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 pyramida