2023 RI H2 Chem Prelims P2 Answers

1 2023 Y6 H2 Chemistry Preliminary Examinations Paper 2 Suggested Solutions 1 (a) The volatilities of the halogens decrease from chlorine to iodine. From C l2 to I2, the electron clouds of the halogens become larger and more polarisable. Hence, more energy is required to overcome the increasing strength of the instantaneous dipole –induced dipole interactions between the halogen molecules down the group, leading to decreasing volatility. (b) (i) X Y Z nucleon number 140 103 37 charge +1 +1 0 (ii) (iii) nucleon number = 103 = 31 + 35 + 37 + 37 There is 1 35Cl atom in Y. Y is PCl2+. Its shape is bent. (c) (i) P(g) + e– ⎯→ P–(g) (ii) 1s 2s 2p 3s 3p (iii)  Nuclear charge increases from A l to C l but shielding effect remains Y X

2 effectively constant. Effective nuclear charge increases. Electrostatic attraction between the nucleus and the incoming electron increases, resulting in an increase in the energy released.  The attraction between the nucleus and incoming electron is weakened due to inter-electronic repulsion between this electron and the one already present in the 3p orbital. Hence, less energy is released as the first electron affinity. 2 (a) Au has a smaller atomic size and thus have more atoms per unit volume . It also has a larger atomic mass than Ba. Hence it has a greater mass per unit volume (i.e. higher density) as compared to Ba. (b) max % of Au in amalgam = 100% − 40% = 60% max mass of Au in 250 g of amalgam = 0.6  250 = 150 g (c) conc of Hg vapour = 1 g of Hg / 1 m3 of air = 1 g of Hg / 1.19 kg of air = 1 g of Hg / 1190 g of air = ( 1 1190 ) g of Hg / 1 g of air = ( 1 1190  106) g of Hg / 106 g of air = 840.3 g of Hg / 106 g of air  840 ppm (d) (i) Water prevents / reduces the vaporisation / evaporation of the collected mercury. (ii) When heated, the mercury in the amalgam vaporises and then condenses on the cooler inner surface of the glass bowl . Then it slides back down to accumulate on the sand to be collected and recovered. (iii) advantage • low cost/convenient to set up because of the use of simple household items or no need special equipment (e.g. metal apparatus) • no mercury-contaminated wastewater produced disadvantage • the mercury collected will mix with the sand and hence be harder to recover • the setup is not as airtight as Method I, i.e. mercury vapour has higher chance to escape back into the atmosphere • when the bowl is removed, mercury can vaporise again whilst collecting the mercury from the sand • mercury poisoning from re -use of cooking pot used in the extraction

3 for cooking purposes (e) (i) O2 + 2H2O + 4e– ⎯→ 4OH– (ii) mol ratio 4Au : 1O2 : 4e– Each Au atom loses 1 e– to form Au+. The oxidation state of Au is +1. (iii) CN– is the ligand in D. The complex anion which is linear would be formed as such : –NC→ Au+ CN–. D is [Au(CN)2]–. (f) (i) Metallic bond is the electrostatic attraction between a lattice of positive ions and delocalised electrons. (ii) In mercury, poor shielding results in the outermost electrons being held so tightly by the nucleus that it is very difficult (or requires a lot of energy) for the outermost electrons to be delocalised. Hence, only weak metallic bonds are formed , resulting in mercury being a liquid and not a solid at room temperature. 3 (a) (i) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0 10 20 30 40 50 [sucrose] / mol dm–3 time / min t1 2 = 12 min t1 2 = 12 min

4 Since half -life is constant at 12 min, order of reaction with respect to sucrose = 1, hence x = 1. (ii) initial rate of reaction = | 0.60 – 0.00 16 – 0 | = 0.0375 mol dm–3 min –1 (iii) Ratio of the rate of reaction of experiment 1 and 2 = (0.0375 / 0.0125) = 3 Comparing expt 2 to 1, when [H+] x 3, initial rate x 3 ⇒ rate ∝ [H+], hence order of reaction with respect to H+ = 1, y = 1. (iv) from (a)(i) and (a)(iii); rate = k[sucrose][H+] from (a)(iii); initial rate = 0.0125 mol dm–3 min–1 (0.0125) = k(0.6)(2.0) k = 1.04 × 10−2 mol–1 dm3 min–1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0 10 20 30 40 50 [sucrose] / mol dm–3 time / min

5 (b) A catalyst provides an alternative pathway of lower activation energy (Ea, catalysed) than the uncatalysed reaction. More molecules possess energy greater than or equal to Ea, catalysed, hence the frequency of effective collision increases, and the rate of reaction increases. A lower Ea, catalysed also results in a larger rate constant. (c) (i) a: –1, b: 3 (ii) (iii) The p-orbital of C l in C overlaps with the π-electron cloud of the C=C bond, causing the lone pair of electrons on Cl atom to delocalise into the C=C bond. This causes the C–Cl bond in C to have a partial double bond character and is strengthened. Hence, more energy is required to break this bond. This results in a higher activation energy for the reaction and therefore a smaller value of the rate constant, k. (iv) The carbocation from the S N1 reaction of compound B is stabilised as the empty p-orbital of the positively charged carbon overlaps with the π- electron cloud of the C=C bond. The π electrons delocalise s into the empty p-orbital dispersing the positive charge on the carbocation, hence stabilising it. For SN2, the presence of bulky Cl atoms bonded to the electron deficient carbon causes the approach of the nucleophile to be sterically hindered. Ea, catalysed kinetic energy no. of particles with a given energy, E total no. of particles with energy ≥ Ea (catalysed reaction) total no. of particles with energy ≥ Ea (uncatalysed reaction) Ea, uncatalysed

6 4 (a) (b) (c) (i) CH3CH2CH2NH2 + H2O ⇌ CH3CH2CH2NH3+ + OH– (ii) pOH = –lg√4.7×10–4×1.3 = 1.606 pH = 14 – 1.606 = 12.4 (iii) nHCl = 0.025 x 1.3 x ½ = 0.01625 mol VHCl = 0.01625 / 0.80 = 0.02031 dm3 = 20.3 cm3 pH = –lg 10–14 4.7×10–4 = 10.67 = 10.7 (iv) When H+ is gradually added to the buffer solution, it reacts with CH3CH2CH2NH2. Therefore, [CH 3CH2CH2NH2] decreases, concentration of CH3CH2CH2NH3+, increases and the position of equilibrium shifts slightly towards the left. Since the ratio of the amine to its conjugate acid remains relatively constant, pH remains unchanged.

7 5 (a) Oxybenzone absorbs radiation over a larger range of wavelengths because the  electrons in oxybenzone are delocalised over a larger number of atoms resulting in greater extent of delocalisation. Hence oxybenzone is a better choice for sunscreen because it absorbs both UVA and UVB radiation whereas compound U absorbs only UVB radiation. (b) (i) (ii) Test: Add 2,4-DNPH to each sample placed in separate test tubes. Observation: For 3-methoxyphenol, no orange ppt formed. For oxybenzone, orange ppt formed. (c) (i) (ii) At high temperatures, the intramolecular H-bonds in oxybenzone can be overcome. Hence oxybenzone would be able to form more extensive H- bonds with H2O, increasing its solubility in water. (d) (e) (i) Zn(s) + H2O(g) ⇌ ZnO(s) + H2 (g) initial amount / mol 2.60 2.60 0 0 change in amount / mol −2.58 −2.58 +2.58 +2.58 equilibrium amount / mol 0.02 0.02 2.58 2.58 partial pressure of H2 at equilibrium = ( 2.58 0.02 + 2.58) x 10 = 9.92 atm β α

8 (e) (ii) Kp = pH2 pH2O = (2.58 2.60) x 10 (0.02 2.60) x 10 = 129 (iii) At lower temperatures, p