Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level. Published

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1 ambridge nternational Examinations ambridge nternational Advanced Subsidiary and Advanced Level HEMSTRY 9701/43 Paper 4 A Level Structured Questions May/June 2016 MARK SHEME Maximum Mark: 100 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. t shows the basis on which Examiners were instructed to award marks. t does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. ambridge will not enter into discussions about these mark schemes. ambridge is publishing the mark schemes for the May/June 2016 series for most ambridge GSE, ambridge nternational A and AS Level components and some ambridge Level components. GSE is the registered trademark of ambridge nternational Examinations. This document consists of 8 printed pages. ULES 2016 [Turn over

2 Page 2 Mark Scheme Syllabus Paper 1 (a) (i) a(h) a 3 + H 2 Ba(H) 2 is soluble, R Ba 3 is insoluble (iii) Mg(H) 2 is insoluble / not very soluble will not form ppt. of Mg 3 (b) carbonates are more stable down the group due to increase in cationic size / radius (causing) less polarisation of 3 2 ion (c) radius of Ni 2+ = nm; radius of a 2+ = nm so Ni 3 decomposes more readily than a 3 [Total: 9] 2 (a) (i) o:...3s 2 3p 6 3d 7 4s 2 o 2+ :...3s 2 3p 6 3d 7 solution starts pink turns blue pink is [o(h 2 ) 6 ] 2+ blue is [ol 4 ] 2 this complex is tetrahedral

3 Page 3 Mark Scheme Syllabus Paper (b) R 3 P R 3 P Ni R 3 P Ni PR 3 R 3 P o PR 3 [Total: 9] 3 (a) K p = {p(s 2 ) (p(h 2 )) 4 } / {(p(h 2 S)) 2 p(h 4 )} units: atm 2 R Pa 2 (b) (i) p(h 2 S) = 196 atm p(h 2 ) = 8 atm K p = (2 8 4 ) / ( ) = (c) (i) S o will be positive, because more gas moles on the RHS / products S o = ( H o G o ) / T = (241 51)/1000 = 0.19 R 190 kj mol 1 K 1 R J mol 1 K 1 (d) G o will become less positive/more negative as T increases, because S o is positive (or T S o is more negative) therefore the reaction becomes more feasible / spontaneous as T increases [2] [Total: 10]

4 Page 4 Mark Scheme Syllabus Paper 4 (a) (i) SP is the EMF / potential of a cell composed of two electrodes (R half cells) under standard conditions (R at 289 K R 1 mol dm 3 ) voltmeter and salt bridge (iii) A is Ag B is Ag + (aq) or AgN 3 (aq) is Pt D is Fe 2+ (aq) and Fe 3+ (aq) (combination of A and B can be reversed with combination of and D) [3] (b) (i) Ag + + Fe 2+ Ag + Fe 3+ E = E o log [Ag + ] = = 0.77 V so E cell = = 0.0 V [Total: 8] 5 (a) (i) pk a = log K a diacids are more acidic than H 3 2 H H 2 group is electron-withdrawing, stabilising the monoanion R H 2 group is electron-withdrawing, weakening the H bond R monoanion is stabilised by H bonding as n increases, the electron withdrawing group is further away from the ionising 2 H group R the (intervening) alkyl groups destabilise the anion (iii) removing H + from an anion is not electrostatically favourable (b) (i) a solution which resists changes in ph when small amounts of H + or H are added

5 Page 5 Mark Scheme Syllabus Paper H 2 H 2 H 2 2 Na + H + H 2 H 2 H 2 2 H + Na + H 2 H 2 H 2 2 Na + NaH Na 2 H 2 H 2 2 Na + H 2 6 (a) (i) 6 H 5 N 2 + 6e + 6H + 6 H 5 NH 2 + 2H H 5 N Hl + 3Sn 2 6 H 5 NH 3 l + 3Snl 4 + 4H 2 [2] [Total: 9] (b) (M r values: 6 H 5 N 2 = H 5 NH 3 l = 129.5) theoretical yield = /123 = 5.26 g percentage yield = /5.26 = 79.8% (80%) (c) (i) 6 H 5 NH 2 = 93 yield of phenylamine = /129.5 = g mass left in water = = g K part = (2.68/50) / (0.336/25) = 3.99 (d) phenylamine is less basic that ethylamine the lone pair on N is delocalised over the ring making it less available for reaction with a proton / δ+ H [2] (e) (i) step 1: HN 2 R (NaN 2 + Hl ) at T 10 step 2: boil / heat in water E is N N (l - ) [Total: 13]

6 Page 6 Mark Scheme Syllabus Paper 7 (a) (i) NH 2 H 3 H H N H H 2 H N H H 2 2 H [2] M r = 233 (b) (i) NH 2 H(H 2 H) 2 F is a D power supply G is the anode R positive electrode is the cathode R negative electrode H is filter paper (R gel) soaked in buffer solution [4] (iii) P is NH 2 H 2 2 or NH 2 H 2 2 H or glycine S is [ala ser gly] ( ) glycine is the smallest, so travels fastest; tripeptide is the largest, so travels slowest (c) (i) heat with H 3 + R heat with H (aq) hydrolysis [Total: 13] 8 (a) H = [2( 580) + 3( 286) + 3( 1438)] [ ( 437) + 3( 814)] = 81 kj mol 1 [2] (b) (i) cis trans R geometrical

7 Page 7 Mark Scheme Syllabus Paper in a complex the d orbitals are split into 2 energy levels colour is due to absorption of light (in visible region) electron promotion to higher orbital absorbs a photon the d d energy gap is different for the two complexes, hence different colours [Total: 7] 9 (a) T is U is N (b) (c) step 1: 6 H 5 l + All 3 (+ heat) step 2: H 3 H 2 l + All 3 (+ heat) step 3: Br 2 + light (or heat) step 4: KN + heat (in ethanol) step 5: H 3 + R H + in H 2 R Hl (aq) etc AND heat / boil / reflux step 1: electrophilic substitution R nucleophilic substitution step 5: hydrolysis R nucleophilic substitution [Total: 9] 10 (a) n(mn 4 ) = / 1000 = mol n( 2 4 H 2 ) = / 2 = (in 25 cm 3 ) = mol in 100 cm 3 M r = = 90 mass of 2 4 H 2 = = g (0.274) percentage = /40 = 0.68% (b) (i) Sl 2 or Pl 5 or Pl 3

8 Page 8 Mark Scheme Syllabus Paper J is H 3 H 3 K is HN NH (c) (i) H 3 at δ 15 H 2 at δ 65 nly one peak, so only one type / environment of atom (d) (i) M is H 2 2 H N is H 3 2 H is H 3 H 3 [3] L is [Total: 13]