NJC 2026 Prelim P3 QP

NJC SH2 Prelim 9476/3/26 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATIONS Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 3 Free Response Candidates answer on Question Paper. Additional Materials: Data Booklet 9476/03 15 Sept 2026 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. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. 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 Section A 1 /16 2 /23 3 /16 Section B (*circle the question you attempted) 4 /20 5 /20 Total /75 This document consists of 32 printed pages (including 3 blank pages).

2 NJC SH2 Prelim 9476/3/26 Section A Answer all the questions in this section in the spaces provided. 1 (a) The ozone molecule, O3, contains a dative covalent bond. (i) Draw a 'dot-and-cross' diagram to show the bonding in a molecule of O3. Include all outer shell electrons in your diagram. [1] (ii) Suggest a value for the bond angle in O3. Explain your answer. ...……………………………………………………………………………………..……… ...……………………………………………………………………………………..……… ...……………………………………………………………………………………..……… ...……………………………………………………………………………………..……… ...……………………………………………………………………………………..…...[2] (iii) Ozone can act as a Lewis base. Draw the structure of the product formed when BF3 reacts with O3. [1]

3 NJC SH2 Prelim 9476/3/26 [Turn over Blank

4 NJC SH2 Prelim 9476/3/26 (b) Atmospheric sulfur species play key roles in both acid rain precursor formation and stratospheric ozone depletion. Sulfuryl chloride, SO2Cl2 is a liquid with low boiling point. When sulfuryl chloride, SO2Cl2, vapour reaches the stratosphere, it decomposes to give sulfur dioxide and chlorine radical. Under high-frequency UV radiation, the mixture undergoes a series of reactions leading to the destruction of ozone, O3, as shown in Table 1.1. Table 1.1 Step 1: One of the S–Cl bonds in an SO2Cl2 molecule undergoes homolytic fission to generate a chlorine radical, Cl•, and a •SO2Cl radical. Step 2: The highly unstable •SO 2Cl radical undergoes further homolytic fission, breaking its remaining S–Cl bond to produce a sulfur dioxide molecule, SO2, and a radical, Cl•. Step 3: The highly reactive Cl• radical collides with an ozone molecule, O3, forming a chlorine monoxide radical, ClO•, and an oxygen molecule, O2. Step 4: The ClO• radical reacts with a free oxygen atom present in the stratosphere, forming another molecule of O2 and the Cl• radical. In the lower atmosphere, NO2 acts as a homogeneous catalyst in converting SO2 to SO3. The first step of the catalytic oxidation is shown in equation 1.1. equation 1.1 SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) ∆H𝑟 = −41.8 kJ mol–1 (i) Define the term homolytic fission. ……………………………………...……………………………………………………….. …………………………………………………..………………………………………..[1]

5 NJC SH2 Prelim 9476/3/26 [Turn over (ii) Use curly arrows to show the movement of electron s in step 1 by completing Figure 1.1. Step 1: O S Cl O Cl Figure 1.1 [1] (iii) Name the reactions for steps 1 and 3. Step 1: ……………………………... Step 3 : ………………………………………… [1] (iv) State the role of the chlorine radical in the overall destruction of ozone and explain your answer by referring to Table 1.1. ……………………………………...……………………………………………………..… ………………………………………...…………………………………………………..… …………………………………………...………………………………………………..… ………………………………………………………………………………………..…..[2]

6 NJC SH2 Prelim 9476/3/26 (v) Figure 1.2 shows how the temperature of the atmosphere varies with height from the ground. height / km temperature / C Figure 1.2 Suggest how the position of equilibrium of equation 1.1 differs at a height of 20 km compared with a height of 50 km from the ground. Explain your answer. ………………………………………………………………………………...…………..… …………………………………………………………………………………...………..… ……………………………………………………………………………………...……..… …………………………………………………………………………………………….[2]

7 NJC SH2 Prelim 9476/3/26 [Turn over (c) 4-acetamidomethylbenzene is formed by the three-step route shown in Figure 1.3. H3C NHCOCH3H3C 4-acetamidomethylbenzene T U step 1 step 2 step 3 Figure 1.3 (i) Draw the structure of each of the organic products, T and U in Figure 1.3. [2] (ii) Give the reagents and conditions needed for steps 1 to 3. Step 1 ………………………………………………………..………….. Step 2 ………………………………………………………..………….. Step 3 ………………………………………………………..………….. [3] [Total : 16]

8 NJC SH2 Prelim 9476/3/26 2 (a) A solution containing hydrazoic acid, HN3(aq), and azide ion, N3– (aq), can act as a buffer. The value of the acid dissociation constant, Ka, for HN3(aq) is 1.8 × 10–5 mol dm–3 at 298 K. (i) Describe how a solution containing HN3(aq) and N3– (aq) can act as a buffer. Include relevant equations to illustrate your answer. ……………………………………………………………………………...……………….. ……………………………………………………………………………...……………….. ……………………………………………………………………………...……………….. ………………………………………………………………………………………………. ………………………………………………………………………………………………. …………………………………………………………………………………………… [2] (ii) A buffer solution is prepared by mixing 50.0 cm 3 of 0.500 mol dm –3 HN3 and 25.0 cm3 of 0.600 mol dm–3 NaOH. Calculate the pH of the buffer solution formed at 298 K. Show your working. [2]

9 NJC SH2 Prelim 9476/3/26 [Turn over (iii) Ethanal reacts readily with HN3 (aq) to form product Q. C N N N H OH C H H H product Q The first step of this mechanism involves the reaction of ethanal with N 3–. The structure of the N3– ion is shown. N NN Suggest the mechanism of this reaction. Include all relevant lone pairs, dipoles, curly arrows and charges. [3]

10 NJC SH2 Prelim 9476/3/26 (b) Burning sodium in air produces a mixture of sodium oxide, Na 2O, and sodium peroxide, Na2O2. The proportion of sodium peroxide in the mixture can be determined by the following method. • The mixture is dissolved in water and the following reaction takes place. Na2O2 + 2H2O → 2 NaOH + H2O2 • The solution is acidified and an excess of potassium iodide solution is added. Iodine is liberated by the following reaction. 2KI + H2O2 +