Chemistry: Model Question Sets and Predicted Questions

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Chemistry: Model Question Sets and Predicted Questions

This chapter does three things. It reads the shape of the Grade 11 Chemistry paper (3 hours, full marks 75, Group A 11\times1, Group B 8\times5, Group C 3\times8) and shows how the 128 teaching hours are spread over the 17 units. It then sets two complete model papers to that grid, each fully solved. Finally it lists twenty-four predicted questions in ranked order, each with the reason it is predicted and a short model answer.

What "model" and "predicted" mean here

The model papers are original practice papers built to the CDC pattern; they are not NEB papers and do not claim to resemble any particular year. A prediction is a judgement, not a leak: nobody outside the board knows the paper. The ranking uses three public things only — the teaching hours of each unit, how standard the question is (a definition, reaction or numerical type that every textbook treats), and whether it is a recurring numerical type. Treat the list as a revision order. No question carries a past-paper year, because Grade 11 papers are not published by NEB.

7.1 Pattern and weight analysis of the Grade 11 Chemistry paper

7.1.1 The grid

Part Questions Marks What to expect
Group A 11 multiple-choice 11\times1 = 11 one mark each; drawn from the whole syllabus
Group B 8 short answer 8\times5 = 40 equations, short theory, one or two numericals
Group C 3 long answer 3\times8 = 24 a manufacturing process, a mechanism or a multi-part numerical

All questions are compulsory. The time per mark is 180/75 = 2.4 minutes: about 12 minutes for a Group B question and 19 for a Group C question.

7.1.2 Where the syllabus hours sit

Area Units Hours Share Expected presence
General and Physical 1–8 48 38 % two or three Group B numericals and a Group C question
Inorganic 9–11 35 27 % two Group B questions of reactions and tests
Organic 12–15 30 23 % two or three Group B questions; mechanisms and conversions
Applied 16–17 15 12 % a manufacturing process, often as Group C

7.1.3 Unit by unit: hours and the form the question usually takes

Unit Hours Typical form Priority
1 Foundation 2 percentage composition, formulae low
2 Stoichiometry 8 mole numericals, limiting reactant, yield, empirical formula very high
3 Atomic structure 8 Bohr, quantum numbers, configurations, uncertainty very high
4 Periodic table 5 trends and their reasons high
5 Bonding and shapes 9 VSEPR, hybridization, hydrogen bond very high
6 Redox 5 oxidation number, balancing, Faraday's laws high
7 States of matter 8 gas laws, Graham, solids high
8 Equilibrium 3 K_c, K_p, Le Chatelier medium
9 Non-metals 22 preparation and properties of N, S, halogens, O, H very high
10 Metals 10 metallurgy, Down's process, alkaline earths high
11 Bio-inorganic 3 roles of metal ions, toxicity medium
12 Basic organic 6 classification, octane number medium
13 Fundamental principles 10 IUPAC, isomerism, Lassaigne, effects very high
14 Hydrocarbons 8 preparation, reactions, Markovnikov very high
15 Aromatic 6 benzene, nitration, Friedel–Crafts high
16 Applied fundamentals 4 stages, batch and continuous medium
17 Manufactures 11 Haber, Ostwald, contact, diaphragm, Solvay very high

Two points follow. The single heaviest unit is non-metals (22 hours), one-sixth of the whole course, so reactions of nitrogen, sulphur and halogens must be known cold. And nothing is safe to skip: Group A is spread over all seventeen units.

7.2 Model Set 1

Time: 3 hours. Full marks: 75. Attempt all the questions. Figures in the margin indicate full marks. Atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, K = 39, Cr = 52, Fe = 56, Ag = 108. This is an original practice paper, built to the CDC pattern.

Group A — Rewrite the correct option of each question in your answer sheet. (11\times1 = 11)

  1. The number of moles in 36\ \text{g} of water is [1]a1b2c18d36
  2. The maximum number of electrons in a 3d subshell is [1]a6b8c10d14
  3. The most electropositive of the following is [1]aLibNacKdCs
  4. The hybridization of nitrogen in ammonia is [1]aspbsp^{2}csp^{3}dsp^{3}d
  5. The oxidation number of chlorine in \text{KClO}_3 is [1]a+1b+3c+5d+7
  6. The law that PV is constant at constant temperature is [1]aCharles' lawbBoyle's lawcAvogadro's lawdGraham's law
  7. For \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3, K_p is related to K_c by [1]aK_p = K_c(RT)^{2}bK_p = K_c(RT)^{-1}cK_p = K_cdK_p = K_c(RT)^{-2}
  8. Sodium is extracted industrially by [1]acarbon reductionbDown's processcthermite processdpoling
  9. The IUPAC name of \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} is [1]apropanalbpropan-1-olcpropan-2-oldpropanoic acid
  10. Ethyne reacts with two molecules of HBr (Markovnikov addition) to give mainly [1]a1,2-dibromoethaneb1,1-dibromoethanecbromoethenedethane
  11. The catalyst used in the contact process is [1]aFebPtc\text{V}_2\text{O}_5dNi

Group B — Short answer questions. (8\times5 = 40)

  1. 4.9\ \text{g} of sulphuric acid is taken. Find the number of moles, the number of molecules, and the number of oxygen atoms in it. [5]
  2. A hydrocarbon contains 85.7\ \% carbon and 14.3\ \% hydrogen, and its molar mass is 56\ \text{g mol}^{-1}. Find its empirical and molecular formulae. [5]
  3. Write the electronic configurations of Fe, \text{Fe}^{2+} and Cr, and give the number of unpaired electrons in each. [5]
  4. State Faraday's first law. For how long must a current of 0.5\ \text{A} be passed through silver nitrate solution to deposit 0.54\ \text{g} of silver? [5]
  5. State Graham's law of diffusion. A gas X takes 40\ \text{s} to diffuse through a porous plug and the same volume of oxygen takes 20\ \text{s}. Find the molar mass of X. [5]
  6. One mole each of \text{H}_2 and \text{I}_2 are heated in a 1\ \text{L} vessel and 1.5 mol of HI is found at equilibrium. Find K_c. [5]
  7. Describe the laboratory preparation of sulphur dioxide and three of its chemical properties. [5]
  8. Write the main product of: (i) propene + HBr, (ii) propene + HBr (peroxide), (iii) ethyne + water (\text{HgSO}_4), (iv) benzene + acetyl chloride (\text{AlCl}_3), (v) sodium acetate heated with soda lime. [5]

Group C — Long answer questions. (3\times8 = 24)

  1. (a) Define oxidation and reduction in terms of electron transfer. [2](b) Balance the equation \text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} + \text{H}^{+} \to \text{Cr}^{3+} + \text{Fe}^{3+} + \text{H}_2\text{O} by the ion–electron method. [4](c) What mass of \text{K}_2\text{Cr}_2\text{O}_7 (M = 294) is needed to oxidize 0.6 mol of \text{Fe}^{2+}? [2] [8]
  2. (a) Describe the preparation of ethyne from calcium carbide and from 1,2-dibromoethane. [3](b) Explain, with equations, the acidic character of ethyne. [2](c) Write the reactions of ethyne with hydrogen, HCl and water. [3] [8]
  3. (a) Describe the manufacture of nitric acid by Ostwald's process, with a flow sheet. [5](b) Write the reactions of dilute and concentrated nitric acid with copper. [3] [8]

7.2.1 Solutions to Model Set 1

Solutions — Group A

1. (b) — 36/18 = 2 mol.

2. (c) — five d orbitals hold ten electrons.

3. (d) — electropositivity rises down Group 1; Cs is the lowest in the list.

4. (c) — three bond pairs and one lone pair give four electron pairs.

5. (c) — 1 + x - 6 = 0 gives x = +5.

6. (b) — Boyle's law.

7. (d) — \Delta n = 2 - 4 = -2.

8. (b) — electrolysis of fused NaCl with \text{CaCl}_2.

9. (b) — three carbons, -\text{OH} on carbon 1.

10. (b) — the second H–Br adds to the carbon already carrying Br, giving \text{CH}_3\text{CHBr}_2.

11. (c) — vanadium pentoxide.

Solutions — Group B (12–19)

12. M(\text{H}_2\text{SO}_4) = 98. Moles = 4.9/98 = 0.05 mol. Molecules = 0.05\times6.022\times10^{23} = 3.01\times10^{22}. Each molecule has 4 oxygen atoms: 4\times3.01\times10^{22} = 1.2\times10^{23} atoms.

13. Moles in 100\ \text{g}: C = 85.7/12 = 7.14; H = 14.3/1 = 14.3. Ratio 1 : 2, so the empirical formula is \text{CH}_2 (mass 14). n = 56/14 = 4; molecular formula \text{C}_4\text{H}_8.

14. Fe (Z = 26): [\text{Ar}]3d^{6}4s^{2}, 4 unpaired electrons. \text{Fe}^{2+}: [\text{Ar}]3d^{6}, 4 unpaired (the 4s electrons leave first). Cr (Z = 24): [\text{Ar}]3d^{5}4s^{1}, 6 unpaired (half-filled d and s are stable).

15. The mass of a substance deposited at an electrode is directly proportional to the quantity of electricity passed, m = ZIt. \text{Ag}^{+} + e^{-} \to \text{Ag}: 108\ \text{g} needs 96\,500\ \text{C}. Q = \dfrac{0.54}{108}\times96\,500 = 482.5\ \text{C}. t = \dfrac{Q}{I} = \dfrac{482.5}{0.5} = 965\ \text{s} (\approx16 min).

16. At constant temperature and pressure the rate of diffusion of a gas is inversely proportional to the square root of its density (or molar mass): \dfrac{r_1}{r_2} = \sqrt{\dfrac{M_2}{M_1}}. For equal volumes the rates are inversely proportional to the times, so \dfrac{r_X}{r_{O_2}} = \dfrac{20}{40} = \sqrt{\dfrac{32}{M_X}}. Squaring, \dfrac{1}{4} = \dfrac{32}{M_X} and M_X = 128\ \text{g mol}^{-1}.

17. \text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}. 1.5 mol HI forms from 0.75 mol each of \text{H}_2 and \text{I}_2, so at equilibrium [\text{H}_2] = [\text{I}_2] = 0.25 and [\text{HI}] = 1.5\ \text{mol L}^{-1}. K_c = \dfrac{[\text{HI}]^{2}}{[\text{H}_2][\text{I}_2]} = \dfrac{2.25}{0.0625} = 36.

18. Preparation: add dilute HCl (or dilute \text{H}_2\text{SO}_4) to sodium sulphite: \text{Na}_2\text{SO}_3 + 2\text{HCl} \to 2\text{NaCl} + \text{SO}_2\uparrow + \text{H}_2\text{O}; the gas is dried with conc. \text{H}_2\text{SO}_4 and collected by upward delivery (heavier than air). Or heat copper with conc. \text{H}_2\text{SO}_4. Properties: (i) acidic: \text{SO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{SO}_3 (turns blue litmus red); (ii) reducing: \text{SO}_2 + \text{Cl}_2 + 2\text{H}_2\text{O} \to \text{H}_2\text{SO}_4 + 2\text{HCl}, and it decolourizes acidified purple \text{KMnO}_4; (iii) oxidizing: \text{SO}_2 + 2\text{H}_2\text{S} \to 3\text{S} + 2\text{H}_2\text{O}; (iv) bleaching of coloured matter by reduction (temporary).

19. (i) 2-bromopropane, \text{CH}_3\text{CHBrCH}_3. (ii) 1-bromopropane, \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}. (iii) acetaldehyde, \text{CH}_3\text{CHO}. (iv) acetophenone, \text{C}_6\text{H}_5\text{COCH}_3. (v) methane, \text{CH}_4 (with \text{Na}_2\text{CO}_3).

Solutions — Group C (20–22)

20. (a) Oxidation is the loss of electrons; reduction is the gain of electrons (OIL RIG). (b) Reduction half: \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6e^{-} \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}. Oxidation half: \text{Fe}^{2+} \to \text{Fe}^{3+} + e^{-}; multiply by 6: 6\text{Fe}^{2+} \to 6\text{Fe}^{3+} + 6e^{-}. Add: \text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^{+} \to 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}. Check charge: -2 + 12 + 14 = +24; 6 + 18 = +24 ✓. (c) 1 mol dichromate oxidizes 6 mol \text{Fe}^{2+}, so 0.6 mol needs 0.1 mol = 0.1\times294 = 29.4\ \text{g}.

21. (a) From carbide: \text{CaC}_2 + 2\text{H}_2\text{O} \to \text{C}_2\text{H}_2 + \text{Ca(OH)}_2. From 1,2-dibromoethane: \text{BrCH}_2\text{CH}_2\text{Br} + 2\text{KOH(alc.)} \to \text{HC}{\equiv}\text{CH} + 2\text{KBr} + 2\text{H}_2\text{O}, heated. (b) The sp carbon holds the bonding electrons more tightly (50 % s character), so the \text{C}-\text{H} bond is polar and the terminal hydrogen is slightly acidic: 2\text{HC}{\equiv}\text{CH} + 2\text{Na} \to 2\text{HC}{\equiv}\text{CNa} + \text{H}_2; and it gives a white precipitate of silver acetylide, \text{AgC}{\equiv}\text{CAg}, with ammoniacal \text{AgNO}_3. (c) With hydrogen: \text{HC}{\equiv}\text{CH} + \text{H}_2 \rightarrow\ [\text{Ni}] \text{CH}_2{=}\text{CH}_2, then \text{CH}_3\text{CH}_3 with excess. With HCl: \text{HC}{\equiv}\text{CH} + \text{HCl} \to \text{CH}_2{=}\text{CHCl} (vinyl chloride); a second HCl gives \text{CH}_3\text{CHCl}_2. With water: \text{HC}{\equiv}\text{CH} + \text{H}_2\text{O} \rightarrow\ [\text{HgSO}_4,\ \text{H}_2\text{SO}_4] \text{CH}_3\text{CHO}.

22. (a) Raw materials: ammonia and air. (i) Catalytic oxidation: a mixture of 1 part \text{NH}_3 to 10 parts air passes over a Pt–Rh gauze at about 800^{\circ}\text{C}: 4\text{NH}_3 + 5\text{O}_2 \to 4\text{NO} + 6\text{H}_2\text{O} (\Delta H = -905\ \text{kJ}; the heat keeps the gauze hot). (ii) Oxidation of NO: the gas is cooled and mixed with more air: 2\text{NO} + \text{O}_2 \to 2\text{NO}_2. (iii) Absorption: \text{NO}_2 goes up an absorption tower against a descending spray of water: 3\text{NO}_2 + \text{H}_2\text{O} \to 2\text{HNO}_3 + \text{NO} (the NO is re-oxidized and recycled), giving 50–60 % acid. Flow sheet: \text{NH}_3 + filtered air \to mixer \to converter (Pt–Rh gauze) \to heat exchanger and cooler \to oxidation chamber \to absorption tower (water) \to nitric acid storage; the tail gas is recycled or scrubbed. (b) Dilute: 3\text{Cu} + 8\text{HNO}_3 \to 3\text{Cu(NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}. Concentrated: \text{Cu} + 4\text{HNO}_3 \to \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}.

7.3 Model Set 2

Time: 3 hours. Full marks: 75. Attempt all the questions. Atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, Cl = 35.5, Ca = 40. R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}; 1\ \text{F} = 96\,500\ \text{C}. This is an original practice paper, built to the CDC pattern.

Group A — Rewrite the correct option of each question in your answer sheet. (11\times1 = 11)

  1. The number of electrons in \text{Mg}^{2+} (Z = 12) is [1]a8b10c12d14
  2. Which has the largest ionic radius? [1]a\text{S}^{2-}b\text{Cl}^{-}c\text{K}^{+}d\text{Ca}^{2+}
  3. The shape of the carbon dioxide molecule is [1]abentblinearctrigonal planardtetrahedral
  4. The charge needed to deposit one mole of aluminium from \text{Al}^{3+} is [1]a96\,500\ \text{C}b193\,000\ \text{C}c289\,500\ \text{C}d386\,000\ \text{C}
  5. The volume of 0.1 mol of a gas at STP is [1]a2.24\ \text{L}b22.4\ \text{L}c4.48\ \text{L}d0.224\ \text{L}
  6. On raising the temperature of an exothermic equilibrium, the equilibrium shifts [1]atowards the productsbtowards the reactantscnot at alldto the side with more moles of gas
  7. Brown fumes are evolved when copper reacts with [1]adilute \text{H}_2\text{SO}_4bdilute HClcconc. \text{HNO}_3dconc. HCl
  8. The formula of bleaching powder is [1]a\text{CaCl}_2b\text{CaOCl}_2c\text{Ca(OH)}_2d\text{CaCO}_3
  9. The general formula of the alkenes is [1]a\text{C}_n\text{H}_{2n+2}b\text{C}_n\text{H}_{2n}c\text{C}_n\text{H}_{2n-2}d\text{C}_n\text{H}_{n}
  10. The reagents for the Friedel–Crafts alkylation of benzene are [1]a\text{CH}_3\text{OH} and \text{H}_2\text{SO}_4b\text{CH}_4 and Nic\text{CH}_3\text{Cl} and anhydrous \text{AlCl}_3d\text{CH}_3\text{COOH} and \text{ZnCl}_2
  11. The raw materials of the Solvay process are [1]aNaOH and \text{CO}_2b\text{Na}_2\text{SO}_4 and cokecNa and \text{CO}_2dbrine, ammonia and limestone

Group B — Short answer questions. (8\times5 = 40)

  1. State the postulates of Dalton's atomic theory and two points on which it has been modified. [5]
  2. 10\ \text{g} of \text{CaCO}_3 reacts with 100\ \text{mL} of 1\ \text{M} HCl: \text{CaCO}_3 + 2\text{HCl} \to \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}. Find the limiting reactant and the volume of \text{CO}_2 at STP. [5]
  3. Define effective nuclear charge. Explain the trends in atomic radius and first ionization energy across the third period. [5]
  4. Explain why \text{H}_2\text{O} is a liquid while \text{H}_2\text{S} is a gas, and why ice floats on water. [5]
  5. A current of 5\ \text{A} is passed for 1 hour through molten NaCl. Find the mass of sodium and the volume of chlorine at STP. [5]
  6. A 10\ \text{L} vessel at 27^{\circ}\text{C} holds 2\ \text{g} of \text{H}_2 and 14\ \text{g} of \text{N}_2. Find the partial pressure of each gas and the total pressure. [5]
  7. Write the preparation and uses of quicklime and bleaching powder, and the reaction by which bleaching powder bleaches. [5]
  8. Explain geometrical isomerism in but-2-ene and optical isomerism in lactic acid. [5]

Group C — Long answer questions. (3\times8 = 24)

  1. (a) State Dalton's law of partial pressures and show that the partial pressure of a gas is p_i = x_iP. [3](b) A mixture of 2.8\ \text{g} of \text{N}_2 and 3.2\ \text{g} of \text{O}_2 is in a 10\ \text{L} vessel at 27^{\circ}\text{C}. Find the total pressure and the partial pressure of each. [3](c) Find the ratio of the rates of diffusion of \text{N}_2 and \text{O}_2. [2] [8]
  2. (a) Describe the mechanism of the addition of HBr to propene and explain Markovnikov's rule in terms of carbocation stability. [4](b) Write the equation for the preparation of propene from propan-2-ol. [2](c) Describe two tests for unsaturation. [2] [8]
  3. (a) List the stages in the development of a new chemical product. [3](b) Compare batch and continuous processing. [3](c) State two measures to reduce the environmental impact of a chemical plant. [2] [8]

7.3.1 Solutions to Model Set 2

Solutions — Group A

1. (b) — 12 - 2 = 10.

2. (a) — the ions are isoelectronic (18 electrons); the lower the nuclear charge, the larger the ion, and \text{S}^{2-} (Z = 16) is the lowest.

3. (b) — two double bonds, no lone pair, 180^{\circ}.

4. (c) — \text{Al}^{3+} + 3e^{-} \to \text{Al}, 3\times96\,500.

5. (a) — 0.1\times22.4.

6. (b) — heat is a product of the forward reaction; adding heat favours the reverse.

7. (c) — \text{NO}_2 is brown.

8. (b) — calcium oxychloride.

9. (b) — one double bond.

10. (c) — alkyl halide with a Lewis acid catalyst.

11. (d) — \text{NaCl} (brine), \text{NH}_3 and \text{CaCO}_3 (for \text{CO}_2 and lime).

Solutions — Group B (12–19)

12. Postulates: (i) matter consists of indivisible atoms; (ii) atoms of one element are identical in mass and properties; (iii) atoms of different elements differ; (iv) compounds form by combination of atoms in simple whole-number ratios; (v) atoms are neither created nor destroyed in a chemical reaction. Modifications: atoms are divisible into subatomic particles; atoms of the same element may differ in mass (isotopes).

13. Moles of \text{CaCO}_3 = 10/100 = 0.10; moles of HCl = 0.1\times1 = 0.10. 0.10 mol \text{CaCO}_3 needs 0.20 mol HCl, so HCl is limiting. 2 mol HCl gives 1 mol \text{CO}_2; so 0.10 mol HCl gives 0.05 mol \text{CO}_2. Volume = 0.05\times22.4 = 1.12\ \text{L}.

14. Effective nuclear charge Z_{eff} = Z - S is the net positive charge felt by an outer electron after shielding by inner electrons. Across Period 3 (Na to Ar), electrons are added to the same shell, so shielding barely changes while Z increases: Z_{eff} rises, the outer electrons are pulled in, and the atomic radius falls (Na 186 pm to Cl 99 pm). The same pull makes the electron harder to remove, so the first ionization energy generally rises, with small dips at Al (the 3p electron is higher in energy than the 3s) and S (pairing repulsion in 3p^{4}).

15. Oxygen is small and highly electronegative, so water molecules form strong hydrogen bonds between \text{O}-\text{H} and the lone pair of a neighbouring oxygen; this holds the molecules together as a liquid at room temperature (100^{\circ}\text{C} boiling point). Sulphur is larger and less electronegative, so \text{H}_2\text{S} has only weak dipole attractions and boils at -60^{\circ}\text{C}. In ice each oxygen is hydrogen-bonded to four others in an open hexagonal lattice with empty space, so ice is less dense than water (0.92\ \text{g cm}^{-3}) and floats.

16. Q = It = 5\times3600 = 18\,000\ \text{C}. Moles of electrons = 18\,000/96\,500 = 0.1865. \text{Na}^{+} + e^{-} \to \text{Na}: mass = 0.1865\times23 = 4.29\ \text{g}. 2\text{Cl}^{-} \to \text{Cl}_2 + 2e^{-}: moles of \text{Cl}_2 = 0.1865/2 = 0.0933; volume = 0.0933\times22.4 = 2.09\ \text{L}.

17. Moles: \text{H}_2 = 2/2 = 1.0; \text{N}_2 = 14/28 = 0.5. T = 300\ \text{K}. p_{\text{H}_2} = \dfrac{nRT}{V} = \dfrac{1.0\times0.0821\times300}{10} = 2.46\ \text{atm}; p_{\text{N}_2} = \dfrac{0.5\times0.0821\times300}{10} = 1.23\ \text{atm}. By Dalton's law the total is 2.46 + 1.23 = 3.69\ \text{atm}.

18. Quicklime: \text{CaCO}_3 \rightarrow\ [1100\ \text{K}] \text{CaO} + \text{CO}_2; uses: making slaked lime and cement, drying agent for ammonia, flux in metallurgy. Bleaching powder: \text{Ca(OH)}_2 + \text{Cl}_2 \to \text{CaOCl}_2 + \text{H}_2\text{O}; uses: bleaching cotton and paper, disinfecting water. Bleaching action: \text{CaOCl}_2 + \text{H}_2\text{SO}_4 \to \text{CaSO}_4 + \text{Cl}_2 + \text{H}_2\text{O}; the chlorine, with water, gives \text{HOCl}, which bleaches by oxidation.

19. Geometrical isomerism: arises from restricted rotation about a double bond when each doubly bonded carbon has two different groups. But-2-ene has cis (both \text{CH}_3 on the same side, boiling point 4^{\circ}\text{C}, polar) and trans (opposite sides, boiling point 1^{\circ}\text{C}, non-polar, more stable) forms. Optical isomerism: arises when a carbon carries four different groups (a chiral carbon). In lactic acid, \text{CH}_3\text{CH(OH)COOH}, the middle carbon holds H, \text{CH}_3, OH and COOH, giving two mirror-image forms (d and l) that rotate plane-polarized light in opposite directions and are non-superposable.

Solutions — Group C (20–22)

20. (a) In a mixture of non-reacting gases the total pressure is the sum of the partial pressures: P = p_1 + p_2 + \cdots. For gas i: p_i = \dfrac{n_iRT}{V} and P = \dfrac{n_{tot}RT}{V}, so \dfrac{p_i}{P} = \dfrac{n_i}{n_{tot}} = x_i and p_i = x_iP. (b) n_{\text{N}_2} = 2.8/28 = 0.1; n_{\text{O}_2} = 3.2/32 = 0.1; n_{tot} = 0.2. P = \dfrac{0.2\times0.0821\times300}{10} = 0.493\ \text{atm}. Each gas has x = 0.5, so p = 0.246\ \text{atm}. (c) \dfrac{r_{\text{N}_2}}{r_{\text{O}_2}} = \sqrt{\dfrac{M_{\text{O}_2}}{M_{\text{N}_2}}} = \sqrt{\dfrac{32}{28}} = 1.07; nitrogen diffuses slightly faster.

21. (a) Step 1: \text{HBr} is polar; the \pi electrons of propene attack \text{H}^{+}, giving a carbocation. H can add to either carbon: adding to \text{CH}_2 gives the secondary carbocation \text{CH}_3\overset{+}{\text{C}}\text{HCH}_3; adding to \text{CH} gives the primary \text{CH}_3\text{CH}_2\overset{+}{\text{C}}\text{H}_2. Two alkyl groups push electrons (+I) and stabilize the secondary ion, so it forms faster. Step 2: \text{Br}^{-} attacks the carbocation to give \text{CH}_3\text{CHBrCH}_3 (2-bromopropane) as the major product. Markovnikov's rule: the negative part of the reagent goes to the carbon with fewer hydrogens. (b) \text{CH}_3\text{CH(OH)CH}_3 \rightarrow\ [\text{conc. H}_2\text{SO}_4,\ \text{heat}] \text{CH}_3\text{CH}{=}\text{CH}_2 + \text{H}_2\text{O}. (c) Bromine water test: an orange solution is decolourized because of addition of \text{Br}_2 across the double bond. Baeyer's test: pink or purple cold alkaline dilute \text{KMnO}_4 is decolourized (a brown precipitate of \text{MnO}_2 may form) as the alkene is oxidized to a diol.

22. (a) (i) Laboratory research: the reaction, catalyst and conditions are found. (ii) Process design and economic evaluation: flow sheet, material and energy balance, cost estimate. (iii) Pilot plant: a small-scale plant tests the process and safety. (iv) Full-scale design, construction and commissioning. (v) Operation with quality control and monitoring. (b) Batch: a fixed charge of reactants is reacted and the product removed before the next charge; flexible, suited to small-volume products (drugs, dyes); higher labour cost per unit. Continuous: feed enters and product leaves without a break; suited to large-volume products (ammonia, sulphuric acid, NaOH); lower cost per tonne and steadier quality, but less flexible and with a high capital cost. (c) Treat effluent before discharge (neutralization, sedimentation, biological treatment); use scrubbers for \text{SO}_2 and \text{NO}_x; recycle by-products and unreacted material; use catalysts and heat exchangers to save energy.

7.4 Predicted Questions — ranked

The ranking is by teaching hours of the unit and how standard the question is. Each prediction states its reason. Tier A are the questions a paper on this syllabus almost has to contain in some form; Tier B are likely; Tier C are possible. These are judgements, never guarantees.

7.4.1 Tier A — revise first

P1. Mole concept numerical with limiting reactant and percentage yield (unit 2, [5]). Why: stoichiometry has 8 hours and a numerical in every model type. Outline: balance; moles of each reactant; the smaller ratio limits; yield = actual/theoretical \times100. P2. Empirical and molecular formula from percentage composition (unit 2, [5]). Why: named in the syllabus, with a clean answer. Outline: divide percentages by atomic masses, divide by the smallest, n = M/M_{emp}. P3. Quantum numbers and electronic configuration (Aufbau, Pauli, Hund) (unit 3, [5]). Why: eight hours and several named rules. Outline: n, l, m_l, m_s ranges; Cr and Cu exceptions; unpaired electrons. P4. Bohr's postulates and the hydrogen spectrum (unit 3, [5\text{ or }8]). Why: a named bullet with a numerical (the Balmer lines). Outline: three postulates; E_n = -13.6/n^{2}\ \text{eV}; \Delta E = h\nu; Lyman, Balmer, Paschen series; defects of the model. P5. VSEPR shapes and hybridization (unit 5, [5\text{ or }8]). Why: the largest of the general chemistry units after stoichiometry (9 h), and the syllabus lists eleven molecules. Outline: table of \text{BeF}_2, \text{BF}_3, \text{CH}_4, \text{NH}_3, \text{H}_2\text{O}, \text{PCl}_5, \text{SF}_6 with pairs, shape and angle. P6. Balancing redox by oxidation number or ion–electron method; Faraday's laws (unit 6, [5]). Why: a standard type with a computed answer. Outline: find O.N. change; balance atoms and charge; m = ZIt. P7. Gas laws numerical; the ideal gas equation and its derivation (unit 7, [5\text{ or }8]). Why: a recurring numerical type. Outline: PV = nRT from Boyle, Charles and Avogadro; R in two units; Graham's and Dalton's laws. P8. K_c, K_p and Le Chatelier's principle (unit 8, [5]). Why: a small unit, but a clean standard question. Outline: K_p = K_c(RT)^{\Delta n}; effects of P, T, concentration and catalyst. P9. Haber's process (unit 17, [8]). Why: the syllabus names principle and flow sheet, and the manufactures unit has 11 hours. Outline: reaction, conditions with reasons, flow sheet, uses. P10. Contact process or Ostwald's process (unit 17, [8]). Why: both named, and sulphuric and nitric acid are the largest-volume acids. Outline: three stages each with reactions, catalyst and temperature. P11. Markovnikov's rule and the peroxide effect (unit 14, [5]). Why: named in the syllabus; a mechanism question. Outline: carbocation stability; HBr only for the peroxide effect. P12. Nitration and Friedel–Crafts reactions of benzene; directing effects (unit 15, [5]). Why: the heart of a six-hour unit. Outline: nitrating mixture and \text{NO}_2^{+}; \text{AlCl}_3 with RX or RCOCl; o,p and m directors.

7.4.2 Tier B — likely

P13. Periodic trends with reasons (unit 4, [5]). Why: five hours, six trends named. Outline: Z_{eff}; radius, IE, EA and EN across and down; exceptions at N/O and Mg/Al. P14. Hydrogen bonding and its consequences (unit 5, [5]). Why: named in the syllabus; applies to water, HF and DNA. Outline: conditions; effect on boiling point; ice floats; examples. P15. Ammonia and nitric acid: preparation and properties (unit 9, [5\text{ or }8]). Why: nitrogen carries 5 hours inside the largest unit. Outline: laboratory preparation, reactions with HCl, \text{Cl}_2, \text{CuSO}_4; \text{HNO}_3 with Cu, Zn, S, C. P16. Ring test; halogens and their displacement reactions (unit 9, [5]). Why: five hours for halogens, one test named. Outline: brown ring equations; \text{Cl}_2 preparation and displacement; halide tests. P17. Metallurgy: calcination, roasting, smelting; Down's process (unit 10, [5\text{ or }8]). Why: a ten-hour unit with named processes. Outline: definitions with equations; Down's cell; refining by poling and electrolysis. P18. IUPAC naming and Lassaigne's test (unit 13, [5]). Why: a ten-hour unit with an exact procedure. Outline: longest chain, lowest locants; sodium extract; Prussian blue, black PbS, curdy white AgCl. P19. Isomerism: chain, position, functional, geometrical and optical (unit 13, [5]). Why: named in the syllabus; answered with structures. Outline: \text{C}_4\text{H}_{10}\text{O} isomers; but-2-ene; lactic acid. P20. Preparation of alkanes and alkenes; the acidity of ethyne (unit 14, [5\text{ or }8]). Why: eight hours with named reactions. Outline: Wurtz, decarboxylation, Kolbe; dehydration of alcohols; \text{AgC}{\equiv}\text{CAg}.

7.4.3 Tier C — possible

P21. Roles of metal ions and toxic metals (unit 11, [5]). Why: the whole content of a three-hour unit. Outline: Na/K pump; Mg in chlorophyll; Fe in haemoglobin; Hg, Pb, As, Cd effects. P22. Ozone: preparation, structure, depletion (unit 9, [5]). Why: named in the syllabus and topical. Outline: silent discharge; resonance structure; CFC radical chain. P23. Diaphragm cell for sodium hydroxide (unit 17, [5\text{ or }8]). Why: named in the syllabus. Outline: electrode reactions; the diaphragm's role; evaporation; uses. P24. Batch against continuous processing; environmental impact (unit 16, [5]). Why: the four-hour applied fundamentals unit has few other standard questions. Outline: comparison table; pollutants and controls.

Chapter summary

  • The Chemistry paper is 75 marks: 11\times1, 8\times5 and 3\times8; General and Physical (48 h) and Inorganic (35 h) make up two-thirds of the teaching time, and non-metals (22 h) is the heaviest unit.
  • Two complete model sets, built to the grid, are solved in full; every equation is balanced and every numerical shows the formula, substitution and unit.
  • Recurring numerical types: mole and limiting reactant; empirical formula; Faraday's laws; gas laws and Dalton's law; K_c and K_p.
  • Recurring reactions and processes: nitrogen, sulphur and halogens; Down's process; Markovnikov and peroxide; nitration and Friedel–Crafts; Haber, Ostwald, contact, diaphragm, Solvay.
  • The twenty-four predicted questions are ranked by unit hours and standardness; they are judgements, not leaks.
  • In the hall: balance every equation and write conditions over the arrow; write the formula before substituting; draw the flow sheet and label it.

Practice questions

Group A — Multiple choice (1 mark each)

  1. The number of atoms in 0.1 mol of \text{CH}_4 is [1]a3.0\times10^{22}b6.0\times10^{22}c3.0\times10^{23}d6.0\times10^{23}
  2. The shape of the \text{PCl}_5 molecule is [1]aoctahedralbsquare pyramidalctrigonal bipyramidaldtetrahedral
  3. For \text{H}_2 + \text{Cl}_2 \rightleftharpoons 2\text{HCl}, increasing the pressure [1]ashifts it rightbshifts it leftchas no effectdlowers K_c
  4. In the Haber process the promoter commonly used is [1]aMobPtc\text{V}_2\text{O}_5dNi
Answers to Group A

1. (c) — each \text{CH}_4 has 5 atoms: 0.1\times5\times6.022\times10^{23} = 3.0\times10^{23}.

2. (c) — five bond pairs, sp^{3}d.

3. (c) — \Delta n = 0.

4. (a) — molybdenum.

Group B — Short answer (5 marks each)

  1. Find the mass of \text{NaOH} needed to neutralize 0.5 mol of \text{H}_2\text{SO}_4. [5]
  2. Find the volume of \text{H}_2 at STP evolved when 6.5\ \text{g} of zinc reacts with dilute HCl (\text{Zn} = 65). [5]
  3. Write the equations for the action of heat on \text{CaCO}_3 and on \text{CaSO}_4\cdot2\text{H}_2\text{O} at 120^{\circ}\text{C}. [5]
  4. Give the reagent and condition for converting(a) ethanol to ethene,(b) benzene to nitrobenzene. [5]
Answers to Group B

1. \text{H}_2\text{SO}_4 + 2\text{NaOH} \to \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}; 0.5 mol acid needs 1.0 mol NaOH = 40\ \text{g}.

2. \text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2; 6.5/65 = 0.1 mol Zn gives 0.1 mol \text{H}_2 = 2.24\ \text{L}.

3. \text{CaCO}_3 \to \text{CaO} + \text{CO}_2 (at about 1100\ \text{K}); \text{CaSO}_4\cdot2\text{H}_2\text{O} \to \text{CaSO}_4\cdot\tfrac{1}{2}\text{H}_2\text{O} + 1\tfrac{1}{2}\text{H}_2\text{O} (plaster of Paris).

4. (a) Conc. \text{H}_2\text{SO}_4 at 170^{\circ}\text{C} (443 K). (b) Conc. \text{HNO}_3 and conc. \text{H}_2\text{SO}_4 at about 323\ \text{K}.

Group C — Long answer (8 marks each)

  1. Describe the manufacture of sulphuric acid by the contact process. [8]
  2. Describe benzene's structure and the evidence for resonance. [8]
Answers to Group C

1. Stages: (i) burn sulphur or roast pyrites to get \text{SO}_2: \text{S} + \text{O}_2 \to \text{SO}_2; purify and dry the gas (impurities poison the catalyst). (ii) Pass \text{SO}_2 with excess air over \text{V}_2\text{O}_5 at 450^{\circ}\text{C} and 1–2 atm: 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3, \Delta H = -197\ \text{kJ}; heat is removed with exchangers to favour the product. (iii) Absorb \text{SO}_3 in conc. \text{H}_2\text{SO}_4 (98 %) to give oleum, \text{H}_2\text{S}_2\text{O}_7. (iv) Dilute the oleum with water to the required strength. Water is not used for direct absorption because it forms a mist of acid. Flow sheet: furnace \to purifier \to dryer \to preheater \to converter \to absorber \to dilution tank.

2. Benzene, \text{C}_6\text{H}_6, is a planar hexagon of six sp^{2} carbons, each bonded to one hydrogen; the six unhybridized p orbitals overlap sideways to give a delocalized \pi cloud of six electrons above and below the ring. Kekulé's two forms (alternating double bonds) are resonance structures; the real molecule is a hybrid. Evidence: all C–C bonds are equal (139\ \text{pm}); only one 1,2-disubstituted product exists; the heat of hydrogenation (-208\ \text{kJ mol}^{-1}) is about 150\ \text{kJ mol}^{-1} less than three times that of cyclohexene (-360); benzene undergoes substitution, not addition, and does not decolourize bromine water or Baeyer's reagent.