2024 Prelims DHS H2 Chem P1 (Ans)

2024 Y6 Preliminary Examination H2 Chemistry 9729 Paper 1 Suggested Solutions © DHS Chemistry Unit Page 1 of 5 Answer Key 1 2 3 4 5 6 7 8 9 10 B D B D B A D C B A 11 12 13 14 15 16 17 18 19 20 C C A D A B D D A C 21 22 23 24 25 26 27 28 29 30 A D C C C B C A B D 1 B P: [Ne]3s23p3 No. of unpaired electrons = 3  1 Ti: [Ar]3d24s2 Ti2+: [Ar]3d2 No. of unpaired electrons = 2  2 V: [Ar]3d34s2 V2+: [Ar]3d3 No. of unpaired electrons = 3  3 Cr: [Ar]3d54s1 Cr3+: [Ar]3d3 No. of unpaired electrons = 3 2 D H1 = 2nd IE of Si = +1580 kJ mol−1 H2 = 2nd IE of Al = +1820 kJ mol−1 H3 = sum of 1st & 2nd IE of Si = 786 + 1580 = +2366 kJ mol−1 order of decreasing enthalpy change: H3 > H2 > H1 3 B 2H2S + 3O2 → 2SO2 + 2H2O CS2 + 3O2 → CO2 + 2SO2 Hence, SO2 : CO2 will be 4 : 1.  Options A & C are incorrect. Both CO 2 and CS 2 are linear around the central C atom so both are non-polar molecules. CS2 has a larger, more polarisable electron cloud than CO2 so CS2 has stronger instantaneous dipole- induced dipole interactions between molecules. The more significant intermolecular forces result in greater deviation of CS2 from ideal behaviour. 4 D  A magnitude of lattice energy  | q+× q- r++ r- | This is a true statement (Ca2+; 0.099 nm, Na+; 0.095 nm). but a larger cationic radius leads to a less exothermic lattice energy so this does not explain the higher melting point of CaO.  B This is a true statement as Ca2+ has a higher charge than Na+ but their ionic radii are similar (Ca2+; 0.099 nm, Na+; 0.095 nm). However, magnitude of lattice energy  | q+× q- r++ r- | so charge density  q r of the cation alone is not representative of the lattice energy of the compound.  C This is a true statement: sum of ionic radii CaO NaF 0.099 + 0.140 = 0.239 nm 0.095 + 0.136 = 0.231 nm However, a larger sum of ionic radii leads to a less exothermic lattice energy so this does not explain the higher melting point of CaO.  D This is a true statement and the magnitude of lattice energy is larger when the magnitude of ionic charges are larger. 5 B molecule structure shape polarity NCl3 N Cl Cl Cl trigonal pyramidal polar HCN H−C≡N linear polar BeCl2 Cl−Be−Cl linear non-polar SOCl2 S Cl Cl O trigonal pyramidal polar Hence, NCl3 and SOCl2 are polar and have the same shape.

Dunman High School 2024 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions © DHS Chemistry Unit Page 2 of 5 6 A Solid X is Na2O which is soluble in water to form the colourless solution of NaOH(aq). On adding CuSO 4(s) to NaOH(aq), pale blue ppt of Cu(OH)2 is formed. Na2O(s) is also soluble in NaOH(aq) as it readily dissolves in water. Solid Y is insoluble in water and could be either Al2O3(s) or SiO 2(s). Since Y is soluble in dilute NaOH(aq), Y is Al2O3(s) which is amphoteric. Solid X (Na2O) is soluble in HC l(aq) as it readily dissolves in water. Solid Y (Al2O3) undergoes acid-base reaction with HCl(aq) to form a colourless solution: Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) 7 D  A As ionic radius increases down the group, charge density and hence polarising power of the cation decreases.  B The reducing (not oxidising) power of the elements increases.  C As polarising power of the cation decreases, the electron cloud of the chloride anion is polarised to a smaller extent so covalent character of the metal chlorides decreases.  D As polarising power of the cation decreases, the electron cloud of the carbonate anion is polarised to a smaller extent and the covalent bonds in the carbonate anion are weakened to a smaller extent. More energy is required to decompose the metal carbonate so thermal stability of the metal carbonates increases. 8 C Cl2 is a stronger oxidising agent than Br 2 so Cl2 will oxidise Br− to Br2 while itself is reduced to Cl−. Cl2(aq) + 2Br−(aq) → Br2(aq) + 2Cl−(aq) The colourless KBr solution turns orange due to the mixture of orange Br2(aq) formed and remaining pale yellow Cl2(aq) so options A & D are incorrect. On addition of AgNO 3(aq), white ppt of AgC l is formed. There is no remaining Br −(aq) ions to form cream ppt of AgBr. Ag+(aq) + Cl−(aq) → AgCl(s) 9 B  1 H2 = 5 × BE(Cl −Cl) = 5 × 244 = +1220 kJ mol−1 H1 = 2 × Hatomisation of P(s) so H1 > 0 since atomisation is an endothermic process. Hence H1 + H2 > +1220 kJ mol−1  2 Note: Reactants and products are in gaseous state when bond energy is used. H3 ≠ 10 × P−Cl bond energy H3 = (2 × Hvaporisation of PCl5(s)) + (10 × P−Cl bond energy) 2P(g) + 10Cl(g) 2PCl5(s) 2 x Hvaporisation of PCl5(s) H3 2PCl5(g) 10 x BE(P Cl)  3 2 mol of PCl5(s) are formed from the constituent elements in their standard states. 10 A S < 0 due to a decrease in number of moles of gas (4 mol to 2 mol). Hence −TS > 0 Since H < 0 and G = H − TS, G < 0 when |−TS| < |H| Hence the reaction is spontaneous only at low temperature. 11 C Let the unknown iodine oxide be IxOy. Given, IxOy : I− : I2 0.02 : 0.2 : 0.12 1 : 10 : 6 In order to balance the number of I atoms, x = 2 Iodine in I2Oy is reduced to I2. Iodide is oxidised to I2. Since [O]: 2I− → I2 + 2e− Hence, number of mol. of electrons lost by 10 mol. of KI = 10 mol = number of mol. of electrons gained by 1 mol. of I2Oy So, mol ratio of I2Oy : e− gained = 1:10 ∴mol ratio of each I in I2Oy : e− gained = 1:5 ∴ the oxidation state of each I in I2Oy is +5 to produce I2.

Dunman High School 2024 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions © DHS Chemistry Unit Page 3 of 5 12 C px ∝ [Cl2] Since it is given that the reaction is zero order w.r.t. Cl2, the rate of reaction should remain unchanged when there is a change to [Cl2]. The gradient of px vs time graph is indicative of the rate of reaction. Hence, the gradient of px vs time graph should remain constant even when px decreases with time due to reaction (downward sloping straight line). 13 A  1 Colorimetry can be used to measure how the light absorbance of the reaction solution changes at regular intervals as the reaction takes place. The light absorbance is proportional to the colour intensity which is in turn proportional to the concentration of the coloured substance. The rate of reaction is directly proportional to the rate of decrease (or increase) in colour intensity of the reactant (or product).  2 Concept: All total volumes were kept constant for all experiments, hence [reactant] ∝ Vreactant. Comparing Expts 1 & 4, when volumes of B and Y were kept constant, while volume of the coloured solution A was doubled, time taken was doubled.  rate of decolourisation remained constant. ∴Order of reaction w.r.t. A is 0. Comparing Expts 1 & 2, when volumes of A and Y were kept constant, while volume of B was doubled, time taken was halved  rate has doubled. ∴Order of reaction w.r.t. B is 1. Comparing Expts 1 & 3, when volumes of A and B were kept constant, while volume of Y was doubled, time taken was halved  rate has doubled. ∴Order of reaction w.r.t. Y is 1. So, rate equation is: rate = k[B][Y].  3 From option 2, order of reaction w.r.t. A is 0. Comparing Expts 4 & 5, since volumes of B and Y were kept constant, while volume of A was halved, time taken should also be halved since rate of decolourisation should remain constant. Hence, the time taken for Expt 5 is 5 s. 14 D Concept: Adding inert gas at constant volume results in an increase in total pressure (due to increase in amount of gaseous particles at a constant volume)  this accounts for the spike observed at X as well as the increase in total