RI Periodic Table II - Answers to Discussion Questions

-1- Raffles Institution Year 6 H2 Chemistry 2025 Tutorial 23: The Periodic Table (II) (Suggested Answers) Q8a ZnCO3(s) ⎯⎯→ ZnO(s) + CO2(g) Q8b The ionic radius of Mg 2+ (0.065 nm) is smaller than that of Zn 2+ (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 CO32– ion in ZnCO3 to a smaller extent. The covalent bonds within the CO32– ion are weakened to a smaller extent as compared to that in MgCO3. 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 XO, YO or ZO, 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. • 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

-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 = 0.01