Mark Scheme for Mock Paper 2

Mark Scheme for Mock Prelim Paper 2 Mark Scheme Abbreviations ; separate marking points (1 marking point is 0.5 marks, unless stated otherwise) R reject I ignore COND mark awarded is conditional on previous marking point OWTTE or words to that effect (accept other ways of expressing the same idea) underline actual word given must be used by candidate (grammatical variants accepted) ( ) the word / phrase in brackets is not required ORA or reverse argument

Question Answer Guidance 1 a i giant metallic structure / lattice; sea of delocalised electrons AND Cs+ cations (arranged in a lattice); electrostatic attraction / electrostatic forces of attraction (known as metallic bonding); between electrons and Cs+ cations; R metallic bonding Candidates are expected to explain the origin of metallic bonding, because the word ‘metallic’ does not give any insight into the bonding present. ii Li has stronger metallic bonding than Cs; because charge density of Li+ cation is higher than that of Cs+ cation; • Li is harder to vapourise than Cs Pb has more valence electrons than Cs; because Pb is in Group 14, while Cs is in Group 1; • Pb has more active electrons than Cs Answers missing either one of the bullet points will be deducted 0.5m. The total number of marks awarded must be non-negative. b i E⦵(Li+/Li) = -3.04 V AND E⦵(Na+/Na) = -2.71 V AND E⦵(K+/K) = -2.92 V;

inconsistent reducing power down the group OR reducing power is approximately invariant / constant OR reducing power remains strong; OWTTE ii number of principal quantum shells increases down the group; valence electron is further AND less attracted to the nucleus; OWTTE 1st ionisation energy decreases down the group (The conclusion must be present. Deduct 0.5m for no conclusion. No marks awarded for this question for wrong conclusion.) The total number of marks awarded must be non-negative. c i Cs + H2O → Cs+ + 0.5 H2 + OH- reaction has Ecell > 0; OWTTE because Group 1 metals have E⦵(X+/X) around -3 V / more negative than -2 V which is more negative than E⦵(H2O/H2) = -0.83 V; OWTTE Cs metal will react with water (spontaneously) to form Cs+ ions; unable to maintain [Cs+] = 1 mol dm-3 OR cannot place Cs in water; The reduction potential of Cs+/Cs is not given in the Data Booklet, so it is the job of the candidate to offer further insight of what they would expect this value to be. ii M1 for amount of electrons ne-=ItF=2.00×48×60×6096500=3.5813 mol A1 for correct answer mass = 3.5813 × 132.9 = 476 g

d Award 1 mark for 3 points. Award 2 marks for 5 points. • down Group 1, the charge density of the cation decreases (because ionic radius increases down the group) • polarising power of the cation weakens / decreases down the group • electron cloud of peroxide ion is less distorted / polarised down the group • more energy required to break the O—O bond in the peroxide ion down the group • decomposition temperature increases down the group 2 a i intermediate B cannot be the intermediate because; the C=C bond in B is non-polar / C atoms have same electronegativity (pi electrons in the C=C bond are localised); so there is no electron deficient / electrophilic site in the C=C bond in B; NaBH4 cannot perform a nucleophilic attack on the C=C bond; OR C=C bond is electron-rich; and will repel / prevent; Both A and B are produced, but the question is asking you which of them is the intermediate. The candidate should argue why B is NOT the intermediate (and why A is). Because A and B are produced, there is no value in arguing why A is produced. Why A can react is also a given, so full credit is given to candidates that argue why

the electron-rich nucleophile NaBH4 from reacting with B further; an explanation on why A is produced will only gain max. 1m B cannot react with NaBH4. ii The following tests are acceptable: To separate test tubes containing A and B Test 1 add 2,4-DNPH; orange precipitate observed for A, but no precipitate is formed for B; Test 2 add KMnO4 AND H2SO4(aq) OR NaOH(aq) (then heat); purple solution remains for A, solution decolourises for B; Test 3 add K2Cr2O7(aq) AND H2SO4(aq) AND heat / warm; orange solution remains for A, orange solution decolourises / turns green for B; Test 4 add Br2(aq) OR Br2(l) OR Br2 in CCl4;

orange solution remains for A, solution decolourises for B; (for aqueous Br2 used) brown solution remains for A, solution decolourises for B; (for pure or organic Br2 used) Test 5 add Na(s) OR any group 1 metal; no effervescence observed for A, effervescence observed for B; iii equal probability of the borohydride nucleophile from attacking the top and bottom of the plane with respect to the trigonal planar geometry of the carbonyl carbon; which forms equal proportions of butan-2-ol enantiomers (and their rotating power cancels each other); iv H2(g), high pressure, Ni (catalyst); (award 1m for this marking point, catalyst must be present) • room temperature condition not needed accept other catalysts such as Pd/C (palladium on carbon), Pt b i electrophilic addition; (1m is awarded)

all partial charges, formal charges indicated; slow step correctly identified; all curly arrows correct; correct final product; ii test tube Y (no marks awarded for writing test tube X as the answer) the pi-electrons in the C=C bond are delocalised in the C=C—C=O system; since O is more electronegative than C, the electron density in the C=C bond decreases OR C=C bond is less electron rich (because electron density will be richer towards O); OH Br Br slow O Br H O Br H H2O Br O /u1D6FF+ /u1D6FF-

but in Y, the pi electrons in the C=C bond is localised OR not delocalised; OWTTE so the C=C bond is more electron-rich in B than that in 3-oxybut-1-ene; and will attack Br2 faster 3 a 1s2 2s2 2p6 3s2 3p6 3d6 ; (1m or 0m awarded) b in an octahedral ligand field; the d-orbitals split; into 2 different energy levels; an electron transitions / is excited / is promoted to a higher-energy level (known as d-d transition); wavelength / frequency of light absorbed; colour seen / observed / reflected / transmitted is complement of colour absorbed; c when a system at dynamic equilibrium experiences a change in conditions, the position of equilibrium will shift so as to reduce OR counteract OR lessen that change; (1m or 0m awarded) R eliminate, stop d aqueous ammonia is basic / contains OH- ; addition of NH3(aq) causes concentration of OH- to increase, causing position of equilibrium 1 to shift right forming green precipitate of Ni(OH)2 ;

further addition of NH3(aq) increases the concentration of NH3 causing the position of equilibrium 2 to shift right forming blue complex of Ni(NH3)62+ ; this decreases the concentration of Ni2+ which causes position of equilibrium 1 to shift left so the green precipitate redissolves (to form Ni2+); e ; cis isomer labelled; COND ; trans isomer labelled; COND R tetrahedral complexes f i Award 1m or 0m for this question. Ni BrBr NC CN 2- Ni CNBr NC Br 2-

ii CO produced is poisonous; and will irreversibly bind with haemoglobin which will deprive the body of vital oxygen; OWTTE R toxic 4 a i condensation OR electrophilic (aromatic) substitution; (award 1m or 0m) ii Hot NaOH(aq) hydrolysis products correct (both carboxylate ion AND alcohol); both phenols deprotonated; Cxx x xOx x