Chemistry: Model Question Sets and Predicted Questions

This chapter does three things. It reads the real papers — the board papers of 2079, 2081, 2082 and 2083 and the model papers of 2078 and 2079, with two further official model sets — and tabulates what the examiner asks, unit by unit. It then sets two complete model papers to the exact grid of the Chemistry paper (3 hours, full marks 75, Group A 11 × 1, Group B 8 × 5, Group C 3 × 8), each fully solved. Finally it lists forty-eight predicted questions, ranked, every one with the reason it is predicted and a short model answer.

7.1 Pattern analysis of the Grade 12 Chemistry papers

7.1.1 The grid, and what has been fixed for years

Part	Questions	Marks	What the examiner does
Group A	11 multiple-choice questions	11 × 1 = 11	one mark each; in 2082 the booklet was given out later in the examination
Group B	Q12 to Q19	8 × 5 = 40	two of the eight carry an internal Or; the five marks are split 1+4, 2+3, 1+1+3 or 1+1+1+2
Group C	Q20 to Q22	3 × 8 = 24	Q20 and Q21 carry an Or; Q22 is four two-mark parts from applied chemistry (six smaller parts in 2079)

All questions are compulsory; the choice is only inside a question, between its two alternatives. The CDC publishes no per-unit grid for Grade 12, so every "weight" below is our reading of the papers. The one official figure is the cognitive-level split printed in the 2079 model paper: knowledge 16 %, understanding 24 %, application 28 % and higher ability 32 % of the 75 marks — so about three-fifths of the paper asks you to use chemistry, not to recite it.

7.1.2 Where the questions sit

The papers of 2081, 2082 and 2083 follow the same order, and the pattern is worth knowing before you open the paper.

Slot	What was set in 2081 / 2082 / 2083
Group A, Q1 to Q4	four general and physical questions: equivalent weight or normality, Ka or pH, order of reaction or half-life, the ranking of E° values or Gibbs–Helmholtz
Group A, Q5 and Q6	two inorganic questions: catalytic metals, coloured ions, the deep blue complex of copper, lithopone, purest form of iron
Group A, Q7 to Q11	five organic questions: basicity order, Tollens' and iodoform tests, the product of a named reaction, the Grignard route
Group B, Q12 and Q13	physical: a cell from two E° values, a heat of formation, a pH calculation, a titration, a rate calculation
Group B, Q14 to Q19	six questions split two inorganic to four organic: a reaction chain, a list of named reactions, an amine, acid or alcohol question, and a pair on copper, mercury, zinc or rusting. The inorganic pair stood at Q14–Q15 in 2081 and at Q18–Q19 in 2082 and 2083
Group C, Q20	one physical question with an Or: volumetric, kinetics or ionic equilibrium
Group C, Q21	one organic question with an Or: a long identification chain
Group C, Q22	applied: cement, polymers, paper, nuclear chemistry, drugs — four two-mark parts

Reading a typical paper, about 33 of the 75 marks are organic, 22 general and physical, 12 inorganic and 8 applied. Each share is close to that area's share of the teaching hours (organic 55 of 128 hours, physical 40, inorganic 20, applied 13), which is the best argument for revising in proportion to hours.

7.1.3 Unit by unit: marks carried, years seen, expected weight

Seen counts the eight papers (2078 model, 2079 model, 2079, 2081, 2082, 2083, model sets A and B) in which the unit supplied at least one question of five or eight marks in Group B or Group C; MCQs are not counted. Expected weight is our estimate from hours and from this count.

#	Unit	Hours	Seen (of 8)	Typical marks when asked	Expected weight
1	Volumetric Analysis	8	8	4–8 (standards, normality equation, titration data)	very high
2	Ionic Equilibrium	10	8	5–8 (pH, Ksp, Ostwald's law)	very high
3	Chemical Kinetics	7	8	5–8 (table of rates, half-life)	very high
4	Thermodynamics	8	8	4–5 (Hess's law, ΔG)	very high
5	Electrochemistry	7	8	4–5 (cell notation and emf)	very high
6	Transition Metals	5	3	3–5 (CFT, d–d transition)	medium
7	Studies of Heavy Metals	15	8	5–10, often two questions	very high
8	Haloalkanes	8	7	4–8 (chloroform, SN, chains)	high
9	Haloarenes	3	4	1–5 (DDT, reactivity)	low
10	Alcohols	7	7	5 (Victor Meyer, ethanol)	very high
11	Phenols	4	3	1–5 (named reactions, acidity)	low
12	Ethers	2	5	1–5 (Williamson, peroxides)	low
13	Aldehydes and Ketones	10	8	5–8 (named-reaction lists, iodoform chain)	very high
14	Carboxylic Acid and its Derivatives	9	6	5–8 (chains through amide, Grignard)	high
15	Nitro Compounds	3	4	3–5 (reduction in media)	medium
16	Amines	7	8	5–8 (Hoffmann, nitrous acid, chains)	very high
17	Organometallic Compounds	2	4	within other chains	medium
18	Chemistry in the service of mankind	4	8	2–4 (polymers, drugs)	high
19	Cement	4	8	2–4 (flow sheet, components, OPC/PPC)	very high
20	Paper and Pulp	3	6	2	medium
21	Nuclear Chemistry	2	5	2	medium

Two things stand out. Nothing in the syllabus is ignored: twelve of the twenty-one units were set in at least seven of the eight papers. And some teaching-hour heavyweights have an uneven record: transition metals and the crystal-field question appear in the model papers but in none of the 2079–2083 board papers at five-mark level, and silver, iron extraction in the blast furnace and calomel have not appeared at all in Group B or C of the eight papers. A gap is not a forecast, but a student with four weeks to spare should cover these after the very-high units, not before.

7.1.4 The numericals and derivations that recur

Type	Forms seen	Papers
Normality and titration	N₁V₁ = N₂V₂ from oxalic acid against KMnO₄ or NaOH; mixtures of acids; Na₂CO₃ against H₂SO₄; molarity and molality from density	2078 model, 2079, 2082, 2083, model sets
Acid–base and solubility	pH of HCl, NaOH, weak acid from Ka; Kw; precipitation test from Ksp; dilution of a base	2078 model, 2079, 2081, 2082, 2083, model set B
Kinetics	order and k from three or four experiments; first-order time; half-life; instantaneous rate expressions	every paper
Thermochemistry	heat of formation from heats of combustion or formation (acetic acid, ethanol, acetylene, CS₂, CH₄)	2078 model, 2079 model, 2081, 2082, 2083, model set A
Electrochemistry	cell notation, anode and cathode, emf, feasibility from two or three E° values	every paper
Derivations	Ostwald's dilution law; N₁V₁ = N₂V₂; first-order k and t½; zero-order t½ ∝ a	2078 model, 2079 model, 2081, 2082, 2083

Definitions asked again and again: primary standard; equivalence point against end point; order against molecularity; rate law; Hess's law; heat of formation and of combustion; standard electrode potential; pH; Portland cement; natural against artificial radioactivity.

Organic items that repeat across papers: Victor Meyer's test (2078 model, 2079 model, 2081, 2083); chloroform with its five reactions (2079 model, 2081, 2082, 2083); Hoffmann's method (2079, 2079 model, 2082, 2083, model set B); an iodoform identification chain (2079 model, 2082, model set A); Reimer–Tiemann and Perkin in the named-reaction lists (2079 model, 2079, 2082, 2083); OPC against PPC (2078 model, 2079 model).

7.2 Model Question Paper 1

Group A — Rewrite the correct option of each question in your answer sheet. [11 × 1 = 11]

1. The equivalent weight of Na₂CO₃ when it is titrated completely with HCl (to CO₂ and water) is [1]a106b53c35.3d26.5
2. A first-order reaction has a rate constant 6.93 × 10⁻³ min⁻¹. Its half-life is [1]a100 minb10 minc0.0048 mind693 min
3. The pH of 0.001 M NaOH solution at 25 °C is [1]a3b11c10d1.4
4. A reaction is spontaneous at all temperatures when [1]aΔH > 0, ΔS < 0bΔH > 0, ΔS > 0cΔH < 0, ΔS > 0dΔH < 0, ΔS < 0
5. The oxidation state of manganese in K₂MnO₄ is [1]a+2b+4c+6d+7
6. The gas that turns acidified K₂Cr₂O₇ solution green is [1]aCO₂bSO₂cNO₂dH₂
7. 2-Bromobutane heated with alcoholic KOH gives mainly [1]abutan-2-olbbut-1-enecbut-2-enedbutane
8. Which test distinguishes acetophenone from benzaldehyde? [1]a2,4-DNP testbsodium bisulphite testciodoform testdFehling's test
9. Which of the following is the strongest acid? [1]aCH₃CH₂COOHbCH₃COOHcClCH₂COOHdC₆H₅OH
10. Nitrobenzene on reduction with Zn dust and NH₄Cl solution gives [1]aanilinebN-phenylhydroxylaminecazobenzenedp-aminophenol
11. Propanamide on heating with Br₂ and NaOH gives [1]apropylaminebethylaminecmethylaminedpropanoic acid

Group B — Short answer questions [8 × 5 = 40]

12. 1.26 g of oxalic acid crystals (H₂C₂O₄·2H₂O) is dissolved in water to make 200 mL of solution.(a) Define a primary standard. Is oxalic acid crystal one? [1+1](b) Find the normality of the solution. [1](c) 20.0 mL of it neutralises 25.0 mL of a NaOH solution. Find the normality and the strength in g L⁻¹ of the NaOH. [2] [5]
13. (a) Define the heat of combustion. [1](b) The heats of combustion of C(s), H₂(g) and CH₄(g) are −393.5, −285.8 and −890.3 kJ mol⁻¹. Calculate the heat of formation of methane. [4] [5]
14. The standard electrode potentials are E°(Zn²⁺/Zn) = −0.76 V and E°(Ag⁺/Ag) = +0.80 V.(a) Define the standard electrode potential. [1](b) Write the cell notation, showing the anode and the cathode. [2](c) Write the cell reaction. [1](d) Calculate the standard emf. [1] [5] Or
15. For the reaction A + B → C the initial rates are: (1) [A] = 0.1, [B] = 0.1 M, rate 2 × 10⁻³ M s⁻¹; (2) [A] = 0.2, [B] = 0.1, rate 8 × 10⁻³; (3) [A] = 0.1, [B] = 0.2, rate 4 × 10⁻³. Find the order with respect to A and B [2], the rate law [1], and the rate constant with its unit [2]. [5]
16. An alkyl halide A (C₃H₇Br) gives B (C₃H₆) with alcoholic KOH. B adds HBr by Markovnikov's rule to give C, an isomer of A. C with aqueous KOH gives D, which on oxidation with acidified K₂Cr₂O₇ gives E. Identify A to E with the reactions. [5]
17. Give an example of a primary and of a tertiary alcohol. How would you apply Victor Meyer's method to distinguish between them? [1+4] [5]
18. A mixture contains ethylamine and diethylamine.(a) Separate them by Hoffmann's method. [3](b) How would you distinguish them by the nitrous acid test? [2] [5] Or
19. Write one chemical equation for each of(a) Clemmensen reduction of propanone,(b) Cannizzaro's reaction of methanal,(c) Perkin condensation,(d) aldol condensation of ethanal,(e) Tollens' test on ethanal. [5]
20. Blue vitriol is an important compound of copper.(a) Give a method for its preparation. [1](b) What happens when it is heated at 100 °C, 230 °C and 720 °C? [2](c) What happens when excess NH₄OH is added to its solution? [1](d) Write its reaction with KI solution. [1] [5]
21. (a) Name the chief ore of mercury and describe its extraction. [2](b) How is the crude mercury purified? [1](c) How is calomel prepared, and what happens when ammonia solution is added to it? [2] [5]

Group C — Long answer questions [3 × 8 = 24]

20. For the reaction 2NO + O₂ → 2NO₂ the initial rates are: (1) [NO] = 0.01, [O₂] = 0.01 M, rate 2.5 × 10⁻⁶ M s⁻¹; (2) [NO] = 0.02, [O₂] = 0.01, rate 1.0 × 10⁻⁵; (3) [NO] = 0.02, [O₂] = 0.02, rate 2.0 × 10⁻⁵; (4) [NO] = 0.03, [O₂] = 0.04, rate R.(a) Differentiate between order and molecularity. [2](b) Write the rate law. [2](c) Calculate the rate constant with its unit. [2](d) Calculate R. [1](e) What is the overall order? [1] [8] Or
21. (a) What is a weak electrolyte? [1](b) Derive Ostwald's dilution law, and obtain α = √(Kb/C) for a weak base. [3](c) Calculate α, [OH⁻] and the pH of 0.05 M NH₄OH (Kb = 1.8 × 10⁻⁵). [4] [8]
22. Benzene (A) is nitrated to give B. B on reduction with Sn/HCl gives C. C on treatment with NaNO₂/HCl at 273 K gives D. D with phenol in alkaline medium gives the dye E, and D on warming with water gives F.(a) Identify B, C, D, E and F with the reactions. [5](b) Name the dye E. [1](c) Why is C a weaker base than ethylamine? [2] [8] Or
23. A compound P (C₃H₆O₂) on hydrolysis gives Q and R. Q turns blue litmus red and reduces Tollens' reagent. R gives a positive iodoform test.(a) Identify P, Q and R. [3](b) What is formed when R is heated with conc. H₂SO₄ at 443 K? [1](c) Write the reaction of Q with Tollens' reagent. [1](d) Convert R into ethanoic acid. [1](e) Compare the acidic strengths of Q and ethanoic acid. [2] [8]
24. (a) What are the raw materials for Portland cement? Why is gypsum added to the clinker? [2](b) Differentiate between the fibrous and non-fibrous raw materials for paper. [2](c) Write the monomers of nylon 6,6 and one use of it. [2](d) Differentiate between nuclear fission and nuclear fusion. [2] [8]

7.3 Solutions to Model Paper 1

Group A — answers

1. (b) — Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ uses two H⁺ per formula unit, so E = 106/2 = 53. Option (a) forgets the n-factor.

2. (a) — t½ = 0.693/k = 0.693/(6.93 × 10⁻³) = 100 min; option (d) multiplies instead of dividing.

3. (b) — [OH⁻] = 10⁻³ M, pOH = 3, pH = 14 − 3 = 11. Option (a) is the pOH.

4. (c) — ΔG = ΔH − TΔS is negative at every T only if ΔH < 0 and ΔS > 0, since then both terms help.

5. (c) — 2(+1) + x + 4(−2) = 0 gives x = +6 (manganate, green).

6. (b) — SO₂ reduces orange Cr₂O₇²⁻ to green Cr³⁺, which is why roasting of sulphide ores is detected this way.

7. (c) — Saytzeff's rule: the more substituted alkene, but-2-ene, is the major product; but-1-ene is minor.

8. (c) — acetophenone has CH₃CO– and gives iodoform; benzaldehyde does not. 2,4-DNP and NaHSO₃ are given by both, and Fehling's test by neither.

9. (c) — the electron-withdrawing chlorine stabilises the carboxylate; phenol (pKa 10) is far weaker than any carboxylic acid.

10. (b) — neutral medium stops reduction at the hydroxylamine; aniline needs acid, azobenzene alkali and p-aminophenol electrolysis in conc. H₂SO₄.

11. (b) — Hofmann bromamide degradation loses the carbonyl carbon: CH₃CH₂CONH₂ → CH₃CH₂NH₂.

7.4 Model Question Paper 2

Group A — Rewrite the correct option of each question in your answer sheet. [11 × 1 = 11]

1. The equivalent weight of KMnO₄ (M = 158) in neutral medium, where it is reduced to MnO₂, is [1]a158b79c52.7d31.6
2. The unit of the rate constant of a zero-order reaction is [1]as⁻¹bmol L⁻¹ s⁻¹cL mol⁻¹ s⁻¹dL² mol⁻² s⁻¹
3. Kw of water at 50 °C is 5.5 × 10⁻¹⁴. The pH of pure water at 50 °C is [1]a7.00b6.63c7.37d6.50
4. The standard electrode potentials of Mg²⁺/Mg, Zn²⁺/Zn, Cu²⁺/Cu and Ag⁺/Ag are −2.37, −0.76, +0.34 and +0.80 V. The strongest reducing agent is [1]aAgbCucZndMg
5. The number of d electrons in Fe³⁺ is [1]a3b4c5d6
6. White vitriol is [1]aCuSO₄·5H₂ObFeSO₄·7H₂OcZnSO₄·7H₂OdNa₂SO₄·10H₂O
7. Benzenediazonium chloride on warming with water gives [1]achlorobenzenebphenolcanilinedbenzene
8. Ethanoyl chloride on reduction with H₂ over Pd–BaSO₄ gives [1]aethanolbethanalcethanoic aciddethane
9. Which of the following is the most basic in aqueous solution? [1]aNH₃bCH₃NH₂c(CH₃)₂NHdC₆H₅NH₂
10. Ethanol heated with conc. H₂SO₄ at 413 K (alcohol in excess) gives [1]aethenebethoxyethanecethanaldethanoic acid
11. CH₃MgBr reacts with CO₂ and the product is hydrolysed. The compound formed is [1]aethanalbethanolcethanoic aciddpropanone

Group B — Short answer questions [8 × 5 = 40]

12. 20.0 mL of 0.02 M KMnO₄ in acidic medium just oxidises 25.0 mL of a FeSO₄ solution.(a) Write the half-reaction of KMnO₄ in acid and its n-factor. [1](b) Find the normality of the FeSO₄ solution. [2](c) Find its strength in g L⁻¹ (M = 152). [1](d) Why is dilute H₂SO₄ and not HCl used? [1] [5]
13. (a) What is a buffer solution? [1](b) 50 mL of 0.2 M ethanoic acid is mixed with 50 mL of 0.1 M sodium ethanoate (pKa = 4.74). Calculate the pH. [2](c) Write the Henderson–Hasselbalch equation and give two applications of buffers. [2] [5]
14. (a) State the second law of thermodynamics. [1](b) For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J mol⁻¹ K⁻¹. Find the minimum temperature at which the decomposition is spontaneous. [2](c) Find ΔG at 298 K and state whether the reaction is spontaneous. [2] [5] Or
15. E°(Mg²⁺/Mg) = −2.37 V and E°(Cu²⁺/Cu) = +0.34 V.(a) Write the cell notation with anode and cathode. [2](b) Write the cell reaction. [1](c) Calculate the emf. [1](d) Can MgSO₄ solution be stored in a copper vessel? Why? [1] [5]
16. An aromatic compound A (C₆H₇N) reacts with NaNO₂/HCl at 273 K to give B. B on warming with water gives C. C with CHCl₃ and NaOH gives D, and B with CuCl/HCl gives E. Identify A to E with the reactions. [5]
17. Write one chemical equation for each of(a) Sandmeyer's reaction,(b) Kolbe's reaction,(c) Hofmann bromamide reaction,(d) Rosenmund reduction,(e) Fittig reaction. [5]
18. A is ethanoic acid. A + PCl₅ → B; B + NH₃ → C; C + Br₂/KOH → D; D + HNO₂ → E. Identify B to E with the reactions. [5] Or
19. Write the products, with reagent and medium, when nitrobenzene is reduced in(a) acidic medium,(b) neutral medium,(c) weakly alkaline medium,(d) strongly alkaline medium and(e) electrolytically in conc. H₂SO₄. [5]
20. (a) Name the chief ore of zinc. [1](b) How is the ore concentrated? [1](c) Write the reactions of roasting and of reduction in the extraction of zinc. [2](d) What happens when zinc is heated with excess NaOH solution? [1] [5]
21. (a) Describe the manufacture of steel by the Basic Oxygen Process: vessel, lining and charge. [3](b) Why is lime added? [1](c) Why is no external fuel needed? [1] [5]

Group C — Long answer questions [3 × 8 = 24]

20. The decomposition of N₂O₅ is first order with k = 6.93 × 10⁻³ s⁻¹.(a) Calculate the half-life. [1](b) Calculate the time for 75 % decomposition. [2](c) What percentage of N₂O₅ is left after 300 s? [2](d) Derive the relation t½ = 0.693/k for a first-order reaction. [3] [8] Or
21. 0.53 g of anhydrous Na₂CO₃ is dissolved to make 100 mL of solution, and 10.0 mL of it requires 12.5 mL of HCl for complete neutralisation.(a) Is Na₂CO₃ a primary standard? Why? [1](b) Find the normality of the HCl. [3](c) Name the indicator and its colour change. [2](d) Differentiate between the equivalence point and the end point. [2] [8]
22. Aniline (A).(a) How is aniline prepared from nitrobenzene? [1](b) What is formed with ethanoyl chloride? [1](c) Write the diazotisation of aniline and the coupling of the product with β-naphthol. [2](d) Write the carbylamine reaction of aniline. [1](e) Why is aniline a weaker base than methylamine? [2] (f) What is formed with bromine water? [1] [8] Or
23. Ethanoic acid (A).(a) Write the Hell–Volhard–Zelinsky reaction of A. [2](b) How is ethanoic anhydride prepared from A? [2](c) How is ethyl ethanoate prepared from A? [2](d) Why are amides amphoteric? [2] [8]
24. (a) Draw the flow sheet for the manufacture of Portland cement. [2](b) Describe the Kraft process for pulping in brief. [2](c) Write the monomers of polystyrene and Bakelite. [2](d) Define radioactivity and state its SI unit. [2] [8]

7.5 Solutions to Model Paper 2

Group A — answers

1. (c) — in neutral medium MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ takes three electrons, so E = 158/3 = 52.7. Option (d) is the acid-medium value.

2. (b) — for zero order the rate has the unit of concentration per time, mol L⁻¹ s⁻¹; s⁻¹ belongs to first order.

3. (b) — [H⁺] = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷, so pH = 6.63; water is still neutral because [H⁺] = [OH⁻].

4. (d) — the most negative E° (−2.37 V) is the strongest reducing agent: Mg.

5. (c) — Fe is [Ar] 3d⁶4s²; Fe³⁺ loses two 4s and one 3d electron, leaving 3d⁵.

6. (c) — ZnSO₄·7H₂O is white vitriol; CuSO₄·5H₂O is blue vitriol and FeSO₄·7H₂O green vitriol.

7. (b) — C₆H₅N₂Cl + H₂O → C₆H₅OH + N₂ + HCl; chlorobenzene needs CuCl/HCl.

8. (b) — the Rosenmund reduction stops at the aldehyde because Pd–BaSO₄ is poisoned.

9. (c) — dimethylamine, a secondary amine, is the strongest base in water; aniline is weakest.

10. (b) — 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O at 413 K; ethene needs 443 K.

11. (c) — CH₃MgBr + CO₂ → CH₃COOMgBr, and hydrolysis gives CH₃COOH.

7.6 Forty-eight predicted questions, ranked

How to read this list. Questions are ranked in three tiers. Tier 1 is the topics that appeared in five or more of the eight papers studied: drill these first. Tier 2 is topics seen in two to four papers, or seen every year in a small form. Tier 3 is the gaps — syllabus topics with a large share of hours that have not yet been set as a five- or eight-mark question in the eight papers, or that appeared only in the model papers. Each entry gives the reason, then a short model answer; the full answers are in the unit-wise chapter (6.1 to 6.21) and the answers there should be learnt in full.

Tier 1 — drill first

1. Distinguish a primary from a secondary alcohol (or from a tertiary) by Victor Meyer's method. Why: set in the 2078 model, 2079 model, 2081 and 2083 papers — four of eight, and in two of the last three board papers (2081 and 2083). Answer: alcohol → (red P/I₂) iodide → (AgNO₂) nitro compound → (HNO₂) then NaOH: primary gives nitrolic acid, blood red; secondary gives pseudonitrole, blue; tertiary gives no nitro compound, colourless.
2. Primary and secondary standards; why oxalic acid and not NaOH. Why: five of eight papers (2078 model, 2079, 2081, 2083, model set B). Answer: a primary standard is pure, stable, non-hygroscopic, of high molar mass and exact formula, so a solution is made by weighing (oxalic acid crystals, anhydrous Na₂CO₃); NaOH absorbs CO₂ and water, so it must be standardised against a primary standard.
3. Titration of oxalic acid against KMnO₄: equivalent weight, normality from burette readings, why dilute H₂SO₄. Why: 2079, 2082 and 2083, each with different data. Answer: E(KMnO₄) = 158/5 = 31.6; N₁V₁ = N₂V₂ with the mean of concordant readings; dilute H₂SO₄ supplies H⁺ and avoids MnO₂; KMnO₄ is the titrant and its own indicator.
4. A table of initial rates: order in each reactant, rate law, k with its unit, a missing rate. Why: every paper; 2079, 2081, 2083 and model set B in full. Answer: compare two experiments with one variable constant; rate = k[A]ᵐ[B]ⁿ; k = rate/([A]ᵐ[B]ⁿ) from one experiment, unit mol^(1−n) L^(n−1) s⁻¹; substitute for the missing value.
5. Derive Ostwald's dilution law and find α for a weak acid. Why: 2079 model, 2083 and model set A (three derivations), plus the numerical in 2082. Answer: Ka = Cα²/(1 − α) ≈ Cα², so α = √(Ka/C); for 0.01 M HCN (Ka = 4.8 × 10⁻¹⁰), α = 2.2 × 10⁻⁴, pH 5.66.
6. pH, pOH and Kw calculations; dilution of a base. Why: 2078 model, 2079, 2081, 2082, model set A. Answer: pH + pOH = 14; [H⁺] = √(KaC) for a weak acid; a base diluted until it is below 10⁻⁷ M OH⁻ tends to pH 7 and never below; Kw rises with temperature.
7. Precipitation test from Ksp. Why: 2078 model, 2083 and model set B. Answer: halve the concentrations after mixing equal volumes; if [A⁺][B⁻] > Ksp the salt precipitates. Example: if the mixture holds 0.01 M Ba²⁺ and 10⁻⁴ M SO₄²⁻, the product is 10⁻⁶ > 10⁻¹⁰ and BaSO₄ precipitates.
8. Hess's law: heat of formation from heats of combustion or formation. Why: seven of eight papers — every one except model set B. Answer: write the formation equation, then add the given equations (reversing and multiplying): ΔH_f = Σ ΔH_c(reactants) − Σ ΔH_c(products) or the formation form; for acetylene from ΔH_c of C, H₂, C₂H₂ it is +226 kJ mol⁻¹.
9. Cell notation, anode and cathode, cell reaction and emf from two E° values. Why: every paper, in three forms: two E° values, a pair with the SHE, and a feasibility question. Answer: the lower E° is the anode; E°cell = E°cathode − E°anode; write M|Mⁿ⁺ ‖ Xᵐ⁺|X with the 1 M and 1 atm conditions.
10. Extraction of copper from copper pyrite, with the blast-furnace reactions for the matte. Why: 2079 model, 2081, 2083 and model set B, with a furnace drawing asked in 2081 and 2083. Answer: flotation → roasting (2CuFeS₂ + O₂ → Cu₂S + 2FeS + SO₂) → smelting (FeO + SiO₂ → FeSiO₃; Cu₂O + FeS → Cu₂S + FeO) → Bessemer (2Cu₂O + Cu₂S → 6Cu + SO₂) → poling and electrolytic refining.
11. Chloroform: preparation from ethanol and the five reactions. Why: 2079 model, 2081, 2082, 2083 and model set B. Answer: ethanol + Cl₂ (from bleaching powder) → ethanal → chloral → + Ca(OH)₂ → CHCl₃; air → COCl₂; Ag → HC≡CH; acetone/KOH → chloretone; conc. HNO₃ → CCl₃NO₂; aniline/KOH → C₆H₅NC.
12. The iodoform chain: X + CH₃MgBr → Y → Z, all giving iodoform. Why: 2079 model, 2082 and model set A, with the 2081 MCQ. Answer: X = ethanal, Y = propan-2-ol, Z = propanone; the CH₃CO– or CH₃CH(OH)– group is oxidised and iodinated to CHI₃; Z with dilute NaOH gives diacetone alcohol.
13. Hoffmann's separation of amines and the nitrous acid test. Why: 2079 model, 2079, 2082, 2083, model set B. Answer: diethyl oxalate makes the primary amine a solid oxamide, the secondary a liquid oxamic ester and leaves the tertiary unchanged; KOH hydrolysis recovers them. HNO₂: primary gives N₂, secondary a yellow nitrosoamine.
14. One equation for each of five named reactions (Tollens', Fehling's, aldol, Cannizzaro, Reimer–Tiemann, Perkin, Williamson, 2,4-DNP). Why: seven of eight papers carry such a list. Answer: write one equation with the reagent over the arrow, for example 2HCHO + NaOH → CH₃OH + HCOONa for Cannizzaro, and C₆H₅CHO + (CH₃CO)₂O → C₆H₅CH=CHCOOH for Perkin.
15. A reaction chain from benzene through nitrobenzene, aniline, benzenediazonium chloride and phenol or an azo dye. Why: 2078 model, 2081, 2082, 2083 and both model sets. Answer: HNO₃/H₂SO₄ → Sn/HCl → NaNO₂/HCl at 273 K → H₂O warm (phenol) or phenol/NaOH (p-hydroxyazobenzene).
16. Portland cement: raw materials, flow sheet, components, role of gypsum, OPC against PPC, quality control. Why: every paper has one. Answer: limestone + clay → grinding → rotary kiln 1400–1500 °C → clinker → + 2–5 % gypsum (retarder) → ball mill; PPC has 15–35 % pozzolana, lower early strength, better durability.

Tier 2 — seen in some papers, or every year in a small form

17. Preparation, properties and uses of corrosive sublimate. Why: set in both 2081 and 2083 and in model set B — three of eight papers, and in each of the last two board sittings that asked about mercury. Answer: HgSO₄ + 2NaCl → HgCl₂↑ + Na₂SO₄; with KI a scarlet precipitate dissolving in excess; with SnCl₂ white then grey; with NH₃ white Hg(NH₂)Cl; antiseptic and Nessler's reagent.
18. Zinc: extraction from blende, the vertical retort, white vitriol, lithopone, why zinc is non-typical. Why: 2079 and 2082, each of them a full five-mark question. Answer: flotation → 2ZnS + 3O₂ → 2ZnO + 2SO₂ → ZnO + C → Zn + CO at 1673 K; ZnSO₄ + BaS → ZnS + BaSO₄ (lithopone); Zn²⁺ is 3d¹⁰.
19. Steel by the open hearth or Basic Oxygen Process; the role of spiegeleisen. Why: the open hearth process was set in 2079; the Basic Oxygen Process has never been asked, so it is the likelier alternative. Answer: BOP: oxygen lance, basic dolomite lining, lime flux, no fuel; OHP: Fe₂O₃ oxidises C, Si, P, Mn; spiegeleisen restores carbon and removes oxygen.
20. Rusting: the electrochemical theory and two methods of prevention. Why: 2082; corrosion has not been repeated since, so it is due. Answer: anode Fe → Fe²⁺ + 2e⁻; cathode O₂ + 4H⁺ + 4e⁻ → 2H₂O; 4Fe²⁺ + O₂ + 4H₂O → 2Fe₂O₃ + 8H⁺, then Fe₂O₃·xH₂O; paint, galvanising, sacrificial anode.
21. The chain A → (SOCl₂) B → (NH₃) C → (Br₂/KOH) D → (HNO₂) E. Why: 2082 and 2081 (Or), and the 2079 model, each through an acid derivative. Answer: propanoic acid → propanoyl chloride → propanamide → ethylamine → ethanol; the Hofmann step loses one carbon.
22. Differentiate order from molecularity. Why: 2079 and 2083; it is also the opening part of the kinetics question. Answer: order is experimental and may be zero or fractional; molecularity is the number of molecules in one elementary step and is a whole number 1, 2 or 3.
23. First-order integrated equation, t½ = 0.693/k and the time for 90 % completion. Why: 2078 model, 2082 (MCQ on t½) and model sets; a derivation is asked in about half the papers. Answer: k = (2.303/t) log[a/(a − x)]; t½ = 0.693/k; t(90 %) = 2.303/k = 3.32 t½.
24. Gibbs–Helmholtz: spontaneity, the temperature at which a reaction turns spontaneous, ΔG and K. Why: model set B, the 2078 model entropy question and the 2082 MCQ; only Hess's law dominates the board papers so far, which leaves ΔG under-tested. Answer: ΔG = ΔH − TΔS; T = ΔH/ΔS at ΔG = 0; ΔG° = −2.303RT log K.
25. Brønsted–Lowry against Lewis acids and bases, with conjugate pairs. Why: 2082 (Or) and the opening part of ionic-equilibrium questions; the syllabus lists both concepts. Answer: B–L acid donates H⁺, Lewis acid accepts an electron pair; NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ has pairs NH₃/NH₄⁺ and H₂O/OH⁻.
26. Reduction of nitrobenzene in acidic, neutral, alkaline and electrolytic media. Why: 2079 model, 2081, 2083 (electrolytic) and the MCQs of 2078 model and 2079 model. Answer: Sn/HCl aniline; Zn/NH₄Cl phenylhydroxylamine; Na₃AsO₃ azoxybenzene; Zn/NaOH azobenzene then hydrazobenzene; electrolytic conc. H₂SO₄ p-aminophenol.
27. Ethanoic acid from CH₃MgBr, and its action with NH₃, SOCl₂, P₂O₅, LiAlH₄. Why: 2083 and model set A. Answer: CH₃MgBr + CO₂ → CH₃COOMgBr → CH₃COOH; then CH₃CONH₂, CH₃COCl, (CH₃CO)₂O and CH₃CH₂OH respectively.
28. Order of acidic strength of substituted acids; the HVZ reaction. Why: 2079 model (2 marks) and a recurring Group A stem. Answer: FCH₂COOH > ClCH₂COOH > C₆H₅COOH > CH₃COOH; the −I effect stabilises the anion; CH₃COOH + Cl₂ → (red P) ClCH₂COOH.
29. Addition against condensation polymers; monomers of nylon 6,6, polystyrene and Bakelite. Why: six of eight papers (2078 model, 2079 model, 2079, 2081, 2082, model set A). Answer: addition: no loss of atoms (polythene); condensation: loss of H₂O or HCl (nylon 6,6 from H₂N(CH₂)₆NH₂ and HOOC(CH₂)₄COOH; Bakelite from phenol and methanal; polystyrene from styrene).
30. Paper: raw materials, fibrous and non-fibrous, paper against pulp, the Kraft process. Why: six of eight papers, always as a two-mark part. Answer: fibrous (wood, bamboo, rags, lokta) supply cellulose; non-fibrous (china clay, rosin, starch, alum) are additives; Kraft: white liquor NaOH + Na₂S at 170 °C dissolves lignin.
31. Natural against artificial radioactivity, with an example of each. Why: 2079 model, 2083 and model set A (the 2081 paper asked for nuclear against chemical reactions). Answer: natural is spontaneous in U, Th, Ra; artificial is induced by bombardment, ²⁷Al + ⁴He → ³⁰P + n, ³⁰P → ³⁰Si + β⁺.
32. Nuclear fusion as a thermonuclear reaction; fission with its chain reaction; nuclear power. Why: 2082 (fusion) and the applied question every year. Answer: fusion needs 10⁷–10⁸ K to overcome repulsion; ²H + ³H → ⁴He + n; fission ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n, controlled by rods in a reactor.

Tier 3 — the gaps, and the rest of the syllabus

33. Crystal field theory: its assumptions and the octahedral d-orbital splitting diagram. Why: set in the 2078 model and model set A but in no board paper of 2079–2083; five teaching hours; a gap. Answer: ligands are point charges; the d-orbitals split into e_g (+0.6 Δo) and t₂g (−0.4 Δo); [Cu(H₂O)₆]²⁺ is blue because the d–d transition absorbs red-orange.
34. What is a d–d transition? List the characteristics of transition metals. Why: the 2079 model; the characteristic properties are also the reasons behind several Group A items (catalysts, green ions). Answer: promotion of an electron between split d-orbitals by visible light; variable oxidation states, coloured ions, complexes, catalysis, paramagnetism.
35. Silver: extraction from argentite by the cyanide process; silver nitrate and silver chloride. Why: never set in Group B or C in the eight papers, despite a full syllabus bullet; a clear gap. Answer: Ag₂S + 4NaCN ⇌ 2Na[Ag(CN)₂] + Na₂S (air removes Na₂S); 2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag; Moebius refining.
36. Iron: extraction in the blast furnace, the zones and the slag. Why: never set in the eight papers (steel and rusting have been set, the extraction has not). Answer: Fe₂O₃ + 3CO → 2Fe + 3CO₂; CaCO₃ → CaO + CO₂; CaO + SiO₂ → CaSiO₃ slag; pig iron is tapped from the hearth.
37. Buffer solutions: Henderson–Hasselbalch equation and pH of an acidic buffer. Why: no five-mark question on buffers in the eight papers although it is a named syllabus bullet. Answer: pH = pKa + log([salt]/[acid]); 0.2 M CH₃COOH + 0.1 M CH₃COONa (pKa 4.74) has pH 4.44.
38. Chlorobenzene: low reactivity towards nucleophiles, DDT, Sandmeyer route. Why: 2078 model, 2082, 2083 (the DDT part); a recurring 2-mark part. Answer: resonance and sp² carbon give the C–Cl bond partial double-bond character; 2C₆H₅Cl + CCl₃CHO → DDT; C₆H₅N₂Cl + CuCl → C₆H₅Cl.
39. Ethers: Williamson's synthesis, the action of HI and the danger of peroxides. Why: 2079, 2081, 2082 and 2083 (as list items). Answer: RONa + R′X → ROR′ + NaX (primary halide); C₂H₅OC₂H₅ + 2HI → 2C₂H₅I + H₂O; peroxides form in air, so keep over iron wire.
40. Phenol: acidity against alcohol and water; Kolbe and Reimer–Tiemann reactions; tests. Why: model set A, 2082 and 2083 (as list items); four teaching hours. Answer: phenoxide is resonance-stabilised, so phenol is more acidic than ethanol but weaker than carbonic acid; FeCl₃ gives violet; Br₂ water gives white 2,4,6-tribromophenol.
41. Grignard reagent with aldehydes and ketones: preparation of 1°, 2° and 3° alcohols. Why: the 2081 MCQ, 2079 model and 2082 chains depend on it. Answer: HCHO gives a primary, another aldehyde a secondary and a ketone a tertiary alcohol, each after hydrolysis.
42. Fermentation of ethanol; absolute, denatured and power alcohol. Why: 2079 (full question); it has not recurred. Answer: C₁₂H₂₂O₁₁ → (invertase) 2C₆H₁₂O₆ → (zymase) 4C₂H₅OH + 4CO₂ overall (with one H₂O taken in); denatured alcohol contains methanol or pyridine.
43. Basic strength of aniline against ammonia and ethylamine; coupling reaction. Why: 2079 model, 2083 (MCQ) and chains in 2082, 2078 model. Answer: ethylamine > ammonia > aniline; in aniline the lone pair is delocalised; C₆H₅N₂Cl + phenol → p-hydroxyazobenzene.
44. Structural formulae of aspirin and DDT; antipyretic against analgesic; habit-forming drugs. Why: 2083 and model set B; the drug part of the service-of-mankind unit has come up twice in three sittings. Answer: aspirin is 2-acetoxybenzoic acid; paracetamol is an antipyretic, aspirin an analgesic; morphine is habit-forming.
45. Radiocarbon dating: age of a sample from its ¹⁴C activity. Why: the unit-21 syllabus names it, and numerical parts are rare in Q22, so it is unseen. Answer: N/N₀ = (½)^(t/5730); for 25 % activity, t = 11 460 years.
46. The standard hydrogen electrode (diagram); the lead storage battery and the H₂–O₂ fuel cell. Why: 2083 (reference electrode), model set B (diagram), and the 2078 model and 2083 MCQs on batteries and fuel cells. Answer: Pt foil in 1 M H⁺ with H₂ at 1 atm, E = 0.00 V; Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; 2H₂ + O₂ → 2H₂O.
47. Indicators and the choice of indicator in acid–base titrations. Why: model sets A and B both open with it as an MCQ. Answer: strong acid–strong base: any; strong acid–weak base (Na₂CO₃): methyl orange; weak acid–strong base: phenolphthalein; weak–weak: no sharp end point.
48. Activation energy, the effect of a catalyst and temperature (Arrhenius equation), energy-profile diagram. Why: the 2079 model MCQ on the effect of temperature and the 2079 MCQ on catalysts; a full question on catalysis has not been set. Answer: a catalyst lowers Ea without changing ΔH; k = Ae^(−Ea/RT); the rate roughly doubles for each 10 °C rise.

Chapter summary

• The grid never changes: 11 MCQs, 8 five-mark and 3 eight-mark questions, all compulsory, with internal alternatives in about two Group B and two Group C questions.
• The paper has a shape. In 2081, 2082 and 2083 Group A ran four physical, two inorganic, five organic questions; Group C ran one physical question (Q20), one organic chain (Q21) and an applied question of four parts (Q22).
• Marks follow teaching hours. Organic chemistry is about 44 % of the marks and 43 % of the hours, physical about 29 % and 31 %, inorganic 16 % and 16 %. Revise in proportion.
• Twelve units have been set in seven or eight of the eight papers — volumetric analysis, ionic equilibrium, kinetics, thermodynamics, electrochemistry, heavy metals, haloalkanes, alcohols, aldehydes and ketones, amines, service of mankind and cement — and these are the Tier 1 and Tier 2 questions above.
• The gaps are real. Crystal field theory, silver, blast-furnace iron, buffers and radiocarbon dating have not been set as five-mark questions in the board papers studied; they are Tier 3 because they are due, not because they are certain.
• Two complete model papers are solved in full; sit each in three hours, mark yourself by the steps in the solution, and then return to the unit-wise chapters for the topics where you lost marks.

Sources and verification