2021 DHSH2 Prelim Paper 1 Worked Solutions

2021 Y6 Preliminary Examination H2 Chemistry 9729 Paper 1 Suggested Solutions ã DHS Chemistry Unit Page 1 of 6 Answer Key 1 2 3 4 5 6 7 8 9 10 C D B D B C C A D C 11 12 13 14 15 16 17 18 19 20 B B C A C A D D B C 21 22 23 24 25 26 27 28 29 30 A B A D C A D B A C 1 C No. of electrons in Re3+75187 = 75 – 3 = 72 No. of neutrons in Re3+75187 = 187 – 75 = 112 Calculate the number of electrons and neutrons in each ion and compare to that of Re3+75187.This ion should have 72 electrons and 112 neutrons. O A No. of electrons in 182W2+ (Z = 74) = 74 – 2 = 72 No. of neutrons in 182W3+ = 182 – 74 = 108 (≠ 112) O B No. of electrons in 186W+ (Z = 74) = 74 – 1 = 73 (≠ 72) No. of neutrons in 186W+ = 186 – 74 = 112 P C No. of electrons in 188Os4+ (Z = 76) = 76 – 4 = 72 No. of neutrons in 188Os4+ = 188 – 76 = 112 O D No. of electrons in 189Os3+ (Z = 76) = 76 – 3 = 73 (≠ 72) No. of neutrons in 189Os3+ = 189 – 76 = 113 (≠ 112) 2 D The largest increase in successive ionisation energies (IE) occur between 9th and 10th IE. This implies that the 10th electron is removed from an inner shell. Since Z ≤ 20, only s and p subshells exists. Hence, each shell can only accommodate up to a maximum of 8 electrons. This would imply that the 2nd to 9th electron also lies in another inner shell. There will be only one electron in the valence shell. This element therefore belongs to Group 1 (either Na or K). O A The element belongs to Group 1 since it has only one electron in the valence shell. O B There is insufficient data to conclude that the element lies in the third period. Since Z ≤ 20, this Group 1 element can be in the fourth period. O C As a Group 1 element, the 6th and 7th electrons are removed from the same p-subshell. P D There is a large electronegativity difference between the Group 1 element and oxygen (Group 16). Hence, the compound formed will be ionic in nature. 3 B O A BeCl2 is linear while SO2 is bent. P B Both CO2 and I3− are linear. O C NO3− is trigonal planar while PH3 is trigonal pyramidal. O D PO43− is tetrahedral while SF4 is see-saw. Be ClCl S OO C OO II I N OO O P H H H P O OO O 3 S F F F F

Dunman High School 2021 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions ã DHS Chemistry Unit Page 2 of 6 4 D O A In giant molecular structures, the lattice of atoms are held together by strong covalent bonds. O B Ionic bonding is non-directional. Each cation is electrostatically attracted to any anion, not necessarily to the anion to which electrons were transferred. O C In giant metallic structures, only valence electrons are delocalised within the metal lattice. P D Simple molecular structures are held together by weak intermolecular forces of attraction such as instantaneous dipole-induced dipole (id-id) interactions, permanent dipole-permanent dipole interactions and hydrogen bonding. 5 B Based on the mole ratio, CH3OH is the limiting reagent. 2CH3OH(g) + 3O2(g) ® 2CO2(g) + 4H2O(g) Initial V / dm3 x 4x 0 0 Change in V / dm3 −x −1.5x +x +2x End V / dm3 0 2.5x x 2x g P 1 From the equation above, mole ratio between CH3OH(g) and O2(g) is 2 : 3. P 2 Partial pressure of CO2, p(CO2), can be calculated using Dalton’s Law of Partial Pressure: p(CO2) = p(CO2) = 0.18P O 3 Assuming these gases behave ideally, the gas particles occupy a negligible volume as compared to the volume of the vessel containing these gases. 6 C O A Electronegativity increases across the period. In Period 3, P precedes S and hence P has a lower electronegativity. Similarly, Arsenic precedes Se in Period 4, As is expected to have a lower electronegativity. O B In Period 3, the highest oxidation state of an element usually corresponds to maximum number of valence electrons. Using this idea, arsenic being a Group 15 element should have a highest oxidation state of +5 instead of +6. P C As is below P in Group 15. Since PCl5 reacts with water to produce HCl as one of its products, the solution is acidic. PCl5(s) + 4H2O(l) ® H3PO4(aq) + 5HCl(aq) O D Oxides of As should behave chemically similar to that of P, which are acidic in nature. They react with water to form acids rather than bases. P4O10(s) + 6H2O(l) ® 4H3PO4(aq) 7 C The Group 2 metal carbonates (MgCO3, CaCO3 and BaCO3) are neutralised by dilute sulfuric acid to form a metal sulfate, water and carbon dioxide. MgSO4 dissolves in water to give a colourless solution while CaSO4 and BaSO4 are sparingly soluble, resulting in formation of white precipitate. From the table of observations given in the question, L can be deduced to be Mg. CaCO3 undergoes thermal decomposition at a lower temperature than BaCO3 since Ca2+ having smaller ionic radius will have a stronger polarising power than Ba2+. The electron cloud of the CO32− ion in CaCO3 would be polarised to a greater extent. CaCO3 would be less thermally stable than BaCO3 and decomposes at a lower temperature. Since K decomposes at a higher temperature than J, K is Ba while J is Ca. 8 A P A Down the group, Group 17 hydrides (H–X) would be more acidic since H–X bond energy decreases and results in decreased bond strength. It would be easier to break the H-X bond and form H+ down the group. O B Oxidising power of the elements decreases down the group as reflected by the less positive E¡ values. The ability of the halogens to be reduced decreases and correspondingly, its ability to oxidise other species decreases. O C Hydrogen halides are colourless gases at room temperature. The colour intensity of the elements increases down the gro O D Down the group, the electron cloud size increases and becomes more polarisable. More energy is required to overcome the stronger instantaneous dipole-induced dipole interactions between non-polar halogen molecules. Boiling point increases and hence volatility decreases. total totall 2 2 (CO ) (CO ) n pp n =´ 2.5 2 x P xx x ´ ++

Dunman High School 2021 Y6 Preliminary Examination – H2 Chemistry 9729/01 Solutions ã DHS Chemistry Unit Page 3 of 6 9 D Since enthalpy change of formation is required to be calculated, simply sum up all the required enthalpies (atomisation, ionisation energy, electron affinity and lattice energy), taking into account mole ratios where appropriate. 10 C O 1 The enthalpy change of hydration of sodium ion is less exothermic than that of magnesium ion. Recall: Calculate the ratio of for Na+ and Mg2+: ⇒for Na+ = =10.5 ⇒for Mg2+ = =30.8 ∵, Mg2+ has a more exothermic enthalpy change of hydration than Na+. P 2 Recall: Lattice energy,. Since Mg2+ has twice the charge and a smaller ionic radius than that of Na+ given the same anionic charge and radius, lattice energy for MgBr2 will be more exothermic than that of NaBr. P 3 A more exothermic enthalpy change of solution suggests higher solubility in water. Since NaBr has a less exothermic enthalpy change of solution (–18 kJ mol–1) than that of MgBr2 (–216 kJ mol–1), NaBr is likely to be less soluble in water than MgBr2. 11 B Oxidation state of S changes from +4 in SO2 to +6 in SO3. Since SO2 ≡ V2O5, there are 2 mol of electrons transferred to 1 mol of V2O5. 12 B The rate equation is derived from the slow step of the mechanism i.e. Rate = k[NO2]2 P 1 The reaction is zero order w.r.t. CO and the change in its partial pressure will not affect the rate. P 2 Since step 1 is the slow step, it has a higher activation energy and hence a smaller rate constant. O 3 Units for k = !"#$% '() )*$+!"#$% '() [-.!]! = mol–1 dm3 s–1 13 C Since reaction is first order with respect to H2O2, rate = k[H2O2] and rate is directly proportional to [H2O2]. The graph of [H2O2] against rate will be an upward sloping straight line passing through the origin. 14 A N2O4(g) ⇌ 2NO2(g) Initial moles / mol 1 0.2 Change in moles / mol –0.24 +0.48 Eqm moles / mol 0.76 0.68 P 1 See ICE table above. O 2 Kc = [