CBSE Class 12 Chemistry Sample Paper 1: marking scheme and solutions
Every answer with its marking scheme, the way an examiner awards the marks. Chapters in this paper: 1 Solutions; 2 Electrochemistry; 3 Chemical Kinetics; 4 The d- and f-Block Elements; 5 Coordination Compounds; 6 Haloalkanes and Haloarenes; 7 Alcohols, Phenols and Ethers; 8 Aldehydes, Ketones and Carboxylic Acids; 9 Amines; 10 Biomolecules.
Section A (16 × 1 mark)
Answer: (d) Order is an experimental quantity and may be zero or fractional; molecularity counts colliding species in an elementary step, so it is a whole number and has no meaning for a complex reaction.
Answer: (b) The lower the standard electrode potential, the stronger the metal as a reducing agent; sodium has the lowest E° of the four.
Answer: (c) Primary amines have two N–H hydrogens for hydrogen bonding and associate most; secondary amines have one; tertiary amines, (a) and (d), have none. So the boiling point order of isomers is 1° > 2° > 3°.
Answer: (b) The molecule is symmetrical: all twelve CH₃ hydrogens are equivalent and the two CH hydrogens are equivalent. Two types of hydrogen give two monochlorides: ClCH₂CH(CH₃)CH(CH₃)₂ and (CH₃)₂CClCH(CH₃)₂.
Answer: (c) Moles of KOH = 2.8/56 = 0.050 mol; volume = 0.250 L; M = 0.050/0.250 = 0.20 M. 0.050 M forgets the volume; 0.0112 divides grams by millilitres.
Answer: (d) Down the group the halogen grows and the C–X bond lengthens, from 139 pm (C–F) to 214 pm (C–I); the bond enthalpy falls accordingly, 452, 351, 293 and 234 kJ mol⁻¹. CH₃–I has the longest bond and the lowest enthalpy.
Answer: (d) Ozonolysis cleaves the C=C and puts C=O on each carbon: (CH₃)₂C= gives (CH₃)₂C=O, propanone; =CH–CH₂CH₃ gives OHC–CH₂CH₃, propanal. Zinc dust prevents oxidation of the aldehyde to the acid.
Answer: (b) Each en is didentate and gives two donor N atoms (2 × 2 = 4); NO₂⁻ and Cl⁻ give one each: 4 + 1 + 1 = 6. Counting ligands instead of donor atoms gives 4.
Answer: (d) Ammonolysis can run on to 2°, 3° amines and the quaternary salt, because each amine formed is itself a nucleophile; a large excess of ammonia makes it far more likely that the halide meets NH₃, so the primary amine is the major product.
Answer: (b) Like dissolves like: non-polar anthracene dissolves in non-polar benzene. NaCl, KCl and sugar are polar or ionic and dissolve in water, not benzene.
Answer: (c) Aquated Mn²⁺ is pink; Ni²⁺ is green, Cr³⁺ violet and Cu²⁺ blue.
Answer: (b) In CH₃CH₂C(CH₃)(OH)CH₂CH₃ the –OH is on a carbon bonded to three other carbons: tertiary. (a) and (d) have –OH on a –CH₂– carbon (primary); in (c) the –OH carbon carries one hydrogen and two carbons (secondary).
Answer: (a) Attack of OH⁻ puts the negative charge on the carbons ortho and para to the site of attack. A nitro group placed there withdraws and stabilises it; at meta it sits on a carbon that never carries the charge, so it has no activating effect. Both are true and R explains A.
Answer: (a) Both are true. The lone-pair donation is the σ part of the M–C bond and the metal-to-π* donation the π part, so R explains A; the two reinforce each other (synergic bonding).
Answer: (b) Both statements are true, but the shape of the chromate ion has nothing to do with K₂Cr₂O₇'s use as a primary standard, which rests on its being a strong oxidant obtained as a pure, weighable crystalline salt.
Answer: (a) Both true, and R explains A: the strongly activating –NH₂ makes all three ortho and para positions react with bromine without a catalyst.
Section B (5 × 2 marks)
ppm = (mass of O₂ / mass of solution) × 10⁶ = (0.016 g / 2000 g) × 10⁶ = 8.0 ppm. The solute is present in trace quantity; as a percentage it would be an awkward 0.0008%.
- 8.0 ppm with working (1)
- ppm is used for a solute in trace quantity (1)
Carnallite is a double salt: in water it dissociates completely into K⁺, Mg²⁺ and Cl⁻. K₄[Fe(CN)₆] is a coordination compound: it gives K⁺ and the complex ion [Fe(CN)₆]⁴⁻, which stays intact, so no free Fe²⁺ or CN⁻ is present.
- carnallite a double salt, giving K⁺, Mg²⁺ and Cl⁻ (1)
- K₄[Fe(CN)₆] a coordination compound, giving only K⁺ and [Fe(CN)₆]⁴⁻ (no free Fe²⁺ or CN⁻) (1)
Step 1: C₆H₅COCH₃ + NH₂NH₂ → C₆H₅C(CH₃)=N–NH₂ + H₂O (the hydrazone). Step 2: C₆H₅C(CH₃)=N–NH₂ → C₆H₅CH₂CH₃ + N₂ on heating with KOH in ethylene glycol.
- hydrazone formation with NH₂NH₂ (1)
- heating with KOH in ethylene glycol gives C₆H₅CH₂CH₃ and N₂ (1)
CH₃COCH₂CH₃ + 4[H] → CH₃CH₂CH₂CH₃ + H₂O, with zinc amalgam and concentrated hydrochloric acid; the C=O becomes CH₂.
- reagents Zn–Hg and conc. HCl (1)
- correct equation: butanone to butane and water (1)
CO + 2H₂ → CH₃OH, over ZnO–Cr₂O₃ at 200–300 atm and 573–673 K (catalytic hydrogenation of carbon monoxide). Methanol is used chiefly to make formaldehyde; it is also a solvent for paints and varnishes.
- equation with ZnO–Cr₂O₃ and high pressure and temperature (1)
- chief use: formaldehyde (1)
t½ = 0.693/k = 0.693/(6.93 × 10⁻³) = 100 min. 87.5% used means 12.5% = ⅛ remains: three half-lives. t = 3 × 100 = 300 min.
- t½ = 100 min (1)
- ⅛ remaining is three half-lives, 300 min (1)
Section C (7 × 3 marks)
(i) C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O (H⁺). (ii) C₂H₅OH + CH₃COCl → CH₃COOC₂H₅ + HCl (pyridine). Reaction (i) is reversible, so removing water shifts the equilibrium towards the ester. In (ii) pyridine, a base, neutralises the HCl formed and so shifts the equilibrium to the right.
- both equations (1)
- (i) reversible; removing water drives it forward (1)
- (ii) pyridine neutralises HCl (1)
Zn(s) + Ni²⁺(aq) → Zn²⁺(aq) + Ni(s), n = 2. E°cell = −0.25 − (−0.76) = 0.51 V. Ecell = E°cell − (0.059/2) log([Zn²⁺]/[Ni²⁺]), so 0.451 = 0.51 − 0.0295 log(0.10/[Ni²⁺]); log(0.10/[Ni²⁺]) = 0.059/0.0295 = 2, so 0.10/[Ni²⁺] = 100 and [Ni²⁺] = 1.0 × 10⁻³ M.
- cell reaction with n = 2 and E°cell = 0.51 V (1)
- Nernst equation with Q = [Zn²⁺]/[Ni²⁺] and log Q = 2 (1)
- [Ni²⁺] = 1.0 × 10⁻³ M (1)
Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s), n = 2. E°cell = 0.34 − (−0.44) = 0.78 V. Ecell = E°cell − (0.059/2) log([Fe²⁺]/[Cu²⁺]), so 0.839 = 0.78 − 0.0295 log(0.0010/[Cu²⁺]); log(0.0010/[Cu²⁺]) = −0.059/0.0295 = −2, so 0.0010/[Cu²⁺] = 0.01 and [Cu²⁺] = 0.10 M.
- cell reaction with n = 2 and E°cell = 0.78 V (1)
- Nernst equation with Q = [Fe²⁺]/[Cu²⁺] and log Q = −2 (1)
- [Cu²⁺] = 0.10 M (1)
Shake the ether solution with dilute HCl: CH₃–C₆H₄–NH₂ + HCl → CH₃–C₆H₄–NH₃⁺Cl⁻. The salt is soluble in water but insoluble in ether, so it passes into the aqueous layer; methylbenzene, not basic, stays in the ether layer. Separate the layers. Add NaOH to the aqueous layer: CH₃–C₆H₄–NH₃⁺Cl⁻ + NaOH → CH₃–C₆H₄–NH₂ + H₂O + NaCl; the free amine separates and can be extracted into fresh ether.
- salt formation with HCl, equation (1)
- salt in water, insoluble in ether; methylbenzene stays in ether (1)
- NaOH frees the amine, equation (1)
(i) n-Pentane: the five carbons are in a straight chain. (ii) An oxime: Y has a carbonyl group. (iii) Oxidation by a mild oxidising agent without loss of carbon: the carbonyl group is an aldehyde group. (iv) A tetraacetate: four –OH groups; since it is stable, they are on different carbon atoms. Y is CHO–(CHOH)₃–CH₂OH, an aldopentose.
- (i) straight chain and (ii) carbonyl group (1)
- (iii) aldehyde group and (iv) four –OH on different carbons (1)
- structure CHO–(CHOH)₃–CH₂OH, aldopentose (1)
(a) Only one Cl⁻ is ionisable, so two Cl⁻ join four NH₃ inside the sphere: [Ir(NH₃)₄Cl₂]Cl. (b) Primary valence 3, the ionisable valence, read today as the oxidation state +3; secondary valence 6, the non-ionisable valence, the coordination number. (c) [Ir(NH₃)₄Cl₂]⁺ is octahedral [MX₂L₄]: two isomers, cis (Cl at 90°) and trans (Cl at 180°).
- [Ir(NH₃)₄Cl₂]Cl (1)
- primary 3 = oxidation state; secondary 6 = coordination number (1)
- two: cis and trans (1)
(i) Br₂ with anhydrous FeBr₃ (or Fe): 1,4-dibromobenzene major, 1,2-dibromobenzene minor. (ii) CH₃COCl with anhydrous AlCl₃: 4-bromoacetophenone, 1-(4-bromophenyl)ethanone, major; the ortho isomer minor. (iii) Bromine withdraws electrons by its –I effect, which outweighs its resonance release and destabilises the carbocation intermediate, so the ring is deactivated; resonance still makes ortho and para the preferred positions.
- 1,4-dibromobenzene, Br₂/FeBr₃ (1)
- 4-bromoacetophenone, CH₃COCl/AlCl₃ (1)
- –I deactivates; resonance directs o,p (1)
Kf = R × M₁ × Tf²/(1000 × Δfus H) = (8.314 × 18 × 273.15²)/(1000 × 6010) = 1.86 K kg mol⁻¹, in agreement with the tabulated 1.86 K kg mol⁻¹.
- correct expression for Kf (1)
- Δfus H in J mol⁻¹ and substitution (1)
- Kf = 1.86 K kg mol⁻¹, agreeing with the table (1)
Section D (2 × 4 marks)
Answer: (a) Ea(forward) = peak − reactants = 130 − 50 = 80 kJ mol⁻¹.
Answer: (a) ΔH = products − reactants = 20 − 50 = −30 kJ mol⁻¹; exothermic.
Uncatalysed: 130 − 20 = 110 kJ mol⁻¹; catalysed: 90 − 20 = 70 kJ mol⁻¹. The forward barrier falls from 80 to 40 kJ mol⁻¹ and the backward from 110 to 70: both are lowered by the same 40 kJ mol⁻¹, so the catalyst speeds both directions alike.
- 110 and 70 kJ mol⁻¹ (1)
- both barriers lowered by the same 40 kJ mol⁻¹ (1)
Both paths start at the same reactant energy (50) and end at the same product energy (20), so ΔH = −30 kJ mol⁻¹ either way. The catalyst lowers the forward and backward barriers by the same 40 kJ mol⁻¹, speeding both reactions equally, so equilibrium is reached sooner but its position, and K, are unchanged.
- same end points, so ΔH unchanged (1)
- both barriers lowered equally, so K unchanged (1)
| Student | Main symptom |
|---|---|
| Ravi | night blindness |
| Meena | bleeding gums |
| Arjun | convulsions |
| Fatima | soft bones that bend under the body's weight (rickets) |
Answer: (a) Bleeding gums are a sign of scurvy, the deficiency disease of vitamin C (ascorbic acid).
Answer: (b) Ravi lacks vitamin A and Fatima vitamin D; both are fat soluble and stored in the liver and adipose tissue. Meena's vitamin C and Arjun's B₆ are water soluble.
Convulsions are caused by deficiency of vitamin B₆ (pyridoxine). Sources (any two): yeast, milk, egg yolk, cereals, grams. It belongs to the B group and is water soluble.
- vitamin B₆ (pyridoxine) (1)
- two sources and water soluble (1)
Fatima lacks vitamin D. Exposure to sunlight is a source that is not a food (fish and egg yolk are food sources). In adults its deficiency causes osteomalacia.
- vitamin D; exposure to sunlight (1)
- osteomalacia in adults (1)
Section E (3 × 5 marks)
(a)(i) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O: manganese(II). (ii) MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻: manganese dioxide. (b) 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. MnO₄⁻ used = 15.0 × 0.020 = 0.30 mmol; Fe²⁺ = 5 × 0.30 = 1.50 mmol; molarity = 1.50/25.0 = 0.060 M. (c) Permanganate oxidises hydrochloric acid to chlorine, so some titrant would be used up by the acid.
- acid half-reaction, Mn²⁺ (1)
- neutral half-reaction, MnO₂ (1)
- balanced Fe²⁺ equation (1)
- 0.060 M from the 5 : 1 ratio (1)
- HCl oxidised to Cl₂ (1)
(a)(i) 2MnO₄⁻ + 5NO₂⁻ + 6H⁺ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O. (ii) 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O. (b) NO₂⁻ = 25.0 × 0.040 = 1.00 mmol; MnO₄⁻ = (2/5) × 1.00 = 0.40 mmol; volume = 0.40/0.020 = 20.0 mL. (c) In acid permanganate is reduced to Mn²⁺ (purple colour discharged). HCl would itself be oxidised to Cl₂, using up titrant.
- nitrite equation (1)
- sulphite equation (1)
- 2 : 5 ratio, 0.40 mmol (1)
- 20.0 mL (1)
- Mn²⁺; HCl oxidised to Cl₂ (1)
(a) First law: the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed. Second law: the amounts of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights (atomic mass ÷ electrons needed to reduce the cation). (b) Q = 5.0 × 3860 = 19 300 C = 0.200 F. Mg²⁺ + 2e⁻ → Mg: 0.100 mol Mg = 2.4 g. 2Cl⁻ → Cl₂ + 2e⁻: 0.100 mol Cl₂ = 2.27 L. (c) Magnesium is a very powerful reducing metal (E° −2.36 V); no suitable chemical reducing agent exists, so it is won by electrochemical reduction of the fused chloride.
- first law (1)
- second law (1)
- Q = 0.200 F and 2.4 g Mg (1)
- 0.100 mol Cl₂ = 2.27 L (1)
- (c) no suitable chemical reducing agent; electrolysis of fused chloride (1)
(a) First law: the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed. Second law: the amounts of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights. (b) Q = 9.65 × 3000 = 28 950 C = 0.300 F. Al³⁺ + 3e⁻ → Al: 0.300/3 = 0.100 mol Al = 2.7 g. (c) E°(Al³⁺|Al) = −1.66 V is far below that of H⁺/H₂ (0.00 V), so water is reduced to H₂ at the cathode before Al³⁺.
- first law (1)
- second law (1)
- Q = 0.300 F (1)
- 0.100 mol Al = 2.7 g (1)
- (c) water reduced in preference to Al³⁺ (1)
(a) 1. H⁺ protonates the carbonyl oxygen of CH₃CH₂COOH, making the carbon more electrophilic. 2. CH₃OH attacks the carbonyl carbon; a tetrahedral intermediate forms. 3. A proton moves from the –O⁺H–CH₃ to an –OH, making –OH₂⁺, a good leaving group. 4. Water leaves, giving the protonated ester. 5. Loss of H⁺ gives CH₃CH₂COOCH₃, methyl propanoate. (b) CH₃CH₂COOH + SOCl₂ → CH₃CH₂COCl + SO₂ + HCl; SO₂ and HCl are gases that escape, so the acid chloride is obtained pure.
- protonation of the carbonyl oxygen (1)
- attack of methanol, tetrahedral intermediate (1)
- proton transfer, loss of water and of H⁺ to methyl propanoate (1)
- SOCl₂ equation (1)
- gaseous by-products escape (1)
(a) CH₃CH₂CH₂COOH + PCl₅ → CH₃CH₂CH₂COCl + POCl₃ + HCl; 3CH₃CH₂CH₂COOH + PCl₃ → 3CH₃CH₂CH₂COCl + H₃PO₃; CH₃CH₂CH₂COOH + SOCl₂ → CH₃CH₂CH₂COCl + SO₂ + HCl. SOCl₂ is preferred: SO₂ and HCl are gases and escape. (b) CH₃CH₂CH₂COOH + C₂H₅OH ⇌ CH₃CH₂CH₂COOC₂H₅ + H₂O; H⁺ protonates the carbonyl oxygen, making the carbonyl carbon more electrophilic towards the alcohol.
- PCl₅ equation (1)
- PCl₃ equation (1)
- SOCl₂ equation and why preferred (1)
- esterification equation (1)
- protonation of carbonyl oxygen activates the carbon (1)