2025 SAJC H1 Chem Prelims P2 Answers

1 ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 1 (STUDENT REVIEW) CANDIDATE NAME CLASS 2 4 S CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 8873/02 02 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. Section A Answer all questions in the spaces provided on the Question Paper. Section B Answer one question. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 25 printed pages (including this cover page). For Examiner’s Use Section A Q1 25 Q2 12 Q3 12 Q4 11 Section B Q5 20 Q6 20 Total 80

2 1 (a) Aluminium oxide , Al2O3, i s a white, crystalline solid with high thermal stability and hardness. (i) Complete the electronic configuration of an aluminium atom. 1s2 …………………………………… [1] 1s2 2s2 2p6 3s2 3p1 (ii) Explain why the first ionisation energy of Al is lower than the first ionisation energy of Mg. [2] In Mg, the first electron is removed from the 3s orbital, whereas for Al, the first electron is removed from the 3p orbital (√). In Al, the valence electron in the 3p orbital experiences additional shielding effect (√) by the 2s electrons. This is so as valence electron of Al in the 3p orbital is further away (√) from the nucleus than the electrons in the 3s orbital. This factor outweighs the effect of increase in nuclear charge from Mg to Al. The valence electron in 3p orbital of Al experiences weaker attraction(√) to the nucleus and requires less energy to be removed. (iii) Draw the ‘dot-and-cross’ diagram of Al2O3. [1] 3+ O x 2- xAl2 3 (iv) The ionic radius of Al3+ is 0.050 nm and Ga3+ is 0.062 nm. Explain the difference in ionic radii between Al3+ and Ga3+. [2] Ga3+ has a higher nuclear charge than A l3+. Ga3+ also has one more filled electronic/electron shells / inner 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 . Ga3+ has a larger ionic size than Al3+.

3 (v) The melting point of Al2O3 is 2072 °C while the melting point of Ga 2O3 is 1720 °C. Explain the difference in melting points of these two oxides. [2] Al2O3 & Ga2O3 have giant ionic lattice structure , with strong electrostatic forces of attraction between oppositely charged ions. The charges of the cations and anions of both oxides are similar. The ionic radius of O2– is the same. |LE| ∝ | 𝑞+ x q− 𝑟++𝑟− | ; (ionic radius) r+ of Al3+ < r+ of Ga3+, more energy is required to overcome the stronger electrostatic forces of attraction between Al3+ & O2− ( or ionic bonding) (√) in Al2O3 than in Ga 2O3, & hence Al2O3 has a higher mpt. (b) (i) Aluminium oxide , Al2O3 and silicon dioxide , SiO2, are both solid ceramic materials with high melting points and are electrical insulators. Explain how differences in their structure and bonding account for their high melting points and poor electrical conductivity. [3] Al2O3 has a giant ionic lattice/structure which is made up of strong electrostatic attraction (or ionic bonding) between the cations and anions / oppositely charged ions . Hence, it is used as an insulator because there is no mobile ions / charge carriers to conduct electricity. The strong ionic bonds require a lot of energy to overcome, hence having a high melting point. SiO2 has a giant covalent structure/lattice , made up of a 3D tetrahedral network of strong covalent bonds between Si and O atoms. Hence, it is used as an insulator because there are no mobile electrons / charge carriers to conduct electricity. The strong covalent bonds require a lot of energy to break, hence, leading to high melting point.

4 (b) (ii) Al2O3 and SiO2 have different acid-base behaviour. With the aid of equations, explain the difference in the acid and base behaviour of these two oxides. [4] Al2O3 is amphoteric /reacts with both acids and bases while SiO2 is acidic/ reacts with bases only: Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H2O (l) Al2O3 (s) + 2NaOH (aq) + 3H2O (l) → 2Na[Al(OH)4] (aq) SiO2 (s) + 2NaOH (conc) →Na2SiO3 (aq) +H2O (l) (c) The pH values of two Period 3 chloride solutions are given below. Compound pH of 1.0 mol dm-3 solution AlCl3 3.0 SiCl4 2.0 [4] Suggest explanations for the pH values of the two chloride solutions. Include appropriate equation(s) in your answer. AlCl3 undergoes partial hydrolysis to form an acidic solution of pH≈ 3. Al3+ has a smaller radius and higher charge (or high charge density) than Na+, which polarises the electron cloud of water molecules , thus weakening O –H bond to release H+ in water. AlCl3 (s) + 6H2O(l) → [Al(H2O)6]3+ (aq) + 3Cl− (aq) [Al(H2O)6]3+ (aq) + H2O(l) [Al(H2O)5(OH)]2+(aq) + H3O+(aq) Si in SiCl4 has empty 3d orbitals to accommodate lone pair of electrons from O of H2O molecules, forming dative bonds. Complete hydrolysis takes place and HC l (aq) is formed. SiCl4 (l) + 4H2O (l) → SiO2.2H2O (s) + 4HCl (aq) Or SiCl4 (l) + 2H2O (l) → SiO2 (s) + 4HCl (aq)

5 (d) Name the shapes and state the bond angles of SiCl4 and PCl3. Explain your answers using the Valence Shell Electronic Pair Repulsion (VSEPR) Theory. Draw diagrams to illustrate your answers. SiCl4 PCl3 diagram shape tetrahedral Trigonal pyramidal bond angle 109.5°/ 110° 107° [6] According to VSEPR Theory, the electron pairs are arranged as far apart to minimise repulsion. (√) SiCl4 has 4 bond pairs and no lone pair(√) while PCl3 has 3 bond pairs and 1 lone pair. (√) Hence, the lone-pair-bond pair repulsion is greater than the bond-bond pair repulsion, (√) the bond angle of PCl3 is smaller. [Total: 25 marks] Si ClCl Cl Cl 109.5o P ClCl Cl 107o

6 2 Manganese can form ions that are bonded to oxygen atoms in different oxidation states. Two examples are MnO42– and MnO4–. (a) Both ions of manganese differ in their oxidation number. State the oxidation of each ion. [2] Oxidation state of Mn in MnO42– : +6 Oxidation state of Mn in MnO4– : +7 (b) To standardise Na2S2O3 solution, 25.0 cm3 of 0.025 mol dm–3 KMnO4 was added to an excess of acidified K I solution. The iodine liberated required 2 2.40 cm 3 of Na2S2O3 for reduction. (i) Use the Data Booklet to write an equation for 1. the reaction between and MnO4– ions and I– 2. the reaction between and I– and S2O32– [2] 16H+ + 2MnO4– + 10I– 2Mn2+ + 5I2 + 8H2O 2S2O32– + I2 S4O62– + 2I– (ii) Calculate the concentration of Na2S2O3, in mol dm–3. [2] Amount of KMnO4 used = 25.0 / 1000 × 0.025 = 6.25 × 10–4 mol Amount of I2 liberated = 5/2 × 6.25× 10–4 = 1.563 × 10–3 mol [1] Amount of S2O32– reacted = 2 × 1.563 × 10–3 = 3.125 × 10–3 mol Concentration of Na 2S2O3 = 3.125 x 10–3 ÷ 22.40 / 1000 = 0.1 395 = 0.140 mol dm–3 (3.s.f.) [1]

7 (c) When K2MnO4 is dissolved in water, the solution appears green. Upon acidification, the green solution , MnO42–, slowly turns into purple MnO4–. This equilibrium is represented by the following equation: 3MnO42– (aq) + 4H+ (aq) 2MnO4– (aq) + MnO2 (s) + 2H2O (l) The equilibrium concentrations are shown in Table 2.1. Table 2.1 Temperature/ °C [MnO42–]/ mol dm–3 [H+]/ mol dm–3 [MnO4–]/ mol dm–3 25 0.010 0.020 0.005 (i) Write the expression for Kc a