MI Prelim H2 Chem P2 Suggested answers for exchange

Class Adm No Candidate Name: 2025 Preliminary Examination Pre-University 3 H2 CHEMISTRY 9729/02 Paper 2 Structured Questions 3 September 2025 2 hours Candidates answer on the Question paper. Additional materials: Data Booklet Suggested answers

2 1 The fire triangle (Fig 1.1), shows the three components necessary for combustion. Disrupting any one of these components can suppress or extinguish a fire. Fig 1.1 Fire retardants are substances added to materials to reduce flammability by disrupting the fire triangle. For Examiners’ Use (a) Magnesium hydroxide, Mg(OH)2, is an example of an inorganic fire retardant. It undergoes thermal decomposition upon heating and releases water vapour in the process. (i) Write a balanced equation for the thermal decomposition of magnesium hydroxide. State symbols are not required. [1] Mg(OH)2 → MgO + H2O (ii) With reference to the components in Fig 1.1, suggest two ways in which this decomposition reaction helps to suppress a fire. [1] The decomposition process is endothermic, which removes the heat component from the fire triangle. Water (vapour) released dilutes the oxygen that is available, reducing the amount of oxygen in the fire triangle. Magnesium oxide forms a protective barrier over the fuel, keeping oxygen and the fuel separated. Any 1 for [1] Aluminium hydroxide, A l(OH)3 is also a good fire retardant. It undergoes thermal decomposition in the same way as magnesium hydroxide. (iii) Explain why t he thermal decomposition temperature of Al(OH)3 is lower than that of Mg(OH)2. [2] Al3+ has a higher charge density than Mg 2+ and polarises the O–H bond and weakens it to a larger extent, hence it decomposes at a lower temperature. [1] higher charge density [1] polarises O–H bond and weakens to a larger extent

3 [Turn over (iv) Hence, explain whether A l(OH)3 would be a better fire retardant compared to Mg(OH)2. [1] Better since the decomposition will occur at the lower temperature and release water vapour earlier in a fire. (b) Halogenated fire retardants often use bromine or chlorine derivatives and they release hydrogen halides or halogen radicals upon decomposition which disrupt free radical chain reactions that sustain combustions. (i) One method of preparing brominated organic compounds is free radical substitution. Draw the mechanism for the formation of bromomethane from methane. [3] Initiation √ √ Propagation √ CH4 + Br• → •CH3 + HBr √ •CH3 + Br2 → CH3Br + Br• √ Termination √ 2 •CH3 → C2H6 √ •CH3 + Br• → CH3Br √ 8 ticks – [3] 5 – 7 ticks – [2] 3 – 4 ticks – [1] Tetrabromobisphenol A (TBBPA) is a common brominated flame retardant. It can be made from bisphenol A (BPA) in a single step reaction, as shown in Fig. 1.2. Fig 1.2 (ii) State the reagent and conditions required to convert BPA to TBBPA. [1] Br2(aq)

4 Tetrachlorobisphenol (TCBPA) is the chlorinated analogue of TBBPA, where the bromine atoms are replaced with chlorine. (iii) Using relevant data from the Data Booklet, explain the difference in strength between C–Cl and C–Br. [2] C–Cl bond: 340 kJ mol–1 C–Br bond: 280 kJ mol–1 [1] for both values The C–Br bond is weaker than the C–Cl bond because the Br atom is larger than the Cl atom, resulting in a longer bond and lower extent of orbital overlap. [1] (iv) Hence, explain whether TCBPA would be a better flame retardant than TBBPA. [1] No it will be less effective because the C-Cl bond requires more energy to break and hence will not produce HCl or Cl radical as easily as TBBPA to suppress the fire. Link bond energy to ease of formation of HX or X radical which suppresses fire. (c) Despite their effectiveness, halogenated fire retardants such at TBBPA are being phased out due to concerns over the release of toxic persistent organic pollutants during fires. Suggest one reason why Mg(OH) 2 is considered a safer alternative to halogenated fire retardants. [1] The products of Mg(OH)2 decomposition (MgO and H2O) are non–toxic. [Total: 13]

5 [Turn over 2 (a) Lead(II) sulfate, PbSO 4, is a sparingly soluble white solid that was historically used as a pigment in paints. For Examiners’ Use (i) Write an expression for the solubility product, Ksp, of PbSO4, stating its units. [1] Ksp = [Pb2+][SO42–], units: mol2 dm–6 (ii) Calculate the solubility of PbSO4 in pure water, given the value of Ksp is 1.60 × 10–8. [1] Let solubility of PbSO4 be x mol dm–3 x2 = 1.60 x 10–8 x = 1.26 x 10–4 mol dm–3 [1] (iii) A common method used by early artists to prepare PbSO 4 was to mix solutions of Na2SO4 and Pb(NO3)2. A solution of 0.0200 mol dm–3 Pb(NO3)2 was mixed with an equal volume of Na2SO4 to prepare PbSO4. Calculate the minimum concentration of Na2SO4 required for PbSO4 to be formed. [2] Upon mixing, [Pb2+] = 0.0200 / 2 = 0.0100 mol dm–3 Hence [SO42–] at equilibrium = 1.60 × 10–8 / 0.0100 = 1.60 × 10–6 mol dm–3 [1] However, before mixing, the concentration is double. Hence [Na2SO4] required = 1.60 × 10–6 × 2 = 3.20 × 10–6 mol dm–3 [1] (b) Due to the increased awareness of the toxicity of lead compounds, titanium dioxide, TiO 2, is now widely used as a non -toxic alternative to PbSO 4 in white pigments by modern artists. (i) Write the full electronic configuration of the titanium ion in TiO2. [1] 1s2 2s2 2p6 3s2 3p6 (ii) Explain what is meant by the term transition element. [1] A transition element is a d–block element which forms one or more stable ions , in compounds, with a partially filled d–subshell. (iii) Explain why TiO2 appears white despite containing a transition element. [2] TiO2 has no electrons in the 3d subshell. [1] Hence no wavelength of light is absorbed [1] and all are reflected hence it appears white. (c) Copper obtained from the extraction of copper ores is often impure and contains small amounts of other metals such as zinc, silver and lead. To obtain high purity copper for electrical use, impure copper is placed at the anode and a pure copper rod is placed at the cathode in an electrolytic cell. The electrolyte used is aqueous copper(II) sulfate.

6 (i) Write the half equation for the reaction occurring at the anode for copper. [1] Cu → Cu2+ + 2 e– (ii) Using relevant data from the Data Booklet, explain how zinc and silver impurities are removed during the purification process. [3] E ⦵ (Zn2+/Zn): –0.76 V E ⦵ (Ag+/Ag): +0.80 V E ⦵ (Cu2+/Cu): +0.34 V Since E ⦵ (Zn2+/Zn) is more negative than E ⦵ (Cu2+/Cu), Zn will be oxidised before Cu and forms Zn2+ in the electrolyte. [1] E ⦵ (Ag+/Ag) is less negative than E ⦵ (Cu2+/Cu), hence it will not be oxidised and be found at the bottom of the anode as Ag metal. [1] At the cathode, since E ⦵ (Zn2+/Zn) is more negative than E ⦵ (Cu2+/Cu), it will not be reduced and remain as Zn2+ in the electrolyte. [1] While the lead impurity was expected to dissolve into the electrolyte as Pb 2+ ions, it was instead found in the anodic sludge as a compound. (iii) State the identity of the compound and explain how it was formed. [2] PbSO4. [1] When Pb2+ is formed, it reacts with the SO 42– present in the electrolyte to form PbSO 4 and hence is deposited in the anodic sludge. [1] [Total: 14]

7 [Turn over 3 Advanced Oxidation Processes (AOPs) are used to degrade organic pollutants in wastewater through the formation of highly reactive hydroxyl radicals (•OH). Two examples of AOP s are the Fenton reaction and the UV/H 2O2 process