2025 SAJC H2 Chem Prelim P2 Answers

1 [TURN OVER ST. ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 2 CANDIDATE NAME CLASS 2 4 S CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 2 September 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your name and class on all the work that you hand in. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of XX printed pages (including this cover page). For Examiner’s Use Q1 21 Q2 7 Q3 16 Q4 12 Q5 19 Total 75

2 [TURN OVER 1 (a) Titanium dioxide, TiO2, is a white solid, which is an amphoteric oxide. In the structure of titanium dioxide, the titanium ion is bonded to six oxide anions. (i) Complete the electronic configuration of a titanium atom. 1s2 …………………………………… [1] 1s2 2s2 2p6 3s2 3p6 3d2 4s2 (ii) Suggest the shape around the titanium ion in titanium dioxide. [1] Octahedral (b) (i) Aluminium oxide is another example of an amphoteric oxide. Write two equations to illustrate the reaction of Al2O3 with an acid and a base of your choice respectively. [2] Al2O3 + 6HCl → 2AlCl3 + 3H2O Al2O3 + 2NaOH + 3H2O → 2Na[Al(OH)4] (ii) The ionic radius of Al3+ is 0.050 nm and Ti2+ is 0.086 nm. Explain the difference in ionic radii between Al3+ and Ti2+. [2] Ti2+ has a higher nuclear charge than Al3+. Ti2+ also has more filled inner shells / electron shells / (principal) quantum shells and the valence electrons are further away from the nucleus and shielding effect increases. The valence electrons are less strongly attracted to the nucleus. Ti2+ has a larger ionic size than Al3+. (c) Titanium(II) chloride is prepared by the thermal decomposition of TiCl3 at 500°C. The reaction is driven by the loss of volatile TiCl4. 2TiCl3 (s) → TiCl2 (s) + TiCl4 (g) (i) State and explain the sign for ∆So. [1] ∆So is positive or >0 as there is an increase in disorderliness / decrease in orderliness / more ways of arranging when gaseous TiCl4 / more gaseous molecule is formed.

3 [TURN OVER (ii) Deduce the sign of the enthalpy change , ∆H, of the thermal decomposition of TiC l3, given that the decomposition is spontaneous only at high temperature. Explain your answer. [1] ∆G = ∆H − T∆S The reaction is spontaneous only at high temperature, hence ∆G is negative at high temperature. Since ∆S is positive, ∆H must be positive. (iii) Explain why TiCl3 forms a violet solution, but TiCl4 forms a colourless solution. [2] TiCl4 does not have any 3d electrons, hence no d-d transition can occur. However, TiCl3 has partially filled 3d subshells / orbitals which split into 2 energy levels in the presence of ligands . When an electron from the lower energy d orbital is promoted to a higher energy d orbital , visible light is absorbed . The complementary colour observed is reflected / colour observed is the complementary of light absorbed. (d) Another transition element that is bonded to oxygen atoms is manganese. Two examples are manganate(VI) ion, MnO42–, and manganate(VII) ion, MnO4–. (i) Given that t he structure of MnO42– is similar to that of SO42–, draw th e ‘dot–and–cross’ diagram of MnO42– and state its bond angle. [2] Mn x x x xxx O OO O xx xx xx xx xx xx xx x x 2- x x 109.5° (ii) Acidified potassium manganate(VII), KMnO 4, and acidified potassium dichromate, K 2Cr2O7, can be used as oxidising agent s in organic reactions. [2]

4 [TURN OVER With reference to relevant Eo values, suggest why KMnO4 is a stronger oxidising agent than K2Cr2O7. MnO4– + 8H+ + 5e– ⇌ Mn2+ + 4H2O Eo = +1.52 V Cr2O72– + 14H+ + 6e– ⇌ 2Cr3+ + 7H2O Eo = +1.33 V Eo(MnO4–/Mn2+) is more positive as compared to Eo(Cr2O72–/Cr3+). Hence MnO4– is easier to be reduced and is a stronger oxidising agent. (e) A, B and C are isomers with the molecular formula, C5H10O, that contains one or two of the following functional groups. • Alkene • Alcohol • Carbonyl Reactions are carried out on A, B and C and the observations are shown in Table 1.1. Table 1.1 with acidified K2Cr2O7 (aq) with acidified KMnO4 (aq) with 2,4–DNPH with Br2 (aq) A orange to green purple to colourless no reaction no reaction B no reaction no reaction orange precipitate no reaction C no reaction purple to colourless no reaction orange to colourless

5 [TURN OVER (i) A is a cyclic compound and does not rotate plane of polarised light. Draw the structure of A. [1] Or OH (ii) B is a symmetrical molecule. Draw the structure of B. [1] (iii) Write the equation for the reaction which occurs when A reacts completely with an excess of acidified potassium dichromate( VI). Use [O] to represent the oxidising agent in the reaction. [1] OH A CH3 CH2 C CH2 CH3 O B

6 [TURN OVER OH + [O] O + H2O Examiner’s Comment: • Common mistakes include: o Writing the molecular formula instead of giving the structural formula of A and the product. o Not balancing the equation with H2O or [O] (iv) State all possible functional groups in C. [1] Alkene and tertiary alcohol (v) C is unable to exhibit stereoisomerism. Draw the structure of C. [1] (f) Ketones can undergo oxidation forming esters through the Baeyer –Villiger oxidation reaction by using peroxycarboxylic acids as shown in the equation below. [2] CH3 C CH3 OH CH CH2 C

7 [TURN OVER O R R1 O R O R1 O O R2 O H peroxycarboxylic acid The first step of the mechanism of the Baeyer –Villiger oxidation reaction involves the nucleophilic attack of the lone pair of electrons on the oxygen atom bonded to the hydrogen atom in the peroxycarboxylic acid to the carbonyl carbon in the ketone. Draw the first step of the mechanism of the Baeyer–Villiger oxidation reaction. Show all relevant dipoles, curly arrows and the structure of the intermediate. O R R1 O O R2 O H O- R1 O+ R H O R2 O [Total: 21] 2 Ozone, O 3, plays a crucial role in the Earth’s atmosphere by absorbing harmful ultraviolet radiation. It is also widely used for its oxidising and disinfecting properties. For example, ozone can be dissolved in ground water or drinking water for disinfection and water quality enhancement. Fig. 2.1 shows one possible structure of O3 (g). O O + O – Fig. 2.1 (a) The overall reaction for the decomposition of ozone can be represented as follows. 2O3 → 3O2 𝛿+ 𝛿–

8 [TURN OVER The rate of decomposition of ozone in ground water, at pH 8, was investigated and the following results were obtained. The reaction is first order with respect to ozone. Fig. 2.2 (i) Define the term order of reaction. [1] It is the power to which the concentration of the reactant is raised in the experimentally determined rate equation. (ii) Use the graph in Fig. 2.2 to show that the overall order of reaction is first order. [1] Since the half-life of the reaction for decomposition of ozone is constant at 1.3 x 10 3 s, the order of the reaction is first order with respect to [ozone]. for correct half-life with working on the graph that shows 2 half-lives t1/2 = 1.3 x 103 s t1/2 = 1