RI 2022 Prelim P2 (Ans)

1 © Raffles Institution 2022 9729/02/S/22 2022 Y6 H2 Chemistry Preliminary Exam Paper 2 – Suggested Solutions 1(a) Amount of iodine reacted = 65 22700  1 3  3 2 = 1.432  10–3 mol Mass of iodine = 1.432  10–3  126.9  2 = 0.363 g Comments: • Students need to recall that the molar volume of a gas at standard temperature and pressure (s.t.p) is 22.7 dm3 mol−1 (the molar volume at s.t.p and r.t.p can also be obtained from the Data Booklet). There is no need to use pV = nRT equation to calculate the amount of HI gas. • Using the given equations, the molar ratio of the species are as follows: amt of I2 : amt of PI3 : amt of HI 3 : 2 1 : 3 3 : 2 : 6 Therefore, amt of I2 : amt of HI 3 : 6 1 : 2 • mass of iodine = mass of I2 molecules = amt of I2  molar mass of I2 1(b)(i) The boiling points increase from HCl to HBr to HI as the size of electron cloud for polarisation increases, resulting in stronger instantaneous dipole -induced dipole (id-id) interactions, and hence requiring more energy to overcome. The boiling point decreases from HF to HCl (or HF has the highest boiling point ) due to the presence of stronger hydrogen bonding between HF molecules which requires more energy to overcome. Comments: • Boiling point is dependent on the amount of energy required to overcome the intermolecular forces of attraction . Hence, the explanation should focus on comparing the strength of attractive forces between molecules. Students should not discuss about halide ions (Cl−, Br−, I−) nor halogen atoms (Cl, Br, I). • The size and ease of polarisation of the electron cloud of HX molecule (X = Cl, Br, I) affects the strength of instantaneous dipole-induced dipole (not permanent dipole-permanent dipole) interactions between the molecules. • The strength of permanent dipole-permanent dipole interactions is not relevant in this question. Some students explained that H-Cl bond is more polar followed by H -Br and then HI. Hence, HC l molecules have the strongest pd -pd interactions followed by HBr and HI. However, the b.p of HCl < HBr < HI. • Students need to revise how hydrogen bonding arises and deduce that only HF molecule can form intermolecular hydrogen bonding.

2 © Raffles Institution 2022 9729/02/S/22 1(b)(ii) Down Group 17, the valence orbital of the halogen atom becomes increasingly diffuse and the orbital overlap between the halogen and hydrogen atoms becomes less effective / electronegativity difference between the halogen and hydrogen decreases, resulting in a decrease in bond polarity. Hence, the H–X bond strength decreases from HCl to HBr to H I and the thermal stability decreases from HCl to HBr to HI. Comments: • Thermal stability is dependent on the strength of H −X