NJC 2025 H1 Chemistry Prelim P2 Ans

1 NJC/H1 Chem Preliminary Examination/02/2025 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 1 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 8873/02 16 September 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your name, registration number and subject class on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all the questions. Section B Answer one question. At the end of the examination, fasten all your work securely together. A1 /13 A2 /13 Paper 2 /80 A3 /6 Paper 1 /30 A4 /11 Weightage A5 /10 Paper 1 (33%) /33 A6 /7 Paper 2 (67%) /67 B6 / B7 /20 Overall The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 23 printed pages and 1 blank page. Section A

2 NJC/H1 Chem Preliminary Examination/02/2025 Answer all the questions in this section in the spaces provided. 1 Boron, carbon, and lithium play critical roles in nuclear, organic, and battery chemistry. Their atomic structure and periodic behaviour provide important insights into chemical properties and bonding patterns. (a) (i) About 0.005% of water molecules consist of an oxygen atom bonded to two atoms of the isotope of hydrogen, deuterium, . The compound formed is deuterium oxide, D2O, and is also known as ‘heavy water’. Define the term isotope. Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. [1] (ii) State the number of subatomic particles present in one molecule of D2O. number of protons number of neutrons number of electrons Protons: 10 (8 from O, 1 from each D)  Neutrons: 12 (8 from O, 2 from each D)  Electrons: 10 (8 from O, 1 from each D) [1] (b) Separate beams of 14C2– and 7Li+ ions are passed through an electric field in the set- up below. The angle of deflection of the 14C2– beam is 7.0 o. Sketch on the diagram below, the paths taken by beams of 7Li+ and 14C2– ions in the presence of an electric field. Label your sketch clearly.

3 NJC/H1 Chem Preliminary Examination/02/2025 [Turn over  1 mark: ¹⁴C²⁻ deflects toward positive plate, curved path labelled (angle = 7°)  1 mark: ⁷Li⁺ deflects toward negative plate, with greater curvature than ¹⁴C² ⁻ (Opposite directions, and lighter ion deflects more.) [2] (c) The element sulfur has four naturally occurring isotopes as shown in Table 1.2. Table 1.2 isotope relative abundance / % 32S 94.93 33S 0.76 34S 4.29 36S 0.02 Use the relative abundance data to calculate the relative atomic mass of sulfur to 4 significant figures. Show your working. Ar = (32.0 𝑥 94.93+(33.0 𝑥 0.76)+(34 𝑥 4.29)+(36 𝑥 0.02) 100 [1] = 32.07 [1] (accept values within + 0.01 if correct method is shown [2] (d) Write the electronic configuration of Cu and explain its anomaly. ● Electronic configuration of Cu: [Ar] 3d¹⁰ 4s¹ [1] ● Explanation: full d-subshell (3d¹⁰) is more stable than partially-filled 3d⁹ 4s² configuration due to exchange energy/stability of half or full subshell [1] [2] (e) Predict and explain the order of first IE of Al, Mg, and S.  Correct order: S > Mg > Al [1]  Explanation: ● S has higher nuclear charge and same shielding, hence greatest nuclear attraction for most loosely held electron.

4 NJC/H1 Chem Preliminary Examination/02/2025 ● Al has electron in 3p orbital → easier to remove due to higher energy level from repulsion from pair electrons in p orbital. Hence most loosely held electron is easier to remove than that of Mg. [2] (f) (i) Draw the dot-and-cross diagram for PCl5. Correct bonding with P in centre forming five single bonds with Cl atoms [1] [1] (ii) Explain why PCl₅ exists but NCl₅ does not, in terms of electronic structure.  P can expand its octet by using empty 3d orbitals → forms 5 bonds [1]  N cannot expand its octet as it has only 2s and 2p orbitals → limited to a maximum of 4 electron pairs (8 electrons) [1] [2] [Total : 13]

5 NJC/H1 Chem Preliminary Examination/02/2025 [Turn over 2 Oxalic acid, found in many leafy vegetables, can be analysed via redox titration using potassium manganate(VII). This question explores thermochemistry and redox stoichiometry in food and materials chemistry. (a) To determine the standard enthalpy change of neutralisation, 60 cm 3 of 0.370 mol dm–3 aqueous ammonia was placed in a polystyrene cup. Constant volume of dilute HCl was added. The mixture was stirred after each addition and the maximum temperature was recorded. The following graph was obtained in Figure 2.1 Fig 2.1 (i) Define the term standard enthalpy change of neutralisation. The energy released when 1 mole of water is formed from the neutralisation of an acid by a base under standard conditions (1 bar, 298 K, solutions at 1 mol dm⁻³). [1] (ii) Determine the maximum temperature change from Figure 2.1. Correctly identifies ΔT = 2.7 °C from the graph. [1] (iii) Calculate the concentration of HCl used in the experiment. Given: ● Volume of HCl used = 22.0 cm³ = 0.0220 dm³ ● Volume of NH₃ = 60.0 cm³ = 0.0600 dm³ ● Concentration of NH₃ = 0.370 mol dm⁻³

6 NJC/H1 Chem Preliminary Examination/02/2025 ● Step 1: Calculate moles of NH₃ added mol NH₃=0.0600×0.370=0.0222 mol Step 2: At equivalence, mol NH₃ = mol HCl mol HCl=0.0222 mol [1] Step 3: Calculate concentration of HCl [HCl]=mol HCl/volume in dm³=0.0222/0.0220=1.01 mol dm⁻³ [1] [2] (iv) Calculate the enthalpy change of neutralisation, ΔHno, for the reaction between HCl and ammonia. Assuming that the density of all aqueous solution is constant at 1.00 g cm –3 and specific heat capacity of the solution is 4.18 J g⁻¹ °C⁻¹. Given: ● Total volume = 60.0 cm³ (NH₃) + 22.0 cm³ (HCl) = 82.0 cm³ ● Assume density = 1.00 g cm⁻³, so mass = 82.0 g ● Specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹ ● Temperature rise (ΔT) = 2.7 °C ● Moles of water formed = moles of NH₃ = 0.0222 mol Step 1: Calculate heat evolved (q) q=mcΔT=82.0×4.18×2.7=925.6J [1] Step 2: Convert to kJ q=0.9256 kJ Step 3: Calculate ΔH⁰ₙₑᵤₜ ΔH=−q/n=−0.9256/0.0222=−41.7 kJ mol⁻¹ [1] [2] (v) The theoretical standard enthalpy change of neutralisation is −57.0 kJ mol −1. Account for the discrepancy in the value calculated in (a)(iv). 1 mark : Heat loss to surroundings / incomplete reaction / inaccurate measurement of temperature / polystyrene cup not perfectly insulated / evaporation or delay in mixing [1] (b) The following investigation was carried out on a sample of oxalic acid, H 2C2O4, extracted from 100 g of spinach (about one serving) using 50.0 cm 3 of an inert organic solvent. The solution is then titrated with acidified potassium manganate(VII), KMnO4. 2MnO4– + 16H+ + 5C2O42– → 10CO2 + 2Mn2+ + 8H2O Solutions used in the experiment: FA 1 The given inert organic solvent containing the oxalic acid was diluted with deionised water to a total volume of 250 cm3 in a volumetric flask.

7 NJC/H1 Chem Preliminary Examination/02/2025 [Turn over FA 2 1.0 mol dm–3 sulfuric acid, H2SO4 FA 3 0.0231 mol dm–3 potassium manganate(VI