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 increases electron density in the benzene ring, mak ing it more reactive towards electrophile / more susceptible towards electrophilic substitution. 1 (d) [1] P: HOCH2CH2CHO [1] R: HOCH2CH2CH2OH (allow ecf from P only if student’s P can be reduced to student’s R using the H2/Pt) [1] Q: HOCH2CO2H (all three structures must fit their given molecular formula)

2022 HCI C2 H2 Chemistry Prelims / Paper 3 1 (e) initiation: → •CH2CH2O• [½] two curly half-arrows ‘fly out’ from one C–O bond, ‘fly to’ the correct atoms [½] correct product radical, • at correct atoms propagation: •CH2CH2O• + → •CH2CH2O–CH2CH2O• [½] correct radical reacts with epoxyethane [½] correct product radical, • at correct atoms 2 (a) (i) No. of moles of thiosulfate needed to react with 25 cm 3 of solution = 12.0/1000 x 0.50 = 0.006 No. of moles of I2 in 25 cm3 of solution = 0.006/2 = 0.003 mol [1] No. of moles of I2 formed from solid NaI = 0.003 x (100/25) = 0.012 mol 2I − + 2e → I2 No. of moles of electrons lost by iodide on reacting with conc H2SO4 = 0.012 x 2 = 0.024 mol No. of moles of electrons gained per mole of sulfuric acid = 0.024 / 0.003 = 8 [1] Oxidation state of sulfur product = +6 – 8 = –2 [1] 2 (a) (ii) Cl2 + 2e– ⇌ 2Cl– Eo = +1.36 V or Eo (Cl2/Cl–)= +1.36 V I2 + 2e– ⇌ 2I– Eo = +0.54 V or Eo (I2/I–)= +0.54 V [1] for quote Eo and comment on more/less positive Eo Iodide is a stronger reducing agent than chloride due to less positive Eo(I2/I–). Iodide is more easily oxidised to iodine and is able to reduce sulfur atom from +6 to a product with a lower oxidation state or chloride is unable to reduce H2SO4. [1] 2 (b) (i) Strong ionic bonds in P require large amounts of energy to overcome, hence P has high melting point and exists as a solid. [1] Weak dispersion forces between molecules of Q require lesser energy to overcome, hence Q has low melting point and exists as a liquid. [1] 2 (b) (ii) P: MgCl2 [1] Q: SiCl4 [1]

2 (b) (iii) P undergoes slight/partial hydrolysis in water [½] Mg(H2O)62+ + H2O ⇌ Mg(H2O)5(OH)+ + H3O+ [1] Q undergoes complete hydrolysis in water [½] SiCl4 + 2H2O → SiO2 + 4HCl [1] 2 (c) (i) Kp = PCl2 PPCl3 / PPCl5 [1] 2 (c) (ii) PCl5(g) ⇌ PCl3(g) + Cl2(g) Initial pressure/kPa 84 + x 0 0 Change in pressure/kPa –x +x +x Eqm pressure/kPa 84 x x Total equilibrium pressure = (84 + 2x) kPa (84 + x) / (84 + 2x) = 0.70 [1] correctly set up initial P/total eqm P = 0.7 84 + x = 58.8 + 1.4x → x = 63 [1] find partial pressure of PCl3 or Cl2 Kp = 632/ 84 = 47.3 kPa or 47300 Pa [1] *must use 84 as equilibrium pressure of PCl5 is given in question. 2 (c) (iii) partial pressure of PCl5 / kPa 84 time / min [1] correct sketch at t1 (sharp increase) + explain: volume reduced hence pressure increased [1] correct sketch between t1 and t2 (increase at decreasing rate) + explain: POE shifts left to lower total pressure by forming fewer gaseous molecules 2 (d) X undergoes oxidation / iodoform reaction with warm alkaline aqueous iodine I3 → X contains CH3CH(OH)− (structure must be drawn out, there is no name for this structure) X undergoes (nucleophilic) substitution reaction with PC l5 → X contains an alcohol group t1 t2

2022 HCI C2 H2 Chemistry Prelims / Paper 3 Y is reduced by tin, conc HC l → Y contains nitrobenzene or Z contains phenylamine. This is followed by an intramolecular nucleophilic substitution reaction between chloroalkane and phenylamine to form Z. 2 marks for any 4 underlined points from above. 1 mark for 2-3 underlined points from above. X: alternative structure: Y: alternative structure: Z: alternative structure: Z must be in 1,2 position otherwise no intramolecular reaction. 3 (a) (i) Nitrogen is in Period two of the Periodic Table whereas phosphorus is in Period three. Phosphorus has energetically accessible 3d subshells to accommodate beyond eight electrons in its valence shell. However, for nitrogen, the next available subshell beyond 2p is 3s, which is not energetically accessible. [1] (credit is awarded for an idea of nitrogen being unable to “expand octet”.)

3 (a) (ii) : 6 : 1 [1] for both 3 (a) (iii) For P–H: head-on overlap between the sp3 orbital P and the 1s orbital of H to form an sp3–1s  bond. [1] [1] correct labels are required to score this credit. 3 (a) (iv) The structures of the conjugate base of H3PO4 and H3PO3 are shown below. H3PO4 has a higher p Ka than H3PO3, implying that the latt