Mark Scheme (Results) Summer GCE Chemistry (6CH05/01)

Transcription

1 Mark Scheme (Results) Summer 05 GCE Chemistry (6CH05/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 05 Publications Code UA04085* All the material in this publication is copyright Pearson Education Ltd 05

3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. Mark schemes will indicate within the table where, and which strands of QWC, are being assessed. The strands are as follows: i) ensure that text is legible and that spelling, punctuation and grammar are accurate so that meaning is clear ii) select and use a form and style of writing appropriate to purpose and to complex subject matter iii) organise information clearly and coherently, using specialist vocabulary when appropriate

4 Section A (multiple choice) Correct Answer Reject Mark C Correct Answer Reject Mark A Correct Answer Reject Mark 3 C Correct Answer Reject Mark 4 B Correct Answer Reject Mark 5 C Correct Answer Reject Mark 6 B Correct Answer Reject Mark 7 B Correct Answer Reject Mark 8 D Correct Answer Reject Mark 9 C Correct Answer Reject Mark 0 D Correct Answer Reject Mark C Correct Answer Reject Mark A

5 Correct Answer Reject Mark 3 D Correct Answer Reject Mark 4 D Correct Answer Reject Mark 5 A Correct Answer Reject Mark 6 D Correct Answer Reject Mark 7 D Correct Answer Reject Mark 8 A Correct Answer Reject Mark 9 B Correct Answer Reject Mark 0 B

6 Section B a(i) Ni: (s s p 6 ) 3s 3p 6 3d 8 4s () Cu: (s s p 6 ) 3s 3p 6 3d 0 4s () capital letters, subscripts for superscripts 4s before 3d Penalise omission of 3s 3p 6 once only if rest is correct a*(ii) First electron removed is from 4s (in both atoms) () Second electron in Cu (is harder to remove so it is) EITHER closer to nucleus/in inner shell less shielded () IGNE Comments about second electron being in full shell/ in a 3d shell/in a 3d orbital Reference to 3d 0 stability a(iii) (attraction on (3d) electrons increases due to) number of protons increasing / nuclear charge increasing IGNE The charge density of the + ions increases Effective nuclear charge

7 b(i) b(ii) Cu + (aq) Cu(s) + Cu + (aq) IGNE Eqm sign for Reverse equation Any equation involving electrons Both white (both) Colourless () COMMENT Ignore states eg solution/precipitate As have 3d 0 / have a full 3d sub-shell /ALL 3d orbitals are full () IGNE Does not have partially filled d orbitals They do not absorb light No d-d transitions occur c (Zinc) does not form a (stable) ion with incompletely/partially filled d orbitals Element has full d shells. d sub-shell for d orbitals The only (stable) ion formed by zinc has full d sub-shell It does not form a (stable) oxidation state with incompletely/partially filled d orbitals Total for = 9 marks

8 (a)(i) 3 Beaker with V electrode in solution containing V + (aq) AND beaker containing V + (aq) and V 3+ (aq) with Pt electrode N.B. Both solution levels must be shown () Labelled salt bridge AND connections to voltmeter Suitable name or formula of salt for label Salts eg NaCl in salt bridge () Ion concentrations = mol dm -3 M for mol dm -3 Concentrations given in one beaker only Salt bridge neither dipping into nor touching solution unless penalised in MP Salt bridge containing an alkali/acid mole of V + and mole of V 3+ () Beaker positions may be reversed Ignore references to temperature and pressure

9 (a)(ii) First mark V 3+ + V 3V + Balanced equation, either direction Eqm sign for e - included IGNE State symbol even if incorrect () Second mark Correct direction If balancing is incorrect or e - included in equation () b(i) ([VO + (aq) + H + (aq)], [V 3+ (aq)+ H O(l)] Pt) ([VO + (aq) + H + (aq)], +.00 [VO + (aq)+ H O(l)] Pt) Sign and value needed

10 (b)(ii) A: (+)0.3 (V) () 6 VO + (may be shown as a product in an overall equation) () EITHER Bubbles / effervescence (of colourless gas) Colour changes (from yellow) to blue TE on negative E cell for stays yellow (from yellow) to green if justified by partial reduction () B: 0.(0) (V) () no change / stays blue () If B=+0. or other positive value allow colour change from blue to green or brown. Violet Stays violet EITHER Consistent use of rule that reaction occurs when E cell is positive Consistent use of rule that no reaction occurs when E cell negative If implied but not stated specifically () c(i) V H O VO + 4H + + 3e Reverse equation unless used to Ox number of V increases by 3, deduce final ox number of Mn decreases by 5 correct equation. Balanced full equation 5V + + 3MnO H + 5VO + +3Mn H O (c)(ii) (35.50 x 0.000/000) = 7.(0)x 0-4 /

11 (c)(iii) In final answer 3 9. scores 3 marks 33. scores marks (ratio inverted) 55.3 scores marks (ratio :) METHOD Mol V + reacting = 7.0x 0-4 x 5/3 =.8333 x 0-3 = mol VO + TE on answer to (c)(ii) () x 3/5 = 4.6x 0-4 Mass NH 4 VO 3 = (.83 x 0-3 x 6.9) = g TE from 4.6 x 0-3 = () % purity = (0.3897x 00/ 0.50) = (9.9333) = 9.% TE from = 33.% () METHOD If 00% pure, moles of NH 4 VO 3 =0.50/6.9=.83 x 0-3 () Mol V + reacting = 7.0x 0-4 x 5/3 =.8333 x 0-3 = mol VO + TE on answer to (c)(ii) () % purity = =.8333 x 0-3 x 00/.83 x 0-3 = 9.% () TE at each step provided that each number used is to at least sf Total for = 7 marks

12 3(a) All carbon to carbon bonds same length/ longer C-C and shorter C=C not present IGNE Just benzene has a delocalised ring Benzene does not have C=C double bonds Any references to shape/ bond angles 3(b)(i) (3 x -8) = -354 (kj mol - ) 3(b)(ii) X (05 354) = 49 (kj mol - ) Benzene Diagram inverted scores 0 Cyclohexane First mark Relative levels with names or formulae () Second mark Value 49 (kj mol - ) + arrow in correct direction double-headed arrow () +49 TE from value in (b)(ii) IGNE 3H if shown / cyclohexene / other arrows/values 3(b)*(iii) The p/pi-/π/6 electrons (of carbon) () are delocalised in benzene (but not in X) ()

13 3(c) Acceptable Answers Reject Mar k First mark: 4 FeBr 3 + Br FeBr 4 + Br + Br-Br + FeBr 3 Br δ+.. Br δ- FeBr 3 () Ignore state symbols even if wrong Second, third and fourth marks: Either Arrow from benzene ring electrons (from inside the hexagon) to Br + / Br δ+ (.. Br δ- FeBr 3 ) () Correctly drawn intermediate with delocalisation covering at least three carbon atoms, but not the carbon atom bonded to the bromine, with the positive charge shown inside the horseshoe Gap in wrong place The bonds to H and Br may be dotted () Arrow from / close to C-H bond to inside the hexagon and H + / HBr as product () Use of Kekulé structure for benzene and intermediate with arrow from C=C double bond to Br + / Br δ+ (.. Br δ- FeBr 3 ) () Correctly drawn intermediate with + charge on the C atom next to the C bonded to H and Br

14 The bonds to H and Br may be dotted () Arrow from / close to C-H bond to bond beside + charged C and H + / HBr as product () Each marking point is independent 3(d)(i) Bromine goes colourless It/the mixture goes from brown to colourless Goes clear Red-brown/ Orange/ yellow/ combinations of these colours Bromine is decolorised () White precipitate/solid forms / Steamy fumes () Red to colourless Bromine is discoloured Effervescence IGNE Antiseptic smell Gets hot 3(d)(ii) + 3Br + 3HBr Organic product with structure shown () Rest of equation correct C 6 H 5 OH or Kekule for phenol () C 6 H 5 OH + 3Br C 6 H Br 3 OH + 3HBr Scores MP only Substitution of Br or Br in any position in balanced equation scores MP only.

15 *3(d)(iii) Lone pair of electrons on oxygen (may be shown on a diagram) and EITHER overlaps with pi cloud Feeds into / donates into / interacts with benzene ring () OH group overlaps Just making it more reactive. Activating benzene ring / increasing electron density of ring / making attack by electrophiles easier () COMMENT Lone pair of electrons on oxygen increases electron density of ring scores () benzene becomes a better nucleophile for MP Total for 3 = 6 marks

16 4(a)(i) 4(a)(ii) Concentrated nitric acid AND concentrated sulfuric acid concentrated nitric and sulfuric acids Concentrated HNO 3 and concentrated H SO 4 Extra reagents To prevent multiple substitutions/ to stop di- or trinitrobenzene forming To stop further substitution (of NO )/ further nitration Further addition of nitro groups IGNE further reaction 4(a)(iii) Tin/ Sn AND concentrated HCl/ concentrated hydrochloric acid Dilute HCl Iron/Fe or Zn/Zinc for tin Conc for concentrated 4(b)(i) C 6 H 5 NH + 3 Cl C 6 H 5 NH 3 Cl 4(b)(ii) Skeletal formula C 6 H 5 for benzene Undisplayed CH 3 Structural formula

17 4(b)(iii) D (transition metal) complex ion Transition metal complex / copper complex IGNE Formulae of ions () F (azo) dye / azo compound / diazo compound diazonium compound molecule for compound () 4b(iv) Benzenediazonium chloride Phenyldiazonium chloride Benzadiazonium chloride Diazonium salt 4b(v) HCl + NaNO Hydrochloric acid + Sodium nitrite / nitrate(iii) alternative cation to Na + IGNE HNO Concentration of HCl HCl + HNO

18 4b(vi) any substitution positions C 6 H (CH 3 ) NH C 6 H 3 (CH 3 ) NH H- C 6 H (CH 3 ) NH Kekule structure Total for 4 = 0 marks 5(a)(i) H + H 3 N C COO CH OH HOCH CH(NH 3 + )CO Charge on NH 3 shown on N atom

19 *5(a)(ii) EITHER At ph 5.68 both NH groups on lysine will be NH + 3 / lysine has an overall positive charge while serine has no overall charge () So only lysine will move to negative terminal / in an electric field () lysine has a greater positive charge (than serine) () Lysine will move faster / further in an electric field () They have different charges so will move different distances/ at different speeds mark max IGNE Answers based on molecular mass *5(b) COMMENT Ignore references to H bonds from amino and carboxyl groups and assume that phenylalanine cannot form H bonds refers to R group in the compound. Serine has an OH (alcoholic) group which can form H bonds with water () Benzene ring in phenylalanine is bulky /is large and non-polar / is hydrophobic / London forces between phenylalanine molecules are significant van der Waals etc for London If MP is not scored, phenylalanine cannot form H bonds with water can score MP () VdW / id-id

20 5(c) H, -H, H- NH CH CO H / NH CH COOH 5(d)(i) (Measure) the amount of / angle of / degree of /extent of rotation (of plane of polarization/ plane polarized light) Just polarimetry IGNE direction they rotate the light 5(d)(ii) (Leu has chiral C but) Ile has chiral C / asymmetric C atoms The R group in Ile contains a chiral C (Leu has optical isomers but Ile has 4 optical isomers) Leu has no chiral C Ile has () more chiral C (than leu) Chiral centres shown on formulae

21 5(e) CON circled with N as part of ring () Fully correct structure () Just N-C circled -CH(CH 3 )NH shown skeletally Dipeptide using NH from alanine and COOH from proline, with correct structure and CONH circled This scores max () 5(f)(i) CH 3 CHO / CH CHOH CH 3 COH (unless with correct structure) Displayed or semi-displayed formula C H 4 O 5(f)(ii) C 8 H 9 NO 4

22 5(f)(iii) Several methods are possible. Amount of O should have been calculated and final answer should be consistent with masses/mols in calculation. There should be evidence of working at each stage for 4 marks 4 MP.64 g CO contain 0.79 g C g H O contain g H.64 g CO contain mol CO g H O contain mol H O () MP Mass O in Q = ( ) = 0. g () % mol/00g ratio C H N O..3 4 MP3 of moles of each element of mol in 00g Allow TE for incorrect mol of H see calculation below () MP4 Ratio stated must follow from numbers in calculation (so empirical formula is consistent with the molecular formula of Q) () Calculation following error in H scores max (3) If mass H is wrongly calculated as g, then mass O will be (TE for MP) () % mol/00g ratio C H N O

23 of moles of each element (TE for MP3) () Ratio stated and empirical formula said not to be consistent (TE for MP4) () 5(f)(iv) Mass of molecular ion in mass spectrum = 303 / any sensible fragment with mass eg 30 / 58 / 45 or smaller fragment () Molecular ion based on formula in f(i) of peaks in nmr = / 3 5 () 4 Total for 5 = 8 marks

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