2022 HCI Prelim P3 Mark Scheme

2022 HCI C2 H2 Chemistry Prelims / Paper 3 HWA CHONG INSTITUTION 2022 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS (PAPER 3) 1 (a) (i) [½] working [½] answer to nearest integer (n must be 65) 63  74.1 100 + n  25.9 100 = 63.5 25.9n = 6350 – 4668.3  n = 1681.7 25.9 = 65 1 (a) (ii) [1] nucleon number is the total number of protons and neutrons in the nucleus of an atom 1 (a) (iii) [1] draw any one of these [1] draw this For each 1m: ½m for drawing the shape ½m for labelling x y z axes AND ‘naming’ that orbital (students who drew, say, the dz2 orbital but named it as dx2 – y2 are penalised) 1 (b) (i) [½] Zn oxidised to Zn2+ AND justify in terms of E(Zn2+/Zn) vs. E(Cu2+/Cu) and E(Au3+/Au) [½] Explain why Zn2+ is not reduced at the cathode in terms of E(Zn2+/Zn) vs. E(Cu2+/Cu) [½] Au is collected as anode sludge / Au falls off the anode (Reject ‘Au remains on the anode’ / ‘remains as anode’ / ‘remains on the Shakudo’) [½] Explain why Au is not oxidised in terms of E(Au3+/Au) vs. E(Cu2+/Cu) Sample write-up:

Zn2+ + 2e− ⇌ Zn E = –0.76 V Cu2+ + 2e− ⇌ Cu E = +0.34 V Au3+ + 3e− ⇌ Au E = +1.52 V At the anode, E(Zn2+/Zn) is the least positive (or most negative), so Zn oxidises to Zn2+ (which enters the solution). E(Au3+/Au) is more positive than E (Cu2+/Cu), therefore Au is not oxidised, Au sinks to the bottom of the container as anode sludge. At the cathode, E(Zn2+/Zn) is less positive (or more negative) than E (Cu2+/Cu), therefore Zn 2+ is not reduced (but remains as Zn2+ in the solution). 1 (b) (ii) Q = n(electrons)  F = n(electrons)  Le = It 0.26 63.52  L  1.6010–19 = 1.10  1260 L = 6.04  1023 [2] Any mistake loses 1m No ½m for this part 1 (c) (i) [1] no electron-rich centres / no region of high electron density / no + carbon / non-polar C–C and C–H bonds 1 (c) (ii) [½] difference in reaction condition : B enzene needs conc . H2SO4 as catalyst while propoxybenzene doesn’t . (Reject discussion of temperature ∵ for benzene, even at higher temperature (55 C), the reaction still doesn’t work if conc. H 2SO4 catalyst isn’t added. The question also never label the temperature for propoxybenzene’s nitration, so students shouldn’t be looking at temperature.) [½] difference in organic products : Benzene only mono -nitration / mono - substitution or draw out the nitrobenzene, while propoxybenzene is tri- substituted. [½] regarding the lone pair on O: Oxygen’s lone pair delocalises over the benzene ring. (Reject: ‘O is electron donating’ / ‘OCH 2CH2CH3 is electron donating’ ∵ that could refer to an inductive effect. ) (Reject: ‘alkyl group electron donating’ – must discuss about the lone pair on O) [½] so what if that lone pair delocalises : This incr