HCI 2026 Prelim P2 Ans

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 1 HWA CHONG INSTITUTION 2026 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION PAPER 2 MARK SCHEME 1 (a) (i) Number of protons: 8 Number of neutrons: 10 Number of electrons: 7 [1] for all correct 1 (a) (ii) Relative atomic mass of O = (16 × 97.1/99.1) + (17 × 1.53/99.1) + (18 × 0.47/99.1) = 16.02 (no units) [2] correct method and final answer in 2 decimal place. 1 (b) (i) Across the period from Sc to Cu, • nuclear charge increases. • electrons are added to the inner 3d subshell, shielding effect increases. • effective nuclear charge relatively constant. • energy requited to remove the outmost electron remains relatively constant. [2] 4 points – 2m, 2 to 3 points – 1m 1 (b) (ii) Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 Cr+: 1s2 2s2 2p6 3s2 3p6 3d5 [1] In second ionisation energy for Cr, the electron is removed from an inner 3d orbital, which is closer to the nucleus. [1]. For the other 2 elements, the electron is removed from 4s orbital. 2 (a) Marking points for why complex are coloured: • presence of ligands in octahedral complex • degenerate 3d orbitals split into 2 energy levels with an energy gap • Co2+ has a partially filled 3d subshell, vacancies present in higher energy level • electrons promoted from lower to higher energy level • absorbs visible light of electromagnetic spectrum • colour observed is complementary to the colour absorbed [3] All 6points – 3m, 4 to 5 points – 2m, 2 to 3 points – 1m Different ligands split the 3d orbitals to different extent / produce energy gaps of different magnitude. [1] 2 (b) (i) Type of reaction: oxidation. Oxidation state of Co changes from +2 to +3. [1] 2 (b) (ii) 4Co3+ + 2H2O → 4Co2+ + O2 + 4H+ [1]

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 2 Eocell = +1.89 – (+1.23) = +0.66 V G = – 4 × 96500 × 0.66 = –255 kJ mol-1 [1] 2 (b) (iii) en molecule forms a complex with Co3+, preventing the redox reaction between Co3+ with water. [1] for any reasonable answer 2 (c) Step 1: liquid Br2 or Br2 in organic solvent [1] Step 2: NH3 in ethanol, heat under pressure [1] 2 (d) A and C are a pair of enantiomers / non-superimposable mirror images. An equimolar mixture of A and C is optically inactive. [1] B has an internal plane of symmetry and is achiral, does not rotate plane polarised light. Hence, the sample will not rotate plane polarized light. [1] 2 (e) [Co(H2O)6]2+ + EDTA4– → [Co(EDTA)]2– + 6H2O [1] S is positive. There are 2 particles before the process and 7 particles after, so there is an increase in no. of particles, more ways to distribute the energy and particles. [1] 3 (a) (i) Orders of reaction with respect to [Cl•] and to [O3] are both 1. [1] The slow/rate-determining step of the mechanism shows one Cl• reacting with one O3. Hence, rate is directly proportional to [Cl•] and to [O3]. [1] (Allow 1m for correct order of O3 identified, and its reason.) 3 (a) (ii) O3(g) + O(g) → 2O2(g) [1] state symbols optional (½ m penalty for not cancelling out common Cl• species on both sides of eqn) 3 (b) (i) homogeneous catalyst [1] (½ m penalty if “homogeneous” not stated) Since Cl• is regenerated in the reaction, [Cl•] remains constant during the experiment, and the rate equation can be simplified to rate = k’[O3] or reaction is pseudo-zero order w.r.t. [Cl•].

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 3 [Cl•] constant [½] + link to overall pseudo-1st order [½] 3 (b) (ii) Half-life = ln2 ÷ (6.69 × 10−3)  104 ms [1] 3 (b) (iii) Every • for [½] • Correct shape for 1st order product-time graph (curve passing thru origin) • Second labelled [O2] is 1.5 times that of the first [O2] • Correctly labelled [O2] values at the correct half-lives (i.e. ½[O2]max ¾[O2]max) (ecf [O2]max value from (a)(ii)) • Correctly labelled half-life values (ecf half-life from (b)(ii)) time / ms 0 [O2(g)] / mol dm−3 2.00 × 10−4 104 208 3.00 × 10−4

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 4 3 (b) (iv) [1] diagram (½ m penalty for each error. Allow legend to be incorporated within explanation with shadings shown.) When temperature is decreased to 193 K, [½] for each • • reactant particles have lower average kinetic energy. • Proportion of particles with KE  activation energy (Ea) decreases (Must be referenced to correctly shaded regions in the Boltzmann curve, either through a legend, or illustrated within the explanation (see sample below) • Frequency of effective collisions decreases. • Since rate of reaction decreases while concentrations of reactants are unchanged, rate constant decreases. 3 (c) (i) HFCs do not release any Cl• and hence do not contribute to the destruction of the ozone layer. [1] (mark negated if mentioned about F• or H• radicals) 3 (c) (ii) HFCs are potent greenhouse gases. [1] idea of greenhouse gases/contribute to global warming. No. of molecules kinetic energy (KE) 193 K proportion of molecules with KE  Ea at 193 K proportion of molecules with KE  Ea at 273 K 273 K 0 Ea +

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 5 3 (d) Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH−(aq) (1) CuS(s) ⇌ Cu2+(aq) + S2−(aq) (2) [Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) (3) [½] quote equilibria for either sparingly soluble salt, i.e. (1) or (2) [½] correct formula for the [Cu(NH3)4(H2O)2]2+ complex (either in eqn or explanation) The formation of [Cu(NH3)4(H2O)2]2+ in (3) causes [Cu2+(aq)] to decrease [½], hence position of equilibrim ( 1) shifts right / ionic product of Cu(OH)2 decreases below its Ksp, so Cu(OH)2 dissolves. [½] However, the ionic product of CuS cannot be lowered sufficiently below its Ksp, as the Ksp of CuS is very much lower than that of Cu(OH)2. Hence, CuS remains insoluble in excess aqueous NH 3. [½] compare IP vs Ksp of CuS; [½] explain why IP cannot become < Ksp for CuS (No marks for merely comparing Ksp values of the two salts) 4 (a) (i) sp3 hybrid orbitals [½] [½] (ii) The negative charge on the CH 3COO– ion is delocalised equally over two highly electronegative oxygen atoms and the negative charge is dispersed. [1] The electron-withdrawing C l atom disperses the negative charge on the CH2ClCOO– ion to a greater extent , hence the CH 2ClCOO– ion is more stabilised compared to the CH3COO– ion and CH2ClCOOH has a lower pKa than CH3COOH. [1] Delocalisation of the negative charge cannot take place for the CH3COOO– ion and hence the CH 3COOO– ion is less stabilised compared to the CH3COO– ion and CH3COOOH has a higher pKa than CH3COOH. [1]

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 6 (iii) CH3COOO– + H2O ⇌ CH3COOOH + OH– (salt hydrolysis) Initial conc / mol dm–3 1.00 Change / mol dm–3 −x +x +x Eqm conc / mol dm–3 1.00 − x x x Kb for (CH3COOO–) = Kw = 10-14 6.31 × 10-9 = 1.58  10–6 [1] Ka for (CH3COOOH) (for correct relationship between Ka, Kb and Kw) Kb = x2 1.00 - x (from ICE table) 1.58  10–6 ≈ x2 1.00 (assuming x <<1.00) [OH–] = x = 1.26  10–3 mol dm–3 [1] for correct relationship between [OH–] and Kb pOH = ‒ lg [OH–] = 2.90 pH = 14−pOH = 11.1 [1] for correct relationship between pH and pOH. (b) Electrophilic addition [1] The secondary carbocation intermediate leading to A is more stable than the primary carbocation intermediate leading to B. Secondary carbocation is more stable, as there are two electron-donating alkyl groups attached to t