June 2019: Paper 3, Unified chemistry

Mark scheme

Examiner's comment

The best candidates explained that bond polarity results from fluorine being more electron negative than carbon. Most candidates were able to explain why a CF4 molecule has no overall dipole in terms of the symmetry of the molecule or dipoles cancelling. Some candidates confused dipoles cancelling with polar bonds cancelling. A significant number of candidates did not attempt to explain bond polarity at all. Candidates are recommended to look closely at the requirements of the question set and to look at the total marks for a short answer question to gauge how many points they need to make.

Notes from the mark scheme and report

• F more electronegative than C
• symmetrical/tetrahedral so dipoles cancel
• DO NOT ALLOW C is more electronegative
• IGNORE 'polar bonds cancel' and 'charges cancel'

Mark scheme

Examiner's comment

Many candidates recognised that the presence of hydrogen or oxygen isotopes could explain the high relative molecular mass of water. The best responses discussed D2O or molecules with other combinations of isotopes including 18O and even 3H. Exemplar 1 shows a good response which explains the relative molecular mass of 20 in terms of the presence of 18O. Many candidates claimed that some water molecules had the formula of H4O, often bringing lone pairs and hydrogen bonds into their explanations, as shown in Exemplar 2. D2O is encountered in proton NMR spectroscopy, so candidates will have encountered it during their studies, but many candidates were unable to link this to the relative molecular mass of 20.

Notes from the mark scheme and report

• molecules contain 2H/deuterium or 18O
• idea of isotopes is critical
• DO NOT ALLOW isotopes of elements other than H and O

Mark scheme

Examiner's comment

This question tested an understanding of ‘partial pressure’ as a concept. Most candidates obtained the correct response of 21,000 Pa or 2.1 × 104 Pa. Scaling proved to be a common error with 2.1 × 10–4 Pa being seen, presumably from dividing, instead of multiplying, 0.21 by 1 × 104. In calculations, candidates are advised to think about whether their answer is sensible, rather than relying just on the answer displayed on the calculator.

Notes from the mark scheme and report

• 0.21 × 1.0 × 10^5 = 2.1 × 10^4 Pa

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

This part required candidates to work out the amount in moles of propane that combusted, followed by a calculation of the volume of CO2. The initial mole calculation required candidates to first convert the volume of 42.0 m3 into dm3 or cm3. A correct calculation shows that 1750 mol of propane was combusted. Many struggled with the unit conversion, with 0.175 and 1.75 × 10–3 often being seen. This value then needed to be multiplied by 3 (from the stoichiometry of the equation) and 44 to produce 231,000 g. Here, ×3 was sometimes omitted. Some candidates ignored RTP in the question and used the ideal gas equation to calculate the moles of propane. This approach was acceptable provided that sensible values had been chosen for the conditions at room temperature and pressure (e.g. 298K, 293K, 100 kPa,101 kPa). Candidates are advised to practice volume unit conversions – important for any calculation involving gases.

Notes from the mark scheme and report

• common errors (1 mark): no conversion of m3 to dm3 (231), cm3/dm3 confusion (0.231), no 3 × for CO2 (77 000)
• ALLOW ideal gas equation with sensible T and p

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This part tested an understanding of pH as a logarithmic scale and the relationship between rates and order. This part discriminated extremely well. A pH of 3 meant that the H+ concentration would be 100 times less than at a pH of 1. Being a first order reaction with respect to H+, the initial rate is simply 100 times less than at a pH of 1: 2.4 × 10–5 mol dm–3 s–1. The higher-ability candidates identified the relationships and often wrote the correct answer on the answer line with no visible working (presumably doing the calculation in their head). Others used their well-learnt equations to calculate the same correct answer. Many candidates found the calculation difficult and 7.2 × 10–3 (3 × (2.4 × 10–3)) was a very common incorrect answer. If candidates had fully scrutinised this answer, they may have realised that a more dilute solution cannot produce a faster rate.

Mark scheme

Examiner's comment

This part required candidates to initially work out the amount in moles of P2O5 in 4.26 g of P2O5, followed by multiplication by 5 (for O) and the Avogadro constant to produce 9.03 × 1022 atoms. The initial moles calculation proved to be a relatively easy mark which most candidates were awarded. Many then struggled, sometimes omitting the ×5 factor to account for the O atoms, producing the common incorrect answer of 1.806 × 1022.

Notes from the mark scheme and report

• minimum 3 SF
• common error: no × 5 (1.806 × 10^22) 1 mark
• ALLOW ECF from incorrect n(P2O5)

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures

Mark scheme

Examiner's comment

This equation presented problems for many candidates, despite the question asking for an equation between carbonate ions and water. An acceptable equation had to be ionic and needed to produce OH– (for the alkaline solution) and either HCO3– or CO2. Many candidates wrote an equation with H+ instead of H2O, with lower ability candidates showing the carbonate ion with the wrong charge as CO3–. Many candidates wrote full equations despite the question asking for an ionic equation. Candidates do need to read the instructions in the question.

Notes from the mark scheme and report

• CO32– + H2O → OH– + HCO3–
• ionic equation required
• IGNORE equations with H+ or H3O+ (question asks for reaction with H2O)
• IGNORE state symbols

Mark scheme

Examiner's comment

Candidates found this part extremely difficult. The question was aimed to stretch and challenge. Many candidates followed on directly from part (a), stating in simple terms that the alkaline solution needed to be neutralised to remove hydroxide ions. However, candidates were expected to recognise that the alkaline conditions would lead to benzoate ions rather than benzoic acid being present in the mixture. The mixture is acidified to protonate the benzoate. The hint in the question was about making the benzoic acid appearing when acid is added.

Notes from the mark scheme and report

• H+ protonates benzoate/carboxylate ion to form benzoic acid
• IGNORE responses only about neutralising the alkali/carbonate

Skills: PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

This part discriminated well with many candidates being able to write a correct equation using their knowledge of the oxidation of alcohols. Mistakes usually resulted in the balancing with either [O] instead of 2[O] or 2H2O instead of H2O. Written equations always need to be checked for the atoms balancing.

Notes from the mark scheme and report

• C6H5CH2OH + 2[O] → C6H5COOH + H2O
• ALLOW molecular formulae

Mark scheme

Examiner's comment

Candidates are well practised with percentage yield calculations with about half obtaining the correct percentage yield of 33.8 or 33.9% to secure all 3 marks. Many were able to secure partial credit for incorrect answers, provided that the working was laid out clearly. Some responses showed a simple percentage of the two masses with no consideration of moles or molar masses. Such a response received no credit.

Notes from the mark scheme and report

• mass from volume × density
• 3 SF
• common errors (2 marks): no use of density (35.2%), ÷ density instead of × density (36.5%)
• ALLOW ECF for each step

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many candidates produced thorough responses, showing that they had encountered recrystallisation as a technique in their practical work. Most candidates were aware that the impure product is dissolved in a minimum volume of hot solvent, although ‘minimum’ was sometimes omitted. The subsequent stages were sometimes incomplete or in the wrong order. Many were aware that the hot solution can be passed through fluted filter paper to remove solid impurities. (This is beyond the specification requirements for A Level but good practice). Most candidates were aware of the need to filter (usually under reduced pressure) but the necessary cooling stage to form the crystals was sometimes omitted. Finally, many responses omitted the need to dry the crystals. Candidates did sometimes dry the crystals by adding an anhydrous salt (e.g. CaCl2 or MgSO4), a clear confusion with drying an organic liquid. Others described the purification of an organic liquid for their response, including use of a separating funnel, drying and distillation.

Notes from the mark scheme and report

• dissolve in the minimum quantity of hot solvent
• cool AND filter AND dry (all three needed)
• DO NOT ALLOW use of a drying agent
• IGNORE hot filtration

Skills: PAG6 Synthesis of an organic solid

Mark scheme

Examiner's comment

This equation proved to be very testing and only a minority of candidates were able to write a balanced equation for this reaction forming Pb, CO2 and H2O. Many different and implausible products were seen such as O2, H2, and CH3OH.

Notes from the mark scheme and report

• 4Pb2O3 + 3CH4 → 8Pb + 3CO2 + 6H2O (ALLOW versions forming CO or C)
• ALLOW multiples
• IGNORE state symbols

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme

Examiner's comment

The mark scheme catered for a wide range of responses that were judged to be worthy of credit. Responses that form part of normal safe laboratory practice, such as keeping bags under benches, tying hair back, etc., were not given. Candidates were expected to identify a precaution for a chemical hazard. As with GCSE, general words to describe hazards, such as harmful and dangerous, were not given. Acceptable chemical hazards would include those encountered in hazard signs such as flammable (for methane) or toxic (for lead or lead compounds). The best responses suggested that naked flames should be kept away from flammable methane or that gloves should be worn when using toxic lead. Other suggested use of a fume cupboard or good ventilation when using flammable or toxic materials. Responses such as ‘face masks’ gained no credit.

Notes from the mark scheme and report

• definite safety issue AND matching precaution (toxic lead compounds – gloves or fume cupboard
• flammable methane – keep away from flame)
• 'harmful'/'dangerous' too vague
• IGNORE safety glasses, lab coat

Mark scheme

Examiner's comment

For one modification, many candidates suggested heating to constant mass. A second modification proved to be elusive for many. Other acceptable modifications included using powdered lead oxide to increase the surface area, bubbling the escape gases through lime water to identify when no more CO2 was being produced in the reaction, and passing an inert gas through the apparatus during cooling to prevent re-oxidation of the lead from oxygen in the air. Some suggestions would have created safety issues, such as working in a closed system. A significant number of candidates suggested collecting the gases evolved and that the reaction was complete when the gas volume no longer increased. Presumably the suggestion came from practical experience of rates experiments involving gas collection. This is unfeasible though when there is a constant flow of methane gas.

Notes from the mark scheme and report

• heat to constant mass
• spread out/powder the oxide
• pass methane or inert gas (not air) while cooling
• excess methane
• IGNORE 'heat for longer' without constant mass
• IGNORE repeats/averages

Skills: PAG1 Moles determination

Mark scheme

Examiner's comment

Most candidates were able to work out the masses of Pb and O atoms in the lead oxide from the supplied experimental results, and to then attempt to find the molar ratio of Pb : O. The 1 : 1.33 ratio was not always identified and a significant number of candidates approximated this to PbO or Pb2O3 instead of scaling up the ratio to 3 : 4 to show that the formula is Pb3O4 .

Notes from the mark scheme and report

• masses Pb 3.132 g and O 0.322 g
• Pb3O4 must come from masses
• NO ECF from incorrect masses

Skills: M0.2 Use ratios, fractions and percentages; PAG1 Moles determination

Mark scheme

Examiner's comment

A good understanding of structure and bonding continues to be difficult for candidates, demonstrated by many explanations seen for the different melting points. Most candidates obtained two relatively easy marks for identifying the giant and simple molecular/covalent structures of SiO2 and CO2 respectively. The explanation proved to be much more difficult as candidates showed some misconceptions. Many identified that CO2 had London forces but their action between molecules was often omitted. Many candidates realised that the forces broken on melting are much stronger in SiO2 than in CO2, but then went on to erroneously compare the strength of London forces or intermolecular forces in both SiO2 and CO2.

Notes from the mark scheme and report

• SiO2 giant covalent, CO2 simple molecular
• London forces in CO2
• covalent bonds in SiO2 stronger than intermolecular forces in CO2
• DO NOT ALLOW intermolecular forces in SiO2, hydrogen bonds or permanent dipoles for CO2
• IGNORE 'more bonds'

Mark scheme

Examiner's comment

Most candidates correctly named the organic compound as 4-chloro-3,5-dimethylphenol. The numbers caused candidates less issues than the need for the ‘di’ prefix to signify two methyl groups in the name.

Notes from the mark scheme and report

• 4-chloro-3,5-dimethylphenol
• look for dimethyl
• DO NOT ALLOW meth or methy

Mark scheme

Examiner's comment

Most candidates predicted that there would be 5 peaks in the 13C NMR spectrum. The common incorrect responses were 3 or 4 peaks, presumably from omitting the C–OH or C–Cl carbon atoms.

Notes from the mark scheme and report

• 5 peaks

Skills: M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

Most candidates identified the phenol functional group but a satisfactory test proved to be far more difficult. Many candidates identified that there would be no effervescence with sodium carbonate but it was common that the pH being less than 7 was often not seen. Neutralisation with sodium hydroxide was often seen but with no observation. Although not in the specification, the reaction with bromine to produce a white precipitate was often seen as an acceptable alternative test for the phenol group.

Notes from the mark scheme and report

• phenol (DO NOT ALLOW alcohol or hydroxide
• IGNORE hydroxyl/OH)
• acidic to indicator/pH < 7 AND no reaction with carbonate
• ALLOW bromine giving white precipitate

Skills: PAG7 Qualitative analysis of organic functional groups

Mark scheme

Examiner's comment

Candidates produced some very good calculations. It was common for candidates to determine the correct Ka value of 1.71 × 10–10 mol dm–3, after first calculating the hydrogen ion and weak acid concentrations. The initial calculation of the hydrogen ion concentration from 10–pH proved to be the easiest mark. The weak acid concentration proved to be more difficult with a significant number unable to determine the correct molar mass of chloroxylenol as 156.5. Most candidates recognised that the molar concentration in mol dm–3 was required from the mass of chloroxylenol in 100 cm3. Almost all candidates divided the mass of 4.80 by their molar mass but the scaling by ×10 was not always successful. Some candidates used the mole equations that they had learnt to obtain the ×10 scaling in a very roundabout way. The final mark was often given by ECF (error carried forward) from an incorrect concentration. The question did also ask for an equation, using molecular formulae, for the acid dissociation. This was sometimes omitted or written using C8H8ClOH as the molecular formula of chloroxylenol, instead of C8H9ClO. Even if the equation had been omitted, the mark could still given from a correct Ka expression using molecular formulae. The usual advice of ‘read and answer the question’ still holds. equation has been written using molecular formulae and all stages of the calculation are clear to see. Most candidates worked out the hydrogen ion concentration as 7.244 × 10–6 mol dm–3 before carrying out the final Ka calculation. This candidate had combined these stages.

Notes from the mark scheme and report

• molecular formulae required in equation (C8H9ClO ⇌ H+ + C8H8ClO–)
• common error: no squaring of [H+] (2.36 × 10–5) 3 of 4 calculation marks
• NO ECF from incorrect formula in equation

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.4 Use calculators to find and use power, exponential and logarithmic functions; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.5 Use logarithms in relation to quantities that range over several orders of magnitude

Mark scheme

Examiner's comment

Most candidates showed one asterisk at the base of the cyclic part of the structure. The most common error was to show two asterisks, the second being on *C(CH3)2OH, despite this carbon not being connected to four different groups.

Notes from the mark scheme and report

• DO NOT ALLOW more than one *
• ALLOW a circle

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

Candidates displayed a good knowledge of the requirements for E/Z isomerism in terms of a C=C double bond and different groups on the carbon atoms of the C=C bond. Many assigned terpineol as the Z isomer explained in terms of the priority groups being on the same side of the C=C bond. Candidates found it difficult to explain why terpineol has only one E/Z isomer. Many candidates thought that the C=C bond could not rotate because it was part of the ring. however, a C=C bond cannot rotate whether it is in a ring or not. Few candidates considered the strain put on the ring if the priority groups (being part of the ring) were to be placed in an E conformation.

Notes from the mark scheme and report

• max 4 from 5 points
• C=C with two different groups on each C
• Z isomer as high priority groups (ring carbons) on same side
• other isomer impossible as ring would be strained – must be linked to the ring
• IGNORE C=C cannot rotate

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects; M4.3 Understand the symmetry of 2-D and 3-D shapes

Mark scheme

Examiner's comment

In this question, candidates were given the opportunity to demonstrate some knowledge of organic reaction in a new context. The choice of reaction was up to the candidate. Most candidates were able to identify the alkene group in terpineol and to suggest a reagent that would react with this functional group. A correct structure for the organic product then usually followed. Although most candidates identified the alcohol group, many struggled with a reagent and resulting product. Although the alcohol –OH group has many reactions, (e.g. elimination, substitution, esterification) many candidates were fixated on oxidation with acidified dichromate despite a tertiary alcohol not being capable of oxidation with this reagent. Some candidates quoted acidified dichromate but then copied the structure of terpineol, stating that there was no reaction, despite the question directing them to select a reagent that would react with their chosen group.

Notes from the mark scheme and report

• functional group MUST be named (alkene
• tertiary alcohol)
• DO NOT ALLOW UV with halogens
• DO NOT ALLOW Cr2O72–/H+ (tertiary alcohol)
• ALLOW addition of HBr/H2O either way across C=C
• reagent must react with that group only

Mark scheme, with the levels and indicative points

Examiner's comment

This question was assessed by level of response (LoR). Candidates were required to process raw experimental results to determine one enthalpy change, and then to determine a second enthalpy change by constructing and using an energy cycle. Levels were determined by the accuracy of the candidates’ processing, calculations and use of the energy cycle. This question discriminated extremely well. Level 3 candidates used the mass of solution as 50.7 g with mc∆T to obtain an energy change of 2861 J. They then divided this value by the moles of CuSO4 that reacted (0.05 mol) to obtain the enthalpy change of –57.22 kJ mol–1. Finally, they constructed an energy cycle which they then used to obtain the second enthalpy change of –65.65 kJ mol–1. Level 2 candidates determined the first enthalpy change but may have used the approximate mass of 50 g for the mass of solution from the experimental method. Their energy cycle was often incorrect or absent, with the second enthalpy change calculated incorrectly. Level 1 candidates often calculated the initial energy change using mc∆T but made little further correct progress. A significant number of lower ability candidates used the solid mass of copper sulfate in their calculation. Overall, mathematical skills were displayed well but some basic errors were made, particularly with subtractions. This may have been the result of mis-keying values into a calculator and believing the answer displayed. An example was the mass of the solution (74.13 – 23.43) being seen as 49.7 rather than 50.7. A significant number of candidates added or subtracted the mass of copper sulfate from 50.7 or 50 for their value of m in mc∆T, using for example 57.98 and 42.72. This limited the level that they could reach. initial energy change using mc∆T, the first enthalpy change using the energy change with the moles of copper sulfate, and the second enthalpy change including a clear energy cycle. Numbers are clearly shown and unrounded until the final values. This is an excellent response.

Notes from the mark scheme and report

• mass of solution 50.70 g (not 50.0) and ∆T 13.5 °C
• n(CuSO4) = 7.98/159.6 = 0.0500 mol
• ∆solH = –57.22 kJ mol–1 with – sign
• ∆rH = ∆solH(CuSO4) – ∆solH(CuSO4•5H2O) = –65.65
• energy cycle expected

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.1 Translate information between graphical, numerical and algebraic forms; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

Just over half the candidates obtained the correct temperature change of 1°C. A common error was a temperature change of 0.5°C, the result of not considering that two temperature readings are made when calculating a temperature change.

Notes from the mark scheme and report

• two readings so 0.2/20 × 100 = 1.0 °C
• IGNORE direction of temperature change

Skills: M1.3 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined; M2.2 Change the subject of an equation; PAG3 Enthalpy determination

Mark scheme

Examiner's comment

This was a testing question. This question discriminated very well but lower ability candidates struggled to score marks. This unfamiliar scenario required candidates to first decide on a strategy for solving the problem. Candidates needed to use the Gibbs’ equation and supplied data to determine the entropy change. They then needed to combine this value with other supplied data to determine the unknown standard entropy. The data provided were all standard, signified by the standard sign. Candidates were expected to use 298 K as the standard temperature to first determine the entropy change. Although many candidates chose 298 K, many different temperatures were seen, with 273 K and 286.5 K (the temperature change in 5a(i)) being common. Some candidates were obviously thrown by the absence of a stated temperature and many omitted T from their Gibbs’ equation, using ∆G = ∆H – ∆S instead. This calculated value for ∆S (correct: –0.133 kJ) then needed to be combined with the supplied entropy data to determine the unknown standard entropy (correct: 156 J) to three significant figures. ECF could be applied during marking but this second calculation first required a unit conversion from kJ to J. Many candidates did not make this conversion. Some candidates made errors combining this information, e.g. incorrect signs, adding rather than subtracting and no unit conversions. Finally, some candidates, successful in their calculation, did not follow the instruction to quote their answer to three significant figures. first in kJ and then, in preparation for the second part of the determination, in J. The ∆S value is combined with the supplied entropy data to obtain the unknown entropy change, shown by ‘x’ in the response. This calculated value is then rounded to the required three significant figures.

Notes from the mark scheme and report

• use of 298 K
• ∆S = (–55.8 – –16.1)/298 = –0.133 kJ K–1 mol–1 (sign required), convert to J
• S = 372.4 – 5 × 69.9 + 133 = 156
• 3 SF
• ALLOW ECF from incorrect T or ∆S

Skills: M0.0 Recognise and make use of appropriate units in calculations; M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

In this part, candidates were expected to use their knowledge and understanding of bond angles and shapes to predict a bond angle of 109.5º and a tetrahedral shape for the thiosulfate ion. About half the candidates answered correctly but an array of other answers were obtained with bond angles ranging from 90º to 180º, and shapes from linear to octahedral.

Notes from the mark scheme and report

• 109.5° AND tetrahedral
• ALLOW 109–110°

Skills: M4.1 Use angles and shapes in regular 2-D and 3-D structures

Mark scheme

Examiner's comment

Candidates were required to combine two thiosulfate ions into a feasible dimer. Many acceptable shapes were allowed with different combinations of O and S atoms and several rings. A significant number of candidates did not seem to know where to start and less than half the candidates were able to draw an acceptable structure.

Notes from the mark scheme and report

• must be 2 atoms in the centre between the 6-bonded S atoms (S–S, O–O or S–O link)
• IGNORE charges
• ALLOW cyclic structures

Skills: M4.2 Visualise and represent 2-D and 3-D forms including 2-D representations of 3-D objects

Mark scheme

Examiner's comment

Most candidates were given 1 or 2 marks for this part. The black precipitate B (Ag2S) was identified correctly more often the orange precipitate A, which was often shown as Fe(OH)2 instead of Fe(OH)3 or Fe(OH)3(H2O)3. Significantly, identification of B required interpretation of new information whereas A required knowledge of transition element chemistry.

Notes from the mark scheme and report

• A = Fe(OH)3 (ALLOW Fe(OH)3(H2O)3)
• B = Ag2S from molar mass 247.9
• IGNORE state symbols

Mark scheme

Examiner's comment

Just over half the candidates identified that the student was incorrect, and that the reaction is not redox. Candidates were expected to provide some evidence to support their statement, in terms of no oxidation number changes. Some candidates claimed that the reaction was not redox because only sulfur changed oxidation number, suggesting a misunderstanding of redox.

Notes from the mark scheme and report

• student incorrect AND no oxidation numbers change (e.g. Fe stays +2)
• ALLOW no electron transfer

Mark scheme

Examiner's comment

Candidates found this equation extremely difficult with only a small number of candidates writing a correct equation. The problem lies with balancing the oxidation numbers and charges. Many wrote an equation with a 1:1 ratio or 1:2 ratio for [Fe(H2O)6]2+ : Cl2. An equation balanced in oxidation number and charge required a 2:1 ratio. When writing equations for redox reactions, candidates are recommended to check that oxidation changes and charges are balanced as well as atoms.

Notes from the mark scheme and report

• 2[Fe(H2O)6]2+ + Cl2 → 2[Fe(H2O)6]3+ + 2Cl–
• must be balanced by charge and oxidation number
• ALLOW multiples

Mark scheme

Examiner's comment

Candidates needed to interpret the information in the flowchart and to use this as the basis for their redox equation. The clue that a yellow solid is a product proved to be very difficult to interpret as being sulfur. The equation then required H2O to be added as the other product. Candidates found this equation difficult and relatively few correct equations were seen. As with 6a(iii), many equations were not balanced by oxidation number or charge. Some candidates omitted this part entirely.

Notes from the mark scheme and report

• 5H2S + 2MnO4– + 6H+ → 2Mn2+ + 5S + 8H2O
• 1 mark for all six correct species or Mn and S species balanced 5:2
• IGNORE extra electrons for 1st mark

Skills: M0.2 Use ratios, fractions and percentages

Mark scheme, with the levels and indicative points

Examiner's comment

This question presented a practical scenario in which candidates were asked to identify 5 unknown chemicals. There are many routes that lead to correct identifications of the unknowns. Most candidates identified G as oxygen and made some headway towards identifying C and D by determining that 9 waters of crystallisation were present in C. The formulae of C and D sometimes followed but many candidates found it difficult to link 9H2O to their formulae. The best responses showed the nitrate ion separately in the formula, e.g. Fe(NO3)3, but many showed the empirical formula instead, e.g. FeN3O9. Gas G (NO2) proved to be the most difficult unknown to identify as it required two pieces of data for its determination. There were some very competent attempts at writing equations, with the decomposition of compound D in Stage 2 to form E, F and G being the most difficult. 1:9 molar ratio of C: H2O. The candidate then writes the correct equation for Stage 1, using NO3 for the nitrate ion. The candidate then identifies E, F and G using a methodical approach with clear working throughout. Finally, the candidate writes the correct equation for the reaction in Stage 2. This is an excellent Level 3 response, given 6/6 marks.

Notes from the mark scheme and report

• n(H2O) = 0.027 so 9H2O
• C = Fe(NO3)3•9H2O, D = Fe(NO3)3, E = Fe2O3 (M 159.6), F = NO2 (0.414/0.00900 = 46), G = O2 (relights glowing splint)
• 2Fe(NO3)3 → Fe2O3 + 6NO2 + 1½O2

Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities