2025 RI H2Chem Prelims P2 Answers

© Raffles Institution 2025 9729/02/S/25 2025 Y6 H2 Chemistry Preliminary Exams Paper 2 – Suggested Solutions 1(a) B6+(g) ⎯→ B7+(g) + e– 1(b) C There is a large jump between the 8th and 9th ionisation energies. This indicates that significantly more energy is needed to remove the 9th electron. Thus this 9th electron is located in an inner electron shell that is nearer to the nucleus, experiencing less shielding and is attracted more strongly by the nucleus. Therefore, there are 8 valence electrons. Hence C belongs to Group 18 of the Periodic Table and is a noble gas. 1(c)(i) Sulfur 1(c)(ii) 1(d) D+: 1s2 2s2 2p6 3s2 3p6 C: 1s2 2s2 2p6 3s2 3p6 D+ and C are isoelectronic species and hence their outermost electrons experience the same shielding effect. However, D+ has a larger nuclear charge than C and so there is stronger electrostatic attraction between the nucleus and the outermost electrons in D+. Hence the 2nd ionisation energy of D is more positive than the 1st ionisation energy of C. 1(e)(i) The electronegativity of an atom in a molecule is a relative measure of its ability to attract bonding electrons. 1(e)(ii) 2.5 1(f)(i) The B–Ha bond is longer. Since two electrons are shared between three atoms in B−H a−B, there is an average of one shared electron per B−H a bond, which is less than the two shared electrons per B–Hb bond. This decrease in electron density between the B and H a atoms causes the B−Ha bond to be weaker and therefore longer. 1(f)(ii) 122 Since the electron density of the B–Hb bond is greater than that of the B–Ha bond, the repulsion between two B–Hb bonds is greater than that of the repulsion between B–Ha and B–Hb, which is greater than that between two B–Ha bonds. Hence Hb–B– Hb bond angle is greater than 109.5.

© Raffles Institution 2025 9729/02/S/25 2(a)(i) or 2(a)(ii) Each carbon atom has an unhybridised p orbital which overlaps side-on with the other p orbitals on adjacent carbon atoms. This continuous side-on overlap of the p-orbitals allows the  electrons to be delocalised and shared equally across all carbon atoms, resulting in all bond lengths to be equal. 2(a)(iii) sp3 Each carbon atom has four bond pairs / regions of electron densities and no lone pairs and exhibits a tetrahedral molecular geometry (shape). Each bond pair must thus be located in an sp3 hybrid orbital. 2(a)(iv) Allotrope F is a non -conductor of electricity while graphite is a conductor of electricity. All four valence electrons on each C atom in allotrope F are used to form σ bonds, and no electrons are available for delocalisation / no mobile charge carrier to conduct electricity. In graphite, three of the valence electrons on each C atom are used to form σ bonds. The last electron on each C atom in graphite delocalises across the whole layer, acting as a mobile charge carrier. Hence, graphite is a conductor of electricity. 2(a)(v) The researcher is incorrect. The two structures are different because Fig. 2.1 Fig. 2.2 These two carbons point in the same direction. These two carbons point in different / opposite directions.

© Raffles Institution 2025 9729/02/S/25 OR Fig. 2.1 Fig. 2.2 There are “rectangular” faces. There are 4-carbon rings on the sides. There are hexagonal faces. There are 6-carbon rings on the sides. OR Fig. 2.1 Fig. 2.2 Chair structures are stacked directly on top of each other. Chair structures are staggered between the layers. 2(b)(i) 2(b)(ii) see-saw / distorted tetrahedral 2(c)(i) ethanolic NaOH and heat 2(c)(ii) 2(c)(iii) SN1 does not occur because N is a 1 bromoalkane and the carbocation formed is not stable enough. The bulky group hinders the backside attack by the OH– nucleophile on the electron-deficient carbon and hence SN2 does not occur.

© Raffles Institution 2025 9729/02/S/25 3(a)(i) Electrophilic Substitution 3(a)(ii) The delocalisation of the six  electrons in the ring structure of benzene causes resonance stabilisation. Addition reactions disrupt the delocalisation of the six  electrons in benzene while substitution reactions restores the resonance stabilisation after temporarily disrupting it. 3(a)(iii) AlCl3 acts as a Lewis acid catalyst in the reaction. 3(a)(iv) With reference to the mechanism, the electrophile is planar about the positively charged carbon in the carbocation. Hence, the  electrons from the benzene ring will attack the positively charged carbon on either side of the plane with equal probability. Hence, a racemic mixture is formed, and the product is unable to rotate plane-polarised light. 3(b)(i) U: 3(b)(ii) V: 3(b)(iii) Free radical substitution reaction may result in produce multiple alternative monosubstituted products / major product is another monosubstituted product / multiple substitutions 3(c)

© Raffles Institution 2025 9729/02/S/25 OR 3(d) Test: To each compound in separate test-tubes, add H2SO4(aq), a few drops of KMnO4 and heat in hot water bath Observation(s): For X, there is decolourisation of purple KMnO4. For Y, there is decolourisation of purple KMnO4 and evolution of a colourless gas (CO2) which gives a white precipitate with limewater. 3(e)(i) 3-bromopent-1-ene 3(e)(ii) 4(a)(i) In 1 m3, mass of air is 1 kg. mass of O3 = 2 x 10–5 100 x 1 = 2 x 10–7 kg = 2 x 10–7 x 103 x 106 = 200 g Hence, concentration of O3 is 200 g m–3 4(a)(ii) PSI of O3 = 200 – 100 235 – 157 (200 – 157) + 100 = 155 (3 s.f.) 4(b)(i) Vehicular emissions are a significant source of NO2. When the lockdown measures were implemented, there were fewer vehicles on the road, and concentration of NO₂ decreased. 4(b)(ii) NO2 is a radical and can react with O3. During the Circuit Breaker, the concentration of NO₂ decreased, leading to less reaction with O3. Therefore, the concentration of O3 increased.

© Raffles Institution 2025 9729/02/S/25 4(c)(i) [O2] / mol dm−3 time / s Draw a tangent at t = 0 s Initial rate = −gradient = −[(0.005 – 0) / (0 − 102)] = 4.90 x 10–5 mol dm-3 s–1 = z 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0 20 40 60 80 100 120 140 160 180 200 (0 , 0.005) (102, 0) 0.005)

© Raffles Institution 2025 9729/02/S/25 4(c)(ii) Comparing experiments 1 and 2, when [O2] x 3, initial rate x 3. [O2] α rate. Hence, reaction is first order with respect to O2. 4(c)(iii) initial rate 0 initial [NO] 4(c)(iv) rate = k[NO]2[O2] 4(c)(v) The half-life of a reaction is the time taken for the concentration of a reactant to decrease to half its initial value. 4(c)(vi) rate = k[NO]2[O2] Using large excess of NO, reaction becomes pseudo first order. rate = k’[O2] where k’ = k[NO]2 t½ = ln 2 k' = ln 2 k[NO]2 Since [NO] in experiment 3 is doubled that of experiment 1, t½ of experiment 3 = 74 22 = 18.5 s 4(c)(vii) N2O2 + O2 ⎯→ 2NO2 4(d)(i) Homogeneous catalysis 4(d)(ii) NO2 + SO2 ⎯→ SO3 + NO (I) NO + ½ O2 ⎯→ NO2 (II) 4(e) n(S2O32−) reacted = 15.60 1000 x 4.00 x 10–4 = 6.24 x 10–6 mol n(I2) formed = 6.24 x 10–6 / 2 = 3.12 x 10–6 mol n(SO2) in 1 m3 sample of air = 3.12 x 10–6 × 5 = 1.56 x 10–5 mol mass in 1 m3 sample of air = 1.56 x 10–5 × 64.1 = 0.00099996 g mass concentration of SO2 = 0.00100 g m–3 (3 s.f.)

© Raffles Institution 2025 9729/02/S/25 5(a) Zn2+ + 4OH– ⎯→ [Zn(OH)4]2– 5(b)(i) At high [OH–], [Zn(OH)4]2– is produced, causing [Zn2+] t