1.	Write the formula and charge of the ion which forms when ammonia, NH3, accepts a proton.
NH4+
(1)

2.	(a)	An aqueous solution of silver nitrate is added to an aqueous solution of potassium iodide in a test–tube.
(i)	State what you would observe.
precipitate/ppt            (1)
NOT goes cloudy
yellow (or reasonable colour)      (1)
NOT cream
(2)
(ii)	Write a balanced equation including state symbols for the reaction that occurs.
Ag+(aq) + I–(aq) ® Agl(s)
OR
AgNO3(aq) + Kl(aq) ® Agl(s) + KNO3(aq)
all formulae correct         (1)
correct balance and state symbols     (1)
Carry forward consequential error from a(i) (–1 max)
If (s) is missing (–1)
Allow one (aq) missing. Do not allow (1)
Equation based on KBr:1 max if correct
(2)
(b)	The reaction in (a) is repeated using potassium bromide instead of potassium iodide.  
The reaction mixture is allowed to stand in sunlight for a few minutes.
(i)	State what you would observe after a few minutes.
Reasonable description of darkening of precipitate,
including descriptions of colour change
ACCEPT purple/green precipitate
ACCEPT mixture darkens/turns purple/green
NOT yellow bromine colour or solution darkens
(1)
(ii)	Suggest a useful application of this reaction.
photography is the most likely answer, but there may be other reasonable acceptable answers
NOT 'to process photographic film' or 'test for bromide ions' or 'photography (development)'
(1)
(iii)	Concentrated ammonia solution is added to the reaction mixture in (b).  
State and explain what you would observe.
Observation precipitate dissolves/disappears or solution clears   (1)
Explanation.
formation of soluble/complex (silver) ion/compound, or correct formula   (1)
(2)

(c)	A few drops of concentrated sulphuric acid are added to some potassium bromide in a test–tube.
	At first a gas is given off which fumes at the mouth of the test–tube and gives dense white fumes with ammonia.
(i)	Name the gas given off.
hydrogen bromide/HBr
(1)
(ii)	Write a balanced equation for the reaction between the gas and ammonia.
HBr + NH3 ® NH4Br 
ACCEPT BrNH4
NOT NH3BrH
Ignore spurious state symbols
(1)
(iii)	After a short time reddish–brown fumes are observed in the test tube.
	Name this gas.
bromine/Br2
(1)
(iv)	During this time a piece of filter paper soaked in acidified potassium dichromate(VI) is held at the end of the test–tube.  The colour changes 
from yellow to green.
	Name the gas which produces this change.
sulphur dioxide/SO2
NOT HBr/H2S/sulphur oxide 
(1)
(v)	Concentrated sulphuric acid has reacted in two ways in these reactions.  
Classify its behaviour.
In reaction (c)(i) 
acid       (1)
ACCEPT acid/base or hydrogen ion/proton donor
NOT hydrogen donor
In reaction (c)(iii)
oxidizing agent/redox    (1)
NOT hydrogen acceptor
(2)
[Total 14 marks]

3.	(a)	An aqueous solution of silver nitrate is added to an aqueous solution of potassium iodide in a test–tube.
(i)	State what you would observe.
precipitate/ppt            (1)
NOT goes cloudy
yellow (or reasonable colour)      (1)
NOT cream
(2)

(ii)	Write a balanced equation including state symbols for the reaction that occurs.
Ag+(aq) + I–(aq) ® Agl(s)
OR
AgNO3(aq) + Kl(aq) ® Agl(s) + KNO3(aq)
all formulae correct         (1)
correct balance and state symbols     (1)
Carry forward consequential error from a(i) (–1 max)
If (s) is missing (–1)
Allow one (aq) missing. Do not allow (1)
Equation based on KBr:1 max if correct
(2)
(b)	The reaction in (a) is repeated using potassium bromide instead of potassium iodide.  
The reaction mixture is allowed to stand in sunlight for a few minutes.
(i)	State what you would observe after a few minutes.
Reasonable description of darkening of precipitate,
including descriptions of colour change
ACCEPT purple/green precipitate
ACCEPT mixture darkens/turns purple/green
NOT yellow bromine colour or solution darkens
(1)
(ii)	Suggest a useful application of this reaction.
photography is the most likely answer, but there may be other reasonable acceptable answers
NOT 'to process photographic film' or 'test for bromide ions' or 'photography (development)'
(1)
(iii)	Concentrated ammonia solution is added to the reaction mixture in (b).  
State and explain what you would observe.
Observation precipitate dissolves/disappears or solution clears   (1)
Explanation.
formation of soluble/complex (silver) ion/compound, or correct formula   (1)
(2)
[Total 8 marks]

4.	A few drops of concentrated sulphuric acid are added to some potassium bromide in a test–tube.
	At first a gas is given off which fumes at the mouth of the test–tube and gives dense white fumes with ammonia.
(i)	Name the gas given off.
hydrogen bromide/HBr
(1)

(ii)	Write a balanced equation for the reaction between the gas and ammonia.
HBr + NH3 ® NH4Br 
ACCEPT BrNH4
NOT NH3BrH
Ignore spurious state symbols
(1)
(iii)	After a short time reddish–brown fumes are observed in the test tube.
	Name this gas.
bromine/Br2
(1)
(iv)	During this time a piece of filter paper soaked in acidified potassium dichromate(VI) is held at the end of the test–tube.  The colour changes 
from yellow to green.
	Name the gas which produces this change.
sulphur dioxide/SO2
NOT HBr/H2S/sulphur oxide 
(1)
(v)	Concentrated sulphuric acid has reacted in two ways in these reactions.  
Classify its behaviour.
In reaction (c)(i) 
acid       (1)
ACCEPT acid/base or hydrogen ion/proton donor
NOT hydrogen donor
In reaction (c)(iii)
oxidizing agent/redox    (1)
NOT hydrogen acceptor
(2)
[Total 6 marks]

5.	(a)	Chloroethene, C2H3C1, is the monomer from which the important plastic poly(chloroethene) is made.  It is often known as PVC.  It can be produced with any degree of flexibility from rigid to pliable.
(i)	Draw the displayed formula of chloroethene.

(1)

(ii)	Draw a ‘dot–and–cross’ diagram of a chloroethene molecule.  You need include only outer shell electrons.

Accept any reasonable electron symbolism that clearly conveys correct arrangement.
Carry forward reasonable consequential error from a(i) double bond    (1)
all other outer shell electrons       (1)
If Cl is replaced by H (1 (max))
Ignore 'rings', right or wrong
(2)
(b)	Use the Hess cycle and the data, at 298 K, to calculate the standard enthalpy change of formation of chloroethene.

(or similar statement of how Hess' law is to be applied)    (1)
 	= 2(–394) + 3/2(–286) – (–1264)
	=+47 kJ mol–1
correct substitution of values to give     (1)
correct evaluation of DH1 with units      (1)
–1 (max) if sign omitted
–1 (max) if number of moles C and H2 missed(*)
(3)
(c)	A possible route for making chloroethene in the laboratory is as follows:

Suggest reagents and conditions for each step.
Step 1 .....
reagent:	chlorine
condition:	no heating/room temperature/no sunlight/dark
	ACCEPT in organic solvent (any one)
Step 2 .
reagent:	potassium hydroxide (in ethanol/sodium hydroxide)
	ignore aqueous
condition:	in ethanol solution or reflux (NOT heat)
	Condition must relate to reagent
(4)
(d)	(i)	Write a balanced equation for the formation of poly(chloroethene) from chloroethene.
n C2H3Cl ® (–CH2–CHCl–)n
ACCEPT (C2H3Cl)n
ACCEPT displayed
Both n's are essential and must be correctly placed 
answer based on ethene (0)
(1)
(ii)	Suggest appropriate conditions for the polymerization reaction.
any two, for one mark each, from:
raised/ high pressure (but not sealed vessel)
raised temperature/warm (accept sensible suggestions for value 30–100 °C) but not high temperature initiator (accept catalyst or named appropriate substance) NOT Ni
Free radical
autoclave
carry across mark if already stated in d(i)
(2)
(iii)	State the type of polymerization reaction involved in the formation of poly(chloroethene).
addition 
ACCEPT free radical if not already stated in part (ii)
(1)

(iv)	Suggest a use for poly(chloroethene), explaining why it is suitable.
For mark to be awarded:
– both suggested use and explanation in terms of desirable property for that use must be given
– the desirable property must be one that reasonably relates to the real properties of PVC
The suggested use does not have to be a known use of PVC
Note that stem states PVC can be produced with any desired degree of flexibility; hence to say that PVC is rigid is not by itself sufficient explanation of any specified use for the mark to be awarded.
In general terms an A level standard of answer is looked for, demonstrating a deeper understanding than the more superficial answers that might be acceptable at GCSE.
(1)
[Total 15 marks]

6.	Chloroethene, C2H3C1, is the monomer from which the important plastic poly(chloroethene) is made.  It is often known as PVC.  It can be produced with any degree of flexibility from rigid to pliable.
(i)	Draw the displayed formula of chloroethene.

(1)
(ii)	Draw a ‘dot–and–cross’ diagram of a chloroethene molecule.  You need include only outer shell electrons.

Accept any reasonable electron symbolism that clearly conveys correct arrangement.
Carry forward reasonable consequential error from a(i) double bond    (1)
all other outer shell electrons       (1)
If Cl is replaced by H (1 (max))
Ignore 'rings', right or wrong
(2)

7.	A possible route for making chloroethene in the laboratory is as follows:

Suggest reagents and conditions for each step.
Step 1 .....
reagent:	chlorine
condition:	no heating/room temperature/no sunlight/dark
	ACCEPT in organic solvent (any one)
Step 2 .
reagent:	potassium hydroxide (in ethanol/sodium hydroxide)
	ignore aqueous
condition:	in ethanol solution or reflux (NOT heat)
	Condition must relate to reagent
(4)

8.	(i)	Write a balanced equation for the formation of poly(chloroethene) from chloroethene.
n C2H3Cl ® (–CH2–CHCl–)n
ACCEPT (C2H3Cl)n
ACCEPT displayed
Both n's are essential and must be correctly placed 
answer based on ethene (0)
(1)
(ii)	Suggest appropriate conditions for the polymerization reaction.
any two, for one mark each, from:
raised/ high pressure (but not sealed vessel)
raised temperature/warm (accept sensible suggestions for value 30–100 °C) but not high temperature initiator (accept catalyst or named appropriate substance) NOT Ni
Free radical
autoclave
carry across mark if already stated in d(i)
(2)
(iii)	State the type of polymerization reaction involved in the formation of poly(chloroethene).
addition 
ACCEPT free radical if not already stated in part (ii)
(1)

(iv)	Suggest a use for poly(chloroethene), explaining why it is suitable.
For mark to be awarded:
– both suggested use and explanation in terms of desirable property for that use must be given
– the desirable property must be one that reasonably relates to the real properties of PVC
The suggested use does not have to be a known use of PVC
Note that stem states PVC can be produced with any desired degree of flexibility; hence to say that PVC is rigid is not by itself sufficient explanation of any specified use for the mark to be awarded.
In general terms an A level standard of answer is looked for, demonstrating a deeper understanding than the more superficial answers that might be acceptable at GCSE.
(1)
[Total 5 marks]

9.	Read the passage on petroleum refining straight through, and then more carefully, in order to answer the following questions.
(a)	Petrol produced by simple crude oil distillation has a low octane rating and is not suitable for modern car engines.
(i)	What technological change led to the need for higher octane rating fuels?
increase in (petrol engine) compression (ratio)/higher pressure
or quote 4: 1 goes to 10:1
(ii)	What is the effect of using a low octane fuel in a modern petrol engine?
pre–ignition/knock/autoignition/pinking
(accept spontaneous ignition)
(2)
(b)	CH3CH2CH2CH2CH2 CH2 CH2 CH3

CH3C(CH3)2C(CH3)2CH3
2,2,3,3–Tetramethylbutane

(i)	Name the two chemicals from the four hydrocarbons above which are isomers of each other.
octane and (2,2,3,3–)tetramethylbutane          
(accept formulae if correct)
(1)
(ii)	Name the hydrocarbon with the lowest spontaneous ignition temperature.
octane (accept formula if correct)          
(1)
(c)	(i)	How has the cost of producing petrol been affected by the reduction in permitted lead content?
'tends to raise cost dramatically'
alternative wordings should convey more than just
'increased', and should imply a marked increase
(1)
(ii)	Why have limits been placed on the maximum concentration of benzene in petrol?
benzene is toxic/carcinogenic/harmful/poisonous/dangerous/hazardous
(Do not allow to 'spray the target')
(1)
(d)	Write a summary in continuous prose, in no more than 100 words, of the chemical processes involved in the manufacture of petrol today.
	You are not asked to summarise the whole passage, nor to include equations in your summary.
	At the end of your summary state the number of words you have used.
(9)
	Credit will be given for answer written in good English, using complete sentences and with correct use of technical words.  Avoid copying long sections from the original text.  Numbers count as one word, as do standard abbreviations and hyphenated words.  Any title you give to your passage does not count in your word total.
	There are penalties for the use of words in excess of 100.
Examiners will need to consider each answer for (i) key points and (ii) style and use of English. Candidates should have recorded their word total at the end of their answer, and this should be checked.
up to 105 words: no penalty
106 – 110 words: –1 
111 – 115 words: –2 
116 – 120 words: –3
and at a rate of – 1 penalty for every 5 words excess thereafter, up to a maximum penalty equal to the number of key points included by the answer.
This penalty should be recorded as a minus number alongside the word total at the end of the answer (but NOT in the margin). Note that words appearing in the title to the summary do not count in the word total. Normally hyphenated words, numbers and chemical formulae count as one word. The question does not ask for equations in the summary, but if included they should be counted in the word total.

MARKING FOR KEY POINTS
One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 7 marks. A tick should be made in the script at which the examiner decided to award each mark. The total marks for key points should be placed in the body of the script at the end of the answer, out of 7.
List of key points; these may occur in any order, and it is not necessary for them to be expressed in these words provided that the sense of each point is conveyed.
1	Crude Oil is separated by fractional distillation...
2	.... into products/fractions of suitable/different boiling point
3	Catalytic cracking is used to convert high boiling point fractions into high octane fuel 
4 	using (crystalline) aluminosilicates/zeolites as catalysts 
Aluminosilicates are used to crack high boiling point fractions
5	 In thermal cracking very heavy residues from (vacuum distillation) are heated briefly to 500°C / high temperature.
6	In catalytic reforming straight chain alkanes (and cycloalkanes) are converted to arenes
7	... by dehydrogenation reactions/and produces hydrogen.
8	The components from these processes are blended to produce a fuel with the right characteristics.
1 mark each key point up to a maximum of 7.
To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
STYLE AND USE OF ENGLISH
These should be impression marked on a scale 2–1–0, and the mark out of 2 should be recorded in the body of the script at the end of the answer.
The aspects to be considered are:
use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
–	articulate expression; the answer should be well–organised in clear, concise English, without ambiguity. It should read fluently, with the links between key points in the original passage maintained, even if the order is different.
–	correct grammar; basic sentence structure, including punctuation, should be consistently correct, without quibbles over the finer points of grammar.
–	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
Good style and use of English, with only infrequent minor faults.        (2) 
Frequent minor or a few major faults in style and use of English.         (1)
Very poor style and use of English.                                                           (0)
 [Total 15 marks]

10.	State the name of the element which is oxidized in the reaction:
Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s)
zinc 
(1)

11.	(a)	Name the type of intermolecular force between water molecules.
hydrogen bonding
(1)
(b)	The H¾O¾H bond angle in water is
104.5° (accept 104° – 105°)  
(1)
[Total 2 marks]

12.	(a)	Pure samples of hydrogen halides, HC1, HBr and HI, are prepared by adding a suitable acid to the corresponding potassium halides.
(i)	Name the acid used.
Phosphoric acid/H3PO4 (1)
Name but incorrect formula (1)
Formula but incorrect name (0)
(1)
(ii)	Draw a diagram of the apparatus you would use.
or diagram similar to Qu. 3 (a)
	conical flasks acceptable
Upward delivery  (1)
Collected over water  (0)
No rubber bung  (1)
Totally enclosed  (0)
(2)
(b)	A test–tube containing hydrogen chloride gas is inverted in water.
Describe and explain what you would observe.
Water rises up inverted test–tube or gas dissolves in water or gas disappears (1)
HCI is (very) soluble in water or reacts with water or forms hydrochloric acid (1)
(2)

(c)	(i)	What would you see when ammonia reacts with hydrogen bromide?
White fumes/white gas/white cloud(s)/white smoke
white solid/white vapour
NOT white precipitate
NOT smoke (1)
(1)
(ii)	What is the formula for the compound formed when ammonia reacts with hydrogen bromide?
NH4Br or BrNH4
No other combination
ignore charges (1)
(1)
(d)	A hot wire is plunged into test–tubes containing each of the hydrogen halides.
(i)	Name the hydrogen halide which decomposes.
hydrogen iodide/HI or hydrogen bromide/HBr
NOT hydrogen chloride/HCI (1)
(1)
(ii)	Describe the appearance of the contents of the test–tube as decomposition takes place.
Brown/purple vapour
brown liquid or brown solid or grey solid or silver solid
black solid
NOT purple solid
colour and state must be given, mark consequentially for HBr in (d)(i)
orange or yellow or brown or red–brown vapour/liquid (1)
(1)
(iii)	Write a balanced equation for the decomposition in (i).

Any balanced multiple of this
mark consequentially from (d) (i) (1)
(1)
(e)	Water companies use chlorine to purify water for domestic use.  Concentrations of chlorine are carefully monitored by testing water samples.
Excess potassium iodide was added to a 1000 cm3 sample of water.
The iodine formed reacted with 14.0 cm3 of 0.000100 M sodium thiosulphate solution.
(i)	Calculate the number of moles of sodium thiosulphate, , Na2S2O3, used in the reaction.

= 1.40 ´ 10–5 (mol) or 1.4 ´ 10–5 (mol)
or 0.000014 (mol)
or figure in any correct form (1)
(1)

(ii)	The ionic equation for the reaction between iodine and sodium thiosulphate is
I2(aq) + 2S2O32– (aq) ® 2I– (aq) + S4O62– (aq)
Calculate the number of moles of iodine molecules, I2, used in the reaction.
accept (e) (i) answer divided by 2
= 7.00 ´ 10–6 (mol) or 7 ´ 10–6
mark consequentially from (e) (i) (1)
(1)
(iii)	Write the equation for the reaction between chlorine molecules, C12, and iodide ions.
Cl2 + 2I– ® I2 + Cl–    or    Cl2 + 2KI ® I2 + 2KCl
ignore state symbols (1)
(1)
(iv)	Write down the number of moles of chlorine in the sample.
7.00 ´ 10–6 mol or answer to (e) (ii) or
answer in (e)(ii) multiplied by 2 (1)
(1)
(v)	Calculate the mass of chlorine molecules in the original sample.
Mass	=	n ´ M
	=	7.00 ´ 10–6 ´ 2 ´ 35.5
	=	497 ´ 10–6g = 4.97 ´ 10–4g or 0.000497
or 0.005 or 5 ´ 10–4
or answer to (iv) multiplied by 71
or if answer in (iv) multiplied by 2 then answer in (iv) ´ 35.5 (1)
(1)
(vi)	The maximum accepted concentration of chlorine in drinking water is 0.5 parts per million by mass.
Show by calculation that the sample of water tested above is acceptable.
You may assume 1000 cm–3 of water has mass 1000 g.
1000g contains 497 ´ 10–6g
106g contains 497 ´ 10–6 ´ 103
i.e. 0.497 ppm or 0.5 ppm
Mark is for the method (´ 1000 or ¸ 1000)
If just answer (0)   (1)
(1)
(vii)	Suggest a reason why the concentration of chlorine in water must not exceed 0.5 ppm.
Chlorine is poisonous/toxic/is a toxin/harmful/irritant/
damaging to respiratory system/irritating to eyes
or unacceptable smell/taste
NOT dangerous (1)
(1)
[Total 17 marks]

13.	Pure samples of hydrogen halides, HC1, HBr and HI, are prepared by adding a suitable acid to the corresponding potassium halides.
(i)	Name the acid used.
Phosphoric acid/H3PO4 (1)
Name but incorrect formula (1)
Formula but incorrect name (0)
(1)
(ii)	Draw a diagram of the apparatus you would use.
or diagram similar to Qu. 3 (a)
	conical flasks acceptable
Upward delivery  (1)
Collected over water  (0)
No rubber bung  (1)
Totally enclosed  (0)
(2)
[Total 3 marks]

14.	A test–tube containing hydrogen chloride gas is inverted in water.
	Describe and explain what you would observe.
Water rises up inverted test–tube or gas dissolves in water or gas disappears (1)
HCI is (very) soluble in water or reacts with water or forms hydrochloric acid (1)
(2)
[Total 2 marks]

15.	(i)	What would you see when ammonia reacts with hydrogen bromide?
White fumes/white gas/white cloud(s)/white smoke
white solid/white vapour
NOT white precipitate
NOT smoke (1)
(1)
(ii)	What is the formula for the compound formed when ammonia reacts with hydrogen bromide?
NH4Br or BrNH4
No other combination
ignore charges (1)
(1)
[Total 2 marks]

16.	A hot wire is plunged into test–tubes containing each of the hydrogen halides.
(i)	Name the hydrogen halide which decomposes.
hydrogen iodide/HI or hydrogen bromide/HBr
NOT hydrogen chloride/HCI (1)
(1)
(ii)	Describe the appearance of the contents of the test–tube as decomposition takes place.
Brown/purple vapour
brown liquid or brown solid or grey solid or silver solid
black solid
NOT purple solid
colour and state must be given, mark consequentially for HBr in (d)(i)
orange or yellow or brown or red–brown vapour/liquid (1)
(1)
(iii)	Write a balanced equation for the decomposition in (i).

Any balanced multiple of this
mark consequentially from (d) (i) (1)
(1)
[Total 3 marks]

17.	Water companies use chlorine to purify water for domestic use.  Concentrations of chlorine are carefully monitored by testing water samples.
Excess potassium iodide was added to a 1000 cm3 sample of water.
The iodine formed reacted with 14.0 cm3 of 0.000100 M sodium thiosulphate solution.
(i)	Calculate the number of moles of sodium thiosulphate, , Na2S2O3, used in the reaction.

= 1.40 ´ 10–5 (mol) or 1.4 ´ 10–5 (mol)
or 0.000014 (mol)
or figure in any correct form (1)
(1)
(ii)	The ionic equation for the reaction between iodine and sodium thiosulphate is
I2(aq) + 2S2O32– (aq) ® 2I– (aq) + S4O62– (aq)
Calculate the number of moles of iodine molecules, I2, used in the reaction.
accept (e) (i) answer divided by 2
= 7.00 ´ 10–6 (mol) or 7 ´ 10–6
mark consequentially from (e) (i) (1)
(1)

(iii)	Write the equation for the reaction between chlorine molecules, C12, and iodide ions.
Cl2 + 2I– ® I2 + Cl–    or    Cl2 + 2KI ® I2 + 2KCl
ignore state symbols (1)
(1)
(iv)	Write down the number of moles of chlorine in the sample.
7.00 ´ 10–6 mol or answer to (e) (ii) or
answer in (e)(ii) multiplied by 2 (1)
(1)
(v)	Calculate the mass of chlorine molecules in the original sample.
Mass	=	n ´ M
	=	7.00 ´ 10–6 ´ 2 ´ 35.5
	=	497 ´ 10–6g = 4.97 ´ 10–4g or 0.000497
or 0.005 or 5 ´ 10–4
or answer to (iv) multiplied by 71
or if answer in (iv) multiplied by 2 then answer in (iv) ´ 35.5 (1)
(1)
(vi)	The maximum accepted concentration of chlorine in drinking water is 0.5 parts per million by mass.
Show by calculation that the sample of water tested above is acceptable.
You may assume 1000 cm–3 of water has mass 1000 g.
1000g contains 497 ´ 10–6g
106g contains 497 ´ 10–6 ´ 103
i.e. 0.497 ppm or 0.5 ppm
Mark is for the method (´ 1000 or ¸ 1000)
If just answer (0)   (1)
(1)
(vii)	Suggest a reason why the concentration of chlorine in water must not exceed 0.5 ppm.
Chlorine is poisonous/toxic/is a toxin/harmful/irritant/
damaging to respiratory system/irritating to eyes
or unacceptable smell/taste
NOT dangerous (1)
(1)
[Total 7 marks]

18.	Read the passage on pages 13 and 14 on ESCA – a Practical Method for Determining charges on Atoms in Molecules – straight through, and then more carefully, in order to answer the following questions.

(a)	(i)	In ammonia, NH3, the ionic character of the bond between nitrogen and hydrogen (N–H) is 18%. State the charge on each hydrogen atom and hence calculate the charge on the nitrogen atom.
charge on each H +0.18 (1)
\ charge on N –0.54 (1)
If + 18 and –54 one mark max
(2)
(ii)	Explain why liquid ammonia is likely to be a good solvent for ionic compounds.
Positive ions will be powerfully attracted to the negative
nitrogen atoms in ammonia molecules or vice versa. (1)
(1)
(b)	Use the results given for the sulphate ion to calculate the charge on the sulphate ion.
charge on sulphate ion = 4(–0.78) + (+1.12)
	= –2.00 or –2
(1)
(c)	Describe the shape of the thiosulphate ion with an appropriate diagram.

accept a diagram which is slightly ambiguous but tetrahedral is written (1)
(1)
(d)	(i)	Using the normal rules, calculate the oxidation number of sulphur in the thiosuiphate ion, S2O32– .
+2 or 2+ or +II (1)
(1)
(ii)	Suggest a reason why oxidation numbers are not always the same as the true electric charge on atoms in molecules or ions consisting of several atoms.
oxidation number represents the charge an atom would bear if all the binding electrons were attributed to the most electronegative element
OR oxidation number does not take intermediate bond types/bond polarisation
into account.
OR oxidation number is not a good guide to charge when atoms form bonds to atoms of the same element.
OR reference to electron delocalisation
(or any reasonable point)
(1)

(e)	Describe the process of electron spectroscopy for chemical analysis (ESCA) and explain how it can be used to give important information about atoms in molecules, in not more than 100 words.
	You are not asked to summarise the whole passage, nor to include equations in your summary. At the end of your summary state the number of words you have used.
Examiners will need to consider each answer for (i) key points and (ii) style and use of English.  Candidates should have recorded their word total at the end of their answer, and this should be checked.
Up to 105 words: no penalty
106–110 words: –1
111–115 words: –2
116–120: –3
and at a rate of –1 penalty for every 5 words in excess thereafter, up to a maximum penalty equal to the number of key points included by the answer.
hyphenated words count as one word
KE counts as two words
EK counts as one word
Eb = hv – Ek counts as one word and cannot be counted as key point
This penalty should be recorded as a minus number alongside the word total at the end of the answer (but NOT in the margin).  Note that words appearing in the title to the summary do not count in the word total.  Normally hyphenated words, numbers and chemical formulae count as one word.  The question does not ask for equations in the summary, but if included they should be counted in the word total.
Marking for key points
One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks.  A tick should be made in the script with the number of the key point above it e.g. ³ at the point at which the examiner decided to award each mark.  The total marks for key points should be placed in the body of the script at the end of the answer, out of 6.
List of key points; these may occur in any order, and it is not necessary for them to be expressed these words provided that the sense of each point is conveyed.
Marking for key points
If deducing in excess of 3 marks for word penalty send script to PE as part of a sample
1	When an atom/molecule absorbs (energy from) X–radiation/X–rays .(1)
2	an (inner/core) electron is expelled (1)
3	The kinetic energy of (expelled) electrons can be measured.(1)
4	using an electron spectrometer .(1)

5	The kinetic energy of the (expelled) electrons depends on their
binding energy. (1)
OR Electron binding energy can be calculated from the kinetic energy of (expelled) electrons
6	which depends on the effective nuclear charge .
(or similar statement)
7	which depends on the (partial) charge on the atom
OR (partial) charges can be calculated
OR there is a linear relation between electron binding energies and partial charges on atoms ….(1)
1 mark each key point up to a maximum of 6.
To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
Style and use of English (2)
These should be impression marked on a scale 2–1–0, and the mark out of 2 should be recorded in the body of the script at the end of the answer.
The aspects to be considered are:
use of technical terms; the answer should be well organised in clear, concise English, without ambiguity.  It should read fluently, with the links between key points in the original passage maintained, even if the order is different.
Correct grammar; basic sentence structure, including punctuation, should be consistently correct, without quibbles over the finer points of grammar.
Legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace with only the occasional difficulty with a word.
Good style and use of English, with only infrequent minor faults. (2)
Frequent minor or a few major faults in style and use of English (1)
very poor style and use of English (0)
(8)
[Total 15 marks]

19.	The table below gives the molar masses and boiling points of three carbon compounds.

(a)	What is the main type of interaction between molecules in
(i)	(Dipole–dipole (interactions) (1).
(1)
(ii)	Van der Waals forces/temporary dipoles/dispersion or London forces/induced dipoles (1)
(1)

(b)	Draw a diagram to illustrate the main interaction between two molecules of butanone.
Dipole–dipole interactions shown (1)
Do NOTpenalise if wrong ketone shown
between atoms or groups with correct d+ and d– charge (1)

Correct C and O labelled d+ and d–
If charges left out (1 max); incorrect atoms (0)
If hydrogen bond in (a) mark consequentially (1 max)
(2)
(c)	Explain why pentane and 2–methylbutane have different boiling points despite having the same molecular formula and molar mass.  You may find drawing a diagram helpful.
2–methylbutane less cylindrical/less area of contact (1)
This may be shown in a diagram.

2–methylbutane	pentane
so weaker forces (1)
(2)
(d)	When equal volumes of butanone and pentane are stirred together the liquids mix and there is a drop in temperature.  When equal volumes of pentane and water are stirred together they separate into two layers when stirring is stopped.
(i)	Why is there a drop in temperature when butanone is mixed with pentane?
Good explanation in general terms (1)
comparison of strength of forces in butanone and mixture (1)
Energy or relative strength of forces should be considered.
(2)
(ii)	Explain why pentane and water do not mix.
Intermolecular forces in water very strong/hydrogen bonds in water (1)
So not enough energy released on mixing to allow bonds to break/pentane–water interactions are too weak to allow mixing. (1)
(2)
[Total 10 marks]

20.	You should to spend no more than 55 minutes on this section
	(a)	Boron, nitrogen and oxygen form fluorides with molecular formulae BF3, NF3 and OF2.
	Draw the shapes you would expect for these molecules, suggesting a value for the bond angle in each case.

NOT < 120 or > 120
Any reasonable "Y" shaped molecule
ACCEPT shape defined by orbitals
	FNF = 107(°) (accept 100 = 108.5°)
ACCEPT

ACCEPT 120 angled diagram if tetrahedral or pyramidal quoted
	FOF = 105(°) (accept 95–107° providing less than NF3
	angle must be > 90
(3)
(b)	(i)	Calculate the standard enthalpy change of atomization of nitrogen trifluoride,
DHt [NF3(g)].  Your answer should include a sign and units.
	Use the Hess cycle, and the data at 298 K:
DHt [NF3(g)]	=	–124.7 kJ mol–1
DHt [½N2(g)]	=	+472.7 kJ mol–1
DHt [½F2(g)]	=	+79.0 kJ mol–1
				
DHt [NF3(g)] = [DHt [½N2(g)] + 3DHt [½F2(g)]] – DHt [NF3(g)]
	= [(+472.7) + (3 × 79.0)]– (–124.7)
	= + 834.4 kJ mol–1  or (+ 834 kJ mol–1)
1 mark for correct application of Hess
1 mark for correct calculation of atomisation of elements
1 mark for answer, sign, units and sig. fig
–834.4 kJ mol–1 (2 marks max)
+834.4 kJ mol–1 but no working (3)
(3)
(ii)	Use your answer to (i) to calculate the bond energy of the N–F bond
E(N–F) = 
= 278.1 kJ mol–1 or (+278 kJ mol–1)
ACCEPT answer to (i) divided by 3
IGNORE sign, sig. fig., units
(1)

(c)	Another fluoride of nitrogen has the formula N2F2.
(i)	Draw a ‘dot–and–cross’ diagram for the electronic structure of N2F2 showing outer shell electrons only.

one  mark for double bond
one mark for the rest
no need to distinguish between dots and crosses
ACCEPT

(2)
(ii)	Draw diagrams to show two possible shapes for the N2F2 molecule

(1)

concentrate on cis/trans, ignore double bonds and bond angles if linear (0)
(iii)	The N–F bond energy in N2F2 is 282.0 kJ mol–1.  Suggest a reason for the differences between this value and the value you calculated for the N–F bond in NF3 in (b)(ii).
bond energies vary with the nature of the other atoms attached
to the atoms participating in a bond
Accept 'bond energy varies with environment/surroundings'
double bond affects the N–F bond strength
reasonable discussion in terms of delocalisation
(1)

(d)	BF3 and NF3 react together readily to give a solid with composition BN3NF3.
(i)	Draw a ‘dot–and 'cross' diagram for the electronic structure of BF3NF3 showing outer shell electrons only.

dots and crosses need not be distinguished (except below)
N must form the dative covalent bond
ALL outer shell electrons must be shown
NOT

(1)
(ii)	What is the type of bond between the nitrogen and boron atoms?
dative (covalent)
If  in (1) then accept covalent
(1)
(e)	Suggest a balanced equation for the reaction of 1 mole of OF2 with 1 mole of water to form two products.
		
OF2 + H2O ® 2HF + O2
1 mark for correct product, 2nd mark for correctly balanced 2nd product
H2O2 + F2 (2)
H2 + F2O2 (2)
H2 + 2FO (1)
NOT HFO or H2F2
(2)
[Total 15 marks]

21.	(a)	This part of the question is about the hydrolysis of halogenoalkanes.
	2 cm3 of ethanol is added to each of three test–tubes.
	Three drops of 1–chlorobutane are added to the first, three drops of 1–bromobutane to the second, and three drops of 1–iodobutane to the third test–tube.
	2cm3 portions of hot aqueous silver nitrate solution are added to each test–tube.
	A precipitate forms immediately in the third test–tube, slowly in the second test–tube.
	A precipitate forms immediately in the third test–tube, slowly in the second test–tube and extremely slowly in the first test–tube.
	In each reaction the precipitate is formed by silver ions. Ag+(aq), reacting with the halide ions formed by hydrolysis of the halogenoalkane.
(i)	Why was ethanol added to each test–tube?
Make halogenoalkane miscible with AgNo3 solution
ACCEPT to dissolve halogenoalkane/acts as a solvent
(1)

(ii)	The same organic product forms in each reaction.
Name this organic product.
butan–1–ol/1–butanol/n–butanol
(1)
(iii)	Complete the equation for the hydrolysis of 1–bromobutane
C4H9Br + H2O ®
C4H9Br + H2O ® C4H9OH + HBr
1 mark for each product
ACCEPT correct molecular formula
(2)
(iv)	What is the colour of the precipitate in the third test–tube?
(pale) yellow
NOT cream/yellow
(1)
(v)	Name the precipitate which forms extremely slowly in the first test–tube and write the ionic equation, including state symbols, for its formation.
silver chloride
Ag+(aq) + CI–(aq) ® AgCI(s)
one mark for state symbols and balanced equation
one mark for formulae
(3)
(vi)	Ammonia solution is added to the precipitate formed in the first test–tube.  Describe and explain what you would observe.
precipitate dissolves/clears/disappears 
a more soluble/stable complex/compound of silver with ammonia forms
accept 'a complex forms'
(2)
(vii)	Explain why the rates of hydrolysis of the three halogenoalkanes are different.
C–I bonds are weaker than (1)
C–CI bonds and break more easily (1)
(Therefore lower activation energy for reaction)
recognise bond energy or strength for first mark
recognise C–I bond weaker/lower energy second mark
CI is more electronegative than I or Br (1)
Forms the carbocation more easily in C–I (1)
(2)

(b)	1–bromobutane reacts with an alcoholic solution of potassium hydroxide at high temperature to form but–1–ene.
(i)	Draw a fully labelled diagram to show an apparatus for carrying out this reaction in the laboratory and collecting the gaseous but–1–ene

(3)
If boiling tube attached to gas syringe must have a reflux condenser between
If no ceramic wool/aluminium oxide/pumice tube must not be horizontal
If flat bottomed flask but rest OK (2 marks)
If apparatus sealed (0)
(ii)	Suggest a chemical test for an alkene such as but–1–ene.  State the colour change you would observe.
bromine (water	or	(acidified) KMnO4/potassium manganate(1)
(yellow) brown to colourless		pink to colourless (brown) (1)
red/orange/yellow or combination	NOT clear 
NOT clear
(2)
[Total 17 marks]

22.	(a)	Persulphate ions, , react with iodide ions in aqueous solution to form iodine and sulphate ions, .
(i)	What is the oxidation number of iodine in
iodine molecules, I2	0
iodine ions I–	–1/1(1)
(ii)	The oxidation number of sulphur is +6 in both sulphate and persulphate ions.
What is the oxidation number of oxygen in

(1)
(iii)	Identify the oxidant (oxidizing agent) in the reaction.
(oxygen in) persulphate
NOT ‘oxygen’
(1)
(iv)	Write a balanced equation for the reaction between persulphate ions and iodide ions.
S2O+ 2I–  ®  2SO + I2
(1)
(b)	The effect of persulphate ion concentration on the rate of this reaction can be measured as follows.
	Some starch solution and a small measured quantity of sodium thiosulphate solution are added to potassium persulphate solution.
	Potassium iodide solution is added and the time measured for the solution to change colour.
	This occurs when all the sodium thiosulphate is used up and the starch reacts with the iodine.
	The reaction is repeated using difference concentrations of potassium persulphate.
(i)	What is the final colour of the reaction mixture?
blue–black/blue/black/purple
(1)
(ii)	Why is the concentration of iodide ions constant until the reaction mixture changes colour?
as fast as I– reacts to form I2 it is reduced to form I– by 
ACCEPT I2 reacts with 
(1)
(c)	The rate of reaction is measured by taking the reciprocal of the time for the colour of the reaction mixture to change.
	Typical results are shown in the table below.

(i)	Plot a graph of 1/time on the vertical axis against concentration of persulphate ions on the horizontal axis.

correctly labelled axes
ACCEPT (s–1), in s–1, (mol dm–3), in mol dm–3,
(10–3 s–1). 1/time × 103 s–1
IGNORE aq
axes increasing from left to right and bottom to top getting more positive, linear scale 3 points out of 4 correctly plotted, straight line drawn with a ruler
(3)
(ii)	Use your graph to deduce the order of the reaction with respect to persulphate ions.
	Justify your answer.
graph is a straight line
\ rate µ [S2O82– (aq)]
\ reaction is order 1
doubling the conc doubles the rate therefore first order (2)
gradient is constant therefore rate is constant therefore zero order (1max) if just 1 order (1 max)
(2)
(d)	(i)	The reaction is first order with respect to iodide ions.  Use this information and your answer to (c)(ii) to write the overall rate equation for the reaction.
rate = k[S2O82– (aq)] [I–(aq)]
If zero order in (ii) then acceptor rate = k[(I–(aq)]
IGNORE state symbols
(1)

(ii)	What are the units of the rate constant for the reaction?
dm3 mol–1 s–1/dm6 mol–2 s–1
ACCEPT s–1 if zero order in (ii)
Mark d(i) and (ii) consequentially from c(ii)
(1)
[Total 13 marks]

23.	Read the passage on BENZENE straight through, and then more carefully, in order to answer the following questions.
(a)	(i)	Draw the displayed formula for benzene

(1)
(ii)	Write a balanced equation for the dehydrogenation of cyclohexane, C6H12, to benzene, C6H6.
C6H12  ®  C6H6 + 3H2
ACCEPT

(1)
(iii)	Suggest ONE reason why platforming might be preferred to using aluminium oxide
cheaper 
lower pressure
(1)
(b)	Classify the following reactions as substitution, elimination or addition reactions.
(i)	The reaction between benzene and chlorine to form
1,2,3,4,5,6–hexachlorocyclohexane
(free radical) addition
(1)
(ii)	The reaction between benzene and chlorethene to form ethylbenzene
substitution
(1)

(c)	Name the chemical used to produce dodecylbenzenesulphonic acid from dodecylbenzene.
fuming sulphuric acid
allow sulphur trioxide/concentrated sulphuric acid
NOT sulphuric acid
(1)
(d)	Write a summary in continuous prose, in no more than 100 words, describing the production of benzene from petroleum.
	You are NOT asked to summarise the whole passage, nor to include equations in your summary.
	At the end of your summary state the number of words you have used.
	Examiners will need to consider each answer for (i) key points and (ii) style and use of English.  Candidates should have recorded their word total at the end of their answer, and this should be checked.
	Up to 105 words: no penalty
106 – 110 words: –1
111 – 116 words: –2
117 – 120 words: –3
	and at a rate of –1 penalty for every 5 words in excess thereafter, up to a maximum penalty equal o the number of key points included by the answer.
	This penalty should be recorded as a minus number alongside the word total at the end of the answer (but NOT in the margin).  Note that words appearing in the title to the summary do not count in the word total.  Normally hyphenated words, numbers and chemical formulae count as one word.  The question does not ask for equations in the summary, but if included they should be counted in the word total.
1.	Petroleum is fractionally distilled to produce naphtha	(1)
2.	Sulphur (Impurities) are removed/sulphur is removed	(1)
3.	Naphtha is then heated to 770K (and then either passed over an)
	aluminium oxide (catalyst) at 40 atmospheres pressure 	(1)
4.	or a platinum (catalyst) at 15 atmospheres/platforming/platformed 	(1)
5.	where hexane is dehydrogenated/hydrogen is removed to cyclohexane.	(1)
6.	The cyclohexane is then dehydrogenated to form benzene, and	(1)
7.	other products which are dissolved in a solvent and (extracted) in
	fractionating towers/distilled/fractionally distilled	(1)
8.	Residual impurities are removed and further fractionation/distillation
	separates benzene from other products	(1)
	(underlined points must be included)

	Marking for key points
	One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 7 marks.  A tick should be made in the script with the number of the key point above it eg3  at the point at which the examiner decided to award each mark.  The total marks for key points should be placed in the body of the script at the end of the answer, out of 7.
	List of key points; these may occur in any order, and it is not necessary for them to be expressed in these words provided that the sense of each point is conveyed.  1 mark each keypoint up to a maximum of 7.
	To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
	Style and use of English
	These should be impression marked on a scale 2–1–0, and the mark out of 2 should be recorded in the body of the script at the end of the answer.
The aspects to be considered are:
•	use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
•	articulate expression, the answer should be well–organised in clear, concise English, without ambiguity.  It should read fluently, with the links between key points in the original passage maintained, even if the order is different.
•	correct grammar; basic sentence structure, including punctuation, should be consistently correct, without quibbles over the finer points of grammar.
•	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
Good style and use of English, with only infrequent minor faults.	(2)
Frequent minor or a few major faults in style and use of English.	(1)
Very poor style and use of English	(0)
	Credit will be given for answers written in good English, using complete sentences and with correct use of technical words.  Avoid copying long sections from the original text.  Numbers count as one word, as do standard abbreviations and hyphenated words.  Any title you give to your passage does not count in your word total.
	There are penalties for use of words in excess of 100.
[Total 15 marks]

24.		The gases hydrogen sulphide, H2S, and sulphur dioxide, SO2 react according to the equation
2H2S(g) + SO2(g) ®  3S(s) + 2H2O(1)
(a)	Explain why water is a liquid but hydrogen sulphide is a gas at room temperature, even though a hydrogen sulphide molecule has almost twice the mass of a water molecule.
Water molecules form (2) hydrogen bonds between its molecules
whereas hydrogen sulphide molecules do not....
MUST mention both substances for first mark
because oxygen is more electronegative than sulphur
or more energy is needed to break these extra bonds
(2)
(b)	Which of the four substances in the equation would you expect to have the lowest entropy at 298 K?
Justify your answer.
Sulphur (1)
because it is a solid whereas the other substances are either
liquids or gases and the particles in a solid are more ordered. (1)
(2)
(c)	(i)	Would you expect DSsystem for the reaction to have a positive or negative value?
	Justify your answer.
DSsystem negative
because (3 moles of) gas turn into a solid and (2 moles of) a liquid, 
and gases have more entropy because there are more ways of arranging 
the particles in a gas.
(2)
(ii)	The reaction is know to be spontaneous at 298 K.  
What can you deduce about DSsurrounding?
DSsurroundings must be positive as DStotal must be positive if a reaction is to occur 
and DSsurroundings must be more positive than DSsystem is negative (1)
mark consequentially from (i)
(2)
(iii)	What can you deduce about DH?
DH must be negative (as DSsurroundings = –DH/T)
ACCEPT exothermic
(1)
[Total 9 marks]

25.	Iodine is only slightly soluble in water, but it is much more soluble in organic solvents and also in aqueous potassium iodide solution.
(a)	What is the colour of iodine
(i)	grey/black
	ACCEPT purple / black or purple/grey
	NOT purple or brown or blue black
(1)

(ii)	pink/red/purple/violet/mauve/lilac
(1)
(iii)	yellow/orange/brown/reddish–brown
(1)
(b)	A solution of iodine in a hydrocarbon solvent was shaken with water until equilibrium was reached.  The mixture was allowed to settle and then 5 cm3 of the hydrocarbon solvent layer and 50 cm3 of the aqueous layer were removed and titrated separately with sodium thiosulphate solution of concentration 0.0100 mol dm–3.
	The following results were obtained.

(i)	Which indicator, if any, would you use for this titration and what would the colour change be at the end point?
starch	or	none
blue–black ® colourless		yellow ® colourless
If colour changes are in wrong order (2 max)
(3)
(ii)	The equation for the reaction between iodine and sodium thiosulphate is
I2(aq) + 2Na2S2O3(aq) ® 2NaI(aq) + Na2S4O6(aq)
	Calculate the concentration of iodine molecules, I2, in mol dm–3 in the hydrocarbon layer.
Organic layer
[I2(CCl4)] = [76 ×  × ] ×  = [7.6 × 10–2  or 0.076] mol dm–3
If correct answer but no working (2)
(2)
(iii)	Use your knowledge of intermolecular forces to explain why iodine dissolves so much more in a hydrocarbon than in water.
Iodine and hydrocarbons are molecular – van der Waals’ forces only
Water is polar with H–bonds, so energy needed to break those before iodine could dissolve
(2)

(iv)	When a solution of iodine in a hydrocarbon solvent is shaken with water the following equilibrium occurs.
I2(hydrocarbon) Û I2(water)
Write down an expression for Kc for this equilibrium and calculate its value.

NOT more than 3 sig fig
mark consequentially provided correct data is used from (ii)
(2)
(v)	Suggest two improvements to this titration procedure which would increase its accuracy.  Assume that the results table is a complete record of all of the practical work carried out.
Any two from
Each titration should be repeated.
Volume of sodium thiosulphate solution used should be measured
to 1 dec. place.
Titration value of 76 cm3 too large to measure accurately in a burette (and so the concentration of the sodium thiosulphate solution should be increased).
The volume of 5 cm3 in a pipette is less accurate than a larger volume.
All liquids should be measured to 0.1 cm3
(2)
(vi)	How would you expect the value of Kc to change when the concentration of iodine molecules, I2, in the hydrocarbon solvent at the start is doubled?
Kc unaffected/constant (by change in initial concentrations)
(1)
(c)	The experiment was repeated using potassium iodide solution instead of water.
The following equilibrium was set up in the aqueous layer:
I2(aq) + I– (aq) Û I–3(aq)
	How would you expect this to affect the volume of sodium thiosulphate solution used in the titration of the aqueous layer?
Justify your answer.
More iodine is likely to dissolve in the aqueous layer if potassium iodide is present so more Na2S2O3 solution used because I3–(aq) is formed, I2 molecules are removed and so by Le Chatelier's Principle more I2 molecules will take their place from the organic layer.
Read it overall, some understanding of equilibrium must be shown for second mark 
(2)
[Total 17 marks]

26.	A solution of iodine in a hydrocarbon solvent was shaken with water until equilibrium was reached.  The mixture was allowed to settle and then 5 cm3 of the hydrocarbon solvent layer and 50 cm3 of the aqueous layer were removed and titrated separately with sodium thiosulphate solution of concentration 0.0100 mol dm–3.
(a)	The following results were obtained.

(i)	Which indicator, if any, would you use for this titration and what would the colour change be at the end point?
starch	or	none
blue–black ® colourless		yellow ® colourless
If colour changes are in wrong order (2 max)
(3)
(ii)	The equation for the reaction between iodine and sodium thiosulphate is
I2(aq) + 2Na2S2O3(aq) ® 2NaI(aq) + Na2S4O6(aq)
	Calculate the concentration of iodine molecules, I2, in mol dm–3 in the hydrocarbon layer.
Organic layer
[I2(CCl4)] = [76 ×  × ] ×  = [7.6 × 10–2  or 0.076] mol dm–3
If correct answer but no working (2)
(2)
(iii)	Use your knowledge of intermolecular forces to explain why iodine dissolves so much more in a hydrocarbon than in water.
Iodine and hydrocarbons are molecular – van der Waals’ forces only
Water is polar with H–bonds, so energy needed to break those before iodine could dissolve
(2)

(iv)	When a solution of iodine in a hydrocarbon solvent is shaken with water the following equilibrium occurs.
I2(hydrocarbon) Û I2(water)
Write down an expression for Kc for this equilibrium and calculate its value.

NOT more than 3 sig fig
mark consequentially provided correct data is used from (ii)
(2)
(v)	Suggest two improvements to this titration procedure which would increase its accuracy.  Assume that the results table is a complete record of all of the practical work carried out.
Any two from
Each titration should be repeated.
Volume of sodium thiosulphate solution used should be measured
to 1 dec. place.
Titration value of 76 cm3 too large to measure accurately in a burette (and so the concentration of the sodium thiosulphate solution should be increased).
The volume of 5 cm3 in a pipette is less accurate than a larger volume.
All liquids should be measured to 0.1 cm3
(2)
(vi)	How would you expect the value of Kc to change when the concentration of iodine molecules, I2, in the hydrocarbon solvent at the start is doubled?
Kc unaffected/constant (by change in initial concentrations)
(1)
(b)	The experiment was repeated using potassium iodide solution instead of water.
The following equilibrium was set up in the aqueous layer:
I2(aq) + I– (aq) Û I–3(aq)
	How would you expect this to affect the volume of sodium thiosulphate solution used in the titration of the aqueous layer?
Justify your answer.
More iodine is likely to dissolve in the aqueous layer if potassium iodide is present so more Na2S2O3 solution used because I3–(aq) is formed, I2 molecules are removed and so by Le Chatelier's Principle more I2 molecules will take their place from the organic layer.
Read it overall, some understanding of equilibrium must be shown for second mark 
(2)
[Total 14 marks]

27.	Trichloromethane, CHCl3, has dipole–dipole interactions between its molecules, as has ethoxyetane, CH3CH2OCH2CH3.  Hexane, C6H14, has only van der Waals interactions.

(a)	In an experiment to measure the strength of the dipole–dipole interactions in trichloromethane, 10g of trichloromethane was mixed with 30g of hexane.
	The dipole–dipole interaction were disrupted and the temperature fell by 1.1 °C.

	Use this information to calculate a value for the strength of the dipole–dipole interactions in 1 mole of trichloromethane.
Enthalpy change
= 10 × 0.98 × 1.1 + 30 × 2.26 × 1.1
= 10.78	+ 74.58
= 83.36 Joules
Moles of trichloromethane (hexane in excess)
= 10/119.4 = 0.08375
Number of moles (1)
Mass × SHC × temp change (1)
for any number of substances
Therefore strength of dipole/dipole interaction

= (+) 1.02 kJ mol–1 (1)
(3)
(b)	When trichloromethane is mixed with ethoxyethane a hydrogen bond is formed between the molecules.  Plan how you would carry out an experiment to measure the strength of this hydrogen bond.
In your account you should include:
•	a displayed formula to show the hydrogen bond between the two molecules;
•	a justification of the quantities of each substance you would use;
•	details of the apparatus you would choose to ensure an accuracy of at least 5%;
•	any risks involved and hence safety precautions you would take;
•	details of how you would calculate the strength of the hydrogen bond from your results and the approximate value you would expect.
	(1)
(1 mark)

Displayed not essential
A good answer is likely to include the following points
1.	suitable choice of quantities that will fit into the reaction vessel (2)
1 mark for justification and 1 mark for working out quantities bonus mark for realisation that excess of one substance is better (+1)
2.	appreciation that the temperature will rise (1)
3.	the need for accurate thermometer (0.5 °C) (1)
4.	concern about heat loss to the apparatus and to the surroundings and suitable description of apparatus, diagram helpful but not essential (2)
5.	realisation of the flammability and toxicity of the reagents and hence justification of safety measures (2)
6.	expected value of + 2 ® + 50 kJ mol–1 (1)
7.	method of calculation (if not awarded in (a)) (1)
8.	measure the steady temperature after stirring/mixing (1)
If whole idea does not work (4 max)
(8)
(c)	Comment on the likely solubility of these three organic substances in water.
Hexane can only form van der Waals bonds so will be insoluble in water. (1)
Ethoxyethane and trichloromethane can both form hydrogen bonds
with water so should be soluble (1)
Not very soluble because of rest of molecule OR
Only likely to be soluble in water if the energy needed to break
the 2–H–bonds per molecule in the water can be regained by
the bonds formed between the molecule and water. (1)
(3)
[Total 15 marks]

28.	Trichloromethane, CHCl3, has dipole–dipole interactions between its molecules, as has ethoxyetane, CH3CH2OCH2CH3.  Hexane, C6H14, has only van der Waals interactions.
	In an experiment to measure the strength of the dipole–dipole interactions in trichloromethane, 10g of trichloromethane was mixed with 30g of hexane.
	The dipole–dipole interaction were disrupted and the temperature fell by 1.1 °C.

	Use this information to calculate a value for the strength of the dipole–dipole interactions in 1 mole of trichloromethane.
Enthalpy change
= 10 × 0.98 × 1.1 + 30 × 2.26 × 1.1
= 10.78	+ 74.58
= 83.36 Joules
Moles of trichloromethane (hexane in excess)
= 10/119.4 = 0.08375
Number of moles (1)
Mass × SHC × temp change (1)
for any number of substances
Therefore strength of dipole/dipole interaction

= (+) 1.02 kJ mol–1 (1)
(3)

29.	When trichloromethane is mixed with ethoxyethane a hydrogen bond is formed between the molecules.  Plan how you would carry out an experiment to measure the strength of this hydrogen bond.
(a)	In your account you should include:
•	a displayed formula to show the hydrogen bond between the two molecules;
•	a justification of the quantities of each substance you would use;
•	details of the apparatus you would choose to ensure an accuracy of at least 5%;
•	any risks involved and hence safety precautions you would take;
•	details of how you would calculate the strength of the hydrogen bond from your results and the approximate value you would expect.
	(1)
(1)
Displayed not essential
A good answer is likely to include the following points
1.	suitable choice of quantities that will fit into the reaction vessel (2)
1 mark for justification and 1 mark for working out quantities bonus mark for realisation that excess of one substance is better (+1)
2.	appreciation that the temperature will rise (1)
3.	the need for accurate thermometer (0.5 °C) (1)
4.	concern about heat loss to the apparatus and to the surroundings and suitable description of apparatus, diagram helpful but not essential (2)
5.	realisation of the flammability and toxicity of the reagents and hence justification of safety measures (2)

6.	expected value of + 2 ® + 50 kJ mol–1 (1)
7.	method of calculation (if not awarded in (a)) (1)
8.	measure the steady temperature after stirring/mixing (1)
If whole idea does not work (4 max)
(8)
(b)	Comment on the likely solubility of these three organic substances in water.
Hexane can only form van der Waals bonds so will be insoluble in water. (1)
Ethoxyethane and trichloromethane can both form hydrogen bonds
with water so should be soluble (1)
Not very soluble because of rest of molecule OR
Only likely to be soluble in water if the energy needed to break
the 2–H–bonds per molecule in the water can be regained by
the bonds formed between the molecule and water. (1)
(3)
[Total 12 marks]

30.	The boiling points and molar masses of the hydrogen halides are given in the table.

(a)	Why do the boiling points increase from hydrogen chloride to hydrogen iodide as molar mass increases?
van der Waals forces increase
(as number of electrons increase)
(1)
(b)	(i)	Name the strong attractive force between hydrogen fluoride molecules which accounts for the unusually high boiling point
hydrogen bond
(1)
(ii)	Draw a diagram showing the attractive force between two hydrogen fluoride molecules.
H–F–––H–F
(1)

31.	(a)	Boron, nitrogen and oxygen form fluorides with molecular formulae BF3, NF3 and OF2.
	Draw the shapes you would expect for these molecules, suggesting a value for the bond angle in each case.
BF3 

not < 120 or > 120
Any reasonable "Y" shaped molecule
ACCEPT shape defined by orbitals

NF3

ACCEPT	FNF = 107(°) (accept 100–108.5°)

OF2
ACCEPT 120 angled diagram if tetrahedral or pyramidal quoted
IGNORE lone pair
 	FOF = 105(°) (accept 95–107° providing less than NF3)
(3)
(b)	(i)	Calculate the standard enthalpy change eof atomization of nitrogen trifluoride,
DH [NF3(g)].  Your answer should include a sign and units.
	Use the Hess cycle, and the data at 298 K:
DH [NF3(g)]	=	–124.7 kJ mol–1
DH [½N2(g)]	=	+472.7 kJ mol–1
DH [½F2(g)]	=	+79.0 kJ mol–1

			
DH [NF3(g)] = [DH [½N2(g)] + 3DH [½F2(g)]] – DH [NF3(g)] (2)
	  = [(+472.7) + 3 × 79.0)]– (–124.7)
	  = + 834.4 kJ mol–1  or (+ 834 kJ mol–1) (1)

1 mark for correct application of Hess
1 mark for correct calculation of atomisation of elements
1 mark for answer, sign, units and sig. fig.
–834.4 kJ mol–1 (2 marks max)
+834.4 kJ mol–1 but no working (3)
(3)
(ii)	Use your answer to (i) to calculate the bond energy of the N–F bond
E(N–F) = 
= + 278.1 kJ mol–1 or (+278 kJ mol–1
ACCEPT answer to (i) divided by 3
IGNORE sign, sig.fig., units
(1)
(c)	Another fluoride of nitrogen has the formula N2F2.
(i)	Draw a 'dot–and–cross' diagram for the electronic structure of N2F2 showing outer shell electrons only.
	
one mark for double bond
one mark for the rest
no need to distinguish between dots and crosses
ACCEPT
	
(2)
(ii)	Draw diagrams to show two possible shapes for the N2F2 molecule

concentrate on cis/trans, ignore double bonds and bond angles
If linear (0)
(1)

(iii)	The N–F bond energy in N2F2 is 282.0 kJ mol–1.  Suggest a reason for the differences between this value and the value you calculated for the N–F bond in NF3 in (b)(ii).
bond energies vary with the nature of the other atoms attached to the atoms participating in a bond
Accept 'bond energy varies with environment/surroundings'
double bond affects the N–F bond strength
reasonable discussion in terms of delocalisation
 (1)
(d)	BF3 and NF3 react together readily to give a solid with composition BN3NF3.
(i)	Draw a 'dot–and 'cross' diagram for the electronic structure of BF3NF3 showing outer shell electrons only.

dots and crosses need not be distinguished (except below)
N must form the dative covalent bond
ALL outer shell electrons must be shown
NOT
 
(1)
(ii)	What is the type of bond between the nitrogen and boron atoms?
dative (covalent)
If  in (i) then accept covalent
(1)
(e)	Suggest a balanced equation for the reaction of 1 mole of OF2 with 1 mole of water to form two products.
		
OF2 + H2 O ® 2HF + O2
1 mark for correct product, 2nd mark for correctly balanced 2nd product
H2O2 + F2 (2)
H2 + F2O2 (2)
H2 + 2FO (1)
NOT HFO or H2F2
(2)
[Total 15 marks]

32.	(a)	This part of the question is about the hydrolysis of halogenoalkanes.
	2 cm3 of ethanol is added to each of three test–tubes.
	Three drops of 1–chlorobutane are added to the first, three drops of 1–bromobutane to the second, and three drops of 1–iodobutane to the third test–tube.
	2cm3 portions of hot aqueous silver nitrate solution are added toe ac test–tube.
	A precipitate forms immediately in the third test–tube, slowly in the second test–tube.
	A precipitate forms immediately in the third test–tube, slowly in the second test–tube and extremely slowly in the first test–tube.
	In each reaction the precipitate is formed by silver ions. Ag+(aq), reacting with the halide ions formed by hydrolysis of the halogenoalkane.
(i)	Why was ethanol added to each test–tube?
Make halogenoalkane miscible with AgNO3 solution
ACCEPT to dissolve halogenoalkane/acts as a solvent
(1)
(ii)	The same organic product forms in each reaction.
Name this organic product.
butan–1–ol/1 butanol/n–butanol
(1)
(iii)	Complete the equation for the hydrolysis of 1–bromobutane
C4H9Br + H2O  ®
C4H9Br + H2O ® C4H9OH + HBr
1 mark for each product
ACCEPT correct molecular formula
(iv)	What is the colour of the precipitate in the third test–tube?
(pale) yellow
NOT cream/yellow
(1)
(v)	Name the precipitate which forms extremely slowly in the first test–tube and write the ionic equation, including state symbols, for its formation.
silver chloride
Ag+(aq) + CI–(aq) ®  Ag(CI(s)
one mark for state symbols and balanced equation
one mark for formulae
 (3)
(vi)	Ammonia solution is added to the precipitate formed in the first test–tube.  Describe and explain what you would observe.
precipitate dissolves/clears/disappears
a more soluble/stable complex/compound of silver with ammonia forms
accept 'a complex forms'
(2)

(vii)	Explain why the rates of hydrolysis of the three halogenoalkanes are different
C–I bonds are weaker than
C_CI bonds and break more easily
(Therefore lower activation energy for reaction)
recognise bond energy or strength for first mark
recognise C–I bond weaker/lower energy second mark
CI is more electronegative than I or Br (1)
forms the carbocation more easily in C–I (1)
(2)
(b)	1–bromobutane reacts with an alcoholic solution of potassium hydroxide at high temperature to form but–1–ene.
(i)	Draw a fully labelled diagram to show an apparatus for carrying out this reaction in the laboratory and collecting the gaseous but–1–ene

If boiling tube attached to gas syringe must have a reflux condenser between
If no ceramic wool/aluminium oxide/pumice tube must not be horizontal
If flat bottomed flask but rest OK (2 marks)
If apparatus sealed (0)
(3)
(ii)	Suggest a chemical test for an alkene such as but–1–ene.  State the colour change you would observe.
bromine (water	or	(acidified) KMnO4/potassium manganate
(yellow) brown to colourless	pink to colourless (brown)
red/orange/yellow or combination	NOT clear
NOT CLEAR
(2)
[Total 17 marks]

33.	(a)	Persulphate ions, S2O82–, react with iodide ions in aqueous solution to form iodine and sulphate ions.  SO42–.
(i)	What is the oxidation number of iodine in
I2	0
I–	–1/1
(1)
(ii)	The oxidation number of sulphur is +6 in both sulphate and persulphate ions.
What is the oxidation number of oxygen in
SO42–.	–2/2S2O82.	=7/4 OR 7/4– OR 14/8 OR 1.75
(1)

(iii)	Identify the oxidant (oxidizing agent) in the reaction.
(oxygen in) persulphate
NOT 'oxygen'
(1)
(iv)	Write a balanced equation for the reaction between persulphate ions and iodide ions.
S2O82– + 2I–  ®  2SO42–  + I2
(1)
(b)	The effect of persulphate ion concentration on the rate of this reaction can be measured as follows.
	Some starch solution and a small measured quantity of sodium thiosulphate solution are added to potassium persulphate solution.
	Potassium iodide solution is added and the time measured for the solution to change colour.
	This occurs when all the sodium thiosulphate is used up and the starch reacts with the iodine.
	The reaction is repeated using difference concentrations of potassium persulphate.
(i)	What is the final colour of the reaction mixture?
blue–black / blue / black / purple
(1)
(ii)	Why is the concentration of iodide ions constant until the reaction mixture changes colour?
as fast as I– reacts to form I2 it is reduced to form I– by S2O32–
ACCEPT 'I2 reacts with S2O32–1
(1)
(c)	The rate of reaction is measured by taking the reciprocal of the time for the colour of the reaction mixture to change.
	Typical results are shown in the table below.

(i)	Plot a graph of 1/time on the vertical axis against concentration of persulphate ions on the horizontal axis.

correctly labelled axes
ACCEPT (s–1), in s–1, (mol dm–3), in mol dm–3,
(10–3 s–1). 1/time × 103 s–1
IGNORE aq
axes increasing from left to right and bottom to top getting more positive,
linear scale
3 points out of 4 correctly plotted, straight line drawn with a ruler
(3)
(ii)	Use your graph to deduce the order of the reaction with respect to persulphate ions.
	Justify your answer.
graph is a straight line
\ rate µ [S2O82– (aq)]
\ reaction is order 1
doubling the conc doubles the rate therefore first order (2)
gradient is constant therefore rate is constant therefore zero order (1max)
if just 1 order (1 max)
(2)

(d)	(i)	The reaction is first order with respect to iodide ions.  use this information and your answer to (c)(ii) tow rite the overall rate equation for the reaction.
rate = k [S2O82– (aq) [I–(aq)]
If zero order in (ii) then acceptor rate = k(I–(aq)]
IGNORE state symbols
(1)
(ii)	What are the units of the rate constant for the reaction?
dm3 mol–1 s–1/dm6 mol–2 s–1
ACCEPT s–1 if zero order in (ii)
Mark d(i) and (ii) consequentially from c(ii)
(1)
[Total 13 marks]

34.	Ionization energies in Kj mol–1 of three elements R, Sand T which are consecutive in atomic number are.

(a)	In which Group of Periodic Table would T be found?  Justify your answer.
Group 1 (1), ‘jump’ between (1) 1st and 2nd IE
ACCEPT S is a noble gas (1) and T follows it (1)
(2)
(b)	Estimate the 8th ionization energy of the element S to 2 significant figures.
14000, 15000
13000, 16000 (1)
IGNORE sig figs and units
(2)
(c)	What type of electron (s, p, etc) is removed when an atom of R is ionized by the removal of 1 electron?
p electron
(1)
(d)	Write an equation with state symbols to represent the reaction accompanied by the second ionization energy of T (use T as the symbol for the element).
T+ (g) ® T2+(g) + e– / T+(g) – e– ® T2+(g)
state symbols essential for 2 marks
1 or 2 incorrect state symbols (1 max)
ACCEPT e for electron
–1 mark per error
(2)

(e)	The sketch graph shows the first ionization energies of R, S and T.  Continue the sketch so as to show the pattern of the first ionization energies of the next three elements of the Periodic Table, U, V and W, assuming that transition elements are not involved.

(3)
(f)	The first six ionization energies of another element M are

Explain why M cannot have an atomic number less than 12.
in Group 2 (1)
not Be (Be only has 4 electrons) (1)
\ must be Mg or higher (1)
OR
There are 2 s–type electrons outermost. (1)
4 electrons which must be part of 8 (1)
Must be at least 2 electrons inside of this (1)
(3)
[Total 13 marks]

35.	(a)	Propane C3H8, is used as a feedstock for making ethene using a cracking process.  
The yield of ethene is about 42%.
(i)	Write an equation for the cracking or propane to form ethene using displayed formulae.

(2)
(ii)	Draw a labelled diagram of an ethene molecule, showing the electron density distribution in thes and pbonds between the carbon atoms.

both correctly drawn but no labels (1 max)
one correctly drawn and both labels (1 max)
sigma bond cannot be represented by a line
(2)
(b)	The cracking of propane also produces propene.

	Calculate the standard enthalpy change for this reaction.  Use the data, at 298 K.
DHf[C3H8(g)] = –104.5 kJ mol–1
DHf [C3H6(g)] = +20.2 kJ mol–1
	Your answer should include a sign and units.
DHt	= DHt[C3H6] – DHt [C3H8]
	= +20.2 – (–104.5)
	= +124.7 (125) kJ mol–1
value (1) sign/unit (1)
if positive sign not shown (–1)
–124.7 kJ mol–1 with some correct reasoning (1 max)
(2)

(c)	Propene can be used to make a number of important chemical products.  The processes involved can be summarized in the diagram:

(i)	Give the displayed formula and name of compound A.
	 (1)
1,2– dibromopropane (1)
comma and hyphen can be omitted
 (2)
(ii)	State the type of reaction occurring and the type of reagent used in the formation of compound A.
addition (1)
electrophile (1)
(2)
(iii)	Give the formulae of compounds B and C.
B 	H2SO4/H3PO4	or 	B H2O (1)
C 	H2O	C AI2O3 (1)
B and C can be reversed
B Hbr
C NaOH/KOH
(2)
(iv)	Write a balanced equation for the formation of poly(propene) for propene in Reaction 1.

1st mark for stoichiometry, n moles of C3H6 ® (C3H6)n
2nd mark for structure showing CH3 group attached to polymer chain
(2)

(v)	Suggest a reagent and condition for Reaction 2.
CI2 (1)
ACCEPT name
UV (light) / sunlight / photoflood light (1)
NOT light
(2)
(vi)	State the type of mechanism in the substitution in Reaction 2.
free radical (substitution)
NOT free radical/nucleophile (0)
(1)
(vii)	Give the systematic name for CH2==CH—CH2—C1
1–chloroprop–2–ene or 3–chloroprop–1–ene or 3–chloropropene
(1)
[Total 18 marks]

36.	(a)	Iodine reacts with hot concentrated sodium hydroxide solution according to the equation:

(i)	What is the oxidation number of iodine in
I2	    0	1®0
NaI 	–1	2®1
NaIO3  	+5	3®2
(2)
(ii)	What is the name given to this type of redox reaction?
disproportionation
(1)
(b)	An experiment was carried out to determine the purity of a sample of sodium iodate, NaIO3, made by the reaction described in (a).
	0.060 g of the sample was dissolved in pure water to make 100cm3 of solution.
	A 10.0 cm3 portion of this solution was taken and added to an excess of acidified potassium iodide solution,  The iodine liberated was titrated with 0.0100M sodium thiosulphate solution.
(i)	What piece of apparatus would you use to measure out the 10.0 cm3 sample?
pipette
(1)
(ii)	Name a suitable indicator to use for this titration and give the colour change expected.
starch (1)
(dark) blue or black or blue/black or black/blue ® colourless (2)
NOT purple or clear
If colours given wrong way round (1 max)
mark independently
ACCEPT no indicator, yellow ® colourless (2 max)
no indicator, colourless ® yellow (1 max)
(3)
(iii)	The volume of sodium thiosulphate solution required in the titration was 16.7 cm3.
	Calculate the number of moles of sodium thiosulphate used.

correct answer no working (1)
(1)
(iv)	The equation for the reaction between iodine, I2, and thiosulphate ions, , is

	Calculate the number of moles of iodine molecules.  I2, which reacted with the sodium thiosulphate solution.

ie 
correct answer no working (1)
(1)
(v)	The equation for the reaction between sodium iodide and sodium iodate, NaIO3, is

Calculate the number of moles of sodium iodate, NaIO3, in the original sample.

ie 
correct answer no working (1)
(1)
(vi)	Calculate the mass of sodium iodate, NaIO3, in the original sample and hence calculate the percentage purity of the sample to 2 significant figures.
(Molar mass of NaIO3 = 198 g mol–1)
0.278 ´ 10–3 × 198 (=0.0551) (1)
ie answer (v) × 198 (1)
% purity = 
	 = 91 –92% (1)
ie   with their correct value to 2 or 3 sig figs (1)
(2)

(c)	When a similar experiment is repeated, making and analysing potassium iodate, KIO3, a higher percentage purity is obtained.  Suggest a reason for this.
Either NaIO3 may contain some water of crystallization
or NaIO3 may be more soluble
or NaI may be less soluble than KI
(or answers expressed in terms of potassium salts)
or I2 is soluble in KI
or KIO3 more stable than NaIO3 (or reversed argument)
NOT potassium more reactive
(1)
(d)	An acidified solution of sodium iodate, NaIO3, is added to a solution of iron(II) sulphate.  Describe and explain what you would expect to see.
A yellow or brown or yellow/brown (solution) forms (1)
as Fe2+ (oxidized) to Fe3+ (1)
(mark independently)
(2)
[Total 15 marks]

37.	Some organic compounds having molecules containing eight carbon atoms are listed below with their molar masses and boiling points.

(a)	(i)	What type of intermolecular force is present in all of the se compounds?
Van der Waals’ (forces)
reasonable spelling
(1)
(ii)	2,2,4–trimethylpentane has a lower boiling point than its isomer, octane.  Suggest a reason for this.
You may find it helpful to draw a diagram.

molecular shape, described or diagrams linked to packing (1)
Explanation why packing influences van der Waals’ forces/boiling point (1)
(2)

(b)	(i)	Suggest ONE reason why octanoic acid has a higher boiling point than octan–1–ol.
stronger dipole/dipole attraction (and correct justification) (1)
ACCEPT stronger van der Waals and correct justification (1)
NOT explanation in terms of hydrogen bonding
(1)
(ii)	Estimate the boiling point of octanal and justify your estimate.
(Molar mass of CH3(CH2)6CHO = 128 g mol–1).
B.p within range 150 – 185 °C; (true value 171 °C) (1)
octanal has van der Waals + dipole/dipole attractions but octane
only has van der Waals (1)
OR octanal has dipole/dipole but octan–1–ol has hydrogen bonds (1)
Mark independently
(2)
(iii)	Would you expect octan–1–ol to dissolve readily in water?
Justify your answer.
Slightly soluble because the –OH group will form H–bonds
with water (1)
The long carbon cain will not interact with water hence 
insoluble / slightly soluble (1)
(2)
[Total 8 marks]

38.	Economic and environmental factors must be considered if anew fuel is to be used.
(i)	Do you think the production of hydrogen forms team is likely to cost–effective?  Give a reason for your answer.
No, because of the large energy expenditure
in raising and decomposing steam ie energy cost
NOT a high temperature / pressure is needed
NOT water is cheap and readily available
(1)
(ii)	State and explain one environmental advantage of hydrogen over fossil fuels.
Fossil fuels produce CO2 greenhouse gas / global warming gas
OR no carbon particles produced
OR SO2 which can lead to acid rain
OR mining and quarrying of fossil fuels disfigures landscape etc.
NOT SO2 is a pollutant
NOT fossil fuels finite resource 
(1)
[Total 2 marks]

39.	Cynogen, (CN)2, behaves like the halogens in its chemical reactions.  Its hydride, hydrogen cyanide, HCN, is similar to the hydrogen halides.
(a)	(i)	Draw ‘dot-and-cross’ diagrams of hydrogen chloride, hydrogen cyanide and cyanogen molecules.  You need include only outer shell electrons.

or similar
ACCEPT

(3)
(ii)	Draw the displayed formula for cyanogen showing the CCN bond angle.
	
Correctly bonded structure (1) eg ACCEPT 
180° bond angle (1) eg
 (1 max)
(2)
(b)	Use the Hess cycle and the data, at 298 K, to calculate the standard enthalpy change of formation of cyanogen, including its sign and unit.
DH [(CN)2(g)]	=	+2121 kJ mol-1
DH [C(graphite)]	=	+717 kJ mol-1
DH [½N2(g)]	=	+473 kJ mol-1

DH1 = DHf [(CN)2(g) + DHat [(CN)2(g)]
(or similar statement applying Hess’ law) (1)
DHat[(CN)2(g)] = 2(717) + 2(473) – 2121
correct numbers and multiples (1)
= +259 kJ mol-1
correct value, sign, unit (1)
If Hess applied incorrectly but correct multiples accepted answers are:
-259 kJ mol-1
-4501 kJ mol-1
+4501 kJ mol-1
(2 max)
If wrong multiples used acceptable answers are:
-931 kJ mol-1
-458 kJ mol-1
-214 kJ mol-1
(2 max)
If wrong in Hess and multiples (0)
(3)
(c)	Potassium cyanide, KCN, reacts with concentrated sulphuric acid in a similar way to potassium iodide.  Suggest THREE possible gaseous products.
HCN (1)
(CN)2 (1)
either one or two from
SO2, H2S, CO2, CO and NH3
to give a total maximum mark of 3
(3)
(d)	Chlorine reacts with cold sodium hydroxide solution in a disproportionation reaction.

(i)	State the oxidation number of chlorine in
C12.		0
NaC1	-1	or 1NaC1O	+1	or 1+	3®2
				2®1
				1,0®0
(2)
(ii)	What is meant by the term disproportionation?
chlorine/an element oxidizes and reduces itself
(chlorine disproportionates is not acceptable)
(1)
(iii)	Suggest a balanced chemical equation for the reaction between cyanogen and cold sodium hydroxide solution.
(CN)2 + 2NaOH ® NaCN + NaCNO + H2O
(1)
(e)	Suggest a reason why there are no experiments on the chemistry of cyanogen in Advanced level chemistry courses.
HCN (may form easily from it which is very) poisonous/toxic or (CN)2
is poisonous/toxic
(1)
[Total 16 marks]

40.	Cynogen, (CN)2, behaves like the halogens in its chemical reactions.  Its hydride, hydrogen cyanide, HCN, is similar to the hydrogen halides.
(i)	Draw ‘dot-and-cross’ diagrams of hydrogen chloride, hydrogen cyanide and cyanogen molecules.  You need include only outer shell electrons.

or similar
ACCEPT

(3)
(ii)	Draw the displayed formula for cyanogen showing the CCN bond angle.
	
Correctly bonded structure (1) eg ACCEPT 
180° bond angle (1) eg
 (1 max)
(2)
[Total 5 marks]

41.	Use the Hess cycle and the data, at 298 K, to calculate the standard enthalpy change of formation of cyanogen, including its sign and unit.
DH [(CN)2(g)]	=	+2121 kJ mol-1
DH [C(graphite)]	=	+717 kJ mol-1
DH [½N2(g)]	=	+473 kJ mol-1

DH1 = DHf [(CN)2(g) + DHat [(CN)2(g)]
(or similar statement applying Hess’ law) (1)
DHat[(CN)2(g)] = 2(717) + 2(473) – 2121
correct numbers and multiples (1)
= +259 kJ mol-1
correct value, sign, unit (1)
If Hess applied incorrectly but correct multiples accepted answers are:
-259 kJ mol-1
-4501 kJ mol-1
+4501 kJ mol-1
(2 max)
If wrong multiples used acceptable answers are:
-931 kJ mol-1
-458 kJ mol-1
-214 kJ mol-1
(2 max)
If wrong in Hess and multiples (0)
(3)

42.	Potassium cyanide, KCN, reacts with concentrated sulphuric acid in a similar way to potassium iodide.  Suggest THREE possible gaseous products.
HCN (1)
(CN)2 (1)
either one or two from
SO2, H2S, CO2, CO and NH3
to give a total maximum mark of 3
(3)

43.	Chlorine reacts with cold sodium hydroxide solution in a disproportionation reaction.

(i)	State the oxidation number of chlorine in
C12.		0
NaC1	-1	or 1NaC1O	+1	or 1+	3®2
				2®1
				1,0®0
(2)
(ii)	What is meant by the term disproportionation?
chlorine/an element oxidizes and reduces itself
(chlorine disproportionates is not acceptable)
(1)
(iii)	Suggest a balanced chemical equation for the reaction between cyanogen and cold sodium hydroxide solution.
(CN)2 + 2NaOH ® NaCN + NaCNO + H2O
(1)
[Total 4 marks]

44.	Suggest a reason why there are no experiments on the chemistry of cyanogen in Advanced level chemistry courses.
HCN (may form easily from it which is very) poisonous/toxic or (CN)2
is poisonous/toxic
(1)

45.	Butane reacts with chlorine, in a free radical chain reaction, to form 1-chlorobutane as one of the products.  The reaction takes place in a number of steps:

(i)	What condition is needed to promote Step 1?
	UV or sunlight
(1)
(ii)	What type of bond breaking occurs in Step 1?
homolytic
(1)
(iii)	Classify the type of reaction step occurring in Steps 2 and  3.
propagation
(1)
(iv)	Suggest an equation for another possible chain termination step.(1 mark)C4H9- + C4H9- ® C8H18
or CI. + CI. ® CI2
(1)

(v)	Give one other example of a reaction with a free radical chain mechanism.
formation of polymers (poly(ethene))
or combustion of hydrocarbons
or reaction of benzene with chlorine
or reactions in the ozone layer
or hydrogen with chlorine
NOT alkanes + halogens
(1)
[Total 5 marks]

46.	Read the passage below on GROWING DIAMONDS straight through, and then more carefully, in order to answer the following questions.
(a)	Give TWO reasons why alchemists tried to make gold from copper, rather than diamonds from graphite.
Any two of the following:
Allotropy 	They did not know diamond and graphite are different forms of the same element
or They did not know diamond and graphite are both made of carbon (1)
Appearance 	gold resembles copper
or diamonds do not look like graphite (1)
hardness 	diamond is hard, graphite is soft (1)
Any two
(2)
(b)	 (i)	State the physical property of diamond which suggested that high pressure would be needed in its synthesis from graphite.
 (1 mark)Density
(1)
(ii)	Explain why high pressure is needed in the synthesis of diamond from graphite
High pressure required to force/squeeze atoms/carbons into close/tighter packing arrangement/closer together
(1)
(iii)	Suggest a reason why high temperatures are needed to make diamond from graphite.
High activation energy of reaction
OR (energy) required to break bonds (before new bonds made)
OR to melt the graphite
(1)
(iv)	Where do many scientists believe that the conditions of pressure and temperature for making diamonds occur naturally?
100 to 200/several hundred miles below the Earth’s surface
(1)
(c)	Explain why large natural diamonds rather than synthetic ones are still used in jewellery.
Time for diamonds to grow makes the process too costly
OR wrong colour for jewellery and not of gem quality
(1)

(d)	Describe in no more than 100 words how diamonds can be made from graphite.

	Cross-section of apparatus used in diamond making. Synthetic diamonds appear within the graphite in the cylindrical cell, which is squeezed to a pressure of 1.0 × 105 atm by hydraulically driven pistons. A strong belt of concentric metal rings keeps the material from sqeezing out the sides.
Examiners will need to consider each answer for (i) key points and (ii) style and use of English.  Candidates should have recorded their word total at the end of their answer, and this should be checked.
Up to 105 words: no penalty
106 – 115 words: -1
116 – 125 words: -2
126 – 135 words: -3
and at a rate of -1 penalty for every 5 words in excess thereafter, up to a maximum penalty equal to the number of key points included by the answer.
This penalty should be recorded as a minus number alongside the word total at the end of the answer (but NOT in the margin).  Note that words appearing in the title to the summary do not count in the word total.  Normally hyphenated words, numbers and chemical formulae count as one word.  The question does not ask for equations in the summary, but if included they should be counted in the word total.
1200-2400 °C = 4 words
FeS = 2 words
105 atmosphere ie number and unit = 2 words

Marking for key points (6)
One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks.  A tick should be made in the script with the number of the key point above it e.g. 3Ö at the point at which the examiner decided to award each mark.  The total marks for key points should be placed in the body of the script at the end of the answer, out of 6.
List of key points; these may occur in any order (but see below), and it is not necessary for them to be expressed in these words provided that the sense of each point is conveyed, but sections in bold are essential.
If deducting in excess of 3 marks for word penalty send script to PE as part of a sample.
1.and high temperature, up to 2500 °C, are needed
(accept 1200 – 2400 °C) (1)
2.To make diamonds from graphite high pressure, 1.0 x 105 (1)
3.Graphite is mixed with a metal which melts below 2500 °C (1)
4.and the mixture heated till the molten metal dissolves the graphite (1)
5.under pressure supplied by pistons (1)
6.and a belt of concentric metal rings (needed to contain the mixture) (1)
1 mark for each key point up to a maximum of 6.
To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
Style and use of English (2 marks)
These should be impression marked on a scale 2-1-0, and the mark out of 2 should be recorded in the body of the script at the end of the answer.
The aspects to be considered are:
•	use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which re involved in the key points.
•	articulate expression, the answer should be well-organised in clear, concise English, without ambiguity.  It should read fluently with the links between key points in the original passage maintained.
•	correct grammar; basic sentence structure, including punctuation, should be consistently correct, without quibbles over the finer points of grammar.
•	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty 
with a word.
•	points must be in a logical order.
Good style and use of English, with only infrequent minor faults. (2)
Frequent minor or a few major faults in style and use of English. (1)
Very poor style and use of English (0)
(8)
(Total 15 marks)

47.	The boiling points of four compounds which have hydrogen bonds between their molecules are shown below:

(a)	Draw a diagram showing the position of a hydrogen bond between two molecules of methanol.
One bond from O to the hydrogen in the -OH group

Allow ethanol formulae
(1)
(b)	By considering the structures and intermolecular forces involved, explain
(i)	why water has a higher boiling point than ethanol,
Water can from 2 H-bonds per molecule (1)
methanol/ethanol only one (1)
Water has stronger H-bonds + justification (1 max)
(2)
(ii)	why methanol has a lower boiling point than ethanol.
Methanol has similar strength hydrogen bonds to ethanol (1)
and has lower van der Waals forces (1)
or
lower van der Waals forces (1) as it has fewer electrons/shorter Hydrocarbon chain/smaller area contact (1) NOT molar mass
(2)
(c)	Butan-1-ol has an isomer named ethoxyethane, C2H5OC2H5
	Explain why there are  no hydrogen bonds between ethoxyethane molecules.
No H-bonds as the hydrogen atoms are not bonded directly to oxygen/another electronegative element
or C-H bonds are not polar.
NOT reference to positive H’s.
(1)
(d)	Iodine is almost insoluble in water but is more soluble in ethoxyethane.
(i)	What type of force is present between molecules of iodine in a crystal?
Van der Waals’ forces or induced dipole
NOT dipole
(1)

(ii)	Explain why iodine is more soluble in ethoxyethane than in water.
Bonds between water molecules are much stronger than between ethoxyethane molecules. (1)
Bonds between ethoxyethane molecules are similar strength to iodine (so it can penetrate lattice). (1)
Iodine can not interact strongly with water (to compensate for bonds broken)
or bonds between iodine molecules are much weaker than between water molecules 
or statement comparing strengths of bonds formed with strength of bond broken (1)
Argument based on polar dissolves in polar but not in non-polar (1 max)
NOT statement of types of bonds or “like dissolves like”
3 comparisons required for 3 marks
ie iodine/ethoxyethane; ethoxyethane/water; iodine/water
(3)
[Total 10 marks]

48.	This question is concerned with the molecular interactions between the polymer chains in various semi-crystalline plastic materials in common use.

(a)	Complete the right hand column to show the strongest type of bonding present between polymer chains.
Low density Poly(ethene van der Waals (1)
High density poly(ethene) van der Waals
Terylene dipole-dipole (1)
Nylon hydrogen bonds (1)
(3)
(b)	(i)	Why are the densities of the two types of poly(ethene) different?
Low density poly(ethene) has short side chain
branching off from the main chain so molecules do not pack as well giving a greater volume and so lower density.
Must be a comparison
(1)

(ii)	Why is the melting temperature of high density poly(ethene) greater than that of low density poly(ethene)?
Stronger (van der Waals) forces in high density poly(ethene) because more surface of the molecules in contact and so more energy is needed to break these stronger bonds and so higher melting temperature
(1)
(c)	Why is the melting temperature of bakelite not given?
Bakelite chars/decomposes/breaks down on heating rather than melting or larger number of cross-links prevents melting
or Bakelite is a thermoset
(1)
(d)	Suggest FOUR further pieces of data that you would like to see included in the table if you wished to evaluate the suitability of various polymers for use as a rope on board a boat.
Any four from
tensile strength (Young’s modulus)
Wet strength
water absorbance
elasticity, (flexural strengths, flexibility)
density
cost
glass temperature
or other suitable properties
NOT solubility or availability
(4)
[Total 10 marks]

49.	(a)	4.50 g of water is evaporated by 10.0 kJ of electrical energy.
(i)	calculate the number of moles of water evaporated.
Number of moles of water = 4.5/18 = 0.250
(1)
(ii)	Calculate the enthalpy change of vaporization per mole of water. 
Your answer should include both sign and units.
DHvap = 10.0/0.250 = + 40.0 kJ mol-1
(1)
(b)	(i)	Draw a diagram to show the principal molecular force between two water molecules.

Correct atoms (1)
linear O-H...O (1)
(2)
(ii)	Explain why the value for the enthalpy change of vaporization of water is much greater than that for the other Group 6 hydrides.
H2O hydrogen bonds much stronger than (van der Waals')
forces in other compounds.
(1)
[Total 5 marks]

50.	(a)	A and D Or names	1
(b)	D and E Or names	1
both needed for each mark

51.	(a)	(i)		2
Ö (includes correct stoichiometry)
Ö (Ignore stoichiometry)
(ii)	lron(III) bromide or iron	1
(b)	(i)	Bromine	1
	Sunlight/ UV light       (1)
NOT light
	Free radical	1
	Substitution        (1)   (4)
(ii)	C2 H4+ HBr ® CH3CH2Br
	Ö	Ö	2
(iii)	Nucleophilic	1
	Substitution       (1)	2
(c)	(i)	Phosphoric acid or H3PO4	1
(ii)	HBr is oxidized
Or conc.H2SO4is an oxidizing agent	1
ACCEPT if Br2 is mentioned (oxidation implied)
(iii)	HBr reacts immediately       (1)
Or mechanistic point e.g. protonation of –OH group
(iv)	Water rises/fills up the test–tube.	1
	HBr is soluble in water	2

52.	(a)	
	1
	\Hat = [4 × 716.7 + 9 × 218.0 + 106.8] – [–52.0]	1
	= + 4987.6 kJ mol–1 [or + 4990, or + 5000]	3

(b)	(i)	

	carbon chain including C–1 bond	1
	iodine       (1)	2
(ii)	overall shape ÖÖ correct multiples
E(C–1) = He – 9E(C–H) – 3E(C–C)	2
	= + 4987.6 – 9 × 413 – 3 × 347
= (+) 229.6 kJ mol–1 (exact) = (+) 230 kJ mol–1     (1)	3
(c)	(i)	Any yellow precipitate unless pre–fixed by cream  (1)
           
Ag+(aq) + 1–(aq) ® Agl(s)
	symbols    (1) states    (1)	3
(ii)	Butan–l–ol	1
(d)	C–1 bond compared to C–Cl bond    (1) is weaker    (1)	2
	[Chlorine is more electronegative than iodine (1 max)]
(e)	(i)	
	1
(ii)	A white precipitate	1
(iii)	Any two pertinent points, eg electron donating effect of methyl
groups and stability of carbocation	2

53.	(a)	(i)	The compound is (energetically)more stable	1
	NOT to complete outer shell electrons
(ii)	Either
Inner shells are full/complete so there is no room for more electrons
NOT inner shells are stable
Or
The ionization energy for an inner electron is too high
Or
The electron held more tightly/harder to remove/attracted
more strongly to the nucleus	1
(iii)	Either a partly filled shell
Or outer shell of electrons that take part in bonding	1

(iv)	Only fluorine and oxygen form bonds
with sufficient energy/sufficiently strong bonds       (1)
to promote electrons in krypton and xenon to a
higher energy level/to break into inner/filled shells    (1)	2
(b)	Either: in metals, electrons are not associated with particular atoms
Or: electrons are free to move/sea of electrons description	1
(c)	All compounds fall inside the triangle as none are purely ionic 
or covalent (or metallic)	1
(d)	Examiners will need to consider each answer for (i) key points and (ii) style
and use of English.  Candidates should have recorded their word total at the
end of their answer, and this should be checked (noting the word count in 
blocks of 20).
	up to 105 words: no penalty
106 – 115 words: 1
116–125words:2
126 – 135 words: 3
	and at a rate of 1 penalty for every 5 words excess thereafter, up to a maximum penalty equal to the number of key points included by the answer.
	This penalty should be recorded as a minus number alongside the word total at the end of the answer– Note that words appearing in the title to the summary do not count in the word total.  Normally hyphenated words, numbers and chemical formulae count as one word.  The question does not ask for equations in the summary, but if included they should be counted in the word total.
1.	There are three types of chemical bond.  They are ionic, covalent 
and metallic.	1
2.	In ionic bonding, electrons are transferred from metal (atoms) to
non–metal (atoms).	1
3.	(Coulombic) attractions between ions (lead to the binding energy 
between ions.)	1
4.	Ionic bonding is non–directional/Covalent bonding is directed 
Both needed	1
5.	In covalent bonding, electrons occupy the (region of) space
between the atoms.	1
6.	The nuclei approach each other more closely (and achieve a
lower energy state.)	1
7.	In metals (bonding) electrons are free to move (throughout the metal)	1
NOT metallic molecule
Any 6 points
[6]

	Marking for key points
	One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks.  A tick should be made in the script with the number of the key point above it eg 3 Ö at the point at which the examiner decided to award each mark.  The total marks for key points should be placed in the body of the script at the end of the answer, out of 6.
	List of key points; these may occur in any order, and it is not necessary for them to be expressed in these words provided that the sense of each point is conveyed.
To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.

	Style and use of English
	These should be impression marked on a scale 2–1–0, and the mark out of 2 should be recorded in the body of the script at the end of the answer.
	The aspects to be considered are:
•	use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
•	articulate expression; the answer should be well–organised in clear, concise English, without ambiguity.  It should read fluently, with the links between key points in the original passage maintained, even if the order is different. 
•	correct grammar; basic sentence structure, including punctuation, should be consistently correct, without quibbles over the finer points of grammar.
•	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word
•	points must be in logical order
	Good style and use of English, with only infrequent minor faults.	2
	Frequent minor or a few major faults in style and use of English.	1
	Very poor style and use of English.	0
[8]

54.	(a)	(i)	They have more electrons/a lot of contact	1
(ii)	Strength or high melting point	1
(iii)	Hydrogen bonds	1
(iv)	O–H bond polarized –O¾ H+        (1)
So H atoms attracted to O in other molecules        (1)	2
(v)	
	than poly(ethene)/poly(ethene) or water/water        (1) 
So insufficient energy available for breaking 
existing intermolecular ‘bonds’       (1)	2
(b)	H2O molecules correctly aligned at edge of crystal        (1)
Na+ and Cl– with correctly orientated H2O molecules
in solution        (1)	2
(c)	(i)
	3
(ii)	–446 + 409 = –37 kJ mol–1	1
(iii)	Temperature rise	1
	More energy released when bonds formed between ions 
and water than needed to break ionic bonds and 
hydrogen bonds.        (1)	2

55.	(a)	(i) and (ii)
Award any of these points to a maximum of 4 marks for part (a) 
The electron density falls to zero between sodium and chloride ions (1)
Mention of discrete ions (1)
No contours go round more than one nucleus/atom (1)
There is significant electron density between the nuclei in the hydrogen molecule.
Any four
	Some contours go round both nuclei (1)
Electrons are transferred to form ions (1)
Electrons are shared in molecules (1)	4
(iii)	H × H or H ·  H	1
(b)	(i)	Electrons are delocalized	1
(ii)	Some electron density contours form a ring around all six carbon atoms/
electrons evenly spread.	1
[7]

56.	(a)	(i)	+2
+4
   0	2
3 ® 2
2 ® 1
1 ® 0
(ii)	Disproportionation (1)
An element/sulphur is oxidized and reduced
OR gains and loses electrons at the same time	2
(b)	(i)	Heat in (boiling tubes) in the same water bath	1

(ii)	Attach a piece of paper marked with a dark ink spot to the outside of a glass beaker (1)
Time for spot to disappear as reaction occurs in the beaker measured with a stop watch (1)
OR
volume of gas (1)
Time to a fixed volume (1)
OR reasonable suggestion	2
(iii)	rate = 	1
(c)	(i)	3.10 × 10–3  –3.40      only acceptable answers	2
(ii)	Labels on axes including units (1)
Axes correct, reasonable scales (one axis must use 8 cm squares) (1)
4/5 points correct/straight line drawn (1)	3
(iii)	Gradient = – 5000 K
(with sign in the range 4200–5400) (1)
EA	 = –R x gradient (1)
	 = –8.31 × 5000
	 = +41.6 kJ mol–1 (in the range 35–45) (1)	3
[16]

57.	(a)	Miners overcome by toxic gases and burned in mine explosions.	1
(b)	A method of detecting the gas was needed (1)
A way of illuminating the mine without igniting/causing explosion the gas was needed. (1)	2
(c)	Charcoal is carbon
mole ratio H:C = 4/1 : 11.5/12 » 4:1	1
(d)	(i)	The strength of the bond between iron and the gases (accept methane or oxygen) is just right.
	Accept thermal conductivity/conducts heat	1
(ii)	Ruthenium Osmium (two Groups either side of iron)
Any transition metal (except Hg and Tc) and Al
NO alloys EXCEPT bronze	1
(e)	By the application of science many men’s lives were saved.	1
(f)	Examiners will need to consider each answer for (i) word total, (ii) key points and (iii) quality of language. Candidates should have recorded their word total at the end of their answer,and this should be checked.
Word Total
	Up to 125 words: no penalty
126 – 135 words: –1
136 – 145 words: –2
146 + words: –3
and at a rate of –1 penalty for every 10 words in excess thereafter, up to a maximum penalty equal to the number of key points included by the answer.

	Words appearing in the title to the summary do not count in the word total. Normally hyphenated words, numbers and chemical formulae count as one word. The question does not ask for equations in the summary, but if included they should be counted in the word total.
	Marking for key points	6
	One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks.
	Key points; may occur in any order, and it is not necessary for them to be expressed these words, provided that the sense of each point is conveyed.
1.	Davy worked out the formula of fire–damp/that fire damp was methane (1)
2.	He showed that when methane is mixed with 6–14 times its volume of air an explosive mixture results (1)
3.	He used a wire mesh cylinder surrounding a candle/flame/lamp.
	OR He used a chemical fire sieve (1)
4.	The mesh diffuses the heat of the lamp (1)
5.	At low concentrations, the size of the flame increases/brightens, at high concentrations the flame is extinguished. (1)
6.	A metal shield was added to prevent the effects of draughts (1)
7.	A further layer of iron/bronze/mesh was added. (1)
8.	Glass windows and a metal windshield were added. (1)
	1 mark for each key point up to a maximum of 6.
	To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
	Quality of language	2
	These should be impression marked on a scale 2–1–0.
	The aspects to be considered are:
•	use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
•	articulate expression; the answer should be well–organised inclear, concise English, without ambiguity. It should read fluently, with the links between key points in the original passage maintained.
•	correct grammar; basic sentence structure, including punctuation, should be consistently correct, without quibbles over the finer points of grammar.
•	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
•	points must be in a logical order.
	Good style and use of English, with only infrequent minor faults. (2)
Frequent minor or a few major faults in style and use of English. (1)
Very poor style and use of English. (0)
[15]

58.	(a)	DHf / kJ mol–1	DHf / kJ mol–1 Cl
	BeCl2	–490.4	–245.2
	BCl3	–427.2	–142.4
	(CCl4	–129.6)	–32.4
	NCl3	+230.1	+76.7
		 (1)	(2)	3
(b)	axes correct (1)
x–axis labelled atomic number (1)
y–axis labelled DHf  / kJ mol–1 Cl (1)	6, 5 ® 3
sensible scale (1)	4, 3 ® 2
points correctly plotted (1)	2 ® 1
straight line drawn (1)	1, 0 ® 0	3
(c)	(i)	As the graph is a straight line
DHf kJ mol–1 Cl increases steadily
X–Cl bond becomes steadily weaker	2
(ii)	Electronegativity increases Be N, (1)
as Cl is so electronegative, X–Cl bond becomes less polar (1)
(Or other reasonable argument)	2
(iii)	LiCl is ionic (1)
these chlorides are essentially covalent (1)	2
(d)	DHf becomes more positive (1)
X–Cl bond becomes weaker (Or reasonable explanation) (2)	3
[15]

59.	(a)	(i)	Too high temperature is needed to boil the oil (1)
so that the oil either burns (or chars) (1) or decomposes (1)	Any two
	2
(ii)	Lower temperature/Maximum temperature (100 °C),
so oil unaffected at this temperature.	1

(b)	(i)	Diagram: suitable container for water and  lemon peel 
OR separate container for generating steam    (1)
Liebig condenser (1)
- must have separate water jacket. If condenser tipped the wrong way then this mark is lost.
Vessel to collect oil     (1)
-1 for sealed system, minor errors, poor drawing, no heat	3
(ii)	Use a pipette/syringe to remove oil OR separating funnel     (1)
Suitable drying agent 
ACCEPT Calcium oxide, calcium chloride, silica gel,
sodium/magnesium sulphate, anhydrous sodium carbonate    (1)
	If solvent extraction suggested:
Extract with hydrocarbon solvent, use drying agent.     (1)
Distil off hydrocarbon solvent.     (1)	2

(c)	(i)	Alkene/C=C	1
(ii)	Alcohol/O-H
	If name and formula of group given, both must be correct for the mark	1
[10]

60.	(a)	(i)	Orange/yellow/red/brown    (1)
pink/purple/mauve     (1)	2
(ii)	Only molecular substances/halogen molecules dissolve in the solvent	1
(iii)	Cl2(aq) + 2Br-(aq) ®  Br2(aq) + 2Cl-(aq)
OR
Cl2(aq) + 2KBr(aq) ®  Br2(aq) + 2KCl(aq)	2
(b)	(i)	U.V. light/sunlight/photoflood	1
(ii)	Substitution (1), free radical (1)	2
(iii)	1,2 – dibromoethane     (1)	2

(c)	 mol  I2  react with   mol Na2S2O3
	1 × 10-4 mol I2 react with 2 × 10-4 mol Na2S2O3
	\ 1 mol I2 reacts with 2 mol Na2S2O3  (1)
	I2(aq) + 2Na2S2O3(aq) ® 2Na I(aq) + Na2S4O6(aq)
	OR   I2(aq) + 2S2O3 2-(aq) ® 2I-(aq) + S4O6 2-(aq)
	formulae (1)  balancing (1)	3
(d)	(i)	Destroy bacteria/microbes/bleaches/disinfectant/antiseptic	1
(ii)	Cl2/Br2 in swimming pools/Cl2 in drinking water
I2 for cuts 
OR
reasonable suggestion	1
[15]

61.	(a)	(i)	
	1

(ii)	C6H10 + H2 ® C6H12
OR
	1
(iii)	Platinum/nickel	1
(iv)	DHt=  DHf  [C6H12(l)] – (DHf [C6H10(l)] (+DHf[H2(g)])) (1)
= -156 –(–38)
= -118 kJ mol-1     (1)	2
(v)	Hardening of vegetable oils/for use in margarine.	1
(b)	(i)	High temperature (1) and high pressure (1)
(150 – 240 °C and 10 – 100 atmospheres)	2
(ii)	3(-118) = -354 kJ mol-1
ACCEPT their (a)(iv) answer × 3	1
(iii)	Any three of the following points:

	Diagram to show delocalization molecular orbitals
above and below the benzene ring
OR
In words to show an understanding of electrons shared by several
carbon atoms/all 6 carbon atoms    (1)
Delocalization of electrons     (1)
makes the benzene molecule more/very stable/more energy
needed to break the bonds     (1)
	All C-C bonds of same length/strength
OR
the length/strength is intermediate between
single and double bonds     (1)	3
(c)	(i)	Substitution  (1),  electrophilic (1)	2
(ii)	Sensible product formula (1)
Balanced equation (1) 
e.g. C6H6 + Br2 ® C6H5Br + HBr	2
[16]

62.	(a)	Particles of gold are trapped in a mineral lattice structure	1

(b)	The mineral mixture acts as an impervious physical barrier/shielding 
most of the gold particles from attack by cyanide ions.	1

(c)	Sulphur in refractory minerals is converted to sulphur dioxide,
and arsenic to arsenic(III) oxide,
which have environmental and …..
economic implications.
	3 points correct (2)
	2 points correct (1)	2
(d)	Fresh water contains dissolved oxygen needed to 
oxidize sulphur minerals.	1
(e)	Recovery of base metals is high	1
(f)	Up to 105 words: no penalty
   106 – 115 words: -1
   116 – 125 words: -2
   126+ words: -3
and at a rate of -1 penalty for every 5 words in excess thereafter, up
to a maximum penalty equal to the number of key points included by
the answer.
1200 – 2400°C = 4 words
FeS = 2 words
105 atmosphere i.e. number and unit = 2 words
marking for key points (7 marks)
	These may occur in any order (but note that the Quality of Language mark may be affected).  It is not necessary for them to be expressed in  these words, provided that the sense of each point is conveyed.
	The Bactech process of extracting gold from its ore
1.	Gold ore is crushed and ground.    (1)
2.	Refractory minerals are separated from the oxide ore/other non-metallic minerals (1)
3.	……by froth flotation to produce a sulphide concentrate     (1)
4.	The concentrate is treated with the thermophilic  bacterium Sulpholobus 
acidocalderius.     (1)
5.	Which catalyses atmospheric oxidation of sulphide minerals    (1)
6.	Helps the products to dissolve/products are water soluble     (1)
7.	The bacteria, concentrate and fresh water are mixed together in stirred tanks   (1)
8.	pH is adjusted/altered by adding lime, then sodium cyanide solution is added.     (1)
9.	... which is reacted with zinc shavings (to produce gold)     (1)
	1 mark for each key point up to a maximum of 7.

	Quality of Language (2 marks)
	There should be impression marked on a scale 2-1-0.
	The aspects to be considered are:
•	Use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
•	Articulate expression; the answer should be well-organised in clear, concise English, without ambiguity.  It should read fluently, with the links between key points in the original passage maintained.
•	Correct grammar; basic sentence structure, including punctuation, should be consistently correct, without quibbles over the finer points of grammar.
•	Legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
•	Points must be in a logical order
Good style and use of English, with only infrequent minor faults (2)
Frequent minor or a few major faults in style and use of English  (1)
Very poor style and use of English     (0)

63.	(a)	(i)		1
(ii)	Hold charged/static rod 
close to stream of liquid from burette    (1)
Stream deflected/attracted     (1)
- Can be shown on diagram	2
(iii)	Stronger/more van der Waals forces (in butanone)    (1)
because more electrons in butanone molecules     (1)	2
(iv)	Because same types of intermolecular forces/same 
number of electrons.    (1)
Similar boiling point expected     (1)	2
(b)	Glucose is soluble in water but cyclohexane is not    (1)
Water can form hydrogen bonds to the –OH groups or internal
O group in glucose but cannot hydrogen bond to cyclohexane     (1)	2
[9]

64.	(a)	(i)	Ka = 	1
(ii)	pH = -log [H+]     (1)
	-log[H+] = 5.1
	\  [H+] = 7.94 × 10-6    (1)
	Ka =  = 6.31 × 10-10  mol dm-3     (1)	3

(iii)	H2BO3- k H+ + HBO32-  could be H2BO3  + H2O k H3O+ + HBO32 (1)
HBO3- k H+ + BO33-    (1)
H2O k  H+ + OH-     (1)	Any two	2
(iv)	Boric acid is weaker than carbonic acid 
OR
too weak to displace CO2     (1)
So is unable to donate protons to CO32- ions     (1) 
Must make a comparison	2
(v)	Na3BO3 /Na2HBO3 /NaH2BO3/NaOH	1
(b)	(i)	Hydrogen bonds	1
(ii)	
	1
[11]

65.	(a)	A   C2H5COCH3 OR CH3CH2COCH3	1
(b)	B
	 (1)
	butanal                  OR             2-methylpropanal (1)	2
(c)	C   CH2=CH-CH2-CH2OH OR alternatives	1
(d)	D and E 

	Any two 
MUST be cyclic
ACCEPT displayed formula	2
(e)	(i)	C, D, and E  all three needed	1
(ii)	e.g. CH2=CH-CH2-CH2OH + Na  ®  CH2=CH-CH2-CH2-O-Na+ + ½H2
	(1)	(1)	2
(f)	C	1

(g)	Al2O3 /H3PO4 /H2SO4/porous pot
Any two, mark independently  (2)
If dilute stated loses mark  (1)

	ALLOW  TE of an alkene structure based on answer in (d)	3
[13]

66.	(a)	(i)	Yes silicon is in group 4 OR in the same group as carbon  (1)
forms 4 bonds
OR has 4 electrons in outermost shell
OR needs 4 more electrons for a full shell   (1)	2
(ii)	Covalent  (1)
Liquid at room temperature   (1)
Reacts with water rather than just dissolving  (1) 	3
(iii)	Filter  (1) 
i.e. reasonable method to separate solid from liquid e.g. decanting
Wash with water  (1) 
i.e. reasonable way to get rid of excess acid
Oven/Heat over a Bunsen flame
OR leave to dry OR put between filter papers (1) 
i.e. reasonable way to dry	3
(b)	Completely ionized/dissociated in solution (at moderate concentrations) 
Must imply a large degree of ionization in aqueous solution   (1)
	HCl + H2O/aq ®  H3O+ + Cl-  (1) 	2
(c)	(i)	(A)    volumetric flask OR graduated flask   (1)
(B)    pipette        (1) 	2
(ii)	Any named indicator 
e.g. methyl orange/red             (1)
red ® yellow                       (1)		2
phenolphthalein	colourless ® pink 
bromothymol blue	yellow ® blue
screened methyl orange	purple ® green
NOT Universal Indicator/ litmus
(iii)	Any two from:
Shaking mixture thoroughly in flask
Repeating titration
Add to water in a closed vessel to prevent escape of HCl
Rinse weighing boat/all washings transferred 	2

(iv)	Moles of NaOH =   	= 3 × 10-3 = 0.003   (1)
	Moles of HCl (in 25.0 cm3) 	= 3 × 10-3    
Moles of HCl in 250 cm3 	= 3 × 10-2
\ Moles of SiCl4  	= ¼ × 3 × 10-2    (1) i.e. 1:4 ratio
	= 7.5 × 10-3
award first two marks if this is seen
	SiCl4 = 170  (1)
	Mass of SiCl4 	= 7.5 × 10-3 × 170
	= 1.275 g
\% Purity 	=  × 100 = 85.0 (%)  (1)	4
[20]

67.	(a)	(i)	KCl + H2SO4 ® HCl + KHSO4       (2)   ACCEPT  ionic
OR 	2KCl  +  H2SO4   ®  2HCl(g)  +  K2SO4
	H3PO4 + KCl  ® HCl + KH2PO4  	1 (out of 2)
H3PO4  + 2KCl   ®  2HCl + K2HPO4 etc.     	(2)
(ii)	It does not oxidize..(1)
the halide to halogen (1)	2
(b)	(i)	HCl is  soluble              (1)
it would rise up the glass tube/to prevent suck-back (1)	2
(ii)	White precipitate                 (1)
Precipitate darkens              (1)	2
(c)	NH3+ HBr   ®  NH4Br	1
(d)	(i)	Purple/yellow/brown/violet	1
(ii)	2H ® I2 + H2	1
(e)	(i)	4KClO3   ®  3KClO4 + KCl        
formulae of KClO3 and KClO4  (1)  KCl and balancing  (1)	2
(ii)	Oxygen/O2	1
[14]

68.	(a)	(i)	Description of asymmetry of electron/charge cloud hence attractive
forces between neighbouring induced dipoles	1
(ii)	NCl3 / chlorine because more electrons	1
(iii)	NF3 because F more electronegative (than Cl)	1
(iv)	Van der Waals forces more significant/greater than permanent
dipole-dipole interactions	1

(b)	(i)	N(g) + 3F(g) in top right-hand box
½ N2(g) + 1 ½ F2(g) in lower box.	1
(ii)	

	Arrows in correct directions and labelled with correct data	2
(iii)	DHοat for [NF3(g)] ® N (g) + 3F (g) = 710 – ( – 125) = (+) 835 (kJ mol –1) (1)
	E (N – F) =  = (+) 278 kJ mol–1 (1)
	Penalise 4 or more SF
Penalise incorrect units	2
[9]

69.	(a)	(i)	Insert hot glass rod / (nichrome) wire into tube of gas / heat over	1
Bunsen flame (1)
(ii)	Violet / purple   vapour / gas / fumes
OR black / brown-black / grey solid (1)	1
(b)	(i)	Forward and reverse reactions occur at the same rate (1)
so there is no change in the proportions of reactants and products (1)	2
(ii)	Proportion of HI expected to become smaller (1)
because reaction will respond to temperature increase by shifting in the endothermic direction (1)	2
(iii)	No change expected because catalysts affect only the rate at
which equilibrium is attained.	1

(c)	

	Products at higher energy level than reactants (1)
	Line going up from reactants to peak (corresponding to transition state),
then down to products (1)
	Second line with lower peak representing the catalysed reaction (1)
Activation energies of uncatalysed and catalysed reactions correctly marked (1)	4
[11]

70.	(a)	(i)	Ticks in Cl2 /KBr and Cl2 /Kl boxes	1
(ii)	Solution / it becomes darker / turns brown / orange / yellow
OR black ppt	1
(iii)	Br2 + 2I– ® I2 + 2Br –	1
(iv)	Test for bromide ions: add silver nitrate (solution) (1)
pale yellow/cream/off-white/ivory ppt (1)
OR
test for iodine: add starch (solution) (1)   OR   hydrocarbon solvent (1)
		blue-black colour    (1)            purple/pink upper layer (1)
Colour must be consistent with reagent.	2
(b)	(i)	Cr/chromium and I/iodine identified (1)
Cr, initially +6, finally +3 (1)
I, initially –1, finally 0 (1)	3
(ii)	Cr2O72– + 14 H+ + 6I– ® 2Cr3+ + 3I2 + 7H2O	1
(c)	(i)	Bromine is toxic / poisonous / corrosive / causes burns	1
(ii)	May leave harmful residues (1)	)
	Harmful vapour could escape (1)	)
	Total opposition to use of chemical pesticides as posing a	) Any one
general health risk (1)	)
Depletes /damages/ harms ozone layer. (1)	)
Build up in the food chain (1)	)	1
[11]

  71.	(a)	It is a mixture / not a single compound	1
(b)	(i)	2,4-dimethylpentane	1
(ii)	C7H16	1
(iii)	More volatile / lower boiling point / vaporises more readily / branched so
doesn’t knock / higher octane number	1
(iv)	Heat / high temperature / ³ 200 °C (1)
Silica / alumina (catalyst) /zeolites (1)	2
(v)	Diagram should show:
	Test tube containing paraffin absorbed on suitable absorbent – (1)
absorbent can be just shown in the diagram
	Aluminium oxide catalyst (1)
Heat catalyst (1)
Recognition of collection of gas over water /gas syringe (1)	4
	Penalties
–1 for poor diagram
(c)	(i)	(CH3)2C = CH2
ACCEPT (CH3)2CCH2	1
(ii)	Elimination	1
(iii)	Potassium hydroxide / KOH / NAQH (1)
Ethanolic / alcoholic solution + heat / reflux (1)	2
[14]

72.	(a)	(i)	Viscosity / it affects ease /extent of penetration	1
(ii)	High viscosity caused by strong intermolecular forces	1
(b)	The proportion / fraction of pores connected to the surface of the
material / accessible to consolidant.	1
(c)	Diagram shows H–bond between any –OH group in cellulose
and an ‘ether oxygen’ in PEG
	1
(d)	Molar mass of each repeat unit (– CH2 – O – CH2 –) = 44	(1)
so number of repeat units per molecule = 1500 = 34 (nearest whole number)(1)
	44	2

(e)	

	Must show an Si atom joined, tetrahedrally, to 4 O
atoms and one of the O atoms must be
joined to two Si atoms	1
(f)	Examiners will consider each answer for (i) key points and (ii) quality of written communication. Candidates should have recorded their word total at the end of their answer, and this should be checked.
Up to 125 words: no penalty
126 – 135 words: –1
136 – 145 words: –2
146 – 155 words: –3
	and at a rate of –1 penalty for every 5 words in excess thereafter, up to a maximum penalty equal to the number of key points included in the answer.
Formulae count as names in the word count
eg Polyethylene glycol = 2 words
		PEG = 1 word
		SIOC = 1 word
		SiOSi = 1 word
	Note that words appearing in the title to the summary do not count in the word total, BUT subheadings do count.
	Marking for Key Points
	One mark awarded for every key point clearly identified in an answer, up to a maximum of 6 marks.
	List of key points: these may be in a different order, and need not be expressed in the wording below provided that the sense of each point is conveyed.
	Key points:
1	Stone objects deteriorate + reason(s)	(1)
2	consolidants are used to restore (structural) strength;	(1)
3	consolidant must penetrate the object and bind it together;	(1)
4	has to be applied as a liquid / solution which then solidifies;	(1)
5	organosilane consolidant applied as a monomer;	(1)
6	which polymerises within the object;	(1)
7	polymer formed resembles silica/silicon dioxide which is very stable:	(1)
8	polymer shrinks back on to pore walls and binds the object together	(1)
Any Six (6)

	Quality of Written Communication	(2)
	These should be impression marked on	 a scale 2–1–0, and the mark out of 2 should be recorded in the body of the script at the end of	 the answer. This mark can not be lost as a result of a word penalty.
	Candidates are expected to:
·	show clarity of expression;
·	construct and present coherent argument;
·	demonstrate effective use of grammar punctuation and spelling.
	The aspects to be considered are:
·	use of technical terms; the answer, should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
·	articulate expression; the answer should be well-organised in clear, concise English, without ambiguity. It should read fluently, with the links between key points in the original maintained.
·	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
·	points must be in a logical order.
	Good style and use of English, with only infrequent minor faults, no use of formulae .(2)
	Frequent minor or a few major faults in style and use of English (1)
	Very poor style and use of English (0)
[15]

73.	(a)	(i)	White / colourless	1
(ii)	Yellow / orange	1
(iii)	2Br – + Cl2 ® Br2 + 2Cl – ACCEPT multiples	1
(iv)	Separate layers – stated or implied (1)
Organic /Hydrocarbon / upper layer coloured orange (1)	2
(b)	(i)	Sulphur / S	)
Bromine / Br	)    (1)
	S, initially –2, finally +1 sign needed (1)
Br,initially 0, finally –1 (1)	3
(ii)	2 × +3 = +6, 6 × –1 = –6
OR   total change in ON of S = +6, total change in ON of Br = –6
OR   Up 6, down 6
OR   6 electrons lost, 6 electrons gained	1
(c)	(i)	Greater van der Waals attractions in HI / iodine (1)
because it has more electrons (1)
Can be from a HBr perspective	2

(ii)	Hydrogen / H bonding in HF (but not in HBr or HI)	1
(iii)	Within range 174 to 195 (actually 188) (K) (1)
Fewer electrons than in HBr (but no hydrogen bonding)
weaker van der Waals forces than in HBr (1)	2
[14]

74.	(a)	(i)	Two reactants form one product	1
(ii)	Substitution reactions occur under these conditions	1
(iii)	The electrons of the double / p bond polarise the Br – Br molecule (1)
	and Br d+ is the electrophile (1)
OR show Br d+ – Br d – attacking in the correct orientation	2
(b)	(i)	
	2
(ii)	No rotation about a C =C double bond
OR only single bonds can rotate	1
(c)	(i)	
	1
(ii)	Low temperature because exothermic reaction (1)
High pressure because fewer molecules of product than of
reactants gases are being converted into liquids (1)	2
[10]

75.	(a)	(i)	
	1
(ii)	180° (1)
Two regions / areas of negative charge / electrons repel so as to
be as far apart as possible (1)	2

(b)	4HCN + 5O2 ® 2H2O + 4CO2 + 2N2	1
(c)	Mark according to quality of argument given.
No mark if answer
•	effectively re–quotes the question without amplification
•	is chemically unsound
•	is nonsensical
	Answer must come to a conclusion and then look......
for EITHER a sophisticated statement about dangers of HCN
OR the usefulness of perspex.
	Not enough to say “No, because it is very toxic” but a developed
point such as ‘Danger of an escape of HCN is an unacceptable risk
to local residents” would score.
	Alternatively, an argument in favour of its use could be based on the
existence of adequate risk assessment and/or procedures for evacuation
in the event of an emergency or need for a safer alternative to glass or
other justification for use of Perspex.	1
(d)	(i)	H(g) + C(g) + N(g) (1)
	½ H2(g) +C(s/graphite) + ½ N2(g) (1)
all formulae correct, but missing state symbols: 1 max	2
(ii)		2

(iii)	E (CºN) = 1408 – 110 – 413
= +885 (kJ mol–1)	1
[10]

76.	(a)	(i)	1,2-dichloroethane	1
(ii)	CH2 = CHCl / CH2CHCl	1
(iii)	e.g. dissolve / bubble HCl in water / absorb in an alkali / condense the
HCl(g)	1
(b)	(i)	Species having unpaired electron	1
(ii)	Action of UV radiation/sunlight / named initiator / photoflood	1

(c)	(i)	Water / OH–	1
(ii)	Unshared / lone pair of electrons on a legitimate nucleophile
based on (c)(i) (1)
	(c)(i) “nucleophile” attacks / forms bond with C of C – Cl (1)	2
(iii)	Chloride ion / Cl–	1
(iv)	Add silver nitrate solution (1)
white ppt (1)	2
[11]

77.	(a)	Reactions producing the smog require / involve sunlight	1
(b)	2CO+2NO ® 2CO2 + N2	1
(c)	If combustion is complete, the only products are carbon dioxide and water
Incomplete combustion gives CO / carbon / soot / unburnt hydrocarbons
NOT SUFFICIENT to say “not all fuel is burned /not fully oxidised”	1
(d)	Higher temperature in conventional engines / lower temperatures used
in lean-burn (or diesel) engines	1
(e)	Iridium / silver/ osmium / ruthenium / gold	1
(f)	Fuel must be of very low sulphur content / unleaded because sulphur / lead
compounds ‘poison’/ affect the catalyst	1
(g)	The catalyst / engine / exhaust system is too cold / isn’t hot enough /
temperature of 150 / 200 / 240 °C needed	1
(h)	Examiners will need to consider each answer for (i) key points and (ii)
quality of written communication. Candidates should have recorded their
word total at the end of their answer, and this should be checked.
Up to 125 words: no penalty
126 – 135 words: –1
136 – 145 words: –2
146 – 155 words: –3
and at a rate of –1 penalty for every 5 words in excess thereafter, up to a
maximum penalty equal to the number of key points included in the answer.
Formulae count as names in the word count
Correctly hyphenated words = 1 word
eg		NO2 = 2 words	but NOx = 1 word
		200 °C = 2 words
		air: fuel = 2 words
	Note that words appearing in the title to the summary do not count in the word total, BUT subheadings do count.

	Marking for key points (6 marks)
	One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks. List of key points: these may be in a different order, and need not be expressed in the wording below provided that the sense of each point is conveyed.
	Key points:
1	converter contains platinum supported on a ceramic honeycomb; (1)
2	uses NOx / oxides of nitrogen to oxidise carbon monoxide and hydrocarbons /the other pollutants; (1)
3	air:fuel proportions controlled by fifting oxygen sensors; (1)
4	linked to electronically / computer controlled fuel injection system; (1)
5	efficiency is improved by including rhodium / palladium / other platinum group metals; (1)
6	diesel-engined vehicles need a filter to remove carbon/soot; (1)
7	Nitrogen oxides are converted to nitrogen dioxide with a platinum catalyst...... (1)
8	.... and the nitrogen dioxide then oxidizes the carbon / soot at 200 °C / low
temperature. (1)
Any six (6)
	Quality of Written Communication (2)
	These should be impression marked on a scale 2–1–0, and the mark out of 2 should be recorded in the body of the script at the end of the answer. This mark can not be lost as a result of a word penalty.
	Candidates are expected to:
·	show clarity of expression;
·	construct and present coherent argument;
·	demonstrate effective use of grammar punctuation and spelling.
	The aspects to be considered are:
·	use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
·	articulate expression; the answer should be well-organised in clear, concise English, without ambiguity. It should read fluently, with the links between key points in the original maintained.
·	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
·	points must be in a logical order.
	Good style and use of English, with only infrequent minor faults, no use of formulae EXCEPT NOx (2)
	Frequent minor or a few major faults in style and use of English (1)
	Very poor style and use of English (0)
[15]

78.	(a)	

	2-chloro-2-methylpropane (1)
	No marks for primary or secondary
halogenoalkane even if both formula
and name are consistent
Must be displayed	2
(b)	(i)	Hydrogen chloride
OR HCl
OR (concentrated) hydrochloric acid NOT dilute hydrochloric acid
OR sodium / potassium chloride and concentrated sulphuric acid/
phosphoric acid	1
(ii)	Substitution (1)
Nucleophilic (1)	2
(c)	(i)	

	2-methylprop-1-ene	2
(ii)	No because each carbon atom joined by the double bond has the
same two groups attached to it/ OWTTE	1
(d)	Reduction / addition hydrogenation (1)
Oxidation / addition (1)	2
(e)	

	Must be consistent with (c)(i)
Must be displayed	1
[11]

79.	(a)	(i)	a particle / species /group with an unpaired electron /OWTTE	1
(ii)		1
(iii)	homolytic	1
(b)	B and C	1
(c)	(i)	C12    +    CH4    ®    CH3Cl    +    HCl	1
(ii)	+242  +  4  +  –339  =  –93 kJ mol–1
(A     +  B +      F)
OR
+4 – 97 = –93 kJ mol–1
(B + C)
Method (1)
answer with units (1)	2
(d)	(i)	-242 kJ mol–1	1
(ii)	Exothermic because a bond has been formed.	1
(e)	Less endothermic (1)
the bond is weaker (1)	2
[11]

80.	(a)	(i)	It would turn red.	1
(ii)	White fumes	1
(iii)	Sulphur dioxide / SO2	1
(b)	(i)	A pale yellow / off-white / cream precipitate / solid	1
(ii)	The precipitate would dissolve / disappear / the mixture goes clear
Both AgCl and AgBr are soluble in concentrated ammonia	2
(c)	(i)	No more of the solid will dissolve (1)
at that temperature (1)	2
(ii)	100 g KBr and 25g KCl (1)
crystals only KBr (1)
solution contains both KBr and KCl (1)	3
[11]

81.	(a)	81 g mol–1	1
(b)	(i)	

	correctly plotted points (1)
smooth curve (1)	2
(ii)	As you go down the group the number of electrons increases. (1)
so the strength of the van der Waals forces increase. (1)	2
(c)	(i)	204 – 210 K	1
(ii)	Hydrogen/H- bonds	1
(iii)	Oxygen is more electronegative than the others (because
the outer electrons are closer to the nucleus)	1
(iv)	ammonia (1)
hydrogen fluoride (1)	2

(d)	Higher surface tension	)
Comparison of density of water and ice ie ice is lighter than water	)
It expands on freezing	)
Higher enthalpy change of vaporization	)   Any two
Shape of snow flakes/ice crystals	)
Higher viscosity	)
Higher heat capacity	)	2
[12]

82.	(a)	(i)		1
(ii)	180°	1
(b)	Increases the flow rate through the system
OR
Increasing pressure increases the rate of the reaction (1)
3.5 moles of gas on LHS but 4 moles of gas on the RHS therefore
increased pressure will favour the reactants (1)	2
(c)	
	1
(d)	(i)	Gold from 0 to (+)l	1
(ii)	Oxygen from 0 to –2	1
(e)	Key points
Examiners will need to consider each answer for (i) key points and (ii) quality of written communication. Candidates should have recorded their word total at the end of their answer, and this should be checked.
Up to 105 words: no penalty
106 – 115 words: –1
116 – 125 words: –2
126 – 135 words: –3
and at a rate of –1 penalty for every 5 words in excess thereafter, up to a maximum penalty equal to the number of key points included in the answer.
	Note that words appearing in the title to the summary do not count in the word total. Normally hyphenated words and numbers count as one word. The question does not ask for equations in the summary, but if included they should be counted in the word total.
If units given = 1 word
HCN = 2 words
90 % = 2 words
HCN (g) = 3 words
KCN(aq) = 3 words
1200 – 1300 °C = 4 words
Eqn 1 = 8 words
Eqn 2 = 6 words
Eqn 3 = 6 words	7

	Marking for key points
	One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks. A tick should be made in the script at which the examiner decided to award each mark eg 3. The total marks for key points should be placed in the body of the script at the end of the answer, out of 6. To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
1	In the Andrussov process ammonia is reacted / mixed with methane )
(or hydrocarbons) and oxygen... / oxygenated	)	(1)
2	and (passed over) a precious OR specified metal catalyst ...	)
e.g. Pt and Rh OR Pt and Ir	) Any four
NOT Pt on its own	)	(1)
3	at a slightly elevated pressure and high temperatures.	)	(1)
4	In the Degussa process a simpler / platinum catalyst system is	)	(1)
used without oxygen
5	To prevent polymerisation stabilisers of strong acids are used (1)
6	Hydrogen cyanide is used to make other compounds such as /methyl methacrylate / Perspex / pharmaceuticals
NOT "speciality chemicals" on its own (1)
7	Sodium cyanide used to extract gold and silver from their ores 	)
ACCEPT extract gold from gold beating rock	) Any one of (1)
	Potassium cyanide used in electroplating / dyestuffs	) these
	Sodium hexacyanoferrate(II) is used to prevent table salt caking	) three
		MAX 6
	Quality of Written Communication (2)
	These should be impression marked on a scale 2–1–0, and the mark out of 2 should be recorded in the body of the script at the end of the answer. This mark can not be lost as a result of a word penalty.
Candidates are expected to:
·	show clarity of expression
·	construct and present coherent argument
·	demonstrate effective use of grammar punctuation and spelling.

	The aspects to be considered are:
·	use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points
·	articulate expression; the answer should be well–organised in clear, concise English, without ambiguity. It should read fluently, with the links between key points in the original maintained.
·	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
·	points must be in a logical order.
	Good style and use of English, with only infrequent minor faults (2)
	Frequent minor or a few major faults in style and use of English (1)
	Very poor style and use of English (0)
[15]

83.	(a)	(i)	C12(aq)/(g) + 2I–(aq) ® 2Cl– (aq) + I2(aq)/(s)
Formulae – including charges (1)
Balancing and all state symbols (1) ALLOW multiples	2
(ii)	Brown/yellow/red
OR any combination of the three eg red-brown.
	NOT “orange” on its own or as a combination of the above	1
(b)	(i)	–1/1–	)
+5/5+	) 	1
(ii)	+4/4+	)
+6/6+	) 	1
(iii)	3SO2(aq) 3SO42– (aq) (1)
3H2O(I) 6H+(aq) (1)
ALLOW multiples	2
(c)	(i)	Separating funnel (1)
upper hydrocarbon layer (1)
pink/purple (1)	3
(ii)	Evaporate / Distil solvent	1
[11]

84.	(a)	(i)	C2H6(g)/(I) ® C2H4(g) + H2(g)
If a state symbol is missing (0)
If (aq) (0)	1
(ii)	At high pressure reaction goes in direction to reduce
pressure/to oppose change by Le Chatelier’s principle (1)
towards side with fewer molecules/moles (1)	2

(b)	Shapes of orbitals between and above carbon

If p orbitals drawn msut show overlapping
	Shapes (1) ACCEPT crescents for p bonds NOT lines for s bond	2
Labels (1)
(c)	Addition of bromine water/solution (1)
from yellow/brown/orange to colourless (1)
OR
acidified potassium manganate(VII) (1)
from pink/purple to colourless (1)	2
(d)	Addition (1)
Elecrophilic/electrophile OR appropriate explanation (1)	2
(e)	(i)	Potassium/sodium hydroxide/KOH/NaOH (1)
NOT hydroxide/OH– ions
Reflux/heat/warm and ethanol/ethanolic/
alcoholic solution (1)	2
(ii)	Van der Waals and (permanent) dipole-(permanent) dipole	1
(f)	(i)	nCH2(=)CHCl ® (–CH2–CHCI–)n
Formulae (1)
Balancing ie n's (1)	2
(ii)	Window frames
doors
floor coverings
electrical insulation
records/LPs
guttering
pipes
‘cling film’
waterproof/mock leather clothing
shiny handbags
vinyl/washable wall paper
furniture covers
table cloths	2
[16]

85.	(a)	(i)
	1
ACCEPT all dots/crosses

(ii)	

	Trigonal pyramid/Tetrahedral/‘Three leg stool’ shape (1) –
must be some attempt at 3D or correct name
107° ALLOW 92-108  (1)	2
(iii)	repulsion between four pairs of electrons gives
tetrahedral shape (1))
Greater repulsion of non-bonding electrons/lone pair
closes down tetrahedral bond angle (1)	2
(b)	(i)	PH3(g) ® P(g) + 3H(g)	1
(ii)	Hess applied (1)
Multiples (1)
Correct answer + 963(.2)/960 kJ mol–1 (1)	3
(iii)	Answer to (ii) divided by 3
+ 321(.1)/320 kJ mol–1	1
[10]

86.	(a)	Phosphine has more electrons	1
(b)	(i)	Hydrogen/H bonds	1
(ii)	

	Correct atoms (1)
Angle 180° /N-H ... N in straight line (1)	2
(c)	First line – soluble, insoluble, insoluble (1)
Second line – soluble, soluble, insoluble (1)	2
(d)	(i)	white fumes/smoke	1
(ii)	NH3 + HCl ® NH4(+)Cl(–)	1
[8]

87.	(a)	Answers should focus on the information in the 2nd sentence.
	Increased leisure time/holidays
Cheap travel (both needed)	1
(b)	Photoageing/premature ageing (1)
Skin cancer (1)	2

(c)	(i)	Breaking of a bond when both electrons go to the same atom
OR
Forming two oppositely charged ions – can be shown by an example	1
(ii)	An entity/species with an unpaired electron/uneven number of
electrons / odd number of electrons	1
(iii)	A chemical which increases the oxidation number/positive
charge/ removes the electron(s)/adds oxygen/removes hydrogen
from something else	1
(d)	up to 105 words: no penalty
106–115 words: –1
116–125 words: –2
126–135 words: –3
and at a rate of –1 penalty for every 5 words excess thereafter, up to a
maximum penalty equal to the number of key points included in the answer.
	Note that words appearing in the title to the summary do not count in the
word total. Normally hyphenated words, numbers, chemical formulae and
the abbreviations UVR, UVA and UVB count as one word. 70 % counts
as two words.
	Marking for key points
To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
	Key points
1	Avoiding going out in the sun /limit sun exposure (1)
2	But is socially unacceptable and impractical (1)
3	Opaque /non-transparent etc clothing /"not flimsy and not see through" (1)
4	Clothing is effective, providing it is dry/not wet (1)
5	Broad/wide-brimmed/large hats
OR
hats do not stop reflected light (1)
6	Suncreams/preparations which reflect the suns rays are effective/reflect
UVR (1)
7	But need fine particles of titanium dioxide to limit reflection of visible light to make them cosmetically acceptable.
8	...Filter preparations/suncreams filter/absorb ultraviolet
radiation/UVA and UVB/UVR (1)
9	but may not absorb all wavelengths/frequencies/ harmful radiations/UVR. (1)
		Mark to a maximum of 7 marks
	Quality of Written Communication
	Candidates are expected to:
·	show clarity of expression;
·	construct and present coherent argument;
·	demonstrate effective use of grammar punctuation and spelling.

	The aspects to be considered are:
·	use of technical terms; the answer should convey a
correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
·	articulate expression; the answer should be well-organised in clear, concise English, without ambiguity. It should read fluently, with the links between key points in the original maintained.
·	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional dificulty with a word.
·	points must be in a logical order.
	Good style and use of English, with only infrequent minor faults, no use of formulae (2)
Frequent minor or a few major faults in style and use of English (1)
Very poor style and use of English (0)
[15]

88.	(a)	Van der Waals/induced dipole-dipole	1
(b)	(i)	Hydrogen/dipole-dipole in propan-1-ol,(but no hydrogen/
dipole-dipole in butane)	1
(ii)	Van der Waals forces in propan-1-ol are stronger
OR reverse argument (1)
because chain is not branched/so more surface contact between molecules)
OR reverse argument (1)	2
(c)	(i)		2

(ii)	109(.5)° for HCH (1)
120 – 124° inclusive for CCO bonds (1)	2
(iii)	(permanent) dipole-dipole	1
(iv)	25 (°C) < T < 75 (°C)	1
(d)	Propanone can form hydrogen bonds with water, (but butane cannot)	1
[11]

89.	(a)	(i)	KCl + H2SO4 ® KHSO4 + HCl
OR
2KCl + H2SO4 ® K2SO4 + 2HCl	1
(ii)	

	Preparation (1 mark)
	Boiling/test tube but not on its side
Pear-shaped flask
Conical flask
	Collection of gas (1 mark)
	Downward delivery (mark lost if delivery tubing less than 1/3 down tube)
	Gas syringe
	Penalties
-1 for seated apparatus
-1 for poor diagram e.g. continuous pieces of apparatus, delivery
    tubing entering side of test tube	2
(b)	(i)	White smoke/solid/fumes NOT gas/vapour/cloud	1
(ii)	HCl(g) + NH3(g) ® NH4Cl(s)	1
(iii)	Ammonium chloride	1
(c)	(i)	Cl in HCl: –1 / 1 –	Cl in Cl2: 0 (1)
O in H2O2: –1 / 1–	O in H2O: –2 / 2– (1)	2
(ii)	hydrochloric acid/hydrogen chloride/chloride ion/ HCl/Cl–
NOT “chlorine” UNLESS “the chlorine in HCl”	1
(iii)	N.B. Read whole thing through.
Look for good use of chemical language eg use of molecule / atom
	O.N. of 2 Cls has increased by (1 × 2=) TWO
O.N. of 2 Os has decreased by (1 × 2=) TWO
	OR
	In terms of electron exchange from one oxidation state to another.
OR
	Can consider total oxidation numbers remaining constant eg “sum of O and Cl
O.N. = –4 on left, sun of O and Cl O.N. = –4 on right too”.	1

(d)	(i)	C12(g)   (+)   ½ O2(g)	1
(ii)	

	ALLOW T. E from (i)	1
(iii)	DHοat [Cl2O(g)] = –80.3 + 492.6 = + 412(.3) (kJ mol–1)
SIGN essential
PENALISE wrong units
Correct answer with no working (1)	1
ALLOW TE from(ii), provided Hess’ Law applied correctly
for their cycle
(iv)	E[Cl –O] (= ½ × (+ 412.3)) = (+)206(.2) (kJ mo1–1)
ALLOW TE from (iii)
ALLOW (+)206.15
ALLOW missing sign	1
[14]

90.	(a)	(i)	Ammonia/NH3 (1)
Ethanolic/alcoholic conditions	)
high temp/	) Any one (1)
high pressure	)	2
(ii)	Substitution (1)
Nucleophilic (1)	2
(b)	(i)	CH3CH(OH)CH2CH3	OR	CH3CH(OH)C2H5	1
(ii)	cream/off-white/ivory (1)
Ag+(aq) + Br–(aq) ® AgBr(s) (1)	2
(iii)	more slowly + some attempt at explanation (1)
C-Cl bond stronger than C-Br bond (1) - must be a comparison
of specific bonds
	2nd mark dependent on 1st	2

(c)	(i)	Potassium/sodium hydroxide (1)
ALLOW formulae
	Ethanolic and heat/reflux (1) - ethanol can be given as a
reagent	2
(ii)	
	1
(iii)	N. B. Read whole answer to see if it makes sense
	Two hydrogen atoms (two identical atoms) are linked to one of
the carbon atoms holding the double bond (so two distinct geometric
isomers not possible)	1
[13]

91.	(a)	

	Missing end links (max 1)
C2H5 could be on C1 and C4 or C2 and C3.	2

(b)	activation energy for catalysed process clearly marked (1)
appropriate curve drawn starting at “reactants” , finishing at “products” (1)
with the maximum above the one for the catalysed reaction (1)	3
(c)	(i)	(catalyst) provides route/pathway with lower activation energy
NOT reach activation energy earlier/faster	1
(ii)	to provide the activation energy necessary/energy to break some
bonds	1
[7]

92.	(a)	(Sweat is a dilute aqueous) solution of sodium chloride and urea,
(and also other metabolic waste products, such as the lactates
produced in muscles)
OR
Is a mixture of water, sodium chloride, urea	1
(b)	(Sweat is produced by the eccrine glands) via emotional, thermal and
sensory stimuli	1

(c)	CH3CH2CH2CH(CH3)CH2COOH
correct answer (2)
correct structure, except CH3 branch in wrong place (1)
structure for a methylheptanoic acid (1)	2
(d)	Antiperspirants were too acidic and irritated the skin/rotted clothes	1
(e)	Al2(OH)2Cl4	1
(f)	e.g. more wasted using aerosol
application more precise with roll-on/ consequences for atmospheric
pollution using aerosols/any other feasible alternative
ACCEPT environmental pollution + qualification	1
(g)	Examiners will need to consider each answer for (i) key points and (ii)
Quality of communication.
[7]
	Word total/penalty
	Candidates should have recorded their word total at the end of their answer, and this should be checked.
	up to 105 words: no penalty
106 – 115 words: –1
116 – 125 words: –2
126 – 135 words: –3
	and at a rate of –1 penalty for every 5 words excess thereafter, up to a maximum penalty equal to the number of key points included by the answer.
	Note that words appearing in the title to the summary do not count in the word total. Normally hyphenated words (such as odour-causing, roll-ons, mid-1970s, zinc-based), numbers and chemical formulae count as one word. The question does not ask for equations in the summary, but if included they should be counted in the word total.
	99 %	= 2 words
RCOOH	= 2 words
BO	= 2 words
Al2(OH)mCln	= 2 words
ACH	= 2 words
1947	= 2 words
C4 – C10	= 3 words
m+n = 6	= 3 words
	Marking for key points (6 marks)
	One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks.
	A tick should be made in the script. Examiners should show the key point being awarded,
 i.e.3 shows key point 3 given.
	List of key points: these may be in a different order, and need not be expressed in the wording below provided that the sense of each point is conveyed.

	Key pt
	Distinction
1	Deodorants act (solely) to reduce BO by killing / eradicating /
destroying the (odour-causing) bacteria. (1)
2	Antiperspirants reduce both odour and wetness
OR
In addition antiperspirants reduce wetness - dependent on key pt 1 (1)
3	Ethanol is the principal antibacterial agent (1)
4	with further activity [OWTTE] derived from some of the added
perfume oils.(1)
5	Aluminium salts are (commonly) used in antiperspirants
(nowadays).......
NOT aluminium chlorohydrates (1)
6	....... and these physically block the eccrine/sweat glands - 
must follow on from “aluminium” (1)	4
	Formulation
7	Deodorants and antiperspirants can be sold as a solution, a
suspension or emulsion (1)
	Application
8	and can be applied in	)
pump sprays	)
roll-ons	)   Any two (1)
sticks	)
aerosols	)
gels	)
		2
	Quality of Written Communication (2 marks)
	These should be impression marked on a scale 2-1-0, and the mark out of 2 should be recorded in the body of the script at the end of the answer. This mark can not be lost as a result of a word penalty.
	Candidates are expected to:
·	show clarity of expression;
·	construct and present coherent argument;
·	demonstrate effective use of grammar punctuation and spelling.

	The aspects to be considered are:
·	use of technical terms; the answer should convey a correct understanding by the writer of the technical terms used in the passage which are involved in the key points.
·	articulate expression; the answer should be well-organised in clear, concise English, without ambiguity. It should read fluently, with the links between key points in the original maintained.
·	legible handwriting; the reader should be able to read the answer without difficulty at normal reading pace, with only the occasional difficulty with a word.
·	points must be in a logical order.
	Good style and use of English, with only infrequent minor faults, no use of formulae (2)
	Frequent minor or a few major faults in style and use of English (1)
	Very poor style and use of English (0)
[15]

93.	(a)	(i)	Hydrogen iodide / HI/ H-I / IH
ALLOW either correct name or formula
NOT hi / ih
NOT “+” or “–” on any formula
If both name and formula given but one is incorrect (1)	1
(ii)	H3PO4 + 3KI → 3HI + K3PO4
OR
H3PO4 + 2KI → 2HI + K2HPO4	ALLOW HK2PO4
OR
H3PO4 + KI → HI + KH2PO4	ALLOW H2KPO4
	Correct formulae (1) ALLOW minor slip in formula
Correctly balanced (1)	2
ALLOW multiples
If overall formula of salt is correct ignore charges, even if incorrect
(iii)	Reagent
silver nitrate (solution) / Ag(+)NO3(–) (1)
If formula given must be correct
NOT Ag+
	Observation
Yellow / pale yellow precipitate/solid/crystals (1)
OR
Add chlorine/bromine with a hydrocarbon solvent to give
purple/pink/violet colour
	2nd mark depends on correct test BUT allow 2nd mark if Ag+
given as reagent	2
(b)	(i)	Purple vapour / gas/ fumes
OR
Brown/black solid/solution/liquid
NOT increase in temperature/fizzing/solid dissolves
	If more than one observation – eg all correct or correct plus
neutral (1) but if correct plus contradictory (0)	1
(ii)	Paper goes black /silvery black / grey/brown/
black ppt forms / darkens
IGNORE original colours
NOT “discoloured”	1
(iii)	Lead(II) sulphide / Pb(II) sulphide
ALLOW Pb(II)S / lead(II) sulphide(II)
NOT lead sulphide (II)
NOT PbS
Brackets not essential	1
(iv)	Elements
iodine / I / I– / iodide and sulphur / S (1)
NOT I2
	Oxidation numbers
iodine oxidised from –1 to 0 (1)
sulphur reduced from (+)6 to –2 (1)
Oxidation numbers must be given in correct order
1 mark for each correct pair of oxidation numbers	3
(c)	(i)	(Standard) sodium thiosulphate / Na2S2O3
ALLOW any named thiosulphate
ACCEPT “thiosulphate ions” / S2O32–
NOT “thiosulphate” on its own	1
(ii)	Starch (1)
	Blue-black /(dark) blue/black to colourless (1)
	ALLOW “no indicator/self-indicating/iodine’” and ‘yellow to
colourless’ 1 (out of 2)
	If indicator left blank (0)
2nd mark is dependent on 1st	2
(iii)	4.8 × 10–4 × 20 = 9.6 × 10–3/0.0096/0.00960 / 0.009600 etc	1
(iv)	Molar mass of Kl = 166
Mass pure KI = 166 × 9.6 x10–3 (1) = 1.59/1.593 / 1.5936 g
ALLOW TE from (iii)	
	% purity = (1.59/1.75) × 100 = 90.9 / 91.1 / 91 / 91.06 / 91.063% (1)
ALLOW sensible TE from molar mass
	OR
Calculation based on moles
	IGNORE SF	2
[17]

94.	(a)	(i)	
	Bonding pairs can be shown horizontally
or vertically in all positions
	Can be all dots/crosses
IGNORE inner shells of electrons if shown
Watch for lone pairs on O
H can be above or below O	1
(ii)	Electron pairs/electron clouds repel allowing bigger angles
in three dimensions
OR
It is a three-dimensional shape being represented in two
dimensions
OR
Explanation of why angles are not 90° in CH3 / are not 180°
in COH (1)
	HCO or HCH = 109° / 109.5°
OR
COH = 103-105° (1)
Angle can be stated rather than marked on the diagram but
must be between two bonds, not between two atoms
	ALLOW “it is tetrahedral not flat/two dimensional”
NOT “Tetrahedral” on its own	2
(b)	(i)	
Arrow is essential
Watch out for arrow direction
	1
(ii)	Shorter/atoms are closer in CO (as multiple bond) (1)
	More electrons / greater electron density (between the two
nuclei) in the bond (1)
2nd mark depends on 1st
	ACCEPT vice versa argument for methanol	2
(c)	(i)	CH3OH(g) → C(g) + 4H(g) + O(g)
Watch out	Watch out
for wrong	for H2
state symbol		1

(ii)	2039 = 3E(C-H) + E(C-O) + E(O-H) (1)
= 3(413) + E(C-O) + 464
	E(C-O)	= 2039 – 1239 – 464
	= (+) 336 (kJ mol–1) (1)
	Correct answer with no working (2)
	If C–H not multiplied by 3, giving (+)1162 (kJ mol–1) 1 max	2
(iii)	

	Balanced cycle with state symbols and data (1)
	Calculation
∆Hfο + 2039 = 1837.9 kJ mol–1
∆Hfο = 1837.9 – 2039 = –201(.1) kJ mol–1
Hess applied correctly with allowance for 4H (1)
– which will give correct sign
	Answer with units (1)
	ALLOW TE from use of 1H or wrong ∆Hat
If 4H not used allow TE from cycle answer = – 855(.1) kJ mol –1
max 2 (out of 3)
If + 336 kJ mol –1 used instead of correct ∆Hat answer = (+) 1501.9 /
(+) 1502 kJ mol –1 max 2 (out of 3)
Penalise same error once ONLY	3
(iv)	More negative as energy is given out when the liquid forms (from
the gas)
OR
More negative as more/stronger intermolecular bonds/forces are made
	IGNORE type of intermolecular bond	1

(v)	

	Hydrogen bond/ dotted line between O in one molecule and
hydroxyl hydrogen in another (1)
	Bond angle 180° (1) – must go across H
2nd mark depends on correct atoms in bond
	ALLOW diagram showing methanol/ethanol
NOT diagram showing methanol/ethanol and water
ALLOW minor slip eg one missing H on a CH3	2
(d)	Temperature
Low temperature as forward reaction is exothermic / reverse reaction
uses heat / endothermic (1)
Pressure
	Pressure high as number of molecules/moles is decreasing / fewer /
3 molecules/moles go to one (1)
If numbers specified must be correct
NOT 2 molecules go to 1
NOT 2 gases go to 1
	Explanation using Le Chatelier is fine BUT not “Le Chatelier” on its own
	ALLOW 1 mark for correct choice of temperature (low) and pressure (high)
with some attempt at explanation	2
[17]

95.	(a)	(i)	1-chloropropane has more electrons than chloroethane (1)
	So van der Waals’ forces (between molecules) stronger/greater
OR
More/greater van der Waals’ forces (1)
	OR reverse argument
	If dipoles are mentioned they must be temporary /induced /
transient / fluctuating / flickering	2
(ii)	Molecules in 2-chloropropane make less contact / pack less well /
can get closer together OWTTE
	ACCEPT annotated diagram
	If the explanation about van der Waals’ forces is given here
allow it in (i) UNLESS incorrect intermolecular force mentioned
in (i)	1

(b)	(i)	Reagent with a lone pair of electrons
OR
Pair of electrons which it can use to make a bond
OR
Reagent which attacks species with a (δ) + charge
	NOT “attacks nucleus” on its own
NOT “species with a negative charge”	1
(ii)	C-l bond is weaker than C-Cl
Must say which bond is weaker	1
(c)	(i)	Use ethanolic KOH/KOH in alcohol/KOH in ethanol/
ethanol as solvent (and raise temperature)	1
(ii)	Elimination (1)
IGNORE comment on what is eliminated
IGNORE qualification eg electrophilic	1
(d)	(i)	

ALLOW C bonded to HSO4 anywhere
ALLOW C2H5+ HSO4–HCCH
	Need not be displayed
NOT “C2H5 + “ on its own	1
(ii)	Electrophilic (addition)
ALLOW electrophile / electrophylic	1
(iii)	React with water / hydrolysis
ALLOW Steam	1
(e)	Cheaper, because ethene comes from crude oil / alkanes / is readily
available / no separation of by-products required OWTTE
OR
Cheaper because halogenoalkanes are expensive to make / do not
occur naturally / need 2 stages in synthesis
OR
Faster because only one stage
OR
Safer because halogenoalkanes more toxic / more irritant / damage
ozone layer or other specific reason.	1
[11]

96.	(a)	Species / particle / entity / group / atom / molecule having unpaired /
odd electron/odd number of electrons
NOT ion/element/chemical
NOT a lone electron / lone pair of electrons
NOT atom with an incomplete shell of electrons
If a diagram is given an explanation is also needed	1
(b)	Homolytic (fission)	1

(c)	UV /UV radiation / UV light / (sun)light / hν / photons
NOT protons
IGNORE references to heat	1
(d)	C2H6 + 2Cl2 → C2H4Cl2 + 2HCl
	Correct dichloroalkane, C2H4Cl2 or displayed (1) balanced with HCl (1)
	C2H6+Cl2 ® C2H4Cl2 + H2	1 (out of 2)
C2H6+Cl2 ® C2H5Cl + HCl	1 (out of 2)	2
(e)	
	ACCEPT “Fl” for “F”
IGNORE brackets and n as long as at least 2 monomer units shown.
MUST have continuation bonds at both ends	1
(f)	
	ALLOW indication that C = O is polar eg circled.
If charges or dipole arrow shown must be correct. H
	Structure of propanone must be correct
d/δ is needed not just “+” and “–”
		1
(g)	Examiners will need to consider each answer for (i) key points and (ii) style
and use of English. Candidates should have recorded their word total at the
end of their answer, and this should be checked.
up to 115 words:	no penalty
116 – 120 words:	–1
121 – 125 words:	–2
126 – 130 words:	–3
and at a rate of –1 penalty for every 5 words excess thereafter, up to a
maximum penalty equal to the number of key points included by the answer.
	Note that words appearing in the title to the summary do not count in the
word total. Normally hyphenated words, numbers and chemical formulae
count as one word. The question does not ask for equations in the summary,
but if included they should be counted in the word total. Do not count
subheadings “Advantages” and “Disadvantages”.
	mono-, di-, tri- and tetra-chlorinated products	6 words
poly(ethene)	1 word
PTFE	1 word
acid- or base-sensitive reagents	4 words
tributyltin	1 word
150 °C	2 words

	Marking for key points
One mark should be awarded for every key point clearly identified in an answer, up to a maximum of 6 marks. A tick should be made in the script at which the examiner decided to award each mark e.g. 3(1). The total marks for key points should be placed in the body of the script at the end of the answer, out of 6.
	To gain the mark for a key point the wording used by the candidate must make clear the essential chemistry of the point.
	Key points
Maximum of 6 marks available.
1	Radical reactions are often unselective / low yielding / difficult to
control/ uncontrollable	(1)
2	Mixtures/products/by-products may be difficult to separate.	(1)
3	Radicals may produce polymers of varying chain lengths / degrees of
branching.	(1)
4	Radicals form under mild conditions / at low temperatures / neutral
conditions	(1)
5	…..so reagents less likely to decompose in radical reactions (than in
ionic reactions).	(1)
6	Radical reactions are not very sensitive to changes in solvent
OR
Radicals are generally not solvated
OR
Changing solvent is not important
	NOT Radicals are neutral	(1)
7	Reactivity of radicals makes them useful in reactions which are
difficult to achieve
OR
Radicals will carry out reactions that ions will not.
NOT “Radicals are highly reactive” UNLESS qualified	(1)
8	Tributyltin radicals/hydride are expensive and neurotoxic.	(1)
	Any six	6

Quality of Written Communication	2
	These should be impression marked on a scale 2-1-0, and the mark out of 2
should be recorded in the body of the script at the end of the answer. This
mark can not be lost as a result of a word penalty.
	Candidates are expected to:
·	show clarity of expression;
·	construct and present coherent argument;
·	demonstrate effective use of grammar punctuation and spelling.
	The aspects to be considered are:
·	use of technical terms; the answer should convey a correct understanding by the
	writer of the technical terms used in the passage which are involved in the key 
	points.
·	articulate expression; the answer should be well-organised in clear, concise
	English, without ambiguity. It should read fluently, with the links between key
	points in the original maintained.
·	legible handwriting; the reader should be able to read the answer without difficulty
	at normal reading pace, with only the occasional difficulty with a word.
·	points must be in a logical order.
	Good style and use of English, with only infrequent minor faults, no use of
formulae (2)
	Frequent minor or a few major faults in style and use of English (1)
	Very poor style and use of English (0)
	NB: The quality of written communication mark cannot be lost through
word penalties.
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