2025 JC2 Prelims H2 Chem Paper 2 QP and Ans TJC

2025 JC2 Prelims H2 P2 (Review) 1 In 1932, the American chemist Linus Pauling developed the most common scale of relative electronegativity (EN) values for the elements. The Pauling EN values of elements can be used to predict the chemical properties of compounds. The EN values of four Period 3 elements are given in Table 1.1. Table 1.1 Element Sodium Aluminium Phosphorus Chlorine Pauling EN value 0.93 1.61 2.19 3.16 l (a) (i) Explain the difference in the Pauling electronegativity values of Na and Cl. [2] Comparing sodium and chlorine, [✓] chlorine has a higher nuclear charge due to larger number of protons, [✓] Screening effect remains approximately constant as electrons are added to the same electronic shell. [✓] As there is stronger a ttraction between the nucleus and the bonding electrons in the outer shell, chlorine has higher [✓] electronegativity, resulting in a larger EN value. 2[✓] = [1] The ionic character of a bond is directly related to the electronegativity difference (EN) between the bonded atoms. Fig. 1.1 shows the plot of percent ionic character against EN for NaCl. Fig. 1.1

(ii) Using the values in Table 1.1, calculate EN for A lCl3 and PC l5 respectively. Hence plot the percent ionic character of A lCl3 and PCl5 on Fig. 1.1. Label your points clearly. [2] EN for AlCl3 = 3.16 – 1.61 = 1.55 [✓] EN for PCl5 = 3.16 – 2.19 = 0.970 [✓] 2[✓] = [1] [1] Correct relative order for percent ionic character for the three compounds (iii) Explain the difference in bonding for NaC l and PCl5 in terms of electronegativity. [2] [1] Ionic bonds exist in NaC l due to large EN (or large difference in electronegativities) between Na and Cl. [1] Covalent bonds exist in PCl5 due to small EN (or small difference in electronegativities) between P and Cl. (b) (i) The polarity of bonds in covalent molecules is also affected by EN. SeF4 and BrF3 are two fluorine-containing molecules. State whether SeF4 or BrF3 contains covalent bonds with a higher ionic character. [1] [1] SeF4 > BrF3 (ii) Using VSEPR theory, predict and explain the shape and bond angles of SeF4. Illustrate the shape of SeF4 with an appropriate diagram. [3] There are [✓] 4 bond pairs and 1 lone pair in the valence shell of Se atom. [✓] To minimise repulsion and maximise stability, the shape of SeF4 is [✓] see-saw.

[✓] Since [✓] lone pair – bond pair repulsion > bond pair – bond pair repulsion, the [✓] bond angles are 88o and 118o (accept other bond angles < 90o and < 120o). 2[✓] = [1] [Total: 10]

2 PCl5 is a Period 3 chloride commonly used as a chlorinating agent and catalyst in making organic compounds. Industrial production of PCl5 involves the reaction of Cl2 with PCl3. Reaction (1) PCl3(g) + Cl2(g) ⇌ PCl5(g) (a) Suggest, with an explanation, how the position of equilibrium and the composition of the equilibrium mixture might change when chlorine is added to the equilibrium system. [2] By Le Chatelier’s Principle, the [✓] position of equilibrium will shift right to [✓] decrease the concentration of Cl2. The equilibrium mixture will contain [✓] more PCl5, less PCl3 and [✓] more Cl2. 2[✓] = [1] x mol of Cl2 gas is added to a 2 dm 3 vessel containing an equilibrium system of 0.4 mol of PCl3, 0.25 mol of Cl2 gas and 1.2 mol of PC l5. The new equilibrium amount of PCl5 is 1.28 mol. (b) (i) Write the Kc expression for reaction (1), giving its units. [1] [1] Kc = [PCl5] [PCl3][Cl2] , mol−1dm3 (ii) Using the information given above, calculate Kc. [1] Kc = [PCl5] [PCl3][Cl2] = (1.2 2 ) (0.4 2 )(0.25 2 ) = 24.0 mol−1dm3 [1] (iii) Hence calculate x, the amount of Cl2 added. [2] PCl3(g) + Cl2(g) ⇌ PCl5(g) Initial amount /mol 0.4 0.25 + x 1.2 Change in amount /mol -y -y +y Equilibrium amount /mol 0.4 – y = 0.32 0.25 + x – y = 0.17 + x 1.28 [1] ICE Table Working y = 1.28 – 1.2 = 0.08 Kc = [PCl5] [PCl3][Cl2] = (1.28 2 ) (0.32 2 )(0.17+𝑥 2 ) = 24 x = 0.163 mol [1] (c) Period 3 oxides follow a similar trend in bonding as the Period 3 chlorides. Describe the action of water on the oxides of sodium, aluminium and phosphorus, write equations for any reactions that occur, and suggest the pH of each solution formed. [3] Na2O [✓] reacts vigorously with water to give a strongly alkaline solution of NaOH(aq) with a [✓] pH = 13. [✓] Na2O + H2O → 2NaOH P4O10 [✓] reacts vigorously with water to give a strongly acidic solution of

H3PO4(aq) with a [✓] pH = 2. [✓] P4O10 + 6H2O → 4H3PO4 [✓] Al2O3 does not react with water due to its high lattice energy. Hence Al2O3 is insoluble in water. [✓] pH = 7 3[✓] = [1] 6[✓] = [2] 8[✓] = [3] [Total: 9]

3 (a) Nitrogen exhibits a range of oxidation numbers in its compounds. A few of such species are NO, N2O, NO2, N2 and NH2OH. (i) Draw the dot-and-cross diagram for NO2. [1] [1] 0.074 g of hydroxylamine, NH 2OH, is dissolved in water. Excess solution of acidified iron(III) salt is added to the dissolved hydroxylamine to form a nitrogen- containing product and iron(II) ions. The iron(II) ions produced requires 44.8 cm3 of 0.02 mol dm-3 acidified potassium manganate(VII) for complete reaction. The reacting ratio of iron(II) ions and manganate(VII) ions is 5:1. (ii) Calculate the amount of iron(III) reacted with hydroxylamine. [1] Amount of manganate used = (44.8/1000) x 0.02 = 0.000896 mol Amount of iron(II) produced = 0.000896 x 5 = 0.00448 mol Fe2+ ≡ Fe3+ Amount of iron(III) reacted with hydroxylamine = 0.00448 mol [1] (iii) Determine the oxidation state of the nitrogen atom in the nitrogen- containing product. Hence deduce the identity of the nitrogen-containing product. [2] Amount of hydroxylamine used = 0.074 / 33 = 0.00224 mol Reduction: Fe3+ + e– → Fe2+ Hence 2Fe3+ ≡ NH2OH ≡ 2e– Oxidation of 1 NH 2OH loses 2e –, so final oxidation state of nitrogen in product = – 1 + 2 = +1 [1 with clear working] [1 with clear working] N2O (iv) Construct a balanced ionic equation for the reaction of NH 2OH with Fe 3+. [1] Fe3+ + e– → Fe2+ (x4) 2NH2OH → N2O + H2O + 4H+ + 4e– (x1) [1] 4Fe3+ +2NH2OH → 4Fe2+ + N2O + H2O + 4H+

(b) Figure 3.1 shows the second ionisation energies of nine consecutive elements A to I with atomic numbers below 20 in the Periodic Table. Labels A to I are not the atomic symbols of the elements. Fig. 3.1 (i) Define the second ionisation energy of element F. [1] [1]The second ionisation energy of element F is the minimum energy required to completely remove one mole of electrons from one mole of ground-state gaseous F+ ions to form 1 mole of gaseous F2+ ions. F+(g) → F2+(g) + e– (ii) Suggest the identity of B. Explain how you arrived at your answer. [2] [1] Large decrease in 2nd ionisation energy from D to E implies that the 2nd electron in E is removed from the outer electron shell. E belongs to Group 2 and B belongs to group 17. [1] B is fluorine (iii) Explain the difference in second ionisation energy between element A and element B. [1] A: s2p4 A+: s2p3 B: s2p5 B+: s2p4 [1] The 2nd electron removed in B is a paired electron which experiences interelectronic repulsion, so less energy is needed to remove the paired electron than the unpaired electron removed in A. [Total: 9]

4 Hydrogen peroxide, H2O2, finds its applications in a diversity of fields as it is considered an environmentally-friendly oxidising agent. (a) (i) Suggest why it is environmentally -friendly to use H 2O2 as an oxidising agent. [1] [1] The product of reduction of H2O2 is water which is clean / non-pollutant / environmentally friendly. Today, most of the world’s hydrogen peroxide is manufactured by the anthraquinone, C14H8O2 process. This process involves the two steps shown below. step 1: O O + H2 OH OH Pd(s) catalyst Anthraquinone Anthra