2019 HCIS IB2 HL Chem Prelim P2 (ans + comments)

2019 IB2 HL Chem P2 (Answers) 1 2019 IB2 HL Chemistry Prelim Exam Paper 2 – Answers 1. 9.74 tonnes of magnesium is need to react with 1.60 x 102 tonnes of impure titanium (IV) oxide, TiO2, ore. All the titanium is completely extracted from the sample of impure ore. The following are reactions involved in the extraction process: Step 1: TiO2 + C + 2Cl2  TiCl4 + CO2 Step 2: TiCl4 + 2Mg  Ti + 2MgCl2 (1 tonne = 106 g) (a) Write an overall equation for extraction of titanium from TiO2. [1] TiO2 + C + 2Cl2 + 2Mg  Ti + CO2 + 2MgCl2 (b) Calculate the amount, in mols, of TiO2 present in the sample of impure ore. [1] nMg nTiO2 = 2 1 9.74 𝑥 106 24.31 nTiO2 = 2 1  nTiO2 = 1 2 x 9.74 𝑥 106 24.31 mol = 200329.08 mol = 2.00 x 105 mol [Note: e.c.f. (a) equation] Some students did not convert tonnes to g, hence, no marks awarded. Some students uses 1.60 x 102 tonnes for calculation. Note that not all 1.60 x 102 tonnes are TiO2. ( ½ for nMg) ( ½ nTiO2) (c) Calculate the mass, in tonnes, of TiO2 in the sample of impure ore. [2] Mass of 200329.08 mol TiO2 = 200329.08 x (47.87 + 2x16.00) (½) = 200329.08 x 79.87 g = 15997961 g = 1.60 x107 g (½) = 1.60 x101 tonnes (accept 16 tonnes) (1) [Note: e.c.f. (a) equation] If student did not convert to g, and get 16 tonnes, only award 1.5. Student must ensure units consistency in their calculation. Some students uses the wrong Ar for Ti. (d) Calculate the percentage by mass of TiO2 in the sample. [1] % by mass of TiO2 in the sample = 1.60 x 101 1.60 x 102 x 100 = 10.0% [Note: e.c.f. (a) equation] Ecf is allowed as long as student uses the answer (in tonnes) calculated in part (c) for calculation of % by mass here.

2019 IB2 HL Chem P2 (Answers) 2 (e) (i) Titanium (IV) oxide, TiO2, is a solid, however titanium (IV) chloride, TiCl4, is a liquid at room temperature. Suggest a reason for this difference between the two compounds, in terms of structure and bonding. [2] TiCl4 is a simple molecular structure, weak intermolecular forces of attraction between the molecules./ Owtte TiO2 is an giant ionic compound, with strong ionic bonds / strong electrostatic forces of attraction between oppositely charged ions/ owtte 5 points : 2 structure, 2 bonding, strength comparison. 2 points stated 0.5, 3 – 1, 4 – 1.5, 5 – full marks (ii) Titanium (IV) chloride is relatively stable in aqueous solutions. However, it dissolves in concentrated hydrochloric acid to form an acidic solution of the hexachlorotitanate (IV) complex ion. Draw the structure of the complex ion, suggest its molecular geometry and bond angle. [2] (show 6 dative covalent bonds  ½ , correct charge on the ion  ½) Geometry: Octahedral (½) Some students wrote square bipyramidal. Bond angle: 90⁰ (½) No other answer allowed (f) Titanium exits in three important oxidation states, +2, +3 and +4. However, zinc exhibits only +2 oxidation state. (i) Write the full electron configurations of Ti2+ and Zn2+ ions. [2] Ti2+ : 1s2 2s2 2p6 3s2 3p6 3d2 Zn2+: 1s2 2s2 2p6 3s2 3p6 3d10 Write atom configuration, fill in 4s before 3d. Followed by removing electrons from 4s, followed by 3d to obtain ion configuration. Ti

(ii) Suggest why titanium has variable oxidation states. [1] Ti has partially filled d–orbitals / incomplete filled d shells(½) this is not at its highest oxidation state/ able to lose multiple electrons form 3d and 4s / energy needed to remove the electrons form the partially filled d shells is almost equal(½), which allows multiple oxidation states. (accept answers OWTTE) (g) Magnesium is a s–block element. (i) Define what is meant by ‘s’. [1] Only s orbital in the highest main energy level contains electrons. [1] Majority of the student did not answer to the question to define s in the term s–block element, and hence give definition of what an s orbital is. (iii) Sketch on the axes below, the first six ionisation energies (IE), in logarithmic scale, of the element, magnesium. [1] Some students drew the first IE trend across the period, instead of successive IE trend. (iv) Explain one salient feature of the graph. [1] – Ionisation energy increases with subsequent electrons being removed.  (½) Reason: – successive electron to be removed experiences a greater nuclear attraction.  (½) OR – A sharp increase in ionisation energy for removal of 3rd electron while the increase in ionisation (½) energies is gradual for the other electrons. Reason: – The 3rd electron is in a lower energy main energy level, it experiences stronger electrostatic forces of attraction towards the nucleu s. Thus greater ionisation energy is required to remove it.  (½) [Note: have to state “lower energy main energy level” and not “inner energy level”] Note: ½ m: increase gradient or IE over 6 electrons ½ m: sharp increase from 2nd to 3rd electron log IE No. of electron removed 1 2 3 4 5 6

2019 IB2 HL Chem P2 (Answers) 4 2. Chlorofluorocarbons, CFCs, have been used previously for a wide range of applications including cleaning solvents and refrigerant. However, they break down easily in the stratosphere that lead to ozone (O3) depletion. (a) The following equations illustrate the process of ozone depletion. CCl2F2 ●CClF2 + Cl●  (1) Cl● + O3  ClO● + O2  (2) ClO● + O●  Cl● + O2  (3) (i) Identify the type of processes shown in the three equations by completing the table below: Process Equation(s) initiation 1 (1) propagation 2,3 (1) termination Concept: initiation: produce first radicals propagation: radicals produce more radicals termination: radicals combine and no more radicals are formed. Hence eqn (3) is not a termination step  common mistake. [2] (ii) Cl in equation (1) is the reactive species in this process. State the name of such species and explain its reactivity. [1] Cl is a (free) radical and it is reactive due to its unpaired electron  Comments: Qn says “explain” so give a reason: Saying it’s highly reactive is not explaining. Also common mistake: “Cl is highly electronegative”: What is the relevance/link? (b) The hydroflurocarbon, F 3CCFH2, is a suitable replacement for CFCs . It is prepared industrially by reacting hydrogen fluoride with Cl2C=CClH. HCl is the by–product. (i) State the IUPAC name of F3CCFH2. [1] 1,1,1,2–tetrafluoroethane Comments: draw out structure if cannot visualise F3CCFH2. Poorly attempted question: only less than 10 students got it right. (ii) Construct a balanced overall equation for this reaction. [1] Cl2C=CClH + 4HF  F3CCFH2 + 3HCl (iii) Explain how HF can act as a nucleophile in this reaction. [1]

The F in the molecule has a lone pair of electrons  that it can donate (to the δ+ charged carbon atom) and form a (covalent) bond. Comments: Qn says “explain how” and italicised the word ‘nucleophile’. So take the cue and explain how a nucleophile reacts. (iv) Draw