HCI 2026 Prelim P1 Ans

2026 HCI C2 H2 Chemistry Preliminary Examination / Paper 1 1 HWA CHONG INSTITUTION 2026 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 B C C C B B D B A D 11 12 13 14 15 16 17 18 19 20 A B C A D C D A D D 21 22 23 24 25 26 27 28 29 30 C A A A D B D B C B Comments 1 B I, II and III are allotropes of carbon. They represent diamond, fullerene and graphite respectively. Out of the three, only fullerene has a clearly defined molecular formula (usually C 60) and exists as discrete molecules . II is simple molecular, while I and III are giant molecular. 2 C The Mr of N 2 and the Mr of C 2H4 are the same (= 28.0). This means that the number of moles of nitrogen is the same as the number of moles of ethene. Hence the initial pressures of nitrogen and ethene in separate vessels of identical volume and temperature, are the same. Consequently after mixing, the partial pressure of nitrogen must be the same as the partial pressure of ethene as they have the same final volume and temperature (Option C is correct). The average Mr of the gaseous mixture would be 28.0 regardless of the mole fractions of nitrogen and ethene , since the Mr of each gas is 28.0 . The average kinetic energy of the gaseous particles should be unchanged upon mixing as there was no change in temperature. The density of the gaseous mixture, mixture, would be half of the sum of the original densities of nitrogen and ethene , as shown below. mixture = (mnitrogen + methene) / 2V = ½ (mnitrogen / V + methene / V) = ½ (nitrogen + ethene) 3 C At point X, pV / RT is lower than that predicted by the general gas equation as the actual volume of the gas is lower than expected due to non -negligible intermolecular forces of attraction between particles. Graph 1 deviates more from ideality than Graph 2, so we may deduce that the temperature for Graph 1 is lower. 4 C A Lewis acid is an electron -pair acceptor while a Lewis base is an electron -pair donor. In option 1, the bromide ion uses a lone pair to form a covalent bond with the ethyl carbocation, producing the product bromoethane. In the context of

2026 HCI C2 H2 Chemistry Preliminary Examination / Paper 1 2 organic reactions, the bromide ion and carbocation are typically regarded as nucleophile and electrophile respectively, but they also fit the description of a Lewis base and Lewis acid respectively. In option 2, two ammonia molecules each uses a lone pair on their nitrogen to form a dative bond with the beryllium of BeCl2, which was electron deficient with only 4 electrons around Be. The beryllium fulfils octet after forming the two dative bonds. The product is BeCl2(NH3)2. All products have Mr>100. 5 B The decreasing volatility of the halogens down Group 17 is familiar to us from the chapter on Chemical Periodicity. This is due to the increasing strength of intermolecular dispersion forces, owing to the increasing number of electrons in the molecule and increasing electron cloud size. Applying a similar trend, the boiling point of BrCl is expected to be lower than that of BrI (option A is wrong). Option C is wrong as regardless of the comparison of boiling point, as we do not break the covalent bond in discrete covalent molecules for a physical change of state. Option D is wrong as Br 2 is a liquid at rtp so BrI can be expected to be a liquid or solid, not a gas (i.e. less volatile). 6 B We can approach this question by writing out the electronic configurations of all the Period 2 elements and check ing whether each element is counted under E, F, neither or both. element configuration E F Li 1s22s1 Be 1s22s2 ✓ B 1s22s22p1 ✓ C 1s22s22p2 ✓ ✓ N 1s22s22p3 ✓ O 1s22s22p4 ✓ F 1s22s22p5 Ne 1s22s22p6 7 D A General Trend: Across the period, nuclear charge increases, shielding effect stays about the same and effective nuclear charge increases. Hence, energy required to remove the outermost electron for the element increases. Generally, 1st IE increases across the period. Dip between Mg and Al: Less energy is required to remove the 3p electron in Al than to remove the 3s electron from Mg. Dip between P and S: Less energy is required to remove the paired 3p electron in S due to the presence of interelectronic repulsion than to remove the unpaired 3p electron from P. B Na, Mg and Al are metals. Hence, there is a ‘sea’ of delocalised electrons that can act as mobile charge carriers. Electrical conductivity increases from Na to Al as the number of valence electrons that are delocalised per atom increases from Na to Al. Si is a metalloid and can conduct electricity, but not as well as Na, Mg and Al.

2026 HCI C2 H2 Chemistry Preliminary Examination / Paper 1 3 P4, S8 and Cl2 are non-metals with simple molecular structures and do not have free moving ions or delocalised electrons to act as mobile charge carriers. C Electronegativity of an atom is a measure of its ability to attract the electrons in a covalent bond to itself. Across the period, nuclear charge increases, shielding effect stays about the same and effective nuclear charge increases. Hence, electronegativity increases. D Na, Mg and Al have giant metallic lattice structures. From Na to Al, charge density of the metal cation increase and the number of delocalised electrons increases. Hence, more energy is required to overcome the stronger metallic bonding in Al than in Na. Hence, melting point increases. Si has a giant covalent structure. A lot of energy is required to break the strong covalent bonds between Si atoms. P4, S8 and Cl2 have simple molecular structures. S8 has the highest melting point as the largest amount of energy is required to overcome the stronger dispersion forces between the largest S8 molecules. 8 B The relevant equations and Eo values obtained from the Data Booklet are summarised below. equation Eo / V MnO4– + 8H+ + 5e– ⇌ Mn2+ + 4H2O +1.52 Fe3+ + e– ⇌ Fe2+ +0.77 C2O42– ⇌ 2CO2 + 2e– not given Since EoMnO4–/Mn2+ is more positive than EoFe3+/Fe2+, Fe2+ will also be oxidised to Fe3+ alongside C2O42–. For every 1 mol of FeC2O4, 3 mol of e– is given out during oxidation. Since 2 mol of MnO4– is used, 10 mol e– will be taken in by MnO4– and these 10 electrons are being contributed by FeC2O4. Since 1 mol of FeC 2O4 contributes 3 mol of e –, divide 10 by 3 and you will get 3.33. So, 3.33 mol of FeC2O4 reacts with 2 mol of MnO4–. 9 A Let the formula of G be CxHyOz. CxHyOz(g) + ___ O2(g) → xCO2(g) + y/2 H2O(l) Vi/cm3 15.0 a + a’ 0.0 Vf/cm3 0.0 a’ 75.0 15.0 + a + a’ – (a’ + 75.0) = 30.0 a = 90.0 cm3

2026 HCI C2 H2 Chemistry Preliminary Examination / Paper 1 4 CxHyOz(g) + ___ O2(g) → xCO2(g) + y/2 H2O(l) Vreacted/cm3 15.0 90.0 75.0 Ratio 1 6 5  x = 5  Answer is A or B. In terms of number of O atoms in the balanced equation, z + 12 = 10 + y/2 If y = 6, z = 1 So the answer is A. 10 D 1. The electrolysis of water generates H2 and O2 gas. Since there is an increase in the amount of gas in the system, there is an increase in entropy. 2. When water is heated from 30°C to 60 °C, the water molecules will have a greater distribution of molecular speeds, thus there is an increase in entropy. 3. When water evaporates, there is a change of state through vaporisation, thus entropy increases. 11 A Comparing experiments 1 and 2, when [H2O2] is held constant and [I−] is increased 1.25 times, the rate is increased 1.25 times, hence order wrt [I−] is 1. Comparing experiments 3 and 1, rate3 / rate1 = k(0.05)m(0.06) / [k(0.02)m(0.04)] = 4.5 × 10−4 / 1.2 × 10−4 m = 1 (order wrt [H2O2])  Statement B is wrong