2021 CJC H2 CHEM Prelim P3 Mark Scheme (with comments)

9729/03 CJC JC2 Preliminary Examination 2021 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/03 Paper 3 Free Response September 2021 2 hours Candidates answer on the question paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class 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. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 24 printed pages. Catholic Junior College JC2 Preliminary Examination Higher 2 MARK SCHEME

2 9729 CJC Preliminary Examination 2021 Section A Answer all the questions in this section. 1 Fluorine is the most electronegative element and forms many interesting compounds. Antimony, Sb, is in Group 15 of the periodic table and forms two covalent fluorides that exist as simple molecules in gas phase, SbF3 and SbF5. Krypton, Kr, is in Group 18 of the periodic table and its first compound discovered is KrF2. (a) Draw 'dot-and-cross' diagrams showing the electrons (outer shells only) in SbF3, SbF5, and KrF2. Use the VSEPR (valence shell electron pair repulsion) theory to predict their shapes and hence state whether the species is polar or non-polar. [8] ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... 3 bond pairs of electrons, 1 lone pair of electrons around Sb, trigonal pyramidal shape (CAO) Polar 5 bond pairs of electrons, no lone pair of electrons around Sb, trigonal bipyramidal shape. (CAO) Non-polar. 2 bond pairs of electrons, 3 lone pair of electrons around Kr, linear shape. (CAO) Non-polar. [1] x3 For each molecule, correct dot & cross (–1 if incorrect valence electrons shown for F, –1 if ‘orbit’ used) [1] x3 correct shape (CAO, –1 if number/type of electron pairs not stated) [2] All 3 correct polar/non-polar stated (CAO, 1m if 2 correct) Diagrams were generally drawn well, though can do better with a relatively larger central atom, so that the electrons are more clearly differentiated. Critically, several missed out lone pairs on F or Kr and mistakenly showed dative or double bonds in KrF2. Majority of the candidates could not score well for predicting shapes, answer as if it was to “state” or “suggest”, hence missed the number and type of electron pairs. On the other hand, some answers were as if it was to “explain” (see 2019/P3/Q1) and stated the full VSEPR principles. Incorrect answers include trigonal planar, tetrahedral and even shapes for 3 bond pairs for KrF2 when clearly 2 bond pairs were drawn. Candidates had should not

3 9729 CJC Preliminary Examination 2021 [Turn over (b) SbF5 can react as shown in the following reaction and is a useful reagent as an exceptionally strong Lewis Acid. SbF5 + F− → [SbF6]− Explain the term Lewis Acid, and suggest the reason why SbF5 is a strong Lewis Acid. [2] ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... (c) SbF3 exists as gaseous molecules at a temperature of 700 K. (i) Calculate the volume of 0.10 mol of an ideal gas at a temperature of 700 K, at a pressure of 1.01 x 105 Pa. [1] (ii) The volume of 0.10 mol of SbF 3 measured at 700 K and 1.01 x 10 5 Pa in a gas syringe was significantly different from your answer in (i). Suggest two possible reasons why this might be the case in terms of the properties of SbF3. [2] ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ................................ ..... ................................ ................................ ................................ ...........................