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 el