DHS 2026 Prelim P1 Ans

2026 Y6 Preliminary Examination H2 Chemistry 9476 Paper 1 Suggested Solutions © DHS Chemistry Unit Page 1 of 6 Answer Key 1 2 3 4 5 6 7 8 9 10 C D A C C D B C C A 11 12 13 14 15 16 17 18 19 20 B A B D C A B D B C 21 22 23 24 25 26 27 28 29 30 D A C D A A B B B D 1 C Both He24 + and H11 + (proton) carry the same charge of +1 but He24 + has a larger mass. Since angle of deflection ∝ q m, with a larger mass, the angle of deflection for He24 + is smaller than that of the proton. Both are positively-charged and hence, will be deflected towards the negatively-charged plate. 2 D electronic configuration of ground state oxygen: 1s2 2s2 2p4  A The occupied orbital of highest energy is 2p which has a dumbbell shape.  B There are two unpaired electrons present in the n = 2 energy level.  C The electrons are distributed across a total of three subshells and five orbitals in total.  D There are equal number of electrons in p orbitals as well as s orbitals; 4 electrons in 1s, 2s and 4 electrons in 2p. 3 A  1 In liquid state, water molecules are not in fixed positions and the shifting of positions/ kinetic motion causes the hydrogen bonds to continually disconnect and reform.  2 During melting, intermolecular hydrogen bonds are disrupted. The covalent O –H bonds within each water molecule remain intact.  3 In the solid state, the two lone pairs of electrons on O atom and the two H atoms in each water molecule result in a regular tetrahedral structure where each water molecule forms four hydrogen bonds to four other water molecules, creating an open structure.  4 Pure water undergoes only slight self - ionisation, so it contains very few mobile ions and conducts electricity only very weakly. 4 C  1 trigonal planar, no net dipole, non-polar  2 trigonal pyramidal, net dipole, polar  3 trigonal bipyramidal, no net dipole, non-polar  4 bent, net dipole, polar 5 C Based on general combustion equation of hydrocarbons, nCxHy : nO2= 1 : (x+ y 4)O2 For combustion of C3H6, nC3H6 : nO2= 1 : (3+ 6 4) = 1 : 4.5 For combustion of C4H6, nC4H6 : nO2= 1 : (4+ 6 4) = 1 : 5.5 Minimum number of moles of O2 molecules needed = 4.5 + 5.5 = 10 6 D Since pV=nRT,  A When T is not at 0 K, p cannot be at 0 Pa so the graph should not pass through the origin. The units on the x-axis of the graph is in C and 0 C is 273 K.  B nR is the gradient of pV against T graph. At 140 C the conversion reaches near completion and n should ≈2 mols, i.e. gradient from – 10 to 140 C should continuously increase smoothly, until the gradient has doubled at 140 C. The gradient should not increase and then decrease as seen in B.

Dunman High School 2026 Y6 Preliminary Examination – H2 Chemistry 9476/01 Solutions © DHS Chemistry Unit Page 2 of 6  C V = R P (nT) with constant R P The change in gradient of V vs T graph should not be sharp and should not only occur at 413K. The gradient should continuously increase smoothly as the dissociation process goes towards near completion from 263 to 413 K, with the gradient beyond 413 K being about double that of 0 to 263 K, since n has almost doubled.  D Since pV = nRT, T= 1 nR pV, so the gradient of T against pV graph is 1 nR . Hence from from 263 to 413 K, as n increases from 1 mol to ≈2 mols, the gradient decreases continuously and smoothly, until it is halved at 413 K. Above 413 K, negligible dissociation occurs, and n remains ≈2, and hence the line continues above 413 K as a straight line. 7 B Fe3+ + e‒ → Fe2+ Given amount of NH3OH+ ions : amount of Fe3+ = 1 : 2, amount of electrons gained by 2 mol Fe3+ = 2 mol = amount of electrons lost by 1 mol of NH3OH+ Initial oxidation number of N in NH3OH+ = x x + 4 + (‒2) = +1 x = ‒1 Hence, final oxidation number of N = –1 + 2 = +1 8 C  A MgCl2 + 6H2O → [Mg(H2O)6]2+ + 2Cl– [Mg(H2O)6]2+ ⇌ [Mg(H2O)5(OH)]+ + H+  B AlCl3 + 6H2O → [Al(H2O)6]3+ + 3Cl– [Al(H2O)6]3+ ⇌ [Al(H2O)5(OH)]2+ + H+  C SiCl4 + 2H2O → SiO2 + 4HCl 0.5 mol 2.0 mol  D PCl5 + 4H2O → H3PO4 + 5HCl 0.5 mol 2.5 mol 9 C  A The melting point of a Group 2 element is dependent on the charge density of the cations. Since the charge density decreases with increasing ionic radius down the group, less energy is required to overcome weaker electrostatic forces of attraction between positively charged cations and sea of delocalised electrons, hence the melting points of Group 2 elements decrease down the group.  B The magnitude of the enthalpy change of hydration of the metal ions will decrease as charge density decrease. Hence, magnitude of the enthalpy change of hydration of the metal ions will decrease down the group.  C The thermal stability of the Group 2 carbonates increases down the group as the polarising power of the metal cations decreases down the group. Hence, more heat energy is required to thermally decompose the carbonates.  D The acidity of the aqueous solution of the chlorides decreases down the group as the polarising power of the metal cations decreases down the group. 10 A  1 Volatility is inversely related to boiling point and since H I has a higher boiling point, H I is less volatile than HBr. However, it is due to HI having a larger, more polarisable electron cloud and hence stronger instantaneous dipole - induced dipole interactions between molecules.  2 Thermal stability increases with bond strength. From the Data Booklet, the bond energy of HF is 562 kJ mol −1 compared to HI with bond energy 299 kJ mol −1. Hence, HF is thermally more stable than HI.  3 The bond length of Cl2 is expected to be shorter than that of Br 2. However, the oxidising power of halogens is related to their affinity for external source of electrons and not due to the bond length. Gradient should not increase and then decrease!

Dunman High School 2026 Y6 Preliminary Examination – H2 Chemistry 9476/01 Solutions © DHS Chemistry Unit Page 3 of 6 11 B Magnesium oxide has the most exothermic lattice energy while sodium bromide has the least exothermic lattice energy. Hence, options A and D are ruled out. |LE| of magnesium bromide  | 2×1 0.065+0.195 | = 7.69 |LE| of sodium oxide  | 1×2 0.095+0.140 | = 8.51 Hence, the lattice energy of magnesium bromide is less exothermic than that of sodium oxide. 12 A  1 Since the temperature decreases upon dissolution, the enthalpy change of dissolution is endothermic. Since the process occurs readily, the Gibbs Free Energy must be negative and therefore, the entropy change must be positive.  2 The decrease in temperature of the water results in the decrease in the temperature of the system. Hence, G will be more positive as the magnitude of TS becomes smaller.  3 Hsol = –LE + [Hhyd(NH4+) + Hhyd(Cl–)] Since the enthalpy change of solution is endothermic (positive), sum of the hydration energies of ammonium ion and chloride ion is more endothermic/ less exothermic than the lattice energy of ammonium chloride. 13 B  A Iodine is not a catalyst as it is not regenerated. The order of reaction with respect to iodine is zero.  B The rate equation is rate = k[CH3COCH3][H+] since the slow step involves one CH3C(OH)CH3+ and [CH3C(OH)CH3+] = Kc[CH3COCH3][H+].  C The rate equation must only show the concentration of reactants and not intermediates.  D Rate constant is only dependent on temperature and the presence of a catalyst. Increase in [H+] will increase the rate of reaction but not the rate constant. 14 D The rate of decomposition is 0.01 (= 0.1 × 0.1) mol dm –3 for every 5 minutes for