RI 2018 Prelim P2 Ans

© Raffles Institution 2018 9729/02/S/18 1 2018 Y6 H2 Chemistry Preliminary Exams Paper 2 (Suggested Solutions) 1(a) X is likely to be in Group 14 as there is a large jump from the 4th to 5th ionisation energies. This shows that the 5th electron is removed from an inner electron shell and requires more energy for removal as it is more strongly attracted to the nucleus. Hence, there are 4 valence electrons in X. Since X has at least 10 electrons but fewer than 20 electrons, X is Si. Comments • Students need to be familiar with the Group number in the Periodic Table given in the Data Booklet. Silicon is a Group 14 (NOT Group 4 or Group IV) element. • Since X has at least 10 electrons, carbon is not a possible answer for X. • Answers need to be specific when referring to the 5th electron being removed from the inner electron shell. Avoid using vague phrases like ‘different electron shell’ or ‘another electron shell’. 1(b) Mn and Fe are first-row transition metals. Although Fe has a higher nuclear charge (greater proton number) than Mn, Fe has more 3d electrons which provide more shielding between the nucleus and the outer 4s shell of electrons. This increase in shielding effect offsets the increase in nuclear charge and hence, the increase in effective nuclear charge from Mn to Fe is minimal, resulting in the first ionisation energies of Mn and Fe to be similar. Comments • Most students were able to identify that Fe has both higher nuclear charge and shielding effect than Mn. However, students need to be clear er in their explanation on why the inc rease in effective nuclear charge from Mn to Fe is minimal. 1(c) Ca2+ has a larger cationic radius than Cu2+, thus Ca2+ has a lower charge density and weaker polarising power. Consequently, there is lower extent of distortion of the electron cloud of the carbonate anion and hence covalent bonds within the carbonate anion are weakened to a lesser extent . Hence, CaCO3 decomposes at a higher temperature. Comments • Most students were able to link the larger cationic radius of Ca 2+ to its lower charge density and weaker polarising power. • Students are required to make the link between the stronger covalent bonds within the carbonate anion of CaCO3 to its higher decomposition temperature. 2(a)(i) Anthracene is a cyclic molecule. All carbon atoms of anthracene are sp2 hybridised and are trigonal planar. The unhybridised p -orbitals of the carbon atoms in anthracene can form a continuous p-orbital overlap, thus the  electrons are delocalised throughout the molecule. Anthracene has 4(3) + 2 = 14  electrons where or n = 3.

© Raffles Institution 2018 9729/02/S/18 2 Comments • Most students correctly identified the hybridisation of the carbon atoms as sp 2 and also stated that the unhybridised p-orbital of the carbon atoms form a continuous overlap allowing the elect rons in the p-orbitals to be delocalised forming the -elect