2022 EJC Prelims Paper 2 (solution with comments)

© EJC [Turn Over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2022 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME CIVICS GROUP 2 1 – INDEX NUMBER CHEMISTRY Paper 2 Structured Questions 9729/02 15 September 2022 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, civics group, index number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue, or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Paper 2 1 / 20 2 / 20 3 / 18 4 / 17 Total / 75 This document consists of 19 printed pages and 1 blank page.

2 © EJC 9729/02/J2PE/22 For Examiner’s Use 1 The carbon family consists of the elements of Group 14. The elements at the top of the group, carbon to germanium, have very different properties from those at the bottom, tin and lead. For instance, Group 14 elements tend to adopt oxidation states of +4 , whereas the heavier elements, such as tin and lead, exhibit the +2 oxidation state due to the inert pair effect. (a) State the valence shell configuration of Group 14 elements. ................................ ................................ ................................ ................................ [1] Comments: This is generally badly attempted. Many erroneous responses include electron in box diagram, +4, p4, s2p2. (b) One of the contributing factors to the inert pair effect is the unexpected increase in the ionisation energies, after lead, down the group. Explain why the 1st ionisation energies are expected to decrease down the group. ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ [2] Comments: Generally well done. Candidates were penalised for mistaking that nuclear charge remains unchanged due to same number of valence electrons instead of number of protons. Down the group, effective nuclear charge actually increases slightly down the group or remains approximately unchanged and does not really aid in the explanation for the decreasing 1st ionisation energies down the group. Hence, the emphasis is on the decreasing electrostatic forces of attraction between the nucleus and the valence electron being further away due to increasing number of shells of electrons. (c) Carbon forms carbide anion, C 2– 2 in calcium carbide while silicon mostly forms Si4+ ions. (i) Draw the dot-and-cross diagram of the carbide anion, C 2– 2 . [1] Comments: Generally well done. Some responses were penalised for giving solely dots only or crosses only. Candidates were reminded the purpose of the use of dots and crosses is to distinguish the valence electrons from different atoms regardless if they are of the same eleme nt. Also, the use of the dots and crosses also differentiates the valence electrons of the carbon atoms from the additional electrons from the −2 charge. ns2 np2 Down the group, nuclear charge increases and shielding effect increases. Valence electrons are further away from the nucleus and this leads to d ecrease in electrostatic forces of attraction between the nucleus and the valence electron.. Less energy needed to remove the valence electron, resulting in decrease in 1 st I.E. down the group.

3 © EJC 9729/02/J2PE/22 [Turn Over For Examiner’s Use (ii) In a particular experimental set-up, a beam of 28Si4+ ions was deflected by an angle of +4.2º. Assuming an identical set of conditions, by what angle will the 12C 2– 2 ions be deflected? [1] Comments: Generally badly done. Many erroneous responses failed to recognise that there are 2, not just 1, carbon atoms in the 12C2– 2 ions. Omission of the negative sign is common. Candidates are reminded of the importance of the sign as it indicates the direction of the deflected beam of ions. (d) Table 1.1 shows that the melting points of the elements of Group 14. Table 1.1 element C Si Ge Sn Pb melting point / C >3550 1410 937 232 327 Carbon, silicon and germanium each form a solid with the same type of structure. Using bonding and structure, suggest why the melting points of these elements decrease from carbon to germanium. ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ [2] Comments: Generally well done. Common mistakes include omission of the bonding and structure, failure to recognise that germanium is of the same bonding and structure as carbon, silicon as stated in the question stem or mistaking carbon, silicon and germanium to be of giant metallic lattice structure. Another common mistake is that of breaking of ionic bonds or intermolecular instantaneous dipole-induced dipole interactions in the giant covalent structure. Several candidates also tried to replicate their response to Question 1(b) here failed to recognise that this question is on chemical bonding rather than atomic structure in Question 1(b). 28Si4+ whose 41 gives an angle of deflection, , of +4.2 º28 7 q m += = +  14.2º 29.4 7 qk k km     =  + = +  =       For 12C 2– 2 whose 21 ,24 12 q m −= = − 1angle of deflection, 29.4 12  =  − = 2.45º− Carbon, silicon and germanium are of giant covalent structure . To melt these elements, strong covalent bonds between the atoms have to be broken. From C to Ge, the orbitals become bigger and more diffuse . The overlap of the orbitals becomes less effective. Bond strength decreases and hence less energy is required for melting.

4 © EJC 9729/02/J2PE/22 For Examiner’s Use (e) Carbon forms many allotropes such as graphite and diamond . Recent scientific research has found that replacing the graphite electrodes with graphene in lithium-ion batteries can extend battery life. (i) Graphene is a single, one atom thick layer of graphite. Describe the hybridisation of the orbitals in, and the bonds between, the carbon atoms within graphene. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ........