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 (standard temperature). The candidate may choose to assume a high temperature since energy is applied for the decomposition. Taking T = 293 K, p = nRTV = 0.0061813 × 8.31 × 2931.7318 × 10-6 = 8.69 × 106 Pa (Note: This pressure is enough to blow off your hand if you held the detonator in your hand.) Only ecf for the 3rd marking point is awarded. volume of a cylinder. (ii) High charge density of Fe3+ centre polarises/weakens the O—H bond on the H2O ligand. [1] (iii) Different ligand results in different splitting parameter (or energy gap). So light of a different colour is absorbed and its complementary colour transmitted is different. [1] (iv) [1] correct expression for Kc (no approximation should be done) [1] correct substitution of values leading to [N3-] 1.40 × 10-6&0.0200 - 1.40 × 10-6'&10-2.72' [N3-] = 0.5088 [N3-] = 0.072207 mol dm-3 (candidates that did not read that a buffer was used would obtain the value 0.0722604 mol dm-3, penalise 1 mark) [1] for calculating amount N3- [1] for calculating mass of Cu(N3)2 Since the amount of free N3- is much more that in the iron(III) complex, we can ignore the amount of N3- in the iron(III) complex. (If the candidate takes into account of the N3- in the iron(III) complex, it is also Candidates are reminded to leave intermediate values to 4 to 5 significant figures. Failure to do so will result in s.f. penalty or truncation errors.

acceptable. This method is only done because it simplifies the calculations by a bit.) Amount of N3- = 0.072207 × 5.001000 = 0.000361035 mol Mass of Cu(N3)2 = 0.000361035 ×63.5 + 14.0 × 62 = 0.0266 g (3 s.f.) 2 (a) [1] for correct reaction with KMnO4 [1] for correct reaction with K2Cr2O7 [1] for explanation that a side reaction occurred with KMnO4 A chemist may want to convert the starting product to the bottom one on the right. Using KMnO4 will result in side-chain oxidation occurring too which produces the unintended product on top. (b) (i) Immediate distillation [1] (ii) R—CH2OH → 2 e- + R—CHO + 2 H+ [1] Ignore state symbols (iii) To separate samples of A and B, add PCC at room temperature The following combinations of tests are accepted: Then add Tollen’s reagent (and warm) [1] No silver mirror is formed for A. [1] Silver mirror is formed for B. [1] OH KMnO4 dil. H2SO4 dil. H2SO4 K2Cr2O7 OH O O OH OH O

Then add Fehling’s reagent (and warm) [1] No precipitate is formed for A. [1] Brick red precipitate is formed for B. [1] Then add I2(aq), NaOH(aq) (and warm) [1] Yellow precipitate is formed for A. [1] No precipitate is formed for B. [1] (c) (i) [1] for each correct step [1] for each correct intermediate No ecf is awarded. For the 3rd step, accept any (rather strong) iodine containing acids like: HIO3, HIO4. Credit for an intermediate and its previous step will be voided if DMP and IBX is used. Alternative answer: [1] excess DMP [1] correct intermediate (Note: either of the 2 alcohol groups will be oxidised. If the C3 alcohol group is NOT oxidised, then DMP must be added again. To guarantee the C3 alcohol group to be oxidised, DMP should be in excess. The word excess should be present to score credit for the first point. Alternatively, the candidate include an additional step where DMP is use to guarantee the oxidation of the alcohol to a ketone.) [1] I2(aq), NaOH(aq), warm/heat [1] correct intermediate (second last intermediate in answer) [1] iodine-containing acid (ii) Oxidation state of iodine in both IBX and DMP is +3. [1] OH OH I2(aq), NaOH(aq) warm O O-Na+ OH O O-Na+ O IBX or DMP HI(aq)O OH O

Geometry of iodine in IBX: See-saw Geometry of iodine in DMP: Square pyramidal [1] for both correct 3 (a) (i) An atom that has four different groups/ligands attached to it (in tetrahedral configuration). [1] (ii) [1] correct structure of B Different melting point AND rotates plane polarised light to a different extent when compared to A. [1] (iii) 26 = 64 stereoisomers [1] (iv) A and B are not enantiomers of each other [1] since only one of the chiral carbons has a different stereochemical configuration. (b) (i) First order because the rate constant has units in the dimension of time inverse. [1] Candidates are expected to write an explanation and deduce the answer based on the dimensions of k. (ii) 0 J mol-1. R and S have same physical properties so their Gibbs’ Free Energy of formation must be the same. [1] (iii) H2SO4(aq), heat with reflux (any strong acid is acceptable) OR NaOH(aq), heat with reflux (any strong alkali is acceptable) [1] (iv) Since [R] = 0 mol dm-3 before the hydrolysis when t = 0 s, ln([S][S]+ = 2k(0) + C A convincing explanation must be offered. CO2H NH2

So C = 0. [1] (v) [1] for substitution of [R] = 0.059[S] into the expression ln([S] + 0.059[S][S]−0.059[S]+ = 2(4×602)k2 k2 = 4.102 × 10-6 s-1 [1] (correct value in SI units) (vi) [1] for substitution of [R] = 0.090[S] into the expressi