2023 DHS H2 Chem Prelim Paper 1 Worked Solutions

2023 Y6 Preliminary Examination H2 Chemistry 9729 Paper 1 Suggested Solutions  DHS Chemistry Unit Page 1 of 6 Answer Key 1 2 3 4 5 6 7 8 9 10 C A B C D C C D A B 11 12 13 14 15 16 17 18 19 20 D B A C C D D B D A 21 22 23 24 25 26 27 28 29 30 B A D C B A A D B C 1 C Since atomic radius of X is larger than that of sulfur, X is the element below sulfur in Group 16 , selenium. Since atomic radius of Y is larger than that of X, Y precedes X in the same period and is arsenic. Its proton number is 33. 2 A Se: [Ar]3d104s24p4 Br: [Ar]3d104s24p5 Kr: [Ar]3d104s24p6 The 7th electron removed from Se is from an inner shell (n = 3) as compared to those removed from Br and Kr (n = 4). The inner shell electron is more strongly attracted to the nucleus and requires more energy for its removal. Kr has one more proton and a higher nuclear charge than Br. Since the additional electron in Kr is added to the same valence shell (n = 4), the valence electrons in Kr experience a shielding effect similar to those in Br. Thus, Kr has a higher effective nuclear charge than Br and more energy is required to remove the 7 th electron from Kr. FYI: element Se Br Kr 7th IE / kJ mol−1 14990 9940 10710 3 B Hexane molecule is non -polar and has id -id interactions between molecules. W ater molecule is polar and has hydrogen bonds between molecules. Hence statement 2 is correct. Hexane and water molecules form weak id -id interactions which do not release sufficient energy to overcome the stronger hydrogen bonds between water molecules. Hence statement 4 is correct. As a result , hexane and water molecules do not mix well and hexane is immiscible with water.  1 Hexane has a larger, more polarisable electron cloud than water, leading to stronger id -id interactions between hexane molecules. This statement is true but it does not help to explain the observation.  3 O–H bond has a larger bond energy than C–H bond so it is a true statement but it does not help to explain the observation as these covalent bonds are not broken or formed when hexane and water are mixed. 4 C  A Ice has a simple molecular structure with hydrogen bonds between water molecules. There are covalent bonds between atoms in the water molecules.  B Iodine, I2, has a simple molecular structure and is a non-polar molecule. There exists id -id interactions between molecules and covalent bond between iodine atoms in a molecule.  C Sodium sulfate, Na 2SO4, has a giant ionic structure with ionic bonds between oppositely charged Na + and SO 4 2− ions. Within the SO 4 2− ion, there are covalent bonds between S and O atoms.  D Copper has a giant metallic structure with metallic bonds between the positively charged metal cations and sea of delocalised electrons. 5 D pV = nRT  Mr = mRT/(pV) Mr If there are equal amounts of Al2Cl6(g) and AlCl3(g), average Mr Since average Mr = 214.9 > 200.25, there are more moles of heavier Al2Cl6(g) than AlCl3(g). Hence > 1

Dunman High School 2023 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions  DHS Chemistry Unit Page 2 of 6 6 C  1 H formation (H2O(l)) and H combustion (H2(g)) are represented by the same equation: H2(g) + ½O2(g) → H2O(l)  2 Bond energy values apply to covalent bonds in the gaseous molecules. H formation (H2O(l)) ≠ [BE(H–H) + BE O=O − 2BE(O–H)] because the enthalpy change of vaporisation of H 2O(l) has not been taken into account. FYI: H formation (H2O(g)) = BE(H–H) + BE O=O − BE O–H)  3 H2O(g) H2(g) + 1/2O2(g) Ho formation(H2O(l)) Ho vaporisation H2O(l) Ho formation(H2O(g)) H formation (H2O(g)) = H formation (H2O(l)) + H vaporisation (H2O(l)) 7 C H solution = −LE + H hydration (Na+(g)) + H hydration (Cl−(g)) 4 = 786 + H hydration (Na+ − H hydration (Na+ = −4 9 kJ l−1 Na(g) Na+(g) + e Na+(aq) + e 1st IE H Hhydration H = 494 + −419 = +75 kJ mol−1 8 D  A Al is a metal which conducts electricity in both the solid and molten state. P exists as P 4, a simple molecule, which does not conduct electricity in any state.  B Al2O3 is an ionic compound with partial covalent character due to the high charge density of the Al3+ ion. The oxide of P is P 4O10 which is a solid at room temperature.  C AlCl3 reacts rapidly (not slowly) with (limited) cold water to produce white fumes of HCl(g): AlCl3(s) + 3H2O(l)  Al(OH)3(s) + 3HCl(g) PCl5 reacts rapidly with (limited) water to produce white fumes of HCl(g): PCl5(s) + H2O(l) → POCl3(l) + 2HCl(g)  D AlCl3 reacts rapidly (not slowly) with (limited) cold water to produce white fumes of HCl(g): AlCl3(s) + 3H2O(l)  Al(OH)3(s) + 3HCl(g) When PCl5 is added to water , a strongly acidic solution is formed: PCl5(s) + 4H2O(l)  H3PO4(aq) + 5HCl(aq) Both observations are correct. 9 A  1 MCO3(s) → MO s + O2(g) where M = Ca, Ba For the same mass of MCO 3, CaCO 3 has a smaller Mr and a larger amount which leads to more moles of CO2(g) formed.  2 |LE| and are constant Since Ca 2+ has a smaller than B a2+ , CaCO3 has a larger magnitude of LE than BaCO3.  3 Since Ca 2+ has a smaller than B a2+ , Ca2+ has a higher charge density and polarising power than Ba2+ .  4 Since Ca 2+ has a higher polarising power than Ba2+ , the CO3 2− electron cloud is more polarised and the covalent bonds are weakened to a greater extent in CaCO 3. Hence CaCO3 requires less energy and a lower temperature for thermal decomposition. 10 B  A No observable change because there is no reaction between NH 3(aq) and the halide ions. No Ag+ ions was added.  B Brown I2(aq) is added in excess to colourless solution containing the halide ions. Hence the colourless solution turns brown.  C Br2 is a stronger oxidising agent than I2. Hence Br2(aq) is reduced to Br −(aq) while I−(aq) is oxidised to brown I2(aq). Since excess Br2(aq) is added, the resulting solution is brown, a mixture of orange Br2(aq) and brown I2(aq).  D White ppt of AgC l and yellow ppt of Ag I will be formed. 11 D  1 No. of mol of N2 molecules No. of N2 molecules  2 No. of mol of MgO2 Each mol of MgO 2 contains 1 mol of Mg 2+ cations and 1 mol of O2 2– anions. No. of moles of ions in 2.43 g of MgO2  3 Sodium carbonate has the formula Na 2CO3. Hence there are 6 mol of atoms in 1 mol of

Dunman High School 2023 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions  DHS Chemistry Unit Page 3 of 6 Na2CO3. 12 B When the hot gaseous product mixture is cooled to room temperature and pressure, water vapour condenses to form liquid water. HC l dissolves readily in water. Only N 2 (g) and O 2 (g) are collected under experimental conditions. Mr of NH4ClO4 No. of mol of NH4ClO4 used No. of mol of gases collected Volume of gases collected FYI: the experimental setup is shown in the diagram below 13 A n(BrO3 −) = (0.02)(0.02) = 0.0004 mol n(NH2OH) = (0.08)(0.01) = 0.0008 mol BrO3 − ≡ 2NH2OH ≡ e−  Every 1 mol of NH2OH lose 3 mol of e–  2O.N. of N in NH OH 3 1 2 1    O.N. of N in product 1 3 2    Product is NO (O.N. of N = +2) 14 C Step 2 is the rate–determining step. Therefore, rate = k’[ H l3][Cl]. Since Cl is an intermediate, it cannot be present in the rate equation and [Cl] is dependent on [Cl2] in step 1. Kc = [ [  [Cl