2025 AMKSS Prelim P2 MS

1 ANG MO KIO SECONDARY SCHOOLPRELIMINARY EXAMINATION 2025SECONDARY FOUR EXPRESSCHEMISTRY [6092]ANSWER SCHEMEPaper 2 Section A: Compulsory Structured Questions [70 Marks]1aP and S11bQ 11cP3U2 11d S 1Total42aCovalent bonding.Bonding between non-metal elements phosphorus and oxygen is covalent in nature.112bP4O6 1(R): P2O3 / O6P42cLiquid X is not able to conduct electricity.X has a simple molecular structure / is a simple molecule and does not have mobile ions or electrons to act as charge carriers to conduct electricity.112dpH of solution Y is less than pH 7. Substance X is acidic oxide as it is a non-metal oxide and will dissolve in / react with water to form an acidic solution.11Total73aBond breaking energy for 4 moles of C-F = 4 x 495 kJ/mol 2

2 = 1980 kJ Bond breaking energy for 2 moles of C-Cl = 2 x 339 kJ/mol = 678 kJ Bond breaking energy for 2 moles of C-H = 2 x 413 kJ/mol = 826 kJ Total bond breaking energy = 1980 + 678 + 826 = 3484 kJ (correct working = 1; correct ans = 1)3bBond forming energy for 4 moles of C-F = 4 x 495 kJ/mol = 1980 kJ Bond forming energy for 1 mole of C=C is 610 kJ Bond forming energy for 2 moles of H-Cl = 2 x 431 kJ/mol = 862 kJ Total bond forming energy = 1980 + 610 + 862 = 3452 kJ (correct working = 1; correct ans = 1)23c∆ H of the reaction = 3484 kJ – 3452 kJ = + 32 kJ1Allow for ecf3dFluorine is more reactive than chlorine. Therefore, the covalent bond formed between C – F is stronger than between C – Cl. Hence, it needs more energy to break the C – F bond compared to C – Cl bond.1Total64a2 NaN3  2 Na + 3 N2 14bMr of NaN3 = 23 + 3(14) = 65Number of moles of NaN3 used = 120 ÷ 65 = 1.84615 mol [1m]Mole ratio of NaN3 : N2 = 2 : 3 = 1.84615 : 2.76923Number of moles of N2 produced = 2.76923 mol [1m]Volume of N2 produced = 2.76923 × 24 dm3 = 66.46154 = 66.5 dm3 [1m]34ci10 Na + 2 KNO3  5 Na2O + K2O + N2 14ciiSodium is a very reactive metal which reacts violently with water and oxygen in air. 14dMethod: titration1

3 Reagents formulae: KOH / K2CO3 and HNO3 1Total85aNaOH and (NH4)2SO4 15bHNO3 and Zn15cCu(NO3)2 and Na2CO3 15dCu(NO3)2 and Zn15eKI and Ag(NO3)2 1Total56aPotassium chloride is an ionic compound.In the solid state, the oppositely charged ions in potassium iodide are fixed in position and cannot move freely to act as charge carries to conduct electricity.116bAdd distilled water to solid potassium chloride to form an aqueous solution / heat potassium chloride until it melts to form a molten liquid. 16c4 OH− (aq)  2 H2O (l) + O2 (g) + 4 e−Effervescence observed. A colourless and odourless gas is given off.OH− ions are preferentially discharged in place of SO42− ions. Each OH− ion loses an electron to form water and oxygen gas. OH− ions are oxidised.1116dCu2+ (aq) + 2 e−  Cu (s)A brown solid is deposited on the electrode.Cu2+ ions are preferentially discharged in place of H+ ions. Cu2+ ion gains two electrons to form Cu metal. Cu2+ ions are reduced.1116eBlue copper(II) sulfate solution turns colourless / pale blue.1Total107afractional distillation17bC8H18 17cThe greater the number of carbon atoms per molecule, the 1

4 higher the melting and boiling points of the alkanes.As the molecular mass / size increases with more carbon atoms per molecule, the intermolecular forces of attraction between the molecules increases. More energy is needed to overcome the stronger attractive forces. 17diEquation: C16H34  C8H18 + 2 C3H6 + C2H4 17diiName: propanoic acid Full structural formula:117ePercentage by mass of carbon in octane= {8(12) ÷ [8(12) + 18(1)]} × 100%= 84.2% Percentage by mass of carbon in hexadecane= {16(12) ÷ [16(12) + 34(1)]} × 100%= 84.95%= 85.0%Octane has a lower percentage by mass of carbon than hexadecane.Octane is more flammable than hexadecane as flammability of alkanes decreases with higher percentage of carbon by mass.111Total108awater18bEsterSweet smelling118c 18dRed-brown aqueous bromine becomes decolourised / turns colourless. 1

5 Compound X is an unsaturated organic compound and will undergo addition reaction with bromine. 18ei (correct amide linkage = 1; two repeating units = 1)28eiiAmide bond / linkage1Total99aCl2 (g) + 2 NaBr (aq)  2 NaCl (aq) + Br2 (aq) (balanced equation = 1; state symbols = 1)29bExperiment 219cIn the first two minutes, the rate of change of absorbance is very fast as there is a high concentration of bromide ions being displaced.The rate of change of absorbance then slows down as the concentration of bromide ions being displaced decreases over time.After 3 minutes, the reaction has come to a stop as all the bromide ions are displaced from the solution.1119dExperiments 1 and 319eiThe absorbance reading of experiment 5 is lower than 0.4. Chlorine is less reactive than fluorine and will not displace fluorine from sodium fluoride. The colour of aqueous chlorine and sodium fluoride is lighter than aqueous bromine.119eiiThe absorbance reading of experiment 6 is higher than 0.4.Chlorine is more reactive than iodine and will displace iodine from sodium iodide to form aqueous iodine. Aqueous iodine is darker in colour than aqueous bromine. 11Total11Section B: Free Response Questions [10 Marks]10aThe decomposition point of potassium chlorate(V) is below its 1(R): Sublimation

6 boiling point / potassium chlorate(V) decomposes at a temperature below its boiling point.10bLiquid state110cHeat the mixture to 400 oC to decompose KClO3 to form KCl.Weigh and reheat the mixture until the mass remains constant.Add excess distilled water to the cooled mixture to dissolve KCl.Filter the mixture to remove CuO as the residue.Heat the filtrate of KCl to saturation.Cool the filtrate to allow crystallisation of KCl.Filter the filtrate to obtain KCl crystals as residue.Wash with distilled water and dry between dry filter papers.(8 points = 3; 6-7 points = 2; 4-5 points= 1)310d+ 5110eThe oxidation state of chlorine decreases from +5 in KClO3 to −1 in KCl. Therefore, chlorine in KClO3 is reduced.The oxidation state of oxygen increases from −2 in KClO3 to 0 in O2. Therefore, oxygen in KClO3 is oxidised.Since there is oxidation and reduction occurring during the decomposition reaction, it is a redox reaction.1110facid-base neutralisation / precipitation reaction(an appropriate equation based on the stated reaction; state symbols not required) 11Total1011aIonisation energy decreases down group 1.Down group 1, the number of electron shells increases. The increase in distance between the outer electron and the nucleus / protons resulted in weaker electrostatic attraction between them and less energy needed to remove an electron. 11111bIonisation energy increases across period 2.Across period 2, the number of protons increases but the number of electron shells is the same.The increase in proton number but with similar distance between protons and outer electrons resulted in stronger electrostatic attraction between them and more energy needed to remove an electron. 11111cThe ionisation energies of Li−6 and Li−7 are the same / similar. 1

7 Ionisation energy is dependent on the electrostatic attraction between protons and electrons, not the number of neutrons. Both isotopes have the same number of protons and electrons.111d NaS(cor