2024 DHS Y6 H1 Prelim Paper 1 Worked Solutions

2024 Y6 Preliminary Examination H1 Chemistry 8873 Paper 1 Suggested Solutions © DHS Chemistry Unit Page 1 of 5 Answer Key 1 2 3 4 5 6 7 8 9 10 C B A D B D B C C B 11 12 13 14 15 16 17 18 19 20 A D C C A B D A D B 21 22 23 24 25 26 27 28 29 30 C A A D C C B A D D 1 C S: 1s22s22p63s23p4 S+: 1s22s22p63s23p3  options A and B are incorrect. By Aufbau principle, the lower energy 3s orbital should be fully filled before occupying the higher energy 3p orbitals  option D is incorrect. Note: By Hund’s rule, the 3p electrons will occupy the three degenerate 3p orbitals singly. 2 B Given the relative masses, V and W are isotopes of sulfur while X and Y are isotopes of iron.  1 Isotopes have the same number of protons (and electrons) but different number of neutrons. Hence, Y (not Y2+) is isoelectronic with X.  2 isotope Y26 56 V16 32 number of neutrons 56 – 26 = 30 32 – 16 = 16  3 relative atomic mass of iron in this sample = (54)(31.4)+(56)(2) 31.4+2 = 54.1  4 V2+ and W2+ have the same charge of 2+ but V2+ has a smaller relative mass than W2+ Since angle of deflection  charge mass , V2+ has a larger angle of deflection than W2+. 3 A All p orbitals have the same dumbbell shape and two lobes per orbital. A 3p orbital is larger than a 2p orbital. A px orbital lies on the x -axis which is perpendicular to a pz orbital which lies on the z-axis  different orientation of the orbitals in space 4 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 5 B  A & D Factors affecting atomic radius down the group are number of electron shells, shielding effect and nuclear charge  options A and D are incorrect.  B Increase in number of filled electron shells increases shielding effect and distance of valence electrons from the nucleus. This helps to explain the larger atomic radius of iodine.  C Increase in number of protons increases nuclear charge and the attraction of the nucleus on the valence electrons. This does not explain why iodine has a larger atomic radius than chlorine. 6 D magnitude of lattice energy  | q+× q- r++ r- | From the Data Booklet, Mg2+; 0.065 nm, Na+; 0.095 nm, O2−; 0.140 nm, F−; 0.136 nm. Magnitude of product of ionic charges: MgO > MgF = Na2O > NaF The sum of ionic radii: MgO (0.205 nm) ≈ MgF (0.201 nm) < NaF (0.231 nm) ≈ Na2O (0.235 nm) Hence MgO has the largest magnitude of lattice energy.

Dunman High School 2024 Y6 Preliminary Examination – H1 Chemistry 8873/01 Solutions © DHS Chemistry Unit Page 2 of 5 7 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. 8 C  A amount of Na2SO4 = 0.08 × 0.1 = 0.008 mol Since one formula unit of Na 2SO4 contains three ions (two Na+ and one SO42−), total amount of ions 0.008 mol of Na2SO4 = 0.008 × 3 = 0.024 mol  B amount of Br2(l) = 4 / (79.9 × 2) = 0.0250 mol  C amount of H2(g) = 500 / 24000 = 0.0208 mol  D amount of H2O = 1 / 18 = 0.055555 mol amount of atoms = 0.055555 × 3 = 0.167 mol 9 C  A A mole of substance is the amount of that substance which contains as many elementary entities as there are carbon atoms in 12 grams of carbon-12. The correct statement should be: One mole of compound is the amount that contains the same number of compounds as there are atoms in 12.000 g of carbon-12. Note: a compound contains at least two elements and each formula unit contains at least two atoms  B Relative isotopic mass = mass of 1 atom of the isotope 1 12 x mass of 1 atom of carbon-12 The correct statement should be: Relative isotopic mass of lithium-7 = mass of 1 atom of the lithium-7 isotope 1 12 x mass of 1 atom of carbon-12  C Relative atomic mass = average mass of 1 atom 1 12 x mass of 1 atom of carbon-12  D Relative molecular mass = mass of 1 molecule 1 12 x mass of 1 atom of carbon-12 The correct statement should be: Relative molecular mass of E = mass of 1 molecule of E 1 12 x mass of 1 atom of carbon-12 10 B amount of Na 2S2O3 needed to react with iodine liberated = 0.010 × 0.0225 = 2.25 × 10−4 mol from the Data Booklet, Cr2O72−  3I2 from the question, 3I2  6S2O32– Hence, Cr2O72−  3I2  6S2O32− Moles of the Cr2O72− ions = 1/6 × (2.25 × 10−4) = 3.75 × 10−5 mol [Cr2O72−] in 25 cm3 solution = 3.75 × 10−5 / 0.025 = 0.00150 mol dm−3 11 A Solid X is Na2O which is soluble in water to form the colourless alkaline solution of NaOH(aq). Note: X cannot be MgO because MgO is only slightly soluble in water so there will be white solid of MgO(s) remaining in the alkaline solution of Mg(OH)2(aq). 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) 12 D  A first ionisation energy: Mg (736 kJ mol−1) < S (1000 kJ mol−1) < P (1060 kJ mol−1)  B electronegativity: increases across the Period so Al < Si < Cl  C melting point: Al > Na > P  D electrical conductivity: S (non-conductor) < Si (semi-conductor) < Na (conductor)

Dunman High School 2024 Y6 Preliminary Examination – H1 Chemistry 8873/01 Solutions © DHS Chemistry Unit Page 3 of 5 13 C AlCl3(s) undergoes partial hydrolysis in water: AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl–(aq) [Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5(OH)]2+(aq) + H3O+(aq) (pH = 3) PCl5(s) undergoes complete hydrolysis in water: PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq) (pH = 2) Hence option D is incorrect. Acidic solutions are formed so option A is incorrect. Both chlorides react rapidly with water, so option B is incorrect. Note: acidic fumes of HCl(g) are observed when the chlorides react with limited cold water: PCl5(s) + H2O(l) → POCl3(l) + 2HCl(g) AlCl3(s) + 3H2O(l) → Al(OH)3(s) + 3HCl(g) NaCl dissolves in water and no hydrolysis occurs so pH of solution formed = 7. NaCl(s) → Na+(aq) + Cl–(aq) 14 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) 15 A The reaction took 10 s for [N 2O5] to decrease from 0.040 to 0.020 mol dm−3, and another 10 s for [N2O5] to further decrease from 0.020 to 0.010 mol dm−3. Hence, the half-life of the reaction is constant at 10 s and the order of reaction with respect to N2O5 is 1. It follows that rate = k[N2O5] and the units of k is s−1. 16 B Comparing experiments 1 & 2, When initial [BrO3−] × 2 while keeping initial [Br−] and [H+] constant, initial rate × 2. Hence order of reaction with respect to BrO3− is 1  options A & D are incorrect. Comparing experiments 1 & 3, When initial [Br−] × 2 and initial [H+] × 3 while keeping initial [BrO3−] constant, initial rate × 18. The factor of 18 can be attributed to 21 × 32 where the order of reaction with respect to Br− and H+ are 1 and 2 respectively. 17 D  1 HNO3 oxidises P 4 to H 3PO4 while itself is reduced to NO2. 