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CBSE Class 12 Chemistry: FULL SYLLABUS MOCK TEST - 4
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): Ethanol molecules are associated through strong intermolecular hydrogen bonding.
Reason (R): The molecularity of an elementary reaction can be zero.
Reason (R): The lone pair of electrons on the nitrogen atom in aniline is delocalized over the benzene ring via resonance.
Reason (R): It is an octahedral complex and possesses a plane of symmetry.
(i) Finkelstein reaction
(ii) Swarts reaction
Sn(s) | Sn2+ (0.050 M) || H+ (0.020 M) | H2(g) (1 bar) | Pt(s)
[Given: E°(Sn2+/Sn) = -0.14 V]
(a) Haloarenes undergo electrophilic substitution reactions slower than benzene.
(b) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(i) Acetone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Benzene to Acetophenone
(i) Nature of the sugar present.
(ii) Type of pyrimidine bases present.
(iii) Main biological function.
Crystal Field Theory (CFT) provides a widely accepted model for understanding the electronic structure of coordination complexes. In an octahedral complex, the ligands approach the central metal ion along the x, y, and z axes. This asymmetric approach causes the five degenerate d-orbitals of the metal to split into two sets: the lower energy t2g set and the higher energy eg set. The energy difference between these two sets is denoted by Δo (crystal field splitting energy). If Δo is less than the pairing energy (P), the electrons prefer to occupy higher energy orbitals before pairing, resulting in high-spin complexes.
(a) Which specific d-orbitals belong to the higher energy eg set in an octahedral field? [1 Mark]
(b) What happens to the color of the complex [Ti(H2O)6]3+ if it is heated to remove all the water molecules? [1 Mark]
(c) On the basis of CFT, write the electronic configuration for a d4 ion if Δo > P. Will this complex be a high-spin or low-spin complex? [2 Marks]
Aldehydes and ketones undergo a variety of nucleophilic addition reactions. However, the presence of an α-hydrogen atom makes them susceptible to condensation reactions. For example, aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali to form β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols), which readily dehydrate to form α,β-unsaturated carbonyl compounds. On the other hand, aldehydes lacking an α-hydrogen undergo self-oxidation and reduction (disproportionation) when heated with concentrated alkali.
(a) Name the disproportionation reaction undergone by aldehydes lacking an α-hydrogen. [1 Mark]
(b) Identify whether benzaldehyde will undergo the aldol condensation reaction or not. Give a reason. [1 Mark]
(c) Write the chemical equation for the cross-aldol condensation product formed between ethanal and benzaldehyde in the presence of dilute NaOH followed by heating. [2 Marks]
(b) Why does the molar conductivity of a weak electrolyte increase steeply on dilution, while that of a strong electrolyte increases only slightly? [2 Marks]
(c) The conductivity of a 0.001028 mol L-1 acetic acid solution is 4.95 × 10-5 S cm-1. Calculate its dissociation constant (Ka) if Λ°m for acetic acid is 390.5 S cm2 mol-1. [2 Marks]
(a) Identify the compounds A, B, and C. [3 Marks]
(b) Write the chemical equations for the following reactions of compound A:
(i) Tollens' test.
(ii) Cannizzaro reaction. [2 Marks]
(i) Transition metals act as excellent catalysts.
(ii) The enthalpies of atomization of transition metals are very high.
(iii) Cu+ ion is not stable in an aqueous solution.
(iv) The lowest oxide of a transition metal is basic, whereas the highest oxide is acidic.
(v) Separation of lanthanoid elements is extremely difficult.
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SOLUTIONS & MARKING SCHEME
1. Answer: (b) Osmotic pressure
Reasoning: Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions.
2. Answer: (b) From anode to cathode
Reasoning: In a galvanic cell, oxidation occurs at the anode, releasing electrons. These electrons flow through the external wire to the cathode, where reduction occurs.
3. Answer: (b) First-order reaction
Reasoning: All natural and artificial radioactive decay processes follow first-order kinetics, where the rate depends only on the first power of the concentration of the reactant.
4. Answer: (c) Fe3+ (Z=26)
Reasoning: Fe3+ has a 3d5 configuration, possessing 5 unpaired electrons, which is the maximum possible, leading to the highest magnetic moment (5.92 BM).
5. Answer: (a) Tetraamminecopper(II) sulphate
Reasoning: Ligand is ammine. Four of them makes 'tetraammine'. Copper is in +2 oxidation state (x + 0 = +2). The counter ion is sulphate.
6. Answer: (b) CH3I > CH3Br > CH3Cl > CH3F
Reasoning: For SN2 reactions, the rate depends heavily on the leaving group ability. Iodine is the largest and weakest base among halides, making it the best leaving group.
7. Answer: (a) Conc. HCl and anhydrous ZnCl2
Reasoning: The Lucas reagent is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride, used to distinguish between 1°, 2°, and 3° alcohols based on turbidity appearance.
8. Answer: (b) Clemmensen reduction
Reasoning: The reduction of aldehydes and ketones directly to alkanes using Zinc amalgam (Zn-Hg) and concentrated HCl is called the Clemmensen reduction.
9. Answer: (a) 3° > 2° > 1° > NH3
Reasoning: In the gaseous phase, there is no solvation effect. The basicity strictly follows the +I (inductive) electron-donating effect of the alkyl groups. More alkyl groups = higher electron density on nitrogen = stronger base.
10. Answer: (b) Cobalt
Reasoning: Vitamin B12 is clinically known as Cyanocobalamin, and it contains the transition metal Cobalt in its complex structure.
11. Answer: (c) 3
Reasoning: Potassium sulphate (K2SO4) dissociates completely in water to yield 2 K+ ions and 1 SO42- ion. Total ions = 3. Hence, the van't Hoff factor (i) = 3.
12. Answer: (c) Two or more
Reasoning: A pseudo-first-order reaction is an apparent first-order reaction where the molecularity is 2 or more (bimolecular or termolecular), but it behaves as a first-order reaction because one of the reactants is in massive excess.
13. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: Dimethyl ether and ethanol are isomers. Ethanol has a much higher boiling point because its -OH group allows for strong intermolecular hydrogen bonding, which is completely absent in the ether.
14. Answer: (c) A is true but R is false.
Reasoning: The order of a reaction is an experimentally derived quantity and can be fractional. However, molecularity is the actual count of colliding molecules in an elementary step and can never be zero.
15. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: Aniline is weaker than ammonia because the lone pair on the nitrogen atom in aniline is involved in resonance with the benzene ring, making it less available to accept a proton.
16. Answer: (c) A is true but R is false.
Reasoning: The complex [Co(en)3]3+ is optically active (Assertion is true). The reason is false because optical activity requires the molecule to be asymmetric (lacking a plane of symmetry), not symmetrical.
17. Solution:
Mass of solvent (water) = 2.5 kg. Molality (m) = 0.25 mol kg-1.
Molality = Moles of solute / Mass of solvent (in kg)
0.25 = Moles of urea / 2.5
Moles of urea = 0.25 × 2.5 = 0.625 mol.
Mass of urea = Moles × Molar mass = 0.625 × 60 = 37.5 g.
18. Solution:
Transition elements form coloured compounds because they have incompletely filled d-orbitals. When ligands approach, the d-orbitals split into two energy levels. Electrons from the lower energy d-orbital absorb specific wavelengths of visible light and jump to the higher energy d-orbital (d-d transition). The transmitted/reflected light shows the complementary colour.
19. Solution:
(i) Finkelstein Reaction: Used to prepare alkyl iodides by halogen exchange in dry acetone.
R-Cl + NaI →(Dry Acetone) R-I + NaCl↓
(ii) Swarts Reaction: Used to prepare alkyl fluorides by heating an alkyl chloride/bromide with metallic fluorides (like AgF or Hg2F2).
R-Br + AgF → R-F + AgBr↓
20. Solution:
Equation: R-CONH2 + Br2 + 4NaOH → R-NH2 + Na2CO3 + 2NaBr + 2H2O.
Synthetic Utility: It is used to "step down" the carbon series, as the primary amine formed has one carbon atom less than the parent amide.
21. Solution:
In an aqueous solution, the carboxyl group (-COOH) of an amino acid loses a proton and the amino group (-NH2) accepts that proton to form a dipolar, neutral ion known as a Zwitterion (+H3N-CH(R)-COO-).
Because it contains both a positive and a negative site, it can react with both acids (via COO-) and bases (via NH3+), thus behaving as an amphoteric substance.
22. Solution:
Given: k = 1.15 × 10-3 s-1, [R]0 = 5 g, [R] = 3 g.
For a first-order reaction: t = (2.303 / k) log ([R]0 / [R])
t = (2.303 / 1.15 × 10-3) log (5 / 3)
t = 2002.6 × (log 5 - log 3) = 2002.6 × (0.699 - 0.477)
t = 2002.6 × 0.222 = 444.6 seconds.
23. Solution:
Cell reaction: Sn + 2H+ → Sn2+ + H2. (n=2).
E°cell = E°Cathode - E°Anode = 0.00 V - (-0.14 V) = +0.14 V.
Nernst Equation: Ecell = E°cell - (0.0591 / 2) log ([Sn2+] / [H+]2)
Ecell = 0.14 - 0.02955 × log (0.050 / (0.020)2)
Ecell = 0.14 - 0.02955 × log (0.050 / 0.0004)
Ecell = 0.14 - 0.02955 × log (125)
Ecell = 0.14 - (0.02955 × 2.097) = 0.14 - 0.062 = 0.078 V.
24. Solution:
Complex: [Co(NH3)6]3+. Oxidation state of Co = +3.
Co3+ configuration = [Ar] 3d6.
NH3 acts as a strong field ligand with Co3+, forcing the pairing of 3d electrons against Hund's rule.
The 6 electrons pair up completely in the three t2g orbitals, leaving two inner 3d orbitals vacant.
Hybridization: d2sp3 (Inner orbital complex).
Geometry: Octahedral.
Magnetic Behaviour: Diamagnetic (Zero unpaired electrons).
25. Solution:
(a) The highly electronegative halogen atom exerts a strong -I effect, withdrawing electron density from the benzene ring and deactivating it toward electrophilic attack.
(b) In chlorobenzene, the C-Cl bond is sp2-sp3, and resonance creates a partial double bond character, shrinking the bond distance. In cyclohexyl chloride, the C-Cl bond is sp3-sp3 (longer bond). Since Dipole Moment = Charge × Distance, the shorter bond and lower charge separation (due to sp2 carbon's high electronegativity) in chlorobenzene lead to a lower dipole moment.
26. Solution:
Step 1: Protonation of ethanol to form oxonium ion.
CH3CH2OH + H+ ⇔ CH3CH2OH2+
Step 2: Formation of carbocation by loss of water (Slow, rate-determining).
CH3CH2OH2+ ⇔ CH3CH2+ + H2O
Step 3: Deprotonation to form ethene.
CH3CH2+ ⇔ CH2=CH2 + H+
27. Solution:
(i) Acetone to Propene:
CH3COCH3 →(LiAlH4) CH3CH(OH)CH3 (Propan-2-ol).
CH3CH(OH)CH3 →(Conc. H2SO4, 443 K) CH3CH=CH2 (Propene).
(ii) Benzoic acid to Benzaldehyde:
C6H5COOH + SOCl2 → C6H5COCl (Benzoyl chloride).
C6H5COCl + H2 →(Pd/BaSO4) C6H5CHO (Benzaldehyde).
(iii) Benzene to Acetophenone:
C6H6 + CH3COCl →(Anhydrous AlCl3) C6H5COCH3 (Acetophenone) + HCl.
28. Solution:
(i) Sugar: DNA has 2-deoxyribose; RNA has ribose.
(ii) Pyrimidines: DNA has Cytosine and Thymine; RNA has Cytosine and Uracil.
(iii) Function: DNA transfers hereditary characteristics; RNA synthesizes proteins.
29. Solution (Case-Based Kinetics):
(a) In octahedral fields, the dx²-y² and dz² orbitals experience more repulsion from ligands as they point directly along the axes. They form the higher energy eg set.
(b) Heating removes the water ligands, leaving the complex without a crystal field. Since splitting (Δo) becomes zero, d-d transitions cannot occur. The complex becomes colourless.
(c) Since Δo > P, the crystal field splitting is strong. The pairing energy is low enough that the 4th electron prefers to pair up in the lower t2g orbital rather than jump to the eg set.
Configuration: t2g4 eg0. It forms a Low-Spin complex.
30. Solution (Case-Based Organics):
(a) The disproportionation reaction is called the Cannizzaro Reaction.
(b) Benzaldehyde will not undergo the aldol condensation because it lacks an alpha-hydrogen atom, which is mandatory to form the reactive enolate ion intermediate.
(c) Cross-Aldol Condensation:
CH3CHO (Ethanal) + C6H5CHO (Benzaldehyde) →(dil. NaOH) C6H5-CH(OH)-CH2-CHO
C6H5-CH(OH)-CH2-CHO →(Δ, -H2O) C6H5-CH=CH-CHO (Cinnamaldehyde).
31. Solution:
(a) Kohlrausch's law: The limiting molar conductivity of an electrolyte can be represented as the sum of the individual limiting molar conductivities of its constituent cations and anions.
(b) For a weak electrolyte, dilution increases the degree of dissociation (α) drastically, resulting in a steep increase in the total number of ions, hence a steep increase in Λm. For strong electrolytes, dissociation is already almost complete at normal concentrations, so Λm increases only slightly due to decreased inter-ionic attractions.
(c) Numerical:
Λm = κ × 1000 / M = (4.95 × 10-5 × 1000) / 0.001028 = 48.15 S cm2 mol-1.
Degree of dissociation (α) = Λm / Λ°m = 48.15 / 390.5 = 0.123.
Dissociation constant Ka = Cα2 / (1-α) = (0.001028 × (0.123)2) / (1 - 0.123)
Ka = 1.55 × 10-5 / 0.877 = 1.77 × 10-5.
32. Solution:
(a) Identification:
- Positive Tollens' and 2,4-DNP: It is an Aldehyde.
- Undergoes Cannizzaro: It lacks an α-hydrogen.
- Formula C7H8O. Matches Benzaldehyde (C6H5CHO). Hence, A is Benzaldehyde.
- Vigorous oxidation gives C7H6O2 (B). B is Benzoic Acid (C6H5COOH).
- C is a hydrocarbon that oxidizes to Benzoic Acid. C must be Toluene (C6H5CH3).
(b) Chemical Equations:
(i) Tollens' test:
C6H5CHO + 2[Ag(NH3)2]+ + 3OH- → C6H5COO- + 2Ag↓ (Silver mirror) + 4NH3 + 2H2O.
(ii) Cannizzaro reaction:
2 C6H5CHO + Conc. NaOH → C6H5CH2OH (Benzyl alcohol) + C6H5COONa (Sodium benzoate).
33. Solution:
(i) Catalysts: Transition metals possess multiple oxidation states (forming unstable intermediates) and offer a large surface area for reactants to adsorb, lowering the activation energy.
(ii) Atomization Enthalpy: High number of unpaired d-electrons results in exceptionally strong interatomic metallic bonding, requiring vast energy to break the lattice.
(iii) Cu+ instability: Cu2+(aq) is much more stable than Cu+(aq) because the extremely high, negative hydration enthalpy of the smaller Cu2+ ion more than compensates for the high second ionization enthalpy required to form it.
(iv) Oxides: In lower oxidation states, the M-O bond is ionic (metals donate electrons/act basic). In highest oxidation states, the high effective nuclear charge highly polarizes the oxygen electron cloud, creating strong covalent bonds that act acidic.
(v) Lanthanoid separation: Due to Lanthanoid Contraction, the ionic radii of consecutive lanthanoids decrease only marginally. This makes their chemical and physical properties practically identical, rendering standard chemical separation techniques ineffective.
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