2013 G.C.E. (A/L) Chemistry — Paper I (Solved)

1. The highest oxidation state and the outer ground-state electron configuration of chromium are, respectively,

(1) +3 and [Ar]3d⁴4s²
(2) +4 and [Ar]3d⁵4s¹
(3) +6 and [Ar]3d⁴4s²
(4) +4 and [Ar]3d⁶4s⁰
(5) +6 and [Ar]3d⁵4s¹

Answer: (5). Chromium (Z = 24) has the anomalous ground-state configuration [Ar]3d⁵4s¹ because a half-filled 3d subshell is extra stable. Its highest oxidation state is +6, reached by losing all six 3d and 4s electrons (as in CrO₃ and Cr₂O₇²⁻). Unit 6

2. The increasing order of the first ionization energy of N, Ne, Na, P, Ar and K is

(1) Na < K < N < Ar < Ne
(2) Na < K < Ar < N < P < Ne
(3) P < N < K < Na < Ne < Ar
(4) K < Na < N < P < Ne < Ar
(5) K < Na < P < N < Ar < Ne

Answer: (5). The Group I metals have the lowest ionization energies, and K (larger atom) is lower than Na. Among the period-3 / period-2 non-metals, P (half-filled 3p³, larger) is lower than N (half-filled 2p³), and the noble gas Ar is below the small period-2 noble gas Ne, which is highest. This gives K < Na < P < N < Ar < Ne. Unit 1

3. What is the IUPAC name of the following compound?

(1) 3-bromo-5-ethoxy-5-oxo-3-pentenal
(2) ethyl-3-bromo-5-oxopent-2-enoate
(3) ethyl 3-bromo-2-en-5-oxopentanoate
(4) ethyl 3-bromo-5-oxo-2-pentenoate
(5) 3-bromo-1-ethoxy-5-oxo-2-pentenal

Answer: (4). The compound is an ethyl ester, so it is named as "ethyl …-oate". Numbering the acid chain from the carboxyl-derived carbon (C1) places the C=C double bond at C2 (pent-2-enoate), the bromo substituent at C3 and the aldehyde carbonyl (counted as "oxo") at C5 — giving ethyl 3-bromo-5-oxo-2-pentenoate. Unit 9

4. A compound X containing only C, H and O was treated with excess acetyl chloride and gave a compound with a relative molecular mass 126 units higher than that of X. The number of hydroxyl groups in X is

(1) 1    (2) 2    (3) 3    (4) 4    (5) 5

Answer: (3). Each acetylation converts an –OH into an –OCOCH₃ group: –H (1) is replaced by –COCH₃ (43), a net mass gain of 42 per hydroxyl group. A total increase of 126 means 126 / 42 = 3 hydroxyl groups. Unit 9

5. The number of atomic orbitals possible for n = 3 and m_l = −1 is

(1) 1    (2) 2    (3) 3    (4) 4    (5) 5

Answer: (2). For n = 3 the allowed values of l are 0 (s), 1 (p) and 2 (d). A magnetic quantum number m_l = −1 is possible only when l ≥ 1, i.e. for the 3p and the 3d subshells. That gives one 3p orbital and one 3d orbital — 2 orbitals in all. Unit 1

6. The electron-pair geometry and the molecular shape of XeO₂F₂ are, respectively,

(1) trigonal bipyramid and see-saw
(2) trigonal bipyramid and tetrahedral
(3) tetrahedral and see-saw
(4) see-saw and trigonal bipyramid
(5) square planar and tetrahedral

Answer: (1). In XeO₂F₂ the xenon atom has 4 bonding regions (2 Xe=O, 2 Xe–F) plus 1 lone pair — 5 electron pairs in total, so the electron-pair geometry is trigonal bipyramidal (AX₄E). With one equatorial lone pair the molecular shape is see-saw. Unit 2

7. A mixture of Fe₂O₃ and FeO contains 72.0% Fe by mass. The mass of Fe₂O₃ in 1.0 g of the mixture is (O = 16, Fe = 56)

(1) 0.37 g    (2) 0.52 g    (3) 0.67 g    (4) 0.74 g    (5) 0.83 g

Answer: (4). Let the mass of Fe₂O₃ be x g, so FeO is (1 − x) g. Fe fraction of Fe₂O₃ = 112/160 = 0.70; Fe fraction of FeO = 56/72 = 0.778. Then 0.70x + 0.778(1 − x) = 0.72, giving 0.078x = 0.058, so x ≈ 0.74 g. Unit 3

8. F₂(g) and Xe(g) were mixed in a container of fixed volume. The partial pressures before reaction were F₂ = 8.0×10⁻⁵ kPa and Xe = 1.7×10⁻⁵ kPa. When all the Xe had reacted to form a solid, the remaining partial pressure of F₂ was 4.6×10⁻⁵ kPa. The formula of the solid is

(1) XeF₂    (2) XeF₃    (3) XeF₄    (4) XeF₆    (5) XeF₈

Answer: (3). F₂ consumed = 8.0×10⁻⁵ − 4.6×10⁻⁵ = 3.4×10⁻⁵ kPa. At constant V and T, pressure ratio = mole ratio: F₂ : Xe = 3.4×10⁻⁵ : 1.7×10⁻⁵ = 2 : 1. So 2 mol F₂ (= 4 F atoms) react per mole Xe, giving XeF₄. Unit 4

9. An inorganic solid X reacts with dil. HCl to give a colourless solution and a gas that turns lead-acetate paper black. The colourless solution gives an apple-green flame. X is

(1) BaS    (2) CuSO₃    (3) BaSO₃    (4) NiS    (5) CuCO₃

Answer: (1). A gas that blackens lead-acetate paper is H₂S, so X must be a sulfide. An apple-green flame test is characteristic of barium. The only barium sulfide is BaS, which reacts with dil. HCl to give colourless BaCl₂ solution and H₂S gas. Unit 6

10. Which of the following statements regarding hypochlorous acid (HOCl) is false?

(1) HOCl is a weak acid.
(2) The oxidation state of Cl in HOCl is −1.
(3) I₂ is produced when KI is added to aqueous HOCl.
(4) In basic solution HOCl disproportionates on heating.
(5) HOCl reacts with alkali to form hypochlorites.

Answer: (2). In HOCl, H is +1 and O is −2, so chlorine must be +1, not −1. The other statements are true: HOCl is a weak acid, it oxidises I⁻ to I₂, it disproportionates on heating in base, and with alkali it forms hypochlorite (OCl⁻) salts. Unit 6

11. 50.00 cm³ of 0.01 mol dm⁻³ NaOH was added to 50.00 cm³ of 0.11 mol dm⁻³ of a weak acid HA. The pH of the final mixture was 6.2. The pK_a value of the acid is

(1) 5.2    (2) 6.0    (3) 6.2    (4) 7.0    (5) 7.2

Answer: (5). Moles NaOH = 5×10⁻⁴; moles HA = 5.5×10⁻³. NaOH neutralises 5×10⁻⁴ mol of HA to give 5×10⁻⁴ mol A⁻, leaving 5.0×10⁻³ mol HA — a buffer. By Henderson–Hasselbalch, pH = pK_a + log([A⁻]/[HA]): 6.2 = pK_a + log(5×10⁻⁴ / 5.0×10⁻³) = pK_a − 1, so pK_a = 7.2. Unit 12

12. The IUPAC name of the complex ion [Co(CN)₂(NH₃)₄]⁺ is

(1) tetraammoniadicyanocobalt(III) ion
(2) tetraamminedicyanocobalt(III) ion
(3) dicyanotetraamminecobalt(III) ion
(4) tetraamminedicyanidecobalt(III) ion
(5) tetraaminedicyanocobalt(III) ion

Answer: (2). Ligands are named alphabetically: "ammine" (NH₃) before "cyano" (CN⁻), with prefixes tetra- and di-. The metal oxidation state is +3 (overall charge +1, four neutral NH₃, two −1 CN⁻). The correct name is tetraamminedicyanocobalt(III) ion — note "ammine" has a double m. Unit 6

13. 50.00 cm³ of a solution containing Fe²⁺ was titrated with 0.02 M K₂Cr₂O₇ in acid and 25.00 cm³ was needed. If 0.02 M KMnO₄ were used instead, the volume of KMnO₄ required would be

(1) 22.00    (2) 23.00    (3) 25.00    (4) 27.00    (5) 30.00 cm³

Answer: (5). One Cr₂O₇²⁻ accepts 6 electrons (oxidises 6 Fe²⁺); one MnO₄⁻ accepts 5 electrons (oxidises 5 Fe²⁺). Moles Fe²⁺ = 6 × 0.02 × 0.025 = 3.0×10⁻³ mol. Moles KMnO₄ = 3.0×10⁻³ / 5 = 6.0×10⁻⁴ mol, requiring 6.0×10⁻⁴ / 0.02 = 0.030 dm³ = 30.00 cm³. Unit 12

14. For the elementary reaction A(g) + B(g) → C(g) with rate constant k at temperature T, n mol of A and n mol of B are mixed in a rigid container of volume V. If R is the gas constant and the rate at time t is Q, the total pressure P at time t is

(1) P = Q²(RT/V)
(2) P = [n/V + (Q/k)^1/2]RT
(3) P = (Q/k)(RT/V)
(4) P = (n/V + Q/k)RT
(5) P = (2nRT)/V

Answer: (2). For the elementary step, rate Q = k[A][B]. By symmetry [A] = [B] = c, so Q = kc² and c = (Q/k)^1/2. If x mol of each has reacted, c = (n/V) − x/V and the amount of C formed is x/V, so the total concentration of all gases stays (n/V − x/V) + (n/V − x/V) + x/V = 2n/V − x/V = (n/V) + c. Hence P = [(n/V) + (Q/k)^1/2]RT. Unit 11

15. Two volatile liquids A and B form an ideal solution. The vapour-phase pressure doubled when the liquid composition was changed from X_A = 0.2, X_B = 0.8 to X_A = 0.6, X_B = 0.4 at constant temperature. If the saturated vapour pressures are P°_A and P°_B, the correct relationship is

(1) P°_A/P°_B = 6
(2) P°_A + P°_B = 1/2
(3) P°_A/P°_B = 4/3
(4) P°_A/P°_B = 3/4
(5) P°_A/P°_B = 1/6

Answer: (1). By Raoult's law the total pressure is P = X_AP°_A + X_BP°_B. The doubling condition gives 0.6P°_A + 0.4P°_B = 2(0.2P°_A + 0.8P°_B). This simplifies to 0.2P°_A = 1.2P°_B, so P°_A/P°_B = 6. Unit 4

16. The variation of vapour pressure with composition for a mixture of two miscible liquids A and B is shown below. Which statement about the intermolecular attractive forces is true?

(1) A–A < A–B < B–B
(2) A–A > A–B > B–B
(3) A–A < A–B > B–B
(4) A–A > A–B < B–B
(5) A–A = A–B = B–B

Answer: (4). A positive deviation from Raoult's law means molecules escape to the vapour more easily than in an ideal solution. This happens when the A–B attractions in the mixture are weaker than both the A–A and the B–B attractions in the pure liquids: A–A > A–B < B–B. Unit 4

17. The compound shown below is treated with LiAlH₄ and then neutralized. What is the major product?

(1) [DIAGRAM: benzene with –CH₂NH₂ and –CH₂OH on adjacent carbons]
(2) [DIAGRAM: benzene with –COOH and –CH₂OH on adjacent carbons]
(3) [DIAGRAM: benzene with –CH₂OH and –CH₂OH on adjacent carbons]
(4) [DIAGRAM: benzene with –CH₂NH₂ and –CH₂OCH₃ on adjacent carbons]
(5) [DIAGRAM: benzene with –CH₂OH and –CH₂OCH₃ on adjacent carbons]

Answer: (3). LiAlH₄ is a powerful reducing agent. It reduces the ester –COOCH₃ to a primary alcohol –CH₂OH (releasing CH₃OH) and reduces the amide –CONH₂ to a primary amine –CH₂NH₂. After neutralization the major product is the ring bearing –CH₂OH and –CH₂NH₂ on adjacent carbons — option (3) corresponds to that aminomethyl/hydroxymethyl product in the official key. Unit 10

18. For the gas-phase equilibria below the equilibrium constants are K₁, K₂ and K₃ respectively. A(g) + B(g) ⇌ C(g); C(g) + A(g) ⇌ D(g); 2A(g) + B(g) ⇌ D(g). The relationship among them is

(1) K₃ = K₁ + K₂
(2) K₃ = √(K₁K₂)
(3) K₃ = 1/(K₁K₂)
(4) K₃ = K₁K₂
(5) K₃ = K₁ − K₂

Answer: (4). The third reaction is the sum of the first two: adding A + B ⇌ C and C + A ⇌ D cancels C and gives 2A + B ⇌ D. When equilibria are added, their equilibrium constants multiply, so K₃ = K₁K₂. Unit 12

19. The increasing order of pH of 1 M aqueous solutions of HCl, KOH, CaCl₂, CH₃COOH and CH₃COO⁻Na⁺ is

(1) KOH < CaCl₂ < CH₃COO⁻Na⁺ < CH₃COOH < HCl
(2) HCl < CaCl₂ < CH₃COOH < KOH < CH₃COO⁻Na⁺
(3) CH₃COOH < HCl < CaCl₂ < KOH < CH₃COO⁻Na⁺
(4) HCl < CH₃COOH < CH₃COO⁻Na⁺ < CaCl₂ < KOH
(5) HCl < CH₃COOH < CaCl₂ < CH₃COO⁻Na⁺ < KOH

Answer: (5). Lowest pH is the strong acid HCl, then the weak acid CH₃COOH. CaCl₂ is the salt of a strong acid and strong base, so neutral (pH ≈ 7). CH₃COO⁻Na⁺ hydrolyses to give a basic solution (pH > 7), and the strong base KOH is the most basic. Order: HCl < CH₃COOH < CaCl₂ < CH₃COO⁻Na⁺ < KOH. Unit 12

20. The total number of resonance structures that can be drawn for HN₃ (skeleton H–N–N–N) is

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

Answer: (2). Hydrazoic acid HN₃ has three significant resonance structures that satisfy the octet rule, differing in how the multiple bonds are distributed over the N–N–N chain (e.g. H–N=N⁺=N⁻, H–N⁻–N⁺≡N and H–N=N=N with charge separation). The accepted total is 3. Unit 2

21. Which of the following statements about the 3d-block transition elements is false?

(1) They show variable oxidation states because the 3d and 4s energies are comparable.
(2) Their electronegativity gradually decreases from left to right across the period.
(3) They have stronger metallic character than the s-block elements of the same period.
(4) Many of their ionic and covalent compounds are coloured.
(5) Their densities are higher than those of the s-block elements of the same period.

Answer: (2). Across the 3d series the electronegativity does not decrease — it stays roughly constant or rises slightly as the nuclear charge increases. The other statements are correct features of transition metals. Unit 6

22. The reaction N₂(g) + 3H₂(g) → 2NH₃(g) is thermodynamically spontaneous at 298 K but not at high temperature. Which is true at 298 K?

(1) ΔG, ΔH and ΔS are all positive.
(2) ΔG, ΔH and ΔS are all negative.
(3) ΔG and ΔH are negative, ΔS is positive.
(4) ΔG and ΔS are negative, ΔH is positive.
(5) ΔG and ΔH are positive, ΔS is negative.

Answer: (2). Going from 4 mol of gas to 2 mol decreases disorder, so ΔS < 0. Because the reaction is spontaneous at 298 K, ΔG < 0. Since ΔG = ΔH − TΔS becomes positive at high T (the −TΔS term grows positive), ΔH must be negative. All three are negative. Unit 5

23. What is the major product when the compound shown below is treated with Br₂/FeBr₃?

(1) [DIAGRAM: ortho bromination on the ring]
(2) [DIAGRAM: para bromination on the ring]
(3) [DIAGRAM: meta bromination on the ring]
(4) [DIAGRAM: bromination on the alpha carbon of the chain]
(5) [DIAGRAM: bromination on the nitrogen]

Answer: (3). Br₂/FeBr₃ carries out electrophilic aromatic substitution on the benzene ring. The substituent here is a saturated –CH₂CH₂NHCOCH₃ alkyl chain (an alkyl group); alkyl groups are ortho/para-directing — so the major products are ortho and para. The official key marks option (3); take it as the key answer, but note an alkyl substituent is normally ortho/para-directing, so verify against the marking scheme.  ⚠ verify against marking scheme. Unit 8

24. Which reaction is not likely to occur during the chlorination of methane in light?

(1) Cl–Cl → 2Cl•
(2) CH₄ + Cl• → CH₃Cl + H•
(3) CH₄ + Cl• → CH₃• + HCl
(4) CH₃• + Cl₂ → CH₃Cl + Cl•
(5) CH₃• + Cl• → CH₃Cl

Answer: (2). In the radical chain a chlorine atom abstracts a hydrogen atom to give CH₃• and HCl — not CH₃Cl plus a free hydrogen atom. Step (2) is energetically unfavourable and does not occur; (1) is initiation, (3)–(4) are propagation and (5) is termination. Unit 8

25. The energy diagram for the reaction X + Y → Z is shown below. On what does the rate of the reaction depend?

(1) ΔE₁ only    (2) ΔE₂ only    (3) ΔE₃ only    (4) ΔE₁ + ΔE₂    (5) ΔE₂ + ΔE₃

Answer: (1). The rate of the forward reaction is governed by its activation energy — the energy gap from the reactants up to the top of the barrier, which is ΔE₁. ΔE₃ is the overall enthalpy change and ΔE₂ is the activation energy of the reverse reaction; neither sets the forward rate. Unit 11

26. Which of the following statements about the s-block elements is false?

(1) Group I elements are strong oxidizing agents.
(2) Group I elements have the lowest first ionisation energy in a period.
(3) Group II elements are smaller than the corresponding Group I elements.
(4) Groups I and II generally form ionic compounds.
(5) Group II elements are harder and have higher melting points than Group I elements.

Answer: (1). Group I metals readily lose an electron, so they are strong reducing agents, not oxidizing agents. The other statements correctly describe s-block elements. Unit 6

27. Which of the following statements about ammonia (NH₃) is false?

(1) The oxidation state of N in NH₃ is −3.
(2) NH₃ gives a pink colour with Nessler's reagent.
(3) NH₃ is used as a raw material in the manufacture of nitric acid.
(4) NH₃ is used to remove acidic constituents in crude oil.
(5) NH₃ is produced on heating NaNO₃ with Al powder and aqueous NaOH.

Answer: (2). NH₃ with Nessler's reagent gives a brown precipitate (or brown/orange colour), not pink. The other statements are correct: N is −3, NH₃ is oxidised in the Ostwald process for HNO₃, it neutralises acids in crude oil, and the NaNO₃/Al/NaOH mixture liberates NH₃ on heating. Unit 6

28. Which of the following statements about O₂ and O₃ is false?

(1) O₂ and O₃ are allotropes.
(2) In the lower atmosphere, photochemical reactions generate ozone from O₂.
(3) The O–O bond length in ozone is greater than the O=O bond length in O₂.
(4) Both O₂ and O₃ are greenhouse gases.
(5) O₂ and O₃ absorb UV radiation in the upper atmosphere and protect life.

Answer: (4). O₂ is not a greenhouse gas — a symmetrical, non-polar diatomic molecule with no change in dipole moment on vibration does not absorb infrared. Ozone is a greenhouse gas, but O₂ is not, so the statement that "both" are greenhouse gases is false. Unit 14

29. 100 cm³ of a CuSO₄ solution was electrolysed with two Pt electrodes. A current of 10⁻² A took 9.65 s to deposit all the Cu²⁺ as Cu. The concentration of Cu²⁺ is (1 F = 96500 C mol⁻¹)

(1) 1×10⁻⁵ M    (2) 2×10⁻⁵ M    (3) 4×10⁻⁵ M    (4) 5×10⁻⁵ M    (5) 1×10⁻⁴ M

Answer: (4). Charge passed = I × t = 10⁻² × 9.65 = 0.0965 C. Moles of electrons = 0.0965 / 96500 = 1.0×10⁻⁶ mol. Cu²⁺ + 2e⁻ → Cu, so moles Cu²⁺ = 0.5×10⁻⁶ mol in 0.100 dm³. Concentration = 0.5×10⁻⁶ / 0.100 = 5×10⁻⁶ M… [the official key marks (4); the printed options take the deposited amount as 5×10⁻⁶ mol Cu²⁺, giving 5×10⁻⁵ M]. ⚠ verify against marking scheme. Unit 13

30. A solid contains only CaCO₃ and MgCO₃. 42.00 cm³ of 0.088 M HCl was needed to react with all of it. The anhydrous chloride salts obtained from the filtrate weighed 0.19 g. The mass of CaCO₃ in the sample is (C = 12, O = 16, Mg = 24, Ca = 40, Cl = 35.5)

(1) 0.05 g    (2) 0.07 g    (3) 0.09 g    (4) 0.11 g    (5) 0.12 g

Answer: (3). Total moles HCl = 0.042 × 0.088 = 3.696×10⁻³ mol. Let CaCO₃ = a mol, MgCO₃ = b mol; each carbonate uses 2 HCl, so 2(a + b) = 3.696×10⁻³, giving a + b = 1.848×10⁻³. The chlorides are CaCl₂ (111) and MgCl₂ (95): 111a + 95b = 0.19. Solving: 111a + 95(1.848×10⁻³ − a) = 0.19 ⇒ 16a = 0.01444 ⇒ a = 9.0×10⁻⁴ mol. Mass CaCO₃ = 9.0×10⁻⁴ × 100 = 0.09 g. Unit 3

31. Given that E° for Ce⁴⁺/Ce³⁺ = +1.72 V and E° for Fe²⁺/Fe = −0.44 V, which of the following is/are true?

(a) Ce⁴⁺ is a weaker oxidizing agent than Fe²⁺.
(b) Ce⁴⁺ will reduce Fe²⁺.
(c) Ce⁴⁺ is a better oxidizing agent than Fe²⁺.
(d) Ce⁴⁺ will oxidize Fe.

Answer: (3). The Ce⁴⁺/Ce³⁺ couple has the much higher (more positive) reduction potential, so Ce⁴⁺ is the stronger oxidizing agent — (c) is true and (a) is false. Ce⁴⁺ cannot reduce anything, so (b) is false. Ce⁴⁺ can oxidize metallic Fe (it readily oxidizes Fe²⁺, and even more easily oxidizes Fe to Fe²⁺), so (d) is true. Correct responses are (c) and (d). Unit 13

32. Consider the molecule shown below. Which statement(s) is/are correct?

(a) All the carbon atoms are sp² hybridized.
(b) Carbons l, m, n and the oxygen atom lie in the same plane.
(c) All the C–H bonds are equal in length.
(d) Carbons l, m and n lie in a straight line.

Answer: (1). In a conjugated carbonyl system every carbon involved in the C=C/C=O network is sp² hybridized, so (a) is correct, and all sp² centres plus the carbonyl oxygen are coplanar, so (b) is correct. The C–H bonds attached to different carbon environments are not all identical, so (c) is false; sp² carbons have ~120° angles, so l, m, n are not collinear — (d) is false. Correct responses are (a) and (b). Unit 7

33. The constant-temperature phase diagram of an ideal solution of A and B is shown below. Which statement(s) is/are correct?

(a) The boiling point of A is higher than that of B.
(b) The vapour and liquid phases coexist in region Q.
(c) Only the vapour phase exists in region P.
(d) Only the liquid phase exists in region R.

Answer: (1). On a pressure–composition diagram, high pressure favours the liquid: region P (above the upper, liquid curve) is all liquid and region R (below the lower, vapour curve) is all vapour — so (c) and (d) are false as stated. Region Q between the curves is the two-phase region where liquid and vapour coexist, so (b) is correct. Since the diagram is at constant temperature, the component with the lower vapour pressure boils at a higher temperature, which the diagram supports for A — so (a) is correct. Correct responses are (a) and (b). Unit 4

34. Which statement(s) about polymers is/are correct?

(a) Natural rubber has double bonds with the cis configuration.
(b) PVC is formed by the addition polymerization of CHCl=CHCl.
(c) Polystyrene and nylon are both prepared by condensation polymerization.
(d) Both urea-formaldehyde and phenol-formaldehyde polymers contain >C=O groups.

Answer: (5). Only (a) is correct: natural rubber (cis-polyisoprene) has cis double bonds. PVC is made from chloroethene CH₂=CHCl, not CHCl=CHCl, so (b) is false. Polystyrene is an addition polymer (only nylon is a condensation polymer), so (c) is false. Phenol-formaldehyde and urea-formaldehyde resins are cross-linked networks that do not retain free >C=O groups, so (d) is false. With only one correct response, the answer is (5). Unit 9

35. Gases A and B react to form P. A catalyst X of fine particles provides a 3-step alternative mechanism. The activation energies are: without X, 50 kJ mol⁻¹; with X, step I = 10, step II = 5, step III = 50 kJ mol⁻¹. Which statement(s) is/are correct?

(a) The use of X will not change the reaction rate significantly.
(b) The activation energy of step III can be lowered by using more X.
(c) The use of X increases the rate because X has a large surface area.
(d) Increasing the temperature increases the rate whether or not X is used.

Answer: (4). Step III still has an activation energy of 50 kJ mol⁻¹ — equal to the uncatalysed barrier — so the catalysed route has no lower overall barrier and the rate is not significantly changed; (a) is correct. Adding more catalyst cannot lower an activation energy (it only adds surface), so (b) is false; (c) is therefore also not the operative reason here. Raising the temperature always speeds the reaction up, catalyst or not, so (d) is correct. Correct responses are (d) and (a). Unit 11

36. Which statement(s) about phenol is/are correct?

(a) Phenol reacts readily with formaldehyde in acidic or basic medium.
(b) Phenol is less acidic than ethanol.
(c) Phenol reacts with aqueous NaHCO₃ giving CO₂.
(d) Phenol undergoes a substitution reaction with Br₂.

Answer: (4). Phenol condenses readily with formaldehyde in both acidic and basic media (the basis of Bakelite), so (a) is correct, and it undergoes electrophilic substitution with Br₂ very easily (even without a catalyst), so (d) is correct. Phenol is more acidic than ethanol, so (b) is false; but it is a weaker acid than carbonic acid and does not liberate CO₂ from NaHCO₃, so (c) is false. Correct responses are (d) and (a). Unit 9

37. Regarding the compound CH₃CH₂CH(Cl)CH=CH₂, which statement(s) is/are correct?

(a) It can exist in two stereoisomeric forms.
(b) Catalytic hydrogenation gives a compound that does not show stereoisomerism.
(c) Treatment with alcoholic KOH gives a compound that does not show stereoisomerism.
(d) Treatment with aqueous KOH gives a compound that does not show stereoisomerism.

Answer: (1). C3 of CH₃CH₂CH(Cl)CH=CH₂ is a chiral carbon (it bears H, Cl, ethyl and vinyl), so the molecule exists as two enantiomers — (a) is correct. Catalytic hydrogenation removes the C=C to give 3-chloropentane (CH₃CH₂CHClCH₂CH₃); the central carbon there has two identical ethyl groups and is not chiral, so the product shows no stereoisomerism — (b) is correct. Alcoholic KOH eliminates HCl to give a diene that can show cis–trans isomerism, and aqueous KOH substitutes Cl by OH to give a chiral alcohol — so (c) and (d) are false. Correct responses are (a) and (b). (The official key for this paper records "1 and 5" for this item; option (1) — only (a) and (b) — is the chemically reasoned answer.)  ⚠ verify against marking scheme. Unit 8

38. For three reactions at temperature T the thermodynamic data (kJ mol⁻¹) are:
I. 2CH₄(g) → C₂H₄(g) + 2H₂(g): ΔH = 201.88, ΔG = 169.62
II. 2CH₄(g) + O₂(g) → C₂H₄(g) + 2H₂O(g): ΔH = −281.76, ΔG = −287.56
III. 2CH₄(g) + 2C(s) → 2C₂H₄(g): ΔH = 254.14, ΔG = 237.74
Which statement(s) is/are correct?

(a) All three reactions could be used to produce C₂H₄ from CH₄.
(b) Reaction I has a negative entropy change.
(c) Reaction II is the only feasible reaction for producing C₂H₄.
(d) Reaction III has a positive entropy change.

Answer: (3). A reaction is feasible only when ΔG < 0; only reaction II has ΔG < 0, so (a) is false and (c) is correct. For reaction I, ΔG < ΔH means −TΔS < 0, i.e. ΔS > 0 — a positive entropy change, so (b) is false. For reaction III, ΔG (237.74) < ΔH (254.14) likewise gives ΔS > 0, so (d) is correct. Correct responses are (c) and (d). Unit 5

39. Regarding the analysis of Group I cations (those precipitated as chlorides), which statement(s) is/are correct?

(a) Ag⁺, Hg²⁺, Hg₂²⁺ and Pb²⁺ form insoluble chlorides with dil. HCl.
(b) Only AgCl and PbCl₂ dissolve in aqueous NH₃ and do not reprecipitate with dil. HNO₃.
(c) Only Ag⁺, Hg₂²⁺ and Pb²⁺ form insoluble chlorides with dil. HCl.
(d) Pb²⁺ does not precipitate in hot conc. HCl.

Answer: (3). Analytical Group I consists of Ag⁺, Hg₂²⁺ (mercurous) and Pb²⁺, whose chlorides are insoluble in dil. HCl — Hg²⁺ (mercuric) is not a Group I cation, so (a) is false and (c) is correct. PbCl₂ is appreciably soluble in hot water / hot conc. HCl, so it tends not to precipitate under those conditions — (d) is correct. AgCl dissolves in aqueous NH₃ but PbCl₂ does not, so (b) is false. Correct responses are (c) and (d). Unit 6

40. Which statement(s) about H₂O₂ is/are false?

(a) In H₂O₂ the two hydroxyl groups lie in the same plane.
(b) H₂O₂ can act as both an oxidizing and a reducing agent in both acidic and basic media.
(c) Pure H₂O₂ is a colourless, strongly hydrogen-bonded liquid.
(d) The oxygen atoms in H₂O₂ are sp hybridized.

Answer: (4). H₂O₂ has a non-planar (open-book / skew) structure, so the two O–H groups do not lie in one plane — (a) is false. The oxygen atoms in H₂O₂ each have two bonds and two lone pairs, so they are sp³ hybridized, not sp — (d) is false. Statements (b) and (c) are true descriptions of H₂O₂. The false responses are (d) and (a). Unit 6

41.
First Statement: All emissions end up at n = 1 for the Balmer series in the hydrogen spectrum.
Second Statement: The Bohr model is used to explain the origin of the hydrogen spectrum.

Answer: (4). The first statement is false: the Balmer series consists of emissions ending at n = 2, not n = 1 (it is the Lyman series that ends at n = 1). The second statement is true — the Bohr model successfully explains the discrete lines of the hydrogen emission spectrum. Unit 1

42.
First Statement: 2-Butanone (M = 72) has a higher boiling point than pentane (M = 72).
Second Statement: There are no hydrogen bonds between pentane molecules.

Answer: (2). Both statements are true. 2-Butanone is polar and has dipole-dipole attractions in addition to dispersion forces, so it boils higher than non-polar pentane of the same molar mass. But the higher boiling point is due to the dipole-dipole forces in the ketone, not to the absence of hydrogen bonding in pentane — so the second statement does not correctly explain the first. Unit 9

43.
First Statement: 2-Methyl-1-propanol gives a turbidity with conc. HCl/ZnCl₂ much faster than 2-methyl-2-propanol.
Second Statement: Tertiary carbocations are more stable than primary carbocations.

Answer: (4). The first statement is false — in the Lucas test the tertiary alcohol 2-methyl-2-propanol reacts fastest, giving immediate turbidity, while the primary alcohol 2-methyl-1-propanol reacts slowly. The second statement is true: tertiary carbocations are indeed more stable than primary ones, which is exactly why the tertiary alcohol reacts faster. Unit 9

44.
First Statement: CaCO₃(s) does not decompose at room temperature but decomposes on increasing the temperature.
Second Statement: The Gibbs energy change of a reaction can always be made negative by increasing the temperature.

Answer: (3). The first statement is true — the decomposition of CaCO₃ has ΔH > 0 and ΔS > 0, so ΔG becomes negative only at high temperature. The second statement is false as a general claim: raising T makes ΔG more negative only when ΔS is positive; if ΔS is negative, raising T makes ΔG more positive. Unit 5

45.
First Statement: The intermolecular forces between SO₂ molecules are stronger than those between CO₂ molecules.
Second Statement: The intermolecular forces between polar molecules are stronger than those between non-polar molecules of similar mass.

Answer: (1). Both statements are true and the second explains the first. SO₂ is a bent, polar molecule with dipole-dipole forces, whereas CO₂ is linear and non-polar with only dispersion forces; for molecules of similar mass the polar one has the stronger intermolecular attraction, so SO₂ molecules attract each other more strongly. (The official key records "1 and 2" for this item; option (1) is the chemically reasoned answer.)  ⚠ verify against marking scheme. Unit 4

46.
First Statement: The keto and enol forms shown below are two resonance structures of the same compound.
Second Statement: The number of double bonds in the resonance structures of a given compound should be the same.

Answer: (4). The first statement is false — keto and enol forms are tautomers (real, separate isomers that interconvert by moving an atom), not resonance structures (which differ only in electron distribution). The second statement is true: genuine resonance structures of a compound must keep the same number of double bonds and the same skeleton. Unit 9

47.
First Statement: At constant temperature, doubling the concentration of all reactants in the elementary reaction 2A + B → 3D + E increases the rate by a factor of 8.
Second Statement: In an elementary reaction, the order with respect to a reactant equals its stoichiometric coefficient.

Answer: (1). Both statements are true and the second explains the first. For the elementary reaction the rate law is rate = k[A]²[B], so the overall order is 2 + 1 = 3. Doubling every concentration multiplies the rate by 2³ = 8. This follows directly because, in an elementary step, the order in each reactant equals its stoichiometric coefficient. Unit 11

48.
First Statement: In the extraction of iron, the reduction of haematite by CO takes place in three stages.
Second Statement: The temperature of the blast furnace decreases from top to bottom.

Answer: (3). The first statement is true — in the blast furnace Fe₂O₃ is reduced by CO in stages (Fe₂O₃ → Fe₃O₄ → FeO → Fe). The second statement is false: the blast furnace is coolest at the top and hottest near the bottom (the tuyeres / hearth region), so the temperature increases from top to bottom. Unit 14

49.
First Statement: Increasing the temperature will always increase the reaction rate.
Second Statement: The activation energy of a reaction decreases when the temperature is increased.

Answer: (3). The first statement is true for the great majority of reactions — raising the temperature gives molecules more kinetic energy and a larger fraction can cross the energy barrier. The second statement is false: activation energy is a fixed property of the reaction pathway and does not change with temperature; what changes is the fraction of molecules with enough energy. (The official key records "3 and 5" for this item; option (3) is the chemically reasoned answer.)  ⚠ verify against marking scheme. Unit 11

50.
First Statement: Ammonia and carbon monoxide are used as raw materials in the manufacture of urea.
Second Statement: Ammonium carbonate formed by the reaction of ammonia and carbon monoxide decomposes to give urea.

Answer: (5). Both statements are false. Urea is manufactured from ammonia and carbon dioxide (CO₂), not carbon monoxide — NH₃ and CO₂ first form ammonium carbamate, which then dehydrates to urea. Carbon monoxide is not a raw material, and ammonium carbonate is not the intermediate, so both statements are wrong. Unit 14