2024 Prelims SAJC H2 Chem P3 (Ans)

1 [TURN OVER ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 2 CANDIDATE NAME CLASS 2 3 S CHEMISTRY Paper 3 Free Response Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/03 9 September 2024 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. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. 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 32 printed pages (including this cover page). For Examiner’s Use Q1 8 Q2 21 Q3 20 Q4 11 Q5 or Q6 20 Total 80

2 [TURN OVER Section A Answer all the questions in this section. 1 (a) (i) Define the term relative atomic mass. [1] Relative atomic mass of an element is the average mass of 1 atom of the element relative to 1/12th the mass of 1 atom of 12C. [1] OR average mass of one atom of an element 1 12 the mass of one atom of carbon-12 [1] OR It refers to mass of one mole of atoms relative/compared to 1/12 (the mass) of 1 mole of 12C or in which 1 mole 12C (atom) has a mass of (exactly) 12g. (ii) Diamond, which is made up of both 12C and 13C atoms, has a relative atomic mass of 12.011. With reference to Table 1.1, calculate the percentage abundance of 13C in diamond. Table 1.1 Relative isotopic mass 12C 12.000 13C 13.003 [1] Let the percentage abundance of 13C be y%. (13.003)(y) + (12.000)(100-y) 100 =12.011 y = 1.096 ≈ 1.10 [1] (b) (i) Define the term enthalpy change of atomisation of C(graphite). [1] It refers to t he enthalpy change/energy required to form one mole of gaseous C atoms from C(graphite). [1] (ii) Some enthalpy change values are shown in Table 1.2. Table 1.2 value / kJ mol–1 enthalpy change of atomisation of C(diamond) +717

3 [TURN OVER enthalpy change of combustion of C(diamond) –395 enthalpy change of combustion of C(graphite) –393 Using the data in Table 1.2, construct a suitable energy cycle to show that the enthalpy change of atomisation of C(graphite) is +719 kJ mol–1. [3] [2] Hatom = -393 + 395 + 717 = +719 kJ mol–1 [1] (iii) Even though graphite and diamond are allotropes of carbon, they have different enthalpy change of atomisation. With reference to hybridisation of the carbon atoms, explain this difference. [2] Carbon atoms in graphite are sp2 hybridised while the carbon atoms in diamond are sp3 hybridised. As sp3 hybrid orbitals have less s character than sp2 hybrid orbitals, C-C in diamond will experience less effective orbital overlap, leading to weaker bond which requires less energy to overcome. Hence, the enthalpy change of atomisation of C(diamond) is less endothermic. [Total: 8]

4 [TURN OVER 2 This question is on the chemistry of main group elements and their compounds. (a) (i) The melting point of beryllium and magnesium are shown in Table 2.1. Table 2.1 melting point / oC beryllium 1287 magnesium 650 Explain, in terms of structure and bonding, the difference in melting point between beryllium and magnesium. [2] Both Be and Mg are giant metallic lattices with strong electrostatic forces of attraction between the respective cations and sea of delocalised electrons / metallic bonds. Be2+ has a smaller cationic radius / higher charge density than Mg 2+ and therefore, the metallic bond in Be is stronger and will require more energy to overcome. As such, Be has a higher melting point. (ii) Explain why beryllium is a weaker reducing agent than magnesium in terms of shielding and nuclear charge. [2] Down the group from Be to Mg, the number of protons increases and hence, nuclear charge increases . As the number of filled electron shells / inner shells increase, the valence electrons are further away from nucleus and experience stronger shielding effect . Therefore, the valence electrons experience weaker nuclear attraction and require less energy to be removed. (iii) Explain why the radius of beryllium ion is smaller than the radius of beryllium atom. [2] Both Be and Be 2+ have the same number of protons and therefore, same nuclear charge. Be2+ has 1 less shell of electrons than Be. Therefore, the valence electrons of Be 2+ will experience lower shielding effect and stronger nuclear attraction.

5 [TURN OVER (iv) Unlike other Group 2 carbonates, beryllium carbonate is unstable as it will quickly decompose to form beryllium oxide and carbon dioxide. On Fig. 2.1, draw the mechanism for the decomposition of beryllium carbonate. Show the relevant lone pair and movement of electron pairs by using curly arrows on the carbonate ion. Be2+ O C-O O- Be2+ O2- + CO2 Fig 2.1 [1] Be2+ O C-O O- Be2+ O2- + CO2 [1] (b) Beryllium can be extracted from ores. An ore containing beryllium is first heated to a very high temperature to form beryllium fluoride, BeF2, before it is dissolved in water. KOH pellets are then added to 1 mol dm -3 of aqueous BeF2 to precipitate Be(OH)2. The pH of the resultant solution is 12.4. Be(OH)2 is sparingly soluble in water. The Ksp of Be(OH)2 is 2.55 x 10–4 at 25 ⁰C. (i) Write an expression for the Ksp of Be(OH)2, stating its units. [1] Ksp = [Be2+][OH-]2 mol3 dm–9 [1] (ii) Calculate the percentage of Be2+ that has been precipitated. [2] pOH = 14 – 12.4 = 1.6 [OH–] = 10–1.6 2.55 x 10–4 = [Be2+][0.025118]2 [Be2+] = 0.40415 mol dm–3 % precipitated = (1 - 0.40415) / 1 x 100% = 59.6% (iii) HCl(aq) and NaOH(aq) are added to separate saturated solutions containing Be(OH)2, which is a white solid. The observations are recorded in Table 2.2.

6 [TURN OVER Table 2.2 procedures observations HCl(aq) is added to Be(OH)2. White solid dissolve s to form colourless solution. NaOH(aq) is added to Be(OH)2. White solid dissolve s to form colourless solution. Explain how the solubility of Be(OH) 2 is affected by the addition of HC l and NaOH. [2] Be(OH)2 ⇌ Be2+ + 2OH– When HCl(aq) is added, [OH–] decreases due to neutralisation and by Le Chatelier’s Principle, position of equilibrium shifts right to produce more OH–, resulting in Be(OH)2 to dissolve. OR Be(OH)2 reacts with HCl via neutralisation to form a soluble salt BeCl2. When NaOH(aq) is added, the white solids dissolve due to the formation of complex ion [Be(OH)4]2–. (iv) In the gaseous state, BeF2 has the same number of σ and π bonds as CO2. Draw the structure of BeF2 in the gaseous state. [1] Be FF [1]

7 [TURN OVER (c) 25.0 cm3 of 0.100 mol dm–3 of benzoic acid, C6H5COOH, was titrated with 0.100 mol dm –3 of KOH. The change in pH was monitored and shown in the graph below. (i) Show that Ka of benzoic acid is 6.61 x 10–5 mol dm–3. [1] [H+] = 10−2.59 = 2.5704 x 10–3 mol dm–3 Ka = (2.5704 x 10–3)2 0.1 = 6.61 x 10–5 mol dm–3 (ii) Calculate the pH of the solution when 12.5 cm3 of KOH was added. [1] When 12.5 cm 3 of KOH was added, [C 6H5COOH] and [C 6H5COO–] were the same. In other words, there is maximum buffering capacity. pH = pKa = –lg(6.61 x 10–5) =4