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CBSE Class 12 Chemistry: FULL SYLLABUS MOCK TEST - 2
General Instructions:
- This question paper contains 33 questions. All questions are compulsory.
- Section A: Q. No. 1 to 16 are objective type questions carrying 1 mark each.
- Section B: Q. No. 17 to 21 are very short answer questions carrying 2 marks each.
- Section C: Q. No. 22 to 28 are short answer questions carrying 3 marks each.
- Section D: Q. No. 29 to 30 are case-based questions carrying 4 marks each.
- Section E: Q. No. 31 to 33 are long answer questions carrying 5 marks each.
- Use of calculators is not allowed. Log tables may be used.
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.
Reason (R): The SN2 reaction occurs in a single step where the nucleophile attacks from the side opposite to the leaving group.
Reason (R): Transition metals have variable oxidation states and can provide a large surface area for the adsorption of reactants.
Reason (R): Elevation of boiling point is a colligative property that depends on the number of particles in the solution.
Reason (R): The -NH2 group is strongly activating and ortho-para directing.
(Kf for acetic acid = 3.9 K kg mol-1)
Mg(s) | Mg2+ (0.130 M) || Ag+ (0.0001 M) | Ag(s)
Given: E°(Mg2+/Mg) = -2.37 V, E°(Ag+/Ag) = +0.80 V.
(a) Actinoid contraction is greater from element to element than lanthanoid contraction.
(b) Zirconium (Zr) and Hafnium (Hf) have almost identical atomic radii.
(c) Zn, Cd, and Hg are not considered transition elements.
(i) Propanal and Propanone
(ii) Phenol and Benzoic acid
(iii) Primary (1°) and Secondary (2°) amines
(i) Reimer-Tiemann reaction
(ii) Kolbe's reaction
(iii) Williamson ether synthesis
Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, and development of a green coating on copper and bronze are some examples of corrosion. Corrosion of iron (commonly known as rusting) occurs in the presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical phenomenon. At a particular spot on an object made of iron, oxidation takes place and that spot behaves as an anode.
(a) Write the half-cell reaction that occurs at the anode during the rusting of iron. [1 Mark]
(b) Write the half-cell reaction that occurs at the cathode during rusting. What is the role of H+ ions? [1 Mark]
(c) Suggest any two methods to prevent the corrosion of iron. Explain the principle of cathodic protection (galvanization). [2 Marks]
Vitamins are organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism. Vitamins are designated by alphabets A, B, C, D, etc. Some of them are further named as sub-groups e.g., B1, B2, B6, B12, etc. Excess of vitamins is also harmful and vitamin pills should not be taken without the advice of a doctor. The term “vitamine” was coined from the word vital + amine since the earlier identified compounds had amino groups.
(a) Which vitamin deficiency causes Scurvy? Classify this vitamin as water-soluble or fat-soluble. [1 Mark]
(b) Why must water-soluble vitamins be supplied regularly in our diet? Which water-soluble vitamin is an exception and can be stored in the body? [1 Mark]
(c) Identify the deficiency diseases associated with Vitamin A and Vitamin D. [2 Marks]
(b) 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol-1. What is the percentage association of acid if it forms a dimer in solution? [3 Marks]
(a) Identify the compounds (A) and (B). [2 Marks]
(b) Write the chemical equations for the following reactions of compound (A):
(i) Reaction with I2 and NaOH (Iodoform test).
(ii) Reaction with Zinc amalgam and concentrated HCl (Clemmensen reduction).
(iii) Reaction with Hydrogen Cyanide (HCN). [3 Marks]
(b) Write balanced chemical equations to show the oxidizing action of acidified KMnO4 on:
(i) Iron(II) ions (Fe2+)
(ii) Oxalate ions (C2O42-) [2 Marks]
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SOLUTIONS & MARKING SCHEME
1. Answer: (c) Ethanol + Acetone
Reasoning: Ethanol molecules are strongly hydrogen-bonded. Addition of acetone breaks these H-bonds, making the A-B interactions weaker than A-A and B-B interactions. This results in a positive deviation from Raoult's law.
2. Answer: (d) cm-1
Reasoning: Cell constant (G*) is defined as l/A (length/area). Units = cm / cm2 = cm-1.
3. Answer: (c) 3
Reasoning: Let rate = k[A]n. If [A] is doubled, Rate' = k(2[A])n = 2n × k[A]n. Given Rate' = 8 × Rate. Therefore, 2n = 8, which means n = 3.
4. Answer: (a) Mn2+
Reasoning: Magnetic moment depends on the number of unpaired electrons. Mn2+ has 5 unpaired electrons (3d5), which is the maximum possible in the 3d series.
5. Answer: (a) +2
Reasoning: Let the oxidation state of Fe be x. Cyanide (CN-) has a charge of -1. Therefore, x + 6(-1) = -4 ⇒ x - 6 = -4 ⇒ x = +2.
6. Answer: (c) 2-Chloro-2-methylpropane
Reasoning: SN1 reactions proceed via a carbocation intermediate. 2-Chloro-2-methylpropane is a tertiary (3°) alkyl halide, which forms a highly stable tertiary carbocation, making it the most reactive.
7. Answer: (a) Benzene
Reasoning: Phenol is reduced to benzene when heated with zinc dust. (C6H5OH + Zn → C6H6 + ZnO).
8. Answer: (b) Chromyl chloride (CrO2Cl2)
Reasoning: The Etard reaction specifically utilizes Chromyl chloride in CS2 to partially oxidize toluene into an intermediate chromium complex, which on hydrolysis yields benzaldehyde.
9. Answer: (a) Benzenesulphonyl chloride
Reasoning: Benzenesulphonyl chloride (C6H5SO2Cl) is Hinsberg's reagent, used to distinguish between 1°, 2°, and 3° amines based on the solubility of the resulting sulphonamides.
10. Answer: (d) Uracil
Reasoning: Both DNA and RNA contain Adenine, Guanine, and Cytosine. However, DNA contains Thymine, whereas RNA contains Uracil instead of Thymine.
11. Answer: (c) Loss of secondary and tertiary structures
Reasoning: Denaturation disrupts the hydrogen bonds and other forces holding the 2° and 3° structures together, causing the protein to unfold. The primary structure (peptide bonds) remains intact.
12. Answer: (c) 2,4,6-Tribromoaniline
Reasoning: The -NH2 group is strongly activating. In an aqueous medium (bromine water), substitution occurs rapidly at all ortho and para positions, yielding a white precipitate of 2,4,6-tribromoaniline.
13. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: SN2 is a concerted, single-step reaction where the nucleophile attacks from the backside of the leaving group, naturally pushing the other three bonds inside out like an umbrella in the wind, resulting in inversion of configuration.
14. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: The catalytic properties of transition metals are primarily due to their ability to adopt multiple oxidation states (forming unstable intermediates) and providing a large surface area for reactants to adsorb.
15. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: Boiling point elevation is a colligative property. NaCl dissociates into 2 ions (i=2), while glucose does not dissociate (i=1). Because NaCl provides twice as many particles in solution, its boiling point elevation is higher.
16. Answer: (b) Both A and R are true but R is not the correct explanation of A.
Reasoning: While it is true that aniline does not undergo Friedel-Crafts and the -NH2 group is activating, the real reason it fails is because the basic aniline forms a salt complex with the Lewis acid catalyst (AlCl3), placing a positive charge on N, which severely deactivates the ring.
17. Solution:
Molar mass of Ascorbic acid (M2) = 176 g mol-1.
Given: ΔTf = 1.5 K, Kf = 3.9 K kg mol-1, Mass of solvent (W1) = 75 g = 0.075 kg.
ΔTf = Kf × [W2 / (M2 × W1)]
1.5 = 3.9 × [W2 / (176 × 0.075)]
W2 = (1.5 × 176 × 0.075) / 3.9 = 5.08 g.
18. Solution:
IUPAC Name: Pentaamminechloridocobalt(III) chloride.
Isomerism: If swapped with [Co(NH3)5(SO4)]Cl, the counter ions inside and outside the coordination sphere are exchanged. This is known as Ionization Isomerism.
19. Solution:
In chloroacetic acid (Cl-CH2COOH), the highly electronegative chlorine atom exerts an electron-withdrawing -I (inductive) effect. This effect withdraws electron density away from the O-H bond, facilitating H+ release and stabilizing the resulting carboxylate ion. Acetic acid (CH3COOH) has an electron-donating methyl group (+I effect) which destabilizes the anion. Hence, chloroacetic acid is a stronger acid.
20. Solution:
Equations: Phthalimide reacts with KOH to form Potassium phthalimide. This reacts with an alkyl halide (R-X) to form N-Alkylphthalimide. Alkaline hydrolysis yields a primary amine (R-NH2).
Why aniline fails: Aniline is an aromatic primary amine. Its preparation would require the nucleophilic substitution of an aryl halide (like chlorobenzene) by the phthalimide anion. Aryl halides are unreactive towards nucleophilic substitution due to the partial double bond character of the C-X bond.
21. Solution:
Globular Proteins: The polypeptide chains are highly folded into a spherical (globe-like) shape. They are generally soluble in water. Example: Insulin or Albumin.
Fibrous Proteins: The polypeptide chains run parallel to each other and are held together by hydrogen/disulphide bonds to form fiber-like structures. They are generally insoluble in water. Example: Keratin or Myosin.
22. Solution:
Reaction: Mg + 2Ag+ → Mg2+ + 2Ag (n = 2).
E°cell = E°Cathode - E°Anode = 0.80 - (-2.37) = 3.17 V.
Using Nernst Equation:
Ecell = E°cell - (0.0591 / 2) log ([Mg2+] / [Ag+]2)
Ecell = 3.17 - 0.02955 × log (0.130 / (10-4)2)
Ecell = 3.17 - 0.02955 × log (0.130 × 108)
Ecell = 3.17 - 0.02955 × log (1.3 × 107)
Ecell = 3.17 - 0.02955 × 7.11 = 3.17 - 0.21 = 2.96 V.
23. Solution:
For 99% completion: [R] = 1% of [R]0 = 1/100.
t99% = (2.303 / k) log (100 / 1) = (2.303 / k) log(102) = (2.303 / k) × 2
For 90% completion: [R] = 10% of [R]0 = 10/100 = 1/10.
t90% = (2.303 / k) log (100 / 10) = (2.303 / k) log(101) = (2.303 / k) × 1
Comparing both: t99% = 2 × t90%. (Hence proved).
24. Solution:
(a) The 5f electrons in actinoids provide even poorer shielding from the increasing nuclear charge than the 4f electrons in lanthanoids, resulting in a more pronounced actinoid contraction.
(b) Zr (period 5) and Hf (period 6) have almost identical radii due to Lanthanoid Contraction, which cancels out the expected size increase down the group.
(c) Zn, Cd, and Hg have completely filled d-orbitals (nd10) in both their ground states and their common oxidation states. Hence, they do not qualify as transition elements.
25. 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+
26. Solution:
(i) Propanal and Propanone: Tollens' test. Propanal (aldehyde) forms a silver mirror with Tollens' reagent; propanone (ketone) does not.
(ii) Phenol and Benzoic acid: NaHCO3 test. Benzoic acid gives brisk effervescence of CO2 gas with aqueous NaHCO3; phenol does not.
(iii) 1° and 2° amines: Carbylamine test. 1° amine produces a foul-smelling isocyanide when heated with CHCl3 and alc. KOH; 2° amine does not.
27. Solution:
(i) Reimer-Tiemann reaction: Phenol reacts with CHCl3 in the presence of aq. NaOH at 340 K to form Salicylaldehyde (2-Hydroxybenzaldehyde).
(ii) Kolbe's reaction: Phenol is converted to sodium phenoxide, which reacts with CO2 at 400 K and 4-7 atm pressure, followed by acidification to yield Salicylic acid (2-Hydroxybenzoic acid).
(iii) Williamson ether synthesis: Reaction of an alkyl halide with sodium alkoxide via SN2 mechanism to yield an ether (R-X + R'-ONa → R-O-R' + NaX).
28. Solution:
Complex: [Ni(CN)4]2-. Oxidation state of Ni = +2.
Ni2+ configuration = [Ar] 3d8.
CN- is a strong field ligand, which forces the pairing of the two unpaired 3d electrons into a single orbital against Hund's rule.
This leaves one inner 3d orbital vacant. It hybridizes with one 4s and two 4p orbitals, resulting in dsp2 hybridization.
Geometry: Square planar.
Magnetic Behaviour: Diamagnetic (since all electrons are paired).
29. Solution (Case-Based Electrochemistry):
(a) Anode Reaction: Fe(s) → Fe2+(aq) + 2e-.
(b) Cathode Reaction: O2(g) + 4H+(aq) + 4e- → 2H2O(l). The H+ ions are available from H2CO3 formed by dissolution of atmospheric CO2 in water, facilitating the reduction of oxygen.
(c) Prevention: Painting, oiling, or Galvanization. Galvanization (Cathodic Protection): Iron is coated with a more reactive metal like Zinc. Zinc acts as a sacrificial anode, oxidizing itself (Zn → Zn2+ + 2e-) to protect the iron object, which functions as the cathode.
30. Solution (Case-Based Biomolecules):
(a) Scurvy is caused by the deficiency of Vitamin C (Ascorbic acid). It is a water-soluble vitamin.
(b) Water-soluble vitamins dissolve easily in water and are readily excreted from the body through urine. They cannot be stored in fatty tissues. The exception is Vitamin B12, which can be stored in the liver.
(c) Vitamin A deficiency causes Night blindness or Xerophthalmia. Vitamin D deficiency causes Rickets in children (or Osteomalacia in adults).
31. Solution:
(a) Derive zero order: Rate = -d[R]/dt = k[R]0 = k. Therefore, d[R] = -k dt. Integrating: [R] = -kt + I. At t=0, [R] = [R]0, so I = [R]0. The integrated equation is [R] = -kt + [R]0.
(b) Numerical: Given W2 = 2g, M2 = 122 g mol-1, W1 = 25g, ΔTf = 1.62 K, Kf = 4.9 K kg mol-1.
ΔTf = i × Kf × [(W2 × 1000) / (M2 × W1)]
1.62 = i × 4.9 × [(2 × 1000) / (122 × 25)] ⇒ 1.62 = i × 3.213 ⇒ i = 0.504.
For dimerization: 2 C6H5COOH ⇔ (C6H5COOH)2.
i = 1 - α/2.
0.504 = 1 - α/2 ⇒ α/2 = 0.496 ⇒ α = 0.992. Percentage Association = 99.2%.
32. Solution:
(a) Identification:
Formula C3H6O. Positive 2,4-DNP indicates carbonyl. Negative Tollens' indicates a ketone. Positive Iodoform indicates a methyl ketone. The only 3-carbon ketone is Propanone (Acetone).
Hence, Compound (A) is Propanone (CH3-CO-CH3).
Compound (B) formed upon Clemmensen reduction is Propane (CH3-CH2-CH3).
(b) Chemical Equations:
(i) Iodoform reaction: CH3COCH3 + 3NaOI → CH3COONa + CHI3↓ (Iodoform) + 2NaOH.
(ii) Clemmensen reduction: CH3COCH3 + 4[H] →(Zn-Hg / Conc. HCl) CH3CH2CH3 (Propane) + H2O.
(iii) Reaction with HCN: CH3COCH3 + HCN → CH3-C(OH)(CN)-CH3 (Acetone cyanohydrin).
33. Solution:
(a) Preparation of KMnO4 from Pyrolusite (MnO2):
Step 1: Fusion of MnO2 with an alkali (KOH) and an oxidizing agent (O2 or KNO3) to form green potassium manganate.
2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O
Step 2: Disproportionation of manganate ions in an acidic medium (or electrolytic oxidation) to form purple potassium permanganate.
3 MnO42- + 4 H+ → 2 MnO4- + MnO2 + 2 H2O.
(b) Oxidizing action of KMnO4 (in acidic medium):
(i) Oxidation of Fe2+ to Fe3+:
MnO4- + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ + 4 H2O
(ii) Oxidation of Oxalate (C2O42-) to CO2:
2 MnO4- + 5 C2O42- + 16 H+ → 2 Mn2+ + 10 CO2 + 8 H2O.
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