MS3£3.00

GENERAL CERTIFICATE OF EDUCATIONTYSTYSGRIF ADDYSG GYFFREDINOL

MARKING SCHEME

CHEMISTRYAS/Advanced

SUMMER 2008

INTRODUCTION

The marking schemes which follow were those used by WJEC for the Summer 2008examination in GCE CHEMISTRY. They were finalised after detailed discussion atexaminers' conferences by all the examiners involved in the assessment. The conferenceswere held shortly after the papers were taken so that reference could be made to the fullrange of candidates' responses, with photocopied scripts forming the basis of discussion.The aim of the conferences was to ensure that the marking schemes were interpreted andapplied in the same way by all examiners.

It is hoped that this information will be of assistance to centres but it is recognised at thesame time that, without the benefit of participation in the examiners' conferences, teachersmay have different views on certain matters of detail or interpretation.

WJEC regrets that it cannot enter into any discussion or correspondence about thesemarking schemes.

1

CHEMISTRY CH1

Section A

1. 20 protons. [1]

2.

2nd electron in line with other second shell electrons (1)10th electron similar value to 11th electron (1) [2]

3. C iodine [1]

4. (a) The mole is the amount of material which contains the same number ofparticles as there are atoms in 12 g of 12C isotope. [1]

(b) C 0.60 [1]

5. B hydrogen [1]

2

6. Dumbbell shape or appropriate diagram [1]

7. (a) Electronegativity is the ability of an atom to attract electrons within a(covalent) bond [1]

(b) Mg and N [1]

Section A Total [10]

3

Section B

8. (a) (i) X is an electron / β particle [1]

(ii) Mass number 206 Symbol Pb 2 × (½) [1]

(iii) 56→ 28→ 14→ 7 picograms so 3 half-lives (1)3 × 138 = 414 days (1) [2](2 marks for correct answer with no working)

(iv) α - particles (have low penetrating power and) are stopped by glass /air / luggage. (1)but when ingested they cause damage by direct contact with organs etc.(1) [2]

(b) PoCl4 will involve covalent bonding [1]

(c) (i) 35Cl and 37Cl 2 × (½) [1]

(ii) 350 [210Po 35Cl4]+

352 [210Po 35Cl337Cl]+

354 [210Po 35Cl237Cl 2]

+

356 [210Po 35Cl 37Cl 3]+

358 [210Po 37Cl4]+

All 5 correct - 2 marks 3 correct - (1)(Allow 1 consequential mark if (c)(i) isotopes are wrong but candidateshows correct formulation of the molecular ions.) [2]

(iii) Ions formed by electron bombardment (1)Ions accelerated and/or collimated by charged plates/electric field (1)Ions passed through a (curved) (electro)magnetic field (1)Ions deflected according to their mass / charge ratio (1)((½) if only mass mentioned)(any three)

(Marks can be obtained by appropriate labelled diagram) [3]

Total [13]

4

9. (a) (i) Cl, Ne, Ar (1) for three, (½) for two.Treat more than three answers as "right –wrong" [1]

(ii) (Only) covalent (½) giant lattice / macromolecular structures (½).Must break strong bonds, requiring considerable energy formelting (1) [2]

(iii) (Monatomic) species with only van der Waals forces betweenatoms (1)

Weak interatomic forces require little energy to break (1)[2]

(iv) Na, Mg, Al have respectively 1, 2 and 3 outer electrons (1)so more electrons involved in the electron sea / delocalised electronsresulting in stronger bonds. (1) [2]

(v) Covalent bonds become weaker going down a group /C-C bonds stronger than Si-Si [1]

(vi) Atoms become larger/have more electrons going down a group so thereare more/stronger van der Waals forces. [1]

(b) Q9(b) is to be assessed for Quality of Written Communication. [L] [4]Candidates must satisfy the following criteria• Selection of a form and style of writing appropriate to the question• Organisation of the relevant information clearly and coherently• Use of legible text with adequate spelling, punctuation and grammar.

Any 8 × (½)

layered structure / macromolecular stucturecovalent bonding within layersinterlinked / fused hexagons within layersdelocalised electrons within layerssideways overlap of p-orbitals produces delocalisationvan der Waals forces between layers / weak forces between layersbonding within layers much stronger than bonding between layerselectrical conductivity (within layers) due to delocalized electronsno / negligible electrical conductivity from one layer to anothercrystals weak along the layer planes / layers easily slip over one anothercrystals strong at right angles to the layershigh melting temperature due to strong covalent bondslow solubility due to strong covalent bonds

(Marks can be obtained by appropriate labeled diagrams) Total [13]

5

10. (a) Large difference in electronegativities between O and H/ water has bond dipoles / is a polar molecule (½) Hδ+−Oδ− on both covalent bonds in water (½)Oδ− attracted to (Ba2+) cation (1)Hδ+ attracted to (OH−) anion (1) [3]

(b) Flame test (½)(Apple / light ) green flame produced (½ )orDissolve in water and add dilute sulphuric acid (½)White precipitate formed (½)

[1]

(c) (i) 315.46 − 171.30 = 144.16 (1)144.16 / 18.02 = 8 so x = 8 (1)(2 marks for correct answer with no working) [2]

(ii) 5.60 / 315.36 = 0.0177(6) mol (1)so concentration = 0.0177(6) × 1000 / 100 = 0.178 mol dm-3 (1) [2]

(iii) Calcium hydroxide will have a lower solubility than barium hydroxideas the Group II hydroxides increase in solubility down the group. [1]

(d) (i) An oxidizing agent accepts electrons (hence oxidizing another species). [1]

(ii) Chlorine, Cl2, is a stronger oxidizing agent than bromine, Br2, as it hasa stronger attraction for electrons / greater electronegativity / oxidisingpower decreases down Group VII. [1]

(iii) 2Cl2 + 2H2O = 4HCl + O2 (1)0 +1 -2 +1 -1 0

Cl is reduced / accepts electron as Ox No decreases (1)O is oxidized as Ox No increases. (1)(must have before and after ox. nos correct) [3]

orCl2 + H2O = HCl + HClO (1)0 +1 -2 +1 -1 +1 +1 -2

One Cl atom reduced (1)The other Cl atom oxidized (1)(must have before and after Cl ox. nos correct) [3]

(iv) 2 × (1)Add silver nitrate (solution)white precipitate formed / ppt soluble in dilute ammonia [2]

Total [16]

6

11. (a) 1s22s22p63s23p3. [1]

(b) (i) P2O5 [1]

(ii) Covalent (½) discrete molecules (½) [1]

(c) (i) 284 g mol-1 [1]

(ii) No moles P4O10 = 28.4 / 284 = 0.100 (½)No moles water = 0.100 × 6 = 0.600 (½)Mass water = 0.600 × 18.02 = 10.8 g (1) [2]

(d) (i) Ca3(PO4)2 [1]

(ii) (Main constituent of) skeletons / bones. [1]

(e) (i)

P Cl

Cl

Cl

xx

ooo

x x

x

oooo

oo

oo

oooo

oooo

oo

(1)

Lone pair: 2 electrons in the outer shell not involved inbonding (1) [2]

(ii) PCl5 trigonal bipyramid (1) PCl3 pyramidal (1) [2]

(Allow unambiguous diagrams)

(iii) Dative covalent / co-ordinate bond (1)Lone pair on P used to form a bond to Al (1) [2]

Total [14]

Section B Total [56]

7

CHEMISTRY CH2

Section A

1. A [1]

2. (i) Sodium hydroxide / potassium hydroxide [1]

(ii) Elimination / dehydrobromination / dehydrohalogenation [1]

3. (i) But-2-ene [1]

(ii) 6 [1]

(iii) 4C(s) + 4H2(g) → C4H8(g) (eqn ½ ; state symbols ½ ) [1]

4. Fossil fuels contain sulphur (½)gives sulphur dioxide (½)this dissolves in / reacts with water (½)producing sulphurous or sulphuric acid (½) [2]

(accept answers based on NO2)

5. H+ + OH− → H2O [1]

6. C [1]

Section A Total [10]

8

Section B

7. (a) (i) Type of catalyst : heterogeneous [1]

(ii) Catalyst / reaction : e.g., nickel in hydrogenation (of alkenes) [1]

(b) (i)22

22

C1pSO

pC1pSO ×=pK [1]

(ii) By Le Chatelier’s principle if more SO2 is added to the system theposition of equilibrium moves to the left to oppose the addition of SO2

(1)The formation of SO2Cl2 is exothermic, therefore the temperaturerises (1) [2]

(c) (i)

[1](ii) (warm) with (aq) NaOH (½)

(acidify) with (dilute) nitric acid (½)add Ag + / AgNO3 (½)white precipitate (½) [2]

(d) CH4 + Cl• → •CH3 + HCl (1)

•CH3 + Cl2 → CH3Cl + Cl•Or •CH3 + •Cl → CH3Cl (1) [2]

(e) The hydrocarbon is ethane / C2H6 (1)

formed by the combination of two methyl radicals / 2 •CH3 → C2H6 (1)[2]

(f) formula of propane C3H8 (½)

Mr 44.08 (½)

∴ 44.08 g → - 2217 (kJ mol-1) sign (½) number (½) [2]

Total [14]

C C C

Cl

C H

H

H H

H

H

H

H

Cl

9

8. (a) (i) Fractional distillation / fractionation [1]

(ii) There is a greater demand for smaller, (more) useful, compounds [1]

(b) (i) C9H20 [1]

(ii) C6H14 [1]

(c) (i) I. A movement of two electrons / electron pair [1]

II. A polarisation of the bromine molecule / a polar covalentbond (1)

OR

Partial positive and negative charges (1) [1]

III. The carbocation B is more difficult to form / too much energyis needed to form the carbocation B / carbocation B is lessstable / 3,4-dibromobut-1-ene cannot be formed from this ion

[1]

(ii) I.

[1]

II. There is restricted rotation about the C=C double bond / eachcarbon atom of the C=C double bond has two different groupsor atoms [1]

(iii) I. )sdm(mol100.3200

10)60.166.1( 1362

−−−

×=×−[1]

II. It would become smaller (1) (must be a comparison) becausethe collision rate between but-1,3-diene molecules wouldbecome less. (1)

[2]

C

H

H

Br C C C

H

H

H

H

Br

10

III.

correct profile (1)Ea correct Ea (1)

energy /kJ mol -1

[2]

extent of reaction

Total [14]

11

9. (a) (i) potassium / sodium dichromate / correct formulae

or

potassium manganate(VII) / correct formulae [1]

(ii) orange to green / purple to colourless [1]

(b) −239 + (−111) + ∆H = −485 (1)∴ ∆H = −485 + 239 + 111 = −135 kJ mol-1 (1) [2]

(c) (i) ∆H = 100 × 4.2 × 7.9 = 3318 J [1]

(ii) Number of moles of acid / H+ / OH− used = 060.01000

0.5020.1 =×(1)

∴ 0.060 mole gives 3318 J

∴ 1.0 mole gives J55300060.0

3318 =

∆Hneutralisation = − (½) 55.3 (½) kJ mol-1 (1)(accept error carried forward) [2]

(d) more concentrated - this is a measure of the quantity of acid dissolved/ contains more mol dm-3 (½)ethanoic acid is the more concentrated acid (½) [1]

stronger - this is a measure of the amount of dissociationof the acid in solution / the concentration ofhydrogen ions / H+ in solution (½)hydrochloric acid is the stronger acid (½)accept suitable pH values / it has a lower pHvalue (½) [1]

(explanation needs to be given to gain credit for the acid)

(e) pentane - boiling temperature is lower than pentan-1-ol (½)as no hydrogen bonding (between molecules)is present (½)

[1]

hexan-1-ol - boiling temperature is higher than pentan-1-ol (½)

as the chain length increases so does the boilingtemperature /explanation in terms of increasedVan der Waals forces (½) [1]

(explanation needs to be given to gain credit for the boiling temperaturestatement)

12

(f) (i)

2,2-dimethylpropan-1-ol [2]

(ii) e.g., other isomers are possible which do not contain an acidgroup / it could be an ester / a molecular formula does not showthat an acid group is present [1]

Total [14]

CH3 C

CH3

CH3

C

H

H

OH

13

Q10 (a) (i)is to be assessed for Quality of Written Communication. [L]Candidates must satisfy the following criteria

• Selection of a form and style of writing which is appropriate tothis question

• Organisation of the relevant information clearly and coherently• Use of legible text with adequate spelling, punctuation and

grammar

10. (a) (i) The process uses nitrogen (½) from the air (½) and hydrogen (½)from natural gas (½).

N2(g) + 3H2(g) � 2NH3(g) (1)

The forward reaction is exothermic (½) and is favoured by lowertemperatures (½).The forward reaction is also favoured by higher pressures (½) asthe number of moles / molecules is less on the right hand side ofthe equation. (½)Using too low a temperature slows the rate of the reaction toomuch (½) and so a catalyst of iron (½) ) is used.Too high a pressure is too expensive / more dangerous (½)Compromise values of 200 atmospheres (½) and 400 oC are used(½) maximum [6]

(accept values between 150 and 250 atmospheres and 350 to 450oC inclusive)

(ii) Any two from

plant costs; cost of raw materials; fuel / energy cost ; cost ofcoolant; transport ; markets; availability of labour; labour costs;effect on local population; waste disposal and otherenvironmental considerations; safety considerations [2]

14

(iii) I. mole070.01000

00.20.35 =×[1]

II. The number of moles of diluted ammonia is 0.070mole (½)

The number of moles of concentrated ammonia is also0.070 mole (½) [1]

(accept 0.070 mole once for (1) )

III. 5.00 cm3 of the concentrated ammonia solution contain0.070 mole ammonia

Concentration = 3dmmol0.1400.5

070.01000 −=×[1]

(iv) I. 214

28 = [1]

II. 60 − 28 − 16 − 12 = 4 [1]

III. Molecular formula of compound Z is C1H4N2O1 [1]

Total [14]

Section B Total [56]

15

CHEMISTRY CH3a

1. (a) (i) Points correctly plotted (2); axis labels (1); suitable scale (1) [4]

(ii) good line (1) [1]

(iii) 1.1×10-4 (within 10%) (1); mol dm-3 s-1 (1) [2]

(b) Working description - sampling and titration, colorimeter, clock, etc. [2]

(c) Runs at different iodide concs (1) while other initial concs constant (1) [2]

(d) Rates de[end strongly on temperature (1) [1]

Total [12]

2. (a) Suitable sketch and labels - calorimeter, insulation, thermometer, stirrer,lid, clock [4]

(b) (i) Good lines (1); ∆T = 9.9 ° (±0.2 for 2; ±0.3 for 1) [3]

(ii) −209,000 J mol-1 (½ for sign, ½ for units, 2 for calc) [3]

(c) Greater surface are gives faster reaction (1) [1]

Total [11]

3. (a) (i) Conc of hydroxide from second titration = 1.00×14.0/25.0= 0.56 mol dm-3 [2]

(ii) Vol acid for carbonate in first titration = 22.0−14.0 = 8.00 cm3 (1)Conc carbonate = 1.00× (8.00/2)/25 = 0.16 mol dm-3 (1) [2]

(b) 1 mol carbonate needs 2 mol HCl (1) [1]

(c) Test solution with BaCl2 to check no more ppt [1]

(d) Filter and wash BaCO3 ppt, dry and weigh. [1]

Total [7]

Total [30]

16

CHEMISTRY CH4

Section A

1. (a) Amide, amine, carboxylic acid. (2) for three, (1) for two correct [2]

(b) Optical Isomerism (1)Chiral centre / asymmetric carbon / four different groups attached to onecarbon (1) [2]

(c) Ionic bonding (½)Zwitterion correctly drawn / explained (H+ from –COOH to –NH2) (1)Strong bonding needs high energy / temperature for melting (½ ) [2]

(d) (1) for each COO− correct. [2]

O COO

NH2

C CH2 CH

O

_ _

(e) (i) Either(Cold) sodium carbonate (solution) / sodium hydrogencarbonate(solution) (1)Asparagine produces effervescence / fizzing / bubbles of gas (½)2-propenamide has no reaction (in the cold) (½)orAdd (a small amount of) bromine (water). (1)Asparagine has no (rapid) reaction (½)2-propenamide rapidly decolourises the bromine (½) [2]

(ii) I. Addition / free radical chain reaction [1]II. Accept any unambiguous form of the structure [1]

NH2

C C

H

C

H

H

OTotal [12]

17

2. (a) (Primary alcohol C5H12O) structures .2 × (1), allowing any unambiguous form of the structures: [2]

CH3CH2CH2CH2CH2OH CH3CH2

CH3

CHCH2OH

(b) (i) Conc sulphuric acid / phosphoric acid / aluminium oxide [1]

(ii) 2 × (1) [2]

CH3 CH CH CH2

CH3

CH3 CH2 C CH2

CH3

(c)

CH3CHBrCH3 CH

CH3

CBrCH3 CH2

CH3

CH3

is optically active is not optically active (1)

so C must be

(1)

B, a primary alcohol, must be

CH2OHCH3 CH

CH3

CH2

(1)

(mark consequentially if an earlier structure is incorrect)

CH3 CH

CH3

CH CH2

18

so A must be

CH2OCH3 CH

CH3

CH2 C CH3

O

3-methylbutyl ethanoate (isoamyl acetate) (1) [4](mark consequentially if the alcohol structure is incorrect).

(d) (i) Ester. [1]

(ii) Heat / warm / reflux (½) with ethanoic acid (½)and conc (½) sulphuric acid (catalyst) (½)

or

room temp (½) ethanoyl chloride (1)dry conditions (½) [2]

Total [12]

19

3. (a) (i) Polyamides [1]

(ii) Condensation/addition-elimination/step polymerisation [1]

(iii) 2 × (1). [2]

NH2 NH2 COOHHOOC

(b) Ethanoyl chloride / CH3COCl (1)(or ethanoic anhydride / (CH3CO)2O (1)C6H5NH2 + CH3COCl → C6H5NHCOCH3 + HCl (1) [2]

(c) (i) Quantisation of energy / Only certain electron energy levels availablewithin an atom (1)

Light given out when an electron falls from a higher level to a lowerlevel (1)

[2]

(ii) [1]

RED VIOLET

(iii) 2 × (1) [2]

655 nm line n=3 to n=2

433 nm line n=5 to n=2

Total [11]

Section A Total [35]

20

Section B

4. (a) (i) NaHCO3 shows P is carboxylic acid / -COOH (1)2,4-DNPH shows carbonyl (½)Iodine in alkali shows Q is methyl carbonyl / CH3CO- (½)((1) for CH3CO- with no explanation) [2]

(ii) 1st spectrum 60 2nd spectrum 582 × (½) [1]

(iii) Carboxylic acid group COOH+ m/e 45 (1)Methylcarbonyl group CH3CO+ m/e 43 (1)

Probable that 1st spectrum is acid (P), 2nd spectrum ismethylcarbonyl (Q) (1)

60 – 45 = 15 must be CH3+ as other group in acid (½)

or58 - 43 = 15 must be CH3

+ in methyl carbonyl

43 – 15 = 28 must be CO+ (½)

Compounds are ethanoic acid (P) and propanone (Q) 2 x (1)(Lose one (½) if no mention at all of positive charges on peak(s))

[6]

(b) (i) 3 ×(½) for splitting + (½) for shift) [3]COOH no splitting 11 ppmCH3 - triplet 0.2 to 2.0 ppm- CH2 – CO- quadruplet 2 to 3 ppm

(ii) Hydrogen bonding to (corresponding OH) dipoles on water. (1)Longer (alkyl) chain length in butanoic acid decreases effectiveness ofhydrogen bonding. (1) [2]

(iii) State the reagent(s) and conditions necessary to convert propanoic acidintoI. Heat (1) (not reflux) with sodalime (1) [2]

II. LiAlH4 (1) in dry ether (½) at room temperature (½) [2]

III. PCl5 / SOCl2 (1) dry conditions (½)at room temperature (½) [2]

Total [20]

21

5. (a) Q5(a) is to be assessed for Quality of Written Communication. [L] [5]Candidates must satisfy the following criteria• Selection of a form and style of writing appropriate to the question• Organisation of the relevant information clearly and coherently• Use of legible text with adequate spelling, punctuation and grammar.

Any of the points obtainable through clearly labeled diagrams.

5 × (1):1. Planar molecule with ring of six carbons2. Each C forms three single covalent bonds (σ-bonds) to 2C and 1Hor

sp2 hybridisation / 120° bond angles / trigonal arrangement about C

3. A π- electron cloud above and below plane of molecule4. π- electrons / double bonds delocalised around the ring / all C-C bonds

equivalent5. Benzene more stable than expected due to the delocalisation

(b) (i) Calculate ∆Ho

for the reaction

6 C-H 6 × 412 = 2472 12 C-H 12 × 412 = 49443 C-C 3 × 346 = 1038 6 C-C 6 × 346 = 20763 C=C 3 × 612 = 1836 7020 (1)3 H-H 3 × 436 = 1308

6654 (1)

∆Ho

= 6654 – 7020 = − 366 kJ mol-1 (1) [3]

(ii) Delocalisation energy of benzene

is the difference between the experimental ∆Ho

reaction for benzene

and the ∆Ho

reaction according to the Kekule structureorThe extra stability in the benzene structure due to electrondelocalisation. (1)

Delocalisation energy = −210 − (−366) = 156 kJ mol-1 (1) [2]

C

C

C

C

C

C

H

H

H

H

H

H

+ 3H2

C

C

C

C

C

C

H

H

H

H

HH

H H

H

HH

H

22

(c)

NO2

+

H

NO2

+NO2+ H+

(½) for cation (½) for intermediate (½) for product(½) for arrow (½) for curly arrow (½) for proton

[3]

(d) (i) For the reaction sequence

C6H5NO2 → C6H5NH2 iron / tin (½) plus hydrochloric acid (½)room temperature or just above / exothermicso needs to be cooled (½)add NaOH (to liberate base) (½)

C6H5NH2 → C6H5N2Cl sodium nitrite (½) plus hydrochloric acid (½)(or nitrous / nitric(III) acid (1))cooled to < 10°C / in ice) (1)

C6H5N2Cl → C6H5OH presence / addition of water (½)allow to warm to room temperature (½)

[5]

(ii) (Neutral) iron(III) chloride (1)gives a purple colour (1)

or

bromine (added slowly) (1)gives a white precipitate (1) [2]

Total [20]

Section B Total [40]

23

CHEMISTRY CH5

Section A

1. (a) (i) Rate = k[O3]2 (1)

Order (2) is the power to which the concentration (of O3) israised (1) [2]

(ii) Rate = 3.4 × 10-5 × (0.023)2 (1)

= 1.8 × 10-8 mol dm-3 s-1 (1) [2]

(Mark consequentially to (i))

(iii) Catalysts provide a different route (1)

of lower activation energy (1) [2]

(iv) Causes skin cancer / damages eyesight / lets in (more) harmfulUV radiation [1]

(b) (i) I. 8 days [1]

II. Tangent at 6 days = 5.5

Tangent at 12 days = 2.75 (1)

As concentration doubles, rate doubles. Therefore firstorder (1)

[2]

(Or compares at least two half-life values (1)

to show that half life is constant. (1))

(ii) I. 5570 × 3 = 16710 years [1]

II. 14C 14N + β [1]

Total [12]

24

2. (a) (i) I2 + 2S2O32− 2I− + S4O6

2− [1]

(ii) I. Cl2 + 2e− 2Cl− [1]

II. S2O32− + 5H2O 2SO4

2−+ 10H+ + 8e− [1]

III. 4Cl2 + S2O32− + 5H2O 8Cl− + 2SO4

2− + 10H+

[1]

(iii) I. Equation A: S +2 to +2.5 (½)

Equation B: S +2 to +6 (½) [1]

II. Chlorine produces greater oxidation number (change) insulphur(therefore is more powerful oxidising agent) [1]

(b) Sodium chloride in food industry, sodium chlorate(V) as weedkiller [1]

(c) (i) Starch solution [1]

(ii) Moles 3232 1088.1

1000

4.20092.0OS −− ×=×= (1)

Moles I2 = 9.4 × 10-4

Moles ClO− = 9.4 × 10-4 (1)

(Moles ClO− in 250 cm3 solution = 9.4 ×10-3)

Moles ClO− in original bleach = 9.4 × 10-3 (1)

Concentration ClO− 376.0025.0

104.9 3

=×=−

mol dm 3− (1) [4]

(d) Add HNO3(aq) followed by AgNO3(aq) (½ each) (1)

White precipitate shows chloride, yellow precipitate shows iodide(½ each) (1) [2]

(Accept adding chlorine / bromine water no change if chloride brownsolution if iodide)

Total [14]

25

3. (a)]CO[H

][H]HCO[

32

3a

+−

=K [1]

(b) pH = −log[H+] [1]

(c)3

27

101.85

][H102.0910.54 −

+−−

××=× (1)

[H+] = 3.98 × 10-8 (1)

pH = −log 3.98 × 10-8 = 7.4 (1) [3]

(d) Keeps pH constant (when small amounts of acid or alkali are added). [1]

(e) Solution contains a large amount of CH3COOH and CH3COO− ions.

(Accept correct equations) (1)

When an acid is added, the CH3COO− ions react with the H+ ions,removing them from solution and keeping the pH constant. (1)

When an alkali is added the CH3COOH reacts with the OH− ions,removing them from solution and keeping the pH constant. (1) [3](Accept answer in terms of H+ ions reacting with OH− ions)

Total [9]

Total Section A [35]

26

Section B

4. (a) (i) 1s22s22p63s23p63d10 (1)

Cu+ has full 3d orbitals. (1) [2]

(ii) I. [Cu(H2O)6]2+ octahedral (1)

blue (1)

[CuCl4]2− tetrahedral (1)

yellow (accept green) [4]

II. Complexes consist of a Cu2+ ion surrounded byligands (1)that form coordinate bonds with it. (1) [2]

(b) (i) Enthalpy change when 1 mole of compound is formed from itselements in their standard states / under standard conditions. [1]

(ii) ∆Hf = ∆HatCu + I.E. Cu + ∆HatF2 + E.A. F + ∆Hlat form CuF2 (1)

Doubling value for forming 2F and 2F− (1)(These marks can be obtained from Born-Haber cycle)

∆Hf CuF2 = 339 + 2705 + 158 – 696 – 3037 (1)

∆Hf CuF2 = _531 kJ mol-1 [4]

(c) (i) HCl, 1 mol dm-3 [1](Accept H+)

(ii) Platinum [1]

(iii) Salt bridge. It allows charged species to flow between thesolutions / allows conductivity without mixing solutions /completes the circuit [1]

(iv) From beaker A to beaker B because the electrons flow to thepositive electrode. [1]

(v) emf = 0.77 – 0.34 = 0.43 V [1]

(d) Iodide (1)

Only one with more negative / less positive standard potential than Fe3+,Fe2+ half-cell (1) [2]

(2nd mark can be obtained from calculation values)

Total [20]

27

Q. 5(a)(i) is to be assessed for the quality of Written Communication.Candidates must satisfy the following criteria to be awarded the marksdesignated as [L].• Selection of a form and style of writing which is appropriate to this

question.• Organisation of the relevant information clearly and coherently.• Use of legible text with accurate spelling, grammar and punctuation.

5. (a) Diamond shows giant covalent (macromolecular)structure. (1)In diamond each carbon atom is joined to four others. The four outerelectrons are all involved in bonding (1) giving tetrahedral structure. (1)In lead there is a giant metallic structure (1) with a regular array of metalions surrounded by a ‘sea’ of delocalised (valence) electrons. (1)The bonding force is the attraction between the positive ions and thedelocalised electrons. (1)Therefore carbon at the top of the group is a non metal, while lead at thebottom of the group is a metal. (1) [5](Accept any 5 points)

(b) Oxidation state II becomes more stable (1)CO2 more stable than CO (½), but PbO more stable than PbO2 (½)Inert pair effect – the two s electrons become more stable as the group isdescended. (1) [3]

(c) (They react by losing an electron.) As group is descended it is easier tolose an electron / ionisation energy decreases (1)since effective nuclear charge decreases / ionisation energy decreases /there is less attraction to the nucleus due to increased shielding (acceptfurther distance from the nucleus.) (1) [2]

(d) White solid disappearing / colourless solution / bubbles (1)pH = 11 – 14 (1) [2]

(e) From ionic to covalent / becomes more covalent across period 3. [1]

(f) NaCl Na+(aq) + Cl−(aq) (1)

pH = 7 (1)

PCl5 + 4H2O H3PO4 + 5HCl (1)

pH = 0 – 3 (1)

Ionic chlorides dissolve in water, covalent chlorides react with water (1) [5]

(g) e.g., Carbon dioxide / phosphorus(V) oxide / sulphur dioxide (acceptformula) (1)Na2CO3 / Na3PO4 / Na2SO3 (1) [2]

Total [20]

Total Section B [40]

28

CHEMISTRY CH6a

Section A

1. (a) (i) 206Pb [1]

(ii) 800 → 400 → 200 → 100 → 50 4 half lives

∴ 4 × 138 = 552 days [1]

(b) (i) Sodium nitrate melts (½) bubbles (½) white solid (½) produced,which is sodium nitrite (½)

Calcium nitrate brown gas (½) which is nitrogen dioxide (½)white solid produced (½) which is calcium oxide(½) condensation (½) bubbles (½)

Accept formulae from equations

maximum [4]

(ii) I UV light decomposes CFCs (½) to give chlorineatoms (½) [1]

II The C-Cl bond is the weakest and this is broken (by UV) [1]

(iii) moles of ozone = 3101.224000

28.8 −×= (1)

moles of compound Z = 4.0 × 10-4

mole ratio 3 : 1 (½)

∴ 3 double bonds in compound Z (½) [2]

Section A Total [10]

29

Section B

2. (a) Ti Cl4 + 4Na → Ti + 4Na Cl

+4 4(−1) 0 0 4(+1) 4(−1)

titanium has changed from +4 to 0 reduction (1)

sodium has changed from 0 to +1 oxidation (1) [2]

(b) (i) 1s 2s 2p 3s 3p 3d 4s [1]

(ii) the (water) ligands split the (3) d orbitals (into higher and lowerlevels) (½)

energy

(½)

the 3d electron (in the ground state) is in the lower level (½)

this absorbs energy from (white) light and is excited to the higherlevel (½)

the energy for this is in the red region of the (visible) spectrum (½)and violet light / remaining light is transmitted / reflected (½) [3]

(c) the purple MnO4− ions become colourless / pale pink Mn2+ ions [1]

(d) Ti has 4 bonding pairs of electrons (and no lone pairs) (½)position of minimum repulsion adopted (½)∴ tetrahedral molecule / correct diagram (1) [2]

(e) e.g., sulphuric acid in esterification [1]

(f) TiCl4 + 2H2O → TiO2 + 4HCl

accept TiCl4 + 4H2O → Ti(OH)4 + 4HCl [1]

30

(g) Colorimeter reading 0.96 ≡ 0.064 g dm-3 (diluted solution) (1)

dilution factor 1:20 (½) 20 × 0.064 = 1.28 g dm-3 (½) (undiluted)

% of titanium in the sample = 1.901.42

1001.28 =×(accept 90) (1) [3]

(h) orange / yellow precipitate [1]

Section B Total [15]

31

Section C

3. (a) (i) Reagent e.g., (aqueous) bromine / acidified potassiummanganate(VII) [1]

Observation yellow brown to colourless / purple to colourless [1]

(ii) I.

C C C C C C C C

H

H

H H

H

H

H

H

H

H

H

H

H

OH

H

H

H

H

*

[1]

II. A racemic mixture contains equimolar quantities of eachenantiomer / optical isomer (1) these would not affect theplane of polarised light (as external compensationoccurs) (1) [2]

(b) ‘Amphoteric’ means that it can behave both as an acid and as a base /dissolves or reacts with both an acid and an alkali. [1]

eg. Pb(OH)2 + 2HCl → PbCl2 + 2H2O [1]

Pb(OH)2 + 2NaOH → ‘NaPbO2’ + 2H2O [1]

(c) (i) Phenylamine is reacted with nitrous acid at < 10 oC (½), to givea (benzene)diazonium compound (½). This is then coupledwith a phenol (½) in alkaline solution (½) to produce the dye. [2]

(ii) Reagent Iron(III) chloride solution [1]Observation Purple solution [1]

(d) Amaranth becauseλ1

f ∝ and therefore smaller wavelength has a higher

frequency [1]

Total [13]

32

4. (a) Hydrogen iodide acts as a reducing agent and reduces sulphuric acid tolower oxidation state sulphur compounds. [1]

Q 4 b (ii) is to be assessed for Quality of Written Communication. [L]Candidates must satisfy the following criteria

• Selection of a form and style of writing which is appropriate to this question• Organisation of the relevant information clearly and coherently• Use of legible text with adequate spelling, punctuation and grammar

(b) (i)

222

[HI]

]I][[H=cK (1) no units (1) [2]

(ii) The concentration of hydrogen iodide is decreasing (1)The concentrations of hydrogen / iodine are (initially zero) and areincreasing (1)

Rate of reaction HI → H2 / I2 is fast at the start (1)The rate of decomposition of hydrogen iodide is slowing (1)Shown by a less negative gradient / angle of slope decreasing (1)The gradient / slope of the hydrogen iodide system is also decreasing (1)

At (dynamic) equilibriumRate of the forward reaction = Rate of the backward reaction (1)The concentration of hydrogen iodide, hydrogen and iodine remainconstant (1)[HI] > [H2] or [I2] (1)[HI] = 2 [H2] / [I2] (1)

maximum [6]

33

(c) either

H2(g) + I2(s) → 2HI(g) ∆H = +50.2 kJ mol-1

2HI(g) → H2(g) + I2(g) ∆H = +12.1 kJ mol-1

Add ∴ I2(s) → I2(g) ∆H = +62.3 kJ mol-1

method (1) answer (1)

or

H2(g) + I2(s) ∆H H2(g) + I2(g)

+50.2 -12.1

2HI(g)

∆H = 50.2 (–12.1) = +62.3 kJ mol-1

method (1) answer (1) [2]

(d) In ethane all the protons are in equivalent environments, therefore a single line/ singlet is seen [1]

Total [12]

Section C Total [25]

GCE Chemistry Marking Scheme (Summer 2008)/LG

WJEC245 Western AvenueCardiff CF5 2YXTel No 029 2026 5000Fax 029 2057 5994E-mail: exams@wjec.co.ukwebsite: www.wjec.co.uk/exams.html

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