JPJC 2026 H2 Chemistry WA1 Answer

2026 JC1 H2 Chemistry / WA1 1 © Jurong Pioneer Junior College JURONG PIONEER JUNIOR COLLEGE 2026 JC1 H2 CHEMISTRY (9476) WEIGHTED ASSESSMENT 1 Atomic Structure, The Mole Concept & Stoichiometry and Chemical Bonding Name : Answers for Students Raw score Class : 26S n Lack 3sf in final ans –1 / NA Date : _______________ Missing/wrong units in final ans –1 / NA Duration: 50 min Final score 25 INSTRUCTIONS TO CANDIDATES Answer ALL questions in the space provided. All numerical answers should be given to 3 significant figures, where appropriate. A Data Booklet is provided. 1 (a) Aluminium, Al, and gallium, Ga, are elements in Group 13. (i) Write the full electronic configuration of a Ga atom. {Ga belongs to Period 4, Group 13  valence shell: 4s2 4p1} Working: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 [1] [1]

2026 JC1 H2 Chemistry / WA1 2 © Jurong Pioneer Junior College (ii) Sketch a graph in Fig ure 1.1 to show the trend in the first seven ionisation energies of an Al atom. Figure 1.1 (✓1) big ‘jump’ between 3rd IE and 4th IE (✓2) successive IEs increase (moderate “jump” between 1st IE and 2nd IE) [1] Note: 13Al : 1s2 2s2 2p6 3s2 3p1 Al has 3 valence electrons. There is a large increase from the 3rd IE to the 4th IE due to a change in quantum shells where the 4th electron is removed from the inner quantum shell closer to the nucleus and is more strongly attracted to the nucleus. There is also a moderate increase from the 1 st IE to the 2 nd IE due to a change in subshells, where the 2nd electron is removed from the 3s subshell and the 1st electron is removed from the 3p subshell. The 3s subshell has a lower energy and is closer to the nucleus than the 3p subshell. The type of electrons removed are as follows: 1st 2nd 3rd 4th 5th 6th 7th 3p 3s 3s 2p 2p 2p 2p x x x x x x x [1] 1st 2nd 3rd 4th 5th 6th 7th

2026 JC1 H2 Chemistry / WA1 3 © Jurong Pioneer Junior College (b) Table 1.1 shows the first ionisation energies for three elements, Mg, Al and Ga. Table 1.1 element Mg Al Ga first ionisation energy / kJ mol–1 736 577 577 (i) Explain why the first ionisation energy of Al is smaller in magnitude than that of Mg. The 3p electron of Al is further away from the nucleus and has a higher energy than the 3s electron of Mg. [1] Hence less energy is required to remove the 3p electron from A l than the 3s electron from Mg. [1] (ii) Explain why the first ionisation energy of Al is the same as that of Ga. Compared to A l, Ga has more quantum shell s of electrons (✓1) and hence a greater shielding effect (✓2), which cancels out (or offsets) the effect of its greater nuclear charge (or greater number of protons) (✓3). Hence, both have a similar net nuclear attraction on their outermost electrons (✓4) and require the same amount of energy to remove the outermost electron. [2] (c) (i) Deduce which ions, Al3+ or O2–, is larger. Explain your answer. Al3+ and O 2– are isoelectronic (or have the same electronic configuration) (✓1) and hence have similar shielding effect (✓2) by inner shell electrons , but O2– has a smaller nuclear charge (or smaller number of protons) (✓3) than Al3+-.  O2– has a weaker net nuclear attraction on its outermost electrons (✓4) than Al3+ and hence O2– is larger (✓5). [2]

2026 JC1 H2 Chemistry / WA1 4 © Jurong Pioneer Junior College (ii) When separate beams of 27Al3+ and 16O2– are passed through the same electric field, the particles are deflected to opposite directions. Given that the angle of deflection of 16O2– is 4.0°, calculate the angle of deflection of 27Al3+. Note: Angle of deflection  charge, q mass, m If the given angle of deflection has no sign, then do not include a sign for the angle of deflection in your answer. o 3+ x27 =2- 4 16 Angle of deflection of 27Al3+, x = 3.56o (or 3.6o) [1] [1] [Total: 8] 2 (a) Acidified potassium manganate( VII) is a common oxidising agent. It can oxidise ClO2– ions, C2O42– ions, Fe2+ ions, and many other reducing agents. (i) Construct a balanced ionic equation for the reaction between ClO2– ions and acidified MnO4– ions, given that ClO2– ions are oxidised to ClO3– ions. Note: Acidified MnO4– is reduced to Mn2+. [O] ClO2– + H2O → ClO3– + 2H+ + 2e– ( 5) [R] MnO4– + 8H+ + 5e– → Mn2+ + 4H2O ( 2) Overall, 5ClO2– + 2MnO4– + 6H+ → 5ClO3– + 2Mn2+ + 3H2O [1] [1]

2026 JC1 H2 Chemistry / WA1 5 © Jurong Pioneer Junior College An impure sample of BaC 2O4 with a mass of 0.550 g is added to 50.0 cm 3 of 0.0200 mol dm–3 acidified MnO4–(aq), which is in excess. A redox reaction occurs in which all the C2O42– ions react completely to form CO2 gas. 5C2O42– + 2MnO4– + 16H+ → 10CO2 + 2Mn2+ + 8H2O The resulting solution, containing unreacted acidified MnO 4–, is then titrated with 0.0500 mol dm–3 Fe2+(aq). The following reaction occurs during this titration: MnO4– + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O 29.40 cm3 of 0.0500 mol dm–3 Fe2+(aq) is required to reach the end-point. (ii) Calculate the amount of MnO4– left after the reaction with BaC2O4. n(Fe2+) reacted with leftover MnO4– = 29.40 0.05001000 = 1.47  10–3 mol n(MnO4–) left after the reaction with BaC2O4 = 1 5  1.47  10–3 = 2.94  10–4 mol [1] [1] (iii) Calculate the amount of MnO4– ions that reacted with all the C2O42– ions in the 0.550 g impure sample. n(MnO4–) reacted with C2O42- =   50 0.02001000 – 2.94  10–4 = 7.06  10–4 mol [1] [1] (iv) Hence, calculate the percentage by mass of BaC2O4 in the 0.550 g impure sample. [Mr of BaC2O4 = 225.3] n(C2O42–) reacted = 52  7.06  10–4 = 1.765  10–3 mol [1] = n(BaC2O4) in 0.550 g sample Mass of BaC2O4 in the sample = 1.765  10–3  225.3 = 0.398 g % mass of BaC2O4 in the sample = 0.398 1000.550 = 72.3 % [1] [2]

2026 JC1 H2 Chemistry / WA1 6 © Jurong Pioneer Junior College (b) Sodium chlorate( I), NaC lO, is the active ingredient in commercial bleach. The concentration of chlorate(I) ions, ClO–(aq), can be determined by adding an excess of hydrogen peroxide to the sample of the bleach solution and measuring the volume of oxygen gas produced. H2O2(aq) + NaClO(aq) → NaCl(aq) + O2(g) + H2O(l) When an excess of aqueous hydrogen peroxide was added to a 5.0 cm3 of a bleach solution, 82 cm3 of oxygen was produced at room temperature and pressure. Calculate the concentration of ClO– ions in this bleach solution. From the Data-Booklet, Vm = 24 dm3 mol-1 at r.t.p.  1 mol of any gas occupies 24 dm3 at r.t.p. n(O2) produced at r.t.p. = 82 24 000 = 3.417  10–3 mol [1] = n(ClO–) in 5 cm3 of bleach solution [ClO–] =  -31000 3.417 105.0 OR    -33.417 10 5.0 1000 = 0.683 mol dm–3 [1] [2]

2026 JC1 H2 Chemistry / WA1 7 © Jurong Pioneer Junior College (c) 40 cm 3 of gaseous hydrocarbon, C xHy, was exploded with 200 cm 3 of excess oxygen. The mixture was completely burnt. The resultant gaseous mixture has a total volume of 140 cm 3. When these products were passed through aqueous sodium hydroxide, the volume decreased by 120 cm3. Given that all volumes were measured at room temperature and pressure, deduce the molecular formula of the hydrocarbon. 2COV = 120 cm3 (✓1) 2O unreactedV = 140 – 120 = 20 cm3 2O reactedV = 200 – 20 = 180 cm3 (✓2) CxHy(g) + (x + y 4 ) O2(g) → xCO2(g) + y 2 H2O(l) Volume / cm3 40 : 180 : 120 - volume ratio