EJC 2020 JC2 Prelim H2 Chemistry Paper 2 Suggested Solutions

© EJC [Turn Over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2020 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME CIVICS GROUP 1 9 – INDEX NUMBER CHEMISTRY Paper 2 Structured Questions 9729/02 02 September 2020 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 / 12 2 / 12 3 / 18 4 / 16 5 / 17 Total / 75 This document consists of 21 printed pages and 1 blank page.

2 © EJC 9729/02/J2PE/20 For Examiner’s Use 1 (a) Nitrogen can form a few hydrides such as NH3 and N2H4. The boiling points of these two hydrides are given in Table 1.1. Table 1.1 compound boiling point / ºC NH3 −33 N2H4 114 Provide two explanations for why N2H4 has a higher boiling point than NH3. ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ [2] N2H4 can form an average of two hydrogen bonds per molecule while NH3 can only form one. More energy is required to break the more extensive hydrogen bonds between N 2H4, hence it has a higher boiling point. N 2H4 also has more electrons, which gives rise to a larger and more polarisable electron cloud, and hence stronger intermolecular in stantaneous dipole -induced dipole interactions than in NH 3. Hence, more energy is required to overcome the intermolecular forces of attraction between N2H4 than C2H4

3 © EJC 9729/02/J2PE/20 [Turn over For Examiner’s Use (b) When NH3 is reacted with BF3 in the gaseous phase, a white solid of NH3BF3 is formed. (i) Draw a dot-and-cross diagram to show the bonding within NH3BF3 and state the shape about the nitrogen atom. shape ................................ ........................ [2] (ii) State the type of reaction that NH 3 has undergone with BF3, and explain why it is possible. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ......................... [2] (c) Both NH3 and PH3 have trigonal pyramidal shape, but the bond angles in NH3 are 107º while those in PH3 are 94º. By considering the electronegativities of the central atoms, suggest why this is so. ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ .... ................................ ................................ ................................ ................................ [2] tetrahedral Lewis acid-base reaction . N atom on NH 3 has a lone pair which can be donated to the electron-deficient B atom on BF 3 which has an empty 2p orbital to achieve a stable noble gas configuration. Nitrogen is more electronegative than phosphorus, and draws the electron density of its bond pairs closer to itself than phosphorus does. The bond pairs in NH 3 are closer to the central atom, leading to greater bond pair-bond pair repulsion. Hence, NH3 has a larger bond angle than PH3.

4 © EJC 9729/02/J2PE/20 For Examiner’s Use (d) Table 1.2 shows the successive ionisation energies of an element A. Table 1.2 ionisation energies / kJ mol−1 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th element A 945 1794 2735 4839 6056 11690 14180 17370 20550 23830 (i) Explain why the successive ionisation energies of an element always increase. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ......................... [1] (ii) Is element A likely to be from the same group as nitrogen? Explain your answer clearly. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ......................... [2] (iii) State and explain how the first ionisation energy of element A will compare to that of the element to its right in the periodic table. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................ ............................. ................................ ................................ ................................