Copy of PLMGSS 2024 4E Chem P2 Ans

1 Answers to Sec 4 Pure Chem Preliminary Exam P2 2024 1(a) The solid dissolves/disappears/becomes smaller. [1] (b) Carbonic acid [1] (c) Barium ethanoate or (CH3COO)2Ba [1] (d) Ba + 2H2O → Ba(OH)2 + H2 [1] (e)(i) CaCO3 decomposes first. CaCO3 has a lower thermal stability as calcium is less reactive than barium. [1] [1] (ii) BaCO3 → BaO + CO2 [1] 2(a) The ionic radius decreases from P to Cl. As you move across the Period from P to Cl, the nuclear charge increases (OR number of protons increases), the electrons are more closely attracted to the nucleus (attraction between protons and electrons increases). [1] [1] (b)(i) Ion chlorate perchlorate hypochlorite chloride formula ClO3- ClO4- ClO- Cl- oxidation state of chlorine +5 +7 +1 -1 2m- all correct 1m-2 or 3 correct (ii) Oxidation state of chlorine increases from +5 in KClO3 to +7 in KClO4. Chlorine is oxidised. (accept if KClO3 oxidised) Oxidation state of chlorine decreases from +5 in KClO3 to -1 in KCl, Chlorine is reduced. (accept if KClO3 reduced) Since chlorine is both oxidised and reduced, it is a disproportionation reaction. [1] [1] (c) test observation addition of a few drops of Br2(aq) Solution remains reddish brown addition of a few drops of AgNO3(aq) White precipitate [1] [1]

2 3(a) Covalent bonding, simple covalent (molecular) structure [1]-both answers (b) [1-bonding electrons between O & P, O electrons] [1-bonding electrons between Cl & P, electrons of Cl] (c) Enthalpy change for bond breaking in reactants = 2(289) + 496 = +1074 kJ Enthalpy change for bond formation in products = - (2 x 592) = -1184 kJ Enthalpy change for reaction = + 1074 – 11 84 = -110 kJ [1] [1] (d) [1- exo + labelling of reactants & products [1-Ea labelling & value] [1-∆H & value] 4(a) Similarity: At the anode, iodine is produced. Difference: At the cathode, for molten magnesium iodide, magnesium is produced while hydrogen is produced for concentrated aqueous magnesium iodide. [1] [1] (b) Colourless solution turns brown. Chlorine which is more reactive than iodine displaces iodine from magnesium iodide, producing the brown iodine. [1] [1] (c) The oxidation state of iodine increases from -1 in I- to 0 in I2. I- is oxidised, hence it is acting as the reducing agent. [1] [1] Ea = +1074kJ ∆H = -110kJ Progress of reaction energy

3 (d)(i) Iodine -123 and iodine-131 have the same number of 7 valence electrons, hence will have similar chemical properties. [1] (ii) 𝑃15 32 → 𝑆16 32 + 𝛽−1 0 [1] 5(a) No of moles of S2O32- = 0.0200 x 20.10/1000 = 0.000402 mol Mole ratio I2: S2O32- = 1: 2 = 0.000402/2 : 0.000402 mol No of moles of I2 = 0.000402/2 = 0.000201 mol [1] (b) Mole ratio I2: Cu2+ = 1: 2 = 0.000201: 0.000201 x 2 No of moles of Cu2+ = 0.000402 mol No of moles of Cu2+ in origin