RI 12a 2025 The Periodic Table 1 tutorial Answers

-1- Raffles Institution Year 5 H2 Chemistry 2025 Tutorial 12a – The Periodic Table 1 Suggested Answers to Practice Questions 8 Answer: D Strategy 1. Observe x-axis (electrical conductivity) The metals (Na, Mg and A l) have high electrical conductivity (lie to the right of the axis) and increases from Na to Mg to Al i.e. Al would be most far right. ⇒ Reject A and B. 2. Observe y-axis (1st ionisation energy) Using data from data booklet, P has the highest first IE ⇒ Reject C. 9 Answer: D A Cl2 molecule has less electrons than P4 molecule. Hence Cl2 has weaker instantaneous dipole-induced dipole interactions and has a lower melting point. B Na has a lower melting point than Mg due to weaker metallic bonding in Na. C S8 molecule has more electrons than P4 molecule. Hence S8 has stronger instantaneous dipole-induced dipole interactions and has a higher melting point. D S8 molecule has more electrons than P4 molecule. Hence S8 has stronger instantaneous dipole-induced dipole interactions and has a higher melting point. SiO 2 is also the only insoluble oxide amongst the three elements. 10 Answer: A W is Silicon. X is Aluminium. Y is either Phosphorus or Sulfur. Z is Sodium. 11 (a) Molecular formula of the vapour: Al2Cl6 Note: 2 molecules of AlCl3 dimerise to form Al2Cl6 which is more stable (as Al is electron deficient in AlCl3). [1+1] (b) When only a few drops of water are added to the solid, steamy white fumes of HC l are evolved and together with the formation of Al(OH)3, a white solid , which is insoluble in water. [1] Al2Cl6(s) + 6H2O(l) → 2Al(OH)3(s) + 6HCl(g) [1] white solid steamy white fumes OR AlCl3(s) + 3H2O(l) → Al(OH)3(s) + 3HCl(g) white solid steamy white fumes When a large amount of water is added to the solid, [Al(H2O)6]3+(aq) is formed. AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl–(aq) colourless solution [Al(H2O)6]3+(aq) undergoes hydrolysis as shown below.

-2- [Al(H2O)6]3+(aq) ⇌ [Al(H2O)5(OH)]2+(aq) + H+(aq) OR [Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5(OH)]2+(aq) + H3O+(aq) [1] acidic Al3+ ions, with high charge density and hence high polarising power, are able to distort the electron cloud of the water molecules , weakening the O –H bond. As a result , the O –H bond undergoes heterolytic fission readily to release H+. [1] Note: • yellow–white solid can be denoted by AlCl3 or Al2Cl6 (c) Na2O(s) + H2O(l) → 2NaOH(aq) [1] Na 2O(s) dissolves to produce NaOH(aq). The litmus solution turns from purple to blue due to the presence of OH– ions. [1] (d) (i) [2] (d) (ii) Mg 3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) [1] . . ( .) ( .) 32 20Amount of Mg N 0 01982 mol 3 24 3 2 14 0 = = + [1] ..2Amount of Mg(OH) 3 0 01982 0 05946 mol= ×= [1] . ( . ( . . )) .2Mass of Mg(OH) 0 05946 24 3 2 16 0 1 0 3 47 g= ×+ += [1] Alternative answer MgO can be accepted: Mg3N2(s) + 3H2O(l) → 3MgO(s) + 2NH3(g) [1] . . ( .) ( .) 32 20Amount of Mg N 0 01982 mol 3 24 3 2 14 0 = = + [1] ..Amount of MgO 3 0 01982 0 05946 mol= ×= [1] . ( . .) .Mass of MgO 0 05946 24 3 16 0 2 40 g= ×+= [1] 12 (a) L: MgCl2 M: Mg(OH)2 [1] (b) [Mg(H2O)6]2+(aq) [Mg(H2O)5(OH)]+(aq) + H+(aq) [1] OR [Mg(H2O)6]2+(aq) + H2O(l) [Mg(H2O)5(OH)]+(aq) + H3O+(aq) Mg2+ ions, with high charge density and hence high polarising power, are able to distort the electron cloud of the water molecules, weakening the O–H bond. As a result, the O– H bond undergoes heterolytic fission readily to release H+. [1] 13 (a) (i) Giant covalent structure / Giant molecular structure [1] (a) (ii) To melt the elements, strong covalent bonds between the atoms have to be broken. From C to Ge, the valence orbitals become more diffuse. The overlap of those orbitals becomes less effective. Strength of covalent bond decreases and hence less energy is required for melting. [2] Mg N3 2 x x x 2+ 3

-3- (b) (i) SiCl4 + 2H2O → SiO2 + 4HCl [1] Alternative answer (this answer is also accepted): SiCl4 + 4H2O → SiO2 .2H2O + 4HCl Or SiCl4 + 4H2O → Si(OH)4 + 4HCl (b) (ii) C–Cl : 340 kJ mol–1 and Si–Cl : 359 kJ mol–1 [1] [Teacher’s note: Based on b(ii), inertness of CCl 4 to water is not due to C -Cl bond being too strong and not broken] (b) (iii) The electron deficient carbon atom in CCl4 is attached to four large chlorine atoms which hinder the approach of the H2O nucleophile. OR C does not have low –lying vacant orbitals to acc ept the lone pair of electrons from water during hydrolysis. [1] 14 First, add HCl(aq) to a sample of the white powder. If the solid is insoluble in acid, it must be the acidic oxide SiO2. If the solid dissolves in acid, add NaOH(aq) to a fresh sample of the white powder. If the solid is insoluble in NaOH(aq), it must be the basic oxide MgO. If the solid dissolves in both acid and base, it must be the amphoteric oxide Al2O3. sample is When HCl(aq) is added, insoluble soluble soluble When NaOH(aq) is added, insoluble soluble Identity SiO2 MgO Al2O3 Equations: MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na+[Al(OH)4]–(aq)