2025 RI Prelims H2 Chemistry P2 suggested solutions

© Raffles Institution 2025 9729/02/S/25 2025 Y6 H2 Chemistry Preliminary Exams Paper 2 – Suggested Solutions 1(a) B6+(g)  B7+(g) + e– Comments:  Generally well done. 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. Comments:  Most students were able to correctly identify C as the noble gas.  A common error was the incorrect use of the term “inner subshell”. A subshell refers to s, p, d, f. The large jump in ionisation energies is due to the change in principal quantum number, n, of the electron shell from which the electron is removed from (from 3 to 2) and not because of the subshell. 1(c)(i) Sulfur Comments:  Generally well done. 1(c)(ii) Comments:  This question was well answered 1(d) D +: 1s 2 2s2 2p6 3s2 3p6 C: 1s 2 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.

© Raffles Institution 2025 9729/02/S/25 Comments:  Most students understood the concept in this question but had missing key terms, used incorrect terms or demonstrated misconception in their answers.  Many students had the misconception that only the inner electrons contribute to the shielding effect. This is not true, because the outer electrons also contribute to shielding effect (though comparably less). Therefore, there is a need to mention that both D+ and C contain the same number of electrons and hence the outermost electron for both D+ and C experience the same shielding effect.  The 2nd I.E. of D refers to removal of the outermost electron from D+ and the 1st I.E. of C refers to the removal of the outermost electron from C. o Some students were not aware that the outermost electron were being removed from D+ and C. o Writing the electronic configuration of D is irrelevant to this question since the 2nd I.E of D refers to the removal of outermost electron from D+ (instead of D). 1(e)(i) The electronegativity of an atom in a molecule is a relative measure of its ability to attract bonding electrons. Comments:  A significant number of students did not specify “bonding” electrons in their answers. 1(e)(ii) 2.5 Comments:  Generally well done. 1(f)(i) The B–Ha bond is longer. Since two electrons are shared between three atoms in B−Ha−B, there is an average of one shared electron per B−Ha bond, which is less than the two shared electrons per B–Hb bond. This decrease in electron density between the B and Ha atoms causes the B−Ha bond to be weaker and therefore longer. Comments:  Students are required to clearly show the comparisons between B−Ha and B−Hb bond. This includes the number of shared electrons, bond strength and hence the bond length.  Students are also required to demonstrate understanding of the term “three- centre two-electron bond” for the B−Ha−B as having an average of one shared electron per B−Ha bond.

© Raffles Institution 2025 9729/02/S/25 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. Comments:  The explanation required in question was challenging for most students.  The concept tested here is the VSEPR theory, but students are required to adapt and apply it to the context of this question. It is analogous to the idea of “repulsion of lone pair-lone pair > lone pair-bond pair > bond pair-bond pair”. However, there are no lone pair of electrons about the B atom, but only bond pairs of B–Ha and B–Hb which differ in their electron density. Hence, students are expected to recognise that the B–Hb bond pair, which has greater electron density, exerts greater repulsion compared to the B–Ha bond pair. As a result, repulsion between B–Hb and B–Hb > B–Ha and B–Hb > B–Ha and B–Ha. 2(a)(i) or Comments:  Generally well done.  Some candidates incorrectly gave as the answer – note that only 10 carbons are present. 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. Comments:  Many students described the  bond involvement when it was explicitly mentioned in the question not to do so.  There is one electron per carbon atom involved in the delocalisation – not one pair of electrons.  Most students were not able to accurately bring out the equal / even distribution of electron density between carbon atoms.  Citing “intermediate” bond length between that of C−C and C=C bond just means the bond length is in between these two. It does not necessarily mean that any two given such bonds are equal in length e.g. 0.4 and 0.6 are both intermediate between 0.3 and 0.8 but they are not equal.

© Raffles Institution 2025 9729/02/S/25  “Partial double bond character” does not mean that any two bonds having such a characteristic are equal in bond length. 2(a)(iii) sp 3 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. Comments:  Hybridisation cannot be determined solely based on the number of bonds since the presence of lone pairs or lone electrons must also be accounted for i.e. the total number of regions of electron densities involved should be clear.  Shape alone is insufficient as well since the same shape can be the outcome for different hybridisations.  Students are reminded to use proper convention in science, sp3 is not an acceptable representation of hybridisation. 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. Comments:  Students are reminded to always answer the question as some students did not clearly state the difference in difference in conductivity.  F is a non-conductor and should not be described as though there is some level of conductivity.  For graphite, electricity is conducted along the layers, not between the layers. 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.