ChemCA.in
CBSE Class 12 Chemistry: FULL SYLLABUS MOCK TEST - 1
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): Elevation in boiling point is directly proportional to the molarity of the solution.
Reason (R): Aniline reacts with AlCl3 (Lewis acid) to form a salt, which strongly deactivates the benzene ring.
Reason (R): D-Glucose exists primarily in a cyclic hemiacetal form.
Reason (R): Polar protic solvents stabilize the carbocation intermediate through solvation.
(Molar mass of Cu = 63.5 g mol-1, 1F = 96500 C)
(Atomic number of Co = 27)
(i) Aldol Condensation (using Ethanal)
(ii) Cannizzaro Reaction (using Benzaldehyde)
(iii) Rosenmund Reduction
(Kb for water = 0.512 K kg mol-1, Molar mass of NaCl = 58.5 g mol-1. Assume complete dissociation).
(b) What are essential amino acids? Give one example.
(i) Phenol to Salicylic acid
(ii) Propanone to Propane
(iii) Aniline to Chlorobenzene
The dependence of the rate of a chemical reaction on temperature is best expressed by the Arrhenius equation, k = A e-Ea/RT. The quantity Ea is the activation energy, which is the minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to the threshold value. A catalyst lowers the activation energy, providing an alternative path for the reaction, thereby increasing the rate of reaction. The plot of ln k vs 1/T gives a straight line.
(a) What does the intercept of the plot of ln k vs 1/T represent? [1 Mark]
(b) Does a catalyst change the enthalpy (ΔH) of the reaction? [1 Mark]
(c) The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the activation energy (Ea) of the reaction. (Given R = 8.314 J K-1 mol-1). [2 Marks]
Carbohydrates are optically active polyhydroxy aldehydes or ketones. Glucose is the most abundant monosaccharide. Its open-chain structure explains many of its reactions, such as the formation of an oxime with hydroxylamine and oxidation to gluconic acid with bromine water. However, it fails to explain why glucose does not give the 2,4-DNP test. This anomaly was resolved by proposing a cyclic hemiacetal structure. Two cyclic forms of glucose exist, known as alpha (α) and beta (β) forms.
(a) What product is formed when glucose is treated with HI upon prolonged heating? What does this prove? [1 Mark]
(b) What are the α and β forms of glucose commonly called? [1 Mark]
(c) Draw the Haworth structure of α-D-Glucopyranose. Why does glucose not give a positive 2,4-DNP test? [2 Marks]
(b) Calculate the EMF of the following cell at 298 K:
Zn(s) | Zn2+ (0.1 M) || Ag+ (0.01 M) | Ag(s)
Given: E°(Zn2+/Zn) = -0.76 V, E°(Ag+/Ag) = +0.80 V. [3 Marks]
(c) What happens to the specific conductivity of a solution upon dilution? [1 Mark]
(a) Identify the compounds A and B. [2 Marks]
(b) Write the chemical equations for the following reactions of A:
(i) Reaction with 2,4-DNP reagent.
(ii) Iodoform reaction.
(iii) Oxidation to B. [3 Marks]
(b) Explain why transition metals:
(i) exhibit variable oxidation states.
(ii) form complex compounds. [2 Marks]
ChemCA.in
SOLUTIONS & MARKING SCHEME
1. Answer: (d) 0.1 M Urea
Reasoning: Freezing point depression (ΔTf) depends on the number of particles (i). NaCl (i=2), BaCl2 (i=3), Al2(SO4)3 (i=5), Urea (i=1). Urea has the lowest depression, meaning it has the highest freezing point.
2. Answer: (d) remains constant for a given cell.
Reasoning: Cell constant (G*) depends only on the physical dimensions of the cell (distance between electrodes 'l' and area of cross-section 'A'). G* = l/A, hence it is independent of the electrolyte or its concentration.
3. Answer: (b) First order
Reasoning: The unit of the rate constant (k) for a first-order reaction is s-1 (time-1), which matches the given unit.
4. Answer: (d) Sc3+
Reasoning: Sc (Z=21) has configuration [Ar] 3d1 4s2. Sc3+ has [Ar] 3d0. Since it lacks unpaired d-electrons, no d-d transition can occur, making it colourless.
5. Answer: (b) 6 and +3
Reasoning: Oxalate (C2O42-) is a bidentate ligand. With 3 oxalates, the coordination number is 3 × 2 = 6. Oxidation state calculation: 3(+1) + x + 3(-2) = 0 ⇒ x = +3.
6. Answer: (a) RX + KOH (aq) → ROH + KX
Reasoning: Aqueous KOH provides OH- which acts as a nucleophile, substituting the halide ion (X-) to form an alcohol.
7. Answer: (c) 2,4,6-Tribromophenol
Reasoning: In aqueous medium (bromine water), phenol is highly activated. Bromination happens at all ortho and para positions simultaneously, yielding a white ppt of 2,4,6-tribromophenol.
8. Answer: (b) Acetophenone
Reasoning: A yellow precipitate of iodoform (CHI3) is given by compounds with the methyl ketone group (CH3-CO-). Acetophenone (C6H5COCH3) contains this group.
9. Answer: (b) (CH3)2NH > CH3NH2 > (CH3)3N > NH3
Reasoning: In aqueous phase, basicity of methylamines is governed by +I effect, steric hindrance, and solvation. This specific balance makes the secondary amine the most basic, followed by primary, tertiary, and ammonia.
10. Answer: (c) n-Hexane
Reasoning: Prolonged heating of glucose with HI causes complete reduction of all functional groups, proving that all six carbons are linked in a straight chain as n-hexane.
11. Answer: (b) Glycosidic linkage
Reasoning: The oxide linkage formed between two monosaccharides by the loss of a water molecule is known as a glycosidic linkage.
12. Answer: (c) Manganese
Reasoning: Manganese (Z=25) has the configuration 3d5 4s2. It can lose all 7 valence electrons to exhibit a maximum oxidation state of +7 (e.g., in KMnO4).
13. Answer: (c) A is true but R is false.
Reasoning: Elevation in boiling point is a colligative property. However, it is directly proportional to molality (m), not molarity. Hence the reason is false.
14. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: Aniline is a Lewis base (lone pair on N) and AlCl3 is a Lewis acid. They react to form an insoluble salt. The positive charge on the nitrogen atom strongly deactivates the ring, halting the Friedel-Crafts reaction.
15. Answer: (d) A is false but R is true.
Reasoning: Assertion is strictly false. D-Glucose does not give the 2,4-DNP test. The Reason is true, as it exists primarily in a cyclic hemiacetal form, locking up the free aldehyde group required for the test.
16. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: SN1 mechanism involves the formation of a carbocation intermediate. Polar protic solvents (like water or alcohols) stabilize this carbocation through highly effective solvation via hydrogen bonding, driving the reaction forward.
17. Solution:
Henry's Law: The partial pressure of a gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution. p = KH × x.
Application: To avoid 'bends', scuba divers use tanks filled with air diluted with Helium. Also, to increase solubility of CO2 in soft drinks, the bottle is sealed under high pressure.
18. Solution:
For a zero-order reaction, k = ([A]0 - [A]) / t.
Given: k = 0.0030 mol L-1 s-1, [A]0 = 0.10 M, [A] = 0.075 M.
t = (0.10 - 0.075) / 0.0030 = 0.025 / 0.0030 = 8.33 seconds.
19. Solution:
Lanthanoid Contraction: The steady and regular decrease in atomic and ionic radii of lanthanoid elements with increasing atomic number due to the poor shielding effect of 4f electrons.
Consequence: The atomic radii of the second and third transition series elements (e.g., Zr and Hf) become nearly identical, making their separation extremely difficult.
20. Solution:
Step 1: Protonation of alkene to form a carbocation by electrophilic attack of H3O+.
CH2=CH2 + H3O+ ⇔ CH3-CH2+ + H2O
Step 2: Nucleophilic attack of water on the carbocation.
CH3-CH2+ + H2O ⇔ CH3-CH2-O+H2
Step 3: Deprotonation to form ethanol.
CH3-CH2-O+H2 + H2O ⇔ CH3-CH2-OH + H3O+
21. Solution:
Hinsberg's Test (Reaction with Benzenesulphonyl chloride, C6H5SO2Cl).
- 1° Amine: Reacts to form N-alkylbenzenesulphonamide, which is soluble in aqueous NaOH (due to the presence of an acidic hydrogen on N).
C6H5SO2Cl + RNH2 → C6H5SO2NHR + HCl
- 2° Amine: Reacts to form N,N-dialkylbenzenesulphonamide, which is insoluble in aqueous NaOH (due to absence of acidic hydrogen).
22. Solution:
Cathode reaction: Cu2+ + 2e- → Cu (n = 2).
Given: I = 1.5 A, t = 10 min = 600 s.
Using Faraday's First Law: m = ZIt = (M / nF) × I × t
m = (63.5 / (2 × 96500)) × 1.5 × 600
m = (63.5 × 900) / 193000 = 57150 / 193000 = 0.296 g of Cu.
23. Solution:
Complex: [CoF6]3-. Oxidation state of Co = +3.
Co3+ configuration = [Ar] 3d6.
F- is a weak field ligand, so pairing of electrons does not occur against Hund's rule.
The 3d electrons remain: ↑↓ | ↑ | ↑ | ↑ | ↑ (4 unpaired electrons).
It utilizes outer 4d orbitals for bonding (one 4s, three 4p, two 4d), resulting in sp3d2 hybridization.
Geometry: Octahedral.
Magnetic Behaviour: Paramagnetic (High Spin complex due to 4 unpaired e-).
24. Solution:
1. Resonance effect: The lone pair on halogen is in conjugation with the benzene ring, imparting partial double bond character to the C-X bond, making it stronger and harder to break.
2. Hybridization: Carbon in haloarenes is sp2 hybridized (more s-character, more electronegative), holding the electron pair more tightly than the sp3 carbon in haloalkanes.
3. Instability of Phenyl cation: In SN1, the phenyl cation formed is highly unstable and cannot be stabilized by resonance. Also, electron-rich nucleophiles face repulsion from the electron-rich arene ring.
25. Solution:
(i) Aldol Condensation:
2 CH3CHO →(dil. NaOH) CH3CH(OH)CH2CHO →(Δ, -H2O) CH3CH=CHCHO (But-2-enal)
(ii) Cannizzaro Reaction:
2 C6H5CHO + Conc. NaOH → C6H5CH2OH (Benzyl alcohol) + C6H5COONa (Sod. benzoate)
(iii) Rosenmund Reduction:
R-COCl + H2 →(Pd/BaSO4) R-CHO + HCl
26. Solution:
Molar mass of NaCl (M2) = 58.5 g mol-1. For complete dissociation of NaCl (Na+ + Cl-), i = 2.
W2 = 15 g, W1 = 250 g = 0.25 kg, Kb = 0.512 K kg mol-1.
ΔTb = i × Kb × m = 2 × 0.512 × (15 / (58.5 × 0.25))
ΔTb = 1.024 × (15 / 14.625) = 1.024 × 1.025 = 1.05 K.
Boiling point of solution = Tb° + ΔTb = 373.15 + 1.05 = 374.20 K.
27. Solution:
(a) Differences between DNA and RNA:
1. DNA contains 2-deoxyribose sugar; RNA contains ribose sugar.
2. DNA contains Thymine as a pyrimidine base; RNA contains Uracil instead.
(b) Essential amino acids: Amino acids which cannot be synthesized in the human body and must be obtained through diet. Example: Valine or Leucine.
28. Solution:
(i) Phenol to Salicylic acid (Kolbe's reaction):
C6H5OH →(NaOH) C6H5ONa →(CO2, H+) o-Hydroxybenzoic acid (Salicylic acid).
(ii) Propanone to Propane (Clemmensen/Wolff-Kishner):
CH3COCH3 →(Zn-Hg / Conc. HCl) CH3CH2CH3 + H2O.
(iii) Aniline to Chlorobenzene (Sandmeyer's reaction):
C6H5NH2 + NaNO2 + HCl (0-5°C) → C6H5N2+Cl-
C6H5N2+Cl- + Cu2Cl2/HCl → C6H5Cl + N2.
29. Solution (Case-Based Kinetics):
(a) The Arrhenius equation is ln k = -Ea/RT + ln A. The intercept of the plot of ln k vs 1/T represents ln A, where A is the Arrhenius factor (or frequency factor).
(b) No, a catalyst does not alter the enthalpy change (ΔH) of the reaction. It only alters the activation energy.
(c) Numerical: Given T1 = 293 K, T2 = 313 K, k2/k1 = 4.
log(k2/k1) = (Ea / 2.303R) × [(T2 - T1) / (T1T2)]
log(4) = (Ea / 19.147) × [20 / (293 × 313)]
0.602 = (Ea / 19.147) × (20 / 91709)
Ea = (0.602 × 19.147 × 91709) / 20 = 52863 J mol-1 = 52.86 kJ mol-1.
30. Solution (Case-Based Biomolecules):
(a) Prolonged heating with HI produces n-hexane. This proves that all six carbon atoms of glucose are linked in a straight chain.
(b) The α and β cyclic forms of glucose are cyclic stereoisomers that differ in configuration only at the hemiacetal carbon (C-1). They are commonly called Anomers.
(c) (Draw Haworth Pyranose structure: 6-membered ring with Oxygen. For α-D-Glucose, the -OH at C-1 points DOWN).
Glucose does not give the 2,4-DNP test because the free aldehyde (-CHO) group is consumed during the formation of the intramolecular cyclic hemiacetal, leaving no free aldehyde available to react with the reagent.
31. Solution:
(a) Kohlrausch's law: The limiting molar conductivity of an electrolyte is the sum of the individual ionic conductivities of its constituent cations and anions.
(b) Nernst Equation: Cell reaction: Zn + 2Ag+ → Zn2+ + 2Ag (n = 2).
E°cell = E°Cathode - E°Anode = 0.80 - (-0.76) = 1.56 V.
Ecell = E°cell - (0.0591 / 2) log ([Zn2+] / [Ag+]2)
Ecell = 1.56 - 0.02955 × log (0.1 / (0.01)2) = 1.56 - 0.02955 × log (1000)
Ecell = 1.56 - (0.02955 × 3) = 1.56 - 0.08865 = 1.47 V.
(c) Specific conductivity (κ) decreases upon dilution because the number of current-carrying ions per unit volume (1 cm3) of the solution decreases.
32. Solution:
(a) Identification:
- Positive 2,4-DNP test means it is a carbonyl compound.
- Negative Tollens' test means it is a Ketone.
- Positive Iodoform test implies it contains a methyl ketone group (CH3-CO-).
- Oxidation yields C7H6O2 (Benzoic acid), indicating a phenyl ring attached to the carbonyl.
Compound A (C8H8O): Acetophenone (C6H5-CO-CH3).
Compound B: Benzoic Acid (C6H5COOH).
(b) Chemical Equations:
(i) With 2,4-DNP: C6H5COCH3 + H2N-NH-C6H3(NO2)2 →(H+) C6H5C(CH3)=N-NH-C6H3(NO2)2 + H2O.
(ii) Iodoform Reaction: C6H5COCH3 + 3NaOI → C6H5COONa + CHI3↓ (Iodoform) + 2NaOH.
(iii) Oxidation: C6H5COCH3 →(CrO3 / H2SO4) C6H5COOH.
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) Transition metals properties:
(i) Variable oxidation states: Because the energy difference between the (n-1)d and ns orbitals is very small, electrons from both energy levels can participate in bond formation.
(ii) Formation of complex compounds: Due to their small ionic sizes, high effective nuclear charge, and the availability of vacant d-orbitals to accept lone pairs of electrons from ligands.
No comments:
Post a Comment