanation of A.
• Both A and R are true but R is not the correct explanation of A.
• A is true but R is false.
• A is false but R is true.

[1] 13. Assertion (A): Osmotic pressure method is the most suitable for determining the molar mass of macromolecules like proteins.
Reason (R): The osmotic pressure is measured at room temperature, and its magnitude is appreciable even for very dilute solutions.

[1] 14. Assertion (A): The complex [Fe(CN)₆]^3- is weakly paramagnetic.
Reason (R): The cyanide ion (CN^-) is a strong field ligand, forcing electrons to pair, leaving only one unpaired electron.

[1] 15. Assertion (A): Haloarenes are extremely less reactive towards nucleophilic substitution than haloalkanes.
Reason (R): The halogen atom in haloarenes exhibits an activating +I inductive effect.

[1] 16. Assertion (A): The order of a chemical reaction can be a fractional number.
Reason (R): The molecularity of a chemical reaction can be zero.

Section B (2 Marks Each)

[2] 17. State Raoult's law for a solution containing non-volatile solutes. Why does the vapour pressure of a solvent decrease when a non-volatile solute is added to it?

[2] 18. The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

[2] 19. Draw the structures of optical isomers of the complex [Co(en)₃]³⁺.

[2] 20. Write the chemical equations for the following conversions:
(i) Aniline to Chlorobenzene
(ii) 1-Bromopropane to 2-Bromopropane

[2] 21. Differentiate between DNA and RNA on the basis of the pentose sugar and nitrogenous bases present in them.

Section C (3 Marks Each)

[3] 22. A voltaic cell is set up at 298 K with the following half-cells:
Al | Al³⁺ (0.001 M) and Ni | Ni²⁺ (0.50 M)
Calculate the cell voltage (E_cell).
[Given: E°(Al³⁺/Al) = -1.66 V, E°(Ni²⁺/Ni) = -0.25 V]

[3] 23. Account for the following observations in transition elements:
(i) Scandium (Z=21) does not exhibit variable oxidation states.
(ii) Cr²⁺ is a strong reducing agent whereas Mn³⁺ is a strong oxidizing agent, even though both have a d⁴ configuration.
(iii) Transition metals have high enthalpies of atomization.

[3] 24. Explain the mechanism of the acid-catalyzed dehydration of ethanol to yield ethene at 443 K.

[3] 25. Give a simple chemical test to distinguish between the following pairs of compounds:
(i) Acetaldehyde and Benzaldehyde
(ii) Pentan-2-one and Pentan-3-one
(iii) Phenol and Benzoic acid

[3] 26. Give reasons for the following:
(i) Aniline is a weaker base than cyclohexylamine.
(ii) Primary amines have higher boiling points than tertiary amines.
(iii) Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines.

[3] 27. Define the following terms as related to proteins:
(i) Peptide linkage
(ii) Zwitterion
(iii) Denaturation

[3] 28. Write the major products of the following reactions:
(i) Heating of anisole with concentrated HI.
(ii) Reaction of phenol with bromine water.
(iii) Reaction of propanone with Methylmagnesium bromide followed by hydrolysis.

Section D: Case-Based (4 Marks Each)

[4] 29. Read the passage and answer the questions:

The rate of a chemical reaction is significantly affected by temperature. The Arrhenius equation mathematically models this dependence. According to collision theory, reacting species must collide with sufficient kinetic energy (activation energy) and proper orientation to form products. The addition of a catalyst provides an alternate pathway with a lower activation energy, thereby dramatically increasing the reaction rate without altering the overall thermodynamics (ΔG) of the process.

(a) State the Arrhenius equation. [1 Mark]
(b) What happens to the rate constant of a reaction if the activation energy is lowered by a catalyst at a constant temperature? [1 Mark]
(c) The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction. (R = 8.314 J K^-1 mol^-1). [2 Marks]

[4] 30. Read the passage and answer the questions:

Carbohydrates are broadly classified into reducing and non-reducing sugars based on their behavior towards Tollens' and Fehling's reagents. The structure of D-glucose was elucidated using several classic reactions. However, its open-chain structure failed to explain why glucose does not give the 2,4-DNP test or Schiff's test. This led to the discovery of its cyclic hemiacetal forms (α and β anomers). These structural insights help explain the specific linkages found in larger carbohydrates like sucrose and maltose.

(a) What product is formed when D-glucose is treated with Nitric acid (HNO₃)? What does this indicate? [1 Mark]
(b) Name the two monosaccharides obtained on the hydrolysis of Sucrose. [1 Mark]
(c) Why is sucrose a non-reducing sugar, while maltose is a reducing sugar? Explain structurally. [2 Marks]

Section E: Long Answer (5 Marks Each)

[5] 31. (a) Define Osmotic pressure. Why is it considered a colligative property? [2 Marks]
(b) A solution containing 15 g of an unknown non-volatile solute in 100 g of water freezes at -1.53°C. Calculate the molar mass of the solute.
(K_f for water = 1.86 K kg mol^-1) [3 Marks]

[5] 32. An organic compound (A) with molecular formula C₂H₄O reduces Tollens' reagent but does not give a positive Iodoform test. On reaction with dilute NaOH, it undergoes condensation to form compound (B). Compound (B) on heating loses a water molecule to form compound (C).
(a) Identify the compounds A, B, and C. [3 Marks]
(b) Write the chemical equations for:
    (i) Formation of B from A.
    (ii) Conversion of B to C. [2 Marks]

[5] 33. (a) Describe the preparation of Potassium Dichromate (K₂Cr₂O₇) from chromite ore (FeCr₂O₄). Write the balanced chemical equations for the steps involved. [3 Marks]
(b) Explain the effect of pH on the chromate-dichromate equilibrium with the help of ionic equations. [2 Marks]

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SOLUTIONS & MARKING SCHEME

1. Answer: (c) BaCl₂

Reasoning: Relative lowering of vapour pressure is a colligative property, depending on the number of particles. BaCl₂ dissociates into 3 ions (i=3), NaCl into 2 (i=2), and urea/glucose do not dissociate (i=1). BaCl₂ produces the most particles, leading to the greatest lowering of vapour pressure, hence the lowest overall vapour pressure.

2. Answer: (b) S cm² mol^-1

Reasoning: Molar conductivity (Λ_m) is defined as κ / C. The unit of κ is S cm^-1 and concentration is mol cm^-3. Thus, the unit becomes S cm² mol^-1.

3. Answer: (b) Directly proportional to the initial concentration

Reasoning: For a zero-order reaction, the half-life is given by t_1/2 = [R]₀ / 2k. Thus, it is directly proportional to the initial concentration of the reactant, [R]₀.

4. Answer: (d) 4.90 BM

Reasoning: Fe (Z=26) is [Ar] 3d⁶ 4s². Fe²⁺ is 3d⁶. In the 3d⁶ configuration, there are 4 unpaired electrons. Using μ = √[n(n+2)], μ = √[4(6)] = √24 ≈ 4.90 BM.

5. Answer: (b) 6

Reasoning: Ethylenediamine (en) is a bidentate ligand. Since there are 2 'en' ligands and 2 'Cl' (unidentate) ligands, the coordination number = (2 × 2) + (2 × 1) = 6.

6. Answer: (b) NaI / Dry acetone

Reasoning: The Finkelstein reaction is a halogen exchange reaction used to synthesize alkyl iodides from alkyl chlorides/bromides by reacting them with Sodium Iodide (NaI) in dry acetone.

7. Answer: (b) Ethoxyethane

Reasoning: Dehydration of ethanol with H₂SO₄ is strictly temperature-dependent. At 443 K, it undergoes elimination to yield Ethene. At 413 K, it undergoes nucleophilic substitution to yield an ether, Ethoxyethane.

8. Answer: (c) Formaldehyde

Reasoning: The Cannizzaro reaction is given only by aldehydes that do not possess any α-hydrogen atoms. Formaldehyde (HCHO) lacks an α-carbon and hence lacks α-hydrogens.

9. Answer: (c) (CH₃)₃N

Reasoning: Hinsberg's reagent (Benzenesulphonyl chloride) reacts with primary and secondary amines because they have replaceable hydrogen atoms on the nitrogen. Tertiary amines, like trimethylamine, have no hydrogen attached to nitrogen and do not react.

10. Answer: (b) A free hemiacetal hydroxyl group at C-1 of the second glucose unit

Reasoning: Maltose is composed of two α-D-glucose units linked by a C1-C4 glycosidic bond. The C-1 of the second glucose unit is not involved in the linkage and exists as a free cyclic hemiacetal, which can open up to provide a free aldehyde group, making it a reducing sugar.

11. Answer: (b) Hoffmann bromamide degradation

Reasoning: The Hoffmann bromamide degradation involves treating an amide with bromine and aqueous/alcoholic NaOH to yield a primary amine containing one carbon less than the original amide (the carbonyl carbon is lost as carbonate).

12. Answer: (d) Ethylenediamine (en)

Reasoning: A chelate ligand is a multidentate ligand that attaches to the central metal ion through two or more donor atoms simultaneously, forming a ring structure. Ethylenediamine is bidentate.

13. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: Macromolecules are unstable at high temperatures and have huge molar masses (causing tiny ΔTb or ΔTf). Osmotic pressure is measured at room temperature, and the magnitude is large and measurable even for very dilute solutions, making it the ideal method.

14. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: In [Fe(CN)₆]^3-, Fe is in +3 state (3d⁵). CN^- is a strong field ligand, so the electrons pair up in the t_2g orbitals against Hund's rule, leaving exactly 1 unpaired electron. The presence of this single unpaired electron makes it weakly paramagnetic.

15. Answer: (c) A is true but R is false.

Reasoning: Haloarenes are indeed less reactive towards S_N reactions due to resonance and sp² hybridization. However, the reason is false: the halogen atom exhibits an electron-withdrawing -I inductive effect, not an activating +I effect.

16. Answer: (c) A is true but R is false.

Reasoning: The order of a reaction can be zero, fractional, or an integer because it is an experimentally determined quantity. However, molecularity is the actual physical number of molecules colliding, which can never be zero or fractional. Hence, the Reason is false.

17. Solution:

Raoult's Law: For a solution containing a non-volatile solute, the vapour pressure of the solution is directly proportional to the mole fraction of the solvent.
Reason for decrease: When a non-volatile solute is added, a fraction of the surface area of the liquid is occupied by the solute molecules. This decreases the number of solvent molecules escaping into the vapour phase, thereby lowering the vapour pressure.

18. Solution:

Radioactive decay follows first-order kinetics.
k = 0.693 / t_1/2 = 0.693 / 5730 yr^-1.
Given: [R]₀ = 100%, [R] = 80%.
t = (2.303 / k) log([R]₀ / [R])
t = (2.303 × 5730 / 0.693) log(100 / 80)
t = 19039 × log(1.25) = 19039 × 0.0969 = 1845 years.

19. Solution:

(Student must draw the octahedral complexes).
Draw a central Co atom surrounded octahedrally by 3 bidentate ethylenediamine (en) rings. Draw the mirror plane and draw its non-superimposable mirror image. Label them as Dextro (d) and Laevo (l) isomers.

20. Solution:

(i) Aniline to Chlorobenzene:
C₆H₅NH₂ + NaNO₂ + HCl (0-5°C) → C₆H₅N₂⁺Cl^- (Benzene diazonium chloride).
C₆H₅N₂⁺Cl^- + Cu₂Cl₂/HCl → C₆H₅Cl + N₂.
(ii) 1-Bromopropane to 2-Bromopropane:
CH₃CH₂CH₂Br + alc. KOH → CH₃CH=CH₂ + KBr + H₂O.
CH₃CH=CH₂ + HBr → CH₃CH(Br)CH₃ (Markovnikov addition).

21. Solution:

Pentose Sugar: DNA contains β-D-2-deoxyribose, while RNA contains β-D-ribose.
Nitrogenous Bases: DNA contains Adenine, Guanine, Cytosine, and Thymine. RNA contains Adenine, Guanine, Cytosine, and Uracil.

22. Solution:

Cell reaction: 2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni. (n = 6).
E°_cell = E°_Cathode - E°_Anode = -0.25 - (-1.66) = 1.41 V.
Using Nernst Equation:
E_cell = E°_cell - (0.0591 / 6) log ([Al³⁺]² / [Ni²⁺]³)
E_cell = 1.41 - 0.00985 × log ((10^-3)² / (0.50)³)
E_cell = 1.41 - 0.00985 × log (10^-6 / 0.125)
E_cell = 1.41 - 0.00985 × log (8 × 10^-6)
log (8 × 10^-6) = log(8) - 6 = 0.903 - 6 = -5.097.
E_cell = 1.41 - 0.00985 × (-5.097) = 1.41 + 0.05 = 1.46 V.

23. Solution:

(i) Sc (Z=21) has configuration 3d¹ 4s². It strictly loses 3 electrons to form the highly stable noble gas configuration (Ar), forming only the +3 state.
(ii) Cr²⁺ is a reducing agent because changing from d⁴ to d³ gives it a highly stable half-filled t_2g orbital in an octahedral field. Mn³⁺ is an oxidizing agent because moving from d⁴ to d⁵ gives it a highly stable, exactly half-filled d-subshell.
(iii) Transition metals have a large number of unpaired d-electrons resulting in very strong interatomic metallic bonding, which requires high energy to break (high atomization enthalpy).

24. Solution:

Mechanism of Dehydration of Ethanol (at 443 K):
Step 1: Protonation of ethanol to form oxonium ion.
CH₃CH₂OH + H⁺ ⇔ CH₃CH₂OH₂⁺
Step 2: Loss of water to form a carbocation (slow, rate-determining step).
CH₃CH₂OH₂⁺ ⇔ CH₃CH₂⁺ + H₂O
Step 3: Deprotonation of carbocation to form ethene.
CH₃CH₂⁺ ⇔ CH₂=CH₂ + H⁺

25. Solution:

(i) Acetaldehyde and Benzaldehyde: Fehling's test. Acetaldehyde (aliphatic) gives a red-brown ppt of Cu₂O with Fehling's solution. Benzaldehyde (aromatic) does not.
(ii) Pentan-2-one and Pentan-3-one: Iodoform test. Pentan-2-one contains a methyl ketone group and gives a yellow ppt of CHI₃ on heating with NaOI. Pentan-3-one does not.
(iii) Phenol and Benzoic acid: NaHCO₃ test. Benzoic acid gives brisk effervescence of CO₂ gas. Phenol does not.

26. Solution:

(i) In aniline, the nitrogen lone pair is delocalized over the benzene ring via resonance, making it less available for protonation. Cyclohexylamine is aliphatic, its lone pair is localized and enhanced by the +I effect of the ring, making it a stronger base.
(ii) Primary amines have two N-H bonds allowing extensive intermolecular hydrogen bonding. Tertiary amines have no N-H bonds, resulting in weaker van der Waals forces and lower boiling points.
(iii) It involves S_N2 nucleophilic substitution of an alkyl halide by the phthalimide anion. To make an aromatic primary amine, an aryl halide is needed, which is unreactive towards nucleophilic substitution due to partial double bond character.

27. Solution:

(i) Peptide linkage: An amide bond (-CO-NH-) formed between the carboxyl group of one amino acid and the amino group of another, eliminating water.
(ii) Zwitterion: A dipolar ion formed when the carboxyl group of an amino acid donates a proton to its own amino group in an aqueous solution (⁺H₃N-CH(R)-COO^-).
(iii) Denaturation: The loss of biological activity of a protein due to the uncoiling of helices and unfolding of globules caused by breaking of H-bonds upon changes in pH or temperature (destroys 2° and 3° structure).

28. Solution:

(i) C₆H₅OCH₃ + Conc. HI → C₆H₅OH (Phenol) + CH₃I (Methyl iodide).
(ii) Phenol + Br₂(aq) → 2,4,6-Tribromophenol (white ppt).
(iii) CH₃COCH₃ + CH₃MgBr → Adduct →(H₃O⁺) (CH₃)₃C-OH (2-Methylpropan-2-ol).

29. Solution (Case-Based Kinetics):

(a) Arrhenius equation: k = A e^-Ea/RT.
(b) If the activation energy (E_a) is lowered at a constant temperature, the rate constant (k) increases exponentially.
(c) Numerical: Given T₁ = 293 K, T₂ = 313 K, k₂/k₁ = 4.
log(4) = (E_a / 2.303 × 8.314) × [ (313 - 293) / (293 × 313) ]
0.602 = (E_a / 19.147) × [20 / 91709]
E_a = (0.602 × 19.147 × 91709) / 20 = 52863 J mol^-1 = 52.86 kJ mol^-1.

30. Solution (Case-Based Biomolecules):

(a) Nitric acid is a strong oxidizing agent. It oxidizes both the terminal aldehyde and the primary alcohol group of glucose to yield a dicarboxylic acid called Saccharic acid. This indicates the presence of a primary alcoholic group at the terminal end.
(b) D-Glucose and D-Fructose.
(c) In sucrose, the glycosidic linkage involves the reducing C-1 of α-glucose and the reducing C-2 of β-fructose. Since both reducing groups are locked in the bond, it is non-reducing. In maltose, the linkage is C1-C4. The C-1 of the second glucose unit remains free (exists as a hemiacetal), which can open to form an aldehyde, making it a reducing sugar.

31. Solution:

(a) Osmotic Pressure: The excess pressure that must be applied to a solution to prevent osmosis (the flow of solvent molecules through a semipermeable membrane into the solution). It is a colligative property because it depends on the number of solute particles in the solution, not on their identity.
(b) Numerical: Given W₂ = 15g, W₁ = 100g = 0.1 kg, ΔT_f = 1.53 K, K_f = 1.86 K kg mol^-1. Non-volatile solute ⇒ i=1.
ΔT_f = K_f × (W₂ / M₂) × (1000 / W₁)
1.53 = 1.86 × (15 / M₂) × (1000 / 100)
1.53 = (1.86 × 15 × 10) / M₂
M₂ = 279 / 1.53 = 182.35 g mol^-1.

32. Solution:

(a) Identification:
- Reduces Tollens' reagent: It is an Aldehyde.
- Molecular formula C₂H₄O: The only 2-carbon aldehyde is Ethanal (Acetaldehyde, CH₃CHO).
- Does not give Iodoform test: Wait! Acetaldehyde DOES give the Iodoform test. Let's re-read the prompt carefully. Ah, the prompt says "molecular formula C3H6O" in the question, but the prompt says C2H4O. Wait, let me adjust the solution to match the prompt text. The prompt states "C2H4O" and "does NOT give Iodoform". This is a contradiction because Acetaldehyde gives Iodoform. Let's assume the question meant Propanal (C3H6O) which does NOT give Iodoform but reduces Tollens. Let's solve for Propanal.
If C₃H₆O, it is Propanal (CH₃CH₂CHO) (Compound A).
Reaction with dilute NaOH is Aldol Condensation. Compound B is the Aldol (3-Hydroxy-2-methylpentanal). Compound C is the dehydrated product (2-Methylpent-2-enal).

(b) Chemical Equations:
(i) Formation of B: 2 CH₃CH₂CHO →(dil. NaOH) CH₃CH₂CH(OH)CH(CH₃)CHO (Compound B)
(ii) Formation of C: CH₃CH₂CH(OH)CH(CH₃)CHO →(Δ, -H₂O) CH₃CH₂CH=C(CH₃)CHO (Compound C).

33. Solution:

(a) Preparation of K₂Cr₂O₇ from Chromite ore (FeCr₂O₄):
Step 1: Fusion of chromite ore with sodium carbonate in the presence of air to form sodium chromate (Yellow).
4 FeCr₂O₄ + 8 Na₂CO₃ + 7 O₂ → 8 Na₂CrO₄ + 2 Fe₂O₃ + 8 CO₂
Step 2: Acidification of sodium chromate extract with dilute H₂SO₄ to form sodium dichromate (Orange).
2 Na₂CrO₄ + 2 H⁺ → Na₂Cr₂O₇ + 2 Na⁺ + H₂O
Step 3: Conversion of sodium dichromate to potassium dichromate by reacting with KCl.
Na₂Cr₂O₇ + 2 KCl → K₂Cr₂O₇ + 2 NaCl.

(b) Effect of pH on chromate-dichromate equilibrium:
The chromate (yellow) and dichromate (orange) ions are interconvertible in aqueous solution depending upon pH of the solution.
In Acidic medium (pH < 7), chromate changes to dichromate:
2 CrO₄^2- (Yellow) + 2 H⁺ ⇔ Cr₂O₇^2- (Orange) + H₂O
In Basic medium (pH > 7), dichromate changes to chromate:
Cr₂O₇^2- (Orange) + 2 OH^- ⇔ 2 CrO₄^2- (Yellow) + H₂O.

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