Chemistry: Unit-wise Objective Question Bank

1. (b) — H₃PO₃ is dibasic, and here both replaceable H are used (Na₂HPO₃), so E = 82/2 = 41. Option (a) forgets to divide; (c) divides by 3 because the formula has three H, but one of them is bonded to P and is not acidic.

2. (a) — 5 g per 100 mL = 50 g L⁻¹; equivalent weight = 106/2 = 53; N = 50/53 = 0.94. Option (c) is the molarity (50/106), which forgets the n-factor, and (d) doubles the n-factor.

3. (d) — Cl⁻ + Ag⁺ → AgCl↓. This is precipitation titration (Mohr's method), with K₂CrO₄ only as the indicator, so K₂CrO₄ alone cannot be the titrant.

4. (a) — Na₂CO₃ is a weak base; at complete neutralisation the solution holds H₂CO₃ (pH about 4), which methyl orange signals. Phenolphthalein changes colour at the half-way stage, NaHCO₃, so it would give a wrong end point.

5. (c) — Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O takes six electrons, so E = 294/6 = 49. Option (b) uses three electrons, the n-factor of dilute-acid reduction of a different oxidant.

6. (d) — MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so E = 158/5 = 31.6. Option (c) is the neutral-medium value (three electrons).

7. (a) — M = (10 × 0.91 × 25)/17 = 227.5/17 = 13.4 mol L⁻¹. Option (d) uses 34 as the molar mass, which is 2 × 17.

8. (b) — equivalents = 100 × 0.5 + 20 × 0.1 + 50 × 0.2 = 50 + 2 + 10 = 62 milli-equivalents in 170 mL, so N = 0.3647 and pH = −log 0.3647 = 0.44. Option (c) is the pH if the volumes were forgotten and the three normalities were simply averaged.

9. (b) — E = 126/2 = 63; N = (3.15/63)/0.5 = 0.05/0.5 = 0.10 N. Option (a) forgets that the volume is 0.5 L, not 1 L.

10. (d) — oxalic acid crystals are pure, stable and of exactly known formula. NaOH absorbs CO₂ and water, HCl is volatile, and KMnO₄ is reduced by traces of organic matter, so all three are secondary standards.

11. (c) — one drop of excess MnO₄⁻ makes the solution permanently pink, so KMnO₄ is a self-indicator. The solution is warmed to about 60–70 °C because the first reaction is slow at room temperature, and dilute H₂SO₄ supplies the H⁺ the half-reaction needs.

12. (b) — N₁V₁ = N₂V₂ gives N₂ = (0.100 × 25.0)/20.0 = 0.125 N. Option (a) inverts the ratio of the volumes.

1. (c) — the acid contributes 10⁻⁸ M H⁺, but water supplies about 9.5 × 10⁻⁸ M more, so [H⁺] = 1.05 × 10⁻⁷ and pH = 6.98. An acid can never give a pH of 8, which is the trap in option (a).

2. (b) — mixing doubles the volume and NaNO₃ is diluted from 1 M to 0.5 M. Neither KCl nor the spectator ions contribute nitrate. Option (c) forgets the dilution.

3. (a) — ionisation of water is endothermic, so Kw rises with temperature (about 3.8 × 10⁻¹³ at 100 °C). Kw = 10⁻¹⁴ is valid only at 25 °C.

4. (b) — [H⁺] from the acid is 2 × 10⁻⁷ and water adds a little more: [H⁺]² − 2 × 10⁻⁷[H⁺] − 10⁻¹⁴ = 0, so [H⁺] = 2.41 × 10⁻⁷ and pH = 6.62. Option (a) ignores the fact that H₂SO₄ gives two H⁺.

5. (b) — s = √(9.3 × 10⁻⁸) = 3.05 × 10⁻⁴ mol L⁻¹; multiplying by M = 100 gives 3.05 × 10⁻² g L⁻¹. Option (a) stops at mol L⁻¹.

6. (c) — A₂B gives 2s of A⁺ and s of B²⁻, so Ksp = (2s)²(s) = 4s³.

7. (a) — α = √(Ka/C) = √(4.8 × 10⁻¹⁰/0.01) = √(4.8 × 10⁻⁸) = 2.2 × 10⁻⁴. Option (c) uses √Ka alone.

8. (a) — Ka ≈ Cα² = 0.1 × (0.0134)² = 1.8 × 10⁻⁵. Option (d) squares only part of the expression.

9. (d) — 10⁻⁶ M OH⁻ becomes 10⁻⁹ M, which is far below the 10⁻⁷ M that water itself supplies, so the solution is essentially neutral: [OH⁻] = 1.005 × 10⁻⁷, pH = 7.002. A base can never be diluted into an acid (option a).

10. (c) — pH = 4.74 + log(0.1/0.2) = 4.74 − 0.30 = 4.44. Option (b) inverts the ratio of salt to acid.

11. (b) — the conjugate base is formed by removing one proton: H₂PO₄⁻ → HPO₄²⁻ + H⁺. Option (a) is the conjugate acid.

12. (a) — after mixing, [Ba²⁺] = [SO₄²⁻] = 1 × 10⁻⁴ M, so Qsp = 10⁻⁸ > 10⁻¹⁰ and BaSO₄ precipitates. Option (b) forgets to halve each concentration.

13. (b) — CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; the salt of a weak acid and a strong base hydrolyses to an alkaline solution.

1. (c) — the rise is 90 °C, that is nine steps of 10 °C, so the factor is 2⁹ = 512. Option (d) comes from counting ten steps instead of nine.

2. (a) — rate = k[A], so [A] = 2.4 × 10⁻⁵ / 3.0 × 10⁻⁵ = 0.8 mol L⁻¹.

3. (a) — rate = k[A]⁰ = k only when the exponent is zero; the rate is then independent of concentration.

4. (b) — only for first order is t½ = 0.693/k, with no concentration term. For zero order t½ ∝ a and for second order t½ ∝ 1/a.

5. (b) — order comes from the experimental rate law, not from the balancing coefficients. Option (c) is the classic error of reading the 2 in 2N₂O₅ as the order.

6. (b) — molecularity counts the molecules colliding in a single elementary step, so it must be a whole number 1, 2 or 3. Order is an experimental quantity and may be zero or fractional.

7. (c) — the general unit is mol^(1−n) L^(n−1) s⁻¹; putting n = 2 gives L mol⁻¹ s⁻¹.

8. (b) — a catalyst opens a lower-energy route, lowering Ea in both directions, but reactants and products are unchanged so ΔH, ΔG and K are untouched.

9. (a) — ln k = ln A − Ea/RT. Option (c) would be the slope of a log₁₀ k plot.

10. (c) — k = 0.693/10 = 0.0693 s⁻¹, and t = (2.303/k) log 10 = 2.303/0.0693 = 33.2 s. It is not simply 5 × t½.

11. (c) — sucrase is an enzyme, a biological catalyst; the substrate must fit its active site.

1. (b) — reverse the second equation and add it to the first: C + O₂ → CO₂ (−x) plus CO₂ → CO + ½O₂ (+y) gives C + ½O₂ → CO, so ΔHf = (y − x) kJ. Option (a) is the trap of adding the two combustion enthalpies instead of subtracting.

2. (b) — spontaneity is decided by the sign of ΔG at the actual conditions. Option (d) is the common trap: an exothermic reaction can still be non-spontaneous if TΔS opposes it strongly enough.

3. (a) — at equilibrium ΔG = 0. ΔG° is a fixed number for the standard state and equals −2.303RT log K, which is zero only in the special case K = 1.

4. (b) — the system absorbs heat, so its enthalpy rises and ΔH > 0. Sign conventions are worth one mark almost every year.

5. (b) — at constant volume no expansion work is done, so all the heat goes into internal energy.

6. (a) — "adiabatic" means no heat crosses the boundary; work may still be done, so ΔU is not zero.

7. (c) — internal energy is a state function and the system returns to its initial state.

8. (a) — Hess's law of constant heat summation, a direct consequence of enthalpy being a state function.

9. (c) — gases have far more positional and thermal disorder than liquids or solids. Dry ice is solid CO₂.

10. (a) — ΔG = ΔH − TΔS is negative for every T when ΔH < 0 and ΔS > 0.

11. (b) — from ΔG° = −2.303RT log K, log K > 0 requires ΔG° < 0.

12. (a) — both are strong acid plus strong base, so the reaction is the same one, H⁺ + OH⁻ → H₂O.

13. (b) — work = −p(ext)ΔV and the external pressure is zero, so no work is done however large the expansion.

1. (c) — the more negative the reduction potential, the more readily the metal is oxidised, so reactivity falls as E° rises: R (−2.71) > P (−0.76) > Q (+0.34) > S (+0.80). Option (a) is the same list read the wrong way round, which is the usual slip.

2. (b) — reducing power increases as E° becomes more negative, so the most negative value heads the list.

3. (b) — discharge is Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O: sulphuric acid is consumed and its density falls, which is why the acid density is used to test the state of charge.

4. (b) — in a galvanic cell electrons are pushed out of the anode, making it the negative terminal. In electrolysis the anode is connected to the positive terminal of the supply; oxidation still happens there, but the sign flips. This sign reversal is the single most-missed point in the unit.

5. (b) — it is an arbitrary zero assigned by agreement, not a measured quantity; all other potentials are measured against it.

6. (a) — E°cell = E°cathode − E°anode = 0.80 − 0.34 = 0.46 V. Option (c) is the Zn–Cu value.

7. (b) — the negative sign makes a positive cell potential correspond to a spontaneous reaction.

8. (b) — from the Nernst equation E = 1.10 − (0.0591/2) log([Zn²⁺]/[Cu²⁺]); raising the product-ion concentration lowers E.

9. (b) — it is a calibrated, easily maintained electrode whose potential (+0.242 V saturated) is known relative to SHE, which is the primary standard.

10. (b) — H₂ and O₂ combine to give only water, which is why fuel cells are attractive for pollution-free power.

11. (c) — electrode potential is an intensive property; it does not scale with the amount of substance, although ΔG° = −nFE° does.

1. (b) — the d subshell of the penultimate shell is being filled while the outermost s holds one or two electrons. Option (c) is the configuration of the f-block.

2. (b) — variable oxidation states let the metal form unstable intermediates, and partly filled d orbitals let reactants adsorb on the surface. Colour (d) is a consequence of the same d orbitals, not the cause of catalysis.

3. (b) — Ni²⁺ is 3d⁸, partly filled, and absorbs in the red, appearing green. Zn²⁺, Sc³⁺ and Ag⁺ are all d¹⁰ or d⁰ and colourless.

4. (c) — manganese, 3d⁵4s², can lose all seven outer electrons, as in KMnO₄ and Mn₂O₇.

5. (c) — Zn²⁺ is 3d¹⁰, with no vacancy for a d–d transition. "All transition-metal ions are coloured" is the misconception this question is built on.

6. (b) — a transition element must have an incompletely filled d subshell in the atom or in a stable ion; zinc has neither.

7. (b) — the three t₂g orbitals lie between the axes and are stabilised by 0.4Δo, while the two e(g) orbitals point at the ligands and are raised by 0.6Δo.

8. (b) — CN⁻ is a strong-field ligand, so the 3d⁸ electrons pair up, freeing one 3d orbital for dsp² hybridisation; the ion is square planar and diamagnetic. With the weak-field Cl⁻ the same metal gives tetrahedral [NiCl₄]²⁻.

1. (b) — froth flotation works on sulphide ores, whose surfaces are wetted by oil rather than water. Haematite (oxide) is concentrated magnetically or by washing.

2. (b) — hot air blown through molten matte oxidises FeS to slag and then Cu₂S to blister copper: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂, then 2Cu₂O + Cu₂S → 6Cu + SO₂.

3. (b) — ammonia first precipitates Cu(OH)₂, which then dissolves in excess ammonia as the deep blue tetraammine complex.

4. (c) — CuFeS₂ supplies most of the world's copper; the others are minor ores.

5. (a) — all five water molecules are lost by 230 °C, leaving white anhydrous CuSO₄. Black CuO needs about 720 °C.

6. (b) — cuprous oxide is red; cupric oxide CuO is black.

7. (c) — zinc blende is the sulphide; ZnCO₃ is calamine and ZnO is zincite.

8. (a) — "white vitriol" is the zinc salt; green vitriol is FeSO₄·7H₂O and blue vitriol is CuSO₄·5H₂O.

9. (b) — zinc oxide, which sublimes as a woolly white mass; it is yellow while hot and white on cooling.

10. (b) — Zn + 2NaOH → Na₂ZnO₂ + H₂↑; zinc reacts with both acids and alkalis because it is amphoteric.

11. (c) — HgS, bright red, roasted in air to give the metal directly.

12. (c) — calomel Hg₂Cl₂ is insoluble and harmless enough to have been used medicinally, while the soluble HgCl₂ is a violent poison. Solubility is what makes the difference.

13. (b) — excess iodide dissolves the scarlet HgI₂ formed first, giving potassium tetraiodomercurate(II), the basis of Nessler's reagent.

14. (a) — limited SnCl₂ reduces Hg(II) only as far as Hg(I), giving white Hg₂Cl₂; excess reduces it to grey metallic mercury.

15. (b) — haematite Fe₂O₃ contains about 70 % iron and is the main commercial source.

16. (c) — limestone decomposes to CaO, which takes up the silica gangue: CaO + SiO₂ → CaSiO₃.

17. (b) — the acid oxidises the surface to a thin, coherent film of Fe₃O₄ which seals the metal; this is why conc. HNO₃ can be carried in iron containers.

18. (d) — wrought iron is the purest commercial form, under 0.25 % carbon; cast iron holds about 4 %.

19. (b) — zinc has the more negative electrode potential, so it is oxidised in preference to iron; the protection continues even where the coating is scratched.

20. (b) — zinc is more electropositive than silver and displaces it: 2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag.

21. (c) — Ag₂CrO₄ is brick-red; AgCl is white and AgI is yellow.

22. (b) — ammonia dissolves the brown Ag₂O first formed, giving the diamminesilver(I) ion that is reduced by aldehydes to a silver mirror.

1. (b) — reduction removes one chlorine: CHCl₃ + 2[H] → CH₂Cl₂ + HCl. Option (d) is the oxidation product formed in air.

2. (d) — SN1 goes through a carbocation, and the tertiary cation is the most stable, so the rate order follows carbocation stability.

3. (a) — SN2 needs the nucleophile to attack the back of the carbon; the less crowded that carbon is, the faster the reaction. The SN1 and SN2 orders are exact opposites, which is precisely what the examiner is testing.

4. (b) — alcoholic KOH favours elimination; aqueous KOH would have given propan-2-ol.

5. (b) — the cyanide ion of KCN, an ionic salt, attacks through carbon, giving the nitrile CH₃CH₂CN, which is propanenitrile.

6. (b) — AgNO₂ is largely covalent, so the nitrite attacks through nitrogen giving the nitroalkane; ionic KNO₂ would have given the nitrite ester.

7. (c) — two ethyl groups join, so the product has 2 × 2 = 4 carbons.

8. (a) — CHCl₃ + HNO₃ → CCl₃NO₂ + H₂O; chloropicrin is a tear gas.

9. (b) — two molecules lose six chlorines to the silver, joining as 2CHCl₃ + 6Ag → C₂H₂ + 6AgCl.

10. (b) — slow photo-oxidation gives carbonyl chloride; the ethanol added destroys any phosgene formed and the full bottle excludes air.

11. (c) — thionyl chloride gives SO₂ and HCl, both gases, so the haloalkane is left pure; this is Darzen's process.

12. (b) — the C–I bond is the weakest and breaks most easily; C–F is the strongest.

13. (b) — a tertiary substrate ionises to a planar carbocation which is attacked equally from both faces.

1. (b) — chloral, CCl₃CHO (trichloroethanal), condenses with two molecules of chlorobenzene.

2. (b) — the lone pair on chlorine is delocalised into the ring, shortening and strengthening the C–Cl bond; the sp² carbon also holds the halogen more tightly.

3. (b) — copper(I) chloride is the Sandmeyer reagent; copper powder alone would make it the Gattermann reaction.

4. (c) — two rings couple in the Fittig reaction.

5. (c) — the Dow process needs severe conditions precisely because the C–Cl bond is so unreactive.

6. (b) — chlorine deactivates the ring but still directs ortho and para through its lone pair; the para isomer predominates.

1. (b) — a primary alcohol RCH₂OH → RCOOH gains one O (16) and loses two H (2), a net +14. An aldehyde would gain 16 and a ketone would not give an acid of the same chain at all.

2. (c) — the tertiary nitroalkane has no α-hydrogen, so it cannot form the nitrolic acid or pseudonitrole that give the colours.

3. (c) — tertiary alcohols form the most stable carbocation and react at once; primary alcohols need heating.

4. (b) — CaO takes up the last 4.4 % of water chemically, which simple distillation cannot remove because the mixture is an azeotrope.

5. (b) — 95.6 % by mass, the constant-boiling mixture with water.

6. (b) — a petrol–ethanol blend used as a motor fuel.

7. (b) — the lower temperature favours intermolecular dehydration to the ether; 443 K gives the alkene.

8. (b) — a secondary alcohol is oxidised to a ketone with no loss of carbon.

9. (c) — the iodoform test needs a CH₃CH(OH)– group (or CH₃CO–). Methanol and propan-1-ol have neither; ethanol is the one primary alcohol that responds.

10. (b) — copper catalyses dehydrogenation, not dehydration; a primary alcohol loses H₂ to give the aldehyde.

11. (b) — phenol gives a violet colour with neutral ferric chloride and ethanol gives none. Sodium reacts with both, so (a) cannot distinguish them.

1. (c) — a carboxylic acid liberates CO₂ with sodium hydrogen carbonate; phenol, being weaker than carbonic acid, does not. Both dissolve in NaOH and both react with sodium, so (a) and (b) fail. This "acidic enough for NaOH but not for NaHCO₃" line is the whole point of the question.

2. (b) — the diazonium group is displaced by water on warming, releasing nitrogen.

3. (b) — chloroform with alkali generates dichlorocarbene, which formylates the ring at the ortho position.

4. (b) — carbon dioxide under pressure gives sodium salicylate, and acidification gives salicylic acid, the raw material for aspirin.

5. (a) — the violet iron(III)–phenoxide complex; this is the standard identification test for a phenolic –OH.

6. (c) — the –OH group activates the ring so strongly that all three ortho/para positions are substituted, even in cold water and without a catalyst.

7. (b) — the carboxylate ion is stabilised by two equivalent oxygens, the phenoxide by the ring only, and the alkoxide not at all.

1. (b) — the phenoxide oxygen attacks the methyl carbon, giving methoxybenzene (anisole); this is the aromatic Williamson synthesis.

2. (a) — the alkoxide is a strong base as well as a nucleophile, so with 2° and 3° halides elimination competes and finally wins.

3. (b) — the iodide attacks the smaller alkyl group and the oxygen leaves as an alcohol; with excess hot HI that alcohol is converted to a second iodoethane.

4. (b) — attack always occurs at the alkyl carbon, never at the sp² ring carbon, so the aryl group keeps the oxygen.

5. (c) — auto-oxidation at the carbon next to oxygen gives peroxides which concentrate in the still and can detonate.

1. (b) — benzene is acylated or oxidised through to benzaldehyde (Y), and the Perkin condensation with ethanoic anhydride and sodium ethanoate gives 3-phenylprop-2-enoic acid, cinnamic acid.

2. (b) — ethanal is the only aldehyde carrying a CH₃CO– group, so it answers to both tests. Propanone gives iodoform but no mirror; methanal and propanal give a mirror but no iodoform.

3. (b) — 2,4-DNP detects the carbonyl group in both aldehydes and ketones, so it cannot distinguish them; only Tollens' can.

4. (c) — aromatic aldehydes do not reduce Fehling's solution, although they do give a silver mirror with Tollens'. This exception is examined almost every year.

5. (b) — free silver plates out on the glass as a mirror; Fehling's gives the red-brown Cu₂O instead.

6. (c) — methanal has no α-hydrogen, so it cannot enolise and instead disproportionates to methanol and sodium formate.

7. (a) — the base removes an α-hydrogen to form the carbanion that attacks the second carbonyl. Cannizzaro is exactly the opposite requirement.

8. (a) — zinc amalgam with concentrated HCl, so it is used for compounds stable to acid; Wolff–Kishner is the basic alternative.

9. (b) — the poisoned palladium catalyst stops the reduction at the aldehyde stage.

10. (a) — cyanide adds across the C=O to give the cyanohydrin CH₃CH(OH)CN.

11. (a) — the sodium bisulphite adduct crystallises out and is decomposed later with dilute acid or alkali to return the pure carbonyl compound.

12. (a) — the double bond breaks in the middle of the chain, giving two CH₃CHO molecules.

13. (a) — a 40 % solution in water, used as a preservative and disinfectant.

14. (b) — six molecules of methanal and four of ammonia give hexamethylenetetramine, used as a urinary antiseptic and in making RDX.

15. (a) — cyanide ion catalyses the coupling of two aromatic aldehyde molecules to give the α-hydroxy ketone benzoin.

16. (b) — the test needs CH₃CO– or CH₃CH(OH)–; propanal (CH₃CH₂CHO) and propan-1-ol have the methyl one carbon too far from the functional group.

1. (b) — C₃H₄O₄ is propanedioic (malonic) acid; heating it loses CO₂ to give ethanoic acid (Y), and P₂O₅ dehydrates two molecules of that to ethanoic anhydride.

2. (d) — three chlorines pull the most electron density away from the carboxylate, stabilising the anion best.

3. (b) — HCOOH is the one acid in which the carboxyl carbon still carries a hydrogen, so it behaves partly as an aldehyde.

4. (c) — only an acid stronger than carbonic acid displaces CO₂ from a hydrogen carbonate; phenol is too weak.

5. (a) — the enol of the acid halide is halogenated next to the carbonyl.

6. (a) — the carbanion from one ester attacks the carbonyl of the other, giving the β-keto ester ethyl 3-oxobutanoate.

7. (b) — the Hofmann bromamide degradation loses the carbonyl carbon, so a two-carbon amide gives a one-carbon amine.

8. (d) — chloride is the best leaving group of the four and the carbonyl carbon is the most electrophilic.

9. (b) — lithium aluminium hydride is powerful enough to take the acid all the way down to the primary alcohol.

10. (b) — the carboxyl group withdraws electrons, deactivating the ring and directing meta.

11. (b) — alkaline hydrolysis gives the salt, which cannot react back with the alcohol, so the reaction goes to completion.

12. (b) — the lone pair on nitrogen gives a weak base and the N–H a weak acid, so amides form salts with both strong acids and sodium.

13. (a) — the nitrile passes through the amide and then the acid, with ammonia released.

14. (b) — the Grignard carbon attacks CO₂, so the chain grows by one carbon.

1. (b) — tin with concentrated hydrochloric acid is the standard acidic reduction, and it takes the nitro group all the way to the amine.

2. (c) — the N-phenylhydroxylamine formed first rearranges in the strongly acidic medium (the Bamberger rearrangement) to p-aminophenol.

3. (b) — a neutral medium is only mildly reducing, so the reduction stops at the hydroxylamine stage.

4. (b) — –NO₂ withdraws electrons from the ortho and para positions, leaving meta as the least deactivated site.

5. (b) — the pseudonitrole gives a blue colour; the primary compound gives the red nitrolic acid salt and the tertiary compound, having no α-hydrogen, gives nothing.

6. (c) — both are C₂H₅NO₂, but one has a C–N bond and the other a C–O–N linkage, so they belong to different functional classes.

1. (d) — in water the order is 2° > 1° > 3° > NH₃, so diethylamine leads. Aniline (b) is by far the weakest, because its lone pair is tied up in the ring.

2. (a) — the carbylamine (isocyanide) test is given only by primary amines; 2° and 3° amines give nothing.

3. (b) — nitrous acid generated in situ at 273–278 K converts the –NH₂ group into –N₂⁺Cl⁻.

4. (a) — the ester reacts differently with each class: a solid oxamide from the 1° amine, a liquid oxamic ester from the 2°, and nothing from the 3°.

5. (c) — the N–H of a secondary amine is nitrosated; there is no hydrogen left for the further steps that release nitrogen.

6. (b) — resonance delocalisation into the benzene ring reduces the electron density on nitrogen, so it accepts a proton much less readily.

7. (a) — the aliphatic diazonium ion is unstable even at low temperature and decomposes at once, releasing nitrogen quantitatively.

8. (c) — the strongly activating –NH₂ group substitutes all three available ortho and para positions without any catalyst.

9. (b) — acetylation reduces the activating power of the nitrogen and blocks oxidation by the nitrating mixture, so the para product can be isolated.

10. (a) — electrophilic attack by the diazonium ion at the para position of phenol gives p-hydroxyazobenzene, an orange dye.

11. (b) — the nitrile carbon is retained, so CH₃CN gives CH₃CH₂NH₂; the Hofmann route would have shortened the chain instead.

1. (b) — K[PtCl₃(C₂H₄)]·H₂O, prepared by Zeise in 1830; the ethene is held side-on by its π electrons.

2. (b) — sodium ethoxide has a Na–O bond, not a Na–C bond, so it is simply a salt. Every other compound listed has a direct metal–carbon bond, including the carbonyl of nickel.

3. (a) — only methanal adds to give RCH₂OH; ethanal would give a secondary alcohol and propanone a tertiary one.

4. (b) — water is an acid towards the carbanion-like carbon, which is protonated to the alkane. This is why the ether and the apparatus must be perfectly dry.

5. (b) — magnesium is far more electropositive than carbon, so the carbon carries a partial negative charge and attacks electrophilic carbonyl carbons.

1. (a) — polytetrafluoroethene is made by the addition polymerisation of CF₂=CF₂; the C–F bonds give it its non-stick, chemically inert surface.

2. (c) — nylon-6,6 loses a molecule of water at every amide link; the other three simply add.

3. (a) — phenol and methanal condense to a cross-linked, three-dimensional thermosetting network.

4. (b) — a butadiene–styrene copolymer with rubber-like elasticity.

5. (a) — indigo is insoluble, so it is reduced in a vat to the soluble leuco form, taken up by the fibre, and re-oxidised in air.

6. (b) — it lowers fever and relieves pain; it kills no microorganisms, so it is not an antibiotic.

7. (b) — 2,4-dichlorophenoxyacetic acid is a selective weedkiller that acts as a synthetic plant hormone.

1. (b) — gypsum reacts with tricalcium aluminate to form a coating of ettringite that delays hydration, giving workers time to place the concrete.

2. (a) — tricalcium silicate hydrates within about a week; C₂S continues for months and supplies the final strength.

3. (d) — about 1673–1773 K, hot enough to part-fuse the charge into clinker nodules.

4. (b) — the pozzolana replaces part of the clinker, lowering the heat of hydration and improving resistance to sulphates.

5. (c) — "calcareous" means lime-supplying; clay and shale are the argillaceous (silica and alumina) component.

6. (c) — Himal Cement at Chobhar, commissioned in 1974.

7. (b) — decarbonation of limestone is complete before the charge reaches the hottest, clinkering zone.

1. (c) — lignin encrusts the fibres and prevents the hydrogen bonds between cellulose chains that give paper its strength. Removing cellulose would destroy the paper.

2. (a) — caustic soda with sodium sulphide dissolves lignin and gives an unusually strong ("kraft" means strength) pulp.

3. (d) — nothing is dissolved away, so almost all the wood is recovered; the price is a weak, dark, short-lived paper.

4. (a) — rosin precipitated by alum coats the fibres and makes the sheet resistant to the spread of aqueous ink.

5. (a) — fillers occupy the spaces between fibres, raising opacity and smoothness, although they slightly reduce strength.

6. (b) — grams per square metre; ordinary copier paper is about 70–80 GSM.

1. (b) — deflection depends on charge-to-mass ratio; the β particle is about 7300 times lighter than the α particle, so it bends far more, and towards the positive plate.

2. (a) — the curie was defined from the activity of one gram of radium-226.

3. (b) — the α particle carries away two protons and two neutrons, so Z falls by 2 and A by 4. Option (c) confuses the two numbers.

4. (b) — cadmium and boron absorb neutrons strongly and so control the chain reaction; graphite and heavy water are moderators, which slow neutrons rather than absorb them.

5. (b) — carbon-14 decays with a half-life of 5730 years, which suits archaeological time scales.