2024 Prelims VJC H2 Chem P2 (Ans)

1 © VJC 2024 9729/02/PRELIM/24 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/02 Paper 2 Structured Questions Candidates answer on the Question Paper. 12 September 2024 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and CT group in the spaces at the top of this page. Write in dark blue or black pen. You may use an 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. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 20 2 / 22 3 / 16 4 / 17 Total / 75 This document consists of 18 printed pages and 2 blank pages

2 © VJC 2024 9729/02/PRELIM/24 1 (a) The Pauling electronegativity values, EN, for the elements in the first two periods of the Periodic Table range from 1.0 to 4.0. The elements X, Y and Z are all in the first two periods of the Periodic Table. Their EN values are shown in Table 1.1. Table 1.1 element EN X 1.0 Y 2.1 Z 4.0 Substances exist with formulae XZ, YZ and Z2. (i) Use Table 1.1 to deduce the group of the Periodic Table that X and Z belong to. X belongs to Group 1 Z belongs to Group 17 [1] 1m: both correct (ii) Suggest the structure and bonding present in XZ. • XZ has a giant ionic structure with ionic bonds (OR electrostatic forces of attraction) between oppositely charged ions. [1] (iii) Arrange XZ, YZ and Z2 in order of increasing melting point. • Z2 < YZ < XZ [1] (b) (i) Write the full electronic configuration of an aluminium atom. • 1s22s22p63s23p1 [1] (ii) Write an equation to show the third ionisation energy of aluminium. • Al2+(g) → Al3+(g) + e– (state symbols) [1] (iii) Explain why the third ionisation energy of aluminium is greater than the first ionisation energy of sodium. • Al2+ has a greater nuclear charge than Na due to greater number of protons. The shielding effect is constant for both Al2+ and Na as they have the same electronic configuration (OR number of electrons) and involve the removal of the single electron in 3s orbital. • Hence, Al2+ has a greater effective nuclear charge and more energy is required to remove the third electron, leading to greater third ionisation energy than the first ionisation energy of sodium. [2]

3 © VJC 2024 9729/02/PRELIM/24 [Turn over (c) Two Period 3 elements, U and V, burn separately in oxygen to form solid oxides. The oxide of U is insoluble in water. The oxide of V dissolves in water to form solution W. Solution W reacts with the oxide of U at room temperature and dissolves it. V has a larger atomic radius than U. (i) Identify U and V. • U is aluminium (OR Al) • V is sodium (OR Na) [2] (ii) Write an equation for the reaction between oxide of U and solution W. • Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq) (no ecf) [1] (d) (i) State the total numbers of protons, neutrons and electrons in the NO3 – ion. Total number of protons: 31 Total number of neutrons: 31 Total number of electrons: 32 [1] (ii) Similar to carbonate salts, c opper(II) nitrate, Cu(NO3)2, and barium nitrate, Ba(NO3)2, decompose when heated. Compare the decomposition temperatures of these two compounds. Explain your answer. • Both Cu2+ and Ba2+ have the same charge but Cu2+ has a smaller radius than Ba2+. Hence, Cu2+ has a greater charge density and higher polarising power than Ba2+. • Cu2+ distorts the electron cloud of NO3– more than Ba 2+, making Cu(NO3)2 less stable to heat as the N–O covalent bond within the NO3– anion is more weakened. Cu(NO3)2 has a lower decomposition temperature than Ba(NO3)2. [2] (iii) Sketch the shape of and label a filled orbital in Cu2+(aq) ion with the highest energy in Fig. 1.1. Fig. 1.1 [1] z y x

4 © VJC 2024 9729/02/PRELIM/24 OR (iv) Compound T, an anhydrous Group 2 bromide, is dissolved in water and titrated against aqueous silver nitrate. A solution containing 0.250 g of T requires 33.65 cm3 of 0.0500 mol dm –3 AgNO3(aq) for complete reaction. Identify T. Show your working. Amount of AgNO3 = 0.0500 ⨯ 33.65 1000 = 1.68 ⨯ 10–3 mol Let T be MBr2. MBr2 + 2AgNO3 → M(NO3)2 + 2AgBr Amount of T • = 1 2 ⨯ 1.68 ⨯ 10–3 = 0.250 ÷ Mr(T) Mr(T) = 297.2 Ar(M) = 297.2 – 2(79.9) = 137.4 • T is BaBr2. [2] (e) HCl is a product of several different reactions. Some of these are shown in Fig. 1.2. Fig. 1.2 (i) Describe the reaction of SiCl4 with water in reaction 1. Write an equation for the reaction and state the pH of the resultant mixture. • SiCl4 reacts violently with water and it undergoes complete hydrolysis in water to give a strongly acidic solution of pH 2 (accept pH 1 -3). A white ppt is also formed. SiCl4(l) + 2H2O(l) → SiO2(s)+ 4HCl(aq) [1]

5 © VJC 2024 9729/02/PRELIM/24 [Turn over In reaction 2, NaCl reacts with concentrated H2SO4 to form HCl and NaHSO4 only. Similarly, NaBr reacts with concentrated H 2SO4 to form HBr and NaHSO 4. HBr reacts further with concentrated H2SO4 to form Br2 and SO2. (ii) State the types of reaction that occur when NaBr reacts with concentrated H2SO4 and when HBr reacts with concentrated H2SO4. reactants type of reaction NaBr and concentrated H2SO4 acid-base HBr and concentrated H2SO4 redox [1] 1m: both correct (iii) Write an equation for the reaction of HBr with concentrated H2SO4. • 2HBr + H2SO4 → Br2 + SO2 + 2H2O [1] (iv) Suggest an explanation for the difference in the reactions when HCl and HBr react with concentrated H2SO4. • HBr is a stronger reducing agent than HCl. HCl is unable to reduce H2SO4 as there is no change in the oxidation state of S in NaHSO 4. HBr is able to reduce H2SO4 as the oxidation state of S changes from +6 to +4 in SO2. [1] [Total: 20]

6 © VJC 2024 9729/02/PRELIM/24 2 (a) The reaction between iron(III) ions, Fe3+(aq), and iodide ions, I–(aq), occurs as shown. Fe3+(aq) + I–(aq) → Fe2+(aq) + 1 2I2(aq) The rate of this reaction can be followed by the change in the I–(aq) concentration as given in the graph in Fig. 2.1. The initial concentration of Fe3+(aq) is 0.150 mol dm–3. Fig. 2.1 (i) Use Fig. 2.1 to calculate the initial rate of reaction and the rate of reaction at 20 s. Show your working on the graph. Initial rate of reaction at 0 s = – 12.0 ⨯ 10−3 − 4.00 ⨯ 10−3 0 − 23.0 • = 3.48 ⨯ 10–4 mol dm–3 s–1 Rate of reaction at 20 s = – 10.3 ⨯ 10−3 − 4.00 ⨯ 10−3 0 − 47.0 • = 1.34 ⨯ 10–4 mol dm–3 s–1 [2] (0, 10.3⨯10–3) (47.0, 4.00⨯10–3) (23.0, 4.00⨯10–3)

7 © VJC 2024 9729/02/PRELIM/24 [Turn over (ii) Hence, determine the order of reaction with respect to the concentration of iodide ions. Show your working. Let n be the order wrt [I–]. initial rate rate at 20 s = 3.48 ⨯ 10−4 1.34 ⨯ 10−4 = ( 12.0 ⨯ 10 –3 7.60 ⨯ 10 –3 )n (ecf) nlg( 12.0 7.60 ) = lg( 3.48 1.34 ) • n