2023 SAJC H2 Chem Prelim P2 (Ans)

ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 2 CANDIDATE NAME CLASS 2 2 S CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 30 August 2023 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 __ printed pages (including this cover page). For Examiner’s Use Q1 9 Q2 21 Q3 17 Q4 9 Q5 19 Total 75

2 1 (a) Table 1.1 lists the number of protons, neutrons and electrons in five different particles. Each particle may be an atom, an anion or a cation. Table 1.1 particle number of protons number of neutrons number of electrons A 17 18 17 B 17 20 17 C 17 20 18 D 18 22 18 E 18 22 17 F 19 20 19 G 19 20 18 (i) Use the information in Table 1.1 to identify • the two particles which are a pair of isotopes of the same element __________________________________________________________ A and B • the uncharged particle with no unpaired electrons __________________________________________________________ D [2] (ii) Deduce which particle is smaller, F or G. Explain your answer. [2] F have the same number of protons as G and hence, same nuclear charge. F has one more electron shell than G, so there is stronger attraction between the nucleus and the remaining electrons / electrons would be closer to the nucleus in G and G would be smaller. (b) Phosphorus pentachloride, PCl5, is a useful reagent in organic synthesis. It exists in equilibrium with PCl3 and chlorine gas as shown below: PCl5 (g) PCl3 (g) + Cl2 (g) ΔH = +124 kJ mol‒1 At equilibrium, at 200 °C and a total pressure of 5 atm, 40% of PC l5 is dissociated. (i) Write an expression for Kp for the above equilibrium. [1] Kp = PPCl3PCl2 PPCl5

3 (ii) Calculate a value for Kp at 200 °C and state its units. [2] PCl5 PCl3 + Cl2 Initial / mol Z 0 0 Change / mol -0.4 z +0.4 z +0.4 z Eqm / mol 0.6 z 0.4 z 0.4 z Total moles = 1.4 z PPCl5 = (0.6 z / 1.4 z) x 5 = 2.143 atm PPCl3 = PCl2 = (0.4 z / 1.4 z) x 5 = 1.429 atm Kp = 0.953 atm (iii) Predict and explain the effect of increasing the temperature on the value of Kp. [2] When temperature increases, the system will favour the (endothermic) forward reaction to absorb the excess heat / decrease the temperature. Hence, rate constant, kforward increases more than rate constant, kbackward / position of equilibrium shifts right. Hence, Kp increases. [Total: 9] 2 (a) Table 2.1 gives data about some physical properties of the elements potassium, calcium and iron. Table 2.1 Property potassium calcium iron relative atomic mass 39.1 40.1 55.8 atomic radius (metallic) / nm 0.243 0.197 0.126 ionic radius (charge) / nm 0.138 (1+) 0.099 (2+) 0.061 (2+) melting point / K 337 1112 1811 density / g cm−3 0.89 1.54 7.87 electrical conductivity / 106 S m−1 0.14 0.298 1.00

4 (i) Explain why the atomic radius of calcium is less than that of potassium. [2] Nuclear charge is higher for calcium than potassium, while shielding effect by inner shells of electrons remains relatively constant. Hence, effective nuclear charge is higher for calcium than potassium. Attraction between the valence electrons and nucleus is stronger / distance between the valence electrons and the nucleus is smaller in calcium than potassium, resulting in smaller atomic radius for calcium. (ii) Use relevant data from Table 2.1 to suggest why the density of calcium is greater than that of potassium. You may assume that both elements have the same packing arrangement of atoms. [2] The relative atomic mass of calcium (40.1) is higher than that of potassium (39.1). The atomic radius of calcium (0.197 nm) is smaller than that of potassium (0.243 nm). Since density is mass per unit volume , with a higher relative atomic mass (quote data) and smaller atomic radius , the calcium atoms are packed more closely and the density is higher. (iii) Describe the structure and bonding in calcium with the aid of a labelled diagram. [2] Calcium has a giant metallic structure . Electrostatic forces of attractions (metallic bonding) exist between the Ca2+ and the sea of delocalised valence electrons. + + + + + + + + + + + + Ca2+ mobile valence electrons

5 (b) The high conductivity of iron is a consequence of its electronic configuration. (i) Complete the electronic configuration of an iron atom and Fe2+ ion. [1] Fe 1s2 2s22p6 3s2 3p6 3d6 4s2 Fe 2+ 1s2 2s22p6 3s2 3p6 3d6 (ii) Despite its high electrical conductivity, iron is rarely used in electrical wires, unlike copper. Apart from its physical properties, suggest a reason why iron is less preferred than copper in electrical wiring. [1] Iron is prone to oxidation / corrosion, especially in the presence of moisture and oxygen. This can lead to the formation of iron oxide (rust), which negatively impacts the conductivity of the wire and can cause it to deteriorate over time. Copper, on the other hand, has excellent corrosion resistance, making it a more durable choice for electrical wiring. (iii) Iron can form octahedral complexes. In an octahedral complex, the d subshell of a transtion metal ion is split into two energy levels. On the Cartesian axes in Fig. 2.1, draw a fully-labelled diagram of one d orbital at the lower energy level in an octahedral complex. [1] Fig. 2.1

6 Lower: any (iv) Describe two ways in which compounds containing Fe2+ ions are different from those containing Ca2+ ions in terms of their chemical behaviour. [2] Compounds containing Fe 2+ ions are more readily reduced than those containing Ca2+ ions. Compounds containing Fe2+ can act as catalysts but not Ca2+. Fe2+ formed coloured complex but not Ca2+ (c) Iron(II) compounds are generally only stable in neutral, non -oxidising conditions. It is difficult to determine the lattice energy of FeO experimentally. (i) Use data from the Data Booklet and relevant data in Table 2.2 to construct a labelled Born-Haber cycle to determine the enthalpy change of lattice energy of FeO(s). Table 2.2 ΔH/kJ mol−1 Enthalpy change of atomisation of Fe(s) +416 Sum of first and second electron affinity of O(g) +657 Enthalpy change of formation of FeO(s) -272 [3]

7 −272 = +416+ ½ (496) + 762 + 1560 + 657 + LE LE = − 3915 kJ mol–1 (ii) State and explain how the lattice energy of FeO(s) compares to the lattice energy of CaO(s). [2] Since anions are the same and Fe2+ and Ca2+ have same charge but Fe2+ has smaller radius, there is stronger ionic bonds/greater attraction between Fe 2+ and O2−. Hence LE of FeO is more exothermic/more negative than that of CaO. (iii) Most naturally occurring samples of iron( II) oxide are found as the mineral wüstite. Wüstite has formula Fe20Ox. It contains both Fe2+ and Fe3+ ions. 90% of the iron is present as Fe2+ and 10% is present as Fe3+. Deduce the value of x. [1] Charge of cations = charge of anions (per unit of Fe20Ox) 20 × [0.9(+2) + 0.1(+3)] = 2x 0 Energy/ k