2025 G.C.E. (A/L) Chemistry — Paper I (Solved)

1. Which of the discoveries given below is Henry Becquerel best known for?

(1) wave-particle dual nature of matter    (2) radioactivity    (3) X-rays    (4) structure of the nucleus    (5) neutrons

Answer: (2). Henri Becquerel discovered radioactivity in 1896 when he found that uranium salts spontaneously emitted penetrating radiation. X-rays were discovered by Röntgen, the neutron by Chadwick, and the wave-particle duality of matter was proposed by de Broglie. Unit 2

2. The maximum number of electrons that can exist in an atom with quantum numbers n = 4 and l = 2 in the ground state is

(1) 3    (2) 4    (3) 6    (4) 10    (5) 14

Answer: (4). The quantum numbers n = 4, l = 2 specify the 4d subshell. A subshell with l = 2 has 2l + 1 = 5 orbitals, and each orbital holds a maximum of 2 electrons, giving 5 × 2 = 10 electrons. Unit 2

3. The examples to best denote H-bonds, ion-induced dipole interactions, dipole-induced dipole interactions and dipole-dipole interactions are respectively given by

(1) 4-nitrophenol, ortho-nitrophenol, HCl/Ar, I₂ in KI(aq), ClF

(2) ortho-nitrophenol, HCl/Ar, I₂ in KI(aq), ClF

(3) para-nitrophenol, ortho-nitrophenol, HCl/Ar, I₂ in KI(aq)

(4) 4-nitrophenol, I₂ in KI(aq), HCl/Ar, ClF

(5) ortho-nitrophenol, ClF, I₂ in KI(aq), HCl/Ar

Answer: (1). Intramolecular H-bonding is best illustrated by 4-nitrophenol (the –OH and –NO₂ groups can H-bond between molecules); an ion-induced dipole interaction occurs in I₂ dissolved in KI(aq) where the I⁻ ion induces a dipole giving I₃⁻; a dipole-induced dipole interaction occurs in an HCl/Ar mixture; and dipole-dipole interaction is shown by the polar molecule ClF. Unit 4

4. The correct order of increasing electronegativities of C in the chemical species CH₂Cl₂, CH₄, COF₂ and CH₂F₂ is

(1) CH₄ < CH₂Cl₂ < CH₂F₂ < COF₂

(2) CH₂Cl₂ < CH₄ < CH₂F₂ < COF₂

(3) CH₂Cl₂ < CH₄ < COF₂ < CH₂F₂

(4) COF₂ < CH₂F₂ < CH₂Cl₂ < CH₄

(5) CH₄ < CH₂F₂ < CH₂Cl₂ < COF₂

Answer: (1). The effective electronegativity of carbon rises as it is bonded to more, and more electronegative, atoms which withdraw electron density and increase its s-character/effective nuclear pull. CH₄ has the least electron withdrawal; CH₂Cl₂ next; CH₂F₂ has more electronegative F atoms; and COF₂, with a C=O double bond plus two F atoms, has the most electron-withdrawing environment. Unit 3

5. At a given temperature, A₂(g) and B₂(g) in a molar ratio of 1:3 were introduced into a closed-rigid container. Then the following reaction occurs.

A₂(g) + 3B₂(g) ⇌ 2AB₃(g)

At equilibrium, total pressure of the system P, partial pressure of AB₃(g) and the equilibrium constant are P_AB₃, P and K_p respectively. At this temperature, if P_AB₃ <<< P, the value of P_AB₃ is

(1) 3^3/2K_p^1/2P²/4    (2) 3^3/2K_p^1/2P²/16    (3) K_p^1/2P²/16    (4) K_p^1/2P²/4    (5) 3^3/2K_p^1/2P²/4

Answer: (2). With a 1:3 starting ratio and P_AB₃ negligible compared with P, the unreacted gases keep the 1:3 ratio, so P_A₂ ≈ P/4 and P_B₂ ≈ 3P/4. Substituting into K_p = P_AB₃²/(P_A₂·P_B₂³) and solving for P_AB₃ gives P_AB₃ = 3^3/2K_p^1/2P²/16. Unit 8

6. A description of the atoms X and Y is given below. Y is heavier than X.

Which of the following could be correct regarding X and Y?

(1) a = 6   b = 6   c = 6   d = 13   e = 14

(2) a = 6   b = 7   c = 6   d = 13   e = 14

(3) a = 6   b = 8   c = 6   d = 13   e = 14

(4) a = 7   b = 7   c = 6   d = 14   e = 13

(5) a = 6   b = 8   c = 7   d = 13   e = 14

Answer: (3). X has 6 electrons; if X is a neutral atom, a = 6 protons, and with 7 neutrons its mass number d = 13. Y has 6 protons (the same element, carbon) so it must be neutral with c = 6 electrons. For Y to be heavier than X (mass > 13) and a valid carbon isotope, b = 8 neutrons gives e = 14. Option (3) satisfies all of these consistently. Unit 2

7. What is the IUPAC name of the compound CH₃–C≡C–C(Cl)(CH₃)–CHO (a hex-2-yne chain bearing a chloro and a methyl substituent on C3, with the aldehyde as C1)?

(1) 4-chloro-5-methyl-2-hexynal

(2) 3-chloro-2-formyl-4-hexyne

(3) 4-chloro-3-formyl-2-hexyne

(4) 2-methyl-3-chloro-4-hexynal

(5) 3-chloro-2-methyl-4-hexynal

Answer: (5). The principal functional group is the aldehyde (–CHO), so the chain is named as a hexynal with the –CHO carbon as C1. Numbering from the –CHO end places the triple bond starting at C4 (hex-4-ynal), a methyl substituent on C2 and a chloro substituent on C3. This gives the name 3-chloro-2-methyl-4-hexynal. Unit 9

8. CaCO₃ and Al₂(CO₃)₃ undergo thermal decomposition as given below.

CaCO₃(s) → CaO(s) + CO₂(g)
Al₂(CO₃)₃(s) → Al₂O₃(s) + 3CO₂(g)

How much CO₂ is formed from Al₂(CO₃)₃ when 3.34 g of an equimolar mixture of CaCO₃ and Al₂(CO₃)₃ is thermally decomposed? (Relative molar mass: CO₂ = 44, CaCO₃ = 100, Al₂(CO₃)₃ = 234)

(1) 0.44 g    (2) 1.32 g    (3) 1.48 g    (4) 1.76 g    (5) 1.88 g

Answer: (2). An equimolar mixture means equal moles x of each. Total mass = 100x + 234x = 334x = 3.34 g, so x = 0.01 mol. Al₂(CO₃)₃ gives 3 mol CO₂ per mole, so CO₂ from it = 3 × 0.01 = 0.03 mol = 0.03 × 44 = 1.32 g. Unit 1

9. Which of the following species is not formed during the chlorination of methane in the presence of light?

(1) ·CH₃    (2) ·CHCl₂    (3) CH₃CH₃    (4) CH₂Cl₂    (5) H⁺

Answer: (5). Photochlorination of methane proceeds by a free-radical chain mechanism. Radicals such as ·CH₃ and ·CHCl₂ form during propagation, CH₃CH₃ (ethane) forms by radical combination in termination, and CH₂Cl₂ is a substitution product. A free proton, H⁺, is never produced — the mechanism involves homolytic, not heterolytic, bond cleavage. Unit 10

10. Consider the unimolecular zero order reaction A → P at temperature T. Which of the following graphs represents the variation of concentration of A with time at temperature T? (Graphs (1)–(5) plot [A] against time; (1) is a straight line of negative slope reaching zero.)

Answer: (5). For a zero-order reaction the rate is constant and independent of [A], so [A] decreases linearly with time until the reactant runs out, after which [A] stays at zero. The correct graph is the straight line of constant negative slope that levels off at zero concentration. Unit 11

11. Identify the species which has a different shape from that of NCO⁻ ion.

(1) NO₂⁺    (2) N₃⁻    (3) XeF₂    (4) I₃⁻    (5) SF₂

Answer: (5). NCO⁻ has 16 valence electrons, is isoelectronic with CO₂, and is linear with no lone pair on the central atom. NO₂⁺, N₃⁻, XeF₂ and I₃⁻ are all linear species too. SF₂ has two lone pairs on sulfur (AX₂E₂) and is therefore bent, so it differs in shape. Unit 3

12. A volume of 25.00 cm³ of 0.02 mol dm⁻³ KIO₃ solution was added to a titration flask. The solution was acidified with dil. H₂SO₄ and 15 cm³ of 0.5 mol dm⁻³ KI solution was added. The liberated I₂ was titrated with a Na₂S₂O₃ solution using starch as the indicator. The volume of Na₂S₂O₃ solution required for the titration was 20.00 cm³. The concentration of the Na₂S₂O₃ solution in mol dm⁻³ is

(1) 0.05    (2) 0.075    (3) 0.10    (4) 0.125    (5) 0.15

Answer: (5). Moles of KIO₃ = 0.025 × 0.02 = 5×10⁻⁴ mol. The reaction IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O liberates 3 × 5×10⁻⁴ = 1.5×10⁻³ mol I₂. Each I₂ reacts with 2 S₂O₃²⁻ (I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻), giving 3×10⁻³ mol thiosulfate in 20.00 cm³. Concentration = 3×10⁻³ / 0.020 = 0.15 mol dm⁻³. Unit 12

13. Enthalpy change, ΔH for the dissolution of NaOH(s) in water at temperature 23 °C is −42 kJ mol⁻¹. An amount of 20 g of NaOH(s) was dissolved in 230 g of water at 23 °C in an insulated container. The specific heat capacity of the resulting solution is 4.2 J g⁻¹ K⁻¹. What is the final temperature of the solution? (Neglect the heat capacity of the container). (H = 1, O = 16, Na = 23)

(1) 20 °C    (2) 23 °C    (3) 42 °C    (4) 43 °C    (5) 44.7 °C

Answer: (4). Moles of NaOH = 20/40 = 0.5 mol, so heat released = 0.5 × 42 = 21 kJ = 21000 J. Total solution mass = 20 + 230 = 250 g. Temperature rise ΔT = q/(mc) = 21000/(250 × 4.2) = 20 K. Final temperature = 23 + 20 = 43 °C. Unit 7

14. The titration curve for a titration of 25.00 cm³ mono-protic base with mono-basic acid obtained at a given temperature is shown below. The curve falls from a high pH plateau through an end point near pH 5 at a titrant volume of 12.50 cm³. Which of the following descriptions is correct for the above titration curve?

(SA = strong acid, SB = strong base, WA = weak acid, WB = weak base, MO = methyl orange, MR = methyl red, Ph = phenolphthalein.)

Answer: (5). The curve starts high (a base in the flask) and the end point lies on the acidic side at about pH 5, which is characteristic of a strong acid being added to a weak base. The salt formed is acidic, so an indicator that changes in the acidic range — methyl red — is suitable. Unit 12

15. Which of the following reaction schemes is not suitable to prepare 1,2-dibromopropane?

(1) CH₃CHClCH₃ —(conc. H₂SO₄, Δ)→ —(Br₂)→ product

(2) CH₃CH(OH)CH₃ —(LiAlH₄/dry ether; then H⁺/H₂O)→ —(excess PBr₃)→ product

(3) CH₃C≡CH —(H₂/Pd-BaSO₄, quinoline)→ —(Br₂)→ product

(4) CH₃CH=CH₂ —(Br₂)→ product (and then H₂/Pt)

(5) CH₃CH₂CH₂Br —(alcoholic KOH)→ —(Br₂)→ product

Answer: (4). 1,2-dibromopropane needs the two Br atoms on adjacent carbons, obtained by adding Br₂ across the C=C of propene. Scheme (2) cannot work: reducing propan-2-ol with LiAlH₄ does nothing useful and PBr₃ would only replace the –OH to give 2-bromopropane, not a vicinal dibromide. (The official key marks (4) as the answer; note schemes that finish with extra hydrogenation, e.g. H₂/Pt after the dibromide, would destroy the product.) Unit 10

16. The reaction H₂(g) + I₂(g) → 2HI(g) proceeds through a fast equilibrium step followed by a slow elementary step. The rate law for the slow elementary step is, rate = k'[H₂(g)][I(g)]². The rate constant is k'. Which of the following gives the rate law of the overall reaction in terms of [H₂(g)] and [I₂(g)]? (k is the rate constant of the overall reaction.)

(1) k[H₂(g)][I₂(g)]    (2) k[H₂(g)][I₂(g)]²    (3) k[H₂(g)]²[I₂(g)]

(4) k[H₂(g)]²[I₂(g)]²    (5) k[H₂(g)][I₂(g)]³

Answer: (1). The fast pre-equilibrium I₂ ⇌ 2I gives [I]² ∝ [I₂]. Substituting [I]² = K[I₂] into the slow-step rate law rate = k'[H₂][I]² gives rate = k'K[H₂][I₂] = k[H₂][I₂]. The overall reaction is therefore first order in each reactant. Unit 11

17. The colours of AgBr(s) and AgI(s) are pale yellow and yellow, respectively. A solution of NaI(aq) was added to a test tube containing AgBr(s). When the contents in the test tube were stirred, its colour turned yellow. The following ideas were proposed to explain this observation.

I.  The reaction, AgBr(s) + I⁻(aq) → AgI(s) + Br⁻(aq) occurs.
II.  K_sp of AgBr(s) > K_sp of AgI(s)
III.  K_sp of AgBr(s) < K_sp of AgI(s)
IV.  Concentration of I⁻(aq) is increased.
V.  Concentration of Br⁻(aq) is decreased.

Which of the above are correct in regard to this observation?

(1) I, III and IV    (2) I, IV and V    (3) I, II and IV    (4) I, II and V    (5) I, III and IV

Answer: (3). The colour change to yellow shows AgI forms, so reaction I occurs. The more insoluble salt precipitates, which requires AgI to have the smaller K_sp, i.e. K_sp(AgBr) > K_sp(AgI) — statement II. Adding NaI increases [I⁻] — statement IV. The conversion releases Br⁻ into solution, so [Br⁻] increases, not decreases; statement V is wrong. Unit 8

18. A small sample of water placed in a beaker at room temperature was found to have evaporated completely after 6 hours. Which of the following descriptions of this process is correct with respect to the water sample?

Answer: (4). Evaporation is endothermic, so ΔH > 0. Liquid → vapour increases disorder, so ΔS > 0. Because the process actually happens spontaneously at room temperature it must be feasible, so ΔG < 0. Unit 7

19. Which of the following statements is incorrect regarding d-block elements?

(1) V₂O₅ is a basic oxide while V₂O₃ is an acidic oxide.

(2) In Mn₃O₄, manganese is present as a mixture of Mn(II) and Mn(III).

(3) When NH₄Cl/NH₄OH are added to an aqueous solution containing Fe³⁺ and Co²⁺, only Fe³⁺ precipitates.

(4) The white precipitate formed on addition of dil. NaOH to ZnCl₂(aq) is insoluble in dil. HCl.

(5) The first five elements of the 3d transition series achieve their maximum oxidation states by losing both 4s and 3d electrons.

Answer: (4). Adding dil. NaOH to ZnCl₂(aq) gives a white precipitate of Zn(OH)₂, which is amphoteric and therefore does dissolve in dil. HCl (and in excess NaOH). The statement that it is insoluble in dil. HCl is incorrect. Unit 14

20. Consider the following reaction in which a benzene ring bearing a –CH(CH₃)–CH₂–CH₂OH side chain (the side chain also carrying an –OH on the ring-attached carbon) is treated with pyridinium chlorochromate (PCC) in CH₂Cl₂. Which is the major product of this reaction? (The five options are drawn structures.)

Answer: (1). PCC is a mild oxidising agent that oxidises a primary alcohol only as far as the aldehyde (it does not over-oxidise to the carboxylic acid) and a secondary alcohol to a ketone, without touching C=C bonds or the aromatic ring. The major product is therefore the structure in which the primary –CH₂OH has become –CHO and the secondary alcohol is left/converted accordingly — option (1). Unit 9

21. Given that Hg(s) → Hg(l), ΔH = 2.4 kJ mol⁻¹ and normal freezing point of Hg(l) = −38 °C, what is the entropy change (J K⁻¹) when 47 g of Hg(l) freezes at the normal freezing point of Hg? (Hg = 200)

(1) 14.84    (2) 2.40    (3) −0.429    (4) −2.40    (5) −14.84

Answer: (4). Freezing is the reverse of melting, so ΔH_freezing = −2.4 kJ mol⁻¹. Moles of Hg = 47/200 = 0.235 mol. T = −38 °C = 235 K. ΔS = q_rev/T = (0.235 × −2400 J)/235 = −564/235 = −2.40 J K⁻¹. Unit 7

22. Which of the following statements regarding E_cell° and the flow of electrons is correct for the galvanic cell made of Ni(s) | Ni²⁺(aq, 1.0 mol dm⁻³) || Cu²⁺(aq, 1.0 mol dm⁻³) | Cu(s), at 25 °C immediately after the electrodes are connected? At 25 °C, E°_Cu²⁺/Cu = 0.34 V and E°_Ni²⁺/Ni = −0.24 V.

(1) Electrons flow from Ni-electrode to Cu-electrode and E_cell° = 0.58 V

(2) Electrons flow from Ni-electrode to Cu-electrode and E_cell° = −0.58 V

(3) Electrons flow from Ni-electrode to Cu-electrode and E_cell° = 0.10 V

(4) Electrons flow from Cu-electrode to Ni-electrode and E_cell° = 0.58 V

(5) Electrons flow from Cu-electrode to Ni-electrode and E_cell° = 0.10 V

Answer: (1). The more negative electrode (Ni) is the anode where oxidation occurs and electrons are released, so electrons flow through the external circuit from the Ni electrode to the Cu electrode. E_cell° = E°_cathode − E°_anode = 0.34 − (−0.24) = 0.58 V. Unit 13

23. Identify the reaction in which the diazonium ion acts as an electrophile. (Each option (1)–(5) shows a benzenediazonium ion C₆H₅N₂⁺Cl⁻ reacting with a different reagent: (1) H₂O with heat, (2) KI, (3) phenol with NaOH, (4) CuCN, (5) H₃PO₂/H₂O.)

Answer: (3). When a diazonium salt reacts with an activated aromatic compound such as phenol in alkaline conditions, the diazonium ion acts as a weak electrophile and attacks the electron-rich ring in an azo-coupling reaction to give a coloured azo dye. In the other reactions the diazonium group is simply displaced (substitution), so it does not act as an electrophile there. Unit 10

24. At the temperature at which a pure liquid will boil when it is heated in an open container, the

(1) average kinetic energy of the liquid is equal to the average kinetic energy of its vapour.

(2) average kinetic energy of the liquid is equal to the molar-entropy of its vapour.

(3) entropy of the liquid is equal to the entropy of its vapour.

(4) entropy of the vapour above the liquid is equal to the entropy of the atmosphere.

(5) vapour pressure of the liquid is equal to the atmospheric pressure above the liquid surface.

Answer: (5). The boiling point is defined as the temperature at which the saturated vapour pressure of the liquid becomes equal to the external (atmospheric) pressure above the liquid surface; only then can bubbles of vapour form throughout the bulk liquid. Unit 5

25. Consider the following information for an electrochemical cell at 25 °C.

Which of the following correctly shows the overall cell reaction, the corresponding E_cell° and the number of electrons transferred?

(1) 3Br₂(l) + 3H₂O(l) ⇌ 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq)  |  −0.460 V  |  5

(2) 6Br₂(l) + 6H₂O(l) ⇌ 10Br⁻(aq) + 2BrO₃⁻(aq) + 12H⁺(aq)  |  0.920 V  |  10

(3) 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) ⇌ 3Br₂(l) + 3H₂O(l)  |  0.460 V  |  10

(4) 3Br₂(l) + 3H₂O(l) ⇌ 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq)  |  −0.920 V  |  10

(5) 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) ⇌ 3Br₂(l) + 3H₂O(l)  |  0.460 V  |  5

Answer: (5). A spontaneous cell needs E_cell° > 0. The BrO₃⁻/Br₂ couple (1.520 V) acts as the cathode (reduction) and the Br₂/Br⁻ couple (1.065 V) is reversed as the anode (oxidation), giving E_cell° = 1.520 − 1.065 = 0.455 ≈ 0.460 V. Balancing the half-equations so electrons cancel, the lowest whole-number electron transfer is 5, and the overall reaction is 5Br⁻ + BrO₃⁻ + 6H⁺ → 3Br₂ + 3H₂O. Unit 13

26. The following redox couples are given in the decreasing order of their reduction potentials: O₂(g)/H₂O(l), Br₂(l)/Br⁻(aq), I₂(s)/I⁻(aq), H⁺(aq)/H₂(g), Cd²⁺(aq)/Cd(s), Fe²⁺(aq)/Fe(s), Zn²⁺(aq)/Zn(s), Al³⁺(aq)/Al(s). Which of the following reactions will not take place spontaneously in an electrochemical cell?

(1) Zn(s) + Cd²⁺(aq) → Cd(s) + Zn²⁺(aq)

(2) 2Al(s) + 3Br₂(l) → 2Al³⁺(aq) + 6Br⁻(aq)

(3) 2H₂(g) + O₂(g) → 2H₂O(l)

(4) H₂(g) + I₂(g) → 2H⁺(aq) + 2I⁻(aq)

(5) 2Al³⁺(aq) + 3Fe(s) → 2Al(s) + 3Fe²⁺(aq)

Answer: (5). A redox reaction is spontaneous when the couple with the higher reduction potential is reduced and the couple with the lower one is oxidised. Al³⁺/Al has a lower reduction potential than Fe²⁺/Fe, so Al³⁺ cannot oxidise Fe; reaction (5) would force the lower-potential couple to be reduced, giving E_cell < 0 — it is non-spontaneous. Unit 13

27. X and Y are two compounds having the molecular formula C₃H₆O₂. The observations when X and Y are reacted with three reagents are given in the following table.

Which of the following pairs of structures could be X and Y respectively?

(1) CH₃CHClCH₃ and CH₃CH₂CH₂CHO

(2) CH₂=CHCHCH₂OH and CH₃CH₂COCH₂OH (a hydroxy-ketone)

(3) CH₃CH₂COOH and CH₃CH(OH)CHO

(4) HOCH₂CH₂CH₂CHO and HOCH₂C(=CH₂)OH

(5) HOCH₂CH=CHCH₂OH and CH₃CH₂CH₂COOH

Answer: (2). Both X and Y give a gas with Na, so both contain an –OH (or –COOH). X decolourises Br₂/H₂O, indicating a C=C double bond, but gives no 2,4-DNP precipitate, so it has no carbonyl — an unsaturated alcohol fits. Y gives a coloured precipitate with 2,4-DNP, so it has a carbonyl group, and gives a gas with Na from its –OH — a hydroxy-ketone fits. Option (2) provides exactly this pair. Unit 9

28. Manganese(III) fluoride can be prepared according to the following reaction.

2MnI₂(s) + 13F₂(g) → 2MnF₃(s) + 4IF₅(l)

If the percentage yield is 80%, what is the mass of MnF₃ obtained from 0.10 moles of MnI₂ are reacted with excess F₂? (F = 19, Mn = 55, I = 127)

(1) 4.48 g    (2) 7.44 g    (3) 8.96 g    (4) 9.20 g    (5) 11.20 g

Answer: (3). From the 2:2 stoichiometry, 0.10 mol MnI₂ would give 0.10 mol MnF₃ at 100% yield. Molar mass of MnF₃ = 55 + 3×19 = 112 g mol⁻¹, so the theoretical mass = 0.10 × 112 = 11.2 g. At 80% yield the actual mass = 0.80 × 11.2 = 8.96 g. Unit 1

29. A(g) + C·D₂(s) ⇌ AD(g) + BB(g) + C(s). In an experiment carried out at 300 K, it was found that 5% of AB(g) converted to AD(g) and the total pressure of the system at equilibrium was 10 atm. The equilibrium constant K_p of the system at 300 K is

(1) (19×19)/21    (2) 10/(19×21)    (3) 0.10/(19×21)    (4) (19×19×0.10)/39    (5) (19×19×0.10)/39

Answer: (2). [Transcribed as printed; this question relies on values defined in the original printed reaction scheme. The official marking scheme gives the answer as option (2), 10/(19×21).] Unit 8

30. The diagram-I below depicts the variation of rates with time of forward and backward reactions of an equilibrium reaction P ⇌ Q at a given temperature. At time t, when an additional amount of P is added to the system the change in rate of the forward reaction is also shown in diagram-I. Which line (A, B, C, D or E) in diagram-II shows the change in the rate of the backward reaction?

(1) A    (2) B    (3) C    (4) D    (5) E

Answer: (2). Adding more P instantly raises the forward rate, which then falls back as P is consumed. The backward rate does not jump at t (the amount of Q is unchanged at that instant) but rises gradually as more Q is produced, until forward and backward rates become equal again at the new equilibrium. The curve that starts unchanged at t and rises smoothly to meet the new forward rate is line B. Unit 8

31. Consider the mechanism of the nitration of bromobenzene with conc. HNO₃/conc. H₂SO₄. Which of the following structure/structures represent/s the ion/ions formed during this reaction? (Options (a)–(d) show arenium-ion (sigma-complex) intermediates with the –NO₂ group and the added H at ortho/meta/para positions relative to Br.)

(1) only (a) and (b)    (2) only (b) and (c)    (3) only (c) and (d)    (4) only (d) and (a)    (5) any other number or combination of responses

Answer: (4). Bromine is an ortho/para-directing substituent in electrophilic aromatic substitution. The arenium-ion intermediates that actually form are those with the nitro group entering the ortho and para positions; the meta intermediate is not favoured. The correct pair is therefore the ortho and para sigma-complexes — option (4), (d) and (a). Unit 10

32. At a given temperature the following equilibrium exists in a closed-rigid container. CO(g) + 2O₂(g) ⇌ 2CO₂(g); ΔH < 0. Which of the following statements is/are correct for this system?

(a) Adding more CO₂(g) at the same temperature increases the amount of CO(g) with a change in the value of the equilibrium constant.

(b) Increasing the temperature of the system increases the amount of CO(g) with a decrease in the value of the equilibrium constant.

(c) Adding more CO(g) at the same temperature increases the amount of CO₂(g) without a change in the value of the equilibrium constant.

(d) Adding more CO₂(g) at the same temperature increases the amount of O₂(g) without a change in the value of the equilibrium constant.

Answer: (2). Using the response code, statements (b) and (c) are correct. Adding CO₂ or CO at constant temperature shifts the position of equilibrium but never changes K, so (a) is wrong. Since the reaction is exothermic, raising the temperature shifts equilibrium backward, increasing CO(g) and lowering K — (b) is correct. Adding CO drives the equilibrium forward, raising CO₂(g) with K unchanged — (c) is correct. Unit 8

33. The following statements refer to industrial processes. Which of them is/are correct?

(a) A catalyst is not required in the manufacture of NH₃ by the Haber-Bosch process.

(b) Citric acid is used in the purification of soap.

(c) In the extraction of Mg by the Dow process, CO₂ is added to the atmosphere only by the thermal decomposition of limestone or dolamite.

(d) HCl is the main by-product in the manufacture of NaOH using the membrane cell method.

Answer: (5). Under the response code the correct selection is "any other number or combination of responses." Statement (a) is false — the Haber-Bosch process uses a finely divided iron catalyst. The remaining statements as printed do not form one of the simple pairings (a)/(b)/(c)/(d) combinations of responses (1)–(4), so the answer is (5). Unit 14

34. Which of the following statements is/are correct regarding s-block elements?

(a) The metallic radii of Group II elements are less than the metallic radii of the corresponding Group I elements.

(b) The first ionization energies of Group II elements are greater than the first ionization energies of the corresponding Group I elements.

(c) The densities of Group II elements are lower than the densities of Group I elements.

(d) Group II elements have weaker metallic bonds than Group I elements.

Answer: (1). Under the response code, statements (a) and (b) are correct. Across a period a Group II element has a higher nuclear charge than the Group I element in the same period, so its atom is smaller (smaller metallic radius) and harder to ionise (higher first ionisation energy). Group II metals are in fact generally denser and have stronger metallic bonding, so (c) and (d) are false. Unit 6

35. An organic compound A when heated with aqueous NaOH, liberates ammonia. The compound formed when A is heated with acidified K₂Cr₂O₇ gives a coloured precipitate with 2,4-dinitrophenylhydrazine. Which of the following could be A? (The five options are drawn structures of benzene rings bearing combinations of –CH₂OH, –CONH₂, –CH₂NH₂, –OH and –CO₂H groups.)

Answer: (2). Liberation of ammonia on heating with aqueous NaOH shows compound A contains an amide (–CONH₂) group. On heating with acidified K₂Cr₂O₇ the structure must yield a carbonyl compound (an aldehyde or ketone) that gives a 2,4-DNP precipitate, which requires an oxidisable secondary — or primary — alcohol elsewhere in the molecule. The structure satisfying both, an amide together with an oxidisable alcohol, is option (2). Unit 9

36. The phase-diagram of a pure substance X is shown below. Point D is the triple point, B is on the solid-liquid boundary and C is on the liquid-gas boundary, with the gas region at low pressure/high temperature. Which of the following statements is/are correct?

(a) The temperature needed to solidify the liquid decreases as the pressure increases.

(b) The freezing point of the liquid is higher than its normal freezing point at pressures above 1.0 atm.

(c) A is the triple point of X.

(d) At a temperature greater than 100 °C, a pressure above 1.0 atm is needed to liquefy the gas.

Answer: (5). Under the response code the answer is "any other number or combination of responses." For a substance whose solid-liquid line slopes the way shown, statements (a) and (b) cannot both belong to a simple pairing, the triple point is D (not A) so (c) is false, while (d) is consistent with needing higher pressure to liquefy a gas above its normal boiling point — the combination does not match responses (1)–(4), so the answer is (5). Unit 5

37. The following statements are/is about atmospheric ozone. Which of them is/are correct?

(a) The highest ozone concentration is at the ground level.

(b) Stratospheric ozone protects human life.

(c) Motor vehicle emissions contribute to the formation of ground level ozone.

(d) In the ozone layer, ozone is formed and destroyed in the presence of UV radiation.

Answer: (5). Statement (a) is false — the highest ozone concentration is in the stratosphere, not at ground level. Statements (b), (c) and (d) are correct: stratospheric ozone shields life from UV, vehicle emissions generate harmful ground-level (tropospheric) ozone, and stratospheric ozone is continually formed and destroyed by UV. Because three statements are correct, the response is (5), "any other combination of responses." Unit 14

38. The enthalpy change, ΔH, for the reaction given below at 25 °C is −126 kJ. 2Na₂O₂(s) + 2H₂O(l) → 4NaOH(aq) + O₂(g). Which of the following statements about the enthalpy change is/are correct when a given amount of Na₂O₂(s) is added to excess of water at 25 °C? (H = 1, O = 16, Na = 23)

(a) 63.0 kJ of energy is released when one mole of Na₂O₂(s) is added.

(b) 31.5 kJ of energy is released when 39 g of Na₂O₂(s) is added.

(c) 63.0 kJ of energy is absorbed when one mole of Na₂O₂(s) is added.

(d) 31.5 kJ of energy is released when 39 g of Na₂O₂(s) is added.

Answer: All. The official marking scheme marks this question "All" — every listed response is accepted as correct. The reaction releases 126 kJ per 2 mol Na₂O₂, i.e. 63.0 kJ per mole; the molar mass of Na₂O₂ is 78 g mol⁻¹, so 39 g is 0.5 mol, releasing 31.5 kJ. Note that statements (b) and (d) are identical as printed, which is why the key accepts "All". Unit 7

39. Which of the following statements is/are correct regarding hydrogen halides?

(a) Hydrogen halides are acidic in water.

(b) The bond dissociation energies of hydrogen halides decrease when going down the group.

(c) The acidic strength of hydrogen halides decreases when going down the group.

(d) The bond length of hydrogen halides decreases when going down the group.

Answer: (1). Under the response code the answer is (1), meaning only (a) and (b) are correct. Hydrogen halides ionise to give acidic solutions in water, and their H–X bond dissociation energies do decrease down the group (HF > HCl > HBr > HI). Acidic strength actually increases down the group and bond length increases down the group, so (c) and (d) are false. Unit 14

40. Which of the following statements is/are correct regarding the production of H₂SO₄ by the contact process?

(a) The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is endothermic.

(b) Fe is the commonly used catalyst for this process.

(c) SO₃ is converted to SO₃ in four catalytic chambers.

(d) Pressures greater than 1 atm are not utilized.

Answer: (3). Under the response code the answer is (3), meaning only (c) and (d) are correct. In the contact process the conversion of SO₂ to SO₃ is carried out over several catalytic beds and very high pressures are not needed because the equilibrium yield is already high. The conversion 2SO₂ + O₂ ⇌ 2SO₃ is exothermic, not endothermic, and the catalyst used is V₂O₅, not Fe — so (a) and (b) are false. Unit 14

41.
First Statement: Among the elements of the second period of the Periodic Table, the electron gain energy of N and Be have a positive value.
Second Statement: Half-filled shells and completely filled shells have a higher stability than other electron configurations.

Answer: (3). The first statement is true: nitrogen (half-filled 2p³) and beryllium (filled 2s²) resist gaining an electron, so their electron gain energies are positive (endothermic). The second statement, that half-filled and fully-filled shells are extra stable, is also a true idea — but it explains nitrogen well and beryllium only loosely; the official key marks this (3), treating the second statement as not a fully correct general explanation here. Unit 6

42.
First Statement: Ketones cannot be prepared from the reaction between esters and Grignard reagents.
Second Statement: Ketones react much faster than esters with Grignard reagents.

Answer: (1). Both statements are true and the second explains the first. When an ester reacts with a Grignard reagent the ketone formed as the first intermediate product is more reactive than the ester itself, so it is attacked again immediately by a second equivalent of Grignard reagent, ending up as a tertiary alcohol. The ketone therefore cannot be isolated. Unit 9

43.
First Statement: In an evacuated closed-rigid container water boils at a temperature below 100 °C.
Second Statement: When the external pressure is low it is easy for water molecules to be released from the liquid phase to the vapour phase.

Answer: (4). The first statement is false as worded — in an evacuated closed-rigid container the vapour pressure builds up until equilibrium is reached, so the liquid does not freely boil. The second statement is true: at low external pressure water boils at a lower temperature because its vapour pressure equals the small external pressure sooner. Unit 5

44.
First Statement: The bond angles of H₂O, H₂S and H₂Se are in the order H₂O > H₂S > H₂Se.
Second Statement: The electronegativity of the central atom of H₂O, H₂S and H₂Se decreases in the order O > S > Se.

Answer: (1). Both statements are true and the second explains the first. As the central atom becomes less electronegative (O > S > Se) the bonding electron pairs lie further from it and closer to the H atoms, so they repel each other less and the bond angle becomes smaller — giving the order H₂O > H₂S > H₂Se. Unit 3

45.
First Statement: Atmospheric water vapour contributes to global warming.
Second Statement: Water vapour is a greenhouse gas.

Answer: (4). The official key marks this (4): the first statement is taken as false at the syllabus level (water vapour is not classed among the greenhouse gases targeted in global-warming discussions in the way CO₂ etc. are), while the second statement — that water vapour is a greenhouse gas — is accepted as true. Unit 14

46.
First Statement: For water, enthalpy of fusion ΔH_fus is less than the enthalpy of vaporization ΔH_vap.
Second Statement: Water molecules move further apart during fusion compared to vaporization.

Answer: (3). The first statement is true: melting ice only loosens the structure, whereas vaporisation must break essentially all the intermolecular hydrogen bonds, so ΔH_vap > ΔH_fus. The second statement is false — molecules move much further apart during vaporisation (liquid → gas), not during fusion. Unit 5

47.
First Statement: Cu(OH)₂(s) at pH = 5 is higher than that at pH = 10.
Second Statement: In acidic solutions OH⁻ gets neutralized.

Answer: (2). Both statements are true but the second does not, on its own, correctly explain the first as the examiners require. Cu(OH)₂ is more soluble at low pH because added H⁺ removes OH⁻ and shifts the dissolution equilibrium; the bare statement that OH⁻ is neutralised in acid is true but is not framed as the complete explanation, so the key marks this (2). Unit 8

48.
First Statement: In the production of Na₂CO₃ industrially, ammonification precedes carbonation.
Second Statement: When CO₂ is passed into an ammoniated (ammonified) brine solution, NH₄)₂CO₃ is produced in high concentration.

Answer: (1). Both statements are true and the second explains the first. In the Solvay process the brine is first saturated with ammonia (ammonification) and only then is CO₂ passed in (carbonation); ammoniating the brine first allows the CO₂ to be absorbed efficiently and lets NaHCO₃ precipitate, which is why ammonification must come before carbonation. Unit 14

49.
First Statement: In acidic media, H₂O₂ can act as an oxidising agent or a reducing agent, depending on the species it reacts with.
Second Statement: In acidic solutions oxygen exhibits the oxidation states −1 and 0.

Answer: (3). The first statement is true: in H₂O₂ oxygen is in the −1 state, an intermediate value, so H₂O₂ can be oxidised (to O₂, oxidation state 0) or reduced (to H₂O, oxidation state −2). The second statement is false as a general claim — oxygen in acidic solution most commonly shows −2; the −1 and 0 values are specific to peroxide/oxygen, so it is not a correct supporting statement. Unit 12

50.
First Statement: In acidic media, H₂O₂ can act as an oxidising agent, depending on the species it reacts with.
Second Statement: Out of all oxidation states oxygen exhibits, 0 is the most stable and abundant.

Answer: (3). The first statement is true — in acidic media H₂O₂ can act as an oxidising agent towards suitable reducing species. The second statement is false: the most common and stable oxidation state of oxygen in its compounds is −2, not 0; oxidation state 0 occurs only in the free element O₂. Unit 12