NJC 2025 H2 Chemistry Prelim P2 Ans

[Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/02 16 September 2025 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. 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 paper clips, highlighters, glue or correction fluid. Answers all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 /16 2 /14 3 /17 4 /16 5 /12 Paper 2 Total /75 Marks Weightings Paper 1 /30 15% Paper 2 /75 30% Overall Percentag e Paper 3 /80 35% Paper 4 /55 20% Grade This document consists of 19 printed pages and 1 blank page.

2 NJC/H2 Chem Prelim/02/2025 Answer all the questions in the spaces provided. 1 925 silver, also known as sterling silver, is an alloy that is commonly used to make jewellery. It consists of 92.5% silver and 7.5% other metals, such as copper, by mass. Over time, the alloy can form a tarnish of Ag2S(s) when it reacts with hydrogen sulfide, as represented by the following equation. 2Ag(s) + H2S(g)  Ag2S (s) + H2(g) (a) (i) Write the full electronic configuration for copper. 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Note: Cu has a special configuration of 3d10 4s1 Common mistakes - Putting 4s1 before 3d10 - Configuration written as 1s2 2s2 2p6 3s2 3p6 3d10 4s1 [1] (ii) State and explain the difference in atomic radii for silver and copper. Ag has one more filled electronic shell than Cu. The distance of its valence electron is further away from nucleus and experience higher shielding effect . This outweighs the higher nuclear charge in Ag . Nuclear attraction for t he outermost electron in Ag is weaker. Thus, atomic radius of Ag is larger. Common mistakes - Students mention shielding effect but did not state the further distance as a factor as well. - Another common mistake is that having more electrons/more inner subshells does not always means a further distance of the valence electrons from the nucleus. - Students mention both nuclear charge and shielding effect being higher for Ag but did not mention that the higher shielding effect and further distance outweighs the higher nuclear charge. - Students also often miss out nuclear attraction for the valence electrons. - Outweigh and not offset/nullified/cancelled out. [2] (b) The Ag 2S tarnish on sterling silver can be removed until only sterling silver remains. A student weighs a tarnished sterling silver sample both before and after removing the Ag2S and records the data in Table 1.1. Table 1.1 mass before Tarnish Removal /g 54.23 mass after Tarnish Removal /g 52.34

3 NJC/H2 Chem Prelim/02/2025 [Turn over Assuming that only Ag 2S (s) is removed, calculate the number of moles of silver atoms removed. Mass of Ag2S removed = 54.23 – 52.34 = 1.89 g. Amount of Ag2S removed = 1.89 2(107.9)+32.1 = 1.89 247.9 = 0.00762 mol 2Ag(s) + H2S(g)  Ag2S (s) + H2(g) Amount of Ag removed = 2 × 0.00762 = 0.0152 mol Common mistakes - Students did not use the mole ratio to find the number of mol of Ag removed - Many students calculated the no of atom instead of no of mole of Ag [1]

4 NJC/H2 Chem Prelim/02/2025 (c) (i) Suggest and explain the relative magnitude of the lattice energy of the silver compounds Ag2S, Ag2O and Ag2Se. |L.E| ∝ | 𝑞+ × 𝑞− 𝑟+ + 𝑟− | circle but don’t penalize if put “=” ● Product of charges (q + × q −) and the cationic radius (r +) for Ag2O, Ag 2S, Ag2Se are the same ● Interionic distance of Ag2O < Ag2S < Ag2Se OR Anion radius of O2 < S2 < Se2 ● Since |L.E| ∝ | 𝑞+ × 𝑞− 𝑟+ + 𝑟− |, magnitude of lattice energy of Ag2O > Ag2S > Ag2Se Common mistakes - Students did not state the |L.E| ∝ | 𝑞+ × 𝑞− 𝑟+ + 𝑟− | in the answer - Students mention atomic radius comparison rather than the ionic radius. [2] (ii) Using relevant data from the Data Booklet and Table 1.2, calculate the lattice energy of Ag2O. Table 1.2 standard enthalpy change of atomisation of silver +285 kJ mol­1 1st electron affinity of oxygen −141 kJ mol­1 2nd electron affinity of oxygen +798 kJ mol­1 standard enthalpy change of formation of Ag2O(s) −31 kJ mol­1

5 NJC/H2 Chem Prelim/02/2025 [Turn over By Hess Law, - 31 = 2 (285) + 1 2 (496) + 2 (731) + (- 141) + 798 + L.E L.E = -2968 = -2970 kJ mol-1 Common mistakes - Axis not drawn - Putting Electron affinity before Ionisation energy in the energy level diagram - Bond energy for the oxygen is left out. - Students forget to x2 to the ionisation energy of Ag but instead include in 2nd I.E for Ag - Students are not familiar with the use of Hess’s law. Many students changed the ∆Hf sign for Ag2O(s) to +31kJmol–1 and 1st EA of O to +141kJmol–1. [4] (d) Rhodium plating is a process used to protect sterling silver from tarnishing. This involves electroplating (depositing) solid rhodium, Rh(s), onto the surface of the metal from an acidified solution of Rh2(SO4)3 (aq). Oxygen gas is produced during this process. One of the half equations involved in this reaction is Rh3+ + 3e ⇌ Rh Eꝋ = +0.76 V

6 NJC/H2 Chem Prelim/02/2025 (i) Write the half equation for the reaction that has resulted in the formation of oxygen gas in this reaction and hence, write the balanced ionic equation for the overall reaction. [O] : 2H2O ⎯→ O2 + 4H+ + 4e- cannot accept reversible arrow. 4Rh3+ (aq) + 6H2O (l)  4Rh (s) + 3O2 (g) + 12H+ (aq) must have state symbol for ionic equation Common mistakes - [O] equation not written with 🡪 - The basic equation is chosen instead. - The balanced equation does not have state symbol [2] (ii) Calculate the value of Eꝋ cell for the overall reaction in part (i). 𝐸𝑐𝑒𝑙𝑙 ꝋ = 𝐸𝑅𝑒𝑑 ꝋ - 𝐸𝑜𝑥ꝋ =+0.76 V − 1.23 V = −0.47 V [1] (iii) Based on your answer to part (ii), explain why this process requires the use of an external power source. 𝐸𝑐𝑒𝑙𝑙 ꝋ is negative, which means the reaction is not energetically feasible and thus energy must be supplied for the reaction to occur. Generally, well done [1] (iv) Calculate the current that must be supplied for 3.5 g of Rh to be plated onto a piece of sterling silver in 3 minutes. At the cathode: [R] Rh3+ + 3e− ⎯→ Rh Amount of Rh deposited = 3.5 102.9 = 0.0340 mol ηe = 3 × 0.0340 = 0.102 mol I × t = ηe × F I = 𝑛𝑒 × 𝐹 𝑡 = 0.102 × 96500 3 × 60 = 54.7A Generally, well done [2] [Total : 16]

7 NJC/H2 Chem Prelim/02/2025 [Turn over 2 (a) Xenon is a noble gas and forms various fluorides with fluorine. Two of these are xenon difluoride, XeF 2, and xenon tetrafluoride, XeF 4, which are crystalline solids with melting points of 140 C and 112 C respectively. (i) Draw the dot-and-cross diagrams for XeF2 and XeF4 and hence state their molecular shapes. XeF2: linear XeF4: square planar [3] Common mistakes - Students missed out drawing the 3 lone pair electrons on F atom. - Students are still unsure of the molecular shapes associated with the different number of lone pairs & bond pairs. (ii) Even thoug