EJC 2022 GCE A-Level H1 Chemistry Paper 1 Suggested Solution

Eunoia Junior College 8873 H1 Chemistry 2022 Paper 1 Suggested Solution 1 Only A and B with 2 protons has a charge of +2 on the nucleus. The mass of a n atom or ions is due primarily to protons and neutrons in the nucleus as an electron is ~ 1 1836 the mass of a proton or neutron. B with 4 nucleons (neutrons and protons) will have a greater mass than A with 3 nucleons.  B 2 A : 22Ti: 1s2 2s2 2p6 3s2 3p6 3d2 4s2 → 22Ti2+: 1s2 2s2 2p6 3s2 3p6 3d2 B : 24Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 C ✓: 26Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 D : 29Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 → 29Cu+: 1s2 2s2 2p6 3s2 3p6 3d10  C 3 Atomic radius decreases across the period due to increasing effective nuclear charge. Excluding Group 18, the element with the smallest atomic radius in its period must be a Group 17 element. For an element to have o nly 2 electrons in its p sub -shell, the element must have a valence electronic configuration of ns2 np2, i.e. in Group 14.  A 4 As is [Ar] 3d10 4s2 4p3 (Period 4, Group 15) Sb is [Kr] 4d10 5s2 5p3 (Period 5, Group 15) Se is [Ar] 3d10 4s2 4p4 (Period 4, Group 16) Te is [Kr] 4d10 5s2 5p4 (Period 5, Group 16) There will be a significant jump betwee n the 3rd and 4th IE for a Group 15 element (A and C) since the 4 th electron is removed from a lower energy ns subshell, while the first 4 electrons are removed from the np subshell for a Group 16 element , showing a more gradual increase (B or D). A period 5 element in the same group will have lower xth IE compared to a period 4 element as the electrons removed are further away from the nucleus.  B 5 1 ✓: Ice adopts an open hexagonal lattice structure due to hydrogen bonding , resulting in ice having a lower density than water. 2 ✓: The high surface tension of water is due to the strong intermolecular hydrogen bonds between water molecules on the surface. 3 : Water can act as an acid since it can donate a H +, and as a base since it can accept a H +, exemplified by its auto-ionisation: H2O(l) + H2O(l)  H3O+(aq) + OH–(aq) 4 : The pH of pure water at 25 ºC is 7.00 due to the auto-ionisation of water: H2O(l) + H2O(l)  H3O+(aq) + OH–(aq) resulting in equal concentration of H+ and OH– of 10–7 mol dm–3 each.  B 6 The Arrhenius de finition of a base is a substance that dissociates to produce hydroxide ions in aqueous solution.  C 7 pH = –lg [H+] [H+] at pH 4 = 10–4 mol dm–3 [H+] at pH 6 = 10–6 mol dm–3 change in [H+] = 6 4 10 1 10 100 − − =  B 8 ( ) ( )2Ba OH 3 1.00 137.3 16.0 1.0 2 5.838 10 mol n − = + +  = ( ) 3 3 2 5.838 10Ba OH 0.01168 mol dm500 1000 − − == Ba(OH)2 → Ba2+ + 2OH– ( ) 3 2OH 2 Ba OH 0.02335 mol dm−−  ==  pOH lg OH 1.63 pH 14 pOH 12.37 −=− = = − =  D 9 A buffer contains a large reserv oir of a weak acid, HA, and its conjugate base, A–. The conjugate base A – is able to remove small amount of H+ added in: A– + H+ → HA The weak acid HA i