2023 RI Prelim H2 Chem Paper 2 Solutions

1 2023 Y6 H2 Chemistry Preliminary Examinations Paper 2 Suggested Solutions 1 (a) The volatilities of the halogens decrease from chlorine to iodine. From C l2 to I2, the electron clouds of the halogens become larger and more polarisable. Hence, more energy is required to overcome the increasing strength of the instantaneous dipole– induced dipole interactions between the halogen molecules down the group, leading to decreasing volatility. Examiner Comments • Volatility is the tendency of a substance to evaporate and is related to the intermolecular forces of attraction. • There were a number of candidates who did not realise that volatility involves intermolecular forces of attraction and not covalent bond strength (volatility does not involve separating the molecule into its atoms). • Terms used should be spelt out in full i.e. , “instantaneous dipole–induced dipole interactions”, not “id-id”. • Some students talked about the trend in boiling point, which was irrelevant in this question. (b) (i) X Y Z nucleon number 140 103 37 charge +1 +1 0 Examiner Comments • Very well done, with a small number of candidates making mistakes most likely due to not reading the question carefully enough. (ii) Examiner Comments • A number of candidates drew the path for Z instead. Do read the question carefully. • The paths drawn should end outside of the “+” and “ –” plates. Please also note that within the electric field, the paths for X and Y are curved, as shown in the above diagram. Y X

2 • It should be very clear to the examiner that Y deflects to a larger extent than X. • A number of candidates drew a longer horizontal line for X before deviation, with a deflection angle similar to that of Y. Please note that the deviation should start at the same spot. (iii) nucleon number = 103 = 31 + 35 + 37 There is 1 35Cl atom in Y. Y is PCl2+. Its shape is bent. Examiner Comments • Note that there is one lone pair on the central atom P. (c) (i) P(g) + e– → P–(g) Examiner Comments • Very well done, except some missed out giving state symbols. Note that state symbols should be shown without being asked whenever an equation is used to reflect the terms in Energetics e.g., ionisation energy, enthalpy change of combustion, enthalpy change of atomisation, etc (ii) 1s 2s 2p 3s 3p Examiner Comments • Very well done, except some candidates gave the arrangement for the P atom instead of the P– ion. (iii)  Nuclear charge increases from A l to Cl but shielding effect remains effectively constant. Effective nuclear charge increases. Electrostatic attraction between the nucleus and the incoming electron increases, resulting in an increase in the energy released. Examiner Comments • Be precise in your answer – “approximately constant” is not the same as “constant”. • The attraction is between the nucleus (not atom) and an incoming electron here, not with a “valence” electron. Note: P atom has 5 valence electrons, and the incoming electron enters the valence shell.

3  The attraction between the nucleus and incoming electron is weakened due to inter-electronic repulsion between this electron and the one already present in the 3p orbital. Hence, less energy is released as the first electron affinity. Examiner Comments • Be very clear with the terms used and do not use them loosely or interchangeably. • A number of candidates were confused and used incorrect terminology: a 3p orbital is not the same as a 3p subshell. • Note: “electrons” cannot be filled up, “orbitals” cannot be paired. • A “partially filled 3p subshell” does not mean it contains one electron in each of the 3p orbitals – such a phrase just means it can contain between 1 to 5 electrons in total. • Some phrases implied that inter -electronic repulsion only exists between the incoming electron and the electron in the 3p orbital – note that repulsion exists between inner electrons, as well as between the single electrons in each of the half-filled 3p orbitals. 2 (a) Au has a smaller atomic size and thus have more atoms per unit volume . It also has a larger atomic mass than Ba. Hence it has a greater mass per unit volume (i.e. higher density) as compared to Ba. Examiner Comments • This question, covering a fundamental concept, was not well done. Many students incorrectly thought that the atomic radius of Au and Ba are similar or did not mention about Au having a larger atomic mass. • It is insufficient to just state that “ Au has a larger mass per unit volume” without comparing the atomic size and atomic mass of the two metals as the question already mentioned that Au has a higher density than Ba. That answer simply paraphrased the question. • Many incorrect and lengthy answers went on to talk about shielding effect, effective nuclear charge, charge density etc , all of which are all unrelated to explaining density, a macro physical property which is defined as mass/volume. (b) max % of Au in amalgam = 100% − 40% = 60% max mass of Au in 250 g of amalgam = 0.6 × 250 = 150 g Examiner Comments • This question was well done. (c) conc of Hg vapour = 1 g of Hg / 1 m3 of air = 1 g of Hg / 1.19 kg of air = 1 g of Hg / 1190 g of air = ( 1 1190 ) g of Hg / 1 g of air = ( 1 1190 × 106) g of Hg / 106 g of air = 840.3 g of Hg / 106 g of air ≈ 840 ppm Examiner Comments • This question was quite well done.

4 (d) (i) Water prevents / reduces the vaporisation / evaporation of the collected mercury. Examiner Comments • This question was well done. • A handful of students incorrectly thought that mercury dissolves in water or react with water. Such answers received no credit. (ii) When heated, the mercury in the amalgam vaporises and then condenses on the cooler inner surface of the glass bowl . Then it slides back down to accumulate on the sand to be collected and recovered. Examiner Comments • This question was well done with most students recognising that the mercury first vaporises upon heating and then condenses. (iii) advantage • low cost/convenient to set up because of the use of simple household items or no need special equipment (e.g. metal apparatus) • no mercury-contaminated wastewater produced disadvantage • the mercury collected will mix with the sand and hence be harder to recover • the setup is not as airtight as Method I, i.e. mercury vapour has higher chance to escape back into the atmosphere • when the bowl is removed, mercury can vaporise again whilst collecting the mercury from the sand • mercury poisoning from re- use of cooking pot used in the extraction for cooking purposes Examiner Comments • For full credit, students should substantiate their answers appropriately. • Most students correctly recognised that method II would be cheaper or only required household items/less specialised equipment which are an advantage for the miners who are often from poverty-stricken communities (this info was given in the question). • Discussion on scale or efficiency of process were not given credit as the diagrams were labelled to be not drawn to scale and hence insufficient info to draw conclusion on scale or efficiency. (e) (i) O2 + 2H2O + 4e– → 4OH– Examiner Comments • Most students were able to obtain the correct half -equation from the Data Booklet . The clue in the question was in the presence of oxygen, in an aqueous solution, to give hydroxide ions. • As the question already mentioned O 2 is reduced, students should use the irreversible arrow instead of the reversible arrow to illustrate reduction reaction.

5 (ii) mol ratio 4Au : 1O2 : 4e– Each Au atom loses 1 e– to form Au+. The oxidation state of Au is +1. Examiner Comments