RI 12b 2025 The Periodic Table 2 Tutorial Answers

-1- RAFFLES INSTITUTION YEAR 5 H2 CHEMISTRY 2025 Tutorial 12b – The Periodic Table 2 Answers to practice questions Q8a ZnCO3(s) → ZnO(s) + CO2(g) Q8b The ionic radius of Mg2+ (0.065 nm) is smaller than that of Zn2+ (0.074 nm). Hence with the same charge, Zn2+ has a lower charge density and hence weaker polarising power than Mg2+. As a result, Zn2+ distorts the electron cloud of CO 32– ion in ZnCO3 to a smaller extent. The covalent bonds within the CO 32– ion are weakened to a smaller extent as compared to that in MgCO 3. Therefore, more heat energy is needed to decompose ZnCO3. Hence ZnCO3 will decompose at a higher temperature than MgCO3. Q9a 2Mg(IO3)2(s) → 2MgO(s) + 2I2(g) + 5O2(g) Q9b Determine the given three iodates(V): Assume one iodate(V) given is Mg(IO3)2. Molar mass of Mg(IO3)2 = 374.1 g mol–1 Amount of Mg(IO3)2 = 2.00 374.1 = 5.346 x 10–3 mol Amount of MgO formed = 5.346 x 10 –3 mol Molar mass of MgO = 40.3 g mol–1 Mass of MgO formed = (5.346 x 10–3)(40.3) = 0.215 g Since the mass of the oxide calculated does not correspond to that of either X O, YO or Z O, Mg( IO3)2 is not among the three given iodates(V). Hence the given iodates(V) are Ca(IO3)2, Sr(IO3)2 and Ba(IO3)2. [Note: Calculation needs to be made for at least one of the iodates(V) and then reasoning can be made to deduce the identities of the given iodates(V).] Comparison of thermal stability of Ca( IO3)2, Sr(IO3)2 and Ba(IO3)2: • The cationic radius increases from Ca2+ to Sr2+ to Ba2+. • This causes the charge density and polarising power of the cations to decrease from Ca2+ to Sr2+ to Ba2+. • Hence the extent of distortion of the electron cloud of the IO3– anion decreases from Ca(IO3)2 to Sr(IO3)2 to Ba(IO3)2. • The extent of weakening of the covalent bonds in IO3– ion also decreases from Ca( IO3)2 to Sr(IO3)2 to Ba(IO3)2.

-2- • Hence the amount of heat energy and temperature required for decomposition increases from Ca(IO3)2 to Sr(IO3)2 to Ba(IO3)2. • Therefore, thermal stability increases in the order: Ca(IO3)2 < Sr(IO3)2 < Ba(IO3) From the graphs, thermal stability of the given iodates(V) increases in the following order: Z(IO3)2 < X(IO3)2 < Y(IO3)2 Hence, Z(IO3)2 is Ca(IO3)2 ⇒ its rate of thermal decomposition is the fastest. X(IO3)2 is Sr(IO3)2 ⇒ its rate of thermal decomposition is slower than Ca(IO3)2. Y( IO3)2 is Ba(IO3)2 ⇒ it does not undergo thermal decomposition. Q10a Ca(CH3CO2)2 → CaCO3 + CH3COCH3 solid M Comments: • The question has clearly stated that propanone is formed. When writing the balanced equation, the structural formula of propanone should be shown, not just the molecular formula (C 3H6O). Q10b The CaCO3 sample (i.e. M) undergoes partial decomposition to give a white residue. This white residue contains both CaCO3 and CaO. Method 1 Amount of CaCO3 formed at 450 oC = Amount of Ca(CH3CO2)2 = 2.0 158.1 = 0.01265 mol Let the amount of CO2 evolved = y mol. Amount/mol CaCO3 → CaO + CO2 Initial 0.01265 - - Change -y +y +y Final 0.01265 - y y y Mass of CaCO 3 in the residue = (0.01265 – y)(100.1) = (1.266 – 100.1y) g Mass of CaO in the residue = 56.1y g mass of CaCO3in the residue mass of CaO in the residue = 70 30 ⇒ 1.266-100.1y 56.1 y = 7 3 1.266 – 100.1y = 130.9y y = 0.005483 Amount of CO2 evolved = 0.005483 mol Mass of CO2 evolved = (0.005483)(44.0) = 0.241 g

-3- Method 2 Mass of CaCO3 formed at 450°C = (2/158.1)(100.1) = 1.266 g CaCO3 → CaO + CO2 mass after partial decomposition / g 1.266 – x – y x y Since mass of white residue = (1.266 – y) g and % of CaO in white residue = 30%, x 0.3 1.266 y =− Since amount of CaO formed = amount of CO2 formed, xy 56.1 44= Solving simultaneous equations, y = mass of CO 2 lost = 0.241 g Method 3 Mass of CaCO3 formed at 450°C = (2/158.1)(100.1) = 1.266 g Let mass of white residue be x g and mass of CO2 evolved by y g. Mass of CaO = (0.3x) g Since CaCO3 → CaO + CO2, amount of CaO = amount of CO2 0.3x y 56.1 44= By conservation of mass, x + y = 1.266 g Solving simultaneous equations, y = 0.241 g Comments: • Candidates should show all essential steps with appropriate units. • Candidates should keep the answers for intermediate steps to 3 or 4 significant figures. • The question stated that the white residue contains 70% by mass of solid M . Many candidates misinterpreted this information given. Wrong interpretation: Mass of CaCO3 in the white residue = 70% of mass of CaCO3 formed initially = 0.70 x 1.266 Correct interpretation: Mass of CaCO3 in the white residue = 70% of mass of white residue = 0.70 x (mass of CaCO3 + mass of CaO) Q10c Ba(CH3CO2)2 will decompose at a higher temperature. Ba 2+ ion has a larger ionic radius as compared to Ca2+ ion resulting in a lower charge density and weaker polarising power. Hence, Ba2+ distorts the electron cloud of the CH 3CO2– ions to a smaller extent and thus weakens the covalent bonds within CH3CO2– ions to a smaller extent as compared to Ca2+ ion. This results in more heat energy and a higher temperature required for decomposition of Ba(CH3CO2)2. Comments: • Reference must be made to the ion (Ba2+), not atom (Ba).

-4- Q11(a)(i) Down Group 17, the value of EӨ for X2(g) + 2e– ⇌ 2X–(aq) becomes less positive and the position of equilibrium of the reduction of X 2 to Xˉ lies increasingly to the left. Hence, down the group, X 2 has less tendency to be reduced and the oxidising power of X2 decreases. Q11(a)(ii) Chlorine, being a stronger oxidising agent, oxidises thiosulfate to sulfate ions, where the average oxidation state of S increases from +2 to +6. Iodine, being a weaker oxidising agent, oxidises thiosulfate to tetrathionate ions , where the average oxidation state of S only increases from +2 to +2.5. (equations are not required by question) S2O32–(aq) + 4Cl2(aq) + 5H2O(l) → 2SO42–(aq) + 8Cl–(aq) + 10H+(aq) 2S2O32–(aq) + I2(aq) → S4O62–(aq) + 2I–(aq) Alternative answer Chlorine, being a stronger oxidising agent, oxidises Fe to Fe3+ (i.e. from Fe to Fe2+, and then from Fe2+ to Fe3+), where the oxidation state of Fe increases from 0 to +3. Iodine, being a weaker oxidising agent, oxidises Fe to Fe2+, where the oxidation state of Fe only increases from 0 to +2. (equations are not required by question) 2Fe(s) + 3Cl2 → 2FeCl3(s) Fe(s) + I2 → FeI2(s) Alternative answer Chlorine, being a stronger oxidising agent, oxidises Br– to Br2 / I– to I2. (equations are not required by question) Cl2 + 2I– → 2Cl– + I2 Note: We can also compare Br 2 and I–. Furthermore, only displacement reactions allow us to compare the relative oxidising power of Cl2, Br2 and I2. On the other hand, reaction with thiosulfate or Fe only allows us to conclude that Cl2 and Br2 are stronger oxidising agent than I2. Q11(b)(i) 2HX(g) → H2(g) + X2(g) (**Reversible arrows are acceptable too since the extent of reaction depends on the type of hydrogen halides.) Q11(b)(ii) The thermal stability of the hydrogen halides decreases down Group 17. Thermal stability is related to the H –X bond strength. The more endothermic bond dissociation energy of the H–X bond, the stronger the H–X bond and that HX is more thermally stable. Down Group 17, there is less effective orbital overlap between the 1s orbital of H and the valence p orbital of the halogen. Also, the difference in electronegativity between H and the halogens also decrease down the group. This leads to a decrease in bond polarity . Hence, the H –X bon