NJC 2026 Prelim P3 Ans

NJC SH2 Prelim 9476/3/26 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATIONS Higher 2 CANDIDATE NAME Answer 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.

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. or (ii) Suggest a value for the bond angle in O3. Explain your answer. O₃:110° ≤ bond angle < 120° [actual 117°] Central oxygen atom has 3 regions of electron density: 2 bonding pairs (one single bond, one double bond) and 1 lone pair. The lone pair -bond pair repulsion is greater than the bond -bond pair repulsion, compressing the bond angle from the ideal 120° / electron geometry as trigonal planar to approximately 117°. (iii) Ozone can act as a Lewis base. Draw the structure of the product formed when BF3 reacts with O3. A dative covalent (co-ordinate) bond is formed. Note: Ozone acts as a Lewis base by donating a lone pair of electrons to the electron-deficient boron atom in BF₃ (Lewis acid)). [1] (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.

3 NJC SH2 Prelim 9476/3/26 [Turn over 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. The even breaking of a covalent bond where the shared pair of electrons is divided equally, with each atom taking one electron to form radicals. (ii) Use curly arrows to show the movement of electrons in step 1 by completing Figure 1.2. (iii) Name the reactions for steps 1 and 3. Step 1: Initiation Step 3: Propagation [1]

4 NJC SH2 Prelim 9476/3/26 (iv) State the role of the chlorine radical in the overall destruction of ozone and explain your answer by referring to Table 1.1. The chlorine radical acts as a catalyst in the overall destruction of ozone. Cl• is consumed in Step 3 when Cl• reacts with O3 to form ClO• and O2, Cl• is regenerated in Step 4 when ClO• reacts with a free oxygen atom to form O2 and Cl•. (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 differs at a height of 20 km compared with a height of 50 km from the ground. Explain your answer. Since temperature at 20 km from ground is colder/lower compared to 50 km, position of equilibrium at 20 km lies more/ farther to right to favour the forward exothermic reaction as to produce some heat/ increase the temperature. OR Since temperature at 50 km from ground is higher/hotter compared to 20 km, position of equilibrium at 50 km lies more/ farther to left to favour the backward endothermic reaction as to absorb some heat/ decrease the temperature.

5 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] NO2H3C NH2H3C T U (ii) Give the reagents and conditions needed for steps 1 to 3. Step 1 conc. HNO3, conc. H2SO4, 30oC Step 2 1. Sn, conc. HCl, heat 2. NaOH(aq) Step 3 anhydrous CH3COCl [Total : 16]

6 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 HN 3(aq) is 1.8 × 10–5 mol dm –3 at 298 K. (i) Describe how a solution containing HN 3(aq) and N 3– (aq) can act as a buffer. Include relevant equations to illustrate your answer. A buffer solution contains a large reservoir of weak acid, HN3 and its conjugate base, N3− . A buffer solution can resist a change in pH when a small amount of H+ or OH− is added to it. [1] When a small amount of H+ is added, N3− + H+ → HN3 When a small amount of OH− is added, HN3 + OH− → N3− + H2O [1 for both equations] Note: In this buffer, HN3 is the weak acid, N3− is the conjugate base Use full arrow for acid-base reaction. (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. When NaOH is added to HN3, acid base reaction takes place. Use ICF table to determine the composition of the final mixture. Amount of OH− = 0.025 dm3 × 0.600 = 0.015 mol Amount of HN3 = 0.05 dm3 × 0.500 = 0.025 mol OH– + HN3 → N3– + H2O Initial / mol 0.015 (L.R.) 0.025 0 ‒ Change / mol -0.015 -0.015 +0.015 ‒ Final / mol 0 0.010 0.015 ‒