HCI 2026 Prelim P3 Ans

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 1 HWA CHONG INSTITUTION 2026 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION PAPER 3 MARK SCHEME 1 (a) Diagram for electrolytic cell used for purification of copper - Battery and CuSO4 as electrolyte [0.5] - Correct position of pure copper at cathode and impure copper at anode [0.5] Cu2+ + 2e− ⇌ Cu +0.34 V Ni2+ + 2e− ⇌ Ni −0.25 V Ag+ + e− ⇌ Ag +0.80 V [1] – all three values, [0.5] – two correct values At the anode, the oxidation of Cu occurs such that Cu(s) dissolves into the electrolyte as Cu2+(aq) ions: Cu(s) → Cu2+(aq) + 2 e− [0.5] Eo(Ni2+/Ni) is more negative than Eo(Cu2+/Cu) so Ni is more readily oxidised than Cu. Hence, Ni(s) will also be oxidised at the anode and dissolves into the electrolyte as Ni2+(aq). [0.5] Eo(Ag+/Ag) is more positive than Eo(Cu2+/Cu) so Ag is less readily oxidised than Cu. Hence, Ag is not oxidised and remains as a solid. Ag(s) eventually sinks to the bottom of the electrolyte, forming the anode sludge. [1] At the cathode, Cu2+ is more readily reduced than Ni2+ as Eo(Cu2+/Cu) is more positive than Eo(Ni2+/Ni). Hence, Cu 2+(aq) is preferentially reduced and deposited at the cathode while Ni2+(aq) remains in the electrolyte. [1] 1 (b) (i) For Cu(OH)2 precipitation: Ksp = 1.60×10–19 1.60×10–19 = [Cu2+] [OH–]2 1.60×10–19 = (0.1) [OH–]2

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 2 [OH–] = 1.2649×10–9  1.26×10–9 mol dm–3 [0.5] For Fe(OH)3 precipitation: Ksp = 2.70×10–39 2.70×10–39 = [Fe3+] [OH–]3 2.70×10–39 = (0.1) [OH–]3 [OH–] = 3.00×10–13 mol dm–3 [0.5] (Since min. [OH–] needed to precipitate Fe(OH)3 is lower,) Fe3+ is the first to precipitate. [1] 1 (b) (ii) 2.70×10–39 = [Fe3+] (1.2649×10–9)3 [Fe3+] remaining = 1.334×10 –12 mol dm–3 [1] (ecf min [OH -] from the chosen ppt that forms first) % of Fe3+ left in solution = 1.334×10−12 0.10 × 100% =1.33×10–9 % (<< 0.1%) [0.5] OR % of Fe3+ precipitated = = 0.10− 1.334×10−12 0.10 × 100%  1100% [0.5] The separation is considered complete. [ 0.5] (ecf from calculated [ ] remaining) 1 (c) (i) 2-methylpentan-2-ol [1] 1 (c) (ii) K2Cr2O7(aq)/ KMnO4(aq), H2SO4(aq), heat. [1] (aq or dilute for at least one reagent) -0.5 for each error e.g. never specify acid, heat under reflux/distillation, Orange dichromate turns green for B [0.5] but remains orange for A. [0.5]] Conc HCl [1] Solution turns cloudy for A [0.5], solution does not turn cloudy B. [0.5] Conc HCl, (anhydrous) ZnCl2, heat [1] Solution turns cloudy almost immediately for A. [0.5] Solution turns cloudy after some time for B. [0.5] 1 (c) (iii) The nucleophile can attack both electron deficient carbons on the epoxide ring. One carbon contains 2 CH 3 substituent groups and is more sterically hindered compared to the other carbon. As the reaction is S n2 like, the nucleophile will attack the less sterically hindered carbon. [1]

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 3 1 (d) (i) test of F observations / results type of reaction functional groups identified in F addition of 2,4- DNPH orange precipitate formed Condensation/ addition- elimination ketone addition of Tollens’ reagent no silver mirror formed addition of alkaline aqueous iodine yellow precipitate formed oxidation -COCH3 [0.5] for each box 1 (d) (ii) E undergoes side chain oxidation to form G. [1] There is only one side chain on the benzene ring. E F G C D Li+ OR Li+ [1] for each compound 2 (a) The quaternary nitrogen/ammonium group does not have a K b value as the charged nitrogen atom does not have a lone pair [0.5] which can accept a proton [0.5], hence cannot act as a base. 2 (b) Let x be the Ar of X. Mr of salt = 1.2635 × Mr of cholamine chloride 103 + x + 1 + x = 1.2635 (103 + x) x = 35.5, the Ar of chlorine [1] 2 (c) (i) Cholamine chloride is less basic and has a smaller Kb value as the positively charged / electronegative nitrogen draws electron density away from the primary amine, making the nitrogen’s lone pair less available to accept a proton. [1]

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 4 Ethylamine is more basic and has a larger Kb value as the electron-donating ethyl/alkyl group increases electron density around the primary amine, making the nitrogen’s lone pair more available to accept a proton. [1] 2 (c) (ii) Kb = ([conjugate acid] / [weak base]) × [OH−] = (0.035/0.965) × [OH−] = 4.37 × 10−4 [OH−] = 0.01205 mol dm−3 [1] pH = 14 – pOH = 14 + lg(0.01205) = 12.1 [1] ecf from [OH−] 2 (d) (i) A buffer solution resists changes in pH on addition of small amounts of acid or base. [1] 2 (d) (ii) Kb = [conjugate acid][OH−]/[cholamine chloride] [cholamine chloride]/[conjugate acid] = [OH−]/Kb = 10−(14-6.80)/9.12 × 10−8 = 0.6918 = 0.692 shown [1] 2 (d) (iii) Number of moles of weak base formed = number of moles of NaOH used = 0.1x Number of moles of salt remaining = 0.0274 – 0.1x [weak base]/[conjugate acid] = [(0.1x)/V] ÷ [(0.0274 – 0.1x)/V] = 0.692 correct formulation using ratio of concentrations of the two components [1] x = 0.1123 = 0.112 (3sf) [1] correct solving for value of x Total volume of T = 0.0274 ÷ 0.0800 = 0.3425 dm3 [0.5] y = 0.3425 – 0.1123 = 0.230 (3sf) [0.5] [ecf from x] 2 (e) (i) [1] 2 (e) (ii) NaOH(aq), heat [1] 2 (e) (iii) [1]

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 5 3 (a) (i) When NH3 is coordinated to Ag+, there are four electron groups / 4 bond pairs (accept electron pairs) around N atom. To minimise electronic repulsion, the four electron groups will arrange themselves (in a tetrahedral geometry) around the N atom. [1] (minimally 4bp + arrange as far apart as possible to minimise repulsion) Since there are four bonding electron pairs around N atom with no lone pair of electrons), the shape around the N atom is tetrahedral [0.5] and the bond angle is 109.5o [0.5]. (minimally tetrahedral and 109.5o) 3 (a) (ii) dz2 dyz [1] each 3 (a) (iii) Since the 2 NH3 ligands approach Ag(I) from the z-axis, (electrons in the) dz2 orbital experience the greatest electrostatic repulsion from the lone pairs on the N atoms / ligands / NH 3, as the lobes of the dz2 orbital point directly towards the ligands. Hence dz2 is raised to the highest energy. The dyz and dxz orbitals have lobes that lie between the z-axis and the x- or y- axis. Their electrons therefore experience some repulsion from the ligands approaching along the z-axis, although less than those in dz2. In contrast, the dxy and dx2–y2 orbitals lie entirely in the xy plane, orthogonal to the direction of approach of the NH3 ligands, and hence their electrons experience the least repulsion. ✅ Ligands approach directly along z-axis ✅ dyz and dxz is in between the lobes of xz and yz ✅ comparison of extent of repulsion (3 points – 2 marks) (2 points – 1 mark) (1 point – 0 mark) 3 (b) (i) Addition polymerisation [1] 3 (b) (ii) [1]

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 6 3 (b) (iii) [1] Correct diagram – optional to label hydrogen bonds & Must have at least two hydrogen bonding interaction from two water molecules with polyHEA. -1/2 for missing dipoles, lone pairs of electrons, wrong molecule eg HO2 Explanation: PolyHEA can form multiple (or extensive ) hydrogen bonding with water. [1] 3 (b) (iv) In polyHEA, the ester groups are present as side chains, while the polymer backbone consists of strong and non-polar carbon-carbon bonds, which are not readily hydrolysed. Hydrolysis therefore does not cleave the polymer chains. [1] ½ mark: strong C-C bond ½ m