2025 H2 Chem Prelim P3 Answers VJC

© VJC 2025 9729/03/PRELIM/2025 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ANSWERS ……………………………………………….………….. …………………………….. CHEMISTRY 9729/03 Paper 3 Free Response 17 September 2025 2 hours Additional Materials: Data Booklet ________________________________________________________________________________________________________________ READ THESE INSTRUCTIONS FIRST Write your name and CT group on this cover page. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 20 2 / 20 3 / 20 4 / 20 5 / 20 Total / 80 This document consists of 23 printed pages and 1 blank page.

2 © VJC 2025 9729/03/PRELIM/2025 Declocalised electrons Metal cations Section A Answer all the questions in this section. 1 (a) Potassium is a highly reactive alkali metal that must be stored under oil, while copper is a much less reactive metal that resists corrosion. Table 1.1 shows the melting points of both metals. Table 1.1 metal Melting point / C K 63.5 Cu 1085 (i) Copper does not react with most dilute acids, unlike potassium. With reference to the standard electrode potentials in the Data Booklet, explain why this is so. [2] Cu2+ + 2e– ⇌ Cu +0.34 V K+ + e– ⇌ K –2.92 V 2H+ + 2e ⇌ H2 0.00V Or Eo (Cu2+|Cu) = +0.34 V Eo (K+|K) = -2.92 V Eo (H+|H2) = 0.00 V Copper is unable to react with dilute acids. For copper, Eocell = (0– 0.34)= –0.34V <0 (not spontaneous) Potassium can react with acid to produce H2 gas. For potassium, Eocell = (0– (–2.92))= 2.92V > 0 (spontaneous) • • Quote & calculation of EƟcell Explanation (ii) Describe with the aid of a labelled diagram the structure of copper at room temperature. [2] • • Copper has a giant metallic structure which is a three-dimensional arrangement of positive copper ions surrounded by delocalised electrons , held together by strong electrostatic forces of attraction. (iii) Suggest why the melting point of copper is significantly higher than that of potassium as shown in Table 1.1. [2] • Potassium can only delocalise its single 4s electron while copper is able to delocalized electrons from both its 4s and 3d subshells.

3 © VJC 2025 9729/03/PRELIM/2025 [Turn over • Copper forms ions with smaller ionic radius with greater charge. This leads to the stronger metallic bonds in copper which required more energy to break. (iv) Fig. 1.1 shows the first ionisation energies for the elements K to Cu. Fig 1.1 Explain why the first ionisation energy remains relatively constant from scandium to copper. [2] • • From Sc to Cu, the number of protons increases, hence the nuclear charge increases. Additional electron enters the penultimate 3d subshell and shielding effect increases slightly. The increase in nuclear charge is partially nullified by the increase in shielding effect leading to insignificant increase in effective nuclear charge across the period. Hence the 1st ionisation energy remains relatively constant. (v) Using your knowledge in the variation of first ionisation energy of the elements from potassium to copper, sketch the trend of the second ionisation energies on Fig 1.1. [1] • Shape decrease from K to Ca AND Ca to Cu gentle upward sloped or horizontal line. (b) Copper is a transition element which form many different complexes. (i) Explain what is meant by the term transition element. [1] • Transition element is a d-block element which forms one or more stable ions with partially filled d subshell. (ii) Copper(II) ions form a coloured complex with mercaptoethylamine, MEA, a bidentate ligand as shown in equation 1.1. equation 1.1 Cu2++ HSCH2CH2NH2 → [Cu(HSCH2CH2NH2)2]2+ MEA With reference to the complex in equation 1, explain what is meant by the term ligand. [1]

4 © VJC 2025 9729/03/PRELIM/2025 • HSCH2CH2NH2 ligand has lone pair of electrons on S and N atom that can be donated to the empty 3d orbitals of Cu 2+ by dative covalent bonds to form complex. (iii) Explain why complexes of copper(II) are usually coloured. [3] • • The degenerate 3d orbitals in Cu2+ octahedral complex is split into 2 different energy levels due to the presence of ligands (d-d splitting). d-d transition took place whereby a 3d electron from the lower energy level is promoted to the upper energy level by absorbing energy from the visible region of the electromagnetic spectrum. • The colour seen is the complement of the colour absorbed. (c) Polymerisation is a process where small molecules called monomers are combined to form larger polymer chains. Polymers are large molecules made of many repeat units. Thiirane, , can undergo free radical ring-opening polymerisation in the presence of a radical initiator, benzoyl peroxide and an excess of dodecanol, CH3(CH2)10CH2OH. Fig. 1.2 shows a proposed mechanism via three reactions. Reaction 1: Homolytic fission of a C –S bond in thiirane, generating a ‘double-ended’ free radical, •CH₂CH₂S•, that initiates a chain reaction. Reaction 2: A chain reaction occurs where the ring opens and units of •CH₂CH₂S• adds to a growing chain. Repeated steps involve production of other free radicals. Reaction 3: Termination with dodecanol to form CH3(CH2)10CH2–O–(CH2CH2S)n–H. Fig 1.2 (i) Use curly arrows to show the movement of electrons which occur in reactions 1 and reaction 2 on Fig. 1.2. [2] • • Reaction 1: Reaction 2:

5 © VJC 2025 9729/03/PRELIM/2025 [Turn over Correct arrows for each step 1 mark. (ii) Suggest why thiirane undergoes the reaction more easily as compared to (CH3)2S. [1] • Three membered ring is unstable due to ring strain. (e) The three compounds in Table 1.1 behave as monoprotic acids in aqueous solution. Table 1.2 name formula MEA NH₂CH₂CH2SH glycine NH₂CH₂COOH ethanolamine NH₂CH₂CH2OH Arrange the compounds in order of increasing acidity. Explain your answer. [3] • • • Glycine < MEA < ethanolamine NH2CH2COOH ⇌ NH2CH₂COO– + H+ NH2CH2CH2OH ⇌ NH2CH2CH₂O– + H+ NH2CH2CH2SH ⇌ NH2CH2CH2S– + H+ Glycine is acidic as the negative charge is dispersed over COO ⁻. Conjugate base (–COO⁻) is stabilised and position of equilibrium lies most to the right favouring formation of more H+. The conjugate bases of ethanolamine and MEA are less stable as negative charge is intensified on O and S atoms by electron donating inductive effect of alkyl group. The negative charge on S atom in MEA is dispersed to a greater extent due to the larger atomic radius as compared to O in ethanolamine. NH 2CH2CH2S– is more stable than NH 2CH2CH₂O–. Hence NH ₂CH₂CH2SH is more acidic than NH₂CH₂CH2OH. 1m for strong acid forms more stable anions, hence greater dissociation. 1m for anions being most stabilised