Answers to Mock Paper 3

Qn. Suggested Solutions Remarks 1 (a) (i) The VSEPR theory states that: • Electron pairs arrange themselves in a way to minimise electron pair-electron pair repulsion [1] • Repulsion between electron pairs is arranged from the strongest to weakest as follows: lone pair-lone pair, bond pair-lone pair, bond pair-bond pair [1] (ii) Nb and Nd: linear Nc: bent [1] for all correct; no marks otherwise Candidates are required to deduce that 168° is close to linear geometry. (iii) [2] for either of the structures: Structure A Structure B (accept its mirror image) Marking points: [1] for ensuring that Na—Nb and Nd—Ne bond has a bond order that is higher than (or at least equal to) the Nb—Nc and Nc—Nd bond. The former bond length is shorter than the latter. For structure A: the bond order for Na—Nb and Nd—Ne is 3 (triple bond), while that of Nb—Nc and Nc—Nd is 1. For structure B: the bond order for Na—Nb and Nb—Nc is 2, Nc—Nd is 1, and Nd—Ne is 3.

[1] for ensuring that the geometry at Nb and Nd is linear (i.e. no lone pairs on Nb and Nd), and Nc is bent (1 or 2 lone pairs) No other resonance structures will be accepted. (iv) If structure A is drawn, Nc has an oxidation state of +1, while Na, Nb, Nd and Ne has an oxidation state of 0. [1] If structure B is drawn, Nb has an oxidation state of +1, while Na, Nc, Nd and Ne has an oxidation state of 0. [1] • If the candidate states that Nd has an oxidation state of +1, this is acceptable as long as the oxidation state of Nb is stated to be 0. (b) (i) [N5+] [N3-] → 4 N2 [1] Accept equations including spectator ions Cs+ and SbF6-, as long as the equation is balanced AND nitrogen gas is one of the products. (ii) The oppositely charged poly-nitrogen species attract each other once its salt dissolved in liquid SO2, which is a kinetically feasible reaction making this reaction fast and uncontrollable. [1] (iii) Identifies CsSbF6 AND amount of CsSbF6 = 0.002438 mol [1], accept 3 s.f. value An excess was produced. [1] (This is because some liquid SO2 was present too.) (iv) [Cs+] = [N3-] = 3.14 mol dm-3 [1] [1] for final answer with working Volume of SO2 = 2.267×10-33.14 = 7.22 × 10-4 dm3 (v) Lowers the solubility because Cs+ present will lower amount of CsSbF6 needed to be dissolved to achieve [Cs+][SbF6-] = Ksp. [1] Accept answers based on discussing the equilibrium: CsSbF6 ⇌ Cs+ + SbF6- Presence of Cs+ shifts the position of equilibrium to the left so less CsSbF6 dissolves. (c) (i) [1] for calculating volume of the covering (with correct units indicated) V = π$7.00×10-32%2× 45.0 × 10-3 = 1.7318 × 10-6 m3 The candidate is expected to know the

[1] calculates amount of N2 produced nN2=0.600207.2 + 14.0 × 6 × 62 = 0.0061813 mol [1] Establishes relationship pV = nRT and solves for p. The candidate may assume the temperature to be a reasonable one. For example, T = 293 K (room temperature), or T = 298 K (standard conditions), or T = 273 K (sta