2021 C1 Promo Teaching Solutions Compiled

2021 HCI C1 H2 Chemistry Promotional Exam / Paper 1 HWA CHONG INSTITUTION 2021 C1 H2 CHEMISTRY PROMOTIONAL EXAM SUGGESTED SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 C C B B D A C A D B 11 12 13 14 15 16 17 18 19 20 D A B C B C B C D A Comments 1 C Since the s and p subshells in inner quantum shells are completely filled, the s and p electrons from the inner quantum shells are all paired. Hence, we only need to consider the valence shell electronic configuration of the particles. Option Explanation 1 Cr: 3d54s1. It contains an unpaired s electron. 2 Ge: 4s24p2. It contains an unpaired p electron. 3 S: 3s23p4. Hence, valence shell electronic configuration of S 2is 3s23p6. It does not contain an unpaired s electron or an unpaired p electron. 4 Sc: 3d14s2. It does not contain an unpaired s electron or an unpai red p electron. General comments: Give answers that are clear, concise, coherent and in legible handwriting. Do not cram your answers or write into the edges of the page. If more space is required, use the additional page at the end of Paper 3, or request for writing paper for Paper 2. Attempt all calculation questions, so that you would be able to move on to subsequent parts. Follow through marks may be awarded (error carried forward) for correct working steps. For Paper 3, answers should always be written on the lines, except when the question specifies answers to be written in the table/figure itself. Avoid the use of ambiguous terms like “it” or “they”, when there are multiple species, groups, and/or interactions in a question. Use terms like C=C carbon / C=O double bond to refer to specific atoms or bonds, especially if there are a few different kinds of carbons / more than one double bond in the molecule. Be careful with the terms atoms, molecules, ions, carbocation, radical, elements, compounds – these differ in meaning and are not interchangeable. For Paper 1, plan your time well so you do not leave any answer unshaded on the OMS.

1 2 C Balancing the mass number and proton number on both sides of the equation, n0 1 + X7 16  C6 14 + H1 3 Total no. of protons present in products = 6 + 1 = 7 = No. protons present in reactants Since a neutron has proton number of zero, X has 7 protons  X is N (identity of an element is based on the no. of protons it possesses)  X contains 16 – 7 = 9 neutrons 3 B Given that Y has at least 19 electrons, the Period number of Y must be 4 or higher. From Period 4 onwards, the d subshell needs to be considered when deducing the electronic configuration of the element from successive ionisation energy data. Options C and D are incorrect. The graph shows a sharp increase from the 15 th to 16th ionisation energy, which indicates that the 16th electron is in an inner quantum shell. If Y is a Period 4 Group 17 element ([Ne]3s 23p63d104s24p5), after removal of the first seven electrons, there should be a gradual increase in energy for the removal of the ten 3d electrons. A similar reasoning applies for why Y cannot be a Group 14 element. A possible electronic configuration for Y may be 1s22s22p63s23p63d54s2, which is that of manganese. After removal of two electrons, a small jump is expected since the third electron is in an inner 3d subshell. After removal of another five electrons, a slightly greater increase occurs from the 7 th to 8th ionisation energies, as the 8 th electron is in an inner 3p subshell. 4 B Cx is bonded to a H atom besides two other C and one C l, hence Cx is tetrahedral with a bond angle of 109.5. Cy is also bonded to four groups, so it has a bond angle of 109.5. Cz has 3 groups of bonding electrons in total, so its bond angle is 120. 5 D BaO2 contains Ba2+ and O22. Each O atom gains one electron from Ba, as denoted by the open circle in the dot -and-cross diagram. The dot -and-cross diagram for options A, B and C shows O2, O2 and O2 respectively. 6 A The Lewis structures of the molecules provided by the question are drawn in the table below, together with an indication of whether a net dipole exists.

2021 HCI C1 H2 Chemistry Promotional Exam / Paper 1 First molecule Net dipole? Second molecule Net dipole? 1 O S O O C O 2 P FF F F B F F 3 Br F F F F F Si F FF F From the above, it is clear that all options are correct. Therefore A is the option which we will pick. 7 C After combining flasks A, B and C, total volume = 7 dm3 At 40C, water vapour condenses to water and is no longer a gas. Using pFVF/TF = p1V1/T1 + p2V2/T2 we will find the new final pressure (pF) after mixing. (pF)(7)/(313) = (2)(1)/(383) + (4)(2)/(383) PF = 1.2 kPa 8 A Since we know that H2S is being oxidised to form elemental sulfur, we would be able to balance the oxidation half equation: H2S  S + 2H+ + 2e The question mentioned that H2S reacted with BrO3– in a 5:2 ratio. When 5 moles of H2S get oxidised, 10 moles of electrons would be given out and transferred to 2 moles of BrO3–. This implies that every mole of BrO 3– would take in 5 moles of electrons as they get reduced. Oxidation state of Br in BrO3– = +5 After taking in 5 moles of electrons, the new oxidation state = +5 – 5 = 0 9 D CxHy + (x + y/4) O2  x CO2 + y/2 H2O Initial 10cm3 100 cm3 Change 10 65 +40 Final 35 cm3 40 cm3 Volume of CO2 = 40 cm3 (CO2 is an acidic gas and reacted with KOH) Volume of unreacted O2 = 75 – 40 = 35 cm3 Volume of O2 reacted = 100 – 35 = 65 cm3 𝑛 𝐶𝑂2 𝑛 𝐶𝑥𝐻𝑦 = 40 10 = 𝑥 1 Hence, x = 4 𝑛 𝑂2 𝑛𝐶𝑥𝐻𝑦 = 65 10 = 4+𝑦/4 1 Hence, y = 10

2 10 B ∆G = ∆H − T∆S Forward osmosis is spontaneous as the separation of water is done via a natural osmotic pressure gradient, without an external pressure. ⇒ ∆G is negative ∆H is negligible ⇒ ∆S must be positive. Although there are fewer ways to rearrange the energy of the water and solute molecules in the fruit juice after being concentrated, the ways to rearrange the energy of the water molecules after passing through the membrane in the diluted saline water and the ions in the diluted saline water more than compensate so there is a net positive change in entropy. 11 D Hsol = −mcT  nMgCl2 The surrounding which absorbed the heat released from dissolution of the solid is the 100 cm3 of deionised water. Hence, m should be the mass of water alone. The change in temperature, T, is the same value whether in C or K. 12 A H+(aq) is a catalyst and hence its concentration is expected to remain constant during the reaction. It is thus not possible to monitor the rate of reaction by following [H+(aq)]. The reaction produces a gaseous product, hence the rate of reaction may be followed by measuring the changes in total mass of the reaction mixture in an open flask (option B), volume of gas produced (option C), or gas pressure (option D). 13 B Total volume of the reaction mixture for expt 2 is half that of expt 1. Multiplying the volume of each reactant and deionized water by 2 for easy comparison with expt 1, the initial concentration of RBr is the same for both expts, while the initial concentration of NaOH is decreased two times for expt 2. Since rate is halv ed for expt 2, the reaction is 1st order for NaOH. Option A is incorrect. In all expts, NaOH was in excess. The rate equation may hence be simplified to an overall pseudo first order reaction, where rate = k’[RBr] with k’= k[NaOH]. In expt 1, t1/2 = ln2  k’ = ln2  k[NaOH] = ln2  k(0.40) = t min In expt 3, t1/2 = ln2  k’ = ln2  k[NaOH] = ln2  k(0.04) = 10t min Option B is correct. Alternatively, c omparing expts 1 and 3, when [RBr] is doubled, half -life should remain at t min since reaction is 1st order for RBr. However, as [NaOH] i