Chemarkschemejan07 -asa2

Mark Scheme January 2007

GCE

GCE Chemistry (8080/9080) [International]

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January 2007 Publications Code xxxx All the material in this publication is copyright © Edexcel Ltd 2006

2

Contents Unit 6241/01

page 1

Unit 6242/01

page 14

Unit 6243/01A (Practical Test)

page 27

Unit 6243/02

page 36

Unit 6244/01

page 46

Unit 6245/01

Page 66

Unit 6246/01A

Page 81

Unit 6246/02

page 90

i

ii

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6241/01 1

(a)

Neutrons 24 12Mg 26 12Mg 24 12Mg2+

Words or numbers

Electrons 12

14 10 3 marks

1 mark each number (b) 1s

2s

2p ↑↓

(i) Mg

1s

2s

↑↓

2p ↑↓

(ii) Cl

↑↓

3s

↑↓

3p

↑↓ 3s

↑↓

↑↓

3p ↑↓

↑↓

↑

Both arrows up or both down

Arrows can be

Numbers

2 marks

NOTES:

1

ACCEPT

EXPECTED ANSWER (c)

Mg(s) + Cl2(g) → MgCl2(s)

Multiples Mg2+(Cl-)2(s)

Formulae (1) State symbols (1) – only if formulae correct or near miss for MgCl2 (e.g. MgCl/Mg2Cl) (d)

(e)

REJECT

MARK

“Mg2++ 2Cl−” for MgCl2 (0 mark) 2 marks

(56.25x70) + (37.50 x 72) + (6.25x74) (1) 100

Use of Ar (0 mark)

= 71 (1) Any unit max 1 2nd mark consequential on fraction provided 70, 72 and 74 used

Answer ≥ 2 SF

4.73 moles (1) 71

Consequential if wrong answer to (d) used.

Just “71” with no working (0 mark) 2 marks

71 used when (d) incorrect

X 30.6 = 2.04 dm3 (1) Answer with no working 1 max

Answer ≥ 2 SF

NOTES:

2

No or incorrect unit of volume (loses 1 mark)

2 marks

ACCEPT

EXPECTED ANSWER

(f)

REJECT

MARK

Type - Metallic(1) Attraction between Mg2+ (1)

Cations/positive ions /magnesium ions

atoms/nuclei/ions “force between” if used instead of “attraction”

Diagram without brackets

Any suggestion of electrons being shared

3 marks

And (surrounding) sea of electrons/delocalised electrons (1)

(g)

Stand alone Ionic (1) oo o o Mg o o oo

2+

2

oo o o o Cl o oo

Mg with no electrons shown ie [Mg]2+

OR

oo o o o Cl o oo

oo o o Mg o o oo

2+

[Mg• ]+

oo o o o Cl o oo

Correct charges and number of ions (1) Correct electronic structures (1) Stand alone

3 marks Total 17 marks

NOTES:

3

ACCEPT

EXPECTED ANSWER

2

(a)

Penalise lack of nucleus/atom once only in (i) and (ii) Penalise use of element each time it occurs (i) The number of protons in the nucleus of an atom (of an element) OR The number of protons in an atom/nucleus (of an element)

REJECT

MARK

the number of protons in an element the number of protons 1 mark

(ii)

(b)

(i)

The number of protons plus the number of neutrons in the nucleus (of an atom) OR number of nucleons (in the nucleus of an atom)

“and neutrons” instead of “plus the number of……”

The number of protons plus neutrons in an element 1 mark

C and L (1) (Group 0 elements) have the highest (first) ionisation energy (of each period) (1)

implied e.g. immediately precedes large drop

high first I.E. filled shell

Stand alone

Group 0 elements have the highest peaks on the graph highest (effective) nuclear charge/highest number of protons in period

NOTES:

4

smallest atom same shielding

2 marks

ACCEPT

EXPECTED ANSWER

(ii)

REJECT

MARK

F (1) Third after noble gas/C (1)

first mini dip after big drop lowest after Group 1/D in same period (1)

Or first element in period with p electron (1)

just “electron in p orbital” just “s orbitals shield p” 2 marks

(In F, e- removed from) p orbital is at a higher energy level than s orbital (in E) Number of protons/atomic number

(iii) Increase in (effective) nuclear charge (1)

Same distance from nucleus

Same shielding OR same number of electrons in inner shell/orbitals (1)

increased size of nucleus

Stand alone

Same number of shells

2 marks

Electrons in same shell Total 8 marks NOTES:

5

ACCEPT

EXPECTED ANSWER

3

(a)

(i)

2Na + O2 →Na2O2 IGNORE state symbols

4Na + O2  2Na2O or

REJECT

NaO 1 mark

multiples (ii)

Ba + 2H2O →Ba(OH)2 + H2 IGNORE state symbols (iii) NaCl + H2SO4 → NaHSO4 + HCl OR 2NaCl + H2SO4 → Na2SO4 + 2HCl IGNORE state symbols (b)

(i)

(ii)

MARK

multiples

Ba + H2O → BaO + H2

multiples

HNaSO4

1 mark

1 mark

Green/pale green/apple green

yellow-green

Red

deep/dark red / carmine/crimson /scarlet

NOTES:

6

Any mention of blue e.g. blue green OR Any other colour

1 mark

Lilac Any mention of lilac e.g. lilac-red OR any other colour

1 mark

ACCEPT

EXPECTED ANSWER

(c)

Electrons (absorb heat/energy) and are promoted to higher energy levels (1)

“excited/go” instead of “promoted”

as they drop back/down (1)

“orbitals/shells” instead of “energy levels”

Emit radiation (of characteristic colour) OR emit light (1) (d)

If any reference to absorption spectra e.g. light absorbed (0)

3 marks Use of atomic numbers 2 max use of “O2” Mr ~ 32 but only if give formula KO2 (for 3 marks)

e.g. Percentage of oxygen = 45.1 (1) O 45.1 16 2.82

MARK

(produce) colours (0)

Percentage oxygen (=45.1) (1) ÷ Ar (1) Empirical formula = KO2 (1)

K 54.9 39 1.41

REJECT

Mole calculation – then inverted, no consequential marking on formula

(1)

3 marks

KO2 (1)

NOTES:

7

ACCEPT

EXPECTED ANSWER

(e)

Sigma: end on overlap between s and s OR s and p OR p and p orbitals

or

REJECT

MARK

or

Overlap of hybrid orbitals for

1 mark

p Pi :sideways overlap between p and p orbitals

One or both explanations wrong but correct diagrams (or vice versa) (1)

1 mark Total 13 marks NOTES:

8

ACCEPT

EXPECTED ANSWER

4

(a)

(i)

Minimum of one shaded blob and one clear blob labelled (1) Labels are: Na+ or sodium ion and Cl− or chloride ion

REJECT

Na and Cl (ie no charge) sodium chlorine

(ii)

MARK

Strong (force of) attraction between (oppositely charged) ions (1)

Held together by strong ionic forces/bonds “attraction” may be implied by “breaking bonds”

a lot of energy needed to separate ions (1)

a lot of energy implies “strong”

Any reference to atoms or molecules Or covalent bonds Or intermolecular forces Or metallic bonds (scores zero)

1 mark

2 marks

All the bonds need to be broken

break ionic bonds break lattice (b)

Giant covalent delocalised e−

Covalent between carbon atoms in plane (1) Van der Waals’ between planes of carbon atoms (1)

Induced dipole/ dispersion/ London forces/temporary dipoles

Names not linked to bonds (max 1) NOTES:

9

2 marks

ACCEPT

EXPECTED ANSWER

(c)

Covalent Label not needed

(d)

Covalent bonds in diamond are shorter than the distance between layers in graphite (1)

REJECT

Giant covalent BUT do not penalise twice

The atoms in diamond are packed closer together (1)

MARK

1 mark

Layers in graphite are far apart (1) 2 marks Total 8 marks

NOTES:

10

ACCEPT

EXPECTED ANSWER

5

(a)

(b)

HF

hydrogen bonding /H bonding (1)

HCl HBr HI

van der Waals’

REJECT

MARK

just “hydrogen”

(1) – all three needed

(The boiling temperature of HF is higher) because the hydrogen bonding between HF molecules is stronger than the intermolecular forces in HCl (1)

Induced dipole/ dispersion/ London/temporary dipole forces any combination H bonding strongest/strong

dipole-dipole

2 marks

Any mention of ions, ionic bonds or covalent bonds (scores 0)

The rise from HCl to HI is because the strength of the van der Waals’ forces (etc) increases (1) Bigger mass/size for 3rd mark

with increase in number of electrons (1)

3 marks Total 5 marks

NOTES:

11

ACCEPT

EXPECTED ANSWER

6

(a)

(i) (ii)

2ClO- + 4H+ + 2e(−) →Cl2 + 2H2O (1) 2Cl− → Cl2 + 2e(−) (1)

Any multiples Any multiples

(b)

ClO- + 2H+ + Cl− → Cl2 + H2O (1) – stand alone

Any multiples

REJECT

1 mark 1 mark

1 mark

not consequential on wrong equation in (a) (c)

MARK

(i)

Cl2 + 2Br− → 2Cl− + Br2 (1) Ignore states

Any multiples

(ii)

Oxidising agent Ignore “displaces” Mark independently of (c)(i)

To oxidise bromide (ions)

1 mark

just “oxidation” 1 mark

NOTES:

12

ACCEPT

EXPECTED ANSWER

(d)

Moles of BCl3 =

Any alternative method e.g 1 mol BCl3 reacts with 3 mol H2O (1) 117.5 g BCl3 reacts with 54 g H2O (1) 12.3 g BCl3 reacts with 54 x 12.3 g H2O 117.5 = 5.65 g (1) Answer ≥ 2 SF

12.3 mol (1) = 0.1046/0.105 117.5

amount of water = 3 x moles BCl3 (1) = 0.3154/0.315 Mass of H2O = moles H2O x 18 = 5.65/5.67(g) (1) Answer = 5.4 (g) or 5 (g)

- from rounding to 1 s.f. max 2

Correct answer with some working scores 3 marks Mass H2O only (1 max)

(e)

REJECT

Hydrogen ions/H+/H3O+/oxonium ions formed (from HCl and H3BO3)(1)

MARK

3 marks

presence of/contains H+ ions

HCl/H3BO3 is an acid

Hydroxonium ions

H+ ions from water just “H+ ions”

1 mark Total 9 marks

NOTES:

13

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6242/01 1.

(a)

(i)

H

H

H

H

H

H C

C

C

C

C H

H

H

H C H H C H

H

C C C

C

C C C

C

H

H

(ii)

CH3 in branches But do not allow bond directly to H i.e.

(1 mark)

CH3

H Br H

H Br H

H C C C H

H C C C H

H O H

H OH H

H

Bond pointing directly to H in OH i.e.

C HO Hs missing from carbons i.e.

Br C C C OH (b)

90 ° bond angles e.g

Isomer 1

H C C

C C

CH3 H

C2H5

OR C C

(1)

Isomer 2

ACCEPT CH3

C2H5 C C H

(1 mark)

CH3 H

C C2H5

(1)

C

14

(2 marks)

ACCEPT

EXPECTED ANSWER If incorrect alkene eg but-2-ene, allow (1) for both cis and trans isomers

15

REJECT

MARK

ACCEPT

EXPECTED ANSWER

(c)

(i)

(ii)

REJECT

MARK

Any answer containing: “free radical” “electrophile”

Nucleophile OR nucleophilic reagent IGNORE type of reaction e.g. substitution addition (free) radical IGNORE type of reaction e.g. substitution addition

(1 mark) Any answer containing: “nucleophile” “electrophile” (1 mark)

(iii) Oxidising agent OR oxidant

oxidises ethanol/alcohol

“oxidation” on its own “reduced by ethanol” any answer containing “electrophile” “nucleophile” “free radical”

(1 mark)

Total 7 marks

16

ACCEPT

EXPECTED ANSWER

2.

(a)

REJECT

Enthalpy/heat/energy change for one mole of a compound/substance/ a product (1) NOT solid/molecule/species/element

“heat released or heat required” unless both mentioned

to be formed from its elements in their standard states (1) ALLOW normal physical state if linked to standard conditions

“natural state” “most stable state”

standard conditions of 1 atm pressure and a stated temperature (298 K) (1)

“room temperature and pressure”

MARK

(3 marks)

“under standard conditions” (b)

(i)

Bonds broken Bonds made N≡N (+)945 6N-H (−)2346 and 3H-H (+)1308 (1) (+)2253 ∆H = 945 +1308 − 2346 = − 93 sign and value (1) ∆H Ο = − 93 = − 46.5 (kJ mol-1) sign and value q on 3rd mark (1) 2

(1)

− 46.5 (kJ mol-1) with working (4) + 46.5 with working max (3) +93 with working max (2) (4 marks)

17

ACCEPT

EXPECTED ANSWER (ii)

REJECT

N2 + (3)H2 Enthalpy

-46.5

H OR

-93

(2)NH3 “Reactants” and “Products” as labels

Correct labelled levels (1) double headed arrow

∆H labelled (1) direction of arrow must agree with thermicity Diagram marks cq on sign and value of ∆H in (b)(i) IGNORE activation energy humps

(2 marks)

(iii) 350-500 oC (1)

any temperature or range within this range favours endothermic reaction more than exothermic so lower yield

higher temperature gives higher rate (1) but a lower yield because reaction is exothermic (1)

(iv)

MARK

OR Lower temperature give higher yield because reaction is exothermic (1) but rate is slower (1) Iron / Fe (1) IGNORE any promoters no effect on yield (1)

(3 marks) cq on sign of ∆Hf in (b)(i) or levels in (ii)

Lower temp favours exothermic reaction

(2 marks)

18

ACCEPT

EXPECTED ANSWER

(v)

(c)

(i)

(ii)

REJECT

rate too slow without catalyst to lower activation at a temp giving a reasonable energy of reaction yield

temp would have to be much higher for a reasonable rate then yield would be too low “lower activation energy” implies reasonable rate OR Allows reaction at a lower temp at a reasonable/fast rate giving a reasonable yield. advantage higher (equilibrium) yield/more NH3 in equilibrium mixture/equilibrium shifts to right (1)

MARK

(1 mark)

Just “more ammonia”

because smaller number of (gaseous) moles/molecules on rhs (1) IGNORE any reference to change in rate disadvantage (plant more) expensive because thicker pipes would be needed

(2 marks)

Stronger or withstand high pressure for thicker Vessel/container/plant /equipment/reaction vessels for pipes

“just more expensive” “just thicker pipes etc” apparatus (1 mark)

OR cost (of energy) for compressing the gases/cost of pump OR Cost of equipment/pressure not justified by higher yield Total 18 marks

19

ACCEPT

EXPECTED ANSWER

3.

(a)

MARK

Step 1 NaOH/KOH/sodium hydroxide/potassium hydroxide (1) ethanol and heat/reflux/heat under reflux/boil/warm (1) condition dependent on correct reagent or hydroxide

(b)

REJECT

Ethanolic/alcoholic/alcohol/ ethanol solution for ethanol

aqueous ethanol

(2 marks)

Step 2 H2 / hydrogen (1) Ni / nickel and heat OR Pt/Pd/platinum/palladium IGNORE reference to heat (1)

(2 marks)

20

ACCEPT

EXPECTED ANSWER

4.

(a)

aluminium oxide/alumina/Al2O3 dissolved in (1)

REJECT

bauxite

molten cryolite or cryolite at temp≥ 800 oC (1)

(2 marks)

(b)

Al3+ + 3e(-0 → Al

(c)

graphite

carbon /C

(d)

C+O2 → CO2

Multiples or half

(e)

MARK

OR 2C +O2 → 2CO OR C + 202- →C02 + 4eOR C +02- → CO + 2emol Al = 1 x106 = 3.7 x 104 (1) 27

(aq) as state symbol

(1 mark)

charcoal

(1 mark)

(1 mark)

mol Al2O3 = ½ mol Al (1) mass Al2O3 = mol x 102 = 1.9 x106 g / 1.9t(1) value and unit required. If atomic numbers used max 2 If mol Al2 = 1x106 (0) 54 mol Al2O3 = mol Al2 (1) mass Al2O3 = 1.9 t (1) OR 54 g Al made from 102 g Al2O3 (1) 1g Al made from 102 = 1.9 g (1) 54 1 t Al made from 1.9 t / 1.9x106 g (1)

(3 marks)

IGNORE s.f. 21

ACCEPT

EXPECTED ANSWER

(f)

(i)

(energy) to keep the electrolyte/alumina molten (1) OR to produce heat energy to maintain temp 800-10000C (1)

(ii)

no (electricity needed for) electrolysis (1) energy only needed to melt A1 (1) OR Low melting point of Al (1) Compared to high melting point/8000C-1000C for electrolyte (1) OR No (electricity needed for) electrolysis (1) Low melting point of Al (1)

REJECT

MARK

(1 mark)

to keep aluminium molten to melt/heat the electrolyte

Cryolite for electrolyte

Bauxite aluminium oxide for electrolyte

(2 marks)

Purification of bauxite not needed Total 11 marks

22

ACCEPT

EXPECTED ANSWER

5.

(a)

(i)

(ii) (b)

(i)

(ii)

(C2H6 + Br2) → C2H5Br + HBr OR multiple substitution e.g. C2H6 + 2Br2 → C2H4Br2 / CH3CHBr2/CH2BrCH2Br + 2HBr C2H6 + 3Br2 → C3H3Br3 + 3HBr etc (C2H4 + Br2) → CH2BrCH2Br

CH3C H2Br or full structural formula

REJECT

MARK

C2H6+3Br2 → 2C +6HBr (1 mark) C2H4Br2

ethane C− H bond and ethene C=C bond (1) ALLOW carbon-carbon if double in type of bond ethane type: σ/sigma and ethene type: π/pi (1) OR mark horizontally

Reject σ and π for ethene

π/pi bond is weaker (than the σ/sigma bond)

π/pi bond requires less energy to break OR π/pi bond has lower bond enthalpy

OR π/pi bond has higher electron density (than the σ/sigma bond)

(1 mark)

(2 marks)

π breaks more easily π bond is weak (1 mark)

π/pi bond has more accessible electron density Total 5 marks

23

ACCEPT

EXPECTED ANSWER

6

REJECT

MARK

(a)

(i)

Axes labelled (1) Y: number/fraction of molecules/particles (with energy E) and X: (kinetic) energy Correct shape (1) starting at origin, and asymptotic to x-axis and not symmetrical

(2 marks) (ii)

line TH with peak to the right of temp T and peak lower than temp T

(1 mark) 24

ACCEPT

EXPECTED ANSWER

REJECT

(iii) vertical line well to the right of both peaks (b)

(i)

MARK

(1 mark)

higher temp gives molecules higher (average kinetic) energy (1) more collisions per unit time

More collisions

so increase in frequency of collisions (1) area (under curve) to right of Ea greater at TH (1)

molecules/particles for collisions

more collisions have a greater energy ≥ Ea OR a greater proportion of collisions have energy ≥ Ea OR more of the collisions are successful OR a greater proportion of the collisions result in reaction /are successful (1) (ii)

“more successful collisions” “increase in frequency of successful collisions”

Energy of collisions

(4 marks)

(1 mark) Total 9 marks

25

ACCEPT

EXPECTED ANSWER

7.

(a)

(i)

alcohol/OH

hydroxyl

REJECT

Hydroxide/OH− Any additional functional group

(ii)

MARK

(1 mark)

full structural formulae

W (CH3)3COH (1)

(2 marks) IGNORE X (CH3)3CCl must be conseq on their W (1) (iii) Butanoic acid / CH3CHeCH2COOH but not if W is butan-1-ol OR (2) methylpropanoic acid/(CH3)2CHCOOH but not if W is 2-methylpropan-1-ol if name and formula given, both must be correct (b)

both isomers (1) CH3CH2CH2Br/ C2H5CH2Br

(1 mark)

full structural formulae and CH3CHBrCH3

H H H H C C C Br H H H H H H H C C C H H Br H

identification of 2-bromo as the major product (1) (2 marks)

26

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6243/01A 1.

(a)

Observation

lilac (1)

Mauve / purple

Other colours

K+ / potassium ion (1)

potassium

Just “K” on its own

(b)

Inference Ignore K Observation

(i)

(any) yellow precipitate (1)

(2 marks)

White / cream

insoluble in ammonia (1)

(ii)

Inference

(c)

Any brown Any yellow

Just “I- / iodide” on its own “Red” alone “Orange” alone

Blue-black (1) Ignore ppte

Black

purple

iodine / I2 (1)

Iodine / I2 from any equation

I- oxidised (to I2) (1)

Cl2 oxidises I- / displaces I2/I-

Inference AgI / Silver iodide (1) [may have inference from first observation alone] Observation brown (solution) (1) Ignore ppte

K+I−

KI only – stand alone mark

27

(3 marks)

(4 marks) Redox alone Cl- oxidises IOr Cl- displaces I2 Incorrect ionic charges

(1 mark) Total 10 marks

ACCEPT

EXPECTED ANSWER (a)

Table 1 Check subtractions and averaging arithmetic, correcting if necessary. All volumes recorded to 0.05 cm3 (1) ALLOW one slip but withhold this mark if any readings are in the wrong boxes. ALLOW 0 as initial volume NOT 50 as initial volume All subtractions correct (1)

[99top RHS of Table 1] Mean titre For correct averaging of chosen values / choosing identical values and for recording the average correct to 2 or 3 dps or to nearest 0.05 cm3 (1) Do not penalise lack of 2nd d.p. in mean if this has been penalised in Table 1. [9 by the mean in space or near the dotted line in paragraph below]

Accuracy If the candidate has made an arithmetical error in the Table 1 volumes used in the mean or in averaging the examiner must calculate a new average. • For an averaging error simply calculate a new value using the candidate’s chosen titres. • If a wrongly subtracted titre has been used in the mean then choose any two identical titres or take an average of the closest two titres.

28

REJECT

MARK

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Calculate the difference between the candidate’s mean titre and that of the examiner or supervisor. Record the difference on the scripts as d = Examiner’s titre = 24.20 cm3 (to be confirmed at standardisation) Award marks for accuracy as follows.

Difference +0.20 +0.30 +0.40 +0.60 +0.80 +1.00 d= Mark 6 5 4 3 2 1 Range Award a mark on the range of titres used by the candidate to calculate the mean. The range(r) is the difference between the outermost titres used to calculate the mean. If the examiner has corrected titres because of incorrect subtraction then award the range mark on the corrected titres used by the examiner to recalculate the mean Range of titres/cm3 r=

+0.20

+0.30

+0.50

Mark

3

2

1

Examiner to show the marks awarded for accuracy and range as d= value

9

6max

r = value

9 3 max

Then the mark out of 12 written in margin. (12 marks) 29

ACCEPT

EXPECTED ANSWER

REJECT

MARK

[Overseas scripts: examiner to write “SR = titre value” on each script] (b)

Penalise sf once only in (i)-(iv) (i)

Mean titre x 0.100 1000

Mark is for answer to ≥ 2sf.

Answer with no working.

(ii)

Answer to (i) 2

Mark is for answer to ≥ 2sf.

Answer with no working. (1 mark)

(iii)

Answer to (ii) x 1000 Mark is for answer to ≥ 2sf. 25

Answer with no working.

(1 mark)

(iv)

6.00 answer to (iii)

Answer with no working.

(1 mark)

(v)

Molar mass Na2CO3 = 106.0 or implied (1)

Mark is for answer to ≥ 2sf.

106

Answer to (iv) − 106.0 18 Expected answer x = 1.0 (1) (Note 1.0 not 10.0) Ignore S.F. (c)

(1 mark)

(2 marks) 1 / Fractional values

Will react with Na2CO3 / will neutralise Na2CO3

Just “more accurate” (1 mark) Total 19 marks

30

ACCEPT

EXPECTED ANSWER

3.

(a)

REJECT

MARK

Table 2 Both weighings recorded in correct spaces to at least 2 dp (1) Weighings correctly subtracted (1)

Mass of D used – allow loss of 2nd dp if O and both dp if both 0 [Ticks to rhs of Table 2] Table 3 Both temperatures recorded in correct spaces (1) Both to at least 1 dp (1) [Ticks to rhs of Table 3] ∆T negative value

∆T correctly calculated (1) The examiner ratio ∆T / mass A = 1.61 (to be confirmed at standardisation) e.g. 3.8g → 6.1OC increase in temperature. For the candidate calculate (mass A x examiner ratio) = expected ∆T Compare candidate’s expected ∆T with the actual ∆T and record the difference between the two as d = on the script. [∆T : ignore dp’s if zero] Award marks for accuracy as follows. Difference + 0.5 + 1.0 d= 94 93 Marks

+ 1.5

+ 2.0

92

91 (9 marks)

31

ACCEPT

EXPECTED ANSWER

(b)

(i)

(ii)

Mass D 106 IGNORE sf

REJECT

MARK

Answer with no working. (1 mark)

50 x 4.18 x ∆T OR 50 x 4.18 x ∆T 1000 Units must be given and must be correct; J or kJ IGNORE sf IGNORE sign both of ∆T and answer

(1 mark)

(iii) Answer to (ii) (1) Answer to (i)

(iv)

Consequential answer in kJ mol-1, sign(negative only) and 2 significant figures (1) [Units need not be written but answer must be in kJ mol-1 All Na2CO3 dissolves with effervescence.

(2 marks) All Na2CO3 dissolves (1 mark) Total 14 marks

32

ACCEPT

EXPECTED ANSWER

4.

(a)

Add hydrochloric acid/HCl(aq)/dilute HCl(1)

Add nitric acid/HNO3(aq)/ dilute HNO3

Observation

REJECT

MARK

“add hydrochloric acid until carbonate destroyed” “HCl”

“until hydrochloric acid in excess”

until no more effervescence (1) Add barium chloride solution to give white precipitate. (1) OR Add barium chloride solution to give white ppte (1)

Add nitric acid/HNO3(aq)/ dilute HNO3

Then add hydrochloric acid/HCl(aq)/dilute HCl (1)

(3 marks)

Observation no more fizzing + white ppte (remains)(1)

33

ACCEPT

EXPECTED ANSWER

(b)

REJECT

MARK

Heat (1) Cool and weigh (1) Repeat (1) allow lack of “cool” here Until constant mass (1) IGNORE any initial weighing OR Heat (1) Methods of collecting CO2 / accurate diagram/method of detecting CO2 e.g. limewater/observing bubbles (1) Measure volume CO2 (1) Heat until constant volume/no more bubbles/limewater not cloudy (1) [If solution at end max (3)]

(4 marks) (Total 7 marks)

34

35

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6243/02 1

(a)

Lilac (flame/colour) Ignore any references to blue glass

(b)

Add nitric acid/HNO3 and silver nitrate (solution)/AgNO3(aq)

(c)

Sulphur dioxide/SO2 (1)

mauve/purple

given in either order If put these two in and then add ammonia allow

HSO3−/hydrogensulphite

Sulphite / sulphate(IV) / SO32-(1)

(d)

White precipitate/solid/suspension

(e)

Aluminium/Al/Devarda’s Alloy (1) ignore any references to foil or powder or turnings and sodium hydroxide (solution)/NaOH((aq))/KOH((aq))(1)

(f)

Any other colour on its own or in combination with lilac

(1 mark) (1 mark)

Error carried forward e.g CO2 Goes milky/cloudy

(2 marks) (1 mark)

given in any order (2 marks)

(red then) bleached/goes white/ goes colourless

(1 mark) Total 8 marks

36

ACCEPT

EXPECTED ANSWER

2

REJECT

MARK

Notes: candidates may achieve answer with two tests. If they carry out more than two tests penalise those tests that are wrong If minor error in the test allow correct observation e.g inaccuracy in formula Test Observation Gas evolved that turns limewater esterification i.e. P Any carbonate or Group cloudy(1) Test: Add alcohol + conc. 1 hydrogencarbonate as H2SO4 (1) solid or in solution OR Obs: smell (1) correct formula for above including anions Fizzing/effervescence/ ions bubbles OR

P

Add magnesium (1)

Gas evolved burns (with ‘pop’) (1)

OR

P

add blue litmus paper add litmus solution add pH indicator paper add universal indicator paper or solution Use pH meter (1)

(Blue litmus) goes red (1) goes red goes yellow/orange/red

Q

Add bromine water(1) Bromine in non-aqueous solvent or stated e.g hexane

(Brown/red-brown/orange solution) decolourised/goes colourless (1)

esterification i.e. Test: Add carboxylic acid + conc. H2SO4 (1) Obs: smell (1)

Q

Add (acidified /alkaline) potassium manganate(VII)/permang anate OR Add(neutral) solution of

(Purple solution goes ) colourless if acidified green if alkaline

If not specified as acidic or alkaline or neutral, accept colourless or brown (ppt) for observation

OR

Fizzing/effervescence/ bubbles

Gas evolved

(2 marks)

pH below 6

37

Add PCl5 Bromine

ACCEPT

EXPECTED ANSWER potassium manganate(VII)/permang anate(1) OR Heat with acidified /H+ dichromate Cr2O72- / CrO42- (solution)

REJECT

MARK

(2 marks) brown (ppt) if neutral (1)

goes green/blue Dichromate paper

Note: If more than one test is given and both tests are reactions that P and Q would show but one test does not distinguish between P and Q, allow 1 mark. Total 4 marks

38

ACCEPT

EXPECTED ANSWER

3

Magnesium ions/ Mg2+ magnesium compound / contains magnesium OR NOT Ca2+ Ba2+ Sr2+ (ALL THREE) or Ca2+ Ba2+ Sr2+ ABSENT Iodine produced /contains iodine OR Z is an Iodide /iodide ions /I-

Be ions/Be2+ / beryllium compound / contains beryllium

MgI2 Consequential marking: Allow cq on metal stated provided it is Group 2 and not barium No cq on halide

BeI2 if Be2+

REJECT

Mg+

MARK

(1 mark)

bromide iodine ions (1 mark) Name e.g. magnesium iodide

(1 mark)

Total 3 marks

39

ACCEPT

EXPECTED ANSWER

4

(a)

(i)

REJECT

MARK

How it works (Liquid boils and) gas/vapour is condensed (in condenser and runs back) (1) Why it is used Reaction slow /reaction has high activation energy /increase rate / for more time/to enable reactants to be heated for a prolonged period (1)

(3 marks)

When using volatile liquids/ to prevent loss of materials / to prevent escape of reactants (and products)/ to minimise loss of reactants (and products)(1) (ii)

Apparatus Flask properly drawn and thermometer and heat (1) Condenser properly drawn with water jacket with correct water flow(1) Set up Top of still head closed and collection end open Thermometer at correct point in neck (still head) Condenser at angle (1) ALL THREE for 1 mark

(3 marks)

Ignore any attempts to draw a fractionation column and a dropping funnel in a side arm. (b)

Use a water bath/electric heater/electric hot plate/sand bath/ oil bath Ignore Keep away from naked flame / use a fume cupboard

Do not use a Bunsen (unless qualified with what should be used)

(1 mark)

Total 7 marks 40

ACCEPT

EXPECTED ANSWER

5

(a)

(from) colourless (to) pink

pale red

REJECT

MARK

(From) clear to……. OR ….to magenta/ purple/cerise (1 mark)

(b)

The first titre is outside the 0.2 (cm3)limit usually set for volumetric analysis

not concordant closest

Very similar

(1 mark)

OR

(c)

the first titre is rough/trial 23.40 cm3

(d)

(i)

(25.00 x 0.110) = 0.00275 mol / 2.75 x 10-3 (1) (1000)

(23.40 x 0.235) = 0.005499 mol / 5.499 x 10-3 (1) (1000) cq on (c) (iii) (Answer (ii) ) (1) (Answer (i) ) (ii)

(iv)

too far out /overshot 23.4

Not accurate

0.0028

0.003 0.0027

(1 mark)

(1 mark)

0.0055 (1 mark)

ie 0.005499 = 2 0.00275 Cq on (i) and (ii) used to at least 2 sig figs. 2 consequential on (iii) as long as rounded to interger and sensible ⋝ .8 rounded up ⋜ .2 rounded down

(1 mark) Allow mark if no answer to (iii) but (i) and (ii) are correct

41

Any number that is not an integer Any number > 4

(1 mark)

ACCEPT

EXPECTED ANSWER

(e)

=

C 32 12 2.67

H 4 1 4

O 64 16 4

(1)

1

1.5

1.5

(1)

2

3

3

REJECT

MARK

Calculation of percentages I II %C= (C = 48 x 100 0.32 × 150 = 4 ) 150 12 = 32 %H= (H = 6 x 100 0.04 × 150 = 6 ) 150 1 =4 %O= (O = 96 x 100 0.64 × 150 = 6 ) 150 16 = 64 All correct (4) 2 correct (2) 1 correct (1)

Empirical formula mass = 75 (1) x 2 (1) Molar formula C4H6O6

(4 marks)

Notes OR These are in the ratio 4 to 6 to 6 (1) which adds up to 150 molar mass(1) Molar formula C4H6O6 OR C1H1.5O1.5 = 37.5 (1) x 4 = 150 (1) so Molar formula C4H6O6

If calculation stops at C2H3O3 may be out of clip send to review. (f)

HO C CH(OH) O

CH(OH)

C

COOHCH(OH)CH(OH)COOH OR

OH

CO2H

HOOCCH(OH)CH(OH)COOH

O

OR HOOCC(OH)2CH2COOH OR 42

(1 mark)

ACCEPT

EXPECTED ANSWER

REJECT

MARK

COOHC(OH)2CH2COOH

(g)

First mark general statement about the larger volume means it is more accurate Second or both marks for justification based on data given First mark (Percentage/relative )error is less with large titre / error minimised/ reduces error(1)

(2 marks)

Second mark calculation (1) Calculation of percentage error e.g 0.05 x 100 = 0.67% 7.5 0.05 x 100 = 0.2% 25 If do both calculations correctly give both marks Total 14 marks

43

ACCEPT

EXPECTED ANSWER

6

(a)

MARK

IGNORE sig figs provided 2 or better in (i) and (ii) (i)

(ii)

∆T=26 °C (1) STAND ALONE Heat change =104x26.0x4.09=11060 J (1) ignore sign at this point The second mark may be appearing in part (ii)

If use 100g answer is 1063(4) And gives 355 as the final answer If use 4 g gives 425.2 and gives 14.2 kJ mol-1 0.03

Moles = 4.00 (1) = 0.02996 133.5 Answer in (i) x 1 moles 1000

(2 marks)

(1)

= -369 (kJ mol-1) (1)

-369 (kJ mol-1) with some -369000 kJ mol-1 does working (3) not score 3rd marking point -1 -369000 J mol (max 2)

Error carried forward if wrong Mr

(b)

REJECT

(3 marks)

Methods based on increasing insulation alone

Record temp of water at intervals add solid (and stir), continue recording temperature (1) (Plot)graph(1) this could be implied Extrapolate back to time of adding solute (to find actual temperature change) (1) Note An annotated sketch graph showing clear time intervals and temperature plots a vertical line at correct point and evidence at what point the solid was added score all three marks

(3 marks)

Total 8 marks 44

ACCEPT

EXPECTED ANSWER

7

Weigh crucible empty and with solid/ find mass of solid / take known mass of solid(1) Heat (,cool) and reweigh (1) Reheat and reweigh/ heat to constant weight / make sure no gas is being evolved (1)

REJECT

MARK

If say take equal amounts do not give first mark since this indicates a misunderstanding of the whole exercise.

to ensure reaction is complete (1) Compare ratio of mass produced (1) mass taken If 106 reaction I 168 If

40 84

reaction II

If

62 168

reaction III

OR

Mass taken (1) mass produced 168 reaction I etc 106

If

Compare ratio of mass of product with mass of reactant (1) If rxn I: mass of product = mass of NaHCO3 x 106 2 x 84 If rxn II: mass of product = mass of NaHCO3 x 40 (1) 84 If rxn III: mass of product = mass of NaHCO3 x 62 2x84

(1)

OR Calculate actual mass of product or mass lost based on a stated mass taken e.g 10 g gives 6.7 or less of 3.3 10g gives 4.8 or loss of 5.3 10g gives 3.7 or loss of 6.3 (1) for calculation Relate answer to which solid taken (1) Could measure volume of gas produced but it breaks down if temp not above 100 oC max 3 ( the first 3 marks) since this is not the question asked

(6 marks)

45

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6244/01 1.

IGNORE s.f. throughout this question (a) Acid Proton or H+ donor Or forms H+ or H3O+ (1) Weak dissociates to a small extent Or ionises to a small extent(1)

(b)

few molecules dissociate Or incomplete dissociation Or partial dissociation

2HCOOH(aq) + Na2CO3(aq) → 2HCOONa(aq) + CO2(g) + H2O(l) Or HCOOH(aq) + Na2CO3(aq) → HCOONa(aq) + NaHCO3(aq)

“not fully dissociated” Or “not dissociated fully” (2 marks)

… → 2HCOONa(aq) + H2CO3(aq) HCO2H for the acid HCO2Na or HCOO-Na+ for salt

Species + balancing (1) State symbols (1) Consequential on correct species

Notes:

46

(2 marks)

ACCEPT

EXPECTED ANSWER

(c)

(i)

REJECT

MARK

Correct acids and conjugate bases in either order ACCEPT HCO2H and HCO2− OR

one acid: HCOOH Conjugate base: HCOO− 1 mark for both

O HC

OH O

HC

O-

other acid: H3O+ Conjugate base: H2O 1 mark for both (ii)

H+ for H3O+ (2 marks)

(Ka) = [HCOO−][H3O+] [HCOOH]

[ H + ] instead of [ H 3 O + ]

Must use square brackets

[HCO2−]

(1 mark)

Notes:

47

and [HCO2H]

ACCEPT

EXPECTED ANSWER

(iii)

[H+]2 = Ka × [HCOOH] OR Ka =

REJECT

MARK

Any correct expression with [H+]2 or correct numbers

[H+]2 [HCOOH]

If [H+] = √(Ka×c) quoted Scores first two marks

OR [H+]2 = 1.60 × 10-4 × 0.100

(1) pH = 4.8 scores (2) as square root has not been taken

[ H + ] = 1.60 × 10 −4 × 0.100 = 4.0 × 10 −3 (mol dm −3 )

(1)

IGNORE sig figs Max 1 if [H+]2 expression incorrect pH = − log 10 [ H + ]

pH

= 2.40

any pH value consequential on [ H + ] , provided pH < 7

(1)

pH = 2.39 (is a rounding error) so no third mark

Alternative method

pKa = 3.80 (1) 1 1 pKa − log[acid ] 2 2 pH = 1.90 − (−0.50)

pH =

(1)

pH = 2.39 (is a rounding error) so no third mark

pH = 2.40 (1)

Notes:

48

(3 marks)

ACCEPT

EXPECTED ANSWER

(d)

(i)

[ H + ] = Ka ×

REJECT

MARK

[acid ] [ salt ]

OR 0.0500 0.200

(1)

= 4.00 × 10 −5 (mol dm −3 )

(1)

[ H + ] = 1.60 × 10 − 4 ×

pH = 4.40

(1)

0.100 0.400

IGNORE sig figs

4.39 (rounding error) so no third mark

OR

⎧[ HCOOH ] ⎫ pH = pKa − log 10 ⎨ (1) − ⎬ ⎩ [ HCOO ] ⎭ ⎧ 0.0500 ⎫ pH = − log 10 (1.60 × 10 − 4 ) − log10 ⎨ ⎬ ⎩ 0.200 ⎭

(1)

0.100 0.400

pH = 3.80 − (−0.60) pH = 4.40

(1)

IGNORE sig figs

4.39 (rounding error) so no third mark (3 marks)

Notes:

49

ACCEPT

EXPECTED ANSWER

(ii)

REJECT

MARK

Addition of H+ ions: HCOO − + H + → HCOOH

If described in terms of HA ⇌H+ + A− shifting to left

(1)

Addition of OH- ions: HCOOH + OH − → HCOO − + H 2 O

Addition of OH- ions: H + + OH − → H 2 O must be followed by more dissociation of HCOOH (to restore [ H + ] )

(1)

If the ionisation of sodium methanoate shown with ⇌ then max (1) out of 2 for above equations

“molecular” equations or equations described in words or notation involving HA, H+ and A− . Just “large reservoir of both HCOOH and HCOO−”

(buffer solution has) high concentrations Or a large reservoir of both HCOOH and HCOO − relative to added H + / OH − (1) (hence virtually no change in [H+] )

(3 marks)

Total 16 Marks

Notes:

50

ACCEPT

EXPECTED ANSWER

2

(a)

IGNORE s.f. throughout this question (i) moles SO2 (10.0 – 9.00) = 1.00 (mol) moles O2 (5.00 – 4.50) = 0.500 (mol) moles SO3 9.00 (mol) all 3 correct → (2) 2 correct → (1) (ii)

REJECT

MARK

Multiples of the stated moles (2 marks)

All three ÷ total number of moles (1) i.e.

Rounding to 1 sig fig

1.00 (= 0.0952) or 2/21 10.5 0.500 = (= 0.0476) or 1/21 10.5 9.00 = (= 0.857) or 18/21 or 6/7 10.5

X SO 2 = XO2 X SO 3

(1 mark)

Mark consequential on (a)(i)

Notes:

51

ACCEPT

EXPECTED ANSWER

(iii)

REJECT

MARK

All three × total pressure (1) i.e. 1.00 × 2.00 10.5 = 0.190(atm)

pSO 2 =

0.500 × 2.00 10.5 = 0.0952(atm)

pO 2 =

9.00 × 2.00 10.5 = 1.71(atm)

pSO3 =

or 4/21

or 2/21

or 36/21 or 12/7 (1 mark)

Mark consequential on (a)(ii) (iv)

Kp =

(1.71) 2 (0.190) 2 × (0.0952)

K p = 851 (1) atm −1 (1)

Answer with units and no working (2) “Correct answers” between 845 and 855 as this covers rounding up etc

Mark consequential on (a)(iii) and (a)(iv)

Notes:

52

Wrong units e.g. mol-1 dm3 (2 marks)

ACCEPT

EXPECTED ANSWER

(b)

(i)

(Kp) decreases

(ii)

(Kp decreases so) fraction / quotient

REJECT

MARK

(1 mark) Any Le Chatelier argument (this prevents access to 1st mark)

p 2 SO3 p 2 SO 2 × pO 2

has to decrease (to equal new kp) (1) so shifts to left hand side (1) – this mark only available if (b)(i) answer was kp decreases.

Shifts to right, even if answer to (b)(i) was kp increases

(as p SO3 decreases whereas p SO2 and p O increase) 2

(2 marks)

NOTES:

53

ACCEPT

EXPECTED ANSWER

(c)

(d)

REJECT

MARK

(i)

No effect/none/zero (effect)

(1 mark)

(ii)

Increases OR more SO3/more sulphur trioxide OR increases amount of SO3/sulphur trioxide

(1 mark)

(i)

No effect/none/zero (effect)

(1 mark)

(ii)

No effect/none/zero (effect)

(1mark) Total 13 marks

Notes:

54

ACCEPT

EXPECTED ANSWER

3

(a)

REJECT

MARK

Compound A

H H H

O

H C C C C H

H H H

OR a branched chain isomer

H H C H H

O

H C

C

H

H

C H

-CH3 as side chain (1)

Compound B

H O H H H C C C C H H

H H

(1)

Penalise “compressed” formula once only e.g. CH3CH2CH2CHO CH3COCH2CH3

-COH for aldehyde

(2 marks) (2 marks)

Notes:

55

ACCEPT

EXPECTED ANSWER

(b)

O2N H

H

CH3CH2CH2 C N N

REJECT

MARK

C3H7 OR C 2 H 5 CH 2 for CH 3 CH 2 CH 2

NO2

H C N

CH=N

linkage (1)

Remainder of the molecule

NO2 groups in wrong position for remainder of molecule mark

(1)

Lack of circle in benzene ring for second mark (2 marks)

Mark consequential on structure given for Compound A in (a). (c)

(i)

Iodoform Or “triodomethane”

triiodomethane (1) H CHI3

or

I

C I

CH3I

I

(2 marks)

(1)

Notes:

56

ACCEPT

EXPECTED ANSWER

(ii)

butan(-)2(-)ol IGNORE punctuation CH 3 CH (OH )C 2 H 5 or

H

REJECT

2(-)butanol

But-2-ol

Or iso-butanol

2-hydroxybutane

MARK

(1)

Or butane-2-ol

H

OH

H

H

C

C

C

C

H

H

H

H

(1)

H

Only penalise if bond is clearly shown pointing to H ie

OH

is OK

OH is wrong (2 marks)

(d)

(i)

(It is not) superimposable on its mirror image OWTTE

Just “four different groups on the same molecule”

Or Does not have a plane of symmetry

OR Just “ (has an) asymmetric C atom”

Or does not have a centre of symmetry (1)

Notes:

57

(1 mark)

ACCEPT

EXPECTED ANSWER

(ii)

REJECT

MARK

Rotate (the) plane (of plane ) polarised (monochromatic) light (equally) in opposite directions (1) OR pass polarised light through sample OR use a polarimeter rotates the plane (equally) in opposite directions (1)

(e)

H 3C

CH2OH C

90 ° formula

C

H

e.g. H

C C

and CH2OH

H C CH3

(1 mark) (1)

C C

C H

1 mark for each 2nd isomer cq if first isomer is carboxylic acid

cis and trans acid scores (1) consequentially

Notes:

58

(2 marks) Total 12 marks

ACCEPT

EXPECTED ANSWER

4

(a)

(i)

REJECT

Wrong halogen or use of “X” (0)

½Br2 → Br (1) state symbols (1) ½ Br2(g) → Br(g) scores only one

e.g. ½Br2(l)→ Br(g) (2) Br2(l)→ 2Br(g) (1) ie for state symbols Br2→ Br (0) (ii)

MARK

(2 marks)

Heat or enthalpy for energy; energy released instead of energy change

Energy change when 1 mol (1)

“energy required”

Just balanced equation e.g. Na+(g) + Cl−(g) → NaCl(s) can score only last two marks

of a solid/crystal/lattice (1) is formed from its (isolated) gaseous ions (1) IGNORE standard states

(3 marks)

Notes:

59

ACCEPT

EXPECTED ANSWER

(b)

(i)

REJECT

MARK

∆Hf = ∆Ha[Mg] + IE1[Mg] + IE2[Mg] + 2∆Ha [chlorine]+ 2EA[Cl] + LE[MgCl2] Or this in words Or 2EA = −(2× +122) −(+1450) −(+736) − (+150) + (−642) − (−2526) (1) = −696(kJ mol −1 ) EA =

(1) cq on first mark

−696 2

= −348(kJ mol −1 ) (1) must ÷2 [some likely outcomes – but working must be shown] -348 scores (3) -696 or -287 or (+) 348 scores (2) -574 or (+) 696 or (+) 287 scores (1) (+) 574 scores (0)

(3 marks)

Notes:

60

ACCEPT

EXPECTED ANSWER

(ii)

(c)

REJECT

MgCl2 has (a degree of ) covalent character (1)

Mention of “atoms” or “molecules” scores (0) for all of (b)(ii)

due to polarisation of the anion (1) (by Mg 2 + cation)

Just “Mg2+ (strongly) polarising”

As group descended, radius of M 2 + (ion) increases OR cation increases (1)

Reverse arguments

Charge on ions remains the same/2+ (1)

Correct formulae of cations for charge mark “charge density decreases” scores one of the first two marks

“size” instead of “radius”

(2 marks)

Mention specifically of atoms (e.g. Mg atoms) or molecules (MgCl2 molecules) scores (0) for all of part (c)

“weaker bonds” OR “weaker bonding”

(down group) weaker forces of attraction between ions (1)

MARK

(3 marks) Total 13 marks

Notes:

61

ACCEPT

EXPECTED ANSWER

5

(a)

Na NaCl

Mg MgCl 2

Al AlCl3 OR Al2Cl6

Si

REJECT

MARK

PCl4+, PCl6−

P PCl3 OR PCl5

All 4 → (2) 3 → (1) None, one or two correct (0) (2 marks)

(b)

(i)

NaCl(s) + aq → Na+(aq) + Cl−(aq) Or NaCl(s) + H2O(l) → Na+(aq) + Cl−(aq)

NaCl(s) + H2O(l) → NaOH(aq) + HCl (aq) (1)

PCl 3 + 3H 2 O → H 3 PO3 + 3HCl

(ii)

OR PCl 5 + H 2 O → POCl 3 + 2 HCl OR PCl 5 + 4 H 2 O → H 3 PO 4 + 5 HCl (1) NaCl: Ionic (so) dissolves (in water) (1) – both needed

(2 marks)

PClx: Covalent (so) reacts (in water) OR hydrolyses (in water) (1) – both needed

(2 marks)

Notes:

62

ACCEPT

EXPECTED ANSWER

(c)

REJECT

MARK

SiCl 4 reacts/hydrolyses, CCl 4 does not (1)

[This must be clearly stated and not just implied] (lone) pair of electrons (from the oxygen atom) in a water molecule (1)

“lone pair” for “(lone) pair of electrons”

cannot form a bond with/be donated to the C atom Or cannot be accepted by C atom (1)

reverse argument for Si atom CCl4 has no d orbitals [see below] Or “CCl4 has no 2d orbital(s)” Just “C has no dorbital(s)”

because C has no available orbital OR no 2d orbitals in C OR C is a small atom surrounded by Cl atoms OR Cl atoms are large and surround C atom (so attack is sterically hindered) (1) Si has (available) 3d orbital(s) (1)

SiCl4 has available 3d orbitals (but penalise this only once) (d)

(i)

PbO 2 + 4 HCl → PbCl 2 + 2 H 2 O + Cl 2

Multiples (1 mark)

Species and balancing (1) (ii)

(5 marks)

+2 (oxidation state) becomes more stable down the group relative to +4

Relative stabilities of Pb and Si oxidation states

MUST have comparison of +2 and +4 oxidation states (1 mark)

Total 13 marks

Notes:

63

ACCEPT

EXPECTED ANSWER

6

(a)

(i)

(ii) (iii)

Ethanenitrile OR Methyl cyanide OR ethanitrile OR ethanonitrile IGNORE any formula

phonetic spelling e.g. ethanenitrille

(Acid) hydrolysis IGNORE word “acid” before hydrolysis

phonetic spelling e.g. hydrolisis

Step 1: any named mineral acid (eg. hydrochloric acid) or formula

Using a named alkali or formula , then acidify Just “HCl” or “H2SO4”

MARK

Ethenenitrile

(1 mark)

Step 2: PCl 5 / SOCl 2 (iv)

REJECT

(1 mark)

Conc H2SO4

PCl3

(2 marks)

Cl2

CH 3 COCl + CH 3 NH 2 → CH 3 CONHCH 3 + HCl

OR CH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3Cl (1)

(1 mark)

Notes:

64

ACCEPT

EXPECTED ANSWER

(b)

REJECT

MARK

L:

H

M:

H

H

H

H

C

C

C

H

H

H

H

H

H

H

H

C

C

C

C

C

H

O-H

H

H

H

H

(1)

H

Allow OH

(1) Only penalise if bond is clearly shown pointing to H atom N: O C

H-O

H

H

H

C

C

C

H

H

H

H

Allow OH

(1) Penalise omissions of H atoms once only If –CH3, –CH2 are used max (2)

− COOH Or − CO2H

(3 marks)

Total 8 marks

Notes:

65

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6245/01 1.

(a)

(i)

(ii)

The sum of the powers to which the concentrations are raised in the rate equation Note if candidates choose to define order with respect to one species or give complete rate equation in terms of powers of ‘x’ and ‘y’ and explain that the order is x or y or x+ y 1 mark

“The sum of the partial/individual orders” if exemplified by a rate equation OR number of species/reactants involved in (up to and including) the rate determining step

The sum of the partial/individual orders” on its own (1 mark)

1st order because rate halves as [A] halves in expt. 1 → 2 or [B] constant (1) 2nd order because rate quadruples / increases by 22 as [B] doubles in expt. 2 → 3 or [A] constant (1) 1 (out of 2) if incomplete or no reasons given

(iii)

(iv)

rate = k[A][B]2 (1) consequential on their orders

(3 marks)

k = 0.0080 (1) mol-2 dm6 s-1 (1) both marks consequential on rate equation IGNORE SF

(2 marks)

(k) increases

Any reference to endothermic reaction scores zero

66

(1 mark)

ACCEPT

EXPECTED ANSWER

(b)

REJECT

MARK

(i)

Shape i.e. start at origin skewed and asymptotic to x-axis

(1 mark) (ii)

Ecat to left of Euncat and both to the right of hump

If draw energy profile could get this mark if the Es are correct and clearly marked on the profile

(1 mark)

(iii) Peak ( more ) to the right (1) Peak lower (1) OR shown on diagram

Any reference to increase in area under graph deduct 1 mark

(iv)

“Haber process” on its own

Manufacture of ammonia (1) Iron (1) MUST be a metal not a compound OR e.g. Hydrogenation of oils (1) Ni/Pt/Pd (1)

(2 marks)

Sulphuric acid manufacture with V2O5

Manufacture of H2 from CH4 (1) Ni (1) Explanation uses d orbitals to bond with reactants( at active sites) (1) – stand alone

Must have a least three oxidation states Variable /more than 1/several oxidation states 67

(3 marks)

ACCEPT

EXPECTED ANSWER (v)

Catalysed k bigger/Higher OR uncatalysed k lower

REJECT

MARK

(1 mark) Total 15 marks

68

ACCEPT

EXPECTED ANSWER

2

(a)

Pt electrode

(1)

chlorine gas at 1 atm

(1)

REJECT

101 kPa

chloride ions at 1.0 md dm-3 (1) IGNORE references to temperature (b)

(i)

(ii)

MARK

(3 marks)

2Cu+(aq) → Cu(s) + Cu2+(aq) IGNORE state symbols

(1 mark) Could argue reverses reaction is not feasible because…..

Ecell = +0.37 V OR E Ο for Cu+/Cu > E Ο for Cu2+/Cu+ (1)

(2 marks)

Is positive (and thus feasible) (1) (iii) (Copper) oxidised from +1 to +2 (1)

A definition of disproportionation alone does not score

and also reduced to zero (1)

(2 marks) OR The Cu+ is oxidised to Cu2+ (1) and Cu+ also reduced to Cu (1)

69

ACCEPT

EXPECTED ANSWER

(c)

(d)

REJECT

(1s2) 2s22p63s23p63d10 OR (1s2)2s22p63s23p64s03d10 OR (1s2) 2s22p63s23p63d104s0 Ignore spaces between items Ignore punctuation (i)

ligand exchange OR ligand substitution deep/dark blue

(ii)

MARK

(1 mark) “Nucleophilic substitution”

(1) Any type of blue that is darker than hydrated Cu(II) ions

(1)

d-orbitals split (in energy) by ligands / become nondegenerate in presence of ligands (1)

“Substitution” on its own OR “deprotonation” (2 marks)

d-sublevel The first mark may be “UV light” awarded provided at some point in the answer it is clear that there a d – orbitals of different energy

absorbs energy(light in visible region) (1) electron is promoted OR electron moves to a higher energy level (1) Any mention of emission of light can only score 1st mark Any implication of electron promotion before absorption of light can only score 1st mark

(3 marks)

(iii) full d subshell / all d orbitals full (1) Therefore d-d transitions impossible / a clear idea that promotion of electrons by absorbing energy is not possible(1) (2 marks) No d orbital splitting max 1 mark 70

ACCEPT

EXPECTED ANSWER (e)

tetrahedral (1) range 109 – 110

o

REJECT

MARK

OR Sq planar (1) 90 ° (1) with comparison with Ni or Pt complex (1)

(1)

4 (bonding) pairs of electrons repel to a position of maximum separation/minimum repulsion (1) Accept diagram to show shape( ignore charges)

Bonds/atoms repelling if say square planar and then argue that 4 pairs of electrons repel as far as possible max 1

(3 marks)

Total 19 marks

71

ACCEPT

EXPECTED ANSWER

3

(a)

All hydrogen nuclei / hydrogens atoms/ protons in same (chemical) environment

(b)

(i)

MARK

(1 mark)

reagent (1) ethanoyl chloride / CH3COCl

Fe

catalyst (1) (anhydrous) aluminium chloride / AlCl3/Al2Cl6 (ii)

REJECT

AlBr3 FeBr3 , FeCl3

electrophilic substitution (1)

(2 marks))

acylation Friedel-Crafts

(1 mark) Any arrows to C of CH3 rather than of CO

(iii)

(1) this could be shown as part of the mechanism …………………………………………………………………….

(1) for arrow

(1) for intermediateneeds +ve charge …………………………………………………………………….. Either:

(1) for arrow from C-H bond OR: 72

ACCEPT

EXPECTED ANSWER

REJECT

MARK

(1) for arrow ……………………………………………………………………… ALTERNATIVE Kekulé

(1)

(1)

(1) Notes: 1st curved arrow from benzene ring of electrons towards C of COCH3 (1) ALLOW the “+” anywhere on COCH3

(4 marks)

Curved arrow from C-H bond back into benzene ring (1) IGNORE if towards the “+”

(c)

(i)

CN− for KCN

HCN (1) + KCN (1) OR KCN (1) + Acid (1) EXCEPT conc H2SO4 OR HCN (1) + Base / alkali(1) OR HCN/KCN (1) pH 5 - 9 (1) Names or formulae can be given

If KCN, HCN and acid max1

(2 marks) 73

ACCEPT

EXPECTED ANSWER

(ii)

nucleophilic addition Both needed

REJECT

MARK

(1 mark)

74

ACCEPT

EXPECTED ANSWER

(iii)

EITHER

+

(1)

(1) for intermediate (1)

OR

(1)

(1) for intermediate

(1)

75

REJECT

MARK

ACCEPT

EXPECTED ANSWER

REJECT

MARK

• •

The intermediate is not consequential on their first step The minus of the cyanide ion can be on either the C or the N • The arrow can start from the minus of –CN in step 1 (but not from the minus of CN-) and can start from the minus of O- in step 2 • The arrow from the bond must not go past the O atom • Lone pairs not essential • Single step addition of HCN scores zero • Autoionisation of C=O can only score the last two marks ie max 2

(d)

(i)

2 enantiomers drawn

NC

C CH 3 OH

C

C6H5

C6H5

(ii)

(3 marks)

H3 C

or

C

C

C CN OH

(1 mark)

(No effect) as ketone planar (1) Attack possible from top or bottom

(1)

Producing racemic/50:50 mixture (of enantiomers) / rotations cancel out (1) no effect could appear here in the answer (e)

No absorption corresponding to C=O / carbonyl OR No absorption around 1700 cm-1

(3 marks)

Peak / band

Ketone group (1 mark) Total 19 marks

76

ACCEPT

EXPECTED ANSWER

4

(a)

(i)

(free) radical substitution

phonetic spelling e.g. radicle

REJECT

MARK

“radical nucleophilic substitution” (1 mark)

(ii)

(b)

UV radiation OR sunlight OR ultraviolet radiation OR UV OR UV light OR white light OR heat

“light” on its own NOT hν NOT strong light (1 mark)

Diagram labelled axes, lozenge and b.pt. values (1)

The curve must not noticeably go above or below the boiling points indicating a max or min on the curve

At least 2 horizontal + 2 vertical tie lines from anywhere except 100% (1) Explanation Vapour richer in more volatile/chloropropane (1) Condense and then reboil (1) Pure chloropropane distilled off / bromopropane left as residue (1) If heat to 46 (or when at 46) all chloropropane boils off then……. scores (0) for explanation

77

(5 marks)

ACCEPT

EXPECTED ANSWER

(c)

heat with NaOH (1)

REJECT

MARK

Methods based on displacement

add excess HNO3 OR acidify with HNO3 (1) add AgNO3 (1) chloro gives white and bromo gives cream ppt (1) white/off white/ pale yellow ppt soluble in dil NH3, cream ppt slightly/partially soluble in dil NH3 , (or soluble in conc NH3) (1) If fail to add NaOH or fail to add HNO3 3 max (d)

(5 marks)

MS shows different m/e values for molecular ion (1) Because molar masses different / or reason why different(1) Hydrogens in same environment in both molecules

Nmr give same number/3 peaks with both (1) OR Nmr shows different chemical shifts (1) Due to different halides (1) In MS molecular ion peak often absent (1)

(3 marks) Must be a statement about both MS and NMR to score 3 marks Total 15 marks

78

ACCEPT

EXPECTED ANSWER

5

(a)

Moles manganate = 0.0239 x 0.2 (1) = 0.00478

REJECT

MARK

Answers that start from the equation and then use it to derive ratio

Moles bromide = 2.46 (1) = 0.0239 103 ratio MnO4− : Br− = 1:5 OR ratio Br− : MnO4− = 5:1 (1)

(b)

MnO4- + 5Br- + 8H+ → Mn2+ + 4H2O + 2.5Br2 species (1) balance (1) If no calculation allow correct equation marks If calculation wrong equation must be consequential on the ratio calculatedfor balance mark

Multiples

(i)

Cannot be oxidised

(ii)

(5 marks)

Not oxidised by manganate(VII)/ does not react with oxidising agents OR Not hydrolysed by acid non-biodegradable therefore fills landfill sites

unreactive (1 mark)

Non-biodegradable therefore persists in environment

toxic gas if burned (1 mark) Total 7 marks

79

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6246/01A 1.

(a)

Observation Violet/pink/amethyst (1) Inference d-block/transition metal (ion) (1)

Pale violet/pale purple/(pale) brown Accept inference from any colour “s” block from colourless/white

Fe3+/ Fe (III) (1)

(b)

(i)

(ii)

Iron(III) OR ferric Chromium (III) ion OR formula Fe(III)/Fe3+ ONLY from yellow Mn2+ from “pink”

Observation Any brown precipitate/ solid (1) Inference Fe(OH)3 / Iron(III) hydroxide (1)

“s” block ions

Cr(VI) from yellow (3 marks)

Foxy-red iron hydroxide Fe(III)/ Fe3+ alone

(2 marks)

ammonium

(2 marks)

Observation (litmus) turns blue (1) Inference ammonia/NH3 (evolved) (1)

(iii)

Purple/green/yellow

Fe3+/iron(III) (1) NH4+ /ammonium (1)

(2 marks)

80

ACCEPT

EXPECTED ANSWER (iv)

MARK

Observation (turns) red (1) Inference Ligand exchange (1)

(v)

REJECT

Observation White precipitate/solid (1) Inference sulphate/SO4 2− (1)

81

(Ligand) substitution

Complex formation

hydrogen sulphate/ HSO4−/ sulphate from any colour ppt

BaSO4/barium sulphate

(2 marks)

(2 marks)

ACCEPT

EXPECTED ANSWER

2

(a)

Observation Glue-like/ nail varnish/remover/fruity smell (1)

sweet

Inference Type : ester (1) Functional group: carboxylic acid/-CO2H/ -COOH (1) Follows “ester” (b)

(i)

REJECT

MARK

pungent

NOT from “pungent” carboxyl (3 marks)

Observation Yellow precipitate/solid (1)

Orange ppt/solid

Red ppt/solid

Inference Ketone (1) Aldehyde (1) “carbonyl” on its own max 1 (ii)

(3 marks)

Observation Remains orange/yellow (1)

No change

Inference Ketone/“not aldehyde” (1) conditional on b(i) inference (c)

“Nothing” No reaction

Redox alone (2 marks)

Observation (pale) yellow precipitate/solid (1) Inference Triiodomethane/Iodoform/ CHI3 (1) Methyl ketone / methyl carbonyl if KETONE given in b(ii) or structure:

Structural inference from CH3I

82

CH3I CH3CH(OH)

(3 marks)

ACCEPT

EXPECTED ANSWER

H

H

O

C

C

REJECT

MARK

CH3—

R H

(d)

All from ppt (i) (M+) marked on spectrum (1) (ii)

Other designations to identify molecular ion

88

(1 mark) H

(iii) H

(1 mark)

− OH

O

C

C

H

O

C

CH3—

O H

fully correct (2) Otherwise: methyl ketone group /Carboxylic acid (1) max

(2 marks) Total 15 marks

83

ACCEPT

EXPECTED ANSWER

3

REJECT

MARK

Table 1 Check subtractions and averaging arithmetic, correcting if necessary. All volumes recorded to 0.05 cm3 Initial reading as 0 ; 0.0 ; 0.00 Allow one slip in recording to 0.05 cm3 but withhold if any readings are in the wrong boxes

All subtractions completed correctly [9 top RHS of Table 1]

Mean titre: for correctly averaging of chosen titres or for choosing identical titre and for recording the mean to 0.05 or two or three dp [9 by the mean]

(1)

Do not penalise bad recording a second time (1)

Accuracy: If the candidate has made an arithmetical error in Table 1 volumes used in the mean or in averaging the examiner must calculate a new average. • For an averaging error simply calculate a new value using the candidate’s chosen titres • If a wrongly subtracted titre has been used in the mean then choose any two identical titres or take an average of the closest two titres. Calculate the difference between the candidate’s mean and that of the examiner or supervisor. Record the difference on the scripts as d = EXAMINER’S TITRE TBC BY THE CENTRE Examiner to write SV = titre value on each spt

84

Award marks for accuracy as follows: d= ±0.20 mark = 5 d= ±0.25 mark = 4 d = ±0.30 mark = 3 d = ± 0.40 mark = 2 d= ±0.50 mark = 1 d > 0.50 mark = 0

ACCEPT

EXPECTED ANSWER

85

REJECT

MARK

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Range: Award a mark on the range of titres used by the candidate to calculate the mean. The range [r] is the difference between the outermost titres used to calculate the mean. If the examiner has corrected titres because of incorrect subtraction then award the range mark on the corrected titres used by the examiner to recalculate the mean.

Award marks for range as follows:

Examiner to show the marks for accuracy and rang as:

r = ± 0.20 mark = 3

d = value [mark] r = value [mark]

r= ± 0.30 mark = 2 r = ±0.40 mark 1 r > 0.4 mark =0

Calculations Final answers to at least 2 sig fig. Penalise once only (b) (i) 0.02 x titre 1000 (ii)

(1 mark)

Answer (i) x 5

(1 mark)

(iii) Answer (ii) x 10

(1 mark)

(iv)

Answer (iii) x 56

(1 mark)

(v)

Answer (iv) x 100 % 9.50 [Calculated 14.28 %]

(1 mark) 86

ACCEPT

EXPECTED ANSWER

4

MARK

solids

Known volumes of silver nitrate and ionic chloride solution 9V (1) named chloride solution 9N (1) Known / stated concentrations OR known /stated concentration of one and excess of the other

REJECT

Hydrochloric acid

9C (1)

Mix and measure temperature change 9T (1) in polystyrene cup (held in beaker) 9P (1) Use of q = mc∆T 9Q (1)

∆H = q 9H (1) moles

(7 marks)

87

88

ACCEPT

EXPECTED ANSWER

REJECT

MARK

Unit 6246/02 1

(a)

(i)

Cu → Cu2+ + 2e(−) (1) Cr2O72- + 14H+ + 6e(−) → 2Cr3+ + 7H2O (1) Cr2O72- + 14H+ + 3Cu → 2Cr3+ + 7H2O + 3Cu2+ (1)

(ii)

(3 marks)

initial moles Cr2O72- = 0.00750 (1) moles Cr2O72- reacted = 0.00750 − 0.00342=0.00408 (1) moles Cu = 3 x 0.00408 = 0.01224 (1) mass Cu = 63.5 x 0.01224 = 0.77724 g (1) % purity = 97.2 % (1) consequential on equation in (i) if >100 % do not award % mark unless commented on If not 3SF loses last mark

(iii)

(5 marks)

Cu(OH)2 / [Cu(H2O)4(OH)2] (1) Cr(OH)3 / [Cr(H2O)3(OH)3] (1)

(iv)

(2 marks)

[Cr(OH)6]3−/ [Cr(H2O)(OH)5]2− / [Cr(H2O)2(OH)4] − (1)

Cr(OH)4 −

green (1)

pale/light/dark/ bright green

89

(2 marks)

ACCEPT

EXPECTED ANSWER

(b)

REJECT

MARK

dissolve in minimum vol of boiling/hot water (1) filter through heated funnel / filter while hot (1) cool and filter (under reduced pressure) (1) (4 marks)

wash in minimum/cold water (and dry) (1)

Total 16 marks

90

ACCEPT

EXPECTED ANSWER

2

(a)

REJECT

MARK

H3CO H3C C O OR

OCH3 H3C C O (b)

(i)

(1 mark)

4 peaks (1) area 3 : 2 : 2 : 3 (1) (can score (2) because 4 peaks clearly implied) 3 peaks area 3 : 4 : 3 (1 out of 2) 3 peaks area 6 : 2 : 2 / 3 : 1 : 1 (1 out of 2)

(c)

(ii)

[COC6H4OCH3]+ / [CH3COC6H4O]+ charge essential

(i)

LiAlH4 / NaBH4 (1)

(2 marks)

(1 mark) H2+Ni/Pt/Pd, OR Na + ethanol

Reduction / nucleophilic addition (1) secondary alcohol (1)

(3 marks)

91

ACCEPT

EXPECTED ANSWER

(ii)

Molecule cannot be superimposed on its mirror image OR C atom to which four different groups are joined (1)

REJECT

MARK

asymmetric C atom OR no plane/centre of symmetry

(1 mark) (d)

(i)

concentrated H2SO4 OR conc. H3PO4 OR Al2O3 OR names (1) (2 marks)

(ii)

dehydration OR elimination (1) has two H atoms/two atoms the same at one end of the double bond (1)

“can be rotated about double bond”

92

(1 mark)

ACCEPT

EXPECTED ANSWER

(e)

(1) for both arrows

REJECT

MARK

(1) for intermediate

(1) for arrow

Notes: • If the wrong carbocation is shown i.e. Br is on the wrong carbon atom, only 1st and 3rd marks are available • Lone pair is not essential but if shown arrow must start from it • allow arrow from negative charge • allow arrow to “+”

93

(3 marks)

ACCEPT

EXPECTED ANSWER

QWC

(f)

REJECT

MARK

A has van der Waals’/dispersion/London forces and dipoledipole forces (1) C has H bonding and van der Waals/dispersion/London and dipole-dipole forces (1) H bonding stronger than van der Waals/dispersion/ London forces and dipole-dipole forces (1) (therefore more energy required)

van der Waals in C is greater than in A because C has more electrons (3 marks)

Penalise lack of dipole-dipole once only

Total 17 marks

94

ACCEPT

EXPECTED ANSWER

3

(a)

(i)

QWC

MARK

working: ((4x 90.4) + (6x −242) ) − (4x −46.2) OR ∆H = Σ∆Hf(Products) – Σ∆Hf(Reactants) (1) -905.6 kJ mol-1 (1) OR - 906 kJ mol-1 Must have the sign and the units. IGNORE SF

(ii)

REJECT

(2 marks)

high temp = high rate (1) more mols > Eact (1)

If endothermic in (i)

high temp = low yield because reaction exothermic (1)

opposite argument

so compromise temp used to balance rate and yield (1) catalyst causes higher rate by alternative route of lower Eact (1)

(b)

(6 marks)

but same yield as speeds up forward and back reactions OR same yield as k unchanged (1) heat change = 50 x 4.18 x 6.5 = 1358.5 J (1) for 0.025 mol

1359 to give – 54.4 or – 54.36 1360 to give – 54.4 only

÷ heat (J or kJ) by 0.025 mol (1) ∆H = −54.3 kJ mol-1 / 54300 J mol-1 value, sign and unit (1)

(3 marks)

IGNORE SF

95

ACCEPT

EXPECTED ANSWER

(c)

(i)

REJECT

MARK

Add NaOH and warm (1) (Damp) red litmus in gas unchanged (showing no NH4+) (1) Add Al OR Devarda’s alloy OR zinc + NaOH and warm; gas evolved turns red litmus blue (shows NO3−) (1)

(ii)

(3 marks) atom gets bigger weakens bonds less

cation same charge OR are all +1 (1) but ionic radius gets bigger (1) so polarises anion less (1)

distorts anion less

(3 marks) Total 17 marks

96

ACCEPT

EXPECTED ANSWER

4

(a)

(i)

inner electrons too

oo

(ii)

o x

F oo

MARK

all dots/crosses

correct diag (only outer electrons needed) i.e.

H

REJECT

o o

(1 mark)

Hδ+ and Fδ- (1) Large electronegativity difference OR forms strong intermolecular bond (1) Through lone pair OR because of small size of (H and) F atom(s) (1) st

(b)

(i)

(3 marks)

rd

Diagram can score 1 and 3 marks HF + H2O ⇌ H3O+ + F− (1) must show that water is there

HF ⇌ H+ + F-

Ka = [H3O+][F−] [HF] OR Ka = [H+][F-] (1) [HF] (2 marks) IGNORE state symbols

97

ACCEPT

EXPECTED ANSWER

(ii)

REJECT

MARK

moles HF at start = 0.1 × 0.025 = 0.0025 (1) moles NaOH = moles F− = 0.12 × 0.01 = 0.0012 (1) moles HF left = 0.0025 − 0.0012 = 0.0013 (1)

÷ both moles by 0.35 (1) ie [HF]eqm = 0.0013 = 0.03714 (mol dm-3) 0.035 [F−]eqm = 0.0012 0.035

= 0.03429 (mol dm-3)

[H+] = ka × [HF] = 0.000562 × 0.03714 [F−] 0.03429 = 0.000609 (mol dm-3) (1) pH = −log [H+]= 3.22 (1)

QWC

(6 marks)

IGNORE SF (iii) Large reservoir of both HF and F- is needed (to absorb both acid and base) (1) (2 marks)

however [F-] is small (so cannot absorb acid) (1)

98

ACCEPT

EXPECTED ANSWER

(c)

REJECT

MARK

(i)

F B F (ii)

F

(1 mark)

from 120 ° Æ 109.5 ° / 109 ° OR changes by 10.5/11 ° (1) Boron (goes from 3) to 4 electron pairs (around atom) (1) (2 marks) Total 17 marks

99

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