RI 2021 Y5 Promo Solutions

1 © Raffles Institution 2021 9729/S/21 2021 Y5 H2 Chemistry Promotion Examination – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C D D D B B D A A A C B C B A 1 Amount of E = 1 24000 mol (Note: 1 mole of gas occupies 24 dm3 at room temperature and pressure.) No. of molecules in 1 cm3 of E = = 1 24000  L = L 24000 No. of atoms in 1 cm3 of E = L 24000  3 = 3L 24000 (Note: E is a triatomic molecule.) 2 Reduction half-equation: SO2 + 4H+ + 4e– ⎯→ S + 2H2O Oxidation half-equation: H2S ⎯→ S + 2H+ + 2e– Overall balanced equation: SO2 + 2H2S ⎯→ 3S + 2H2O After reaction is completed, the mixture contains unreacted H2S. Hence H2S is present in excess and SO2 is the limiting reagent  Options B and C are incorrect. When 5 mol of H2S reacts with 1 mol of SO2, SO2 is the limiting reagent. 3 mol of H2S will remain unreacted and 3 mol of S will be formed. 3 Since the biggest jump in ionisation energy is from the 5th to 6th ionisation energy, element M has five valence electrons. Since element M is in Group 15, it reacts with chlorine to form the covalent compound MCl3. 4 Rb+, Br− and Kr are isoelectronic species (i.e. with same total number of electrons)  their outermost e− experience the same shielding effect. However, nuclear charge of Br− < Kr < Rb+. Hence, effective nuclear charge of Br− < Kr < Rb+. Electrostatic attraction between the nucleus and the outermost e− of Br− < Kr < Rb+. Energy required to remove the outermost e− increases in the order: H2 < H3 < H1

2 © Raffles Institution 2021 9729/S/21 5 molecule structure shape bond dipoles cancel out? polar / non-polar SF4 see-saw No polar XeF2 linear Yes non-polar CH2F2 tetrahedral No polar SO2 bent No polar 6 Statement 1 Incorrect. In the aldehyde group ( ) of butanal, the hydrogen atom is bonded to carbon (not oxygen). Hence there is no intermolecular hydrogen bonding in butanal. Only intermolecular instantaneous dipole- induced dipole and permanent dipole -permanent dipole interactions exist in butanal. Recall the criteria for hydrogen bonding: • A hydrogen atom covalently bonded to F, O and N. • A lone pair of electrons on a F, O and N atom in a neighboring molecule bearing a − which can attract the + charge on the H atom. Statement 2 Incorrect. All three compounds are simple molecules. Although C=C bond is stronger and requires more energy to break compared to C−C bond, boiling of simple molecules involves overcoming intermolecular forces of attraction, NOT covalent bonds. Statement 3 Correct. Butanal and 2-methylpropanal differs only in their hydrocarbon chain. Both compounds have the same aldehyde functional group and the same number of electrons (as they have the same molecular formula). Butanal is a straight-chain aldehyde, which has greater surface area for intermolecular interactions as compared to its branched isomer, 2-